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https://www.aqua-calc.com/one-to-one/density/gram-per-metric-tablespoon/stone-per-cubic-centimeter/1
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# 1 gram per metric tablespoon in stones per cubic centimeter
## g/metric tbsp to st/cm³ unit converter of density
1 gram per metric tablespoon [g/metric tbsp] = 1 × 10-5 stone per cubic centimeter [st/cm³]
### grams per metric tablespoon to stones per cubic centimeter density conversion cards
• 1
through
25
grams per metric tablespoon
• 1 g/metric tbsp to st/cm³ = 1 × 10-5 st/cm³
• 2 g/metric tbsp to st/cm³ = 2 × 10-5 st/cm³
• 3 g/metric tbsp to st/cm³ = 3 × 10-5 st/cm³
• 4 g/metric tbsp to st/cm³ = 4 × 10-5 st/cm³
• 5 g/metric tbsp to st/cm³ = 5 × 10-5 st/cm³
• 6 g/metric tbsp to st/cm³ = 6 × 10-5 st/cm³
• 7 g/metric tbsp to st/cm³ = 7 × 10-5 st/cm³
• 8 g/metric tbsp to st/cm³ = 8 × 10-5 st/cm³
• 9 g/metric tbsp to st/cm³ = 9 × 10-5 st/cm³
• 10 g/metric tbsp to st/cm³ = 0.0001 st/cm³
• 11 g/metric tbsp to st/cm³ = 0.00012 st/cm³
• 12 g/metric tbsp to st/cm³ = 0.00013 st/cm³
• 13 g/metric tbsp to st/cm³ = 0.00014 st/cm³
• 14 g/metric tbsp to st/cm³ = 0.00015 st/cm³
• 15 g/metric tbsp to st/cm³ = 0.00016 st/cm³
• 16 g/metric tbsp to st/cm³ = 0.00017 st/cm³
• 17 g/metric tbsp to st/cm³ = 0.00018 st/cm³
• 18 g/metric tbsp to st/cm³ = 0.00019 st/cm³
• 19 g/metric tbsp to st/cm³ = 0.0002 st/cm³
• 20 g/metric tbsp to st/cm³ = 0.00021 st/cm³
• 21 g/metric tbsp to st/cm³ = 0.00022 st/cm³
• 22 g/metric tbsp to st/cm³ = 0.00023 st/cm³
• 23 g/metric tbsp to st/cm³ = 0.00024 st/cm³
• 24 g/metric tbsp to st/cm³ = 0.00025 st/cm³
• 25 g/metric tbsp to st/cm³ = 0.00026 st/cm³
• 26
through
50
grams per metric tablespoon
• 26 g/metric tbsp to st/cm³ = 0.00027 st/cm³
• 27 g/metric tbsp to st/cm³ = 0.00028 st/cm³
• 28 g/metric tbsp to st/cm³ = 0.00029 st/cm³
• 29 g/metric tbsp to st/cm³ = 0.0003 st/cm³
• 30 g/metric tbsp to st/cm³ = 0.00031 st/cm³
• 31 g/metric tbsp to st/cm³ = 0.00033 st/cm³
• 32 g/metric tbsp to st/cm³ = 0.00034 st/cm³
• 33 g/metric tbsp to st/cm³ = 0.00035 st/cm³
• 34 g/metric tbsp to st/cm³ = 0.00036 st/cm³
• 35 g/metric tbsp to st/cm³ = 0.00037 st/cm³
• 36 g/metric tbsp to st/cm³ = 0.00038 st/cm³
• 37 g/metric tbsp to st/cm³ = 0.00039 st/cm³
• 38 g/metric tbsp to st/cm³ = 0.0004 st/cm³
• 39 g/metric tbsp to st/cm³ = 0.00041 st/cm³
• 40 g/metric tbsp to st/cm³ = 0.00042 st/cm³
• 41 g/metric tbsp to st/cm³ = 0.00043 st/cm³
• 42 g/metric tbsp to st/cm³ = 0.00044 st/cm³
• 43 g/metric tbsp to st/cm³ = 0.00045 st/cm³
• 44 g/metric tbsp to st/cm³ = 0.00046 st/cm³
• 45 g/metric tbsp to st/cm³ = 0.00047 st/cm³
• 46 g/metric tbsp to st/cm³ = 0.00048 st/cm³
• 47 g/metric tbsp to st/cm³ = 0.00049 st/cm³
• 48 g/metric tbsp to st/cm³ = 0.0005 st/cm³
• 49 g/metric tbsp to st/cm³ = 0.00051 st/cm³
• 50 g/metric tbsp to st/cm³ = 0.00052 st/cm³
• 51
through
75
grams per metric tablespoon
• 51 g/metric tbsp to st/cm³ = 0.00054 st/cm³
• 52 g/metric tbsp to st/cm³ = 0.00055 st/cm³
• 53 g/metric tbsp to st/cm³ = 0.00056 st/cm³
• 54 g/metric tbsp to st/cm³ = 0.00057 st/cm³
• 55 g/metric tbsp to st/cm³ = 0.00058 st/cm³
• 56 g/metric tbsp to st/cm³ = 0.00059 st/cm³
• 57 g/metric tbsp to st/cm³ = 0.0006 st/cm³
• 58 g/metric tbsp to st/cm³ = 0.00061 st/cm³
• 59 g/metric tbsp to st/cm³ = 0.00062 st/cm³
• 60 g/metric tbsp to st/cm³ = 0.00063 st/cm³
• 61 g/metric tbsp to st/cm³ = 0.00064 st/cm³
• 62 g/metric tbsp to st/cm³ = 0.00065 st/cm³
• 63 g/metric tbsp to st/cm³ = 0.00066 st/cm³
• 64 g/metric tbsp to st/cm³ = 0.00067 st/cm³
• 65 g/metric tbsp to st/cm³ = 0.00068 st/cm³
• 66 g/metric tbsp to st/cm³ = 0.00069 st/cm³
• 67 g/metric tbsp to st/cm³ = 0.0007 st/cm³
• 68 g/metric tbsp to st/cm³ = 0.00071 st/cm³
• 69 g/metric tbsp to st/cm³ = 0.00072 st/cm³
• 70 g/metric tbsp to st/cm³ = 0.00073 st/cm³
• 71 g/metric tbsp to st/cm³ = 0.00075 st/cm³
• 72 g/metric tbsp to st/cm³ = 0.00076 st/cm³
• 73 g/metric tbsp to st/cm³ = 0.00077 st/cm³
• 74 g/metric tbsp to st/cm³ = 0.00078 st/cm³
• 75 g/metric tbsp to st/cm³ = 0.00079 st/cm³
• 76
through
100
grams per metric tablespoon
• 76 g/metric tbsp to st/cm³ = 0.0008 st/cm³
• 77 g/metric tbsp to st/cm³ = 0.00081 st/cm³
• 78 g/metric tbsp to st/cm³ = 0.00082 st/cm³
• 79 g/metric tbsp to st/cm³ = 0.00083 st/cm³
• 80 g/metric tbsp to st/cm³ = 0.00084 st/cm³
• 81 g/metric tbsp to st/cm³ = 0.00085 st/cm³
• 82 g/metric tbsp to st/cm³ = 0.00086 st/cm³
• 83 g/metric tbsp to st/cm³ = 0.00087 st/cm³
• 84 g/metric tbsp to st/cm³ = 0.00088 st/cm³
• 85 g/metric tbsp to st/cm³ = 0.00089 st/cm³
• 86 g/metric tbsp to st/cm³ = 0.0009 st/cm³
• 87 g/metric tbsp to st/cm³ = 0.00091 st/cm³
• 88 g/metric tbsp to st/cm³ = 0.00092 st/cm³
• 89 g/metric tbsp to st/cm³ = 0.00093 st/cm³
• 90 g/metric tbsp to st/cm³ = 0.00094 st/cm³
• 91 g/metric tbsp to st/cm³ = 0.00096 st/cm³
• 92 g/metric tbsp to st/cm³ = 0.00097 st/cm³
• 93 g/metric tbsp to st/cm³ = 0.00098 st/cm³
• 94 g/metric tbsp to st/cm³ = 0.00099 st/cm³
• 95 g/metric tbsp to st/cm³ = 0.001 st/cm³
• 96 g/metric tbsp to st/cm³ = 0.00101 st/cm³
• 97 g/metric tbsp to st/cm³ = 0.00102 st/cm³
• 98 g/metric tbsp to st/cm³ = 0.00103 st/cm³
• 99 g/metric tbsp to st/cm³ = 0.00104 st/cm³
• 100 g/metric tbsp to st/cm³ = 0.00105 st/cm³
#### Foods, Nutrients and Calories
PINTO BEANS, UPC: 041512136689 weigh(s) 274.74 gram per (metric cup) or 9.17 ounce per (US cup), and contain(s) 85 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
HERR'S, KETTLE COOKED POTATO CHIPS, HONEY SRIRACHA, UPC: 072600033382 contain(s) 500 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Vitamin B-12, added, foods low in Vitamin B-12, added, and Recommended Dietary Allowances (RDAs) for Vitamin B12
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Ivory Coast weighs 1 281.48 kg/m³ (80.00018 lb/ft³) with specific gravity of 1.28148 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Magnesium phosphate dibasic trihydrate [MgHPO4 ⋅ 3H2O] weighs 2 123 kg/m³ (132.53456 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-438A, liquid (R438A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
Lentor is an obsolete non-metric measurement unit of kinematic viscosity.
The surface density of a two-dimensional object, also known as area or areal density, is defined as the mass of the object per unit of area.
t/l to kg/US qt conversion table, t/l to kg/US qt unit converter or convert between all units of density measurement.
#### Calculators
Calculate gravel and sand coverage in a rectangular aquarium
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CC-MAIN-2019-43
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=tensor/entermetric
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tensor(deprecated)/entermetric - Help
tensor
entermetric
facility for user input of coordinate variables and covariant metric tensor components.
Calling Sequence entermetric( 'g', 'coords' )
Parameters
g - covariant metric tensor (symmetric rank 2 tensor_type) coords - list of coordinate variables
Description
Important: The tensor package has been deprecated. Use the superseding packages DifferentialGeometry and Physics instead.
• The function entermetric( g, coords ) prompts the user for the dimension of the manifold, the coordinate variables, and the components of the covariant metric tensor with respect to the natural basis (that is, the line element), and produces the coordinates list and covariant metric tensor.
• The coordinate variables are returned as a list of names through the output parameter coords. The covariant metric tensor is returned as a rank 2 symmetric tensor_type through the output parameter g. The parameters g and coords must be unassigned names.
• Since diagonal metrics are common, the user is asked to specify if the metric is diagonal or not. If the metric is diagonal, only the diagonal entries are required to be input. If the metric is not diagonal, the entries in the "upper triangle" of the metric components array are required to be input. In both cases, the returned metric components make use of Maple's symmetric indexing function.
• The user is required to end each line of input with a semicolon (";") since each input value is read in as a Maple statement. Long expressions may be broken over more than one line provided the expression contains only one semicolon located at the end of the expression (like regular maple statements).
• After the user has finished inputting the dimension, coordinates, and metric components, entermetric will display them back to the user for confirmation.
Examples
Important: The tensor package has been deprecated. Use the superseding packages DifferentialGeometry and Physics instead.
> $\mathrm{with}\left(\mathrm{tensor}\right):$
Use entermetric to input the Schwarzschild metric.
> $\mathrm{entermetric}\left(g,\mathrm{coord}\right)$
${\mathrm{The coordinate variables are :}}$
${\mathrm{x1}}{=}{t}$
${\mathrm{x2}}{=}{r}$
${\mathrm{x3}}{=}{\mathrm{θ}}$
${\mathrm{x4}}{=}{\mathrm{φ}}$
${}$
${\mathrm{The Covariant Metric}}$
${\mathrm{non-zero components :}}$
${\mathrm{cov_g11}}{=}{1}{-}\frac{{2}{}{M}}{{r}}$
${\mathrm{cov_g22}}{=}{-}\frac{{1}}{{1}{-}\frac{{2}{}{M}}{{r}}}$
${\mathrm{cov_g33}}{=}{-}{{r}}^{{2}}$
${\mathrm{cov_g44}}{=}{-}{{r}}^{{2}}{}{{\mathrm{sin}}{}\left({\mathrm{θ}}\right)}^{{2}}$ (1)
Confirm the results once again:
> $\mathrm{coord}$
$\left[{t}{,}{r}{,}{\mathrm{θ}}{,}{\mathrm{φ}}\right]$ (2)
> $\mathrm{eval}\left(g\right)$
${\mathrm{table}}{}\left(\left[{\mathrm{index_char}}{=}\left[{-}{1}{,}{-}{1}\right]{,}{\mathrm{compts}}{=}{\mathrm{array}}{}\left({\mathrm{symmetric}}{,}{\mathrm{sparse}}{,}{1}{..}{4}{,}{1}{..}{4}{,}\left[\left({2}{,}{2}\right){=}{-}\frac{{1}}{{1}{-}\frac{{2}{}{M}}{{r}}}{,}\left({1}{,}{1}\right){=}{1}{-}\frac{{2}{}{M}}{{r}}{,}\left({3}{,}{3}\right){=}{-}{{r}}^{{2}}{,}\left({4}{,}{4}\right){=}{-}{{r}}^{{2}}{}{{\mathrm{sin}}{}\left({\mathrm{θ}}\right)}^{{2}}\right]\right)\right]\right)$ (3)
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| 2.65625
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https://www.techwhiff.com/issue/which-did-you-include-in-your-answer-demand-higher--485041
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# Which did you include in your answer? Demand higher wages join forces with other employees try to get into a skilled trade try to find a better job
###### Question:
Which did you include in your answer? Demand higher wages join forces with other employees try to get into a skilled trade try to find a better job
### Annibale Carracci was one of the founders of: A. the Renaissance Revival movement. B. the philosophy of art. C. the Italian Mannerist school of painting D. the Bolognese art academy Please help as soon as possible! I really appreciate your guys' help with everything so far!!! :) i think this question is worth 20 points, btw
Annibale Carracci was one of the founders of: A. the Renaissance Revival movement. B. the philosophy of art. C. the Italian Mannerist school of painting D. the Bolognese art academy Please help as soon as possible! I really appreciate your guys' help with everything so far!!! :) i think this questio...
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### 1. 11/20 x 4/14 2. 7/18 x 11/25 3. 3/15 x 3/7 4. 2/4 x 3/6 5. 1/2 x 2/5 6. 1/3 x 2/9 Fractions and please show work if possible
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### ). how adolescents in 27 countries understand, support, and practice human rights
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### I will add you as BRAINLIEST if you can get it right! Someone Plz help!!! Triangles ABC and DEF are similar. Part 1: Find the lenght of segment DF (rounded to the nearest hundreth) _____ units. Part 2: Find the lenght of segment EF (rounded to the nearest hundreth) _____units.
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### If f(x) = |x + 6|, find f(–10). –16 –4 4 16
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### The height h(n) of a bouncing ball is an exponential function of the number n of bounces. One ball is dropped and on the first bounce reaches a height of 6 feet. On the second bounce it reaches a height of 4 feet. PLEASE HURRY
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### The habit of thinking of a people is an essential part of their culture. How does this observation explain the relationship between different cultures and corresponding civilizations
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### What is g in this equation " G+10=20"
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### Sam read 75 pages of a new mystery novel in 2 hrs. if the book contains 350 pages and he always reads the same rate, how long will it take him to read the entire novel
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### Mason brought a tablecloth that measured 8 feet by 4 feet. What was the area of the tablecloth? PLEASE ANSWER QUICK QUICK WILL EARN 44 BADGE IF NOT, I WILL GIVE YOU A BIG RATING AND THUMBS UP ANSWER PLEASE!
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| 3.078125
| 3
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CC-MAIN-2022-33
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latest
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en
| 0.902867
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https://teachingobjects.net/questions/1893-please-help-quickly-first-to-answer-gets-bainliest-which-expression-uses-the-greatest-common-factor.html
| 1,627,632,910,000,000,000
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hellokitty1
2015-10-31 17:22:30
Please help quickly first to answer gets bainliest Which expression uses the greatest common factor and the distributive property to rewrite the sum 32 + 80? A.16(2) + 80 B. 2(16 + 80) C.16(2 + 5) D.8(4 + 10)
Evan259
2015-10-31 19:29:01
C. 16 ( 2 + 5) is your answer hope this helps
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| 2.875
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CC-MAIN-2021-31
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latest
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en
| 0.687449
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http://blog.mathsbank.co.uk/2011/06/
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## Tuesday, 28 June 2011
### Happy Tau Day
Happy Tau Day!
Happy what day? Tau is a mathematical constant, whose value is 6.28... . Is that ringing any bells? Correct:
OK, so tau being 6.28 explains why 28th June is Tau Day (blame the Americans - they write their dates backwards). But why do we need a new mathematical constant, especially one that is simply double another one?
Some maths teachers and academics have been in favour of using tau instead of pi in maths teaching, particularly in early years. It's because 2 pi seems to crop up a lot, probably more often than a single pi, particularly in geometry and trigonometry. For example, the circumference of a circle is given by:
If tau were widely adopted, this would be replaced by:
Personally, I'm not convinced. Tau would certainly be useful in a number of formulae and mathematical solutions. But I think students would end up using a half of tau just as often as they currently use 2 pi.
And if we went to a system where both constants were in use, would this not just add to the confusion, rather than alleviate it?
Finally there are thousands of years of pi tradition. The ancient Greeks obsessed over pi, just as much as modern mathematicians do.
What do you think of tau?
## Thursday, 9 June 2011
### Exam Standards Are Slipping (That's The Papers, Not the Candidates)
You may be used to the idea of going into exams to find questions you find impossible to answer.
If you sat the OCR Decision Maths 1 exam on 26th May you will certainly have found yourself in this situation; one of the questions had no solution.
This embarrassing slip-up was the first of five errors on AS exam papers this summer. The other mistakes came to light on:
• an AQA Business Studies paper, in which there was not enough information to answer the question;
• an Edexcel Biology multiple choice exam, in which none of the answers given were correct;
• an AQA Geography paper, in which the flow of a river was incorrectly labelled at one point;
• an AQA Computing AS Level paper, in which an arrow was shorter than it should have been.
Reports are also circulating of mistakes on a CCEA business studies GCSE and one more, undisclosed paper.
Jim Sinclair, Director of the Joint Council for Qualifications, said that procedures were in place to cope with these errors.
In the case of the Decision Maths, the question was worth 8 marks out of 72, or 11% of the total marks awarded for the paper. The exam board OCR have decided not to discount the question from the paper, since it was worth such a large percentage of the marks. Instead, they will award marks for correct working, which will reward pupils who spent some of their time attempting to answer the question.
But many students are unhappy with this proposal. Facebook groups have been formed to demand a complete rerun of the exam. Some students think the time they wasted will jeopardise their chances of them getting the grades required for their university applications.
More details about the impossible D1 question are here.
Our advice to examination candidates has always been: if you can't do a question on the paper, leave it and come back to it if you have time at the end. It may be time to modify this advice: leave it; it may not be possible anyway.
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http://www.chegg.com/homework-help/heating-ventilating-and-air-conditioning-6th-edition-chapter-14-solutions-9780471470151
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Solutions
Heating Ventilating and Air Conditioning Analysis and Design
# Heating Ventilating and Air Conditioning Analysis and Design (6th Edition)Solutions for Chapter 14
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 9
Cross flow heat exchanger with air as cold fluid and water as hot fluid.
• Step 2 of 9
a) Calculate the Log Mean Temperature Difference (LMTD) by using the below relation for single pass cross flow heat exchanger with both fluids unmixed;
The Log Mean Temperature Difference (LMTD),
Here, is the exit temperature of water , is the inlet temperature of air, is the inlet temperature of water, and is the exit temperature of air.
Substitute for , at standard conditions for , for , for in the above equation.
Therefore , the Log Mean Temperature Difference (LMTD),value is
• Step 3 of 9
Determine the correction factor using below relations for single pass cross flow heat exchanger with both fluids unmixed;
The parameter P,
Substitute, at standard conditions for , for , and for
The parameter R,
Substitute, for , at standard conditions for , for , and for
From the chart “correction factor plot for single-pass cross-flow exchanger (both fluids unmixed)”, take the value of correction factor forand,
Correction factor,
Therefore, the correction factor F is
• Step 4 of 9
b)
Determine fluid capacity rate of air;
The fluid capacity rate of cold fluid air is
…... (1)
Here, mass flow rate of air, Q is the volume flow rate of air and is the density of air
All air properties should be evaluated at bulk mean temperature of the air.
Bulk mean temperature of air
Substitute, at standard conditions for , for
Obtain the fluid properties from the saturated tables at .
Density of air,
Specific heat of air,
Substitute for, for , for in the equation (1).
Therefore, the fluid capacity rate of air, is
• Step 5 of 9
Determine fluid capacity rate of water.
The actual heat transfer rate of air and water are given by
Here, is the fluid capacity rate of hot fluid water,
Substitute for, for , at standard conditions for , for , for
Therefore, the fluid capacity rate of water, is
• Step 6 of 9
c) Determine flow rate of water;
The fluid capacity rate of water is
…... (2)
All water properties should be evaluated at bulk mean temperature of the water.
Bulk mean temperature of water.
Substitute, for , for
Obtain the fluid properties from the saturated tables at .
Density of water,
Specific heat of water,
Substitute for, forand for in the equation (2)
Convert flow rate of water in gpm
Therefore, the flow rate of water, Q is
• Step 7 of 9
d) Determine the overall conductance using the below equation;
The heat transfer rate from one fluid to the other in the heat exchanger is
Here, A is the surface area of heat exchanger, and UA is the Overall conductance,
Substitute for, for , for , 0.983 for, for LMTD in the above equation
Therefore, overall conductance is
• Step 8 of 9
a) Determine the number of heat transfer units (NTU) using below relation ;
Number of Transfer Units (NTU) is defined as
Here minimum fluid capacity rate,
The fluid capacity rate of air
The fluid capacity rate of water is
Therefore
Substitute for, for in the below equation.
Therefore, NTU is
• Step 9 of 9
b) Determine effectiveness of cross flow heat exchanger :
The effectiveness of cross flow heat exchanger is defined as
Since, we get
Substitute for , for , for in the above equation.
Therefore, effectiveness of the heat exchanger is
Corresponding Textbook
Heating Ventilating and Air Conditioning Analysis and Design | 6th Edition
9780471470151ISBN-13: 0471470155ISBN: Authors:
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Standard red. potential (E₀)
V
Temperature
°F
Electrons transferred
mol
Activity (reduced form)
Activity (oxidized form)
Reduction potential (E)
V
# Nernst Equation Calculator
By Bogna Haponiuk
This Nernst equation calculator presents the fundamental formula of electrochemistry - the Nernst equation, also known as the cell potential equation. It is a formula that allows you to calculate the reduction potential of a half-cell or full cell reaction. If you don't know what is the reduction potential, don't worry - we will explain all terms, describe in detail how to calculate the cell potential, and finish with a Nernst equation example that best presents its application.
## What is the reduction potential?
Reduction potential of a half-cell or full cell reaction is also called the redox potential or oxidation/reduction potential. It measures the tendency of molecules (or atoms, ions etc.) to acquire electrons and hence be reduced. This value is measured in volts (V) - the same units that are used by our Ohm's law calculator.
Why exactly is it called oxidation/reduction potential? Oxidation occurs when electrons are removed - for example when a free radical steals an electron from a cell. Reduction, on the other hand, means receiving or gaining electrons, for instance when an antioxidant donates an electron to a free radical.
What does it mean in terms of reduction potential? A solution with a higher potential will have a tendency to gain electrons (be reduced), and a solution with a lower potential - to lose electrons (be oxidized). Note that a high reduction potential doesn't mean that the reaction will occur - the reaction still requires some activation energy to be supplied.
It is difficult to measure the absolute potential of a solution. That's why the reduction potentials are usually defined relative to a reference electrode.
The standard reduction potential is the redox potential measured under standard conditions: 25°C, activity equal to 1 per ion, and pressure of 1 bar per gas participating in the reaction. The standard reduction potential is defined relative to a standard hydrogen electrode(SHE), which is arbitrarily given a potential of 0 volts.
## What is the cell potential equation?
The Nernst equation (cell potential equation) relates the reduction potential to the standard electrode potential, temperature, and activities of molecules. Activities can be substituted by concentrations for an approximate result. For a half-cell or full cell reaction,
`E = E₀ - RT/zF * ln([red]/[ox])`
where:
• E is the reduction potential, expressed in volts (V);
• E₀ is the standard reduction potential, also expressed in volts (V);
• R is the gas constant, equal to 8.314 J/(K * mol);
• T is the temperature at which the reaction would occur, measured in Kelvins (K);
• z is the number of moles of electrons transferred in the reaction (mol);
• F is the Faraday constant, equal to the number of coulombs per mole of electrons (96 485.3 C/mol);
• [red] is the chemical activity of the molecule (atom, ion...) in the reduced form. It can be substituted by concentration;
• [ox] is the chemical activity of the molecule (atom, ion...) in the oxidized form. It also can be substituted by concentration.
## How to calculate reduction potential: a Nernst equation example
We will use the Nernst equation calculator to find the reduction potential of a cell basing on the following reactions:
`Mg → Mg2+ + 2e-`, where E₀ = +2.38 V
`Pb2+ + 2e- → Pb`, where E₀ = -0.13 V
1. First, we need to write down the total reaction and calculate the total standard redox potential:
`Pb2+(aq) + Mg(s) → Mg2+(aq) + Pb(s)`
`E₀(cell) = 2.38 V + -0.13 V = 2.25 V`
1. The next step is to determine the temperature and number of moles of electrons. We can assume the temperature to be equal to 25°C. 2 moles of electrons were transferred in this reaction.
2. Now, we need to decide which molecules are oxidized and which reduced. The ones that gain electrons - that is, lead (Pb) molecules - are reduced. The magnesium (Mg) molecules are oxidized.
3. The last step before before performing calculations is to determine the proportion of activity or concentration. If we know the concentration of Pb molecules to be 0.200 M and of magnesium to be 0.020 M, the the concentration proportion is `[red]/[ox] = [Mg2+]/[Pb2+] = 0.020/0.200 = 0.1`.
4. Now we can combine all of these results and introduce them into the Nernst equation calculator. The final value of reduction potential is equal to 2.28 V.
Bogna Haponiuk
## Get the widget!
Nernst Equation Calculator can be embedded on your website to enrich the content you wrote and make it easier for your visitors to understand your message.
It is free, awesome and will keep people coming back!
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How Do You Calculate the Median of a Set of Numbers?
Calculate the median of a set by putting the numbers in order from least to greatest and finding the middle number, when there is an odd number of values in the set. When there is an even number of values, find the median by averaging the two middle values.
1. Put the set in order from least to greatest
Order the values in the set from least to greatest by putting the smallest value first and arranging the values so that each following value is greater than the previous one. For example, if your set consists of the scores on a particular test with the values of 100, 95, 90, 92, 81, 84, 74, 33, 45, 50, 67, and 79, then set them in order from least to greatest: 33, 45, 50, 67, 74, 79, 81, 84, 90, 92, 95, 100.
2. Find the middle number
If there is an odd number of values, locate the middle number. This is the median. In the example from Step 1, there is an even number of values so there is no middle number.
3. Divide to find the average of the two middle numbers
If there is an even number of values, find the average of the two middle numbers. In the example from Step 1, there is an even number of values. The two middle numbers are 79 and 81. Find the average of 79 and 81 by adding them together and dividing by 2. Therefore, 79 + 81 = 160. 160 / 2 = 80, so the median of the set is 80.
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# 1.1 Patterns and Inductive Reasoning Goal Standards reasoning.
```Untitled.notebook
February 01, 2008
1.1 Patterns and Inductive Reasoning
Goal: Find and describe patterns and use inductive
reasoning.
Standards:
1.0 Students demonstrate understanding by
identifying and giving examples of undefined terms,
axioms, theorems, and inductive and deductive
reasoning.
3.0 Students construct and judge the validity of a
logical argument and give counterexamples to
disprove a statement.
Untitled.notebook
February 01, 2008
A conjecture is an unproven statement that is
based on observations.
Inductive reasoning is a process that involves
looking for patterns and making conjectures.
A counterexample is an example that shows a
conjecture is false.
Untitled.notebook
February 01, 2008
Example 1 Describing a Visual Pattern
Sketch the next figure in the pattern.
1st. term
➦ CW
2nd. term
Next term
3rd. term
➦
➦
➦
Each figure looks like the one before it
except that it has rotated 90°. The next figure
will have the smaller circle in the lower-left
quarter of the bigger circle.
Untitled.notebook
February 01, 2008
1. Sketch the next figure in the pattern.
1st. term
2nd. term
3rd. term
Next term
Untitled.notebook
February 01, 2008
Example 2 Describing a Number Pattern
Describe a pattern in the sequence of numbers.
Predict the next number.
a. 5, 3, 1, -1, ....____
These are consecutive odd numbers, but listed
backwards starting with 5. The next number is
-3.
b. 1, -4, 9, -16,... ____
These numbers look like consecutive perfect
squares, except that every other is negative.
The next number is 25.
Untitled.notebook
February 01, 2008
Example 2 Describing a Number Pattern
Describe a pattern in the sequence of numbers.
Predict the next number.
c.
1
2
, 1
4
,
1
8
, ....____
Each number is
1
2
times the previous number.
The next number is
1
16
.
Untitled.notebook
February 01, 2008
Describe a pattern in the sequence of numbers.
Predict the next number.
2. 1, 2, 6, 24, ____
Untitled.notebook
February 01, 2008
Describe a pattern in the sequence of numbers.
Predict the next number.
3. 0, 3, 8, 15, 24, ____
Untitled.notebook
February 01, 2008
Example 3 Making a Conjecture
Complete the conjecture.
Conjecture: The product of two consecutive
even integers is divisible by ______________
List some specific examples and look for a
pattern.
2(4) = 8 = 8(1)
4(6) = 24 = 8(3)
6(8) = 48 = 8(6)
8(10) = 80 = 8(10)
10(12) = 120 = 8(15) 12(14) = 168 = 8(21)
Conjecture: The product of two consecutive
even integers is divisible by 8
Untitled.notebook
February 01, 2008
4. Complete the conjecture based on the
pattern you observe in the specific cases.
Conjecture: For any two numbers a and b, the
product of (a + b) and (a - b) is always equal to
_____________________
(a + b)(a - b) = a2 - b2
2
2
(2 + 1)(2 - 1) = 3 = 2 - 1
(3 + 2)(3 - 2) = 5 = 32 - 22
(4 + 3)(4 - 3) = 7 = 42 - 32
2 2
(5 + 4)(5 - 4) = 9 = 5 - 4
Untitled.notebook
February 01, 2008
Example 4 Finding a Counterexample
Show the conjecture is false by finding a
counterexample.
Conjecture: All odd numbers are prime.
The conjecture is false. Here is a
counterexample: The number 9 is odd and is a
composite number, not a prime number.
Untitled.notebook
February 01, 2008
Show the conjecture is false by finding a
counterexample.
5. The square if the sum of two numbers is
equal to the sum of the squares of the two
numbers. that is, (a + b)2 = a2 + b2
```
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http://shuttleworthforcongress.org/online-poker-free/3-to-1-odds-payout-in-craps-when-do-us-savings
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3 to 1 odds payout in craps when do us savings
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A further problem is that the Outcome doesn't have a fixed payout in Craps. This will alter the with Variable Odds. This will lead us to refactor Outcome. The other belongs to dice totalling 3 or 11, with odds of 3: 1. We'll address this .. Real savings of effort is the only justification for disrupting the schedule. Architect.
Here are the odds for all the possible numbers that can be rolled in craps: 4 and 10 can be made of 3 separate dice combinations each and offer odds of 1:12. The jackpot sequence has a relatively remote possibility of occurrence and, correspondingly, a high payoff. Given the new generalization, RandomEventwe can rework the. Real savings of effort is the only justification. My question 8 player games wii "does the house edge change at all when playing a strategy of pass line with full odds and making a maximum of two come bets with maximum odds? The Grand Victoria Casino in Elgin, Illinois offers a promotion called "Craps for Cash.
3 to 1 odds payout in craps when do us savings - 888 poker
Transportation: Airport- International arrivals- how long is the arrival process? Craps is nothing if not a communal game, and part of its enduring appeal in casinos spanning the globe is a certain sense of collective congratulations when the dice are rolling the right way. This entangles Roulette and Craps around a feature that is really a. Understanding dice probability is a central component of succeeding at, and even enjoying, the game of craps. We will present two alternative designs paths: minimal rework, and a. Day trips: Springs Preserve Visiting further afield... A put bet is like a come bet made after a point has already been established.
Learn How to Play Craps and Win Part 2
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arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# On a perfect matching in a random bipartite digraph with average out-degree below two.
Michal Karoński, Ed Overman and Boris Pittel
September 14, 2020
September 14, 2020
###### Abstract.
Existence of a perfect matching in a random bipartite digraph with bipartition , , is studied. The graph is generated in two rounds of random selections of a potential matching partner such that the average number of selections made by each vertex overall is below . More precisely, in the first round each vertex chooses a potential mate uniformly at random, and independently of all vertices. Given a fixed integer , a vertex is classified as unpopular if it has been chosen by at most vertices from the other side. Each unpopular vertex makes yet another uniform/independent selection of a potential mate. The expected number of selections made by a generic vertex , i.e. its out-degree, is asymptotic to . Aided by Matlab software, we prove that for , whence for all , the resulting bipartite graph has a perfect matching a.a.s. (asymptotically almost surely). On the other hand, for a.a.s. a perfect matching does not exist. This is a thorough revision of the joint paper (JCT(B) 88 (2003), 1-16) by the first author and the third author.
###### Key words and phrases:
bipartite graphs, perfect matchings, random, asymptotics
###### 2010 Mathematics Subject Classification:
05C05, 05C07, 05C30, 05C80, 60C05
## 1. Introduction and main result
A standard model of a random bipartite (di)graph with bipartition , , is generated by each vertex making uniformly random, independent selections of a potential match from the other side. By computing the expected number of perfect matchings, Walkup [7] proved that asymptotically almost surely (a.a.s.) the graph has no perfect matching. In fact, Meir and Moon [6] (cf. Frieze [2]) earlier proved that the maximum matching number of is a.a.s. about . It is not much more difficult to show, using Hall’s Marriage Lemma, that a.a.s. the graph does have a perfect matching. Remarkably, Walkup managed to show that a.a.s. so does the graph . Frieze [2] was able to prove an analogous result for a non-bipartite graph using Tutte’s criterion.
In this paper we study existence of a perfect matching in a bipartite random graph which is sandwiched between and . is generated in two rounds of random selections of a potential match by every vertex . Specifically, in the first round each vertex selects a vertex from the opposite side uniformly at random, and independently of all other vertices. We call a vertex “unpopular” if it has been selected by at most vertices. (This definition depends on the value : the larger the larger the set of unpopular vertices.) Each unpopular vertex makes yet another uniformly random, and independent selection of a vertex from the other side. (If , then effectively all the vertices select uniformly at random and independently two vertices from the other side, so .) The number of vertices that have selected a given vertex is distributed binomially with trials and success probability ; thus it is in the limit. It follows that the expected out-degree of a generic vertex in is
1+P(Poisson(1)≤m)=1+e−1m∑j=01j!↑2,m→∞.
Thus the average out-degree of a vertex in is strictly between and . Loosely, we can interpret as , . In the joint paper [5] the first author and the third author stated and gave a proof of the claim: a.a.s. (i.e. ) has a perfect matching. Recently Michael Anastos and Alan Frieze [1] pointed out a simple oversight in the proof. We realized that the oversight invalidated the claim. A thorough revision of the method in [5] has allowed us to prove
###### Theorem 1.1.
Let .
(1) P(Bn,m has a perfect matching)≥1−O(n−cm+o(1)), cm:=1−1.51+(m+1)(1+e−1∑j≤m1/j!−log2),
where
(2) P(Bn,m is connected)=1−O(n−cm+o(1)).
Let the reader beware that our rigorous proof techniques produced an explicit function , , , such that, to complete the proof, needed to be proved negative for all . This is because the minimum is the best achievable upper bound for the scaled logarithm of the expected number of Hall’s subgraphs of a given size that are present if there is no perfect matching. By upper-bounding , we checked negativity “manually” for small , but deferred to Matlab algorithmic software to handle the remaining ’s.
As for , Matlab revealed that for all, but very small . Of course, positivity of this minimum even for all does not imply almost sure non-existence of a perfect matching. However this numerical evidence prodded us to try and prove that, contrary to our long-held belief, existence of a perfect matching is indeed highly unlikely. Using the necessity part of Hall’s Lemma we prove the opposite of the claim in [5]:
###### Theorem 1.2.
P(Bn,0 has no perfect matching)=1−O(n−1/2+o(1)).
For the proof itself, Matlab assistance was not required.
## 2. Proof of Theorem 1.1.
Let . Part 1. If the graph has no perfect matching, than by Hall’s Marriage Lemma there exist a set of row vertices (elements of ), or of column vertices (elements of ), such that , where is the set of neighbors of in . We call such “bad” pairs. We focus on the minimal pairs, minimal in a sense that there is no such that is a bad pair. For a minimal bad pair , it is necessary that (i) ; (ii) ; (iii) every vertex in has at least two neighbors in . Let , , denote the expected number of the minimal bad pairs , with . We need to prove that . By symmetry,
(2.1) Enk≤2(nk)(nk−1)Pnk;
here is the probability that and satisfy the conditions (ii) and (iii). In fact we broaden the condition (iii) a bit, replacing it with (iii’): in two rounds of selections every vertex in at least twice either selected a vertex in or was selected by a vertex in .
### 2.1. Case k≤n1/2.
On the event in question, let ( resp.) be the number of columns in (the number of rows in resp.) that selected rows in (columns in resp.) in the first round. Then the number of unpopular rows in (unpopular columns in resp.) is at least ( resp.). (Indeed, every popular row in is selected by at least columns out of columns.) and are independent binomials with parameters and respectively. Therefore
Explanation The first line of the RHS is the probability that the first round choices made by rows from (columns from resp.) are columns from (rows from resp.). The second line (third line resp.) is an upper bound for the conditional probability that the second round choices made by unpopular rows from (unpopular columns from resp.) are still columns from (rows from resp.). Further, the first sum equals
(k−1n)k⋅[1+kn⋅(k−1n)−1/(m+1)−kn]k−1
The second sum equals
(1−kn)n−k+1⋅[(1−k−1n)(1−kn)−1/(m+1)+k−1n]n−k=(1−kn)n−k+1(1+k(m+1)n+O(k2/n2)))n−k=(1−kn)n−k+1exp(km+1+O(k2/n)).
Therefore
Pnk ≤(k−1n)2k(1−kn)2(n−k+1)exp(km+1+O(kmm+1+k2/n)) =(kn)2kexp(−k2m+1m+1+O(kmm+1+k2/n)).
Consequently we have
(2.2) Enk=O(kn(nk)2(kn)2kexp(−k2m+1m+1+O(kmm+1+k2/n))=n−1exp(km+1+O(kmm+1+k2/n)).
In particular, for , , so that
(2.3) P(∃ a bad pair (K,L):k≤(m+1−ε)logn)=O(n−ε/(m+1)).
### 2.2. Case k∈[n1/2,(n+1)/2].
We write .
: first round choices of the rows from are among the columns in , and for every “unpopular” row in (i. e. a receiver of at most first-round proposals), its second round choice is still one of the columns from .
: first round choices of the columns from are among the rows in , and, for every column unpopular among the rows in in the first round, ’s second round choice—in case it is unpopular among rows in too—would still be a row in .
: overall, every column vertex from has taken part, as a proposer or a “proposee”, in at least two contacts with the row vertices from .
Clearly the second round choices of rows from are irrelevant for the events and . Let denote the (muti)graph, with labeled edges, induced by the two rounds of selections by the row set , and the first round selections by the column set . Let be the graph induced by the second round choices by the column set . Then is independent of , where is the number of first round selections of column by rows in , and the distribution of conditioned on is the same no matter what the marginal distribution of is. In the selection process is distributed multinomially, with independent trials, each having equally likely outcomes. The Poissonization device yields that , where and are independent copies of .
Introduce the probability measure defined on the space of triples by
Then for all . By switching to we gain independence of at the expense of the factor. In particular, . So we turn to upper-bounding . To this end, we claim first that
(2.4) P∗(B) =(1−t)n−k+1[1−f(t)pm(t)]n−k+1 pm(t) =m∑ℓ=0(1−t)ℓℓ!,t:=kn,f(t):=te−1+t.
The first factor is the probability that none of columns from selects a row from in the first round, and the second factor is the probability no column , such that would select a vertex in in the second round. Here we used the independence of under and
P∗(Xj≤m)=m∑ℓ=0e−1+t(1−t)ℓℓ!,t=kn.
So we need to estimate the probability of , conditioned on the event : every column in selects a row in , and—if it is unpopular among those rows—would select such a row again. Let stand for the full description of selections by rows from in both rounds, and by columns from in first round, compatible with .
Let us specify, in four items, a generic value of , being a partial description of : (1) let be the set of columns from whose first round choice rows are in ; (2) let be the set of the rows each selected by at least columns, and (3) so that is the set of unpopular rows in . Denote ; evidently , . For , let be the number of columns from which selected row . (4) To finish description of , for , let be the number of rows in whose first round selection is the column , and let be the number of unpopular rows, those from , whose second round selection is column . On the event , we have and .
Let , the event that column has at least two contacts with rows in . For , we have
(2.5) pj(bj,βj) =I{bj+βj≥2}+I{bj=1,βj=0}f(t)pm−1(t) +I{bj=0,βj=1}pm(t)f(t),f(t):=te−1+t,t=k/n.
Explanation. For , the first choice of column is a row in . Suppose . holds if , making unpopular and allowing a second round selection, that happens to be a row in , an event of probability
(e−1+tm−1∑ℓ=0(1−t)ℓ/ℓ!)⋅t=pm−1(t)f(t).
Suppose . This time holds if and, again, ’s second selection is a row in , an event of probability .
For , (’s first choice is a row in ), the counterpart of (2.5) is
(2.6) pj(bj,βj)=I{bj+βj≥1}+I{bj+βj=0}pm(t)f(t).
Conditioned on , the events are independent, so that
P∗(C∣∣B∩{T(S)=T})=P∗(k⋂j−1Fj∣∣B∩{T(S)=T})=k∏j=1pj(bj,βj).
The RHS is the explicit function of , and we need to compute its expected value to obtain . Denoting , , the resulting formula is
(2.7) Pnk=∑0≤u≤v≤k−1Pnk(u,v), ×∑∑jbj=k,∑jβj=k−u(k→b)n−k(k−u→β)n−k+u∏jpj(bj,βj).
(In the last product we can assume that .) Given and , the range of is non-empty only if .
Explanation. We (1) select columns from , and note that the probability that each of the remaining columns selects a row in is ; (2) partition the chosen columns into an ordered sequence of sets of cardinalities , with exactly s above , so that columns select row , with overall probability ; (3) partition rows in (the set of unpopular rows in , i.e. those chosen by at most columns resp.) into an ordered sequence of subsets of cardinalities ( resp.), so that each column is selected by rows in the first round (by unpopular rows in the second round resp.), with overall probability ; (4) add the contributions coming from each triple of partitions weighted with the factors .
To find a tractable upper bound for we will use the generating functions and the Chernoff-type bound: for non-negative sequence , and , we have , where ; analogous inequality holds for multivariate generating functions with non-negative coefficients. Needless to say, this approach is contingent on availability of an explicit formula for .
The bottom sum does not depend on . By symmetry, we have
(2.8) ∑∑iai=v|{i:ai>m}|=u(v→a)=v!(ku)∑a1,…,au>mau+1,…,ak≤m1→a! =v!(ku)[xv](∑a>mxaa!)u⋅(∑a≤mxaa!)k−u≤v!(ku)x−vexpum+1(x)⋅qk−um(x), exps(x):=∑τ≥sxτ/τ!,qs(x):=∑τ≤sxτ/τ!.
Similarly, by (2.5)-(2.6), the bottom sum in (2.7) equals
∑∑jbj=k,∑jβj=k−u(k→b)(k−u→β)∏jpj(bj,βj)=k!(k−u)![ykzk−u]∏j⎛⎝∑b,βybzβb!β!pj(b,β)⎞⎠.
Denoting , the last product equals
(∑b,βybzβb!β!(I{bj+βj≥2}+I{bj=1,βj=0}pm−1(t)f(t) +I{bj=0,βj=1}pm(t)f(t)))k−1−v ×⎛⎝∑b,βybzβb!β!(I{bj+βj≥1}+I{bj+βj=0}pm(t)f(t))⎞⎠v =(∑s≥2(y+z)ss!+ypm−1(t)f(t)+zpm(t)f(t))k−1−v ×(∑s≥1(y+z)ss!+pm(t)f(t))v =(exp1(η)+pm(t)f(t))v⋅(exp2(η)+ypm−1(t)f(t)+zpm(t)f(t))k−1−v.
Therefore we have
(2.9)
Combining (2.7), (2.8) and (2.9), and recalling the bound , we conclude that
(2.10) Pnk ≤(k!)2(1−t)k−1n2kykzkqkm(x)(exp2(η)+(ypm−1(t)+zpm(t))f(t))k−1 ×∑0≤u(m+1)≤v≤k−1(k−1)vu!ξuζv,ξ:=nzexpm+1(x)qm(x), ζ :=gnx(1−t),g=g(t;y,z):=exp1(η)+pm(t)f(t)exp2(η)+(ypm−1(t)+zpm(t))f(t).
Crucially, the sequence of the sums in (2.10) has a simple (exponential) generating function. Indeed if , then we have
∑k≥1wk−1(k−1)!⎛⎝∑0≤u(m+1)≤v≤k−1(k−1)vu!ξuζv⎞⎠ =∑0≤u(m+1)≤vξuζvu!∑k−1≥vwk−1(k−1)v(k−1)!=ew∑0≤u(m+1)≤vξuζvwvu! =ew∑u≥0(ξ(ζw)m+1)uu!∑j≥0(ζw)j=exp(w+ξ(ζw)m+1)1−ζw.
Therefore we obtain: for all , such that ,
(2.11) Pnk≤Qnk(R):=(k!)2(1−t)k−1(exp2(η)+(ypm−1(t)+zpm(t))f(t))k−1n2kykzk ×(k−1)!qkm(x)wk−1⋅exp(w+ξ(ζw)m+1)1−ζw;(η=y+z).
At the price of , yet to be chosen, we got rid of the multi-fold summation. Denoting , we have . The first line expression and the second line expression in (2.11) are respectively of orders
kgexp1(η)+pm(t)f(t)⋅(1−t)k(kne)2k(exp1(η)+pm(t)f(t)yzg)k; k1/2ρt(tqm(x)eρ)k⋅exp[n(ρ+zexpm+1(x)qm(x)⋅(gρ(1−t)x)m+1)]1−gρ(1−t)x,
uniformly for all admissible ; we used (2.10) for both expressions. So, denoting , (2.11) becomes
(2.12) Pnk= O⎛⎜ ⎜⎝k3/2ρgt(exp1(η)+pm(t)f(t))⋅exp(nHn,m(t;r))⎞⎟ ⎟⎠, Hn,m(t;r):= 2tlogte+tlog(1−t)+tlog(exp2(η)+(ypm−1(t)+zpm(t))f(t)yz) +tlog(tqm(x)eρ)+ρ+zexpm+1(x)qm(x)⋅(gρx(1−t))m+1 −n−1log(1−gρx(1−t)).
Recall that and is given by (2.4). Therefore
(2.13) Pnk=O(n1/2PnkP∗(B))=O(n1/2Q∗nk[(1−t)(1−pm(t)f(t))]n−k),
and, by (2.1),
(2.14) Enk=O(kn(nk)2Pnk).
Collecting the estimates (2.12)-(2.14), we arrive at
(2.15a) Enk= O((nk)1/2ρgt(exp1(η)+pm(t)f(t))⋅exp(nHn,m(t;r)), (2.15b) Hn,m(t;r):= −2t+(1−t)log1−pm(t)f(t)1−t+tlog(1−t) +tlog(exp2(η)+(ypm−1(t)+zpm(t))f(t)yz)+tlog(tqm(x)eρ) +ρ+zexpm+1(x)qm(x)⋅(gρx(1−t))m+1−n−1log(1−gρx(1−t)).
where the in equation (2.15b) is not the same as the in (2.12) because of the inclusion of terms from equations (2.1) and (2.4). We already proved (see (2.3)) that , if . So the task is to establish existence of the tuple for every such that is negative enough to out-power the front factor in (2.15a), so that the sum of over the remaining ’s will go to zero as well. It is beneficial to start earlier, with , i.e. with .
### 2.3. Small t’s.
Our focus is on , but for comparison we include here as well. Intuitively, for small ’s the search for the sub-optimal ought to be done by narrowing down the field of candidates. After some tinkering with we chose :
(2.16) x(t)=atσ,y(t)=b1tσ,z(t)=b2tσ,ρ(t)=ct,
with the parameters to be determined. Then, calculating upper bounds for the various terms in (2.15b),
−2t+(1−t)log1−pm(t)f(t)1−t+tlog(1−t)=−t(1+e−1pm(0))+O(t2); tlog(exp2(η)+(ypm−1(t)+zpm(t))f(t)yz)=tlog(b1+b2)2t2+O(t1+σ)b1b2t=tlog(b1+b)22b1b2+O(t1+σ); tlog(tqm(x)eρ)+ρ=tlog1+O(tσ)ec+ct=t(log1ec+c)+O(t1+σ); zexpm+1(x)qm(x)⋅(gρx(1−t))m+1=⎧⎪⎨⎪⎩2b2cb1+b2t+O(t1+σ),m=0,O(t2−σ),m≥1; 1−gρx(1−t)=1−(1+O(tσ))2ct1−2σa(b1+b2)=1−Θ(t1−2σ);
since is the argument of the -function in (2.15b), we need to choose . Combining the estimates we obtain:
Hn,m(t;r(t)) =−γm(b,c)t+O(t1+σ)+Θ(n−1t1−2σ), γm(b,c) −2b2cb1+b2⋅I(m=0).
It follows that is continuous, but not differentiable at . So depends on and only, while determines the behavior of the remainders. With some calculus it follows that, for , attains its maximum at , while for the maximum is attained at . Explicitly,
(2.17)
, and , exceeding by a factor, while . (In this regard the case is drastically different from the case .) Therefore
(2.18) Hn,m(t)=minrHn,m(t;r)≤−γmt+O(t1+σ)+Θ(n−1t1−2σ).
Consequently, for every , and sufficiently large, the function is negative for , where , and is chosen sufficiently small.
The front factor by in (2.15a) is of order . So it follows from (2.15a) and (2.18) that for
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# Semiconductor
Main Article
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This editable Main Article is under development and subject to a disclaimer.
A semiconductor is a substance with electrical conductivity intermediate between metals and insulators. Most commonly these materials are solids.
## Density of states
(CC) Image: John R. Brews
Calculated density of states for crystalline silicon.
In liquid and solid materials where atoms are in close proximity to one another, the energy levels available to electrons fall into bands separated by energy gaps. The density of energy levels per unit energy as a function of energy might look as in the figure. The occupancy of this density of states by electrons relates to a semiconducting behavior for this material as follows:
1. Bands of allowed energy are separated by an energy gap. Those above the gap are called conduction band energy levels, and those below are valence band energy levels.
2. At low temperatures, electrons occupy all the energy levels below the gap, and none above the gap.
3. The energy gap is small enough that at normal temperatures some electrons can acquire thermal energy sufficient to occupy a few conduction band energy levels above the gap, leaving vacant levels (holes) in the valence band energy levels below the gap.
4. The conduction band energy levels, and possibly the higher-energy hole energy levels too, correspond to states spatially extended through large regions of the material and are not localized. That spatial extension means that conduction-band electrons (and possibly valence-band holes too) can travel easily (conduct) by moving between adjacent levels in response to an applied field.
## Dopant impurities
(PD) Image: John R. Brews
Occupancy comparison between n-type, intrinsic and p-type semiconductors
The number of electrons and holes present in a semiconductor can be adjusted by introducing impurities into the semiconductor that alter the balance, allowing the hole and electron densities to differ. The added impurities that control the occupancy of energy levels are called dopants. The impurities may either increase the number of electrons in the conduction band (called donor impurities), or increase the number of holes in the valence band (called acceptor impurities). Semiconductors in which electrons dominate are called n-type, those where holes dominate conduction are called p-type, and those that do not have impurities and exhibit equal hole and electron concentrations are called intrinsic.
The figure shows the Fermi function on the left at a non-zero temperature for the intrinsic case (equal numbers of holes and electrons), the p-type case with holes, and the n-type case with electrons. The horizontal dashed lines indicate how the half-occupancy line is shifted by impurities. Impurities that act as donors, move the Fermi half-occupancy level closer to the conduction band edge, while impurities called acceptors move the Fermi half-occupancy level closer to the valence band.
In the case where the impurity substitutes for a host atom in the semiconductor lattice, the explanation as to why impurities behave as donors or as acceptors is based upon the valence of the substituting atom. The idea is that the substitute must try to emulate a host atom in the lattice by approximating its electron distribution. Thus, acceptor impurities have fewer valence electrons than the host atoms and must "borrow" an electron from the host for this masquerade, becoming negatively charged, while donor impurities have "extra" valence electrons, and must yield an electron to the host and become positively charged.
For example, if a column III impurity like B, Al, Ga or In, is added to silicon, these trivalent atoms grab an electron because they have only three electrons to bond with neighboring atoms, while a silicon atom has four. Similarly, addition of column V atoms, like P, As, Sb, to silicon, which have five electrons in their outer orbitals require only four for bonding, and donate the fifth electron.
This picture is complicated in a compound semiconductor like GaAs (a III-V semiconductor) that has more than one type of atom in the lattice. So, for example, a column VI atom like S, Se and Te can act as a donor if it occupies the site of the column V As host atom, and column II atoms like Be, Mg and Zn can act as acceptors if they replace Ga host atoms. On the other hand, a column IV atom like Si can act as either a donor or an acceptor depending upon whether Si replaces Ga or As.
Where an electron is borrowed, the occupancy of the valence band must be decreased. That means the Fermi occupancy level moves lower, closer to the valence band edge. Conversely, for donors, the occupancy of the conduction band must increase, so the Fermi occupancy level moves closer to the conduction band edge.
## Field effect
(CC) Image: John R. Brews
Field effect: Top panels: An applied voltage bends bands, depleting holes from surface (left). The charge inducing the bending is balanced by a layer of negative acceptor-ion charge (right). Bottom panel: A larger applied voltage further depletes holes but conduction band lowers enough in energy to populate an inversion layer.
In a metal the electron density that responds to applied fields is so large that an external electric field can penetrate only a very short distance into the material. However, in a semiconductor the lower density of electrons (and possibly holes) that can respond to an applied field is sufficiently small that the field can penetrate quite far into the material. This field penetration alters the conductivity of the semiconductor near its surface, and is called the field effect. The field effect underlies the operation of the Schottky diode and of field effect transistors, notably the MOSFET the JFET and the MESFET.[1]
The change is surface conductance is caused because the applied field alters the energy levels available to electrons to considerable depths from the surface, and that in turn changes the occupancy of the energy levels in the surface region. A typical treatment of such effects is based upon a band bending diagram showing the positions in energy of the band edges as a function of depth into the material.
An example band-bending diagram is shown in the figure for a two-layer structure consisting of an insulator as left-hand layer and a semiconductor as right-hand layer. An example of such a structure is the MOS capacitor, a two terminal structure made up of a metal gate contact, a semiconductor body (such as silicon) with a body contact, and an intervening insulating layer (such as silicon dioxide, hence the designation O). The left panels show the lowest energy level of the conduction band and the highest energy level of the valence band. These levels are "bent" by the application of a positive voltage V. By convention, the energy of electrons is shown, so a positive voltage penetrating the surface lowers the conduction edge. A dashed line depicts the occupancy situation: below this Fermi level the states are occupied and above it they are empty, assuming zero temperature. At operating temperatures, however, some electrons populate the conduction band, and some vacancies (holes) populate the valence band.
The example in the figure shows the Fermi occupancy level in the bulk material beyond the range of the applied field as lying close to the valence band edge. This position for the occupancy level is arranged by introducing impurities into the semiconductor. In this case the impurities are so-called acceptors which soak up electrons from the valence band becoming negatively charged, immobile ions embedded in the semiconductor material. The removed electrons are drawn from the valence band levels, leaving vacancies or holes in the valence band. Charge neutrality prevails in the field-free region because a negative acceptor ion creates a positive deficiency in the host material: a hole is the absence of an electron, it behaves like a positive charge. Where no field is present, neutrality is achieved because the negative acceptor ions exactly balance the positive holes.
Next the band bending is described. A positive charge is placed on the left face of the insulator (for example using a metal "gate" electrode). In the insulator there are no charges so the electric field is constant, leading to a linear change of voltage in this material. As a result, the insulator conduction and valence bands are therefore straight lines in the figure, separated by the large insulator energy gap.
In the semiconductor at the smaller voltage shown in the top panel, the positive charge placed on the left face of the insulator lowers the energy of the valence band edge. Consequently, these states are fully occupied out to a so-called depletion depth where the bulk occupancy reestablishes itself because the field cannot penetrate further. Because the valence band levels near the surface are fully occupied due to the lowering of these levels, only the immobile negative acceptor-ion charges are present near the surface, which becomes an electrically insulating region without holes (the depletion layer). Thus, field penetration is arrested when the exposed negative acceptor ion charge balances the positive charge placed on the insulator surface: the depletion layer adjusts its depth enough to make the net negative acceptor ion charge balance the positive charge on the gate.
The conduction band edge also is lowered, increasing electron occupancy of these states, but at low voltages this increase is not significant. At larger applied voltages, however, as in the bottom panel, the conduction band edge is lowered sufficiently to cause significant population of these levels in a narrow surface layer, called an inversion layer because the electrons are opposite in polarity to the holes originally populating the semiconductor. This onset of electron charge in the inversion layer becomes very significant at an applied threshold voltage, and once the applied voltage exceeds this value charge neutrality is achieved almost entirely by addition of electrons to the inversion layer rather than by an increase in acceptor ion charge by expansion of the depletion layer. Further field penetration into the semiconductor is arrested at this point, as the electron density increases exponentially with band-bending beyond the threshold voltage, effectively pinning the depletion layer depth at its value at threshold voltage.
## Notes
1. The acrynoms stand for Metal Oxide Semiconductor Field Effect Transistor, Junction Field Effect Transistor, and MEtal Semiconductor Field Effect Transistor. For a discussion see, for example, M K Achuthan K N Bhat (2007). “Chapter 10: Metal semiconductor contacts: Metal semiconductor and junction field effect transistors”, Fundamentals of semiconductor devices. Tata McGraw-Hill, pp. 475 ff. ISBN 007061220X.
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nLab Hopf fibration
bundles
Examples and Applications
Topology
topology
algebraic topology
Examples
Manifolds and cobordisms
manifolds and cobordisms
Contents
Idea
The Hopf fibration (named after Heinz Hopf) is a canonical nontrivial circle bundle over the 2-sphere whose total space is the 3-sphere.
$S^1 \hookrightarrow S^3 \to S^2$
Definition
Homotopy-theoretic characterization
The Eilenberg-MacLane space $K(\mathbb{Z},2) \simeq B S^1$ is the classifying space for circle group principal bundles. By its very nature, it has a single nontrivial homotopy group, the second, and this is isomorphic to the group of integers
$\pi_2(K(\mathbb{Z},2)) \simeq \mathbb{Z} \,.$
This means that there is, up to homotopy, a canonical (up to sign), continuous map from the 2-sphere
$\phi : S^2 \to K(\mathbb{Z},2) \,,$
such that $[\phi] \in \pi_2(K(\mathbb{Z},2)) = \pm 1 \in \mathbb{Z}$.
As any map into $K(\mathbb{Z},2)$ this classifies a circle group principal bundle over its domain. This is the Hopf fibration, fitting into the long fiber sequence
$\array{ S^1 &\hookrightarrow& S^3 \\ && \downarrow \\ && S^2 &\stackrel{\phi}{\to}& B S^1 \simeq K(\mathbb{Z},2) } \,.$
In other words, the Hopf fibration is the $U(1)$-bundle with unit first Chern class on $S^2$.
Explicit model
An explicit topological space presenting the Hopf fibration may be obtained as follows.
Identify
$S^3 \simeq \{(z_0, z_1) \in \mathbb{C}\times \mathbb{C} \,|\, {|z_0|}^2 + {|z_1|}^2 = 1\}$
and
$S^2 \simeq \mathbb{C P}^1 \simeq \mathbb{C} \sqcup \{\infty\}$
Then the continuous function $S^3 \to S^2$ defined by
$(z_0, z_1) \mapsto \frac{z_0}{z_1}$
gives the Hopf fibration. (Thus, the Hopf fibration is a circle bundle naturally associated with the canonical line bundle.) Alternatively, if we use
$S^2 \simeq \{(z, x) \in \mathbb{C} \times \mathbb{R} \,|\, {|z|}^2 + x^2 = 1\} \,.$
and identify this presentation of the 2-sphere with the complex projective line via stereographic projection, the Hopf fibration is identified with the map $S^3 \to S^2$ given by sending
$(z_0, z_1) \mapsto (2 z_0 z_1^* , {|z_0|}^2 - {|z_1|}^2).$
Variations
For each of the normed division algebras over $\mathbb{R}$,
$A = \mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O},$
there is a corresponding Hopf fibration of Hopf invariant one. The total space of the fibration is the space of pairs $(\alpha, \beta) \in A^2$ of unit norm: ${|\alpha|}^2 + {|\beta|}^2 = 1$. These gives spheres of dimension 1, 3, 7, and 15 respectively. The base space of the fibration is projective 1-space $\mathbb{P}^1(A)$, giving spheres of dimension 1, 2, 4, and 8, respectively. In each case, the Hopf fibration is a map
$S^{2^n - 1} \to S^{2^{n-1}}$
($n = 1, 2, 3, 4$) which sends the pair $(\alpha, \beta)$ to $\alpha/\beta$.
Applications
Magnetic monopoles
When line bundles are regarded as models for the topological structure underlying the electromagnetic field the Hopf fibration is often called “the magnetic monopole”. We may think of the $S^2$ homotopically as being the 3-dimensional Cartesian space with origin removed $\mathbb{R}^3 - \{0\}$ and think of this as being 3-dimensional physical space with a unit point magnetic charge at the origin removed. The corresponding electromagnetic field away from the origin is given by a connection on the corresponding Hopf fibration bundle.
K-theory
In complex K-theory, the Hopf fibration represents a class $H$ which generates the cohomology ring $K_U(S^2)$, and satisfying the relation $H^2 = 2 \cdot H - 1$, or $(H-1)^2 = 0$. (So in particular $H$ has an inverse $H^{-1} = 2- H$, see at Bott generator.)
A succinct formulation of Bott periodicity for complex K-theory is that for a space $X$ whose homotopy type is that of a CW-complex, we have
$K(S^2 \times X) \cong K(S^2) \otimes K(X)$
(It would be interesting to see whether this can be proved by internalizing the (classically easy) calculation for $K(S^2)$ to the topos of sheaves over $X$.)
The Hopf fibrations over other normed division algebras also figure in the more complicated case of real K-theory? $K_O$: they can be used to provide generators for the non-zero homotopy groups $\pi_n(B O)$ for the classifying space of the stable orthogonal group, which are periodic of period 8 (not coincidentally, 8 is the dimension of the largest normed division algebra $\mathbb{O}$). [To be followed up on.]
Revised on November 14, 2013 11:45:21 by Urs Schreiber (188.200.54.65)
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Jeremy Grifski
Posted on
# Fizz Buzz in Every Language
I noticed some folks on here were sharing some coding challenges, and I thought it might be fun to share some of my own (and shamelessly promote my Sample Programs project in the process).
## The Challenge
Fizz Buzz is that notorious interview question almost everyone in the industry knows. Ironically, I don't know of anyone who has every been asked it. Regardless, it's a fun problem to tackle because it involves a little bit of everything: loops, conditions, math, etc. And, there are countless ways to solve it.
For the purposes of this challenge, here are the rules:
Write a program that prints the numbers 1 to 100. However, for multiples of three, print “Fizz” instead of the number. Meanwhile, for multiples of five, print “Buzz” instead of the number. For numbers which are multiples of both three and five, print “FizzBuzz”
Here's a segment of the output:
``````1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
``````
If you're feeling festive, you can swap the terms for whatever you like (i.e. EasterEgg, AprilShowers, etc.). In any case, the solution should be executable in any language of your choice. The more obscure the language, the better!
## The Solution
I'll kick off the challenge with my favorite solution to the problem in Java:
``````public class FizzBuzz {
public static void main(String[] args) {
for (int i = 1; i < 101; i++) {
String output = "";
if (i % 3 == 0) {
output += "Fizz";
}
if (i % 5 == 0) {
output += "Buzz";
}
if (output.isEmpty()) {
output += i;
}
System.out.println(output);
}
}
}
``````
If you're curious how this solution works, Stuart Irwin wrote an excellent article about it. Along with this solution, I've been collecting several others in a repo called Sample Programs. After you drop your solution below, you should see if its missing from the collection. We'd love your contribution!
Stephanie Handsteiner
CSS
``````ol {
list-style-type: inside;
}
li:nth-child(3n), li:nth-child(5n) {
list-style-type: none;
}
li:nth-child(3n):before {
content: 'Fizz';
}
li:nth-child(5n):after {
content: 'Buzz';
}
``````
Jeremy Grifski
Haha I love that you can do this in CSS.
Isabella Muerte
There's several ways to do it in C++ :)
### The Classic
``````#include <cstdio>
int main () {
for (auto i = 1; i < 101; ++i) {
if (i % 3 == 0) { std::printf("Fizz"); }
if (i % 5 == 0) { std::printf("Buzz"); }
std::printf(" %d\n", i);
}
}
``````
### The "GOSH, MOM, IT'S GENERIC"
``````#include <cstdio>
struct number {
number (int x) : x(x) { }
auto operator * () const { return this->x; }
bool operator == (int y) { return this->x == y; }
bool operator != (int y) { return this->x != y; }
number& operator ++ () { ++this->x; return *this; }
number operator ++ (int) { return this->x++; }
int x;
};
struct range {
range (int start, int stop) :
start(start),
stop(stop)
{ }
auto begin () const { return number(this->start); }
auto end () const { return this->stop; }
private:
int start;
int stop;
};
int main () {
for (auto i : range(1, 101)) {
if (i % 3 == 0) { std::printf("Fizz"); }
if (i % 5 == 0) { std::printf("Buzz"); }
std::printf(" %d\n", i);
}
}
``````
### The "Calculated At Compile Time (stare into madness edition)"
``````#include <type_traits>
#include <utility>
#include <cstdio>
template <char... Args>
struct as_string { const char data[sizeof... (Args)] = { Args... }; };
using fizz = as_string<'f', 'i', 'z', 'z'>;
using buzz = as_string<'b', 'u', 'z', 'z'>;
using newline = as_string<'\n'>;
using null = as_string<'\0'>;
struct fizzbuzz: fizz, buzz { };
template <class T> struct type_identity { using type = T; };
template <class T>
constexpr auto digits (T x) {
auto digits = 0;
while (x) {
x /= 10;
digits++;
}
return digits;
}
template <auto Size, auto X, char... Args>
struct to_string : to_string<Size - 1, X / 10, '0' + X % 10, Args...> { };
template <auto X, char... Args>
struct to_string<1, X, Args...> : type_identity<as_string<'0' + X, Args...>> { };
template <unsigned int X>
using int_to_string = typename to_string<digits(X), X>::type;
template <auto X, class Fizzable = std::bool_constant<X % 3 == 0>, class Buzzable = std::bool_constant<X % 5 == 0>> struct check;
template <auto X> struct check<X, std::false_type, std::false_type> : int_to_string<X> { };
template <auto X> struct check<X, std::true_type, std::false_type> : fizz, int_to_string<X> { };
template <auto X> struct check<X, std::false_type, std::true_type> : buzz, int_to_string<X> { };
template <auto X> struct check<X, std::true_type, std::true_type> : fizzbuzz, int_to_string<X> { };
template <auto X> struct calc : check<X>, newline { };
template <auto... X> struct print : calc<X>..., null { };
template <auto... Is>
print<(Is + 1)...> deduce (std::index_sequence<Is...>);
template <auto N>
using result = decltype(deduce(std::make_index_sequence<N>()));
static constexpr result<100> fb{};
int main() {
std::puts(reinterpret_cast<const char *>(&fb));
return 0;
}
``````
nepeckman
I raise you the compile time calculated fizzbuzz in Nim ;)
``````proc fizzbuzz(): seq[string] =
result = @[]
for i in 1..100:
if i mod 15 == 0: "FizzBuzz"
elif i mod 5 == 0: "Buzz"
elif i mod 3 == 0: "Fizz"
else: \$i
)
const compileTimeValue = fizzbuzz()
echo compileTimeValue
``````
Isabella Muerte
C++20 isn't available yet, but we're a bit closer to this approach. :)
Jeremy Grifski
Interested in adding this to the collection? 😁
Jeremy Grifski
That last one scares me. Haha I’m not sure I’d know where to start reading it.
Thanks for jumping into the challenge with some great additions. 😃
Theofanis Despoudis
Casey Brooks • Edited
number 3 is the Cthulu of FizzBuzz
Terrance Lackie
These actually fail the FizzBuzz challenge because they print numbers that are divisible by 3 and/or 5. The challenge specifically states to print Fizz, Buzz, or FizzBuzz INSTEAD of the number.
Avalander • Edited
Here's my implementation in Racket
``````(define (mult-15? n)
(and (mult-5? n)
(mult-3? n)))
(define (mult-5? n)
(= (modulo n 5) 0))
(define (mult-3? n)
(= (modulo n 3) 0))
(define (fizzbuzz n)
(cond
[(mult-15? n) "FizzBuzz"]
[(mult-5? n) "Buzz"]
[(mult-3? n) "Fizz"]
[else n]))
(define (print-list xs)
(map displayln xs))
(print-list (map fizzbuzz (range 1 101)))
``````
The `print-list` function is a bit redundant, since `(map fizzbuzz (range 1 101))` will already print the resulting list to the console.
David Wickes
Ah come on @avalander - surely you should've written a FizzBuzz DSL in Racket? 😉
Avalander
I should, but I don't know enough Racket for that yet 😅
Jeremy Grifski
Great stuff! I like racket a lot, but I haven't written any code in it myself. The fact that there are so many dialects of it is pretty cool to me.
Avalander
Thanks! I've just started learning it myself. I can recommend the book Realm of Racket if you want to give it a shot.
Sql anybody?
``````DECLARE @i INT = 100;
;WITH E1(N) AS (SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1), -- 1*10^1 or 10 rows
E2(N) AS (SELECT 1 FROM E1 a, E1 b), -- 1*10^2 or 100 rows
E4(N) AS (SELECT 1 FROM E2 a, E2 b), -- 1*10^4 or 10,000 rows
E8(N) AS (SELECT 1 FROM E4 a, E4 b), -- 1*10^8 or 100,000,000 rows
cteTally(N) AS (SELECT TOP (@i) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E8)
SELECT CASE WHEN N % 15 = 0 THEN 'fizzbuzz'
WHEN N % 5 = 0 THEN 'buzz'
WHEN N % 3 = 0 THEN 'fizz'
ELSE CONVERT(VARCHAR(10),N) END
FROM cteTally
``````
`
Templar++
Because using a cursor is too mainstream? :)
Avalander
Since nobody has done Javascript yet, here's a crazy implementation.
``````const worder = (predicate, patterner) => (prev, n) =>
predicate(prev, n)
? patterner(prev, n)
: prev
const isDivisible = d => (_, n) =>
n % d === 0
const isEmpty = s =>
s.length === 0
const append = x => (prev) => prev + x
const setNumber = (_, n) => n
const fizzer = worder(isDivisible(3), append('fizz'))
const buzzer = worder(isDivisible(5), append('buzz'))
const numberer = worder(isEmpty, setNumber)
const reducer = (...worders) => n =>
worders.reduce(
(prev, w) => w(prev, n),
''
)
const fizzbuzzer = reducer(
fizzer,
buzzer,
numberer
)
for (let i = 0; i <= 100; i++) {
console.log(fizzbuzzer(i))
}
``````
Consider how easy it is to extend to print `'fazz'` for multiples of 7.
``````const fazzer = worder(isDivisible(7), append('fazz'))
const fizzbuzzfazzer = reducer(
fizzer,
buzzer,
fazzer,
numberer
)
``````
Jeremy Grifski
I appreciate the commitment to the obscure. Haha these are great.
Avalander
That's the whole point of the exercise, right? :D
Jeremy Grifski
Oh absolutely! Got any code golf solutions?
Avalander • Edited
Hmm... the best I can come up with right now is 85 chars. Nothing really clever, just sacrificed readability for space.
``````let i=0;while(i++<101){console.log(i%15==0?'fizzbuzz':i%5==0?'buzz':i%3==0?'fizz':i)}
``````
Another fun one, albeit longer, is this.
``````console.log(new Array(101)
.fill(1)
.map((_, i) =>
i % 15 == 0 ? 'fizzbuzz' :
i % 3 == 0 ? 'fizz' :
i % 5 == 0 ? 'buzz' :
i
)
.slice(1)
.join('\n'))
``````
Niels Bom
Elixir
``````1..100
|> Enum.map(fn
n when rem(n, 3) == 0 and rem(n, 5) == 0 -> "FizzBuzz"
n when rem(n, 3) == 0 -> "Fizz"
n when rem(n, 5) == 0 -> "Buzz"
n -> Integer.to_string(n)
end)
|> Enum.each(&IO.puts/1)
``````
Björn Grunde
Here is another example using as a Module and using pattern matching. It looks hilarious :P
``````defmodule FizzBuzz do
def run(num) when num >= 1 and num <= 100 do:
fizzbuzz(num, rem(n, 3), rem(n, 5))
run(num - 1)
end
def run(0), do: {:ok, "Done"}
def run(_) do: {:error, "Something went wrong"}
defp fizzbuzz(_, 0, 0), do: IO.puts "Fizzbuzz"
defp fizzbuzz(_, 0, _), do: IO.puts "Fizz"
defp fizzbuzz(_, _, 0), do: IO.puts "Buzz"
defp fizzbuzz(n, _, _), do: n |> to_string |> IO.puts
end
# Example
FizzBuzz.run(55)
``````
Jeremy Grifski
Ooh, this would make a nice addition to the repo.
Niels Bom
Alejandro Figueroa • Edited
Didn't see a C# solution, so here's mine (with some linq love):
``````using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace FizzBuzz
{
static class Program
{
static void Main(string[] args)
{
var buzzMap = new Dictionary<int, string>
{
{ 3, "Fizz" },
{ 5, "Buzz" },
};
Func<int, string> buzzer = i =>
{
var buzzed = buzzMap.Keys
.Where(k => i % k == 0)
.Select(k => buzzMap[k]);
return buzzed.Any() ? buzzed.Aggregate((prev, next) => prev += next) : null;
};
var output = Enumerable.Range(1, 100)
.Select(i => buzzer(i) ?? i.ToString())
.Aggregate(new StringBuilder(), (sb, s) => sb.AppendLine(s))
.ToString();
Console.WriteLine(output);
}
}
}
``````
David Wickes
Here's some fun from the world of Common Lisp
#### Inoffensive version
``````(defun divides (divisor dividend)
(= (mod divisor dividend) 0))
(defun fizzbuzz (n)
(let ((fizz (when (divides n 3) "Fizz"))
(buzz (when (divides n 5) "Buzz")))
(if (or fizz buzz)
(concatenate 'string fizz buzz)
(write-to-string n))))
``````
#### Offensive FizzBuzz Builder Macro
For when you want to define your own custom fizzbuzzer. Nice and easy to extend.
``````(defun divides (divisor dividend)
(= (mod divisor dividend) 0))
(defmacro define-fizzbuzzer (name &body pairs)
`(defun ,name (n)
(let ((result nil))
(dolist (p ',pairs)
(let ((test-passed? (if (typep (first p) 'integer)
(divides n (first p))
(funcall (eval (first p)) n))))
(when test-passed? (push (second p) result))))
(if (null result)
(write-to-string n)
(apply #'concatenate 'string (nreverse result))))))
(define-fizzbuzzer fizz-buzz
(3 "Fizz")
(5 "Buzz"))
(define-fizzbuzzer fizz-buzz-bazz
(3 "Fizz")
(5 "Buzz")
(7 "Bazz"))
``````
#### Most Offensive Macrogeddon Too Hot for TV Version
For when you want to define your own custom matcher logic for each word in your custom fizzbuzzer. Still lets you put a single number in for 'divides by' logic.
``````(defmacro define-fizzbuzzer (name &body pairs)
(let ((ps (clean-args pairs)))
`(defun ,name (n)
(let ((result nil))
(dolist (p ',ps)
(let ((test-passed? (funcall (eval (first p)) n)))
(when test-passed? (push (second p) result))))
(if (null result)
(write-to-string n)
(apply #'concatenate 'string (nreverse result)))))))
(defun clean-args (pairs)
(mapcar #'clean-pair pairs))
(defun clean-pair (pair)
(cond ((typep (first pair) 'integer)
(list '#'(lambda (n) (divides n (first pair))) (second pair)))
((eq (caar pair) 'function)
pair)
(t (error "first element of pair must be either an integer or a function"))))
(defun prime? (n)
"evaluates to t when n is prime. Highly inefficient"
(loop for x from 2 to (round n 2) never (divides n x)))
(defun has-a-nine-in-it? (n)
"does the decimal representation of this number have a 9 in it?"
(some #'(lambda (c) (char= c #\9)) (write-to-string n)))
(define-fizzbuzzer fizz-buzz
(3 "Fizz")
(5 "Buzz"))
(define-fizzbuzzer fazz-bazz
(#'prime? "Fazz")
(#'has-a-nine-in-it? "Bazz"))
``````
Jeremy Grifski
Bring on the parentheses!
John Neiberger
Here's a shot at an Erlang version, but I barely know Erlang.
``````
-module(fizzbuzz).
-export([start/0]).
fizzbuzz(X) when X rem 3 == 0, X rem 5 /= 0 ->
fizz;
fizzbuzz(X) when X rem 5 == 0, X rem 3 /= 0 ->
buzz;
fizzbuzz(X) when X rem 3 == 0, X rem 5 == 0 ->
fizzbuzz;
fizzbuzz(X) ->
X.
start() ->
Result = [fizzbuzz(X) || X <- lists:seq(1,100)],
Result.
``````
`````` print('\n'.join(['FizzBuzz' if x % 15 == 0 else 'Fizz' if x % 3 == 0 else 'Buzz' if x % 5 == 0 else str(x) for x in range(1,101)]))
``````
Jeremy Grifski • Edited
Is this a Python solution using a massive list comprehension?! I like it.
Cécile Lebleu
I know this must be very basic, but it's the first time I'm ever doing this.
Here's my implementation using JavaScript, printed into an HTML element :D
``````<p id="target"></p>
``````
``````let target = document.getElementById("target");
for(let i = 1; i <= 100; i++) {
if( i % 5 == 0 && i % 3 == 0 ) {
target.innerHTML += "FizzBuzz<br>";
} else if( i % 5 == 0 ) {
target.innerHTML += "Fizz<br>";
} else if( i % 3 == 0) {
target.innerHTML += "Buzz<br>";
} else {
target.innerHTML += i + "<br>";
}
}
``````
Aniket Bhattacharyea
Here's APL.
``````{(‘FizzBuzz’ ‘Fizz’ ‘Buzz’,⍵)[(0=15 3 5|⍵)⍳1]}¨⍳100
``````
To understand how it works, read this
Jeremy Grifski
Love it! Short and sweet. I haven't gotten around to learning any APL, and the repo shows it.
Aniket Bhattacharyea
Yep. I saw there wasn't an APL in the repo
Subramanian 😎
Here's a solution in Go.
``````package main
import (
"fmt"
)
func main() {
for num := 1; num < 101; num++ {
if num % 3 != 0 && num % 5 != 0 {
fmt.Println(num)
continue
}
if num % 3 == 0 {
fmt.Print("Fizz")
}
if num % 5 == 0 {
fmt.Print("Buzz")
}
fmt.Println()
}
}
``````
Christopher McClellan
Rather than post my own, I’m just going to link you to the fizzbuzz tag on Code Review. I swear I’ve reviewed this one in every language, and that includes LOLCODE.
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# Appendix: Uniform distribution as the optimum illumination
We will formally prove with the model presented in section 13.1.5 that for cells whose grid-line series resistance is negligible, when the local concentration distribution is assumed to vary slowly between the grid-lines, the distribution providing maximum cell efficiency is the uniform one.
Let us consider the cell model presented in section 13.1.5. In the trivial case in which the parameter rs is negligible (i. e. J(C, V)rs ^ VT for any C < CMAX and any V < Voc), any local concentration distribution f (C) produces the same efficiency (equation (13.11) does not depend on f (C)).
Assuming now that rs is not negligible, consider two concentration distributions with the same average concentration (C): one uniform distribution, i. e. fU(C) = S(C – (C)), where 8 denotes the Dirac-delta, and another nonuniform one, with probability density function fNU(C). From equations (13.5) and (13.11) we can write:
( V + JU(V)rs
Iu(V) = (C)IL, lsun-I0exPl——————– with /и = Ac/и (13.21)
, ,„T, {V f°° {Jwj(C, V)rs f
/nu (V) = (C) /L, lsun – /о exp I — 1 J exp I ————- ——— 1 /nu (C) dC
(13.22)
and:
/nu(V) ^ Jnu(C, V)fNU(C)dC = Ac < Jnu(V)). (13.23)
J0
We shall prove that for every value of V, ZU(V) > ZNU( V) and, thus, PU(V) = VIu(V ))VInu (V) = Pnu(V ).
From the properties of the exponential function, we can write
exp(x) > exp(x0)(1 + (x – X0)) (13.24)
for any value of x, and the equality is only fulfilled at x = x0. Let us apply this property to x = JNUrs/ VT and x0 = Jurs/ VT, we can obtain, first, from equation (13.22):
As the second factor is positive, we get Iu(V) > Inu(V).
Updated: August 23, 2015 — 3:17 am
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# How to write a ratio as a rate
The total number of mugs made per day goes in the numerator. When using decimals, your denominator should be 1: Be careful not to confuse the units of measurement.
The rule of thumb for a healthy acid test index is 1. We generally separate the two numbers in the ratio with a colon: When we express ratios in words, we use the word "to"--we say "the ratio of something to something else. Character Strengths and Positive Youth Development.
The risk difference in this study is 70 per vitamin users over ten years. We can just put a negative sign in front of the variable. From ancient scriptures to the latest science, gratitude is known to be good for us and those around us.
They are further sub-divided into 10 ratios as seen in the diagram below. These can be small e. Subjects who underwent incidental appendectomy were 4.
The pottery store can make coffee mugs in an 8 hour day. What is included in other assets. This action is easy to do yet its benefits have been scientifically proven.
Give the unknown quantity the name n. Which of the following would be the best interpretation of this risk ratio. Average Rate of Speed The average rate of speed for a trip is the total distance traveled divided by the total time of the trip.
Also, note that the operating income has dropped significantly in Children who excel with this lesson are ready from more complicated problems.
You can conclude that the per minute lap unit rate is 0. To complete this you must write a ratio. Again, turn into easier problem: Jack wants to compare the number of boys in his class to the number of girls.
Keep it up - try keeping it up for another couple of weeks at least. Consider an example from The Nurses' Health Study. It raises our overall satisfaction with life and helps us have an overall positive outlook.
By doing this we start to notice what goes right as well as wrong in our lives. How much would 10 pens cost. A ratio is a comparison of two numbers; a ratio of 5 to 2 also written 5: Why is this so. What is the rate amount per pizza.
When one of the four numbers in a proportion is unknown, cross products may be used to find the unknown number. Problems with ratios in them usually require good word problem-solving skills. IN!
IN+ OUT Product Folder Sample & Buy Technical Documents Tools & Software Support & Community LMK, LMKA, LM, LMA, LMK, LMKA, LM LM, LMA.
A rate is a special ratio in which the two terms are in different units. For example, if a ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. The first term of the ratio is measured in cents; the second term in ounces. You can write this rate as 69¢/12 ounces or 69¢ ounces.
The ratio can consequently be expressed as fractions or as a decimal. in decimals is A rate is a little bit different than the ratio, it is a special ratio. It is a comparison of measurements that have different units, like cents and grams.
A unit rate is a rate with a denominator of 1.
When a ratio ends up with units (or dimensions) on it, the ratio may also be referred to as a "rate". In the case of the exercise above, the rate was the distance covered per unit-volume of fuel. Conversion factors are simplified ratios, so they might be covered around the same time that you're studying ratios and proportions.
Definitions of the indicators. Adult literacy rate - Percentage of persons aged 15 and over who can read and write. Gross primary or secondary school enrolment ratio - The number of children enrolled in a level (primary or secondary), regardless of age, divided by the population of the age group that officially corresponds to the same level.
Net primary school enrolment ratio - The number of. Jun 01, · How to Calculate Inventory Turnover. In this Article: Article Summary Finding the Inventory Turnover Ratio Mastering the Equation Community Q&A Inventory turnover is a way of measuring how many times a business sells its stock of inventory in a given time period.
How to write a ratio as a rate
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Ratios and Proportions - Ratios - In Depth
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# 1538315 (number)
1,538,315 (one million five hundred thirty-eight thousand three hundred fifteen) is an odd seven-digits composite number following 1538314 and preceding 1538316. In scientific notation, it is written as 1.538315 × 106. The sum of its digits is 26. It has a total of 3 prime factors and 8 positive divisors. There are 1,225,792 positive integers (up to 1538315) that are relatively prime to 1538315.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 26
• Digital Root 8
## Name
Short name 1 million 538 thousand 315 one million five hundred thirty-eight thousand three hundred fifteen
## Notation
Scientific notation 1.538315 × 106 1.538315 × 106
## Prime Factorization of 1538315
Prime Factorization 5 × 359 × 857
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 1538315 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,538,315 is 5 × 359 × 857. Since it has a total of 3 prime factors, 1,538,315 is a composite number.
## Divisors of 1538315
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 1.85328e+06 Sum of all the positive divisors of n s(n) 314965 Sum of the proper positive divisors of n A(n) 231660 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1240.29 Returns the nth root of the product of n divisors H(n) 6.6404 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,538,315 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 1,538,315) is 1,853,280, the average is 231,660.
## Other Arithmetic Functions (n = 1538315)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1225792 Total number of positive integers not greater than n that are coprime to n λ(n) 153224 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 116577 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 1,225,792 positive integers (less than 1,538,315) that are coprime with 1,538,315. And there are approximately 116,577 prime numbers less than or equal to 1,538,315.
## Divisibility of 1538315
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 2 3 8
The number 1,538,315 is divisible by 5.
## Classification of 1538315
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (1538315)
Base System Value
2 Binary 101110111100100001011
3 Ternary 2220011011122
4 Quaternary 11313210023
5 Quinary 343211230
6 Senary 52545455
8 Octal 5674413
10 Decimal 1538315
12 Duodecimal 62228b
20 Vigesimal 9c5ff
36 Base36 wyyz
## Basic calculations (n = 1538315)
### Multiplication
n×y
n×2 3076630 4614945 6153260 7691575
### Division
n÷y
n÷2 769158 512772 384579 307663
### Exponentiation
ny
n2 2366413039225 3640288674435405875 5599910672214101388600625 8614426585727035377605170446875
### Nth Root
y√n
2√n 1240.29 115.438 35.2177 17.2746
## 1538315 as geometric shapes
### Circle
Diameter 3.07663e+06 9.66552e+06 7.43431e+12
### Sphere
Volume 1.52484e+19 2.97372e+13 9.66552e+06
### Square
Length = n
Perimeter 6.15326e+06 2.36641e+12 2.17551e+06
### Cube
Length = n
Surface area 1.41985e+13 3.64029e+18 2.66444e+06
### Equilateral Triangle
Length = n
Perimeter 4.61494e+06 1.02469e+12 1.33222e+06
### Triangular Pyramid
Length = n
Surface area 4.09875e+12 4.29012e+17 1.25603e+06
## Cryptographic Hash Functions
md5 fa415b9c4343528dffc8ad94f2824510 db44b50fa6a18810aa8320136a417c2944c6a828 e784c798046f7130dc1521f2fc7ec06380aca95ea893cb266d561acec324859e 48f6ddb34f3d786ec5af52393ce076dec79bffc4d5bc251a3eb0abae6f2cbdc42fdac7e8f92dd11fcca3c8bbc051a3b00b4ee3d86fc41bbba21c05726556934a 100a8a831ca7c74818bd427b49f02158bd6b871d
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http://www.soulphysics.org/2010/06/cpt-intuitive-approach/
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# CPT: The ‘Intuitive’ Approach
hdfc personal loan emi payment by credit card Khriplovich and Lamoreaux (1997, §2) suggest a very interesting argument that CPT provides the correct notion of “complete reversal” in physics.
payday loans without a lender The background assumption is that “complete reversal” should have effect of reversing the sign of 4-vectors in spacetime. David Malament, for example, has suggested that time reversal in classical electrodynamics should have this effect on enter timelike vectors. The proposal here is that “complete” motion reversal to have this effect on http://acf.ch/wp/?m=fast-auto-and-payday-loans-corporate-office all vectors (timelike, spacelike, and null).
payday loan union city ga Clearly, time reversal T on its own is not enough for this — it doesn’t reverse spacelike vectors. Parity reversal P isn’t either — it doesn’t reverse timelike vectors.
source url What about PT? After all, flipping about two axes is equivalent to a rotation. Shouldn’t that be enough to reverse all four vectors? As it turns out, it’s not enough, at least when it comes to 4-current ja. Since both P and T fix charge density and reverse current, we have:
PT ja = PT (p, http://cfpaldomoro.it/?m=quick-short-term-loans-sa j) = P (p, payday loans in augusta ga no credit check -j) = (p, go site j).
To reverse current, we need an operator C that sends particles to antiparticles, and thus sending ja to -ja. Thus, to get “total” motion reversal in a world with current, we need the CPT operator.
http://condadotravel.com/?q=borrower-paid-compensation-va-loans What I like about this thinking is that it depends crucially on the kind of matter fields in play. It’s only in the presence of 4-currents that PT is not enough to completely reverse motion. But similarly, the discovery of additional exotic matter fields might someday imply that CPT is not enough to reverse motion, either.
secured loan approval process student loan interest calculator nz Update: Wolfgang reports news about evidence for CPT-violation in a recent Fermilab experiment.
Soul Physics is authored by Bryan W. Roberts. Thanks for subscribing.
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## 8 thoughts on “CPT: The ‘Intuitive’ Approach”
1. wolfgang
I always wonder what is so special about the electromagnetic interaction that its charge shows up in such a fundamental operation like CPT.
(Why not change the coloring of quarks?)
2. Bryan
Actually I think the E&M charge isn’t special. All charges get reversed by C, including color. That’s why the operation is equivalent to exchanging particles with antiparticles.
3. Bryan
The quantum numbers that do seem special to me are mass and spin. Both are the charges corresponding to a conserved current. But neither are reversed by C. Why?
Interestingly, these are exactly the quantum numbers arising from the irreducible representations of the homogeneous Poincaré group. Does that have anything to do with it? I’m not sure…
4. wolfgang
>> All charges get reversed by C
you are right.
mass is the only special ‘charge’, but mass is obviously a special case…
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# My Java DP solution beats 93.83%
• ``````public boolean wordBreak(String s, Set<String> wordDict) {
int maxWord = getMax(wordDict);
int len = s.length();
boolean[] dp = new boolean[len + 1];
dp[0] = true;
for (int i = 1; i <= len; i ++) {
int start = Math.max(1, i - maxWord);
for (int j = start; j <= i; j++) {
if (dp[j - 1] && wordDict.contains(s.substring(j - 1, i))) {
dp[i] = true;
break;
}
}
}
return dp[len];
}
private int getMax(Set<String> wordDict) {
int max = 0;
for (String str : wordDict) {
max = Math.max(max, str.length());
}
return max;
}``````
• @biedengle2
Ah, you are trying to apply some heuristic to always pick try the longest word in the diction as the potential candidate.
However, for the time complexity wise, I think the time complexity still O(stringlen^2), is that true?
• @kun5 Not really, if the strings in wordDict is short (<r) then the running time is r*s.length()
• @kun5 unfortunately the time complexity is O(n^3) where n is the word length. because substring takes O(n)
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.
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All new and solved questions in Advanced Physics category
##### Gaussian Cube Electric Field
An electric field given by = 6.0· - 3.9(y2 + 3.5)· pierces a Gaussian cube of edge length 3.2 m and positioned as shown in Fig. 23-56. (The magnitude is in newtonsper coulomb and the position x is in meters.) What net charge (in Coulombs) is enclosed by the Gaussian cube?...
##### A nonconducting spherical shell
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly through its volume between its inner andouter surfaces. The volume charge density ? is the charge per unit volume, with the unit coulomb per cubic meter. For this shell ? = b/r w...
##### Problem 4
An archer shoots an arrow with a velocity of 30 m/s at an angle of 20 degrees with respect to the horizontal. An assistant standing on the level ground 30 m downrangefrom the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the ...
##### Homework 4, Item 1
A road race is taking place along the track shown in the figure (Figure 1) . All of the cars are moving at constant speeds. The car at point F is traveling along astraight section of the track, whereas all the other cars are moving along curved segments of the track. Let be the velocity of the car a...
##### A charge of +1nC and a dipole with charges +9nC and -9nC separated by .3mm contribute to a net...
A charge of +1nC and a dipole with charges +9nC and -9nC separated by .3mm contribute to a net electric field at location A. A -3 nC charge is placed, at rest, atpoint A. Determine what external electric field is needed in order to keep the -3 nC charge in place. (Note: there is a q+ = +9nC, and q- ...
##### Electric field on the axis of a washer
A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 2 cm. The washer has a uniform chargedistribution and a total charge of 8 nC. What is the electric field on the axis of the washer at a distance of 32 cm from the center of the washer?(Enter the...
##### Help
Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly hump") and had it installed. Suppose a 1800-kg car passes over a hump in a roadway that follows the arc of a circle of radius 19.6 m (a) If the car travels at 25.0 k...
##### Carnot Engine
A Carnot engine receives 250 kJ s-1 of heat from a heat-source reservoir at 525?C and rejects heat to a heat-sink reservoir at 50?C. What are the power developed andthe heat rejected?...
##### The image distance in this case is...?
A lens of focal length 20 cm has an object of height 3.0 cm placed 100 cm in front of it. The imagedistance in this case isA) 17 cm. B) 33 cm. C) -25 cm. D) 20 cm. E) 25 cm.Please show work!...
##### HW 6 Item 13
A loudspeaker of mass 22.0{rm kg} is suspended a distance of h = 2.00{rm m} below the ceiling by two cables that make equal angles with the ceiling. Each cable has alength of l = 3.10{rm m} . What is the tension T in each of the cables?...
##### Help
One end of a cord is fixed and a small 0.250-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.50 m as shown in thefigure below. When ? = 19.0°, the speed of the object is 7.10 m/s. At this instant, find each of the following. (a) the tension i...
##### Lens Image. Will Rate for answer with work and explanation.
An object is placed 100 cm in front of a lens of focal length 50 cm. A lens of focal length -20 cm isplaced 90 cm beyond the first lens. The final image is locatedA) 10 cm past the second lens.B) 20 cm in front of the second lens.C) 20 cm past the second lens.D) 10 cm in front of the second lens.E) ...
##### Central Force Problem
A central force potential frequently encountered in nuclear physics is the rectangular well, defined by the potentialV=0 r>a=-Vo r<aShow that the scattering produced by such a potential on classical mechanics is identical with the refraction of light rays by a sphere od radius a and relativein...
##### Tension question
An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at themiddle of the rope (Figure 1) . The rope will break if the tension in it exceeds 2.00×104 , and our hero's mass is 83.4 . If the ang...
##### Electricity and Magnetism Question
A charge q at the origin is at the back corner of an imaginary cube of edge length a. What is the electric flux through the shaded region? Use direct integration...
##### Capacitors
1.)In the figure , each capacitance is 6.3 , and each capacitance is 4.2 . (a)Compute the equivalent capacitance of the network between pointsa and b. (b)Compute the charge on the three capacitors nearest a and b when = 410 .(Q1and Q2) (c)With 410 across a and b, compute ....
##### Introduction to Mechanics 5.7 by Kleppner and Kolenkow
When the flattening of the Earth at the poles is taken into account, it is found that the gravitational potential energy of a mass m a distance r from the centerof the earth is approximately U = -GMe/r[1-5.4 x 10-4(Re/r)2(3cos2θ-1)],where θ is measured from the pole.Show that there is a ...
##### Gauge transformation , electrodynamics
for a constant uniform electric and magnetic fields E=E0 and B=B0, it is possible to choose a gauge such that the scalar potential φ andvector potential A are given by1 φ=0 , A=1/2(B0xr)2 φ=-E0.r , A=1/2(B0xr)3 φ=-E0.r, A=04 φ=0 ,A=-E0tplease explain the correct answer,here φ...
##### 5.15 - True/False
In the picture below, the two wires carry current i1 and i2, respectively, with positive current to the right. The charge q ispositive and has velocity v to the right.If i1 = - i2, then the wires repel each other.If v = 0, then the force on q is zero.If i2 = - i1 then the force on q equals zero.If i...
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Algebra Tutorials!
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Registered: 23.04.2003
From: Girona, Catalunya (Spain)
Posted: Friday 05th of Jan 15:23 Is that so? It is easy to purchase this program. What’s more , you have nothing to lose. The program comes with a money-back assurance . It is available here t: http://www.rational-equations.com/linear-equations-in-two-variables-1.html. I am confident that you will simply enjoy it. Let me know if there is something more that you would like to know from me.
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https://www.176iot.com/2-switches-one-light-wiring-diagram/
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# 2 Switches One Light Wiring Diagram
When wiring two switches to operate one light, a two way switch circuit is created. This type of circuit uses two three-way switches to control the operation of one light from two separate locations. A two way switch circuit allows you to turn on and off a light from either end of the circuit. This type of wiring is sometimes referred to as a “traveler” system because the electricity can travel either way up or down the circuit.
The primary components in a two switch light wiring diagram include two three-way switches, one neutral wire, one hot wire, and one ground wire. Two three-way switches are used to control the operation of one light from two separate locations. The neutral wire connects the hot wire to the light, while the hot wire is connected to one of the three-way switches. The ground wire is connected to both switches and can provide a secure electrical connection between the switches and the light.
In order to connect the two switches to the light, the two three way switches must be wired together and then the hot wire from the light should be connected to the common terminal of one of the switches. Then, the neutral wire should be connected to the load terminal of one of the switches and finally the ground wire should be connected to the ground terminal of both switches. Once all of these connections have been made, the light should be powered when either switch is activated. Two switches one light wiring diagrams can be found online or in wiring books for further guidance.
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http://www.bath.ac.uk/catalogues/2019-2020/ar/AR10313.html
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Student Records
Programme & Unit Catalogues
## AR10313: Structures 1A
Owning Department/School: Department of Architecture & Civil Engineering
Credits: 6 [equivalent to 12 CATS credits]
Notional Study Hours: 120
Level: Certificate (FHEQ level 4)
Period:
Semester 1
Assessment Summary: CW 25%, EX 75%
Assessment Detail:
• Class Test (CW 10%)
• Truss Design Coursework (CW 15%)
• Examination (EX 75% - Qualifying Mark: 40)
Supplementary Assessment:
AR10313 Reassessment (where allowed by programme regulations)
Requisites:
Description: Aims:
To make students aware of the role played by structure in the design and building process. To introduce the concepts of statics and load carrying mechanisms, sufficient for an elementary appraisal of structures. To familiarise students with different types of structural materials and assemblies.
Learning Outcomes:
On successful completion of the unit, the student will be able to design a simple structure and identify and calculate the forces within it.
Skills:
An understanding of Statics and an ability to apply the principles in the context of a design problem. An ability to analyse statically determinate structures and to estimate appropriate member sizes for permissible stress states.
Content:
Stable structures and structural mechanisms. Newton's laws; static equilibrium and free body diagrams. The concepts of forces and moments in structural members. Equilibrium of loads, forces and moments in simple structures. Introduction to load carrying action of trusses, beams, arches, cables and columns. The concepts of stress, section sizes and shapes. Pin-jointed trusses: triangles of forces, resolving at joints and method of sections; physical behaviour and structural form and efficiency. Direct stresses and strains; Young's Modulus. Beams and free body diagrams, bending moments and shear forces. Bending stresses in beams, section shape and structural efficiency; web action and the concept of shear stresses. Overall efficiency of beams and simple bridges. Stability concepts. Hanging chains and funicular shapes; simple suspension systems. Three pin arches and goal-post portal frames. The above topics concentrate on a broad overview of structural concepts and will be supported by laboratory demonstrations and tutorial classes emphasising the relation between structural and architectural concepts, structural safety and examples of structural failures.
Programme availability:
#### AR10313 is Compulsory on the following programmes:
Department of Architecture & Civil Engineering
• UEAR-ANB08 : BSc(Hons) Architecture with Thin sandwich placement(s) (Year 1)
Notes: This unit catalogue is applicable for the 2019/20 academic year only. Students continuing their studies into 2020/21 and beyond should not assume that this unit will be available in future years in the format displayed here for 2019/20. Programmes and units are subject to change in accordance with normal University procedures. Availability of units will be subject to constraints such as staff availability, minimum and maximum group sizes, and timetabling factors as well as a student's ability to meet any pre-requisite rules. Undergraduates: Find out more about these and other important University terms and conditions here. Postgraduates: Find out more about these and other important University terms and conditions here.
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https://ask.sagemath.org/questions/23967/revisions/
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# Revision history [back]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3') sage: eq1=A3(P1-P3)==A3r3u3^2-(A2r2u2^2+A1r1u1^2) sage: eq5a=solve([eq2],r3) sage: eq5b=u3^2+2g/(g-1)(P3/(eq5a[0].rhs()))-2C==0 sage: eq6=solve([eq5b],u3) sage: eq7a=eq1/A3 sage: eq7b=eq7a.subs(-A3r3u3^2 == -(A2u2r2+A1r1u1)u3) sage: eq7b P1 - P3 == -(A1r1u1^2 + A2r2u2^2 + (A3P3g + sqrt(A3^2P3^2g^2 + 2((g^2 - 2g + 1)A1^2r1^2u1^2 + 2(g^2 - 2g + 1)A1A2r1r2u1u2 + (g^2 - 2g + 1)A2^2r2^2u2^2)C))(A1r1u1 + A2r2u2)/(A1(g - 1)r1u1 + A2(g - 1)r2u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs()) #this equation should have multiple answers+_, but it does not have multiple sage: eq7=solve([eq7c],P3)
sage: eq7 [P3 == (A1r1u1^2 + A2r2u2^2 + A3P1 - (A1r1u1^2 + A2r2u2^2 + A3P1)g - sqrt(A3^2P3^2g^2 + 2((g^2 - 2g + 1)A1^2r1^2u1^2 + 2(g^2 - 2g + 1)A1A2r1r2u1u2 + (g^2 - 2g + 1)A2^2r2^2u2^2)*C))/A3]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3(P1-P3)==A3r3u3^2-(A2r2u2^2+A1r1u1^2)
eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2g/(g-1)(P3/(eq5a[0].rhs()))-2C==0
eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3r3u3^2 eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2u2r2+A1r1u1)u3)
-(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1r1u1^2 -(A1*r1*u1^2 + A2r2u2^2 A2*r2*u2^2 + (A3P3g (A3*P3*g + sqrt(A3^2P3^2g^2 sqrt(A3^2*P3^2*g^2 + 2((g^2 2*((g^2 - 2g 2*g + 1)A1^2r1^2u1^2 1)*A1^2*r1^2*u1^2 + 2(g^2 2*(g^2 - 2g 2*g + 1)A1A2r1r2u1u2 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2g 2*g + 1)A2^2r2^2u2^2)C))(A1r1u1 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2r2u2)/(A1(g A2*r2*u2)/(A1*(g - 1)r1u1 1)*r1*u1 + A2(g A2*(g - 1)r2u2))/A31)*r2*u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs()) #this equation should have multiple answers+_, but it does not have multiple
sage: eq7=solve([eq7c],P3)eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1r1u1^2 (A1*r1*u1^2 + A2r2u2^2 A2*r2*u2^2 + A3P1 A3*P1 - (A1r1u1^2 (A1*r1*u1^2 + A2r2u2^2 A2*r2*u2^2 + A3P1)g A3*P1)*g - sqrt(A3^2P3^2g^2 sqrt(A3^2***P3**^2*g^2 + 2((g^2 2*((g^2 - 2g 2*g + 1)A1^2r1^2u1^2 1)*A1^2*r1^2*u1^2 + 2(g^2 2*(g^2 - 2g 2*g + 1)A1A2r1r2u1u2 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2g 2*g + 1)A2^2r2^2u2^2)*C))/A3]1)*A2^2*r2^2*u2^2)*C))/A3]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1*r1*u1^2 + A2*r2*u2^2 + (A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2*r2*u2)/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs()) #this equation should have multiple answers+_, but it does not have multiple
sage: eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1 - (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1)*g - sqrt(A3^2***P3**^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/A3]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1*r1*u1^2 + A2*r2*u2^2 + (A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2*r2*u2)/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs())
sage: eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1 - (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1)*g - sqrt(A3^2***P3**^2*g^2 sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/A3]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side?side (see last equation after the solve in the beginning of sqrt?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1*r1*u1^2 + A2*r2*u2^2 + (A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2*r2*u2)/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs())
sage: eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1 - (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1)*g - sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/A3]
### Solve for variable but variable is still in answer
I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side (see last equation after the solve in the beginning of sqrt?
sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq2=A3*u3*r3==A2*u2*r2+A1*r1*u1
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1*r1*u1^2 + A2*r2*u2^2 + (A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2*r2*u2)/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2))/A3
sage: eq7c=eq7b.subs(u3==eq6[0].rhs())
sage: eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1 - (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1)*g - sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/A3]
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### Servo How To
A servo is a motor that can rotate in 180 degrees by sending a pulse every 20 microseconds. For a detailed description of servos, check out the servo tutorial put together by the Seattle Robotics Club. There are three wires on a servo the most common colors are: black, red, and yellow. To hook it up to Arduino, attach the black wire to the ground (GND or negative), the red wire into +6V (or +5V, the positive voltage), and the yellow wire to a pin with PWM (Pulse Width Modulation) with a 220 Ohm resistor. Sometimes there is a white wire instead of a yellow wire.
It is important to note that this servo works on digital signals (on vs. off) rather than analog which is why it has to be in PWM.
This is a basic example of arduino programming (check this for more information on how servos work with arduino) that allows the servo to move from 50 degrees to 120 degrees and back:
int servoPin = 3; //Connects the yellow wire to pin 3 int myAngle; //This sets the variable for your servo angleint pulseWidth; // servoPulse function variablevoid setup (){ pinMode(servoPin, OUTPUT); //sets pin 3 as output}void servoPulse(int servoPin, int myAngle){ pulseWidth = (myAngle * 10) + 600; //This Procedure translates myAngle to the correct delay in order to make digitalWrite(servoPin, HIGH); //the servo move to the correct set angle. delayMicroseconds(pulseWidth); digitalWrite(servoPin, LOW);}void loop(){ for (myAngle=50; myAngle<=120; myAngle++) //The angle starts at 50 degrees and increases by increments //of one until it reaches 120 { servoPulse(servoPin, myAngle); //Calls the servoPulse Procedure and sets pin and angle delay(20); //refreshes cycle } for (myAngle=120; myAngle>=50; myAngle--) //The angle starts at 120 degrees and decreases by increments //of one until it reaches 50 { servoPulse(servoPin, myAngle); //sets pin and angle delay(20); //refreshes cycle }}// Code sample adapted from the Arduino Notebook.
USING A PUSH BUTTON TO INITIALIZE SERVO ROTATION
Essentially, this uses a lot of the same programing as the above example. However, you have to incorporate an "if...else" loop. In the following example, the servo is programed to return to its initial position after the push button is released.
int servoPin = 3;int myAngle;int pulseWidth;int button = 2; //sets the button to pin2int val = 0; //sets the value to 0void setup (){ pinMode(servoPin, OUTPUT); pinMode(button, INPUT); //sets the push button as an input Serial.begin (9600); //sets serial to 9600 bits per second}void servoPulse(int servoPin, int myAngle){ pulseWidth = (myAngle * 10) + 600; digitalWrite(servoPin, HIGH); delayMicroseconds(pulseWidth); digitalWrite(servoPin, LOW);}void loop(){ val = digitalRead (button); //sets the value to the state of the button if (val==HIGH){ //if the push button is pressed proceed with the for loop for (myAngle = 30; myAngle <= 150; myAngle++) { val = digitalRead (button); //Reads the button between every angle change if (val==LOW) //if the button is released, the angle goes back to 30 { myAngle = 30; } servoPulse (servoPin, myAngle); delay (20); } for (myAngle = 150; myAngle >=30; myAngle--) //If the button is not released, the loop returns to 30 and continues { servoPulse (servoPin, myAngle); delay (20); } }}
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# Degrees of Freedom Formula
It is the number of values that remains during the final calculation of a statistic that is expected to vary. In simple terms, these are the date used in a calculation. The degrees of freedom can be calculated to help ensure the statistical validity of chi-square tests, t-tests, and even the more advanced f-tests. Degrees of freedom is commonly abbreviated as ‘df’. Below mentioned is a list of formulas. The number of degrees of freedom refers to the number of independent observations in a sample minus the number of population parameters that must be estimated from sample data.
#### One Sample T Test Formula
$\LARGE DF=n-1$
#### Two Sample T Test Formula
$\LARGE DF=n_{1}+n_{2}-2$
#### Simple Linear Regression Formula
$\LARGE DF=n-2$
#### Chi Square Goodness of Fit Test Formula
$\LARGE DF=k-1$
#### Chi Square Test for Homogeneity Formula
$\LARGE DF=(r-1)(c-1)$
### Solved Examples
Question 1: Find the degree of freedom for given sequence:
x = 2, 8, 3, 6, 4, 2, 9, 5
Solution:
Given n= 8
$therefore$ DF = n-1
DF = 8-1
DF = 7
Question 2: Find the degree of freedom for given sequence:
x = 12, 17, 19, 15, 25, 26
y = 18, 21, 32, 43
Solution:
Given:
n1 = 6
n2 = 4
Here, there are 2 sequences, so we need to apply
DF = n1 – n2 – 2
DF = 6 -4 -2
DF =0
Related Formulas Radical Formula Sphere Formula Perimeter of a Trapezoid Formula Sample Mean Formula Function Formulas Sampling Error Formula Riemann Sum Formula Magnitude of a Vector Formula
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## Price to Cash Flow Ratio
Price to cash flow (P/CF) is the profitability ratio which is used to compare the price of the company to the underlying cash flow. This is the valuation ratio which uses for the indication of the worth of the company base on the cash flow generated by the company.
## Definition: What is Price to Cash Flow Ratio?
In the company or firm P/CF ratio adjust for all the noncash item and it provides the underlying cash picture generated by the business. Price to Cashflow ratio compares the company’s cash flow to its market value to demonstrate if the valuation justified or not.
If the company has a low P to CF ratio then the firm’s potential will be an undervaluation. Now, look that how to calculate the price to cash flow ratio formula.
## Formula
The price to cash flow ratio can be calculated by dividing the price per share by the operating cash flow per share.
#### Price to cash flow ratio = Price per share/Operating cash flow per share.
For the calculation of price to cash flow ratio formula, we can use the market cap in the formula as
#### Price to Cashflow Ratio = Marketing Cash/Operating cashflow
In the annual report of the cash flow statement operating cash flow is mentioned from which we can take and put the value in the above formula
OCF can be calculated by the following formula as
OCF = Net income + Depreciation + Amortization + Change in WC + other non cash item
The price which the share of stock is traded in the open market is called Market Cap.
## Example
At the end of the year company Z financial statement taken which has the number in the following table. In the year 2 and year, the operating cash flow of the company increase from 7 to 9 but in these 2 years price to cash flow ratio not change which is 1.7x. P/CF ratio unchanged means the increase in the share by the similar proportion.
It means that the investors of company A pay 2 dollars for every 1 dollar cash flow in the year 1. In the year 3 investors not pay 2 dollars more but pat 1.7 dollars for every 1 dollar cash flow.
## Analysis and Interpretation
In the investment industry price to cash ratio is the most important multiple. The analyst needs this ratio to find the valuation of business with respect to cash, the business generates from the underlying operations. From this ratio, we can compare different companies in the same industry.
The analyst needs to examine the short term cash position of the company. They ensure that the management did not set boost adjustment for short term position.
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Estimation of CARA Preferences and Positive Mathematical Programming
Open Journal of Statistics
Vol.08 No.01(2018), Article ID:82011,13 pages
10.4236/ojs.2018.81001
Estimation of CARA Preferences and Positive Mathematical Programming
Quirino Paris
Department of Agricultural and Resource Economics, University of California, Davis, CA, USA
Received: December 6, 2017; Accepted: January 23, 2018; Published: January 26, 2018
ABSTRACT
The purpose of this paper is to combine the estimation of output price risk and positive mathematical programming (PMP). It reconciles the risk programming presented by Freund with a consistent estimate of the constant absolute risk aversion (CARA) coefficient. It extends the PMP approach to calibration of realized production outputs and observed input prices. The results of this specification include 1) uniqueness of the calibrating solution, 2) elimination of the tautological calibration constraints typical of the original PMP procedure, 3) equivalence between a phase I calibrating solution and a solution obtained by combining phase I and phase II of the traditional PMP procedure. In this extended PMP framework, the cost function specification involves output quantities and input prices―contrary to the myopic cost function of the traditional PMP approach. This extension allows for a phase III calibrating model that replaces the usual linear technology with relations corresponding to Shephard lemma (in the primal constraints) and the marginal cost function (in the dual constraints). An empirical example with a sample of farms producing four crops illustrates the novel procedure.
Keywords:
CARA Coefficient, Chance-Constrained Approach, Positive Mathematical Programming, Solution Uniqueness, Calibrating Model
1. Introduction
The treatment of risk in a mathematical programming setting has interested researchers for several decades. It all began with Markowitz [1] who presented the problem of portfolio selection in a mean-variance framework. Freund [2] discussed a quadratic programming approach to deal with output price risk in a mean-variance specification of revenue. Hazell [3] followed with a linear programming minimization of total absolute deviation (MOTAD) of income. Hazell justified his proposal by citing the difficult access―at that time―to a quadratic programming software necessary to solve the mean-variance model.
When dealing with risk, the major issue involves the decision of how to characterize the risk preferences of an economic agent. Pratt [4] proposed a general way to characterize absolute risk aversion―known as the Arrow-Pratt measure of risk―which is defined as the negative ratio of the second derivative to the first derivative of a utility function of wealth. The utility function of an economic agent exhibits decreasing, constant or increasing risk aversion if the Arrow-Pratt risk aversion is decreasing, constant or increasing as a function of wealth. Very often, economists have chosen a negative exponential utility function of wealth $U\left(w\right)=1-\mathrm{exp}\left[-\varphi w\right]$ that exhibits a constant absolute risk aversion (CARA) coefficient $\varphi >0$ . This is the utility function selected also by Freund to represents North Carolina farmers’ preferences. It remains to decide how to estimate the CARA parameter. Freund wrote (Source: [2] , 258). “The estimation of the risk aversion constant $\varphi$ is a purely subjective task, and any chosen value is exceedingly difficult to defend.” Fortunately, the task of estimating $\varphi$ can be made defensible by adopting a chance-constrained approach as presented in section 3.
The objective of this paper, therefore, is twofold: 1) to estimate the CARA parameter in an empirical way that is consistent with the available sample information; 2) to combine the CARA risk analysis with a positive mathematical programming (PMP) approach that uses all the available information.
2. Freund Risk Programming
Freund [2] assumed a $\left(J×1\right)$ random vector of market output prices, $\stackrel{˜}{p}$ , distributed as a normal random vector variable $\stackrel{˜}{p}\sim N\left[E\left(\stackrel{˜}{p}\right),{\Sigma }_{p}\right]$ . He assumed that farmers’ preferences toward risk were characterized by a negative exponential utility function $U\left(\stackrel{˜}{r}\right)=1-\mathrm{exp}\left[-\varphi \stackrel{˜}{r}\right]$ , with CARA coefficient $\varphi >0$ and
random revenue $\stackrel{˜}{r}$ . Finally, Freund assumed that farmers made decisions by maximizing their expected utility subject to a non-random linear technology A and a known quantity $\left(I×1\right)$ vector of limiting inputs $b$ . Given these assumptions, expected utility corresponds to the following integral
$\begin{array}{c}EU\left(\stackrel{˜}{r}\right)=1-\mathrm{exp}\left[-\varphi \left\{E\left(\stackrel{˜}{r}\right)-\frac{\varphi }{2}\mathrm{var}\left(\stackrel{˜}{r}\right)\right\}\right]\\ =1-\mathrm{exp}\left[-\varphi \left\{E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x\right\}\right]\end{array}$ (1)
where $x\ge 0$ is a $\left(J×1\right)$ vector of decision variables, $E{\left(\stackrel{˜}{p}\right)}^{\prime }x$ is expected
revenue, $\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x$ is the risk premium and $\left(E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x\right)$ is the cer-
tainty equivalent (CE) of the risky prospect. Maximization of the certainty equivalent corresponds to the maximization of the expected utility in Equation (1). Therefore, primal and dual specifications of farmer’s risk programming under this CARA model are stated as
Primal $\mathrm{max}CE=E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x$ (2)
subject to $\text{Demand}\le \text{Supply}$
$Ax\le b$ (3)
Dual $\mathrm{min}TC={b}^{\prime }y+\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x$ (4)
subject to $MC\ge MR$
${A}^{\prime }y+\varphi {\Sigma }_{p}x\ge E\left(\stackrel{˜}{p}\right)$ (5)
where TC is total cost, MC is marginal cost, MR is marginal revenue, $y\ge 0$ is the $\left(I×1\right)$ vector of input shadow prices, the primal constraints represent the technological relations between limiting input and output levels, while the dual constraints express the equilibrium relations between marginal cost and marginal revenue of producing and selling outputs. Marginal cost has two components: ${A}^{\prime }y$ is the marginal cost associated with the production technology and fixed limiting inputs; $\varphi {\Sigma }_{p}x$ is the marginal risk premium, that is, the marginal cost associated with farmer’s awareness of operating in the face of risky output prices.
The solution of the risk problem stated in relations (2)-(5) requires either a priori knowledge of the CARA parameter $\varphi$ or a procedure to estimate it simultaneously with the optimal output levels $x$ and input shadow prices $y$ .
3. Chance Constrained Risky Revenue
With some probability, a farmer may survive unfavorable events such as total revenue being less than total cost. Charnes and Cooper [5] proposed a useful approach to deal with this case. Consider the following probabilistic proposition:
$Prob\left({\stackrel{˜}{p}}^{\prime }x\le {y}^{\prime }Ax\right)\le 1-\beta$ (6)
where the probability that uncertain (risky) total revenue ${\stackrel{˜}{p}}^{\prime }x$ be less than or equal to certain total cost ${y}^{\prime }Ax$ should be smaller than or equal to $1-\beta$ . Intuitively, for how many years could a farmer survive while operating in the red? As an example, say once every twenty years. In this case, we could estimate the probability $1-\beta =1/20=0.05$ .
To derive a deterministic equivalent of relation (6) it is convenient to standardize the random variable ${\stackrel{˜}{p}}^{\prime }x$ by subtracting its expected value $E{\left(\stackrel{˜}{p}\right)}^{\prime }x$ and dividing it by the corresponding standard deviation ${\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}$ :
$\begin{array}{l}Prob\left({\stackrel{˜}{p}}^{\prime }x\le {y}^{\prime }Ax\right)\le 1-\beta \\ Prob\left(\frac{{\stackrel{˜}{p}}^{\prime }x-E{\left(\stackrel{˜}{p}\right)}^{\prime }x}{{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}}\le \frac{{y}^{\prime }Ax-E{\left(\stackrel{˜}{p}\right)}^{\prime }x}{{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}}\right)\le 1-\beta \\ Prob\left(\tau \le \frac{{y}^{\prime }Ax-E{\left(\stackrel{˜}{p}\right)}^{\prime }x}{{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}}\right)\le 1-\beta \\ Prob\left(E{\left(\stackrel{˜}{p}\right)}^{\prime }x+\tau {\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}\le {y}^{\prime }Ax\right)\le 1-\beta \end{array}$ (7)
By choosing a value of the standard normal random variable $\tau$ , say $\tau =\overline{\tau }$ , that corresponds to probability $1-\beta$ , the deterministic equivalent of relation (6) assumes the specification
$E{\left(\stackrel{˜}{p}\right)}^{\prime }x+\overline{\tau }{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}\le {y}^{\prime }Ax.$ (8)
There remains to establish a relation between the $\overline{\tau }$ parameter and the CARA coefficient $\varphi$ . This relation is obtained by subtracting the complementary slackness condition of the dual constraints (5) from relation (8):
$\begin{array}{l}E{\left(\stackrel{˜}{p}\right)}^{\prime }x+\overline{\tau }{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}\le {y}^{\prime }Ax\\ -\left[E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\varphi {x}^{\prime }{\Sigma }_{p}x={y}^{\prime }Ax\right].\end{array}$ (9)
With simplification, relation (9) corresponds to
$\overline{\tau }/{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}+\varphi \le 0.$ (10)
Relation (10) defines the CARA parameter $\varphi$ simultaneously with the decision variables $x$ , once the value of $\overline{\tau }$ is selected by the researcher. As an example, if the survival probability is determined to be $1-\beta =0.05$ , the one tail value of the normal random variable is $\overline{\tau }=-1.645$ .
The solution of the risky output price problem―a la Freund―is finally achieved by solving the following set of relations (using the linear complementarity problem (LCP) approach, for example)
dual constraints $\varphi {\Sigma }_{p}x+{A}^{\prime }y\ge E\left(\stackrel{˜}{p}\right)$ (11)
primal constraints $Ax\le b,\text{\hspace{0.17em}}x\ge 0,\text{}y\ge 0$ (12)
chance constraint $\overline{\tau }/{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}+\varphi =0$ (13)
and the associated complementary slackness conditions. This programming framework resolves the dilemma posed by Freund as to the difficulty of “defending any chosen value of the risk aversion constant $\varphi$ .”
4. CARA and Positive Mathematical Programming
Good empirical research requires the use of all the available information. When dealing with a sample of farms, for example, the most accessible piece of information consists in the output levels of crop activities realized in the previous production cycle. Such information is the end result of a decision-making process by an entrepreneur facing technological and market environments. Under the assumption that this economic agent attempted to maximize profit (minimizing cost), the realized (observed) output levels incorporate information about marginal cost and marginal revenue as the fundamental components of his opportunity costs. The research challenge is to unpack the marginal costs hidden in those observed output levels. Another readily available piece of information regards the price of limiting inputs. For example, a farmer has a pretty good idea about the price of his land. Even if his measure is imprecise, the price of land known to him can be used to anchor the model. We assume, therefore, that the output levels of a previous production cycle are observed (measured), ${x}^{obs}$ , as are the prices of limiting inputs, ${y}^{obs}$ . These pieces of information define calibration constraints that take on the following structure
$x={x}^{obs}+h$ (14)
$y={y}^{obs}+u$ (15)
where $h$ and $u$ are unrestricted deviations. This specification of the calibration constraints admits that the observed quantities and prices may be measured with error by either overstating or understating them. The choice approach to deal with $h$ and $u$ , therefore, is to minimize the sum of squared deviations weighed by appropriate weight matrices, say diagonal W and V, respectively. The necessity of introducing matrices W and V is justified by the different nature of the measurement units involving $h$ and $u$ ―in constraints (14) and (15)―and the corresponding dual variables that are indicated with vector variables $\lambda$ and $\phi$ , respectively. Constraint (14) is defined in terms of quantity units and, therefore, the dual variable $\lambda$ is defined in price units, say dollars. The self-duality of the least-squares approach [6] dictates that the matrix W mediates between the deviation vector $h$ and the dual vector $\lambda$ to establish the equation $\lambda =Wh$ , as demonstrated in the following discussion:
$\mathrm{min}LS={h}^{\prime }Wh/2$
subject to $x={x}^{obs}+h$ dual variable $\lambda$
with Lagrange function corresponding to
$L={h}^{\prime }Wh/2+{\lambda }^{\prime }\left(x-{x}^{obs}-h\right)$
and the first order condition
$\frac{\partial L}{\partial h}=Wh-\lambda =0$ . (16)
Therefore, $\lambda =Wh$ , as asserted. Since $\lambda$ is measured in price units and $h$ is measured in quantity units, an appropriate choice for the diagonal terms of the W matrix corresponds to the expected output prices. Analogous discussion involves the deviations $u$ and the corresponding dual variable $\phi$ . In this case, the least-squares relation turns out to be $\phi =Vu$ . Since $u$ is measured in input price units and $\phi$ is measured in quantity units, an appropriate choice of the diagonal terms of the V matrix is $\left({b}_{i}/{y}_{i}^{obs}\right)$ . Notice that the self-duality of the least-squares method allows for the elimination of vector variables $\lambda$ and $\phi$ from the model to be solved, as shown in the following intermediate step whose goal is the derivation of the dual constraint:
$\mathrm{max}CE=E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x$
subject to $Ax\le b$
$x={x}^{obs}+h$
with Lagrange function
$L=E{\left(\stackrel{˜}{p}\right)}^{\prime }x-\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x+{y}^{\prime }\left(b-Ax\right)+{\lambda }^{\prime }\left({x}^{obs}+h-x\right)$
and Karush-Kuhn-Tucker (KKT) condition
$\frac{\partial L}{\partial x}=E\left(\stackrel{˜}{p}\right)-\varphi {\Sigma }_{p}x-{A}^{\prime }y-\lambda \le 0$
but since $\lambda =Wh$ under a least-squares approach, the final specification of the dual constraint takes on the following structure
$\varphi {\Sigma }_{p}x+{A}^{\prime }y+Wh\ge E\left(\stackrel{˜}{p}\right)$ . (17)
The left-hand-side of relation (17) represents the total marginal cost of producing output $x$ under technological and risky output price conditions. Analogous discussion involves the primal constraint of the following dual problem
$\mathrm{min}TC={b}^{\prime }y+\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x$
subject to $\varphi {\Sigma }_{p}x+{A}^{\prime }y+Wh\ge E\left( p ˜ \right)$
$y={y}^{obs}+u$
with Lagrange function
$L={b}^{\prime }y+\frac{\varphi }{2}{x}^{\prime }{\Sigma }_{p}x+{x}^{\prime }\left(E\left(\stackrel{˜}{p}\right)-\varphi {\Sigma }_{p}x-{A}^{\prime }y-Wh\right)+{\phi }^{\prime }\left(y-{y}^{obs}-u\right)$
and KKT condition
$\frac{\partial L}{\partial y}=b-Ax+\phi \ge 0$
but since $\phi =Vu$ , the primal constraint assumes the following structure
$Ax\le b+Vu$ . (18)
Finally, phase I model of the PMP approach under a CARA specification of output price uncertainty can be stated as a weighted least-squares problem of finding nonnegative vectors $x$ and $y$ such that
$\mathrm{min}LS={h}^{\prime }Wh/2+{u}^{\prime }Vu/2$ (19)
subject to
$Ax\le b+Vu$ primal constraints (20)
$\varphi {\Sigma }_{p}x+{A}^{\prime }y+Wh\ge E\left(\stackrel{˜}{p}\right)$ dual constraints (21)
$x={x}^{obs}+h$ calibration constraints (22)
$y={y}^{obs}+u$ calibration constraints (23)
${y}^{\prime }\left(b+Vu-Ax\right)=0$ primal CSC (24)
${x}^{\prime }\left(\varphi {\Sigma }_{p}x+{A}^{\prime }y+Wh-E\left(\stackrel{˜}{p}\right)\right)=0$ dual CSC (25)
$\overline{\tau }/{\left({x}^{\prime }{\Sigma }_{p}x\right)}^{1/2}+\varphi =0$ chance constraint (26)
where CSC stands for complementary slackness conditions.
The solution of model (19)-(26) produces unique least-squares estimates of output quantities ${x}^{*}$ and input shadow prices ${y}^{*}$ that are as close as possible to the observed information ${x}^{obs}$ and ${y}^{obs}$ . This is the meaning of calibration in the novel PMP approach. Furthermore, the estimates of output quantities and shadow prices maximize the certainty equivalent corresponding to expected utility under a CARA specification of risky output prices.
5. Estimation of a Cost Function―Phase II of PMP
Phase II of the PMP approach estimates a cost function. The specification of such a function follows the familiar theoretical properties: it is non-decreasing in output quantities and input prices; it is concave and homogeneous of degree one in input prices. The following specification meets all these properties:
$C\left(x,y\right)=\left({f}^{\prime }x\right)\left({g}^{\prime }y\right)+\left({g}^{\prime }y\right){x}^{\prime }Qx/2+\left({f}^{\prime }x\right)\left[{\left({y}^{1/2}\right)}^{\prime }G{y}^{1/2}\right]$ (27)
where Q is a symmetric positive definite matrix of dimensions $\left(J×J\right)$ . The
term $\left[{\left({y}^{1/2}\right)}^{\prime }G{y}^{1/2}\right]$ follows a generalized Leontief specification. The $\left(I×I\right)$
G matrix has elements ${G}_{i,ii}={G}_{ii,i}\ge 0,i\ne ii,i,ii=1,\cdots ,I$ . The diagonal elements ${G}_{i,i}$ can take on either positive or negative values. The components of vectors $f$ and $g$ are free to take on any value as long as ${f}^{\prime }x>0$ and ${g}^{\prime }y>0$ . The reason for introducing a term like $\left({f}^{\prime }x\right)\left({g}^{\prime }y\right)$ is to add flexibility to the cost function.
The marginal cost function assumes the following specification
$\frac{\partial C}{\partial x}=\left({g}^{\prime }y\right)f+\left({g}^{\prime }y\right)Qx+f\left[{\left({y}^{1/2}\right)}^{\prime }G{y}^{1/2}\right]$ (28)
while Shephard lemma is stated as
$\frac{\partial C}{\partial y}=\left({f}^{\prime }x\right)g+g\left({x}^{\prime }Qx\right)/2+\left({f}^{\prime }x\right)\left[\Delta \left({y}^{-1/2}\right)G{y}^{1/2}\right]$ (29)
where the $\Delta$ matrix is diagonal with elements $\left({y}_{i}^{-1/2}\right)$ .
The estimation of the cost function is performed by combining the elements of phase I and phase II and using all the information for N farms in a weighted least-squares problem:
$\mathrm{min}LS=\sum _{n=1}^{N}{{h}^{\prime }}_{n}{W}_{n}{h}_{n}/2+\sum _{n=1}^{N}{{u}^{\prime }}_{n}{V}_{n}{u}_{n}/2$ (30)
subject to
${A}_{n}{x}_{n}\le {b}_{n}+{V}_{n}{u}_{n}$ primal constraints (31)
${\varphi }_{n}{\Sigma }_{p}{x}_{n}+{{A}^{\prime }}_{n}{y}_{n}+{W}_{n}{h}_{n}\ge E\left({\stackrel{˜}{p}}_{n}\right)$ dual constraints (32)
${x}_{n}={x}_{n}^{obs}+{h}_{n}$ calibration constraints (33)
${y}_{n}={y}_{n}^{obs}+{u}_{n}$ calibration constraints (34)
${{y}^{\prime }}_{n}\left({b}_{n}+{V}_{n}{u}_{n}-{A}_{n}{x}_{n}\right)=0$ primal CSC (35)
${{x}^{\prime }}_{n}\left({\varphi }_{n}{\Sigma }_{p}{x}_{n}+{{A}^{\prime }}_{n}{y}_{n}+{W}_{n}{h}_{n}-E\left({\stackrel{˜}{p}}_{n}\right)\right)=0$ dual CSC (36)
$-1.645/{\left({{x}^{\prime }}_{n}{\Sigma }_{p}{x}_{n}\right)}^{1/2}+{\varphi }_{n}=0$ chance constraint (37)
$\left({{g}^{\prime }}_{n}{y}_{n}\right){f}_{n}+\left({{g}^{\prime }}_{n}{y}_{n}\right)Q{x}_{n}+{f}_{n}\left[{\left({y}_{n}^{1/2}\right)}^{\prime }G{y}_{n}^{1/2}\right]={\varphi }_{n}{\Sigma }_{p}{x}_{n}+{{A}^{\prime }}_{n}{y}_{n}+{W}_{n}{h}_{n}$
marginal cost function (38)
$\left({{f}^{\prime }}_{n}{x}_{n}\right){g}_{n}+{g}_{n}\left({{x}^{\prime }}_{n}Q{x}_{n}\right)/2+\left({{f}^{\prime }}_{n}{x}_{n}\right)\left[\Delta \left\{{y}_{n}^{-1/2}\right\}G{y}_{n}^{1/2}\right]={A}_{n}{x}_{n}$
Shephard lemma (39)
$Q=LD{L}^{\prime }$ Cholesky factorization (40)
$Q{Q}^{-1}=I$ positive definiteness (41)
where L is a unit lower triangular matrix and D is a diagonal matrix with elements ${D}_{j,j}\ge 0$ . The Cholesky factorization guarantees symmetry and positive semidefiniteness of the Q matrix.
The solution of problem (30)-(41) produces least-squares estimates of all unknown variables and parameters, namely ${\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n},{\stackrel{^}{h}}_{n},{\stackrel{^}{u}}_{n},{\stackrel{^}{\varphi }}_{n},{\stackrel{^}{f}}_{n},{\stackrel{^}{g}}_{n},\stackrel{^}{Q},\stackrel{^}{G}$ . In particular, the optimal quantity levels ${\stackrel{^}{x}}_{n}$ , input shadow prices ${\stackrel{^}{y}}_{n}$ and CARA coefficient ${\stackrel{^}{\varphi }}_{n}$ are identical to ${x}^{*}$ , ${y}^{*}$ and ${\varphi }^{*}$ of phase I.
6. Calibrating Model―Phase III of PMP
Using estimates of the cost function parameters, ${\stackrel{^}{f}}_{n},{\stackrel{^}{g}}_{n},\stackrel{^}{Q},\stackrel{^}{G}$ , it is possible to set up a calibrating model―without calibration constraints―that reproduces output levels and shadow input prices that are identical to those obtained with model (30)-(41). This equivalence is achieved because Shephard lemma is equal to the demand for inputs, $Ax$ , as stated in the primal constraint (3) of the CARA risk model. Furthermore, the marginal cost function is equal to the dual constraints (5) of the same problem. In other words, the equivalence between the solution of the calibrating model and the solution of model (30)-(41) reveals the operation of unpacking the information contained in the observed quantities ${x}^{obs}$ and prices ${y}^{obs}$ in the form of effective marginal cost and input demand, respectively.
A calibrating linear complementarity problem for the n-th farm, therefore, can be stated as
$\mathrm{min}CS{C}_{n}={{y}^{\prime }}_{n}z{p}_{n}+{{x}^{\prime }}_{n}z{d}_{n}=0$ (42)
subject to
$\left({\stackrel{^}{{f}^{\prime }}}_{n}{x}_{n}\right){\stackrel{^}{g}}_{n}+{\stackrel{^}{g}}_{n}\left({{x}^{\prime }}_{n}\stackrel{^}{Q}{x}_{n}\right)/2+\left({\stackrel{^}{{f}^{\prime }}}_{n}{x}_{n}\right)\left[\Delta \left\{{y}_{n}^{-1/2}\right\}\stackrel{^}{G}{y}_{n}^{1/2}\right]+z{p}_{n}={b}_{n}+{V}_{n}{\stackrel{^}{u}}_{n}$ (43)
$\left({\stackrel{^}{{g}^{\prime }}}_{n}{y}_{n}\right){\stackrel{^}{f}}_{n}+\left({\stackrel{^}{{g}^{\prime }}}_{n}{y}_{n}\right)\stackrel{^}{Q}{x}_{n}+{\stackrel{^}{f}}_{n}\left[{\left({y}_{n}^{1/2}\right)}^{\prime }\stackrel{^}{G}{y}_{n}^{1/2}\right]=E\left({\stackrel{˜}{p}}_{n}\right)+z{d}_{n}$ (44)
where $z{p}_{n}$ and $z{d}_{n}$ are primal and dual slack variables, respectively.
The solution of model (42)-(44) produces estimates of the output quantities ${x}_{n}$ and shadow input prices ${y}_{n}$ that are identical to the corresponding solutions obtained in solving the phase II model, ${\stackrel{^}{x}}_{n}$ , ${\stackrel{^}{y}}_{n}$ . These estimates are as close as possible to the observed counterparts ${x}_{n}^{obs}$ and ${y}_{n}^{obs}$ . This is no surprise: the PMP process has transferred the same amount of information from the calibration constraints to the cost function while revealing the marginal cost levels and the input shadow prices that presumably influenced the economic agent in making the output and price decisions observed in ${x}_{n}^{obs}$ and ${y}_{n}^{obs}$ . Model (42)-(44) can now be used to evaluate a series of policy scenarios that may consider changes in expected output prices, changes in the quantity of limiting inputs, the introduction of crop subsidies and other analyses.
7. Empirical Example
The PMP procedure discussed in previous sections was applied to a sample of fourteen farms producing four crops (sugar beets, soft wheat, corn and barley). Land is the only limiting input. Given the large amount of information involved in this example, only the quantities of observed output levels and land prices are reported in Table 1.
In any computation of nonlinear models scaling of the original information series is of crucial importance for obtaining a feasible solution. The observed outputs are measured in hundred pound units. The land prices are measured in thousand dollars per acre.
Table 2 presents the optimal quantities of the crop activities and the optimal land prices obtained from the solution of model (30)-(41).
The discrepancy between the observed (Table 1) and the optimal quantities and prices (Table 2) is rather miniscule as reported in Table 3. The specification of the calibration constraints proposed in this paper is similar to a statistical specification of a regression function with non-zero residuals terms. It avoids the tautological specification of the original PMP procedure [7] that concerned only output quantity levels and assumed that $x\le {x}^{obs}\left(1+\epsilon \right)$ , where $\epsilon$ is a user-determined small positive number. In the context of this paper, an analogous specification of the calibration constraints involving both output quantities and input prices would result in an infeasible solution.
The CARA coefficients of the fourteen farms are presented in Table 4.
The CARA coefficient f is measured in 1/$units, as can be determined by examining the certainty equivalent in equation (2). Its reciprocal is measured in$ units and is called the risk tolerance coefficient. From table 4 and the fact that the certainty equivalent in this sample of farms is measured in 1000 dollar’s units, the risk tolerance varies from $50,000 to$190,000.
Table 5 presents the estimate of the cost function Q matrix. The estimate of $G=-30.136728$ .
The estimates of parameters $f$ and $g$ of the cost function are presented in Table 6.
Table 1. Observed output quantities and land input prices.
Table 2. Optimal output quantities and land shadow prices from model (30)-(41).
The parameters $f$ and $g$ can be interpreted as the individual farm deviations from the sample marginal cost function and the sample Shephard lemma, respectively. The conditions ${f}^{\prime }x>0$ and ${g}^{\prime }y>0$ are satisfied for all farms.
Table 3. Percentage difference between observed and estimated quantities and land prices.
Table 4. Estimated CARA coefficients.
Table 5. Estimate of the cost function Q matrix.
Table 6. Estimates of the f and g parameters of the cost function.
8. Conclusions
The extension of a PMP approach to include also the calibration of dual variables around observed limiting input prices has required a modification of the notion of calibration itself as proposed in the original PMP procedure by Howitt [7] . In that seminal paper, calibration means that optimal output levels, say ${x}^{*}$ , are identically equal to the observed output levels ${x}^{obs}$ (up to a user-deter- mined but very small $\epsilon$ number). The research reported in this paper found that the simultaneous calibration of output levels and limiting input prices―as specified in Equations (14) and (15)―can be achieved only in a statistical manner analogous to a statistical regression analysis where the error terms are minimized by least-squares estimation. In other words, if the traditional specification of the calibration constraints were formulated also for the limiting input prices, say $x\le {x}^{obs}\left(1+\epsilon \right)$ and $y\le {y}^{obs}\left(1+\epsilon \right)$ , an infeasible solution of the programming problem would occur.
A useful consequence of specifying the calibration constraints as in equations (14) and (15), coupled with the adoption of a least-squares procedure to minimize the deviations $h$ and $u$ , is that the calibrating solution $\stackrel{^}{x}$ and $\stackrel{^}{y}$ is unique. This extension of a PMP procedure was associated with the treatment of risky output prices according to a famous paper by Freund [2] . In that paper, Freund did not know how to estimate the CARA parameter of the selected utility function. In this paper, a chance-constrained relation involving random revenue is introduced to allow the derivation of a functional relation that ties the CARA parameter to the decision variables of the entrepreneur operating under a risky price environment.
Another methodological advantage of extending the calibration to the limiting input prices concerns the specification of a complete cost function. In the traditional PMP approach, a cost function involved only the output levels and ignored any input price. In this paper, a complete cost function is specified that satisfies all the theoretical properties.
An empirical example involving fourteen farms, four crops and one limiting input confirms that the proposed PMP procedure is feasible without excessive computational burden. In general, however, not every farm produces all the sample crops. This means that, in reality, the matrix of observed output levels contains some zero observations. When this probable event occurs, the proposed PMP procedure can easily accommodate the zero observations with minimal adjustments. It is sufficient to restate the calibration constraints in two parts: one part dealing with the positive output levels and the second part dealing with the zero levels. The rest of the estimation procedure applies without modification.
While the calibrating solution is unique, this cannot be said―in this numerical example―for the estimated parameters of the cost function. To obtain a unique solution of parameters $Q,G,f$ and $g$ it is necessary to have access to at least two observations per farm. In that case, the marginal cost function and Shephard lemma will admit corresponding residuals that must be minimized according to a second level least-squares criterion. Future research will attempt to extend the PMP approach to the estimation of general risk preferences where economic agents can be decreasingly risk averters, as wealth increases.
Cite this paper
Paris, Q. (2018) Estimation of CARA Preferences and Positive Mathematical Programming. Open Jour- nal of Statistics, 8, 1-13. https://doi.org/10.4236/ojs.2018.81001
References
1. 1. Markowitz, H. (1952) Portfolio Selection. The Journal of Finance, 7, 77-91.
2. 2. Freund, R.J. (1956) The Introduction of Risk into a Programming Model. Econometrica, 24,253-263. https://doi.org/10.2307/1911630
3. 3. Hazell, P.B.R. (1971) A Linear Alternative to Quadratic and Semivariance Programming for Farm Planning Under Uncertainty. American Journal of Agricultural Economics, 53, 53-62. https://doi.org/10.2307/3180297
4. 4. Pratt, J.W. (1964) Risk Aversion in the Small and in the Large. Econometrica, 77, 122-136. https://doi.org/10.2307/1913738
5. 5. Charnes, A. and Cooper, W.W. (1959) Chance Constrained Programming. Management Science, 6, 73-79. https://doi.org/10.1287/mnsc.6.1.73
6. 6. Paris, Q. (2015) The Dual of the Least-Squares Method. Open Journal of Statistics, 5, 658-664. https://doi.org/10.4236/ojs.2015.57067
7. 7. Howitt, R.E. (1995) Positive Mathematical Programming. American Journal of Agricultural Economics, 77, 329-342. https://doi.org/10.2307/1243543
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## ◂Math Worksheets and Study Guides Eighth Grade. Linear equations
### The resources above correspond to the standards listed below:
#### California Content Standards
CA.CC.8.EE. Expressions and Equations
Understand the connections between proportional relationships, lines, and linear equations.
8.EE.5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
8.EE.6. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
CA.CC.8.F. Functions
Define, evaluate, and compare functions.
8.F.3. Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s^2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
Use functions to model relationships between quantities.
8.F.4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
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# Length - Mathematics Form 1 Notes
## Introduction
• Length is the distance between two points. The SI unit of length is metres. Conversion of units of length.
1 kilometer (km) = 1000metres
1 hectometer (hm) = 100metres
1 decameter (Dm) =10 metres
1 decimeter (dm) = 1/10 metres
1 centimeter (cm) = 1/100 metres
1 millimeter (mm) = 1/1000 metres
• The following prefixes are often used when referring to length:
Mega – 1000 000
Kilo – 1000
Hecto – 100
Deca – 1 0
Deci – 1/10
Centi – 1/100
Milli – 1/1000
Micro – 1/1000 000
## Significant Figures
• The accuracy with which we state or write a measurement may depend on its relative size. It would be unrealistic to state the distance between towns A and B as 158.27 km. a more reasonable figure is 158 km. 158.27 km is the distance expressed to 5 significant figures and 158 km to 3 significant figures.
Example
Express each of the following numbers to 5, 4, 3, 2, and 1 significant figures:
1. 906 315
2. 0.08564
3. 40.0089
4. 15 600
Solution
number 5 s.f 4 s.f. 3 s.f. 2 s.f 1 s.f. (a) 906 315 906 320 906 300 906 000 91 0 000 900 000 (b) 0.085641 0.085641 0.08564 0.0856 0.085 0.09 (c) 40.0089 40.009 40.01 40.0 40 40 (d) 156 000 156 000 156 000 156 000 160 000 200 000
The above example show how we would round off a measurement to a given number of significant figures
Zero may not be a significant. For example:
1. 0.085 - zero is not significant therefore, 0.085 is a two- significant figure.
2. 2.30 – zero is significant. Therefore 2.30 is a three-significant figure.
3. 5000 – zero may or may not be significant figure. Therefore, 5 000 to three significant figure is 500 (zero after 5 is significant). To one significant figure is 5 000. Zero after 5 is not significant.
4. 31.805 Or 305 – zero is significant, therefore 31 .805 is five significant figure. 305 is three significant figure.
## Perimeter
• The perimeter of a plane is the total length of its boundaries. Perimeter is a length and is therefore expressed in the same units as length.
### Square Shapes
• Its perimeter is 5+5+5+5= 2(5+5)
=2(10)
=20cm
Hence 5 x 4 = 20
So perimeter of a square = Sides x 4
### Rectangular Shapes
Figure 12.2 is a rectangle of length 5cm and breadth 3cm.
Its perimeter is 5+3+5+3 =2(5+3)cm
= 2 x 8
= 16 cm
Hence perimeter of a rectangle p=2(L+ W)
### Triangular Shapes
• To find the perimeter of a triangle add all the three sides.
Perimeter = (a + b + c) units, where a, b and c are the lengths of the sides of the triangle.
### The Circle
The circumference of a circle = 2πr or πD
Example
1. Find the circumference of a circle of a radius 7cm.
2. The circumference of a bicycle wheel is 1 40 cm. find its radius.
Solution
1. C= πd
=22/7 x 7
=44 cm
2. C=πd
=2πr
=2x22/7xr
=140 ÷ 44/7
=22.27 cm
#### Length of an Arc
• An arc of a circle is part of its circumference. Figure 1 2.1 0 (a) shows two arcs AMB and ANB. Arc AMB, which is less than half the circumference of the circle, is called the minor arc, while arc ANB, which is greater half of the circumference is called the major arc. An arc which is half the circumference of the circle is called a semicircle.
Example
An arc of a circle subtends an angle 60 at the centre of the circle. Find the length of the arc if the radius of the circle is 42 cm. (π=22/7).
Solution
The length, l, of the arc is given by:
L =θ/360 x 2πr.
θ=60, r=42 cm
Therefore, L =60/360 x2 x 22/7 x 42
= 44 cm
Example
The length of an arc of a circle is 62.8 cm. find the radius of the circle if the arc subtends an angle 144 at the centre, (take π=3.142).
Solution
L =θ/360 x 2πr = 62.8 and θ= 144
Therefore,
144/360 x 2 x 3.142 x r = 62.8
R= 62.8 x 360/144 x 2 x 3.142
=24.98 cm
Example
Find the angle subtended at the centre of a circle by an arc of length 11 cm if the radius of the circle is 21 cm.
Solution
L= θ/360 x 2 xπr =11 cm and r =21 m
11 =θ/360 x 2 x 22/7 x 21
Thus,
θ = 11 x 360 x 7
2 x 22 x 21
= 300
## Past KCSE Questions on the Topic
1. Two coils which are made by winding copper wire of different gauges and length have the same mass. The first coil is made by winding 270 metres of wire with cross sectional diameter 2.8mm while the second coil is made by winding a certain length of wire with cross-sectional diameter 2.1 mm. Find the length of wire in the second coil .
2. The figure below represents a model of a hut with HG = GF = 10cm and FB = 6cm. The four slanting edges of the roof are each 12cm long.
Calculate
1. Length DF.
2. Angle VHF
1. The length of the projection of line VH on the plane EFGH.
1. The height of the model hut.
2. The length VH.
2. The angle DF makes with the plane ABCD.
3. A square floor is fitted with rectangular tiles of perimeters 220 cm. each row (tile length wise) carries 20 less tiles than each column (tiles breadth wise). If the length of the floor is 9.6m. Calculate:
1. The dimensions of the tiles
2. The number of tiles needed
3. The cost of fitting the tiles, if tiles are sold in dozens at sh. 1 500 per dozen and the labour cost is sh. 3000
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CHEM& 161 PLACEMENT EXAM PREPARATION PROBLEMS
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1 CHEM& 161 PLACEMENT EXAM PREPARATION PROBLEMS CHEM& 161 is a course that requires prerequisite knowledge in math and chemistry. The purpose of the placement exam is to ensure that you are entering at a level that will allow you to succeed in CHEM& 161 and beyond. You will be provided a periodic table (similar to the one on the last page). You are permitted to use a calculator, but you must bring one to use. No cell phones will be allowed. These problems reflect the nature of the exam in terms of content, but the exam will be multiple choice (whereas the problems below are not). The level of difficulty of these problems are similar to those on the placement exam. For the purposes of practicing, do this exam twice. You should first use it as a guide for studying and reviewing material. Show all of your work and check it using the answer key. Then wait a sufficient period (so as not to remember the answers) and do these problems in exam mode no help, 60 minutes timed, no interruptions. Grade your exam and see where you need to review. 1. Calculate: Without a calculator, determine: log(1000) =? Convert 62 µm into Mm and express as scientific notation. 3. x 2 3.4x = 0. Solve for x. 4. Solve for n 1 given the following expression: x 1 y 1 n 1 m 1 = x 2y 2 n 2 m 2 5. Convert 16.5 miles per gallon into km per liter. 1km = miles 1 gallon = liters 6. Convert 2.15x10-4 L into rundlets. 1 cm 3 = 1.47x10-5 rundlets. 7. Given the following data points, sketch a graph that presents the data. Draw a best fit line and determine the slope and y-intercept for the graph. Then write the equation of the line. x y
2 8. Report the following measurement with the appropriate number of significant figures. 9. I have 3.4 moles of Mg(CN) 2. How many total moles of carbon atoms do I have? 10. Balance the following reaction: C 8 H 18 (l) + O 2 (g) ---> CO 2 (g) + H 2 O(l) 11. The density of gold is 19.1 g/cm 3. What is the mass of 16 liters of solid gold? 12. How many electrons are found in one mole of the ion F -? 13. How many protons are in the radioisotope Iodine-131? 14. Name me: SiO What is the ground state electron configuration of sodium? 16. What is the volume of a sphere, with a radius of 43 pm? 17. How much, in total, do 15 atoms of iron weigh, in grams? 18. I have 14 grams of solid magnesium chloride. How many moles of chlorine atoms are contained in this solid? grams of natural gas burns in the presence of excess oxygen. How much CO 2 (g) will be produced as a result of this burning? The balanced chemical equation is given below: CH 4 (g) + 2O 2 (g) ---> CO 2 (g) + 2H 2 O(l)
3 20. A substance X has a mass of 2.50 pounds and a volume of mm 3. What is its density in g/cm 3? (453.6 grams = 1 lb) 21. What is the volume (ml) of ethanol required for a mass of 7.55x10-1 kg? The density of ethanol is g/ml at 25 C What is the number of (a) protons, (b) neutrons, and (c) electrons in 30Zn 2+? 23. How many molecules of water are present in 1 pound (453.6 g) of water? 24. How many moles of methane (CH 4 ) are required to produce 18 moles of H 2 O based on the following balanced chemical equation for its combustion? CH 4 (g) + 2O 2 (g) ---> CO 2 (g) + 2H 2 O(l) 25. What is the correct chemical formula for an ionic compound that contains only calcium ions (Ca 2+ ) and nitrate ions (NO 3 - )? 26. How many milligrams is 4x10-6 kg? 27. A Toyota Prius has a fuel tank that can hold about 11.9 gallons. How many liters is this? (1 liter = gallons) 28. The temperature of a nice warm day in Europe is 30 C. What is this temperature in F? 29. What is the symbol for an ion with 8 protons, 10 neutrons, and 10 electrons? Use A/Z notation (see #23). 30. An unknown isotope, X, has the symbol X. What element is X? 31. What is the chemical formula for lead (II) phosphate?
4 32. If I have grams of Na 2 CO 3, how many moles of Na 2 CO 3 is this? 33. Balance the following chemical equation: AgNO 3 (aq) + CuCl 2 (aq) ---> Cu(NO 3 ) 2 (aq) + AgCl (s) 34. What is the average atomic mass of two hypothetical isotopes with the following isotopic masses and natural abundances: isotope 1 ( amu, 36.34%) and isotope 2 ( amu, 63.66%) 35. Draw an acceptable Lewis structure for the following molecular (covalent) compounds, including all lone pairs. (a) water, H 2 O (b) ammonia, NH 3 (c) carbon dioxide, CO 2 (d) cyanide ion, CN How many moles of nitric acid (HNO 3 ) is obtained from ml of a 3.00 M solution of HNO 3? 37. Convert 4.5 x 10 3 in 3 into cm 3. (1.00 in = 2.54 cm) 38. You have a cylinder with a volume of 750 ml that is 30. cm in height. What is the diameter of its base in cm? (1 ml = 1 cm 3 ) 39. A medical doctor gives the order to administer dopamine at a rate of 3.0 mcg / kg min (mcg is the abbreviation for microgram in a medical context). The dopamine is supplied as a mixture of 400. mg dopamine in 250. ml of a dopamine solution. The patient weighs 73 kg. What is the infusion rate of the dopamine into her body (in units of ml/hour)?
5 Periodic Table of the Elements 1A 8A 1 1 H A 3A 4A 5A 6A 7A 2 He Li Be B C N O F Ne Na Mg B 4B 5B 6B 7B 8B 1B 2B 13 Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (98) 44 Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb B i Po (210) 85 At (210) 86 Rn (222) 7 87 Fr (223) 88 Ra (226) 89 Ac** (227) 104 Rf (261) 105 Db (262) 106 Sg (263) 107 Bh (262) 108 Hs (265) 109 Mt (266) 110 Und (269) 111 Une (272) 112 Unn (277) *Lanthanides 58 Ce Pr Nd Pm (147) 62 Sm Eu Gd Tb Dy Ho Er Tm Yb Lu **Actinides 90 Th Pa (231) 92 U Np (237) 94 Pu (242) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (249) 99 Es (254) 100 Fm (253) 101 Md (256) 102 No (254) 103 Lw (257)
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Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers
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# Stanford GSB Calling all applicants for Class of 2016!
Author Message
Intern
Joined: 03 Mar 2013
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
27 Oct 2013, 22:18
lm11 wrote:
Hi
Anyone receive an interview Invite for 2014 Round 1,yet?
Don't think so, the adcom email specifically mentioned "no invites before Oct 28". So, we shall expect to see at least some today.
Kudos [?]: 2 [0], given: 1
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 05:04
It's going to be very interesting and nail-biting weeks for all of us, starting today!!
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 06:47
Anyone know what time they usually start sending them?
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 08:22
I would guess, based on other schools, somewhere around 12-2pm PDT.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 10:02
Interview season has begun... good luck everybody!
PS. Has anyone's status changed from "submitted" to "under review" yet?
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 10:13
apesanteur wrote:
Interview season has begun... good luck everybody!
PS. Has anyone's status changed from "submitted" to "under review" yet?
Mine says "currently under review" - I'm in the US. No interview notification via email or applyyourself.
Last edited by Holland1234 on 28 Oct 2013, 11:41, edited 2 times in total.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 10:25
as interview emails trickle in can people please include your geography when you post? from some other posts it sounds like that's relevant in determining how applications are prioritized (or anyone has insight into how interviews are triaged, that would be interesting to see)
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 11:03
This month is going to be rough...
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 11:12
anonapplicant wrote:
as interview emails trickle in can people please include your geography when you post? from some other posts it sounds like that's relevant in determining how applications are prioritized (or anyone has insight into how interviews are triaged, that would be interesting to see)
Also perhaps when you submitted. Some places suggest submitting as early as possible - would be interesting to see if that has any impact (and implies they start reviewing as soon as they receive the apps).
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 11:25
I assume if they go out after 12, it will be after 12 PDT.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 11:38
my status has changed to "YOUR APPLICATION IS CURRENTLY UNDER REVIEW"
Does this mean anything?
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 11:38
My status changed to: YOUR APPLICATION IS CURRENTLY UNDER REVIEW.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 11:39
Mine just switched to "YOUR APPLICATION IS CURRENTLY UNDER REVIEW."
No decision yet though. For reference: I'm in Chicago.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 11:42
My app status says " your application is currently under review". Does this mean anything at all??
Posted from my mobile device
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 11:43
I guess it's happened to every Applicant then!
Posted from my mobile device
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 11:45
YOUR APPLICATION IS CURRENTLY UNDER REVIEW.
whats that people?
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
### Show Tags
28 Oct 2013, 11:48
"Approximately two weeks after the application deadline, all applications are updated to "Your application is currently under review."
Got that from the Stanford website
wish it meant good news!
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 12:08
2
KUDOS
My iPhone is at 9% battery because of my excessive refreshing today. I wish I could see a chart of applicants' productivity at work on decisions days.
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 13:32
Kitkat6 wrote:
My iPhone is at 9% battery because of my excessive refreshing today. I wish I could see a chart of applicants' productivity at work on decisions days.
work? productivity? ha!
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink]
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28 Oct 2013, 13:41
Anyone get news yet?
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Re: Stanford GSB Calling all applicants for Class of 2016! [#permalink] 28 Oct 2013, 13:41
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# Stanford GSB Calling all applicants for Class of 2016!
Moderators: ydmuley, mvictor
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## bugs in neb code
Python framework for working with atomic configurations, potentials, and algorithms.
Moderator: moderators
ezanardi
Posts: 1
Joined: Mon Apr 01, 2019 1:31 pm
### bugs in neb code
I've found two bugs in neb code
1) In tsase/neb/util.py vproj(v1,v2) it has to divide by the square of the modulus of v2 ,
but currently it divides by the modulus of v2. A patch to fix that bug (just changing vmag to vmag2) is:
[code]
diff a/tsase/neb/util.py b/tsase/neb/util.py
--- a/tsase/neb/util.py
+++ b/tsase/neb/util.py
@@ -61,7 +61,7 @@ def vproj(v1, v2):
Parameters:
v1, v2: numpy vectors
"""
- mag2 = vmag(v2)
+ mag2 = vmag2(v2)
if mag2 == 0:
printf("Can't project onto a zero vector", ERR)
return v1
[/code]
2) In both tsase/neb/pssneb.py and tsase/neb/ssneb.py forces calculation, when using dneb modification,
it has to multiply fsdneb by 2/pi * atan( |Fperp|²/|Fsperp|² ) but currently it multiplies by
2/pi * atan( |Fperp|/|Fsperp| ) (no squares). A patch to fix that bug (just changing vmag to vmag2) is:
[code]
diff a/tsase/neb/pssneb.py b/tsase/neb/pssneb.py
--- a/tsase/neb/pssneb.py
+++ b/tsase/neb/pssneb.py
@@ -334,8 +334,8 @@ class pssneb:
# New dneb where dneb force converges with (What?!)
if not self.dnebOrg:
- FperpSQ = vmag(self.path[i].fPerp)
- FsperpSQ = vmag(self.path[i].fsperp)
+ FperpSQ = vmag2(self.path[i].fPerp)
+ FsperpSQ = vmag2(self.path[i].fsperp)
if FsperpSQ > 0:
self.path[i].fsdneb *= 2.0 / pi * atan(FperpSQ / \
FsperpSQ)
diff a/tsase/neb/ssneb.py b/tsase/neb/ssneb.py
--- a/tsase/neb/ssneb.py
+++ b/tsase/neb/ssneb.py
@@ -409,8 +409,8 @@ class ssneb:
# dneb modification so that it will converge
if not self.dnebOrg:
- FperpSQ = vmag(self.path[i].fPerp)
- FsperpSQ = vmag(self.path[i].fsperp)
+ FperpSQ = vmag2(self.path[i].fPerp)
+ FsperpSQ = vmag2(self.path[i].fsperp)
if FsperpSQ > 0:
self.path[i].fsdneb *= 2.0/pi*atan(FperpSQ/FsperpSQ)
[/code]
graeme
Posts: 1998
Joined: Tue Apr 26, 2005 4:25 am
Contact:
### Re: bugs in neb code
Wow, thank you very much! The error in the projection algorithm is particularly worrisome. Anyway, again, thank you for the fix. I've updated the code.
xph
Posts: 39
Joined: Tue Mar 13, 2012 9:33 pm
### Re: bugs in neb code
I was worried in the beginning, but later found out that all the "vproj" calls are projecting forces on to a unit vector (tangent direction), except in dneb. So dneb was totally messed up, but the regular (ss)neb was good.
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# Math
An appliance store sells three brands of TV sets, brands A, B, and C. The profit per set is \$30 for brand A, \$50 for brand B, and \$60 for brand C. The total warehouse space allotted to all brands is sufficient for 600 sets, and the inventory is delivered only once per month. At least 100 customers per month will demand brand A, at least 50 will demand brand B, and at least 200 will demand brand B or brand C. How can the appliance store satisfy all those constraints and earn maximum profit?
Delores, I assume you are working n dimensional Simplex algorithm problems. frankly, it has been too long since I set one up, and would probably waste your time. Give this a day or so, we have a math person (Roger) who is a whiz, and if he checks in, may be able to set one of these up for you.
Ok but it's kind of an emergency here. I am so confused with these word problems. I am working witn The Simplex Method: Maximization and Minimization. I can do the problems if they are just the equations, however, the word problems just confuse me with all the wording. So if you know ANYONE who may be of help, PLEEEEEASE ask them. Thanks for taking the time to respond.
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Quantum Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > QLE Home > Th. List > govar GIF version
Theorem govar 896
Description: Lemma for converting n-variable Godowski equations to 2n-variable equations. (Contributed by NM, 19-Nov-1999.)
Hypotheses
Ref Expression
govar.1 ab
govar.2 bc
Assertion
Ref Expression
govar ((ab) ∩ (a2 c)) ≤ (bc)
Proof of Theorem govar
StepHypRef Expression
1 df-i2 45 . . . 4 (a2 c) = (c ∪ (ac ))
21lan 77 . . 3 ((ab) ∩ (a2 c)) = ((ab) ∩ (c ∪ (ac )))
3 ax-a2 31 . . . . 5 (ab) = (ba)
43ran 78 . . . 4 ((ab) ∩ (c ∪ (ac ))) = ((ba) ∩ (c ∪ (ac )))
5 govar.2 . . . . . . . 8 bc
65lecom 180 . . . . . . 7 b C c
76comcom7 460 . . . . . 6 b C c
8 govar.1 . . . . . . . . . . 11 ab
98lecom 180 . . . . . . . . . 10 a C b
109comcom7 460 . . . . . . . . 9 a C b
1110comcom 453 . . . . . . . 8 b C a
1211comcom2 183 . . . . . . 7 b C a
1312, 6com2an 484 . . . . . 6 b C (ac )
147, 13com2or 483 . . . . 5 b C (c ∪ (ac ))
1514, 11fh2r 474 . . . 4 ((ba) ∩ (c ∪ (ac ))) = ((b ∩ (c ∪ (ac ))) ∪ (a ∩ (c ∪ (ac ))))
164, 15ax-r2 36 . . 3 ((ab) ∩ (c ∪ (ac ))) = ((b ∩ (c ∪ (ac ))) ∪ (a ∩ (c ∪ (ac ))))
17 coman1 185 . . . . . . 7 (ac ) C a
1817comcom7 460 . . . . . 6 (ac ) C a
19 coman2 186 . . . . . . 7 (ac ) C c
2019comcom7 460 . . . . . 6 (ac ) C c
2118, 20fh2c 477 . . . . 5 (a ∩ (c ∪ (ac ))) = ((ac) ∪ (a ∩ (ac )))
22 dff 101 . . . . . . . . 9 0 = (aa )
2322ran 78 . . . . . . . 8 (0 ∩ c ) = ((aa ) ∩ c )
2423ax-r1 35 . . . . . . 7 ((aa ) ∩ c ) = (0 ∩ c )
25 anass 76 . . . . . . 7 ((aa ) ∩ c ) = (a ∩ (ac ))
26 an0r 109 . . . . . . 7 (0 ∩ c ) = 0
2724, 25, 263tr2 64 . . . . . 6 (a ∩ (ac )) = 0
2827lor 70 . . . . 5 ((ac) ∪ (a ∩ (ac ))) = ((ac) ∪ 0)
29 or0 102 . . . . 5 ((ac) ∪ 0) = (ac)
3021, 28, 293tr 65 . . . 4 (a ∩ (c ∪ (ac ))) = (ac)
3130lor 70 . . 3 ((b ∩ (c ∪ (ac ))) ∪ (a ∩ (c ∪ (ac )))) = ((b ∩ (c ∪ (ac ))) ∪ (ac))
322, 16, 313tr 65 . 2 ((ab) ∩ (a2 c)) = ((b ∩ (c ∪ (ac ))) ∪ (ac))
33 lea 160 . . 3 (b ∩ (c ∪ (ac ))) ≤ b
34 lear 161 . . 3 (ac) ≤ c
3533, 34le2or 168 . 2 ((b ∩ (c ∪ (ac ))) ∪ (ac)) ≤ (bc)
3632, 35bltr 138 1 ((ab) ∩ (a2 c)) ≤ (bc)
Colors of variables: term Syntax hints: ≤ wle 2 ⊥ wn 4 ∪ wo 6 ∩ wa 7 0wf 9 →2 wi2 13 This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a3 32 ax-a4 33 ax-a5 34 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 ax-r3 439 This theorem depends on definitions: df-b 39 df-a 40 df-t 41 df-f 42 df-i2 45 df-le1 130 df-le2 131 df-c1 132 df-c2 133 This theorem is referenced by: gon2n 898
Copyright terms: Public domain W3C validator
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# Thread: Prove if series converges or diverges
1. ## Prove if series converges or diverges
$\sum\,\frac{n-1}{n^{2}}$
I'm having a brain fart, I don't know if this converges or not.
2. Originally Posted by Pinkk
$\sum\,\frac{n-1}{n^{2}}$
I'm having a brain fart, I don't know if this converges or not.
$\sum_{n=1}^{\infty}\frac{n-1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n^2}$
The first sum is the harmonic series and diverges to infintiy and the 2nd sums to $\frac{\pi^2}{6}$ is I remember correctly so the sum diverges to infinity.
3. Originally Posted by Pinkk
$\sum\,\frac{n-1}{n^{2}}$
I'm having a brain fart, I don't know if this converges or not.
limit comparison with the known divergent series $\sum \frac{1}{n}$ ...
$\lim_{n \to \infty} \frac{\frac{n-1}{n^2}}{\frac{1}{n}}$
$\lim_{n \to \infty} \frac{n-1}{n^2} \cdot \frac{n}{1} = 1$
series diverges
4. Hmm, that seems obvious enough, but since I'm in a real analysis course, I have to show this a bit more formally; it was never proven in my class that the sum/difference of a divergent and a convergent series is divergent.
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Area of . (Give your answer correct to 3 significant figures.) (leave your answers in term of π). Now, Perimeter of the given figure = Length of arc AQO + Length of arc APB + OB. by doing A seg. Radius of semicircle P AQ = 21. . Add all radii and then multiply by $\pi$: Since the perimeter of a half-circle is just $\pi$ times the diameter, the perimeter of the shaded region equals the perimeter of the whole big circle. So we have 100 minus 9 pi is the area of the shaded region. So, the perimeter of a semicircle is 1/2 (πd) + d or πr + 2r, where r is the radius. Perimeter of shaded region = Circumference of semicircle P S R + Circumference of semicircle RT Q + Circumference of semicircle P AQ. 11$\pi cm^2$ C. 15$\pi cm^2$ D. 19$\pi cm^2$ E. None of the above. So if we take pi times diameter, uh, and then we can divide that by two. Answer: Perimeter of the shaded fighure is 14 π + 14cm. Solution: Area of shaded region = Area of sector AOC – Area of sector OBD. Perimeter of the semi circle = 2 π R / 2. How to rewrite mathematics constructively. 12.21, if radii of the two concentric circles with centre 0 are 7 cm and 14 cm respectively and ∠AOC = 40°. D. 198cm. The formula for circumference is. Question 34: In the adjoining figure, and is mid-point of . Want to improve this question? Radius of semicircle RT Q= 21. . In figure there are three semicircles, A, B and C having diameter 3 c m each, and another semicircle E having a circle D with diameter 4. The area of the shaded region will be the sum of the areas of both rectangles. Example 1: If the perimeter of a semi-circular protractor is 66 cm, find the diameter of the protractor (Take π = 22/7). A circle with center touches all the three circles. perimeter of circle = 2πr (or πd, where d-> diameter) perimeter of shaded region = (sum of perimeter of all circles) / 2. how many circles are there ? Change ), You are commenting using your Google account. Threfore, Perimeter of shaded region = π×5+π ×1.5+π×3.5 = … Update the question so it's on-topic for Mathematics Stack Exchange. Area of large semi circle with diameter . note: The radius of the circle is 4. b. [closed]. or Circumference of the Circle = diameter of the circle × π; Circumference of a semi-circle = $$\frac { 2\pi r }{ 2 }$$ = πr and the perimeter of a semi-circular shape = (π + 2) r units. ? As their are the 4 semicircle the formula will be 4πr. perimeter of circle = 2πr (or πd, where d-> diameter), perimeter of shaded region = (sum of perimeter of all circles) / 2. The other has a diameter of 30cm. a. What is the best way to play a chord larger than your hand? How to find the shaded region as illustrated by a circle inscribed in a square. Perimeter semicircle = π * r + 2 * r = r * (π + 2) or. What is the Perimeter of shaded region in semicircle if four small semicircles have radii of 1,2,3,4 respectively? RS is 7 cm. = r ( (33+7)/7). Write down the perimeter of the shaded shape. Question 30. Give your answer correct to 3 significant figures. Let the radius of the semi-circle APB be r. ⇒ The radius of the semi-circle AQO =r/2. Perimeter semicircle = π * r + d, where d is the semicircle diameter. Change ), You are commenting using your Facebook account. c. 40 $\pi$ E. None of the above. Since it's a semi circle. Submit B. So what's the area of a circle with radius 3? Were the Beacons of Gondor real or animated? You can draw out the shaded area to see what is required. $Circumference\, of\, circle=\pi d=\pi\,\times 8=25.12cm$ $Half\, of\, circumference = 12.56cm$ C. 196cm. = Asec.-A triangle. ×7 = 3.5 cm. Calculate: (i) the area of the shaded region (ii)the cost of painting the shaded region at … Why does this current not match my multimeter? The figure shows a semi-circle inscribed in a rectangle. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. The diameter (d) of the circle goes from one side of the circle to the other, through the centre. Is the heat from a flame mainly radiation or convection? =r (3/2xx22/7+1). Circumference is the perimeter of a circle. A. Use π= 3.14 so their radii will be 21cm. ( Log Out / ( Log Out / Find Perimeter of shaded region in semicircle. How does a bank lend your money while you have constant access to it? In Exercises 15-18, Ind the perimeter of the shaded region. Since AB=BC=CD=ED. The circumference of a semicircle is not equal to half of the circle circumference - you need to add also the diameter of the semicircle, as another boundary was created: Perimeter semicircle = (circumference circle / 2) + 2 * radius. The shaded shape below is made from 7 hexagon tiles. And now the perimeter of the 4 small semi circles. What I've done: I know you have to find the area of the small unshaded seg. Which have a diameter of 42cm . Find the area and perimeter of shaded region. Then you need to find the area of the semi-circle created by the shaded … The perimeter of a semicircle is half the original circle's circumference, C, plus the diameter, d. Since the semicircle includes a straight side, its diameter, we cannot describe the distance around the shape as the circumference of a semicircle; it is a perimeter. 1 semi-circle has a diameter of 20cm. 10 $\pi$ 194cm. It only takes a minute to sign up. This becomes six pie, which is about 18.85 units, so 26 plus 18.85 is about 44 0.85. Find the perimeter of the D shaded region in Figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. Does Kasardevi, India, have an enormous geomagnetic field because of the Van Allen Belt? Question: Find the perimeter of the shaded area? The perimeter of a circle is referred to as the 'circumference'. Question 134188: Find the area and perimeter of the shaded region. Can we get rid of all illnesses by a year of Total Extreme Quarantine? The perimeter of a semicircle is half of the circumference plus the diameter. Find the perimeter of the shaded area. =r (40/7)cm. Geometry Area and Perimeter – HW#72 Find the area of the shaded region in each of the following figures. The diameter of the smallest circle is 4cm. =pixxr/2+pixxr+r. Its A because (1 + 2 + 3 + 4) = 10 hence 10pi is perimeter. 20 $\pi$ unix command to print the numbers after "=". Prev. But because there are two of them were also gonna multiply it by two. As the perimeter of a circle is 2πr or πd. Question 17. =r (3/2pi+1). Therefore shaded area . For all questions, assume that things that look like squares are squares, things that look like circles are circles, etc. Why do we not observe a greater Casimir force than we do? So their perimeter of them is 2πr. Step 1: Break the diagram to understand what is required to find the perimeter of the shaded area. The diagram shows a semi-circle inside a rectangle of length 140 m. The semi-circle touches the rectangle at P, Q and R. 140 m P Q R Not to scale Calculate the perimeter of the shaded region. The big circle has radius $10$, hence the answer is $\color{red}{20\pi}$. You can draw out the shaded area to see what is required. How can I find the perimeter of the shaded part? So add the perimeters of the smaller semicircles and of the largest one to get that of the shaded region. Perimeter of the semi circle. d. 60 $\pi$. Find the area of the shaded region in Fig. Calculate the perimeter of the shaded shape and explain how you did it. How can I defeat a Minecraft zombie that picked up my weapon and armor? ( Log Out / Change ), Perimeter of the second semi circle = 2 π R, Perimeter of 2 semi circles + one straight line, Perimeter of 2 semi circles + one straight line = 14 π + 7 cm, Perimeter of 2 semi circles + 2 straight line, Perimeter of 2 semi circles + one straight line = 14 π + 7 cm + 7 cm. [Use π = $$\frac{22}{7}$$] (2011OD) Solution: Perimeter of the shaded region = Circumference of circle + AD + BC = 2πr + 14 + 14 = 2 × $$\frac{22}{7}$$ × 7 + 28 ….[. The radius (r) of a circle is the distance from the centre to the outside of the circle. [5] Is it always one nozzle per combustion chamber and one combustion chamber per nozzle? rev 2021.1.21.38376, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. How to disable OneNote from starting automatically? Not getting the correct asymptotic behaviour when sending a small parameter to zero. How to tell if a song is tuned in half-step down, Protection against an aboleths enslave ability. Now, area of shaded region = Area of the semi-circle – Area of ARPQ. So the radius is 3. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If I'm the CEO and largest shareholder of a public company, would taking anything from my office be considered as a theft? Do RGB cubic-coordinate and HSL cylindrical-coordinate systems both support same colors? How much did J. Robert Oppenheimer get paid while overseeing the Manhattan Project? - Mathematics Question By default show hide Solutions cm . Radius of the semicircle AC AND CE are equal. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Find the area of the shaded parts. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. 2semicircle =a circle. Watch how to calculate the combined area of a semi-circle which is attached to the end of a rectangle.Visit https://maisonetmath.com for more practice. Perimeter of the shaded region=Perimeter of bigger semi-circle+2 (Perimeter of smaller semi-circle)= πD +2(πd)= 722 ×14+2( 722 ×7)= 44+2×22= 44+44= 88cm. 2*22/7*42=264 cm². Perimeter of semi circle πr. That is wrong, you are not adding the larger semicircle. ... A high school track is shaped like a rectangle with a semi-circle (half a circle) on each end. r = $$\frac{14}{2}$$ = 7 cm = 44 + 28 = 72 cm. Change ), You are commenting using your Twitter account. Length of RS is the same as length of TU. Find the area of the shaded region. Find the diameter of the smallest circle is 4cm. Area of large semi circle with diameter . If the radius of the semi-circle is 4 cm, find the area of the shaded region. S and T are the centers of each of the semi-circles. This shaded shape is made using two semi-circles. Semi-circles are drawn on and as diameters. In the Given Figure, Psr, Rtq and Paq Are Three Semicircles of Diameter 10 Cm, 3 Cm and 7 Cm Respectively. 5 cm are shown. And that area is going to be equivalent to the area of one circle with a radius of 3. 7$\pi cm^2$ B. Hint: The perimeter of a semicircle (only the curved part) is $\frac{2\pi r}2=\pi r$. So it's going to be 3 times 3, which is 9, times pi-- 9 pi. The circumference, C, of a circle = 2πr or πd where r is the radius and d is the diameter of the circle. Footnote: Only those workings in the highlighted box are required to be written in your answer, the rest are just for you to understand how to do this question. Well, the formula for area of a circle is pi r squared, or r squared pi. ( Log Out / = 2 X π X 7cm / 2 = 7 π. Perimeter of 2 semi circles. 1: Area of Circle: 49 The radius of each semi-circle must be half the side of the square. Question 16. Perimeter Of A Circle With Examples. ×3 cm = 1.5cm. Step 1: Break the diagram to understand what is required to find the perimeter of the shaded area. Where the shaded region is added to the perimeter. Find the Perimeter of Shaded Region. (See Example 4 . Trilogy in the 80’s about space travel to another world. Perimeter of Semicircle The perimeter of a semicircle is the sum of the half of the circumference of the circle and diameter. Find the Perimeter of the Shaded Region in the Figure, If Abcd is a Square of Side 14 Cm and Apb and Cpd Are Semicircles. Perimeter of shaded figure = 2 × Perimeter contributed by 1 semicircle + 4 × Perimeter contributed by 1 quadrant + Perimeter contributed by rectangle = (2 × 11) + (4 × 4.4) + 7 = 22 + 17.6 + 7 = 46.6 cm Question: In the figure below, RU is a straight line. Comment dit-on "What's wrong with you?" Therefore, the radius is 5. All illnesses by a year of Total Extreme Quarantine up my weapon and?. The perimeter of the circle to the other, through the centre I! 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# [R] Evaluating a function with another function
Guenther, Cameron Cameron.Guenther at MyFWC.com
Thu Apr 6 22:53:51 CEST 2006
```Hello all,
I hope someone can help me with this.
I have function that calculates two values based on input data. A
simple example follows:
test<-function(x,s,rangit=seq(0,10,1))
{
rangit<-rangit
y<-vector()
p<-vector()
for(i in 0:length(rangit)){
y[i]<-x+s[1]+rangit[i]
p[i]<-x+s[2]+rangit[i]
}
return(data.frame(rangit,y,p))
}
And returns the following:
rangit y p
1 0 2 3
2 1 3 4
3 2 4 5
4 3 5 6
5 4 6 7
6 5 7 8
7 6 8 9
8 7 9 10
9 8 10 11
10 9 11 12
11 10 12 13
Which is what I want. The part I am having trouble with is that I want
to write another function that will evaluate the previous function at
multiple levels of x, where the levels of x may not always be the same
length
I have tried several options but so far have not been able to figure it
out.
I tried:
testexp<-function(x,s,rangit){
out<-data.frame()
for (j in 1:length(x)){
out[j]<-test(x[j],s)
}
return(out)
}
Q2<-testexp(x=c(1:4),s=c(1,2))
But that returns a warning with no values. I have also tried various
other methods with no success.
Basically what I want the output to look like is
Out[1] Out[2] Out[3] Out[4]
Rangit y p rangit y p rangit y p rangit y p
0 2 3 0 3 4 0 4 5 0 5 6
1 3 4 1 4 5 1 5 6 1 6 7
2 4 5 2 5 6 2 6 7 2 7 8
3 . . . . . . . . . . .
4 . . . . . . . . . . .
5 . . . . . . . . . . .
6 . . . . . . . . . . .
7 . . . . . . . . . . .
8 . . . . . . . . . . .
9 . . . . . . . . . . .
10 . . . . . . . . . . .
Cameron Guenther, Ph.D.
Associate Research Scientist
FWC/FWRI, Marine Fisheries Research
100 8th Avenue S.E.
St. Petersburg, FL 33701
(727)896-8626 Ext. 4305
cameron.guenther at myfwc.com
```
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http://www.jiskha.com/display.cgi?id=1369598714
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# Math
posted by on .
1. Fantasia Florist Shop purchases an order of imported roses with a list price of \$2,375 less trade discounts of 15/20/20. What is the dollar amount of the trade discount?
2. Using your answer to the question above, what is the net dollar amount of that rose order?
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| 2.53125
| 3
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CC-MAIN-2017-22
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latest
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en
| 0.908837
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http://www.aimsciences.org/journals/displayArticles.jsp?paperID=6194
| 1,561,574,393,000,000,000
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crawl-data/CC-MAIN-2019-26/segments/1560628000414.26/warc/CC-MAIN-20190626174622-20190626200622-00089.warc.gz
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American Institute of Mathematical Sciences
November 2011, 30(4): 1083-1093. doi: 10.3934/dcds.2011.30.1083
Radial symmetry of solutions for some integral systems of Wolff type
1 Department of Mathematics, Yeshiva University, New York, NY 10033 2 Department of Applied Mathematics, University of Colorado at Boulder
Received March 2010 Revised August 2010 Published May 2011
We consider the fully nonlinear integral systems involving Wolff potentials:
$\u(x) = W_{\beta, \gamma}(v^q)(x)$, $\x \in R^n$;
$\v(x) = W_{\beta, \gamma} (u^p)(x)$, $\x \in R^n$;
(1)
where
$\W_{\beta,\gamma} (f)(x) = \int_0^{\infty}$ $[ \frac{\int_{B_t(x)} f(y) dy}{t^{n-\beta\gamma}} ]^{\frac{1}{\gamma-1}} \frac{d t}{t}.$
After modifying and refining our techniques on the method of moving planes in integral forms, we obtain radial symmetry and monotonicity for the positive solutions to systems (1).
This system includes many known systems as special cases, in particular, when $\beta = \frac{\alpha}{2}$ and $\gamma = 2$, system (1) reduces to
$\u(x) = \int_{R^{n}} \frac{1}{|x-y|^{n-\alpha}} v(y)^q dy$, $\ x \in R^n$,
$v(x) = \int_{R^{n}} \frac{1}{|x-y|^{n-\alpha}} u(y)^p dy$, $\ x \in R^n$.
(2)
The solutions $(u,v)$ of (2) are critical points of the functional associated with the well-known Hardy-Littlewood-Sobolev inequality. We can show that (2) is equivalent to a system of semi-linear elliptic PDEs
$(-\Delta)^{\alpha/2} u = v^q$, in $R^n$,
$(-\Delta)^{\alpha/2} v = u^p$, in $R^n$
(3)
which comprises the well-known Lane-Emden system and Yamabe equation.
Citation: Wenxiong Chen, Congming Li. Radial symmetry of solutions for some integral systems of Wolff type. Discrete & Continuous Dynamical Systems - A, 2011, 30 (4) : 1083-1093. doi: 10.3934/dcds.2011.30.1083
References:
[1] L. Caffarelli, B. Gidas and J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth,, C. P. A. M., XLII (1989), 271. [2] W. Chen, C. Jin, C. Li and C. Lim, Weighted Hardy-Littlewood-Sobolev inequalities and system of integral equations,, Disc. Cont. Dyn. Sys., 2005 (): 164. [3] W. Chen and C. Li, Regularity of Solutions for a system of integral equations,, Comm. Pure and Appl. Anal., 4 (2005), 1. [4] W. Chen and C. Li, The best constant in some weighted Hardy-Littlewood-Sobolev inequality,, Proc. AMS, 136 (2008), 955. [5] W. Chen and C. Li, Classification of positive solutions for nonlinear differential and integral systems with critical exponents,, Acta Mathematica Scientia, 4 (2009), 949. doi: 10.1016/S0252-9602(09)60079-5. [6] W. Chen and C. Li, An integral system and the Lane-Emden conjecture,, Disc. Cont. Dyn. Sys., 4 (2009), 1167. [7] W. Chen and C. Li, Classification of solutions of some nonlinear elliptic equations,, Duke Math. J., 63 (1991), 615. doi: 10.1215/S0012-7094-91-06325-8. [8] W. Chen and C. Li, Super polyharmonic property of solutions for PDE systems and its applications, , submitted to Trans. AMS, (2011). [9] W. Chen and C. Li, "Methods on Nonlinear Elliptic Equations,", AIMS Book Series on Diff. Equa. & Dyn. Sys., 4 (2010). [10] C. Ma, W. Chen, and C. Li, Regularity of solutions for an integral system of Wolff type,, Advances of Math, 3 (2011), 2676. doi: 10.1016/j.aim.2010.07.020. [11] W. Chen, C. Li and B. Ou, Classification of solutions for an integral equation,, Comm. Pure Appl. Math., LLVIII (2005), 1. [12] W. Chen, C. Li and B. Ou, Qualitative properties of solutions for an integral equation,, Disc. Cont. Dyn. Sys., 12 (2005), 347. [13] W. Chen, C. Li and B. Ou, Classification of solutions for a system of integral equations,, Comm. PDEs, 30 (2005), 59. doi: 10.1081/PDE-200044445. [14] B. Gidas, W. Ni and L. Nirenberg, Symmetry of positive solutions of nonlinear elliptic equations in $R^n,$, Advances in Mathematics, 7a (1981). [15] F. Hang, On the integral systems related to Hardy-Littlewood-Sobolev inequality,, Math Res Lett, (2007). [16] F. Hang, X. Wang and X. Yan, An integral equation in conformal geometry,, Ann. H. Poincare Nonl. Anal., 26 (2009), 1. doi: 10.1016/j.anihpc.2007.03.006. [17] G. Hardy and J. Littelwood, Some properties of fractional integrals I,, Math. Zeitschr., 27 (1928), 565. doi: 10.1007/BF01171116. [18] C. Jin and C. Li, Symmetry of solutions to some systems of integral equations,, Proc. AMS, 134 (2006), 1661. [19] T. Kilpelaiinen and J. Maly, The Wiener test and potential estimates for quasilinear elliptic equations,, Acta Math., 172 (1994), 137. doi: 10.1007/BF02392793. [20] C. Li, Local asymptotic symmetry of singular solutions to nonlinear elliptic equations,, Invent. Math., 123 (1996), 221. [21] E. Lieb, Sharp constants in the Hardy-Littlewood-Sobolev and related inequalities,, Annals of Math, 118 (1983), 349. doi: 10.2307/2007032. [22] S. Liu, Regularity, symmetry, and uniqueness of some integral type quasilinear equations,, Nonlinear Analysis: Theory, 71 (2009), 1796. [23] Y. Y. Li, Remarks on some conformally invariant integral equations: The method of moving spheres,, J. Euro. Math. Soc., 6 (2004), 153. doi: 10.4171/JEMS/6. [24] C. Li and J. Lim, The singularity analysis of solutions to some integral equations,, Comm. Pure Appl. Anal., 6 (2007), 453. doi: 10.3934/cpaa.2007.6.453. [25] C. Li and L. Ma, Uniqueness of positive bound states to Shrödinger systems with critical exponents,, SIAM J. of Appl. Anal., 40 (2008), 1049. doi: 10.1137/080712301. [26] C. Liu and S. Qiao, Symmetry and monotonicity for a system of integral equations,, Comm. Pure Appl. Anal., 6 (2009), 1925. doi: 10.3934/cpaa.2009.8.1925. [27] D. Li and R. Zhuo, An integral equation on half space,, Proc. AMS, 138 (2010), 2779. [28] L. Ma and D. Chen, A Liouville type theorem for an integral system,, Comm. Pure Appl. Anal., 5 (2006), 855. doi: 10.3934/cpaa.2006.5.855. [29] L. Ma and D. Chen, Radial symmetry and monotonicity for an integral equation,, J. Math. Anal. Appl., 2 (2008), 943. doi: 10.1016/j.jmaa.2007.12.064. [30] L. Ma and L. Zhao, Classification of positive solitary solutions of the nonlinear Choquard equation,, Arch. Rat. Mech. Anal., 2 (2010), 455. doi: 10.1007/s00205-008-0208-3. [31] N. Phuc and I. Verbitsky, Quasilinear and Hessian equations of Lane-Emden type,, Annals of Math., 168 (2008), 859. doi: 10.4007/annals.2008.168.859. [32] S. Sobolev, On a theorem of functional analysis,, Mat. Sb. (N.S.), 4 (1938), 471. [33] N. Trudinger and X. Wang, Hessian measure II,, Annals of Math., 150 (1999), 579. doi: 10.2307/121089. [34] J. Wei and X. Xu, Classification of solutions of higher order conformally invariant equations,, Math. Ann., 313 (1999), 207. doi: 10.1007/s002080050258.
show all references
References:
[1] L. Caffarelli, B. Gidas and J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth,, C. P. A. M., XLII (1989), 271. [2] W. Chen, C. Jin, C. Li and C. Lim, Weighted Hardy-Littlewood-Sobolev inequalities and system of integral equations,, Disc. Cont. Dyn. Sys., 2005 (): 164. [3] W. Chen and C. Li, Regularity of Solutions for a system of integral equations,, Comm. Pure and Appl. Anal., 4 (2005), 1. [4] W. Chen and C. Li, The best constant in some weighted Hardy-Littlewood-Sobolev inequality,, Proc. AMS, 136 (2008), 955. [5] W. Chen and C. Li, Classification of positive solutions for nonlinear differential and integral systems with critical exponents,, Acta Mathematica Scientia, 4 (2009), 949. doi: 10.1016/S0252-9602(09)60079-5. [6] W. Chen and C. Li, An integral system and the Lane-Emden conjecture,, Disc. Cont. Dyn. Sys., 4 (2009), 1167. [7] W. Chen and C. Li, Classification of solutions of some nonlinear elliptic equations,, Duke Math. J., 63 (1991), 615. doi: 10.1215/S0012-7094-91-06325-8. [8] W. Chen and C. Li, Super polyharmonic property of solutions for PDE systems and its applications, , submitted to Trans. AMS, (2011). [9] W. Chen and C. Li, "Methods on Nonlinear Elliptic Equations,", AIMS Book Series on Diff. Equa. & Dyn. Sys., 4 (2010). [10] C. Ma, W. Chen, and C. Li, Regularity of solutions for an integral system of Wolff type,, Advances of Math, 3 (2011), 2676. doi: 10.1016/j.aim.2010.07.020. [11] W. Chen, C. Li and B. Ou, Classification of solutions for an integral equation,, Comm. Pure Appl. Math., LLVIII (2005), 1. [12] W. Chen, C. Li and B. Ou, Qualitative properties of solutions for an integral equation,, Disc. Cont. Dyn. Sys., 12 (2005), 347. [13] W. Chen, C. Li and B. Ou, Classification of solutions for a system of integral equations,, Comm. PDEs, 30 (2005), 59. doi: 10.1081/PDE-200044445. [14] B. Gidas, W. Ni and L. Nirenberg, Symmetry of positive solutions of nonlinear elliptic equations in $R^n,$, Advances in Mathematics, 7a (1981). [15] F. Hang, On the integral systems related to Hardy-Littlewood-Sobolev inequality,, Math Res Lett, (2007). [16] F. Hang, X. Wang and X. Yan, An integral equation in conformal geometry,, Ann. H. Poincare Nonl. Anal., 26 (2009), 1. doi: 10.1016/j.anihpc.2007.03.006. [17] G. Hardy and J. Littelwood, Some properties of fractional integrals I,, Math. Zeitschr., 27 (1928), 565. doi: 10.1007/BF01171116. [18] C. Jin and C. Li, Symmetry of solutions to some systems of integral equations,, Proc. AMS, 134 (2006), 1661. [19] T. Kilpelaiinen and J. Maly, The Wiener test and potential estimates for quasilinear elliptic equations,, Acta Math., 172 (1994), 137. doi: 10.1007/BF02392793. [20] C. Li, Local asymptotic symmetry of singular solutions to nonlinear elliptic equations,, Invent. Math., 123 (1996), 221. [21] E. Lieb, Sharp constants in the Hardy-Littlewood-Sobolev and related inequalities,, Annals of Math, 118 (1983), 349. doi: 10.2307/2007032. [22] S. Liu, Regularity, symmetry, and uniqueness of some integral type quasilinear equations,, Nonlinear Analysis: Theory, 71 (2009), 1796. [23] Y. Y. Li, Remarks on some conformally invariant integral equations: The method of moving spheres,, J. Euro. Math. Soc., 6 (2004), 153. doi: 10.4171/JEMS/6. [24] C. Li and J. Lim, The singularity analysis of solutions to some integral equations,, Comm. Pure Appl. Anal., 6 (2007), 453. doi: 10.3934/cpaa.2007.6.453. [25] C. Li and L. Ma, Uniqueness of positive bound states to Shrödinger systems with critical exponents,, SIAM J. of Appl. Anal., 40 (2008), 1049. doi: 10.1137/080712301. [26] C. Liu and S. Qiao, Symmetry and monotonicity for a system of integral equations,, Comm. Pure Appl. Anal., 6 (2009), 1925. doi: 10.3934/cpaa.2009.8.1925. [27] D. Li and R. Zhuo, An integral equation on half space,, Proc. AMS, 138 (2010), 2779. [28] L. Ma and D. Chen, A Liouville type theorem for an integral system,, Comm. Pure Appl. Anal., 5 (2006), 855. doi: 10.3934/cpaa.2006.5.855. [29] L. Ma and D. Chen, Radial symmetry and monotonicity for an integral equation,, J. Math. Anal. Appl., 2 (2008), 943. doi: 10.1016/j.jmaa.2007.12.064. [30] L. Ma and L. Zhao, Classification of positive solitary solutions of the nonlinear Choquard equation,, Arch. Rat. Mech. Anal., 2 (2010), 455. doi: 10.1007/s00205-008-0208-3. [31] N. Phuc and I. Verbitsky, Quasilinear and Hessian equations of Lane-Emden type,, Annals of Math., 168 (2008), 859. doi: 10.4007/annals.2008.168.859. [32] S. Sobolev, On a theorem of functional analysis,, Mat. Sb. (N.S.), 4 (1938), 471. [33] N. Trudinger and X. Wang, Hessian measure II,, Annals of Math., 150 (1999), 579. doi: 10.2307/121089. [34] J. Wei and X. Xu, Classification of solutions of higher order conformally invariant equations,, Math. Ann., 313 (1999), 207. doi: 10.1007/s002080050258.
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# Sudoku recognizer¶
#### Alejandro Hernández Cordero (GitHub ID: ahcorde) ¶
This notebook details how to recognize a sudoku puzzle from a picture. We'll make use of simple image processing algorithms (edge detection, thresholds... ) and character recognition using the K-Nearest Neighbors (KNN) algorithm. It is a very simple but effective algorithm for solving multi-class classification problems. This puzzle matrix is a 9x9 array of known numbers 1-9 or 0s where the number is unknown
This ipython notebook is divided into two parts. The first part is related to computer vision and the second part is related to machine learning. To complete this task it is necessary to use a computer vision library, in this case we are going to use OpenCV library.
## Introduction¶
The first thing we need to do is to identify the puzzle. We have some challenges: The lines are not perfect, the black grid lines have similar color as a lot of elements in the image or the small squares are difficult to extract.
To simplify the problem we assumed: the puzzle is the biggest square in the image and the puzzle will be orientated reasonably correctly.
In the next picture you can see a sudoku puzzle in a newspaper. To load the image it's used imread.
In [1]:
%matplotlib inline
import pylab as pl
#import Opencv library
try:
import cv2
except ImportError:
print "You must have OpenCV installed"
#Show Images
_=pl.imshow(image_sudoku_original)
_=pl.axis("off")
## Step 1: Segmenting the Sudoku¶
Once we have read the image we have to detect the lines. For this task, It used an adaptativeThreshold to extract the edge of the image (for each pixel in the image take the average value of the surrounding area). This function accepts a gray scale image, with the function cvtColor we change the color space (from RGB to gray scale). In this picture it's possible to see letters, lines or numbers. Light pixels are the paper and dark pixels are the ink.
The adaptativeThreshold function takes as first argument the input image, the second argument returns the binary image, the third argument is the non-zero value assigned to the pixels for which the condition is satisfied. The fourth is the adaptive thresholding algorithm, the fifth argument is the threshold type, the next argument is size of the pixel neighborhood and finally the last argument is a constant that is subtracted from the mean or weight mean.
In [2]:
#gray image
image_sudoku_gray = cv2.cvtColor(image_sudoku_original,cv2.COLOR_BGR2GRAY)
#show image
_=pl.imshow(thresh, cmap=pl.gray())
_=pl.axis("off")
We have to find the connected contours in the image (using the function findContours). These function returns a vector with the corners of the contours. We'll find the sukodu in this contours.
The assumption is that the sudoku puzzle has 4 sides and it's convex. Checking the number of the contour is equal to four and using the funcion of OpenCV isContourConvex to check if the square is convex. We obtain the possible candidates.
We need to filter the candidates using the assumption: the sudoku puzzle is the biggest square in the image. We calculate the area of the possible contours (contourArea). The biggest square is the sudoku puzzle.
In [3]:
#find the countours
contours0,hierarchy = cv2.findContours( thresh,
cv2.RETR_LIST,
cv2.CHAIN_APPROX_SIMPLE)
#size of the image (height, width)
h, w = image_sudoku_original.shape[:2]
#copy the original image to show the posible candidate
image_sudoku_candidates = image_sudoku_original.copy()
#biggest rectangle
size_rectangle_max = 0;
for i in range(len(contours0)):
#aproximate countours to polygons
approximation = cv2.approxPolyDP(contours0[i], 4, True)
#has the polygon 4 sides?
if(not (len (approximation)==4)):
continue;
#is the polygon convex ?
if(not cv2.isContourConvex(approximation) ):
continue;
#area of the polygon
size_rectangle = cv2.contourArea(approximation)
#store the biggest
if size_rectangle> size_rectangle_max:
size_rectangle_max = size_rectangle
big_rectangle = approximation
In the image below it's possible to see the 4 sides of the sudoku puzzle in red
In [4]:
#show the best candidate
approximation = big_rectangle
for i in range(len(approximation)):
cv2.line(image_sudoku_candidates,
(big_rectangle[(i%4)][0][0], big_rectangle[(i%4)][0][1]),
(big_rectangle[((i+1)%4)][0][0], big_rectangle[((i+1)%4)][0][1]),
(255, 0, 0), 2)
#show image
_=pl.imshow(image_sudoku_candidates, cmap=pl.gray())
_=pl.axis("off")
Now we have the sudoku puzzle segmented. We have got the corner points of the puzzle. It's currently not really usable for much. The sudoku puzzle is a bit distorted. It's necessary to correct the skewed perspective of the image. We need a way to mapping from the puzzle in the original picture back into a square. Where each corner of the sudoku puzzle corresponds to a corner on the a new image.
We use a transformation that will map one arbitrary 2D quadrilateral into another. We can use a perspective transformation:
$$X = \frac{ax + by + c}{gx + hy +1}$$$$Y = \frac{dx + ey + f}{gx + hy +1}$$
This perspective transformation maps a point $(x, y)$ in one quadrilateral into a new $(X, Y)$ in another quadrilateral. These two equations contain 8 unknowns, but we have 8 values. (the corners $x$ and $y$ coordinates of the puzzle). Solving these equations gives us the $a,b,c,d,e,f,g,h$ which provide us with a mapping to get our puzzle out nice and straight.
The OpenCV function getperspectivetransform resolved the perspective transformation. Calculates a perspective transform from four pairs of the corresponding points, where the first parameter are the coordinates of quadrangle vertices in the source image and the second parameter are the coordinates of the corresponding quadrangle vertices in the destination image.
To applies a perspective transformation to an image we used the function warpPerspective. This function transforms the source image using the specified matrix. We obtains the image below.
We need to sort the corner of the sudoku puzzle and then associate each point with the new image dimension.
In [5]:
import numpy as np
IMAGE_WIDHT = 16
IMAGE_HEIGHT = 16
SUDOKU_SIZE= 9
N_MIN_ACTVE_PIXELS = 10
#sort the corners to remap the image
def getOuterPoints(rcCorners):
ar = [];
ar.append(rcCorners[0,0,:]);
ar.append(rcCorners[1,0,:]);
ar.append(rcCorners[2,0,:]);
ar.append(rcCorners[3,0,:]);
x_sum = sum(rcCorners[x, 0, 0] for x in range(len(rcCorners)) ) / len(rcCorners)
y_sum = sum(rcCorners[x, 0, 1] for x in range(len(rcCorners)) ) / len(rcCorners)
def algo(v):
return (math.atan2(v[0] - x_sum, v[1] - y_sum)
+ 2 * math.pi) % 2*math.pi
ar.sort(key=algo)
return ( ar[3], ar[0], ar[1], ar[2])
The dataset images have 16x16 pixels. The size of the new image will be 144x144, because we have to divide each row and col by 9 and we have to return the same size of the images.
In [6]:
#point to remap
points1 = np.array([
np.array([0.0,0.0] ,np.float32) + np.array([144,0], np.float32),
np.array([0.0,0.0] ,np.float32),
np.array([0.0,0.0] ,np.float32) + np.array([0.0,144], np.float32),
np.array([0.0,0.0] ,np.float32) + np.array([144,144], np.float32),
],np.float32)
outerPoints = getOuterPoints(approximation)
points2 = np.array(outerPoints,np.float32)
#Transformation matrix
pers = cv2.getPerspectiveTransform(points2, points1 );
#remap the image
warp = cv2.warpPerspective(image_sudoku_original, pers, (SUDOKU_SIZE*IMAGE_HEIGHT, SUDOKU_SIZE*IMAGE_WIDHT));
warp_gray = cv2.cvtColor(warp, cv2.COLOR_BGR2GRAY)
#show image
_=pl.imshow(warp_gray, cmap=pl.gray())
_=pl.axis("off")
we've now got an undistorted square Sudoku puzzle
## Step 2: Segmenting the numbers¶
After remapping, we can divide the puzzle into a 9×9 grid. Each cell in the grid will correspond (approximately) to a cell on the puzzle. The correspondence might not be perfect, but it should be good enough.
Now, we know the size of the bounding box. We also know the size of the image. We can easily center the image. We create a new image that’s the same size as the original.
Sending images directly to k-Nearest Neighbors is not a good idea. A little bit of work on the image can increase accuracy. Here, we’ll center the actual digit in the image. The dataset has been made in such a way, all digits are centered by their bounding box.
To remove the noise around the number (lines in general) we delete all the pixels outside the center of the image from a radius.
In [7]:
def extract_number(x, y):
#square -> position x-y
im_number = warp_gray[x*IMAGE_HEIGHT:(x+1)*IMAGE_HEIGHT][:, y*IMAGE_WIDHT:(y+1)*IMAGE_WIDHT]
#threshold
#delete active pixel in a radius (from center)
for i in range(im_number.shape[0]):
for j in range(im_number.shape[1]):
dist_center = math.sqrt( (IMAGE_WIDHT/2 - i)**2 + (IMAGE_HEIGHT/2 - j)**2);
if dist_center > 6:
im_number_thresh[i,j] = 0;
n_active_pixels = cv2.countNonZero(im_number_thresh)
return [im_number, im_number_thresh, n_active_pixels]
Sometimes after remove the noise of the images can appear other particles close to the number. Because of this, we find the biggest bounding box in the small squared. We make the bounding box a little more bigger to improve the matching with the dataset.
In [8]:
def find_biggest_bounding_box(im_number_thresh):
contour,hierarchy = cv2.findContours(im_number_thresh.copy(),
cv2.RETR_CCOMP,
cv2.CHAIN_APPROX_SIMPLE)
biggest_bound_rect = [];
bound_rect_max_size = 0;
for i in range(len(contour)):
bound_rect = cv2.boundingRect(contour[i])
size_bound_rect = bound_rect[2]*bound_rect[3]
if size_bound_rect > bound_rect_max_size:
bound_rect_max_size = size_bound_rect
biggest_bound_rect = bound_rect
#bounding box a little more bigger
x_b, y_b, w, h = biggest_bound_rect;
x_b= x_b-1;
y_b= y_b-1;
w = w+2;
h = h+2;
return [x_b, y_b, w, h]
Now we have to separate all the numbers from the grid. First we extract the numbers. If we have more active pixels than a threshold, we find the biggest bounding box in the image if the size of this bounding box is bigger than 0, we store this number in a matrix. In another case we store a matrix of zeros.
In [9]:
import math
import numpy as np
#sudoku representation
sudoku = np.zeros(shape=(9*9,IMAGE_WIDHT*IMAGE_HEIGHT))
def Recognize_number( x, y):
"""
Recognize the number in the rectangle
"""
#extract the number (small squares)
[im_number, im_number_thresh, n_active_pixels] = extract_number(x, y)
if n_active_pixels> N_MIN_ACTVE_PIXELS:
[x_b, y_b, w, h] = find_biggest_bounding_box(im_number_thresh)
number = im_t[y_b:y_b+h, x_b:x_b+w]
if number.shape[0]*number.shape[1]>0:
number = cv2.resize(number, (IMAGE_WIDHT, IMAGE_HEIGHT), interpolation=cv2.INTER_LINEAR)
ret,number2 = cv2.threshold(number, 127, 255, 0)
number = number2.reshape(1, IMAGE_WIDHT*IMAGE_HEIGHT)
sudoku[x*9+y, :] = number;
return 1
else:
sudoku[x*9+y, :] = np.zeros(shape=(1, IMAGE_WIDHT*IMAGE_HEIGHT));
return 0
Here you can see the segmented numbers in a grid
In [10]:
index_subplot=0
n_numbers=0
indexes_numbers = []
for i in range(SUDOKU_SIZE):
for j in range(SUDOKU_SIZE):
if Recognize_number(i, j)==1:
if (n_numbers%5)==0:
index_subplot=index_subplot+1
indexes_numbers.insert(n_numbers, i*9+j)
n_numbers=n_numbers+1
#create subfigures
f,axarr= pl.subplots(index_subplot,5)
width = 0;
for i in range(len(indexes_numbers)):
ind = indexes_numbers[i]
if (i%5)==0 and i!=0:
width=width+1
axarr[i%5, width].imshow(cv2.resize(sudoku[ind, :].reshape(IMAGE_WIDHT,IMAGE_HEIGHT), (IMAGE_WIDHT*5,IMAGE_HEIGHT*5)), cmap=pl.gray())
axarr[i%5, width].axis("off")
## Step 3: Recognizing digits (Machine learning)¶
In machine learning, classification is the problem of identifying to which of a set of categories a new observation belongs. Classification is an example of the more general problem of pattern recognition. Which is the assignment of some sort of output value to a given input value. The k-Nearest Neighbors algorithm (or KNN) is a non-parametric method used for classification and regression.
In SHOGUN, you can use CKNN to perform k-Nearest Neighbors algorithm, it's a non-parametric method used for classification and regression. To construct a KNN machine, you must choose the hyper-parameter K and a distance function. Usually, we simply use the standard CEuclideanDistance, but in general, any subclass of CDistance could be used.
We need to load the training data. but the dataset values are between -1 and 1. We need to normalize the data between 0-255.
In [11]:
from modshogun import MulticlassLabels,RealFeatures
from modshogun import KNN, EuclideanDistance
from numpy import random
Xall = mat['data']
#normalize between 0-255
normalize = np.zeros((Xall.shape[1], Xall.shape[0],1), np.uint8)
normalize= cv2.normalize(Xall, normalize, 0, 255, cv2.NORM_MINMAX, cv2.CV_8UC1);
Xall = np.asmatrix(normalize, np.float64)
Yall = np.array(mat['label'].squeeze(), dtype=np.double)
Yall = Yall - 1
#random position to train knn
random.seed(0)
subset = random.permutation(len(Yall))
test_images = Xall[:, subset[:5000]]
test_labels = Yall[subset[:5000]]
We initialize the knn, set up the features and configure the kind of distance that we want to use (in this case Euclidean distance).
In [12]:
k=10
labels_numbers = MulticlassLabels(test_labels)
feats = RealFeatures(test_images)
dist = EuclideanDistance()
knn = KNN(k, dist, labels_numbers)
_=knn.train(feats)
Now we have to take all the segmented numbers and returns what digit it is.
In [13]:
sudoku_puzzle = np.zeros(shape=(9,9))
#Predict the number
for i in range(SUDOKU_SIZE):
for j in range(SUDOKU_SIZE):
#has the image some active pixel?
n_active_pixels = cv2.countNonZero(np.asmatrix(sudoku[i*9+j, :]))
if n_active_pixels>20 :
feats_test = RealFeatures( np.asmatrix(sudoku[i*9+j, :], np.float64).T)
sudoku_puzzle[i,j] = knn.apply_multiclass(feats_test)[0]
else:
sudoku_puzzle[i,j] = 0
print sudoku_puzzle
#show image
_=pl.imshow(image_sudoku_original)
_=pl.axis("off")
[[ 0. 3. 0. 4. 0. 0. 0. 0. 0.]
[ 4. 0. 0. 0. 0. 0. 0. 0. 2.]
[ 2. 0. 0. 5. 0. 0. 7. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 2. 0. 0. 0. 4. 0. 0. 0.]
[ 3. 0. 0. 0. 0. 0. 0. 4. 0.]
[ 0. 0. 0. 0. 7. 0. 0. 3. 0.]
[ 0. 4. 0. 2. 0. 0. 0. 0. 8.]
[ 7. 0. 0. 0. 0. 0. 4. 0. 0.]]
The recognition isn't very accurate because the training data is for handwritten text.
The accuracy of the process could be improved. In this example we have used a holistic method, all the image is introduced in the classifier. But adding new features to the classifier, for example the number of holes, number of straight lines or length of the lines, could improve the accuracy of the classifier.
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BEST FRUSTRATION RELIEF IDEAS???? : The B-School Application
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# BEST FRUSTRATION RELIEF IDEAS????
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03 Jan 2010, 18:18
I am about go nuts writing essays and submitting apps and editing information and creating documents and copying transcripts. Does anyone have any ideas on how to get instantaneous relief from application frustration?!?!?!!?
------v
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03 Jan 2010, 18:35
Sleep for some time.
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03 Jan 2010, 18:41
1
KUDOS
oh man, if only I won't get banned from typing the first thing on my mind
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 18:42
sadly i've turned to eating...this plus my workout time being cutback in lieu of apps has made me not feel so sexy. this process is taxing on all fronts.
_________________
My story of an average chick who stumbled into the 700+ club
My 2009-2010 Application Decisions
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 19:18
I honestly do not even want to apply to apply to bschool this cycle anymore. I feel like my essays would be much better next year. I just decided to apply to bschool after doing well on my gmat. Not really thinking things out. I do not even really like the programs I am applying to anymore.
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 19:36
I hear you. Working on the Chicago Powerpoint, for example, has made me want to trash the entire application altogether. I feel discouraged about all my essays. Haven't been to the gym in weeks. I'm trying to chalk it all up to end-of-round burn-out, though. Just a little longer, y'all! We can do this!
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 19:48
2012dreams: hope you are staying healthy and eating up to your avatar
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 20:23
DITTO!! I dont want to read my essays anymore and I still need to finish some :S :S :S.
MY MOTIVATION IS THAT THIS REALLY IS THE LAST STEP! After the gmat studying, the research, the preparation, etc etc etc... this is the last step!!!!
Listen to an upbeat song! Dance a little! http://www.youtube.com/watch?v=5iBqhMfyoPU&feature=fvst
... and continue working. Remember there are probably A LOT of people exactly like us right now.
GOOD LUCK! GO GO GO!
Manager
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 21:18
Aenigma wrote:
I hear you. Working on the Chicago Powerpoint, for example, has made me want to trash the entire application altogether. I feel discouraged about all my essays. Haven't been to the gym in weeks. I'm trying to chalk it all up to end-of-round burn-out, though. Just a little longer, y'all! We can do this!
Hilarious. The PPT was actually what made me post this but after getting 90% done, I think that they look great and I am encouraged to go make my essays better because of it. Anyways, I think that I was in a slump as I am sure many of you have been/are in. I am on a roll now so no moping for the time being.
Don't worry though, I will be back on here when the red bull wears off.
Manager
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 21:19
justniky wrote:
DITTO!! I dont want to read my essays anymore and I still need to finish some :S :S :S.
MY MOTIVATION IS THAT THIS REALLY IS THE LAST STEP! After the gmat studying, the research, the preparation, etc etc etc... this is the last step!!!!
Listen to an upbeat song! Dance a little! http://www.youtube.com/watch?v=5iBqhMfyoPU&feature=fvst
... and continue working. Remember there are probably A LOT of people exactly like us right now.
GOOD LUCK! GO GO GO!
Manager
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Posts: 101
Schools: HBS, Stanford, Haas, Booth, Columbia
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 21:21
Forgot the topic for a second.
My Tip for frustration is:
BOXING!
I went and punched my friends bag and it really helped. I am exhausted now but at least I won't demolish my macbook or submit my application prematurely out of want to rid myself of the burdens of completion.
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 22:11
asimov wrote:
oh man, if only I won't get banned from typing the first thing on my mind
Ha ha! couldn't think of getting up from the table while working on my essays let alone climbing on the bed!
My suggestion - Pump iron like crazy and then have a swedish massage.
Unplugged
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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03 Jan 2010, 22:12
aj1545 wrote:
Aenigma wrote:
I hear you. Working on the Chicago Powerpoint, for example, has made me want to trash the entire application altogether. I feel discouraged about all my essays. Haven't been to the gym in weeks. I'm trying to chalk it all up to end-of-round burn-out, though. Just a little longer, y'all! We can do this!
Hilarious. The PPT was actually what made me post this but after getting 90% done, I think that they look great and I am encouraged to go make my essays better because of it. Anyways, I think that I was in a slump as I am sure many of you have been/are in. I am on a roll now so no moping for the time being.
Don't worry though, I will be back on here when the red bull wears off.
Good for you, and good luck!
My PPT looks like a rabid chihuahua ate my mom's scrapbook of all my travels and accomplishments... and then vomited it onto a pastel-colored middle-school cafeteria tray. If a toddler scribbled on it in crayon it would probably look better.
Sigh.
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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04 Jan 2010, 00:23
Aenigma wrote:
aj1545 wrote:
Aenigma wrote:
I hear you. Working on the Chicago Powerpoint, for example, has made me want to trash the entire application altogether. I feel discouraged about all my essays. Haven't been to the gym in weeks. I'm trying to chalk it all up to end-of-round burn-out, though. Just a little longer, y'all! We can do this!
Hilarious. The PPT was actually what made me post this but after getting 90% done, I think that they look great and I am encouraged to go make my essays better because of it. Anyways, I think that I was in a slump as I am sure many of you have been/are in. I am on a roll now so no moping for the time being.
Don't worry though, I will be back on here when the red bull wears off.
Good for you, and good luck!
My PPT looks like a rabid chihuahua ate my mom's scrapbook of all my travels and accomplishments... and then vomited it onto a pastel-colored middle-school cafeteria tray. If a toddler scribbled on it in crayon it would probably look better.
Sigh.
That description was hilarious. I am pretty sure that I know exactly what it looks like. I'm sure you will edit a few things and it will go from crayola to art deco in no time. There always is a fine line between crazy and great.
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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04 Jan 2010, 07:06
asimov wrote:
oh man, if only I won't get banned from typing the first thing on my mind
hhahahaha....
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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04 Jan 2010, 07:06
Aenigma wrote:
aj1545 wrote:
Aenigma wrote:
I hear you. Working on the Chicago Powerpoint, for example, has made me want to trash the entire application altogether. I feel discouraged about all my essays. Haven't been to the gym in weeks. I'm trying to chalk it all up to end-of-round burn-out, though. Just a little longer, y'all! We can do this!
Hilarious. The PPT was actually what made me post this but after getting 90% done, I think that they look great and I am encouraged to go make my essays better because of it. Anyways, I think that I was in a slump as I am sure many of you have been/are in. I am on a roll now so no moping for the time being.
Don't worry though, I will be back on here when the red bull wears off.
Good for you, and good luck!
My PPT looks like a rabid chihuahua ate my mom's scrapbook of all my travels and accomplishments... and then vomited it onto a pastel-colored middle-school cafeteria tray. If a toddler scribbled on it in crayon it would probably look better.
Sigh.
That sounds *awesome*
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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04 Jan 2010, 07:11
justniky wrote:
DITTO!! I dont want to read my essays anymore and I still need to finish some :S :S :S.
MY MOTIVATION IS THAT THIS REALLY IS THE LAST STEP! After the gmat studying, the research, the preparation, etc etc etc... this is the last step!!!!
Listen to an upbeat song! Dance a little! http://www.youtube.com/watch?v=5iBqhMfyoPU&feature=fvst
... and continue working. Remember there are probably A LOT of people exactly like us right now.
GOOD LUCK! GO GO GO!
I see you and raise you : http://www.youtube.com/watch?v=VCbO-3hjPok
Always calms me down.
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Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink]
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04 Jan 2010, 07:17
Not particularly creative but I have found that drinking a couple glasses of red wine helps immensely
Although it renders me sleepy and useless for essay-writing so it's not recommended for a night when you need to buckle down.
Re: BEST FRUSTRATION RELIEF IDEAS???? [#permalink] 04 Jan 2010, 07:17
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# What is the interpretation of a glm coefficient on a dependent variable that has a % interpretation
I have a dependent variable that takes on values between 0 and 1, including 0 and 1. The variable signifies a proportion (0 = nothing, 1 = all). I am running a model of the type:
model<- glm(y~x, family=quasibinomial(link = "logit"), data=data)
Now my output looks as follows:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.6510322 0.1614316 10.227 < 0.0000000000000002 ***
x -0.8501458 0.3192826 -2.663 0.007758 **
Doing exp(coef(model)) gives a value for x of 1.97.
My question now is, how do I interpret this coefficient with respect to my dependent variable that is proportion?
Should I interpret it as, an increase in x by 1, (1.97 - 1 = 0.97), decreases the chance that y is one with 97% (which would be kind of annoying because I also want information about the values between zero and one).
Or should I interpret it as, an increase in x by 1 decreases y by 97% (so essentially doubles the value of y)
Or even something different from this?
# EDIT I:
Isabella asked me to provide some more information about the nature of my dependent variable (and I am also posting additional information I found since posting this question, see EDIT II).
My dependent variable, has a continuous range between 0-100% (I scaled it down for this approach) and its distribution looks like the picture below. Although the answer scale is continuous, there is quite a lot of concentration around certain integers (and especially 100%).
I have posted related questions on what model to use here and whether it is possible to demean the dependent variable here.
# EDIT II:
Finally, I found this really nice Stata video on fractional regression (the dependent variable is a proportion including 0 and 1) and this really nice blog on fractional regression, and its counterparts in R, which shows that fractional regression is essentially simply a glm with family quasibinomial. The video shows how the coefficients of the fractional regression (or glm) can be interpreted with the margins command in Stata. I am currently trying to figure out how to achieve the same thing in R (see this post), which now has been answered.
• I would have thought that $1.97$ is the odds ratio: For each unit increase in x, the odds that $y=1$ increase by a factor of $1.97$. You could also convert the odds to a probabiltiy and plot it over the range of x. Apr 14, 2021 at 14:39
• @COOLSerdash Thank you for you comment! Your interpretation is essentially the same as my first guess right (increase by 97% = by a factor of 1.97)? Could you perhaps elaborate on how I can achieve your second comment: "You could also convert the odds to a probabiltiy and plot it over the range of x. " ?
– Tom
Apr 14, 2021 at 14:42
• Your interpretation was slightly off: It's not y that increases by a factor of 1.97 but the odds. Otherwise you'd get probabilities >1, which is not possible. For the plotting, I'd use a package, such as visreg. Then you could use the following code: visreg(model, "x", scale = "response") Apr 14, 2021 at 14:44
• I think the confusion is from the fact that I listed two alternative "possible" explanations. In any case thanks, I will check out the package.
– Tom
Apr 14, 2021 at 14:49
Another fun thread! @COOLSerdash is one of my favourites on this forum!
Tom, when you model an outcome variable expressed as a proportion, you have to be a bit careful with your modelling, as I'll explain below.
In statistics, we tend to think of a proportion as either discrete or continuous. How you model discrete proportions is generally different from how you model continuous proportions.
An example of discrete proportion: proportion of correct answers for an exam with 10 questions. If a student answered correctly 5 of the 10 questions, the discrete proportion of correct answers for that student would be 5/10 or 0.50. Clearly, this type of proportion is the ratio of two discrete counts, hence its labelling as discrete.
An example of continuous proportion: proportion of a study site covered with grass. If a study site had an area of 10.2 squared km and only 5.4 squared km of that area would be covered with grass, then the proportion of the study site covered with grass would amount to 5.4/10.2 = 0.53. This type of proportion is the ratio of two continuous quantities, hence its labelling as continuous.
When the outcome variable in a regression modelling setting takes values that are discrete proportions (and we know the numerator and denominator counts used to obtain those proportions), we use binomial regression modelling (or variations thereof) to relate it to the predictor variables of interest .
When the outcome variable takes values that are continuous proportions, we use beta regression modelling (or variations thereof).
From your post, it's not clear what type of proportions you are dealing with. Once you clarify that, we can proceed to the next step.
EDIT
Thanks for your edits and the excellent resources on fractional regression you shared, Tom.
The concentration you noticed at certain values in your response variable is referred to as inflation in statistical jargon. If you convert your outcome variable to a (continuous) proportion, you can see that you are spoiled for choice in terms of modelling options.
If you only had inflation at 0 and/or 1, you could have used zero and/or one-inflated beta regression (as available in the gamlss package of R, say). But you have inflation at other values in between 0 and 1, so I don't think beta regression is a viable option.
This leaves you with the choice of fractional regression (aka quasibinomial regression with a logit link). (You could also try the mixed effects modelling route with an observation level random effect.)
This is essentially your original modelling option, so we have come full circle. The question now is how you interpret the results.
Personally, I always like to think first about what we are modelling before proceeding with the interpretation. In your case, you are modelling the logit-transformed expected proportion at a given x as a function of x. Something like this:
logit(expected proportion) = beta0 + beta1*x
The expected proportion is the (true) average proportion across all units/subjects in your target population having the same value of x.
If we denote the expected proportion via ep for convenience, the logit transformation is in effect:
logit(ep) = log(ep/(1-ep))
These types of models with a logit link can be interpreted on multiple scales.
On the logit scale, you could just say things like:
Each 1-unit increase in the value of x is associated with a change of b1 points on the logit-transformed value of the expected proportion.
Here, b1 is the estimated value of beta1 from the data. (People talk about points as being units on a logit scale.)
The "odds" scale would be the next possible scale, except that the odds terminology makes more sense with a probability rather than an expected proportion. But that is the scale you would find yourself working with if you exponentiated the value of b1. In other words, exp(b1) would give you the multiplicative factor by which the ratio ep/(1-ep) would change when the value of x increases by one unit. My own view is that ep/(1-ep) does not represent odds per say because ep is not a probability, it is a continuous proportion. So I would just talk about the ratio ep/(1-ep) without calling it odds.
Note that, if you compute (exp(b1) - 1)*100%, you get the % change in the value of the ratio ep/(1-ep) associated with a 1-unit increase in the value of x.
The last possible scale for interpretation of the effect of x is the scale of the expected proportion itself. One can show that:
ep = exp(beta0 + beta1x)/(1 + exp(beta0 + beta1x))
If you plug in the estimated values of beta0 (i.e., b1) and beta1 (i.e., b1) from your model summary output for the quasibinomial regression with a logit link, you get to see how the estimated value of ep (expected proportion) varies with the values of x and can easily visualize that. Of course, on the expected proportion scale, x has a nonlinear effect so you can just qualitatively describe it or just mention whether your study provides evidence that x has a positive nonlinear effect on the expected proportion (if b1 > 0; p-value for testing H0: beta1 = 0 vs Ha: beta1 != 0 reasonably small) or a negative nonlinear effect on the expected proportion (if b1 < 0; p-value for testing H0: beta1 = 0 vs Ha: beta1 != 0 reasonably small).
• Thank you for taking the time to answer my question. I will make an edit, explaining a little bit in more detail what my dependent variable looks like. I also found some info helping with the interpretation that I will add as well.
– Tom
Apr 15, 2021 at 7:03
• +1. I'm not an expert on this, but even with continuous proportions, where the denominator is not known, we can use the "usual" logistic regression (potentially with robust standard errors or quasilikelihood). I don't know if you have access to Stata, but fracreg gives exactly the same result as glm using robust standard errors. If there are 0s or 1s in the data, beta regression unfortunately can't be used. Otherwise, it's certainly a good alternative. Apr 15, 2021 at 7:56
• Thank you very, very much for your edit. I appreciate it enormously! I will have to go through it a couple of times to completely get it, but I will. If I can be so shameful: I just posted this question and I thought you might find that interesting as well: stats.stackexchange.com/questions/519852/… . It would be absolutely amazing if you would be willing to have a quick look (but obviously don't feel obliged.
– Tom
Apr 15, 2021 at 16:19
• You’re welcome, Tom! Ha!Ha! Only the first answer is free. 🤣 I already feel guilty for stepping over @COOLSerdash’s territory and taking over. But this thread is so much fun, so I couldn’t help myself. 😜 Apr 15, 2021 at 17:29
A model with a logit link function assumes a linear relationship between the logit-transformed variable (log-odds when we speak about logit transformed probabilities) and the predictors.
(Note that this assumption of linearity might be false and it can lead to the piranha problem when we extrapolate too far or with too many combined multiple effects)
Below you see an example image of the relationship with the logit link function.
The image is an example of the fraction of the UK Covid variant as function of time. This is modelled with a logistic function. Then the logit transformed data (the log-odds) are considered linear. (In this particular example you see that the model is not perfect and it is not exactly a straight line)
In the upper image it is the fraction on the y-axis. In the lower image it is the same data and fit but expressed as log-odds on the y-axis.
So the fit with a logit function has a linear interpretation in terms of logit transformed data. The coefficients from the output relate to the line in the graph with the logit transformed data on the y-axis.
In terms of the fraction the relationship is non-linear (the upper graph). At different points an increase in the parameter will have a different increase in the fraction. The coefficients will not have a clear/intuitive interpretation in terms of the fraction.
• Thank you very much for your answer! I really appreciate it (heel erg bedankt ;)). Is there any way you could link the terminology a little bit more to my code (not so much to my example, but more to what is what). I notice that I am a bit unfamiliar with the log-odds/logistic terminology (and maybe so might others reading this). Is the original glm the fractional interpretation and the exp() the log odds?
– Tom
Apr 15, 2021 at 16:39
• There really are no “odds” in your example, Tom, because you are modelling a continuous proportion which cannot be conceived as being a “probability”. If you were modelling a discrete proportion, then the “odds” terminology would make sense. In my EDIT to my original answer, I explained that ep/(1-ep) is your version of “odds” but you’re better off to just call it the ratio of the expected proportion to (1 - expected proportion), because that’s really all it is. It is not “odds” unless the expected proportion ep could be conceived as as a “probability”. Apr 16, 2021 at 14:41
• Being careful with terminology is import in a modelling context so that people can clearly understand what it is that you are modelling. Apr 16, 2021 at 14:44
• @Isabella discrete proportion can have a proportion as the parameter of the underlying distribution. For instance the binomial distribution has the parameter $p$ that relates to log odds $\log p/(1-p)$. But I agree that odds is not be a correct terminology here (I will see how to correct it). That is because the underlying variable is not really a proportion or probability. It seems to be a ranking between 0 and 10 which is neither a discrete proportion. Apr 17, 2021 at 6:31
• @Tom I have removed the log-odds out of my answer. Log-odds are a special case of a logit-transformed variable when the original variable represents a probability (for instance the parameter $p$ in a binomial distribution). Apr 17, 2021 at 6:37
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https://www.teacherspayteachers.com/Product/Factoring-Quadratics-Boom-Cards-leading-coefficient-not-1-Digital-Activity-3680878
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Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
Solutions 3-1. (a) Distance hiked = b + c km. b km (b) Displacement is a vector representing Paul’s change in position. Drawing a diagram of Paul’s trip we can see that displacement –c km his displacement is b + (–c) km east = (b –c) km east. (c) Distance = 5 km + 2 km = 7 km; Displacement = (5 km – 2 km) east = 3 km east. 3-2. (a) From v
d t
v
x . t
x . We want the answer in m/s so we’ll need to convert 30 km to meters and 8 min t to seconds: x 30,000 m m 30.0 km 1000 30, 000 m; 8.0 min 160mins 480 s. Then v 63 m . 1 km s t 480 s Alternatively, we can do the conversions within the equation: 1000 m x 30.0 km 1 km v 63 m . s t 8.0 min 160mins
(b) v
In mi/h: 1 mi h 30.0 km 1.61 18.6 mi; 8.0 min 601min 0.133 h. Then v km
x 18.6 mi 140 mi . h t 0.133 h
1 mi x 30.0 km 60 min x 30.0 km 1.61 km 1 mi mi Or, v 1 h 1.61 km 140 h . Or, v 140 mi . 1 h h t t 8.0 min 8.0 min 60 min
There is usually more than one way to approach a problem and arrive at the correct answer! d L v . t t L 24.0 m (b) v 40 ms . t 0.60 s
3-3. (a) From v
d x v . t t x 0.30 m (b) v 30 ms . t 0.010 s
3-4. (a) From v
d 2 r . t t 2 r 2 (400m) 63 ms . (b) v t 40s
3-5. (a) v
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d t
3-6. (a) t = ? From v (b) t
t
h . v
h 508 m 34 s. v 15 ms
(c) Yes. At the beginning of the ride the elevator has to speed up from rest, and at the end of the ride the elevator has to slow down. These slower portions of the ride produce an average speed lower than the peak speed. 3-7. (a) t = ? Begin by getting consistent units. Convert 100.0 yards to meters using the conversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft. d d 91.4 m 3ft 0.3048 m t . Then 100.0 yards 1yard 91.4 m. From v 1ft t v v d 91.4 m (b) t 15 s. v 6.0 ms
d d L t . t v c L 1.00 m 3.33 10-9s 3.33 ns. (This is 3 13 billionths of a second!) (b) t v 3.00 10 8 ms
3-8. (a) t = ? From v
d d vt. t (b) First, we need a consistent set of units. Since speed is in m/s let’s convert minutes to seconds:
3-9. (a) d = ? From v 60 s
5.0 min 1min 300 s. Then d v t 7.5 ms 300s 2300 m.
3-10. (a) v
v0 vf v . 2 2
(b) d ? From v (c) d
d vt d vt . t 2 (1.5s) 1.5 m.
2.0 ms vt 2 2
3-11. (a) d ? From v
d t
vt v v 0 v d vt 0 f t t . 2 2 2
12 ms (8.0s) vt 48 m. (b) d 2 2
3-12. (a) d ? From v
d t
vt v v 0 v d vt 0 f t t . 2 2 2
(b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is in 27.8 ms (8.0 s) vt km 1h 1000 m m 110 m. seconds). 100.0 h 3600 s 1 km 27.8 s . Then, d 2 2
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v v2 v1 . t t 15 km 25 km . Since our time is in seconds we need to convert km to (b) v 40 km h h h h
3-13. (a) a
m s
:
m v 6.94 s 0.35 m2 . s t 20 s Alternatively, we can express the speeds in m/s first and then do the calculation: 11.1 ms 4.17 ms 1hr 1000 m m km 1hr 1000 m m 15 km 4.17 and 40 11.1 . Then a 0.35 m2 . hr 3600 s 1 km s hr 3600 s 1 km s s 20s 1hr m 25 km 3600 1000 6.94 ms . Then a h s 1 km
v v2 v1 . t t (b) To make the speed units consistent with the time unit we’ll need v in m/s:
3-14. (a) a
1hr m v v2 v1 20.0 km 5.0 km 15.0 km 3600 1000 4.17 ms . Then a h h h s 1 km
m v2 v1 4.17 s 0.417 m2 . s t 10.0 s
An alternative is to convert the speeds to m/s first: 1hr m 1hr m v1 5.0 km 3600 1000 1.4 ms ; v2 20.0 km 3600 1000 5.56 ms . h s 1 km h s 1 km
Then a (c) d vt
5.56 ms 1.4 ms v2 v1 0.42 m2 . s t 10.0 s
1.4 ms 5.56 ms v1 v2 t 10.0 s 35 m. Or, 2 2
d v1t 12 at 2 1.4 ms (10.0 s) 12 0.42 m2 (10.0 s)2 35 m. s
v vf v0 0 v v . t t t t m v 26 s (b) a 1.3 m2 . s t 20s
3-15. (a) a
(c) d ? From v
d t
26 ms 0 ms v v d vt 0 f t 20 s 260 m. 2 2
Or, d v0t 12 at 2 26 ms (20 s) 12 1.3 m2 (20 s)2 260 m. s
(d) d = ? Lonnie travels at a constant speed of 26 m/s before applying the brakes, so
d vt 26 ms (1.5 s) 39 m. v vf v0 0 v v . t t t t m v 72 s 6.0 m2 . (b) a s t 12 s
3-16. (a) a
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(c) d ? From v
72 ms 0 ms v v d vt 0 f t (12 s) 430 m. 2 2
d t
Or, d v0t 12 at 2 72 ms (12 s) 12 6.0 m2 (12 s)2 430 m.
3-17. (a) t ? From v (b) t
d t
t
s
d L 2L v v . f 0 v v
2
2L 2(1.4 m) 0.19 s. v 15.0 ms
v v v 3-18. (a) v 0 f . 2 2 350 ms (b) v 175 ms . Note that the length of the barrel isn’t needed—yet!
2
(c) From v
d t
t
d L 0.40 m 0.0023 s 2.3 ms. v v 175 ms
3-19. (a) From v
d t
v v v v d vt 0 f t = 0 t. 2 2
25 ms 11 ms v v t = (b) d 0 (7.8 s) 140 m. 2 2
3-20. (a) v = ? There’s a time t between frames of
1 s, 24
so v=
d t
24 1 x. (That’s 24x per s s x
1 24
second.)
(b) v 24 1s x 24 1s (0.15 m) 3.6 ms . 3-21. (a) a = ? Since time is not a part of the problem we can use the formula vf 2 v0 2 2ad and
v2 solve for acceleration a. Then, with v0= 0 and d = x, a . 2x
1.8 10 7 ms v2 (b) a 2x 2(0.10 m)
1.6 10 2
15 m s2
.
1.8 10 7 ms 0 ms vf v0 1.1 10-8 s 11 ns. (c) t ? From vf v0 at t a 1.6 1015 m2 s
Or, from v d t
© Paul G. Hewitt and Phillip R. Wolf
d L 2L 2(0.10 m) -8 t v v 1.1 10 s. 0 f v (v 0) 1.8 10 7 m s 2
3-4
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d v0 vf 2d t with v0 0 vf . t 2 t
3-22. (a) vf =? From v
(b) af =? From d v0 t 12 at 2 with v0 0 d 12 at 2 (c) vf
a
2d t2
.
2d 2(402 m) 2d 2(402 m) 181 ms ; a 2 40.6 m2 . 2 s t 4.45 s t (4.45 s)
3-23. (a) d=? From v
v v v V d vt 0 f t = t. 2 2
d t
110 ms 250 ms v V t = (b) d (3.5 s) 630 m. 2 2
3-24. (a) t ? Let's choose upward to be the positive direction. v v 0v v From vf v0 at with vf 0 and a g t f 0 . a g g (b) t
m v 32 s 3.3 s. g 9.8 m2 s
m 2
2 32 s d v v v 0 v v (c) d ? From v d vt 0 f t 52 m. 2 2 g 2g 2 9.8 m t 2
9.8 (3.3 s) s
We get the same result with d v0t at 1 2
2
(3.3 s) 32 ms
3-25. (a) v0 = ? When the potato hits the ground y = 0. From d v0t 12 at 2
(b) v0 12 gt
1 2
y v0t 12 gt 2
9.8 (12 s) 59 m s2
0 t v0 12 gt
m . In s
1 2
2
m s2
52 m.
v0 12 gt.
1 km s 1 mi 3600 130 mi . mi/h, 59 ms 1000 m 1.61 km 1h h
3-26. (a) t = ? Choose downward to be the positive direction. From From d v0 t 12 at 2 with v0 0, a g and d h h 12 gt 2
(b) t
t
2h . g
2h 2(25m) 2.26 s 2.3 s . g 9.8 m2 s
. (c) vf vo at 0 gt 9.8 m2 (2.26 s) 22 m s s
2 2 m m Or, from 2ad vf v0 with a g, d h, and v0 0 vf 2gh 2 9.8 s2 (25 m) 22 s .
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3-27. (a) v0 = ? Let’s call upward the positive direction. Since the trajectory is symmetric, vf = –v0. gt Then from vf v0 at, with a g v0 v0 gt 2v0 gt v0 . 2
9.8 m2 (4.0s) gt s (b) v0 20 ms . 2 2
20 ms d v v v d vt 0 f t 0 t (2.0 s) 20 m. 2 2 t 2 We use t = 2.0 s because we are only considering the time to the highest point rather than the whole trip up and down.
(c) d ? From v
3-28. (a) v0 = ? Let’s call upward the positive direction. Since no time is given, use vf 2 v0 2 2ad with a = –g, vf = 0 at the top, and d = (y – 2 m). v02 2(g)(y 2m) v0 2g(y 2 m).
(b) v0 2g(y 2 m ) 2 9.8 m2 (20 m 2 m) 18.8 ms 19 ms . s
3-29. (a) Taking upward to be the positive direction, from 2ad vf2 v02 with a g and d h vf v02 2gh. So on the way up
vf v02 2gh.
(b) From above, on the way down vf v02 2gh, same magnitude but opposite direction as (a). (c) From a
vf v0 t
t
2 2 vf v0 v0 2gh v0 v0 v0 2gh . a g g
(d) vf v02 2gh 16 ms
2
2 9.8 m2 (8.5 m) 9.5 ms . t s
m m vf v0 9.5 s 16 s 2.6 s. a 9.8 m2 s
3-30. (a) vf = ? Taking upward to be the positive direction, from 2ad vf2 v02 with a g and d h vf v02 2gh. The displacement d is negative because upward direction was taken to be positive, and the water balloon ends up below the initial position. The final velocity is negative because the water balloon is heading downward (in the negative direction) when it lands. 2 2 vf v0 vf v0 v0 2gh v0 v0 v0 2gh (b) t = ? From a t . t a g g (c) vf = ? Still taking upward to be the positive direction, from
2ad vf2 v02 with initial velocity = –v0 , a g and d h vf2 v02 2gh vf v02 2gh. We take the negative square root because the balloon is going downward. Note that the final velocity is the same whether the balloon is thrown straight up or straight down with initial speed v0.
© Paul G. Hewitt and Phillip R. Wolf
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5.0 ms 2 9.8 sm (11.8 m) 16 ms for the balloon whether it is 2
(d) vf v02 2gh
2
tossed upward or downward. For the balloon tossed upward, 16 ms 5 ms v v t f 0 2.1 s. a 9.8 m2 s
3-31. (a) Call downward the positive direction, origin at the top. From d v0 t 12 at 2 with a g, d ² y h h v0 t 12 gt 2 From the general form of the quadratic formula x
g , 2
a
b v0 , and c h, which gives t
12 gt 2 v0 t h 0.
b b 2 4ac we identify 2a
v0 v0 2 4
(h) v g 2
0
g To get a positive value for the time we take the positive root, and get
t
v0 + v0 2 + 2gh
v0 2 2gh g
.
.
g
(b) From 2ad vf2 v02 with initial velocity v0 , a g and d h vf2 v02 2gh vf v02 2gh.
v v 2 2gh 0 v0 2 + 2gh . Or you could start with vf v0 at v0 g 0 g
(c) t
v0 v0 2 2gh g
t2
2
9.8 sm2
0.58 s. ;
2
2
3-32. (a) From d v0 t 12 at 2
2(d v0t)
3.2 ms 2 2 9.8 sm (3.5 m)
3.2 ms 2 9.8 sm (3.5 m) 8.9 ms
vf v02 2gh
(b) a
3.2 ms
a
2(d v0 t)
t2 2 120 m 13 ms ·5.0 s
(5.0 s)2
.
4.4 m . s2
(c) vf v0 at 13 ms 4.4 sm2 (5s) 35 ms . 1 km s 1 mi 3600 78 mi . This is probably not a safe speed for driving in (d) 35 ms 1000 m 1.61 km 1h h
an environment that would have a traffic light! v v 2x v0 . 3-33. (a) From x vt 0 f t vf 2 t x x vf v0 (2 t v0 ) v0 2 t 2v0 x v a 2 2 0 . (b) t t t t t
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2x 2(95m) v0 13 ms 3.0 ms . t 11.9s m m 95m 13 ms vf v0 3.0 s 13 s x v m a 2 2 0 2 0.84 or a 0.84 m2 . 2 2 s s t t 11.9s t 11.9s (11.9s)
(c) vf
3-34. (a) From 2ad vf2 v02 with d L vf v02 2aL. This is Rita’s speed at the bottom of the hill. To get her time to cross the highway: From v
d
(b) t
v02 2aL
25m
2
2 3.0 ms
1.5 sm2
(85m)
d t
t
d d . vf v02 2aL
1.54s.
3-35. (a) Since v0 is upward, call upward the positive direction and put the origin at the ground. Then From d v0 t 12 at 2 with a g, d ² y h h v0 t 12 gt 2 12 gt 2 v0 t h 0. From the general form of the quadratic formula x
a
g , 2
b v0 , and c h, which gives t
b b 2 4ac we identify 2a
v0 v0 2 4 g
(h) v g 2
0
v0 2 2gh g
.
(b) From 2ad vf2 v02 with a g and d h vf2 v02 2gh vf v02 2gh. (c) t
v0 v0 2 2gh g
22 ms
22 ms 2 2 9.8 sm (14.7m) 2
9.8 sm2
0.82 s or 3.67 s. So
Anthony has to have the ball leave his had either 0.82s or 3.67s before midnight. The first time corresponds to the rock hitting the bell on the rock’s way up, and the second time is for the rock hitting the bell on the way down. vf v02 2gh
22 ms 2 9.8 sm (14.7m) 14 ms . 2
2
3-31. (a) v1 = ? The rocket starts at rest and after time t1 it has velocity v1 and has risen to a height h1. Taking upward to be the positive direction, from vf v0 at with v0 0 v1 at1 . (b) h1 = ? From d v0 t 12 at 2 with h1 d and v0 0
h1 12 at12 .
(c) h2 = ? For this stage of the problem the rocket has initial velocity v1, vf = 0, a = –g and the distance risen d = h2. vf2 v02 0 v12 v12 (at1 )2 a 2 t12 h2 . 2a 2(g) 2g 2g 2g (d) tadditional = ? To get the additional rise time of the rocket: From v v v v 0 v1 at1 a f 0 t additional f 0 . t a g g (e) The maximum height of the rocket is the sum of the answers from (a) and (b) = a2t 2 hmax h1 h2 12 at12 1 12 at12 1 ag . 2g From 2ad vf2 v02
d
© Paul G. Hewitt and Phillip R. Wolf
3-8
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(f) tfalling = ? Keeping upward as the positive direction, now v0 = 0, a = –g and d = –hmax. From d v0 t 12 at 2
hmax 12 (g)t 2
2 12 at12 1 ag g
2hmax g
t falling
(g) t total t1 t additional t falling t1
at12 (g a) g
2
a(g + a)
at1 t a(g + a) 1 . g g
120 (1.70 s) a t h 2g 2 9.8
(h) vruns out of fuel v1 at1 120 m2 (1.70 s) 204 ms ; h1 12 at12 s
m s2
2 2 1
2
2
1 2
t1 g
120 (1.70 s) 173 m. m s2
2
2
2123 m.
m s2
m at1 120 s2 (1.70 s) 20.8 s. g 9.8 m2 s
hmax 173 m + 2123 m 2296 m 2300 m.
t falling
2hmax 2(2300 m) 21.7 s. g 9.8 m2 s
t total t1 tadditional tfalling 1.7 s 20.8 s 21.7 s 44.2 s. total distance xx 2x 1.14 x . t total time t 0.75t 1.75t 140 km (b) v 1.14 x 1.14 80 km . hr t 2 hr
3-32. v
total distance d . From v d vt. total time t dwalk d jog vwalk t walk v jog t jog v(30 min) 2v(30 min) 3v(30 min) So v 1.5 v. t walk t jog t walk t jog 30 min 30 min 2(30 min)
3-33. (a) v
(b) v 1.5v 1.5 1.0 ms 1.5 ms . s 5400 m = 5.4 km. (c) dto cabin vt total v (t walk t jog ) 1.5 ms (30 min +30 min) 160 min
total distance d . From v d vt. total time t d dfast vslowtslow vfast t fast v(1 h) 4v(1 h) 5v(1 h) So v slow 2.5 v. tslow t fast tslow t fast 1 h 1 h 2h
3-34. (a) v
(b) v 2.5v 2.5 25 km 63 km . h h
© Paul G. Hewitt and Phillip R. Wolf
3-9
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total distance d d x . From v t . total time t v v d1 d2 2x 2x 2 So v v v 2 1 t1 t 2 x x x v1 v1 v1v2 v1 v2 1 2
3-35. (a) v
v1v2 1.5v 2 v(1.5v) 2 2 2.5v 1.2v. Note that the average velocity is biased toward 1.5v v v2 v1 the lower speed since you spend more time driving at the lower speed than the higher speed.
2
(b) v 1.2v 1.2 28 km 34 km . h h 3-36. (a) dAtti =? From VAtti
dAtti t
dAtti Vt. The time that Atti runs = the time that Judy
x x V walks, which is t . So dAtti V x. v v v
4.5 ms V (150 m) 450 m. (b) X x m v 1.5 s
3-37. v
d 3m 2 ms . t 1.5 s
3-38. h = ? Call upward the positive direction. From vf 2 v0 2 2ad with d h, vf 0 and a g
2
14.7 ms v 2 v0 2 v0 2 v0 2 h f 11 m. 2a 2(g) 2g 2 9.8 m 2
s
d 3-39. d ? From v t
0 27.5 ms v0 vf d vt t (8.0 s) 110 m. 2 2
3-40. t ? Let's take down as the positive direction. From d v0 t 12 at 2 with v0 0 and a g d 12 gt 2 t
2d 2(16 m) 1.8 s. g 9.8 m2 s
3-41. a
m m v vf v0 12 s 0 s 4 m2 . s t t 3s
3-42. a
m m v vf v0 75 s 0 s 30 m2 . s t t 2.5 s
3-43. d = ? With v0 0, d v0t 12 at 2 becomes d 12 at 2
© Paul G. Hewitt and Phillip R. Wolf
1 2
2.0 (8.0 s) 64 m. m s2
3-10
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
2
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3-44. a ? With v0 0, d v0 t 12 at 2 becomes d 12 at 2
a
3-45. d = ? With v0 0, d v0t 12 at 2 becomes d 12 at 2
1 2
2d t
2
2(5.0 m) (2.0 s)2
2.5 m2 . s
3.5 (5.5 s) 53 m. m s2
2
3-46. v0 = ? Here we’ll take upwards to be the positive direction, with a =g and vf = 0.
From vf 2 v0 2 2ad v0 2 vf 2 2(g)d v0 2gd 2 9.8 m2 (3.0 m) 7.7 ms . s
3-47. t=? We can calculate the time for the ball to reach its maximum height (where the velocity will be zero) and multiply by two to get its total time in the air. Here we’ll take upward to be the positive direction, with a =g. m vf v0 vf v0 v0 v0 18 s From a t 1.84 s. This is the time to reach t a g g 9.8 m2 s
the maximum height. The total trip will take 2 1.84 s = 3.7 s, which is less than 4 s. Alternatively, this can be done in one step with by recognizing that since the trajectory is symmetric vf = –v0. Then from vf v0 at, with a g v0 v0 gt 2v0 gt
m 2v0 2 18 s t 3.7 s. g 9.8 m2 s
3-48. v0 = ? Since she throws and catches the ball at the same height, vf v0 . Calling upward the positive direction, a = –g.
9.8 m2 (3.0 s) gt s From vf v0 at v0 v0 (g)t 2v0 gt v0 15 ms . 2 2
3-49. For a ball dropped with v0 = 0 and a = +g (taking downward to be the positive direction), dfallen, 1st
v0t 12 at 2
second
1 2
9.8 (1 s) 4.9 m. At the beginning of the 2 2
m s2
we have v0 = 9.8 m/s so dfallen, 2nd
ratio
second
nd
second
v0t 12 at 2 9.8 ms (1 s) 12 9.8 m2 (1 s)2 14.7 m. The s
dfallen, 2nd
second
dfallen, 1st
second
14.7 m 3. More generally, the distance fallen from rest in a time 4.9 m
t is d gt . in the next time interval t the distance fallen is 1 2
2
dfrom time t to 2t v0 t 12 at 2 (gt)t 12 gt 2 23 gt 2 . The ratios of these two distances is
dfrom time t to 2t dfrom rest in time t
© Paul G. Hewitt and Phillip R. Wolf
3 gt 2 2 1 gt 2 2
3.
3-11
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3-50. h ? Call upward the positive direction. From vf 2 v0 2 2ad with d h, vf 0 and a g
1,000 ms vf 2 v0 2 v0 2 v0 2 h 2a 2(g) 2g 2 9.8 m2
2 51, 000 m > 50 km.
s
3-51. h ? With d h, v0 22
m , s
a –g and t 3.5 s, d v0 t 12 at 2 becomes
h (22 ms )(3.5 s) 12 9.8 m2 (3.5 s)2 17 m
3-52. t =? From v
d t
3-53. t ? From a
t
d 65 m 5.0 s. v 13 ms
v vf v0 t t
3-54. (a) t ? From v
d t
s
t
t
m m vf v0 28 s 0 s 4.0 s. a 7.0 m2 s
d d 2d v v . f 0 v v
2
(b) a = ? With v0 0 and vf v, vf 2 – v0 2 2ad becomes a
2
v2 . 2d
28 ms 2d 2(140 m) v2 10 s; a 2.8 m2 . (c) t s v 2d 2(140 m) 28 ms
0 ms 25 ms d v0 vf t (5.0 s) 63 m. 3-55. d ? From v d vt 2 t 2 1 km d d 2462 mi 0.621 mi 3-56. t = ? From v t 8.5 min. t v 28,000 km 1 h 60 min h
3-57. a ? With vf 0, vf 2 – v0 2 2ad becomes
mi 1 km 1000 m 1h v0 2 220 h 0.621 mi 1 km 3600 s a 2d 2(800 m)
3-58. From d v0 t 12 at 2 x
t
2 –6.05 m –6 m . s2
s2
12 at 2 v0 t d 0. From the general form of the quadratic formula
b b 2 4ac we identify a 12 a, b v0 , and c h, which gives 2a
v0 v0 2 4 a
12 a(d) v0
the positive root, which gives us t
© Paul G. Hewitt and Phillip R. Wolf
45 ms
. To get a positive answer for t we take
45 ms 2 2 3.2 sm (440m) 2
3.2 sm2
3-12
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
7.7 s.
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
3-59. v0 = ? The candy bar just clears the top of the balcony with height 4.2m + 1.1m = 5.3 m. With vf 0, vf 2 – v0 2 2ad with v0 and ² y positive and a g
v0 2 vf 2 2(g)h
v0 2gh 2 9.8 m2 (5.3m) 10.19 ms 10.2 ms . The total time is the time for the s
way to the top of the balcony rail plus the time to fall 1.1 m to the floor of the balcony. 2d 2(5.3 m) t up ? From d vf t 12 at 2 with vf 0 and a g d 12 (g)t 2 t up 1.04 s. g 9.8 sm2
t down ? From d v0t 12 at 2 with v0 0, a g and d ² y h h 12 (g)t 2
t down
2h 2(1.1 m) 0 g 9.8 sm2
So t total tup tdown 1.040 s 0.47 s 1.51s. An alternative route is: Since v0 is upward, call upward the positive direction and put the origin at the ground. Then From d v0 t 12 at 2 with a g, d ² y 4.2m d v0 t 12 gt 2 12 gt 2 v0 t d 0. From the general form of the quadratic formula x
a
g , 2
b v0 , and c d, which gives t
10.19 ms
10.19 ms 2 2 9.8 sm (4.2m) 2
9.8 sm2
b b 2 4ac we identify 2a
v0 v0 2 4 g
(d) v g 2
0
v0 2 2gd g
0.57 s or 1.51s. The first answer
corresponds to the candy reaching 4.2 m but not having gone over the top balcony rail yet. The second answer is the one we want, where the candy has topped the rail and arrives 4.2 m above the ground. 3-58. Consider the subway trip as having three parts—a speeding up part, a constant speed part, and a slowing down part. dtotal dspeeding up dconstant speed dslowing down . For dspeeding up , v0 0, a 1.5 m2 and t 12 s, so d v0t 12 at 2 s
For dconstant speed vt. From the speeding up part we had v0 0,
1 2
1.5 (12 s) 108 m.
m s2 a 1.5 m2 and s
so v v0 at 1.5 m2 (12 s) 18 ms and so d 18 ms (38 s) 684 m s
2
t 12 s
For dslowing down , vf 0, a –1.5 m2 and t 12 s, so d vf t 12 at 2 12 -1.5 m2 (12 s)2 108 m. s
s
So dtotal dspeeding up dconstant speed dslowing down 108 m 684 m 108 m 900 m. 3-59. One way to approach this is to use Phil’s average speed to find how far he has run during the time it takes for Mala to finish the race. 100.0 m d From v dPhil vPhilt Mala (12.8 s) 94.1 m. Since Phil has only t 13.6 s traveled 94.1 m when Mala crosses the finish line, he is behind by 100 m 94.1 m 5.9 m 6 m. © Paul G. Hewitt and Phillip R. Wolf
3-13
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3-60. t = ? The time for Terrence to land from his maximum height is the same as the time it takes for him to rise to his maximum height. Let’s consider the time for him to land from a height of 0.6 m. Taking down as the positive direction: From d v0 t 12 at 2 with v0 0 and a g d 12 gt 2 t
2d 2(0.6 m) 0.35 s. g 9.8 m2 s
His total time in the air would be twice this amount, 0.7 s. 3-61. v
d 1 mi 80 mi . h 1h t 45 s 3600 s
total distance d d . If we call the distance she drives d, then from v t . total time t v dthere dback 2d 2d 2 v v So v v v 2 there back d d back there t there t back vback vthere d 1 1
3-62. v
vthere
vthere
vback
vback
v
there vback
2400 km 2 40 km 60 km h 48 km . Note that the average velocity is biased h h 2 2 h km km 60 h 40 h 100 kmh
toward the lower speed since Norma spends more time driving at the lower speed than at the higher speed.
© Paul G. Hewitt and Phillip R. Wolf
3-14
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
Conceptual Physics 12th Edition Hewitt Solutions Manual
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
Conceptual Physics 12th Edition Hewitt Solutions Manual
Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual
| 14,978
| 26,027
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| 4
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CC-MAIN-2018-22
|
latest
|
en
| 0.695587
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https://www.lmfdb.org/L/ModularForm/GL2/Q/holomorphic/4032/2/j/e/
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Properties
Degree 32 Conductor $2^{96} \cdot 3^{32} \cdot 7^{16}$ Sign $1$ Motivic weight 1 Primitive no Self-dual yes Analytic rank 0
Origins of factors
Dirichlet series
L(s) = 1 − 40·25-s − 8·49-s − 80·73-s − 112·97-s + 136·121-s + 127-s + 131-s + 137-s + 139-s + 149-s + 151-s + 157-s + 163-s + 167-s + 128·169-s + 173-s + 179-s + 181-s + 191-s + 193-s + 197-s + 199-s + 211-s + 223-s + 227-s + 229-s + 233-s + ⋯
L(s) = 1 − 8·25-s − 8/7·49-s − 9.36·73-s − 11.3·97-s + 12.3·121-s + 0.0887·127-s + 0.0873·131-s + 0.0854·137-s + 0.0848·139-s + 0.0819·149-s + 0.0813·151-s + 0.0798·157-s + 0.0783·163-s + 0.0773·167-s + 9.84·169-s + 0.0760·173-s + 0.0747·179-s + 0.0743·181-s + 0.0723·191-s + 0.0719·193-s + 0.0712·197-s + 0.0708·199-s + 0.0688·211-s + 0.0669·223-s + 0.0663·227-s + 0.0660·229-s + 0.0655·233-s + ⋯
Functional equation
\begin{aligned} \Lambda(s)=\mathstrut &\left(2^{96} \cdot 3^{32} \cdot 7^{16}\right)^{s/2} \, \Gamma_{\C}(s)^{16} \, L(s)\cr =\mathstrut & \,\Lambda(2-s) \end{aligned}
\begin{aligned} \Lambda(s)=\mathstrut &\left(2^{96} \cdot 3^{32} \cdot 7^{16}\right)^{s/2} \, \Gamma_{\C}(s+1/2)^{16} \, L(s)\cr =\mathstrut & \,\Lambda(1-s) \end{aligned}
Invariants
$$d$$ = $$32$$ $$N$$ = $$2^{96} \cdot 3^{32} \cdot 7^{16}$$ $$\varepsilon$$ = $1$ motivic weight = $$1$$ character : induced by $\chi_{4032} (1, \cdot )$ primitive : no self-dual : yes analytic rank = 0 Selberg data = $(32,\ 2^{96} \cdot 3^{32} \cdot 7^{16} ,\ ( \ : [1/2]^{16} ),\ 1 )$ $L(1)$ $\approx$ $7.578069941e-5$ $L(\frac12)$ $\approx$ $7.578069941e-5$ $L(\frac{3}{2})$ not available $L(1)$ not available
Euler product
$L(s) = \prod_{p \text{ prime}} F_p(p^{-s})^{-1}$ where, for $p \notin \{2,\;3,\;7\}$, $$F_p$$ is a polynomial of degree 32. If $p \in \{2,\;3,\;7\}$, then $F_p$ is a polynomial of degree at most 31.
$p$$F_p$
bad2 $$1$$
3 $$1$$
7 $$( 1 + T^{2} )^{8}$$
good5 $$( 1 + 2 p T^{2} + 54 T^{4} + 2 p^{3} T^{6} + p^{4} T^{8} )^{4}$$
11 $$( 1 - 34 T^{2} + 510 T^{4} - 34 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
13 $$( 1 - 32 T^{2} + 510 T^{4} - 32 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
17 $$( 1 - 22 T^{2} + 174 T^{4} - 22 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
19 $$( 1 + 10 T^{2} + p^{2} T^{4} )^{8}$$
23 $$( 1 + 70 T^{2} + 2262 T^{4} + 70 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
29 $$( 1 - 8 T^{2} + 354 T^{4} - 8 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
31 $$( 1 + 22 T^{2} + p^{2} T^{4} )^{8}$$
37 $$( 1 - 80 T^{2} + 3582 T^{4} - 80 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
41 $$( 1 + 26 T^{2} + 3342 T^{4} + 26 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
43 $$( 1 + 20 T^{2} - 1578 T^{4} + 20 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
47 $$( 1 + 62 T^{2} + p^{2} T^{4} )^{8}$$
53 $$( 1 + 136 T^{2} + 8898 T^{4} + 136 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
59 $$( 1 - p T^{2} )^{16}$$
61 $$( 1 - 14 T + p T^{2} )^{8}( 1 + 14 T + p T^{2} )^{8}$$
67 $$( 1 + 200 T^{2} + 18222 T^{4} + 200 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
71 $$( 1 + 254 T^{2} + 26022 T^{4} + 254 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
73 $$( 1 + 10 T + 150 T^{2} + 10 p T^{3} + p^{2} T^{4} )^{8}$$
79 $$( 1 + 80 T^{2} + 7278 T^{4} + 80 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
83 $$( 1 - 4 T^{2} + 5382 T^{4} - 4 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
89 $$( 1 - 166 T^{2} + 22542 T^{4} - 166 p^{2} T^{6} + p^{4} T^{8} )^{4}$$
97 $$( 1 + 14 T + 222 T^{2} + 14 p T^{3} + p^{2} T^{4} )^{8}$$
\begin{aligned} L(s) = \prod_p \ \prod_{j=1}^{32} (1 - \alpha_{j,p}\, p^{-s})^{-1} \end{aligned}
| 1,853
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| 2.984375
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# Since John Locke acknowledged authorship of his political
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Joined: 01 Aug 2006
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Since John Locke acknowledged authorship of his political [#permalink]
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17 Jan 2009, 14:55
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Since John Locke acknowledged authorship of his political works only in a codicil to his will, the period during which the Second Treatise on Government was written has been established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, what is much more difficult to determine are the personal reason Locke wrote the Treatise, the changes he might have made to his first version, and the extent to which the published version coheres with Locke's intentions.
a. has been established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, what is much more difficult to determine are
b. has been established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, what is much more difficult to determine is
c. have been established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, what is much more difficult to determine is
d. have been established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, what is much more difficult to determine are
e. are established through a close analysis of Locke's reported activities near the time of publication; however, because no original manuscript has been found, that which is much more difficult to determine is
This one is throwing me for a loop. What does the crowd think?
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Re: SC - John Locke [#permalink]
### Show Tags
17 Jan 2009, 15:19
Between A and B.
determine is or determine are.
Typically most courses tell you to inverse the subject here and pick "are".
I believe that " what is much more difficult to determine" can be treated as
the thing (that is much more difficult to determine) is the reason. I believe I read
this from one of the OG answer choices or in another discussion, I can check my log.
Therefore, "what is much more difficult to determine is A, B & C." is correct usage.
Hence B.
Intern
Joined: 01 Aug 2006
Posts: 9
Re: SC - John Locke [#permalink]
### Show Tags
18 Jan 2009, 13:14
Thanks guys, OA was B. I picked A and was having difficulty wrapping my head around "is A, B, and C."
Ryan
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Re: SC - John Locke [#permalink] 18 Jan 2009, 13:14
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# M-file for exercise!!!
4 visualizzazioni (ultimi 30 giorni)
evi il 5 Dic 2013
Modificato: Andreas Goser il 5 Dic 2013
Hi!I have to create an m-file for an exercise and I wanted to ask you what is the difference between a function and a script.Which of them is better??
##### 2 CommentiMostra NessunoNascondi Nessuno
per isakson il 5 Dic 2013
the debugger and the profiler works better with functions
evi il 5 Dic 2013
So,do you think that the error message I get has to do with the debugger and the profiler??
Accedi per commentare.
### Risposte (2)
sixwwwwww il 5 Dic 2013
If you want to re-use some operation again and again then function is better and if you just wanted to start writing codes in MATLAB then script is better at first. Better or not depends upon functionality usually
##### 8 CommentiMostra 6 commenti meno recentiNascondi 6 commenti meno recenti
evi il 5 Dic 2013
Modificato: evi il 5 Dic 2013
The function is like that:
function[]=test(x_0,dimension,e,maxit) H=hilb(dimension); A=H; for j=1:dimension b(j)=sum(A(j,:)); exact(j)=1; end b=b'; exact=exact'; P=diag(diag(A)); F=A-P; x=-P\F*x_0+P\b; end
and the script is like that:
dimension=input('Give dimension: ');
H=hilb(n);
A=H;
e=input('Give e: ');
maxit=input('Give maxit: ');
x_0=input('Give x_0: ');
for j=1:dimension
b(j)=sum(A(j,:));
exact(j)=1;
end
b=b';
exact=exact';
P=diag(diag(A));
F=A-P;
x=-P\F*x_0+P\b;
When I run the function for dimension=50,e: 0.00001,maxit: 50,x_0: zeros(50,1),I get a result,but when I run the script I get this message: ??? Error using ==> mtimes Inner matrix dimensions must agree.
Error in ==> ll at 15 x=-P\F*x_0+P\b
Why does this happen???What have I done wrong???
sixwwwwww il 5 Dic 2013
I am getting result both from your script and function. Following are the inputs: For function
test(zeros(50, 1), 50, 0.00001, 50)
For script
Give dimension: 50
Give e: 0.00001
Give maxit: 50
Give x_0: zeros(50, 1)
then why you dont get correct result. Can you check what inputs you give to both your function and script?
Accedi per commentare.
Andreas Goser il 5 Dic 2013
Modificato: Andreas Goser il 5 Dic 2013
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
### Categorie
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Question
# (1 pt) Cicely invests 3600 dollars in an account paying an effective rate of interest of...
(1 pt) Cicely invests 3600 dollars in an account paying an effective rate of interest of 5 percent. Two years later, she deposits an additional 1650 dollars. If there are no other transactions, how long will it take (from the time of the first investment) for her account balance to reach 8600 dollars? (Assume simple interest between compoundings.)
Answer = (blank) years and (blank) days.
(Note: your answer for the number of years should be a whole number, while your answer for the number of days should be given to at least 3 decimal places.)
Can use excel as long as answer is correct
Future value = Present value*(1+r)^n
where n = number of years
r = rate of return
after 2 years amount = 3600*(1.05)^2
= 3969
after 2 years she deposits additional 1650
total amount after 2 years = 3969+1650 = 5619
8600 = 5619*(1+5%)^n
(1.05)^n = 1.5302
by applying log on both sides
n log(1.05) = log(1.5302)
n = log(1.5302) / log(1.05)
n = 8.723244 years
total = 8.723244+2 = 10.723244 years
1 year = 365 days
0.723244 years = 0.723244*365 = 263.984days
Answer = 10 Years and 263.984 days
#### Earn Coins
Coins can be redeemed for fabulous gifts.
##### Need Online Homework Help?
Most questions answered within 1 hours.
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# ECE 313 ExamsSummer 2016
## Quizzes
Tip: The quizzes represent early checkpoints on your road to learning how to solve problems for this course, which in some cases includes devising notation for word problems. You will be tested over the same material again on the midterm and final exams, without benefit of knowing ahead of time the list of possible problems. So to use your time most efficiently, read the assigned material in the notes, paying special attention to the examples. Then work out the assigned problems on your own, looking at the answers only if you are truly stuck. Start early; don't wait until just before the quiz. If you work the problems yourself, you will be familiar enough with the problems to do well on the quizzes without wasting your time memorizing solutions you don't understand. And you will be in a great position for the exams.
## Midterm Exams
• Review I: Wednesday, July 6th, 6-8pm, 3013 ECEB
• Midterm Exam I: Thursday, July 7, 8.45-10pm
Location: 1013 ECEB
Midterm I and its solution .
• Review II: TBA Wednesday, July 20th, 6-8pm, 3013 ECEB
• Midterm Exam II: Thursday, July 21, 8.45-10p.m.
Location: 1013 ECEB
Midterm II and its solution .
The topics covered in Exam 1 are exactly the ones in the course notes up until (and including) section 2.11, except for Sections 1.5 and 2.9 (those will not be included). Unlike the quizzes, the exam will have problems that are not almost identical to other problems seen during the semester. You should know the meanings, forms, means, and variances for the key discrete distributions. . Exam 2 covers problems, SAQ, lectures, and reading for from the course notes up until (and including) section 3.10, except for Sections 1.5, 2.9, 2.12.3, 2.12.4, 3.8.3 and 3.9 (those will not be included), with emphasis on Chapter 3. Unlike the quizzes, the exam will have problems that are not almost identical to other problems seen during the semester. You should know the meanings, forms, means, and variances for the key discrete and continuous distributions. The final will cover problems, lectures, and reading for all quizzes, covered in the notes through the end of the notes except for Sections 1.5, 2.9, 2.12.3, 2.12.4, 3.8.3 and 3.9 (those will not be included).
You should take the exams in the exam room you have been assigned. Please make sure that you bring your IDs with you, and that you show up at least 15 minutes earlier, in order to avoid last minute problems (such as not being able to find the room).
You may bring one 8.5" by 11" sheet of notes to the midterm exams; and two 8.5" by 11" sheets of notes to the final exam. Both sides of the sheets can be used. The notes can be typed in font size 10 or larger, or written in handwriting of equivalent size or larger. The exams are closed-book and closed-notes otherwise. Electronic devices (calculators, cellphones, pagers, laptops, headphones, etc.) are neither necessary nor permitted.
You can find copies of old midterm exams and final exams by going to the web pages of previous offerings of ECE 313.
The midterm exam dates were announced before the summer session began, and the time for them was announced on June 19, after consulting with the students so that there would be no conflicts. Therefore, you are expected to arrange your schedules to make the exams.
If you miss a midterm exam, the following procedures apply: To receive an excused absence, you must either arrange your absence in advance with your instructor (i.e., prior to the absence), or complete an Excused Absence Form at the Undergraduate College Office, Room 207 Engineering Hall, indicating that you missed the midterm exam and the reason for the absence. This form must be signed by a physician or medical official for a medical excuse, or by the Office of the Dean of Students (Emergency Dean, 610 E. John Street, 3330050) for a personal excuse due to personal illness, family emergencies, or other uncontrollable circumstances. Present the completed form in person to your section instructor as soon as possible after you return. Scores on midterms due to excused absences will not be made up. Your midterm score for an excused absence will be the weighted average of the other midterm score and final exam score. An unexcused absence from a midterm will be counted as a 0.
## Final Exam
Saturday, August 6, 8-11 a.m.
Location: ECEB 3020.
Conflict: Thursday, August 4, 7-10 p.m.
Location: ECEB 1013.
Final Review: Wednesday, August 4th, 6-8pm, 3013 ECEB
Final Exam and its solution
Conflict Final Exam and its solution
Two two-sided 8.5" by 11" sheets of notes are allowed at the final, with font size no smaller than 10 pt or equivalent handwriting. Bring a picture ID. No calculators.
The topics covered in the exam are exactly the ones in the course notes up until (and including) section 4.11, except for Sections 1.5, 2.9, 2.12.3, 2.12.4, 3.8.3, 3.9, 4.7 and 4.10 (those will not be included), with emphasis on Chapter 4. Unlike the quizzes, the exam will have problems that are not almost identical to other problems seen during the semester. You should know the meanings, forms, means, and variances for the key discrete and continuous distributions.
A conflict exam will be offered on Thursday, August 4 at 7pm in room ECEB 1013. You do not need to actually have a conflict to take this exam BUT you do need to tell me so enough exams are printed. If you show up without telling me first, there is a chance you will not get an exam. Most of you have indicated whether you are taking the regular or the conflict exam, and you can find that information in COMPASS. If you have not indicated a preference, you are currently assigned to the regular exam on Saturday.
If for some reason of emergency such as severe illness you are not able to take the final exam at the required time, you will need to obtain a written excuse from the Office of the Dean of Students.
A table of values of the standard Gaussian CDF will be supplied to you if it is needed on an exam.
See Previous semester's web pages for old exams and solutions.
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#jsDisabledContent { display:none; } My Account | Register | Help
# Scale (ratio)
Article Id: WHEBN0000049526
Reproduction Date:
Title: Scale (ratio) Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Scale (ratio)
The scale ratio of a model represents the proportional ratio of a linear dimension of the model to the same feature of the original. Examples include a 3-dimensional scale model of a building or the scale drawings of the elevations or plans of a building.[1] In such cases the scale is dimensionless and exact throughout the model or drawing.
The scale can be expressed in four ways: in words (a lexical scale), as a ratio, as a fraction and as a graphical (bar) scale. Thus on an architect's drawing one might read
'one centimetre to one metre' or 1:100 or 1/100
and a bar scale would also normally appear on the drawing.
Da Vinci's Vitruvian Man illustrates the ratios of the dimensions of the human body; a human figure is often used to illustrate the scale of architectural or engineering drawings.
## General representation
In general a representation may involve more than one scale at the same time. For example, a drawing showing a new road in elevation might use different horizontal and vertical scales. An elevation of a bridge might be annotated with arrows with a length proportional to a force loading, as in 1 cm to 1000 newtons: this is an example of a dimensional scale. A weather map at some scale may be annotated with wind arrows at a dimensional scale of 1 cm to 20 mph.
Map scales require careful discussion. A town plan may be constructed as an exact scale drawing, but for larger areas a map projection is necessary and no projection can represent the Earth's surface at a uniform scale. In general the scale of a projection depends on position and direction. The variation of scale may be considerable in small scale maps which may cover the globe. In large scale maps of small areas the variation of scale may be insignificant for most purposes but it is always present. The scale of a map projection must be interpreted as a nominal scale. (The usage large and small in relation to map scales relates to their expressions as fractions. The fraction 1/10,000 used for a local map is much larger than 1/100,000,000 used for a global map. There is no fixed dividing line between small and large scales.)
## Mathematics
In mathematics, the idea of geometric scaling can be generalized. The scale between two mathematical objects need not be a fixed ratio but may vary in some systematic way; this is part of mathematical projection, which generally defines a point by point relationship between two mathematical objects. (Generally, these may be mathematical sets and may not represent geometric objects.)
## References
1. ^ Scale Ratio
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
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# R/nDep2sam.R In PracTools: Tools for Designing and Weighting Survey Samples
#### Documented in nDep2sam
```nDep2sam <- function(S2x, S2y, g, r, rho, alt, del, sig.level=0.05, pow=0.80){
# check type of test
alt.ok <- alt %in% c("one.sided", "two.sided")
if (!alt.ok)
stop("alt must be either 'one.sided' or 'two.sided'.\n ")
else {
if (alt == "one.sided")
za <- qnorm(1-sig.level)
if (alt=="two.sided")
za <- qnorm(1-sig.level/2)
}
# check ranges of parameters
if (g<0 | g>1) stop("g must be in [0,1].\n")
if (r<0) stop("r must be positive.\n")
if (rho<0 | rho>1) stop("rho must be in [0,1].\n")
if (pow<0 | pow>1) stop("pow must be in [0,1].\n")
# compute sample size in group 1
zb <- qnorm(1-pow)
n1 <- (S2x + r*S2y - 2*g*r*rho*sqrt(S2x*S2y)) / del^2 * (za-zb)^2
METHOD <- "Two-sample comparison of means\n Sample size calculation for overlapping samples"
structure(list(n1 = ceiling(n1),
n2 = ceiling(n1/r),
S2x.S2y = c(S2x, S2y),
delta = del,
gamma = g,
r = r,
rho = rho,
alt = alt,
sig.level = sig.level,
power = pow,
method = METHOD
), class="power.htest")
}
```
## Try the PracTools package in your browser
Any scripts or data that you put into this service are public.
PracTools documentation built on Aug. 17, 2022, 5:06 p.m.
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A coin is tossed once. Write its sample space.
Asked by Sakshi | 1 year ago | 77
##### Solution :-
Given: A coin is tossed once.
We know that, the coin is tossed only once.
Then, there are two probabilities either Head (H) or Tail (T).
So,
S = {H, T}
The sample space is {H, T}
Answered by Aaryan | 1 year ago
### Related Questions
#### One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?
One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?
#### The probability that a student will pass the final examination in both English and Hindi is 0.5
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?
#### A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.
A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.
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# What is the capacity of the human brain?
What would a reasonable estimate be of the data capacity of the human brain expressed in bytes?
Paul Reber, professor of psychology at Northwestern University, stated that
The human brain consists of about one billion neurons. Each neuron forms about 1,000 connections to other neurons, amounting to more than a trillion connections. If each neuron could only help store a single memory, running out of space would be a problem. You might have only a few gigabytes of storage space, similar to the space in an iPod or a USB flash drive. Yet neurons combine so that each one helps with many memories at a time, exponentially increasing the brain’s memory storage capacity to something closer to around 2.5 petabytes.
1PB = 1,000,000,000,000,000 B = 1015 bytes = 1000 TB (terabytes).
1TB = 1,000,000,000,000 bytes = 1012 bytes = 1000 GB (gigabytes). So
2.5PB = 2.5 million gigabytes
For comparison, if your brain worked like a digital video recorder in a television, 2.5 petabytes would be enough to hold three million hours of TV shows. You would have to leave the TV running continuously for more than 300 years to use up all that storage.
The same article states that although it is stated that the brain is capable of storing around 2.5PB of information,
The brain’s exact storage capacity for memories is difficult to calculate. First, we do not know how to measure the size of a memory. Second, certain memories involve more details and thus take up more space; other memories are forgotten and thus free up space. Additionally, some information is just not worth remembering in the first place.
• Sounds like pseudoscience to me Commented Jun 10, 2017 at 19:43
• @Charlie A lot of assumptions we cannot get proof, but they are made explicit, so I believe it is fine and 100% in line with what the OP requested. Commented Oct 16, 2018 at 8:18
• All of these "capacity estimates" are bunk. It could easily be hundreds or thousands of orders of magnitude off. The brain does not store data using bits and bytes, and although information theory does allow us to measure brain capacity in bits, actually calculating that is far beyond us. After all, you're saying that the brain is capable of existing in 2^(2.5×10^18×8) functionally distinct states, i.e. that it can be described in 2.5×10^18×8 bits. RAM only ever contains 1s and 0s, so there's a very, very specific number of functional states a computer can exist in. The brain is far different. Commented Jun 16, 2019 at 2:27
• "one billion neurons” I think that’s a huge understatement I am sure its more like 86 billion. Commented Sep 7, 2023 at 12:01
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W.S 發問於 科學及數學數學 · 1 十年前
# maths f.4 問題
maria takes 5 days more to do a job than John. Together, they can do the job in 6 days.How long does it take each to do the job alone??
ans: Maria: 15 days John: 10 days,,
### 3 個解答
• 1 十年前
最愛解答
Let x be the days that John needs to do the job,
then x + 5 is the day that Maria takes.
From the given condition,
6[1/x + 1/(x + 5)] = 1 ← (Since they need 6 days to work together to do the job.)
6[(x + 5) + x]/x(x + 5) = 1
6[(x + 5) + x]/x(x + 5) = 1
12x + 30 = x2 + 5x
x2 – 7x – 30 = 0
(x - 10)(x + 3) = 0
x = 10 or -3 (rejected)
So, John takes 10 days to finish the job.
Maria takes 15 days to finish the job.
資料來源： Myself~~~
• 1 十年前
Let the time it takes Maria to do the job alone be x days and that of John be y days.
x = y+ 5
1/x + 1/y = 1/6
Then 用代入法就計到答案la~
• 1 十年前
Let m and j be the number of days maria and john takes to do the job alone respectively
since maria takes 5 days more to do a job than john,
equation 1: m=j+5
the proportion of job each of them can do in one day time is 1/m and 1/j respectively,
thus they finished 1/6 of the work each day together
therefore,
equation 2: 1/m+1/j=1/6
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# HEEELP!! need assistance figuring out Firebox-to-Cook Chamber Opening
I need some help, for the life of me I cant wrap my head around figuring out the size of the Firebox-to-Cook Chamber Opening,
I have read the tutorial, and it made my brain hurt, I have found the Circle calculator and am not sure which measurements to enter, can someone help me with this?
I am using a 40 pound propane tank, the internet says it holds 9.4-9.5 gallons, I figured out there are 231 cubic inches in a gallon. the feldon calculator is right on that, My fire box is 10x10 but I cant figure out the Firebox-to-Cook Chamber Opening and on top of that I have read
the size needs to a bit larger than the calculation doe to friction and using a smaller CC.
should the opening on the other end of the baffle be about the same size? thanks in advance.
### SmokingMeatForums.com Top Picks
This tutorial explains stuff better.... It will make you a better RF smoker than the one you are using....
http://www.smokingmeatforums.com/a/reverse-flow-smoker-how-to-calculate-build-tutorial
You have to be careful with propane volumes. Some tanks are labled at a lower volume for safety. A gallon of propane may or may not be equivalent to a gallon of water. You may be best to just to measure it and directly calculate the volume. Just Google vollume of a cylinder or volume of propane tank. There are some good calcs out there.
RG
Quote:
Originally Posted by DaveOmak
This tutorial explains stuff better.... It will make you a better RF smoker than the one you are using....
http://www.smokingmeatforums.com/a/reverse-flow-smoker-how-to-calculate-build-tutorial
yep, but as I clearly mentioned in my original post, I have already read this,,, a few times. I cant get the math to work right in my head. and some of you claim to be able to look right at it and see the numbers naturaly, so how about some help.
It's written in a step by step.... calculate the first number, write it down... do the 2nd, write it down.... as you go down the page, the previous numbers will fit into the calculations....
Calculations for a standard design, reverse flow smoker..
Volume of the Cook Chamber.... Use the Inside Diameter of the tank...
Diameter X Diameter X 0.7854 X Length = Volume in cubic inches / 231 = Volume in gallons
16 x 16 x 0.7854 x 30" = 6032 cu. in. / 231 = 26 gallons
Volume in cubic inches X 0.004 = FB/CC opening in square inches
6032 x 0.004 = 24 sq. in.
Volume in cubic inches X 0.004 = Area under the RF plate in square inches
24 sq. in.
Volume in cubic inches X 0.004 = Area required at the end of the RF plate in square inches
24 sq. in.
Volume in cubic inches X 0.33 = minimum volume of the Fire Box
6032 x 0.33 = 1990 sq. in.
Don't get confused by the volume of calculations, and just slowly work down the tutorial, it should fall into place....
If you still have trouble, put your final numbers in here and I will walk through them...
Quote:
Originally Posted by DaveOmak
This tutorial explains stuff better.... It will make you a better RF smoker than the one you are using....
http://www.smokingmeatforums.com/a/reverse-flow-smoker-how-to-calculate-build-tutorial
Quote:
Originally Posted by 1961shasta
yep, but as I clearly mentioned in my original post, I have already read this,,, a few times. I cant get the math to work right in my head. and some of you claim to be able to look right at it and see the numbers naturaly, so how about some help,
Do as Dave says and read the tutorial again. Ask specific questions about your numbers, but ask about them one at a time. When you clear one hurdle in your head go to the next. Trust me, it worked for me. You can read my thread here http://www.smokingmeatforums.com/t/171029/new-guy-and-this-will-be-my-1st-reverse-flow-smoker-build
and it might help. I ask all the same questions as you and maybe more.
Sorry guys, I get stumped right out of the gate, (Diameter X Diameter X 0.7854 X Length), how can I do this is my ends are rounded? according the feldon calculator my cubic inches will be 2201.43 cubic inches, using 9.54 gallons as the volume I found for a 40 pound propane tank online. so can I use this instead of calculating Diameter X Diameter X 0.7854 X Length?? since that gives you cubic inches, (assuming the calculator is correct with that measurements).
Use a "best guess" scenario...... this ain't rocket science.... Measure from 1/2 length of the dome to 1/2 length of the other dome..... or measure from where the straight side ends to the end of the dome... Guess where 1/2 of the volume of the dome would be to the other 1/2 the volume of the dome would be... Or, fill it with water 1, 5 gallon bucket at a time and count the buckets.... or fill it with water and measure the water as you dump it out....
well using my best guess, the tank is 9.54 gallons, which equals, 2203.74 inches, x.004 = 8.814 cubic inches for the FB to CC opening,
I plugged in Chord AB and segment height Ed into the circle calculator and the closest I can get is 9.083 radius, thats the size of the hole in cubic inches right?
thats 1 inch difference from the feldon calculator which had me at 8 inches
What is the length of your tank.... What is the diameter of your tank...... You seem to be not following the steps.
++++++++++++++++++++++++++++++++++++
For calculating the FB/CC opening.... You are calculating for the GREEN area... that will be the cut-out area in the FB that mates to the CC....
The green area is a segment. It is called a "Segment Area"... That area is the same as the FB/CC opening....
.. ..
When you open the "Circle Calculator" link, Click on the "bullet" to the left of Radius & Segment Height ED....
Circle Calculator
Click on the 2 variables you know
......Radius and Central Angle...........................Radius & Chord AB
[X] Radius & Segment Height ED.....................Radius & Apothem OE
......................................................Radius & Arc AB
......Chord AB & Segment Height ED................Chord AB & Apothem OE
......Segment Height ED & Apothem OE...........Chord AB & Arc AB
You should have opened the circle calculator and a bunch of boxes will appear below..
and a box called [CALCULATE]..
Enter the radius in the box so marked.... If you tank is 24" OD, and has a 3/8" thick steel wall, the ID of the tank 23.25".... the radius is 11.625"..... these numbers need to be accurate if you want stuff to fit... ALSO, take into account the thickness of the FB steel when cutting out the tank...
Enter a guess for the Segment Height ED... I'd start with 6.0
Click on [CALCULATE]....
The other boxes will fill with numbers....
The [Segment Area] box is what you area going to compare with the FB/CC opening in square inches...
If the FB/CC opening number is smaller, change the number in the Segment Height ED box to 5.... continue changing that number until the [Segment Area] matches the FB/CC number.....
Now look at the colored circles above.... Segment height ED is how tall the green area is and corresponds to the area to be cut out.....
Also, [Chord AB] corresponds to the width of the RF plate... NOTE... for ease in fitment, the FB should be at least as wide as the RF plate..
well I cant fill it with water, I already have the door cut and hinges on it. its total height is 25.25 inches with 23/4 inches of dome on each end, so subtract one of the 23/4 inches give me 22.5 inches by 12.3 inches, not sure how thick the tank is looks like over a 16th, so best guess would be 22.5 in x 11.9 inches, )Diameter [11.9] X Diameter [11.9] X 0.7854 X Length 22.5 = Volume) =2502 cubic inches,
which calculates to me needing 10 cubic inches of opening between the FB and CC.
Edited by 1961shasta - 3/27/16 at 7:39pm
according to my previous post, this radius gives me 10 cubic inches between the FB and CC right? I used a 10 in radius because that is the width of the tank where I want my baffle to be at (3 inches up from the bottom) where the tank is 10 inches wide.
Edited by 1961shasta - 3/27/16 at 8:10pm
If the diameter of the tank is 11.9", the radius is 5.95" or 6" is good enough.... You don't put the baffle somewhere..... You calculate for it...
2500 ci. x 0.004 = 10 si I'd use 22 si under the RF plate and the FB/CC opening
2500 x .33 = 825+ ci FB
2500 x .001 = 2.5+ si .. 3" below and 2" above is what I'd do...
2500 x 0.022 = 55 ci.. 55/.7854/3/3=8'" tall... Knowing small exhaust stacks don't suck well, I'd use a 3" stack about 32" tall..
So your 3" guess is right on... good guess.... .....
That 3" will be the green area or ED on the colored picture... That is the shape of the cut out for the FB/CC also....
The reason for enlarging certain areas is friction loss... A long narrow passage creates a lot of friction and when you have a "natural draft" unit, friction is a killer....
I don't know how to copy and paste the circle calculator... Put 6" radius and 3" segment height ED in it and those above numbers are what should come up....
Quote:
Originally Posted by DaveOmak
3" below and 2" above is what I'd do..
So what you are saying is my top cut should be 3 inches below the center of the tank, and my bottom cut will be 2 inches from the floor or bottom of the tank?
how wide does the top cut need to be, I dont under stand the circle calculator, from what I am reading the top cut should be 10.39 inches. right?
First of all, have the tutorial open on a page you can reference...
Look at the picture.... the red circle is the CC...... ED is the segment height..3"...... once you measure up 3" from the bottom of the CC, and draw a line across the CC, it should be Chord AB which should be 10.39" long.... Then you will REMOVE the GREEN area from your CC and slide the FB up to that HOLE you just cut into the CC... Mark the cutout on the FB and remove that area so when it's all welded together, you will end up with the FB/CC opening...
2500 x .001 = 2.5+ si .. 3" below and 2" above is what I'd do...
The above has NOTHING to do with the CC/FB opening.... If you look at the calculation steps, it refers to the air inlets for the FB...
Volume in cubic inches X 0.001 = FB air inlets in square inches...
Recommended upper and lower air inlets... Upper air inlet directly across from the FB/CC opening to facilitate moving heat from the FB to the CC, and insuring good air flow through the CC.. The lower air inlet should be situated at or below the fuel grate.... The two air inlets can share the designated square inches of opening.... approx. 20% upper and 80% lower...
OK, I can finally see it now, thanks for all the help!!!, looks like I got a small issue, my fire box is only 10 inches not quite 10.39, I finished my fire box today, well almost, still have to put a vent in the side.
Quote:
Originally Posted by DaveOmak
2500 x .001 = 2.5+ si .. 3" below and 2" above is what I'd do...
The above has NOTHING to do with the CC/FB opening.... If you look at the calculation steps, it refers to the air inlets for the FB...
on another note about the FB vents, the guy who gave me the steel cut me a 3 opening vent into the door without me asking with his CNC plasma cutter, according to what I have read it was too big, and with my small box it was almost going to draw air in under the grate, which I read produces ash on your food, so I blocked most of the bottom portion of 2 openings off when I was adding some bracing to the inside of the door, it's on 1/8th inch. and I will put a 3 inch in the side below the grate.
and with my small box it was almost going to draw air in under the grate, which I read produces ash on your food,
Well, That may have happened ONCE to someone but it's a bunch of BS.... doesn't work like that... maybe from a gust of wind but not from operating the smoker.... OR someone was thinking that would happen but it won't... Ribwizzard has built hundreds of smokers and has his air inlets in the bottom of the FB.... He has a business selling smoked meats etc..... It's amazing how stupid ideas get spread around and people believe them.... and it's amazing how folks look for information on how to build a smoker, and think the information I give them is WRONG... as if I want them to build a smoker that covers their food with ash....
I'm sorry you listened to that particular idiot that gave you misinformation ... hundreds of smokers have been built as pictured below... The drawing on the right, is from a reputable forum.... the picture on the left is from a member on this forum...
.. ..
Be sure to let me know, in the future, where I'm deliberately trying to steer you wrong so your smoker is a failure...
I didn't mean it that way, and what you described is right and what I am doing, what I had read said if you put a vent in the front door to feed the fire and push smoke into the CC, you don't want it too low or if will push ash into the CC. but you are right, not everything people post is correct. thanks for all the help, I am almost there. unfortunately it will be about a week before I have time to continue working on it.
Another thing I noticed was the expanded metal in your fire box. I would build an angle iron frame and put the expanded metal in if. If left like it is when it heats up it bow and sag, and you will be replacing it.
Here are a couple of FB doors we have made
This is my smoker, we added the top vents after the fact Really Makes a Difference
Here is one showing the FB to CC opening and the baffle (Heat Deflector) plate
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• HEEELP!! need assistance figuring out Firebox-to-Cook Chamber Opening
### SmokingMeatForums.com Top Picks
SmokingMeatForums.com › Forums › Smoking Supplies & Equipment › Smoker Builds › Reverse Flow › HEEELP!! need assistance figuring out Firebox-to-Cook Chamber Opening
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# Question 8ee4f
Jan 8, 2018
The wavelength is 102.5 nm.
#### Explanation:
The Rydberg equation gives the wavelength λ for the transition:
color(blue)(bar(ul(|color(white)(a/a) 1/λ = R(1/n_1^2 -1/n_2^2)color(white)(a/a)|)))" "
where
$R =$ the Rydberg constant (1.0974 × 10^7color(white)(l) "m"^"-1") and
${n}_{1}$ and ${n}_{2}$ are the numbers of the energy levels such that ${n}_{1} < {n}_{2}$
The second line in the Lyman series corresponds to a transition from ($n = 3$)
to ($n = 1$).
1/λ = (1.0974 × 10^7color(white)(l) "m"^"-1")(1/1 - 1/9) = (1.0974 × 10^7color(white)(l) "m"^"-1")× 8/9
= 9.755 × 10^"6"color(white)(l)"m"^"-1"
λ = 1/(9.755 × 10^"6"color(white)(l)"m"^"-1") = 1.025 × 10^"-7"color(white)(l)"m" = "102.5 nm"#
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# Re: [R] Sum(Random Numbers)=100
From: Kenn Konstabel <lebatsnok_at_gmail.com>
Date: Wed, 09 Jul 2008 00:11:05 +0300
On Tue, Jul 8, 2008 at 9:53 AM, Shubha Vishwanath Karanth < shubhak_at_ambaresearch.com> wrote:
> ...actually I need to allocate certain amount of money (here I mentioned
> it as 100) to a randomly selected stocks(50 stocks)... i.e., 100 being
> divided among 50 stocks and preferably all are integer allocations(i.e.,
> 5 8 56 12 etc without any decimals)...
so perhaps you can reformulate your problem: instead of generating random numbers with their sum constrained to be 100, it could be dividing 100 units randomly between 50 "bins", which is, essentially, random sampling with replacement
stocks <- 1:50
money <- 100
allocations <- sample(stocks, money, replace=TRUE) # here you can add a prob argument to sample() # especially if you expect the results to be something like " 5 8 56 12 etc "
allocations <- table(factor(allocations, levels=stocks)) # or, equivalently, colSums(outer(allocations, stocks, "=="))
I don't know if this solves your problem but at least it's guaranteed to sum to 100 and give you only integer values.
Kenn
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# On the definition of small categories in SGA4
L
#### LOCOAS
##### Guest
LOCOAS Asks: On the definition of small categories in SGA4
We assume ZFC+U. A category is an ordered pair $(\operatorname{Ob} \mathcal{C},\operatorname{Mor} \mathcal{C},\operatorname{dom},\operatorname{codom},e,∘)$ of sets (not classes) and maps satifying some conditions.
Let $\mathbb{U}$ be a Grothendieck universe. An element of $\mathbb{U}$ is called a $\mathbb{U}$-set. A set is called $\mathbb{U}$-small if it is isomorphic to a $\mathbb{U}$-set. In the following, we suppose that $\mathbb{N} \in \mathbb{U}$.
In SGA4, a category $\mathcal{C}$ is called $\mathbb{U}$-small if $(\operatorname{Ob} \mathcal{C},\operatorname{Mor} \mathcal{C},\operatorname{dom},\operatorname{codom},e,∘)$ is $\mathbb{U}$-small as a set (if my understanding is correct). However, I don't see this definition working well. For any set $a$ and $b$, an ordered pair $(a,b)$ is always $\mathbb{U}$-small since $(a,b)=\{\{a\},\{a,b\} \}$ is a set consisting of exactly two elements, which is isomophic to $2:=\{\emptyset,\{\emptyset\}\} \in \mathbb{U}$. Thus, $\mathbb{U}$-smallness imposes nothing on categories. In particular, it is not equivalent to $\operatorname{Ob} \mathcal{C}$ and $\operatorname{Mor} \mathcal{C}$ are $\mathbb{U}$-small.
I think I am mistaken somewhere, where is it?
SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.
#### Encoding of categorical variables to reduce the effect of erroneous labels
aby Asks: Encoding of categorical variables to reduce the effect of erroneous labels
I have a structured dataset containing (nominal) categorical variables encoded as labels, let's say a feature includes labels from 1 to 20. Some of the labels in that feature could just be errors, that should not be present in the dataset and are not known priorly.
I wonder if there is an encoding method for that feature, such that the effect of erroneous labels will be mitigated. One-hot encoding of the labels before a ML task will create a dimension for each label, that could lead noisy labels to have a more dominating effect on the dataset.
In case of feature hashing, it's not that easy to determine the number of output features for each variable, therefore I think it wouldn't be reasonable to proceed with it.
Would a compression method such as PCA after having one-hot matrix (sparse) work well in this case? The labels would be represented by continuous values, although this could lead to an information loss for the correct labels besides noisy labels. But eventually noisy labels will not take up a dimension in the dataset, which is better.
I also believe that applying Fourier compression on one-hot matrices considering them as black and white images would be overcomplicated and nonsense for a tabular feature that frequencies do not matter.
What approach should I follow?
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#### How can I do a train test split for an unbalanced panel data set in Python?
Andrew Asks: How can I do a train test split for an unbalanced panel data set in Python?
I have an unbalanced, panel pandas data frame.
I would like to split this data into a training set and a testing set. Python's train_test_split method will not work because it does a random split, and so, it will likely places observations from t + 1 into the training set, and observations from t into the test set.
Which, of course, makes no sense, because the future cannot predict the past.
TimeSeriesSplit will also not work because this function does not take into consideration the panel dimension of my data set.
Is there an easy way to do a train test split on unbalanced panel data sets? This split should (1) take into consideration the panel dimension of the data set, and (2) place earlier observations in the training set and later observations in the testing set.
SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.
#### [Solved] Extracting the Data from Shape File for Data Comparison
• Mathan GIS
• Geography
• Replies: 0
Mathan GIS Asks: Extracting the Data from Shape File for Data Comparison
I have 2 shape files called Existing.shp and Updated.shp with following columns ObjectID,DateTime,X,Y,Location,AOIName,CreatedBy,EditedBy
Existing.shp will be in the Web applications and Updated.shp will be udated once in 10 days from various persons. I have to compare the data with Existing.shp and Updated.shp to identify the duplicate data.
Means Comparing Existing.shp to Updated.shp and remove the duplicate data in Updated.shp, Because Existing.shp is clean and processed by the application.
Steps:
1. Download the Existing.shp from the Application and Store it in my local folder [D:\DataCompare\Existing.shp].
2.Receive the Updated.shp from various persons and store it in[D:\DataCompare\Updated.shp].
3.Compare Existing.shp and Updated.shp and Remove Duplicate Data in Updated.shp
4.Insert the new Data to Existing.shp and Upload to Application.
Environment:
1. Visual Studio 2017 Community/C#/Windows Desktop Forms
2. Installed ESRI.ArcGISRunTime
Query:
1.How to read the data in C# from the Existing.shp and Updated.shp like CreatedBy,Date.
1. How to load in store it in C#/Dictionary with specific Coulmns. So I can Compare with Date/CreatedBy
Once the data are loaded in Dictionary, I can perform other conditions.
I tried :
// Open the shapefile ShapefileFeatureTable myShapefile = await ShapefileFeatureTable.OpenAsync(filepath);
// Create a feature layer to display the shapefile FeatureLayer newFeatureLayer = new FeatureLayer(myShapefile);
But I am unable to access the data Row wise. Am I doing right?
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#### [Solved] How to interpret GPSDestBearing and GPSImgDirection?
• msm1089
• Geography
• Replies: 0
msm1089 Asks: How to interpret GPSDestBearing and GPSImgDirection?
I need to take some photos capturing the GPS coords and direction of camera. I'm using an iPhone 8+ and this is the relevant output I get from Acute Photo EXIF Viewer:
Code:
EXIF GPS Tags
GPSDestBearing:74041/49085
GPSDestBearingRef:M
GPSImgDirection:74041/49085
GPSImgDirectionRef:M
I have searched for about an hour to find a way to convert the bearing/direction into degrees, and every source said the numbers should be between 0-359.99 (i.e. degrees from N, clockwise). So what does 74041/49085 mean and how can it be converted into degrees?
I understand the Ref tags have M for Magnetic North. I need the direction using North the same way that Google Maps does, which is True North. Assuming I can get the direction as degrees from Magnetic North, how would I convert them to degrees from True North?
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#### I don't understand ions. Do they have a taste?
• олеся
• Chemistry
• Replies: 0
олеся Asks: I don't understand ions. Do they have a taste?
1. Do ions define the taste of the compound they form? Blood tastes metallic because hemoglobin contains ions of iron, but like... I'm sure it doesn't always work like that
2. Do they have physical properties like taste when they're on their own? Say I dissolve NaCl in water, then it's just Na+ and Cl- floating around, they're not really a compound anymore, are they? But we would feel the taste
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#### Why do we treat particles as standing waves in QP?
• QuantumGenius
• Physics
• Replies: 0
QuantumGenius Asks: Why do we treat particles as standing waves in QP?
The Quantum Physics course I am taking starts with the Classical Wave Equation and a statement that we treat quantum particles as standing waves.
The explanation they give is that most of the time particles are bound by some kind of potential and that the solution for the Particle in a Box problem is a standing wave.
The way I understood it is that it comes naturally from the Classical Wave Equation, that it is the only solution with such boundary conditions. But then I looked at a guitar string in slow motion and it didn't behave like a standing wave! After some googling I found out that its behavior is called a pulse.
Now I am at a loss. Why do we only consider standing waves and discard other solutions? Are there some restrictions that should be applied that I am not aware of?
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#### QFT in frw spacetime
• Obama2020
• Physics
• Replies: 0
Obama2020 Asks: QFT in frw spacetime
The metric for frw spacetime is $ds^2$ = $a^2(dn^2 - dx^2)$ where $dn$ is the conformal time differential form. The Klein Gordon equation in curved spacetime is $(\frac{1}{g^{1/2}}\partial_{\mu}(g^{1/2}g^{\mu\nu}\partial_{\nu}) + m^2)\phi = 0$
From this one can obtain the Klein Gordon equation for frw spacetime
$\ddot{\phi} + 2\frac{\dot{a}}{a}\dot{\phi} - \Delta \phi + m^2a^2\phi = 0$ (mukhanov 64)
How do we derive this equation from the above equation?
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# Torque, equilibrium and calculating angle of forces
## Homework Statement
I just did two equilibrium/torque problems and it seems like they contradict each other. I understand the basic steps here, but I'm baffled about what I perceive to be a contradiction. Here are the problems:
A 75 kg window cleaner uses a 10 kg ladder that is 5 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked window and climbs the ladder. He is 3 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what is the angle (relative to the horizontal) of the force of the ground on the ladder?
The second question entails a horizontal beam 3 m long, of weight 500 N, suspended horizontally. On the left end, it is hinged to a wall. The right end has a cable that runs to the wall, a distance of D above the hinge point. (The wall, cable and beam form three sides of a triangle). The least tension that will snap the cable is 1200 N. It asks what value of D corresponds to that tension.
## Homework Equations
Horizontal forces are in equilibrium, as are vertical forces and torques.
## The Attempt at a Solution
I'm confused here because in the first problem, the angle of the force of the ground on the ladder is NOT the same as the angle of the ladder and the ground. The angle between the ladder and ground is 60 degrees, whereas the answer to number 1 (and therefore the angle of the force relative to the ground) is 71 degrees.
HOWEVER, in the second problem, it seems like you find the answer by calculating the vertical component of the force of the cable on the beam (it ends up being 250 N), then using the sine trig relationship to find the angle of the force of the cable on the beam (it ends up being 12 degrees). You then assume that this angle is the same as the angle between the beam and the cable, and use the tangent trig relationship to relate 12 degrees, 3 meters (the length of the beam) and D. This yields a D of 0.64 m, which is the answer. My question is, why do we assume that the angle of the force and the angle between the ladder and ground is different whereas, in the second case, we assume that that the angle of force and the angle between the beam and the cable are the same.
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1. Jun 25, 2010
### A B C
1. The problem statement, all variables and given/known data
The moon is observed using a telescope that has an objective lens with a focal length of 3.0 m and an eyepiece with a focal length of 7.5 cm. What is the angular diameter of the moon if the earth-moon distance is 3.85 × 108 m and the diameter of the moon is 3.48 × 106 m?
2. Relevant equations
0 = ho / do
3. The attempt at a solution
0 : the angular size.
ho: the high of the object
do: the distance of the object
2. Jun 25, 2010
### collinsmark
Hello A B C,
What do you think?
3. Jun 25, 2010
### A B C
I think it's 0.36 rad, but I'm not sure
4. Jun 25, 2010
### collinsmark
Fair enough. But if you want us to confirm, you'll have to show us your work.
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# Niece child relationship
### Niece and nephew - Wikipedia
Niece or nephew 4. 5. 6. 7. Gener- ation. Granduncle or -aunt. Grandchildren. Children. YOU. Page 2. This chart shows the blood relationship of “YOU” to. Of course, many of the terms used to express family relationships are well-known and used But what is the relationship of your child to your great-grandniece?. Your cousin's child is NOT your second cousin. So what do you call your cousin's child? What Relationship is Your Cousin's Child? Share; Pin; Email.
All the great-grandkids in the family who are two generations away are "second cousins" to one another, all the great-great-grandkids are "third cousins" to each other, etc.
It's sort of like finding a common denominator--you start with the broadest set of "cousins" you can find in a shared generation, and then count off how much one side is out of balance. As an example, if you and I are in the same generation as second cousins, then you're a second cousin "once-removed" to my kids, since you're in a different generation than they are, by one.
I'm the same thing to your kids, for the same reason in reverse. Our collective set of kids would all be "third cousins" to one another, since they're all once again evenly distant from the great-grandparents that you and I share, bringing the whole thing back into balance.
Hope that's not too complicated--it's actually the type of thing you could explain in 2 secs with some diagrams. Your sister's kids are related to her by 0. Your niece's son is related to her by 0. Your son is related to you by 0. He's related to her son by his relation to her 0. Or equivalently, by your relation to your niece's son, 0.
### How is my neice's baby related to me? And to my son? - family | Ask MetaFilter
Again, assuming a random mating population and no inbreeding, and diploidy -- humans are diploid; for haplodiploid insects, the equations are different, and so are the consequences for kinship and altruism. Hamilton, your inclusive fitness is your own reproductive success plus the reproductive success of your kin times the coefficient of your relation to those kin.
So your niece having a child increases your fitness one-fourth as much as you having a child; equally, if she has four kids, that increases your inclusive fitness as much as you having one kid of your own. A fore-bearer named Japhet had several children including a son named Japhet. When his first wife died, he remarried and had another son.
## Niece and nephew
His second wife also insisted on naming her son Japhet after his father. The boys were only about seven years apart in age.
Although currently the two terms are frequently used interchangeably, there is a difference between a genealogy and a family history. A genealogy starts with one ancestor, most often the original immigrant to the United States, and traces all his descendants to the present time.
If that ancestor arrived on these shores in colonial times, you can imagine the hundreds and hundreds of descendants he now probably has and what a mammoth task it must be to find even half of them! A family history starts with yourself or your children and moves back through your two parents, four grandparents, eight great-grandparents, sixteen great-great-grandparents, thirtytwo great-grandparents, etc.
Most genealogy courses offered today concentrate mostly on techniques for a family history rather than a genealogy.
## Cousins Chart: Understanding Your Family Relationships
However, many of the methods can be used for both. While this is handy for common speech, it can be frustrating to genealogist or family history researcher.
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## calculating critical speed in a ball mill
#### Calculating Critical Speed Of Ball Mills - alanglover.za
Ball mill critical speed formula derivation YouTube · ball mill critical speed speed calculator derivation of critical speed of ball The formula for calculating the critical speed of ball mill design pdf Crusher Price aggregate crushing plant Ball Mill Operation And Maintenance, if you want to konw ball mill design,used second hand Ball .
#### importance of critical speed of the mill
critical speed of mill - wildpeppersf. Axial transport in dry ball mills - CFD- critical speed of mill,12 Dec 2003, in the front to allow the mill internals to be seen The fill level was 30% by volume and the mill speed was 75% of the critical speed required to centrifuge particl The charge consisted of rocks and balls, with balls mak.TECHNICAL NOTES 8 GRINDING R P King - Mineral .
#### To Calculate Critical Speed Of Ball Mill Practical
The "Critical Speed" for a grinding mill is defined as the formula to calculate critical speed in ball mill » critical speed calculation formula of ball mill . grinding mill critical speed calaulator – Grinding Mill China . grinding mill critical speed .
#### What it is the optimun speed for a ball mill .
Oct 19, 2006· Congratulations on your ball mill! You'll want to further reduce your motor speed by 3:1 to mill more efficiently. Right now, your RPM is just slightly under the speed where the media won't do any work. The optimum speed for your jar to turn is around 90 RPM like Frozentech said.
#### SAGMILLING.COM .:. Mill Critical Speed Determination
Mill Critical Speed Determination. The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where balls will not fall away from the mill's shell. Enter the mill diameter inside the shell (excluding liners).
#### SAGMILLING.COM .:. Mill Critical Speed Determination
Mill Critical Speed Determination. The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where balls will not fall away from the mill's shell.
#### derivation to calculate critical speed of ball mill
Calculate critical speed. 3.10 Derivation of equation of critical speed of Ball mill and its calculations. 3k. Get Price Dynamics of Balls and Liquid in a Ball Mill - Department of .
#### Mill Speed - Critical Speed - Paul O. Abbe
The ideal mill speed is usually somewhere between 55% to 75% of critical speed. Critical Mill Speed. Critical Speed (left) is the speed at which the outer layer of media centrifuges against the wall. Second Critical Speed (middle) is the speed at which the second layer of media centrifuges inside the first layer. nth Critical speed (right) is the speed at which the nth layer of media centrifuges inside the n-1 layer …
#### How can I determine the best RPM for Dry Ball Milling .
Note that the critical speed of the mill is calculated as follows: Critical speed (in rpm) = 42.3/sqrt(D - d) with D the diameter of the mill in meters and d the diameter of the largest grinding ball you will use for experiment (also expressed in meters) Then again, a cataracting motion produces a coarser grind.
#### Mill Speed - Critical Speed
Mill Speed - Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product.
#### Ball Mill Parameter Selection – Power, Rotate Speed, Steel .
2.2 Rotation Speed Calculation of Ball Mill Critical Speed_ When the ball mill cylinder is rotated, there is no relative slip between the grinding medium and the cylinder wall, and it just starts to run in a state of rotation with the cylinder of the mill. This instantaneous speed of the mill is as follows:
#### TECHNICAL NOTES 8 GRINDING R. P. King
Figures 8.5 for the popular mill types. 3 c is the mill speed measured as a fraction of the critical speed. More reliable models for the prediction of the power drawn by ball, semi-autogenous and fully autogenous mills have been developed by Morrell and by Austin. (Morrell, S. Power draw of wet tumbling mills .
#### how to calculate critical speed of ball mill
Home / calculate critical speed of ball mill calculate critical speed of ball mill 360 Degree Rotation Lab Grinder For Cocoa Nibs - Buy Lab Grinder Adjustable Revolution Speed: The lab ball mills are best choice for your lab's grinding and research of new materials The following promises are valid within the warranty period which is .
#### Mill Critical Speed Calculation
Effect of Mill Speed on the Energy Input In this experiment the overall motion of the assembly of 62 balls of two different sizes was studied. The mill was rotated at 50, 62, 75 and 90% of the critical speed. Six lifter bars of rectangular cross-section were used at equal spacing.
#### How to calculate critical speed of ball mill - Henan .
calculate critical speed of ball mill ugcnetnicin Ball Mill Critical Speed Mineral Processing 20181110 A Ball Mill Critical Speed actually ball rod AG or SAG is the speed at which the centrifugal forces equal gravitational forces at the mill shells inside 247 online how to calculate critical speed of a ball mill.
#### What Is Critical Speed Of Ball Mill - curacyte.eu
Ball mill critical speed formula derivation YouTube. Nov 10, 2016 · Ball Mill Calculation Critical Speed ball mill critical speed example One example is the large ball mill, critical speed calculation formula of ball mill size calculations Ball Mill Instruction Manual then in the final formula.
#### calculating critical speed of ball mills
Critical Speed Of Ball Mill Calculation India - calculator for ball mill critical speed, Chapter 7.Tubular Ball Mills – Scribd … of the critical speed and a ball charge of 45% of the mill …Mill length, % charge or ball loading, Mill speed, Mill type.
#### critical speed of the ball mill - nerohairsalon
The theoretical critical speed of a ball mill is the speed at which the et price ball mill operating speed mechanical operations solved problems ep 11, 2014 in a ball mill of diameter 2000 mm, 100 mm dia steel balls are being used for grindingalculations the critical speed of ball mill is given by,et price.
#### Ball Mills - Mine Engineer.Com
The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed. Ball Mills are generally used to grind material 1/4 inch and finer, down to the particle size of 20 to 75 microns.
#### how to calculate critical speed of a ball mill - Exodus .
how to calculate critical speed of a ball mill,Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed but most mills operate at speeds between 65 and 79 percent of critical speed Rod mills speed should be limited to a maximum of 70 of critical speed and preferably should be in the 60 to 68 percent critical speed range
#### Formula Calculates The Critical Speed Of A Ball Mill .
critical speed of ball mill formulae. formula calculates the critical speed of a ball mill Founded in 1997, mobile crusher and formula critical speed ball mill Mining Ball Mills Mine Engineer.Com This formula calculates the critical speed of any ball mill.
#### Ball Mills - Mine Engineer.Com
This formula calculates the critical speed of any ball mill. Most ball mills operate most efficiently between 65% and 75% of their critical speed. Photo of a 10 Ft diameter by 32 Ft long ball mill in a Cement Plant.
#### how do i calculate the critical speed of a mill
OVERVIEW ON THE GRINDING MILLS AND THEIR DUAL . - ABB. SAG mill, ball mill, ring gear, dual pinion, drive systems, variable speed ... Figure 7 shows one of the synchronous motors installed on Ball Mill 1. . mill controller measures cascading of the material, by a decreasing torque, before the critical.
#### Ball Mill Critical Speed - Mineral Processing & Metallurgy
Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed, but most mills operate at speeds between 65 and 79 percent of critical speed. Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent …
#### SAGMILLING.COM .:. tools
Mill Critical Speed Calculation Estimates the critical speed of a grinding mill of a given diameter given the mill inside diameter and liner thickness. If given a measured mill rotation (RPM), then the mill`s fraction of the critical speed is given
#### Calculating Critical Speed Of Ball Mills Gulin Machinery
Critical speed ball mill 350 for wet grinding of critical speed mill mathematical and utility software one to calculate belt length for an learn morealculate correct ball mill speed - gulin machineryritical speed of ball mill calculation indiacement popular qa for calculate correct ball mill speed.
#### to calculate critical speed of ball mill practical
20 Jun 2012 . Ball mill rotates at a constant speed, causing the mutual impact and friction of . As per the experience from practical production, when the model of ball mill is . the rotating speed of the ball mill becomes the very critical factor to the . the vector controlling calculation; the force of the ball mill is separated to.
#### calculator for ball mill critical speed - oalebakkershoes
The critical speed of the mill, &c, is defined as the speed at which a single ball will just . Figure 8.3 Simplified calculation of the torque required .. Rod and ball mills in Mular AL and Bhappu R B Editors Mineral Processing Plant Design.
#### formula for critical speed of a ball mill
Ball Mill Critical Speed.Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed, but most mills operate at speeds between 65 and 79 percent of critical speed. Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent critical speed range. Pebble
#### Formula For Critical Velocity Of Ball Mill Pdf
The "Critical Speed" for a grinding mill is defined as the formula to calculate critical speed in ball mill . Get Price And Support Online; Ball Mills - Mine Engineer.Com. This formula calculates the critical speed of any ball mill. Most ball mills operate most efficiently between 65% and 75% of their critical speed.
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Main Page - Lesson Deals - Play Technique
Previous - Lessons - Next
Bidding Conventions
LESSON 24
Alternative Hand Valuations
Introduction Losing Trick Count - Own hand (LTC) Losing Trick Count - Partner's hand (LTC) Trick value Shortage Points (TSP) The Banzai Point Count (BPC) Long Suit Trial bid Deals 89 to 92 Quiz - Answers - Review Recent developments Bidding Guide : advBG-17 - advBG-20 advBG-21 - advBG-22
(
Down - Top)
BR 24.1 - Introduction
Consider Hand 1 below. Your Partner opens with 1S.
What is your standard response ? 4S of course showing 6-10 (Bidding Guide p.4). You have less than 9 HCPs, 4 card trump support and an unbalanced hand.
Counting the the total points of the hand it does not quite add up : 8 HCP + 3 SP = 11 points.
Obviously the combined effect of side suit shortage and an abundance of trumping power makes the hand stronger than the points indicate.
Now look at Hand 2 below.
After Partner's opening bid of 1S you obviously have sufficient points together for a Game contract. But you have two useless 3 card suits in your hand. If Partner has the same in one of these suits, and not much protection in the other, a Game contract may be doomed from the start.
These type of common examples show up the inadequacies of assessing the distributional values of a hand by means of the 5-3-1 shortage points system only.
In recent years an additional new valuation system has been introduced that, in combination with the point count system, is used by many players. It assesses the trick taking potential of the combined hands for trump contracts only. It is called the Losing Trick Count (LTC).
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BR 24.2 - Losing Trick Count - Own Hand
The Losing Trick Count (LTC) is a valuation method for trump contracts only. You can use it after a trump fit has been found.
Count your losers in the top three cards in each suit. Cards beyond these are considered winners.
1. Ace = 1 winner
2. King = 1 winner
3. Queen = half a winner, or a whole winner when supported by a second Honour
4. Void = no loser (therefore represents 3 winners)
5. Singleton = 1 loser (therefore represents 2 winners)
6. Doubleton = 2 losers (therefore represents 1 winners)
Combining above Honour and shortage counts : an A x or K x doubleton counts for 1 loser. A K doubleton is no loser at all, so does a singleton Ace. K Q doubleton is 1 loser as is K Q x, and so on.
The maximum number of losers you can hold in your hand is 4 x 3 losers = 12 losers. The same applies to the hand of your Partner, therefore :
24 - (your losers + Partner's losers) = the likely number of tricks you will make
This means that if you and Partner have 14 losers in the combined hand you are likely to have 10 winning trick, enough for a Game contract in a Major suit. With 12 losers a Slam is a very good proposition.
Let us have a look at how the two hands from the previous chapter are valued using this method.
Hand 1 is valued at 7 losers. This is the same as the strength of an average minimum opening hand.
No wonder then that a Game contract with this supporting hand is going to be a good prospect.
Hand 2 in comparison, despite its much higher point count contains 8 losers, a full loser more than the average opening hand.
If for arguments sake we take away the Club Queen in Hand 1, reducing its strength to 6 HCPs only, it is still equal in loser count (now 8) to Hand 2, which now has more than twice the number of HCPs.
The above comparison makes it very clear that the distribution pattern of a hand has a huge influence on its trick taking potential, much more than expressed by the 1-3-5 shortage point valuation.
It is essential to always remember that you must have a good trump fit, as much of the trick taking potential is based on being able to take advantage (by ruffing) of the shortages in a hand.
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BR 24.3 - Losing Trick Count - Partner's Hand
Valuating your own hand with the LTC method is easy, but how about assessing Partner's hand ?
Here you need to rely on his bidding.
Approximate valuations for standard opening bids and responses are :
• 6-10 pts : Single raise or 1NT Response = 8-9 losers
• 6-9 HCP : Game raise to Partner's 1H or 1S = 7-8 losers
• 11+ pts : New suit at the 2 level Response = 8 losers or less
• 13+ pts : Jump raise = 7 losers or less
• 13-15 pts : Minimum Opening bid or 2NT Response = 7 losers
• 16-18 pts : Strong Opening bid or 3NT Response = 6 losers
• 19-21 pts : Maximum 1 Opening bid = 5 losers
• 22+ pts : Strong 2C, 2NT Opening bid = 4 losers or less
Various other bids are :
• 9-15 pts : Overcall at the 1 level = 8 losers or less
• 11-15 pts : Overcall at the 2 level = 7 losers or less
• 12+ pts : Takeout Double = 7 losers or less
• 6-10 HCP : Weak 2 Opening, Weak Jump overcall = 7-8 losers
• 6-10 HCP : Preemptive 3 Opening bid = 7 losers (not vulnerable) or 6 losers (vulnerable)
• 6-10 HCP : Preemptive 4 Opening bid = 6 losers (not vulnerable) or 5 losers (vulnerable)
• 8-12 HCP : Michaels Cue bid, Unusual 2NT = 5-7 losers
The key totals you need to remember are 14 losers or less required for a Game contract, while 12 losers or less are required for a Slam.
Most importantly however remember that all LTC calculations are only valid when you have a good (8 card or better) trump fit. The best approach is then to use both the LTC and the point count system side by side in your overall assessment of the game.
For a good understanding and effective use of the LTC I recommend Ron Klinger's excellent book on the subject "The Modern Losing Trick Count".
The valuation of your hand (and of the potential of your and Partner's combined holdings) should never be based on just one assessment method alone.
1. Always use the LTC in combination with the Point count system and your own judgement and creative insight of your hand.
2. And always assess your hand in the context of what you know about your Partner's hand.
A frequently heard lament at the bridge table of less experienced players is "But Partner, I could not support you, I had too many losers", or "Partner, I just did not have the point to go on." These are clear signs that the player concerned was looking at his (or her) hand in isolation and not in the context of Partner's holdings.
Comment Be aware that using the LTC on its own can produce a too conservative assessment. For (although this is not stated in any of the textbooks I have come across) the LTC is based on a 12-card hand. The fact that the 13th card is not counted as a loser does not mean (as many appear to believe) that it is a winner. The 13th card in both your and Partner's hand are simply disregarded altogether. The 13th card can be a loser in one hand and a winner in the other a loser in both hands or a winner in both hands A simple analysis of say 100 deals (as I have done for the deals in this course) bears this out. For about one third of the Deals the LTC is spot on (or in a few cases over overvalued), while in the remaining two thirds of the Deals the actual tricks made were 1 or 2 tricks more than the LTC assessment. I therefore suggest that, whenever your hand contains "extra value" (such as a few 10s and 9s, or a strong long suit), you subtract the total assess loser in the combined hands from 25 rather than the prescribed 24. I have included this alternative in the Bidding Guide, BG-15.
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BR 24.4 - Trick Value Shortage Points
Shortage in one suit has several effects on the trick taking value of the whole hand.
1Take a perfectly balanced hand with distribution of 4-3-3-3 and convert this into a hand containing a void. To do this we remove the three cards from one suit. But as the whole hand must consist of 13 cards, three cards must be added to one or more of the other 3 suits, increasing their lengths. Therefore the least unbalanced hand we can create this way has a distribution of 5-4-4-0. In practical terms this means that the chance of making extra tricks with the small cards from a long suit has increased. 2 Having a void in a suit effectively cuts out all 10 High Card Points in that suit.Partner may hold some of these points, but the overall effect would be negative for the Opponents.With a singleton, only the Ace will win a trick, but probably none of the other Honours in the suit (eliminating 6 HCPs). With a doubleton only 3 HCPs are eliminated. 3The shortage enables Declarer to control the suit with his trumps, and when the shortage occurs in the hand with the lesser number of trumps it can provide extra tricks by ruffing.Extra ruffing tricks are also created in both hands in a cross ruff or a dummy reversal play.
In view of the above considerations the Losing Trick Count valuation of voids, singletons and doubletons appears well justified.
Let us now consider the point count system.
To make a Game trump contract in a major suit requires 26 points. This means that each of the 10 winning tricks requires on average 2.6 points.
Therefore, combining the point count system with the the losing trick count :
• If a void is valued at no loser and in consequence represents 3 winning tricks, its true point count value must be 3 x 2.6 = 7.8 points
• Likewise, if a singleton is valued at one loser and in consequence represents 2 winning tricks, its true point count value must be 2 x 2.6 = 5.2 points
• And if a doubleton is valued at two losers and in consequence represents 1 winning trick, its true point count value must be 1 x 2.6 = 2.6 points
This is a very simple approach which I use myself to get a better point count value for all unbalanced hands and appears to be much more realistic in its results than the traditional shortage point valuation.
I call it the Trick Value Shortage Points valuation method (TSP).
The only variation to the above valuation concerns the doubleton
The 2.6 points valuation for a doubleton is fair enough. However before you can effectively consummate this single winner it is necessary to first lose 2 tricks to the opposition. In a Game contract this represents at least 65% of your permissible loss.
Two useless doubletons in a hand are potentially what I call the kiss of death. You may well be down one trick already before you have managed to gain the lead.
Count the doubleton therefore always for 1 point only, unless the doubleton includes the Ace. If so count it for 2 points.
With an abundance of trumps (9 or more in the combined hands) you may add one additional TSP to the singleton and void.
With a 5-3 trump fit and shortage in the long trump suited hand (usually Declarer's hand) it is wise to deduct one TSP of the singleton or void TSP value.
Assessing Hand 1 (after Partner has opened 1S) in terms of TSPs one reaches a more realistic valuation of 8 HCP + 6 TSP = 14 points total. The extra 1 TSP being awarded for the 9 card trump fit.
In general be cautious when you hold a balanced or semibalanced hand with one or two useless doubletons, or worse, a useless 3 card suit.
With an unbalanced hand be aggressive in your approach.
I have included an alternative page 1 for the Intermediate and Advanced Bidding Guides which includes the TSP count. You can use this one instead of the Basic page if you wish.
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BR 24.5 - The Banzai Point Count
Statistical analyses published by Richard Cowan in 1987 and David "Banzai" Jackson & Ron Klinger in 2010 (Better Balanced Bidding) have revealed two deficiencies of the Milton Point Count (A=4 - K=3 - Q=2 - J=1), universally used for the past 90 years.
1. The MPC does not include the 10s in its valuation
2. The MPC substantially overvalues Aces and Kings in balanced hands
They propose the Banzai Point Count as an accurate representation of the statistical findings, to be used by Intermediate-Advanced players.
Banzai Point Count : A=5 K=4 Q=3 J=2 10=1
This increases the total of High Card Points in a whole pack of cards from 40 to 60.
In addition a 5-card suit in a balanced hand (5332) is valued at an additional 2 points.
A minimum Opening can be made with 18 Banzai Points (BPs) and a minimum Response with 9 BPs. The minimum strength for a Game contract is 37 BPs. For further details see advBG-21.
Obviously the Banzai Point Count is superior to Extended Milton. However it will probably take years before this new system (first published in 2010) is fully in mainstream use. For the time being Extended Milton (as now used throughout this online Jazclass Bridge Course) appears therefore the better option, especially for Beginners and most Intermediate players.
Extended Milton : A=4 K=3 Q=2 J=1 10=½
This increases the total of High Card Points in a whole pack of cards from the original 40 to 42.
However I recommend that Intermediate players have at least a working knowledge of the Banzai Point Count, so that in borderline decisions of balanced hands it can be used as an additional valuation check.
Generally it is also useful to be aware that balanced hands containing mainly Aces and Kings are overvalued in point counts and those including several Jacks and/or 10s are undervalued.
With hand Combinations containing at least one unbalanced hand it is best to use a combination of Extended Milton, The Losing Trick Count and Quick Tricks for your valuation assessment.
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BR 24.6 - Long Suit Trial bid
Consider Hand 3 shown below.
You opened with 1S and Partner raised to 2S. He therefore has trump support and 6-10 points. You, after revaluating your hand hold 17 points or 18, when counting TSPs. What next ?
Standard procedure with your hand is to bid 3S, inviting Partner to bid 4 with a maximum (8-10 pts) and Pass with a minimum (6-7 pts).
However if Partner has maximum points, but no values in Diamonds, you will lose 3 tricks right from the start in Diamonds and may well go down one or two.
This is where the long suit trial bid (advBG-17) becomes very useful.
When a major suit is single raised to the 2 level (by either Partner) the bid of a new suit is a long suit trial bid, asking Partner for his losers in the bid suit.
To make a long suit trial bid you should have 3 or 4 cards in the suit and preferably 3 but definitely at least 2 losers. It also should hold none or at the most one of the top three Honour cards (A, K or Q).
Typical long suit trial bid suits are : 8 5 3 or A 9 4 2 or Q 9 3 or K 7 6 4
Partner may not Pass but replies :
For Hand 3 the bidding would therefore go : 1S - 2S - 3D - Partner ?
Partner will bid 4S with no or 1 loser in Diamonds even with minimum points, or with 2 losers and a maximum. He will bid 3S with 2+ losers and a minimum or 3 losers and maximum points.
Here follow two typical cases :
Case A Opener (W) ♠ - A Q 9 6 3 ♥ - A 8 6 4 ♦ - A 8 ♣ - Q 7 Responder (E) ♠ - K J 8 2 ♥ - K 7 ♦ - 7 5 3 ♣ - 9 8 5 2 Comment East has only 1 loser in the Trial suit (Hearts). He therefore bids Game in the agreed trump suit. Bidding : (W) 1S - 2S - 3H - 4S - Pass
Case B Opener (W) ♠ - A Q 9 6 3 ♥ - A 8 6 4 ♦ - A 8 ♣ - Q 7 Responder (E) ♠ - K J 8 2 ♥ - 9 5 3 ♦ - K 7 ♣ - 9 8 5 2 Comment The two hands are identical, except that East's Heart and Diamond holdings are reversed. East now has 3 losers in the trial suit and must therefore sign off in 3S ! Bidding : (W) 1S - 2S - 3H - 3S - Pass
The long suit trial bid can also be made by the Responder.
Partner opened the bidding with 1C, you holding Hand 4 responded with 1S and Partner raised to 2S.
What do you do next ?
Same dilemma, Game is likely, but if Partner has 3 losers in Diamonds (opposite your unprotected Queen) the Game contract may well go down. Once again the long suit trial bid can come to the rescue.
1C - 1S - 2S - 3D - Partner ?
Whatever bid Partner makes, your contract will be more secure for it.
Cue bidding the Enemy suit
A long suit trial bid can also be made after an Enemy Overcall as a Cue bid of the Enemy suit.
For example :
1S - (2D) - 2S - (Pass) - 3D
3D is still a long suit trial bid, to which Partner must respond in the normal prescribed manner.
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BR 24.7 - Deal 89 - 92
Deals 89 - 92 are examples of bidding as outlined in this lesson.
BR 24.8 - Quiz 24 - Answers - Review
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Definition and delimitation of regression model
An embarrassingly simple question -- but it seems it has not been asked on Cross Validated before:
1. What is the definition of a regression model?
Also a support question,
1. What is not a regression model?
With regards to the latter, I am interested in tricky examples where the answer is not immediately obvious. For example,
• What about latent variable models (e.g. ARIMA or GARCH)?
I would say that "regression model" is a kind of meta-concept, in the sense that you will not find a definition of "regression model", but more concrete concepts such as "linear regression", "non-linear regression", "robust regression" and so on. This in the same way as in mathemathics we usually do not define "number", but "natural number", "integers", "real number", "p-adic number" and so on, and if somebody will want to include the quaternions among numbers so be it! it doesn't really matter, what matters is what definitions is used by the book/paper you are reading at the moment.
Definitions are tools, and essentialism, that is discussing what is the essence of ..., what a word really means, are seldom worthwhile.
So, what distinguishes a "regression model" from other kinds of statistical models? Mostly, that there is a response variable, which you want to model as influenced by (or determined by) some set of predictor variables. We are not interested in influence the other direction, and we are not interested in relationships among the predictor variables. Mostly, we take the predictor variables as given, and treat them as constants in the model, not as random variables.
The relationship mentioned above can be linear or nonlinear, specified in a parametric or nonparametric way, and so on.
To delineate from other models we better have a look at some other words often taken to denote something different for "regression models", like "errors in variables", when we accept the possibility of measurement errors in the predictor variables. That could well be included in my description of "regression model" above, but is often taken as an alternative model.
Also, what is meant might vary among fields, see What is the difference between conditioning on regressors vs. treating them as fixed?
To repeat: what matters is the definition used by the authors you are reading now, and not some metaphysics about what it "really is".
• I agree with the essence of your answer. My question was motivated by having encountered statements about regression models that left me wondering what the statement really applies to (and what it does not apply to). Of course, now you could say, "use your best judgement and check the details carefully", but sometimes I might wish to reject the hypothesized statement right away saying that it is not true in general (perhaps true only in a very specific case). Then I need a definition to refer to. There are of course more such situations where having a precise definition is useful. – Richard Hardy Mar 16 '16 at 13:12
• Then you shouls ask specific questions about those uses you have encountered, with references. – kjetil b halvorsen Mar 16 '16 at 13:18
• I don't intend to be picky, but think about it: someone asks you what you are doing, you say "I am analyzing/forecasting/testing [something] using regression models." -- "What is a regression model?" -- (Silence). Or a situation in an introductory econometrics class: "Professor, what is a regression model?" -- (No answer). I think these are very natural questions, so it would be nice to have an answer. – Richard Hardy Mar 16 '16 at 19:24
• Yes, it would be nice to have an answer, but I am not sure there is one canonical answer all can agree about. I got a very different idea of regression from a statistical book such as Seber: "Linear Regression Analysis" as from a text in econometrics. But some ideas all can agree about. I guess it is really a family of models. Then we can ask what is the common core of all this models. – kjetil b halvorsen Mar 16 '16 at 20:44
In regression case we have some random variables $Y$ and $X_1,\dots,X_k$. The variables have some unknown distribution and complicated covariance structure. We simplify this problem to focusing solely on conditional distribution, or more precisely on conditional expectation of $Y$ given the other variables. We simplify it to
$$\mu = E(y|x_1,\dots,x_k) = f(x_1,\dots,x_k)$$
Where $f$ is a function of predictors that can take different forms (linear, non-linear) depending on particular regression model and $\mu$ is a mean of some distribution when thinking of regression models in terms of generalized linear models. In GLM's $\mu$ can be location of Poisson, Binomial, Gamma etc. distributions. With $L_1$ regularized regression it is a location of Laplace distribution, for robust model minimizing Huber loss so called Huber density is used. In case of quartile regression we focus on other feature of distribution, we estimate $\mu$ that is distribution's quartile rather then expected value.
So instead of looking on full joint distribution, we focus on conditional distribution of $Y$. This simplification is a key feature of regression models.
• Thanks. Intuition doesn't hurt, although I am looking for a more formal definition that I could throw at someone who asked me, So what is a regression model anyway? and then tried to pick on details. – Richard Hardy May 6 '16 at 16:47
• @RichardHardy I think that this is the key feature of regression models that is shared by all of them. – Tim May 6 '16 at 16:49
• I think this answer is a correct and useful approach, but it needs to be generalized so it can apply to situations commonly thought of as "regression" (including GLMs, multiplicative errors, regression with transformed responses, quantile regression, and so on). More broadly, a regression model specifies one or more properties of the entire distribution of the response $y$ in terms of the values of the regressors (within specific ranges, random or fixed). In particular, it can go far beyond merely specifying the expectation or assuming an additive error. – whuber May 6 '16 at 18:54
Some thoughts based on the literature:
F. Hayashi in Chapter 1 of his classic graduate textbook "Econometrics" (2000) states that the following assumptions comprise the classical linear regression model:
1. Linearity
2. Strict exogeneity
3. No multicollinearity
4. Spherical error variance
5. "Fixed" regressors
Wooldridge in Chapter 2 of his classic introductory econometrics textbook "Introductory Econometrics: A Modern Approach" (2012) states that the following equation defines the simple linear regression model:
$$y=\beta_0+\beta_1 x+u.$$
Greene in Chapter 2 of his popular econometrics textbook "Econometric Analysis" (2011) states
The classical linear regression model consists of a set of assumptions about how a data set will be produced by an underlying “data-generating process.”
and subsequently gives a list of assumptions similar to that of Hayashi's.
A curiosity relevant to the last bullet point of the OP: Bollerslev "Generalized autoregressive conditional heterosedasticity" (1986) includes a phrase "the GARCH regression model" in the title of section 5 and also in the first sentence of that section. So the father of the GARCH model did not mind calling GARCH a regression model.
• Your three references are all restricted to the linear regression model, but your question is wider than that. (Thus, using this as argument in your answer to another post, which I presume spawned the interest in this issue, is I think not completely valid.) If you'd say that latent variable models are not regression models, then using the immediate connection with measurement errors, regression models with measurement errors would no longer be regression models. Seems odd to me. Wiki just says that a reg model relates indep vars to dep in the sense that $Y\approx f(X, \beta)$. – hejseb Sep 23 '15 at 18:23
• True, my examples are for linear regression models; that is what I was able to find in reliable sources such as these textbooks that are widely used and have become classic. I do not trust Wikipedia that much for statistical and econometric questions. Anyway, even in Wikipedia there is a chapter "Underlying assumptions" that is similar to what I have cited from the textbooks. Regarding the other post, could you post the relevant part of your comment there so that I could respond there? In this post I did not say anything about latent variable models, but it's good to hear you opinion. – Richard Hardy Sep 23 '15 at 18:35
• Why point 3, "no multicollinearity"? I have never seen that used as an assumption in the proof of some result! – kjetil b halvorsen Mar 16 '16 at 13:00
• @kjetilbhalvorsen, please do not hold me responsible over what is written in a textbook which I am not the author of. But thanks for the comment, of course, and even more for the answer! – Richard Hardy Mar 16 '16 at 13:07
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# 132126 (number)
132,126 (one hundred thirty-two thousand one hundred twenty-six) is an even six-digits composite number following 132125 and preceding 132127. In scientific notation, it is written as 1.32126 × 105. The sum of its digits is 15. It has a total of 5 prime factors and 24 positive divisors. There are 41,040 positive integers (up to 132126) that are relatively prime to 132126.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 15
• Digital Root 6
## Name
Short name 132 thousand 126 one hundred thirty-two thousand one hundred twenty-six
## Notation
Scientific notation 1.32126 × 105 132.126 × 103
## Prime Factorization of 132126
Prime Factorization 2 × 3 × 192 × 61
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 6954 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 132,126 is 2 × 3 × 192 × 61. Since it has a total of 5 prime factors, 132,126 is a composite number.
## Divisors of 132126
24 divisors
Even divisors 12 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 283464 Sum of all the positive divisors of n s(n) 151338 Sum of the proper positive divisors of n A(n) 11811 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 363.491 Returns the nth root of the product of n divisors H(n) 11.1867 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 132,126 can be divided by 24 positive divisors (out of which 12 are even, and 12 are odd). The sum of these divisors (counting 132,126) is 283,464, the average is 11,811.
## Other Arithmetic Functions (n = 132126)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 41040 Total number of positive integers not greater than n that are coprime to n λ(n) 3420 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12308 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 41,040 positive integers (less than 132,126) that are coprime with 132,126. And there are approximately 12,308 prime numbers less than or equal to 132,126.
## Divisibility of 132126
m n mod m 2 3 4 5 6 7 8 9 0 0 2 1 0 1 6 6
The number 132,126 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
## Base conversion (132126)
Base System Value
2 Binary 100000010000011110
3 Ternary 20201020120
4 Quaternary 200100132
5 Quinary 13212001
6 Senary 2455410
8 Octal 402036
10 Decimal 132126
12 Duodecimal 64566
20 Vigesimal ga66
36 Base36 2ty6
## Basic calculations (n = 132126)
### Multiplication
n×y
n×2 264252 396378 528504 660630
### Division
n÷y
n÷2 66063 44042 33031.5 26425.2
### Exponentiation
ny
n2 17457279876 2306560560896376 304756620668994575376 40266273262511577266129376
### Nth Root
y√n
2√n 363.491 50.9326 19.0655 10.573
## 132126 as geometric shapes
### Circle
Diameter 264252 830172 5.48437e+10
### Sphere
Volume 9.6617e+15 2.19375e+11 830172
### Square
Length = n
Perimeter 528504 1.74573e+10 186854
### Cube
Length = n
Surface area 1.04744e+11 2.30656e+15 228849
### Equilateral Triangle
Length = n
Perimeter 396378 7.55922e+09 114424
### Triangular Pyramid
Length = n
Surface area 3.02369e+10 2.71831e+14 107880
## Cryptographic Hash Functions
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0
# What is the greatest common factor of 1234 and 78?
Updated: 4/28/2022
Wiki User
13y ago
The Greatest Common Factor (GCF) is: 2
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The greatest common factor of 78 , 41 = 1
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The greatest common factor of 12 and 78 is 6.
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The greatest common factor of 78 and 120 is 6.
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The GCF is 13.The Greatest Common Factor (GCF) is: 13
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comon factor will be 2that is wrong the GCF or (greatest common factor) for 52 and 78 is 13
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The greatest common factor of the numbers 78, 91 and 156 is 13.
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Here is a visualization of several branches of the Lambert W function on the complex plane:
This function is defined implicitly as the inverse of the nonlinear transcendental equation
$W(z) eW(z) =z$
An extremely comprehensive overview of this function, including details of evaluation and applications, is available here.
Evaluation of the function is accomplished using standard inversion algorithms, such as Newton’s method, but one needs a good starting point in order to reach all branches. To find this, first take a logarithm of both sides of the defining equation:
$W(z) =logz -logW(z) ≈logz +f(z)$
The Lambert function can thus be approximated by the natural logarithm, which explains the similarity of its appearance to that function. Putting the right-hand side into the defining equation and ignoring the additive term f in comparison to the logarithm on the resulting left-hand side,
$(logz+f) zef ≈zeflogz =z f=−loglogz$
which gives the approximation
$W(z) ≈logz -loglogz$
This form can be justified by putting it into a slightly rearranged defining equation,
$W(z) =ze −W(z) ≈ze−logz +loglogz =logz$
so that the second term of the approximation allows one to recover the first term.
All that remains now is to specify the branches of the logarithms. This can be done by keeping the second outer logarithm on the principal branch and letting the index of the Lambert function set the branch of the other functions. That is,
$Wn(z) ≈logz +2πin -log(logz +2πin)$
Starting from this complex point allows one to determine the value of the Lambert function for all branches, with the exception of the principal branch W0 in the vicinity of the origin. Since the function is zero there, starting from that value reaches most of the region around the origin, apart from some instability around the negative real axis.
Here is a comparison of the approximation (in red) to the final evaluated result:
While the Lambert function is similar to the natural logarithm, the branches of its imaginary part are not equally spaced as for the latter. This can be seen by visualizing multiple branches of the imaginary part at the same time:
The Lambert function has a complex structure with respect to argument similar to that of the natural logarithm, but reaches its asymptotic coloring noticeably faster than that function.
Uploaded 2020.11.24 — Updated 2020.12.25 analyticphysics.com
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# Calculating Areas and Distances using Integrals
Contributed by:
In this section, we will learn that:
We get the same special type of limit in trying to find the area under a curve or a distance traveled.
1. 5
2. In Chapter 2, we used the tangent
and velocity problems to introduce
the derivative—the central idea in
differential calculus.
3. In much the same way, this chapter starts
with the area and distance problems and
uses them to formulate the idea of
a definite integral—the basic concept of
integral calculus.
4. In Chapters 6 and 8, we will see how to use
the integral to solve problems concerning:
Volumes
Lengths of curves
Population predictions
Cardiac output
Forces on a dam
Work
Consumer surplus
Baseball
5. There is a connection between integral
calculus and differential calculus.
The Fundamental Theorem of Calculus (FTC)
relates the integral to the derivative.
We will see in this chapter that it greatly simplifies
the solution of many problems.
6. 5.1
Areas and Distances
In this section, we will learn that:
We get the same special type of limit in trying to find
the area under a curve or a distance traveled.
7. AREA PROBLEM
We begin by attempting to solve
the area problem:
Find the area of the region S that lies
under the curve y = f(x) from a to b.
8. AREA PROBLEM
This means that S,
illustrated here,
is bounded by:
The graph of a continuous
function f [where f(x) ≥ 0]
The vertical lines x = a and x
=b
The x-axis
9. AREA PROBLEM
In trying to solve the area problem,
What is the meaning of the word area?
10. AREA PROBLEM
The question is easy to answer
for regions with straight sides.
11. For a rectangle, the
area is defined as:
The product of the
length and the width
12. The area of a
triangle is:
Half the base times the
height
13. The area of a polygon
is found by:
Dividing it into triangles
of the triangles
14. AREA PROBLEM
However, it isn’t so easy to find the area
of a region with curved sides.
We all have an intuitive idea of what the area
of a region is.
Part of the area problem, though, is to make this
intuitive idea precise by giving an exact definition
of area.
15. AREA PROBLEM
Recall that, in defining a tangent, we first
approximated the slope of the tangent line
by slopes of secant lines and then we took
the limit of these approximations.
We pursue a similar idea for areas.
16. AREA PROBLEM
We first approximate the region S by
rectangles and then we take the limit of
the areas of these rectangles as we increase
the number of rectangles.
The following example illustrates the procedure.
17. AREA PROBLEM Example 1
Use rectangles to
estimate the area under
the parabola y = x2 from
0 to 1, the parabolic
region S illustrated here.
18. AREA PROBLEM Example 1
We first notice that the
area of S must be
somewhere between 0
and 1, because S
is contained in a square
with side length 1.
However, we can
certainly do better
than that.
19. AREA PROBLEM Example 1
Suppose we divide S into
four strips
S1, S2, S3, and S4 by
drawing the vertical lines
x = ¼, x = ½, and x = ¾.
20. AREA PROBLEM Example 1
We can approximate
each strip by a rectangle
whose base is the same
as the strip and whose
height is the same as the
right edge
of the strip.
21. AREA PROBLEM Example 1
In other words, the
heights of these
rectangles are the values
of the function f(x) = x2
at the right endpoints of
the subintervals
[0, ¼],[¼, ½], [½, ¾],
and [¾, 1].
22. AREA PROBLEM Example 1
Each rectangle has
width ¼ and
the heights are (¼)2,
(½)2, (¾)2, and 12.
23. AREA PROBLEM Example 1
If we let R4 be the sum of the areas
of these approximating rectangles,
we get:
1 2 1 2 3 2 2
R4 1
4 4 1
4 2
1
4 4 1
1
4
15
32
0.46875
24. AREA PROBLEM Example 1
We see the area A of
S is less than R4.
So, A < 0.46875
25. AREA PROBLEM Example 1
rectangles in this figure,
we could use the
smaller rectangles in
the next figure.
26. AREA PROBLEM Example 1
Here, the heights are
the values of f at
the left endpoints of the
The leftmost rectangle
has collapsed because
its height is 0.
27. AREA PROBLEM Example 1
The sum of the areas of these approximating
rectangles is:
2 1 2 1 2 3 2
L4 0
1
4
1
4 4 1
4 2 1
4 4
327
0.21875
28. AREA PROBLEM Example 1
We see the area of S is
larger than L4.
So, we have lower and
upper estimates for A:
0.21875 < A < 0.46875
29. AREA PROBLEM Example 1
We can repeat this
procedure with a larger
number of strips.
30. AREA PROBLEM Example 1
The figure shows what
happens when
we divide the region S
into eight strips of equal
31. AREA PROBLEM Example 1
By computing the sum of the areas of
the smaller rectangles (L8) and the sum of
the areas of the larger rectangles (R8),
we obtain better lower and upper estimates
for A:
0.2734375 < A < 0.3984375
32. AREA PROBLEM Example 1
So, one possible answer to the
question is to say that:
The true area of S lies somewhere
between 0.2734375 and 0.3984375
33. AREA PROBLEM Example 1
We could obtain better
estimates by increasing
the number of strips.
34. AREA PROBLEM Example 1
The table shows the
results of similar
calculations (with a
computer) using n
rectangles, whose
heights are found with
left endpoints (Ln)
or right endpoints
35. AREA PROBLEM Example 1
In particular, we see
that by using:
50 strips, the area lies
between 0.3234 and 0.3434
1000 strips, we narrow it
down even more—A lies
between 0.3328335 and
0.3338335
36. AREA PROBLEM Example 1
A good estimate is
obtained by averaging
these numbers:
A≈
37. AREA PROBLEM
From the values in the
table, it looks as if Rn
is approaching 1/3 as n
We confirm this in
the next example.
38. AREA PROBLEM Example 2
For the region S in Example 1, show that
the sum of the areas of the upper
approximating rectangles approaches 1/3,
that is,
lim Rn 13
n
39. AREA PROBLEM Example 2
Rn is the sum of the
areas of the n rectangles.
Each rectangle has width
1/n and the heights are
the values of the function
f(x) = x2 at the points
1/n, 2/n, 3/n, …, n/n.
That is, the heights are
(1/n)2, (2/n)2, (3/n)2, …,
(n/n)2.
40. AREA PROBLEM Example 2
2 2 2 2
11 1 2 1 3 1 n
Rn ...
n n n n n n n n
1 1 2 2 2 2
2 (1 2 3 ... n )
n n
1 2 2 2 2
3 (1 2 3 ... n )
n
41. AREA PROBLEM E. g. 2—Formula 1
Here, we need the formula for the sum of
the squares of the first n positive integers:
2 2 2 n(n 1)(2n 1)
2
1 2 3 ... n
6
Perhaps you have seen this formula before.
It is proved in Example 5 in Appendix E.
42. AREA PROBLEM Example 2
Putting Formula 1 into our expression
for Rn, we get:
1 n(n 1)(2n 1)
Rn 3
n 6
(n 1)(2n 1)
2
6n
43. AREA PROBLEM Example 2
(n 1)(2n 1)
So, we have: lim Rn lim
n n 6n 2
1 n 1 2n 1
lim
n 6
n n
1 1 1
lim 1 2
n 6
n n
1
12
6
1
3
44. AREA PROBLEM
It can be shown that the lower
approximating sums also approach 1/3,
that is,
lim Ln 1
3
n
45. AREA PROBLEM
From this figure, it
appears that, as n
increases, Rn becomes a
better and better
approximation to the
area of S.
46. AREA PROBLEM
From this figure too, it
appears that, as n
increases, Ln becomes a
better and better
approximations to the
area of S.
47. AREA PROBLEM
Thus, we define the area A to be the limit of
the sums of the areas of the approximating
rectangles, that is,
A lim Rn lim Ln 13
n n
48. AREA PROBLEM
Let’s apply the idea of
Examples 1 and 2
to the more general
region S of the earlier
49. AREA PROBLEM
We start by
subdividing S into n
S1, S2, …., Sn of equal
50. AREA PROBLEM
The width of the interval [a, b] is b – a.
So, the width of each of the n strips is:
b a
x
n
51. AREA PROBLEM
These strips divide the interval [a, b] into n
[x0, x1], [x1, x2], [x2, x3], . . . , [xn-1, xn]
where x0 = a and xn = b.
52. AREA PROBLEM
The right endpoints of the subintervals are:
x1 = a + ∆x,
x2 = a + 2 ∆x,
x3 = a + 3 ∆x,
.
.
.
53. AREA PROBLEM
Let’s approximate the i th
strip Si by
a rectangle with width ∆x
and height f(xi), which is
the value of f at the right
Then, the area of the i th
rectangle is f(xi)∆x.
54. AREA PROBLEM
What we think of
intuitively as the area of
is approximated by the
sum of the areas of these
rectangles: Rn = f(x1) ∆x
+ f(x2) ∆x + … + f(xn) ∆x
55. AREA PROBLEM
Here, we show this
approximation for
n = 2, 4, 8, and 12.
56. AREA PROBLEM
Notice that this
approximation appears to
become better and better
as the number of strips
that is, as n → ∞.
57. AREA PROBLEM
Therefore, we define
the area A of the region S
as follows.
58. AREA PROBLEM Definition 2
The area A of the region S that lies
under the graph of the continuous function f
is the limit of the sum of the areas of
approximating rectangles:
A lim Rn
n
lim[ f ( x1 )x f ( x2 )x ... f ( xn )x]
n
59. AREA PROBLEM
It can be proved that the limit in
Definition 2 always exists—since
we are assuming that f is continuous.
60. AREA PROBLEM Equation 3
It can also be shown that we get the same
value if we use left endpoints:
A lim Ln
n
lim[ f ( x0 )x f ( x1 )x ... f ( xn 1 )x]
n
61. SAMPLE POINTS
In fact, instead of using left endpoints or right
endpoints, we could take the height of the i th
rectangle to be the value of f at any number xi*
in the i th subinterval [xi - 1, xi].
We call the numbers xi*, x2*, . . ., xn*
the sample points.
62. AREA PROBLEM
The figure shows
approximating rectangles
when the sample points
are not chosen to be
63. AREA PROBLEM Equation 4
Thus, a more general expression for
the area of S is:
A lim[ f ( x1*) x f ( x2 *)x ... f ( xn *) x]
n
64. SIGMA NOTATION
We often use sigma notation to write sums
with many terms more compactly.
For instance,
n
f ( x )x f ( x )x f ( x )x ... f ( x )x
i 1
i 1 2 n
65. AREA PROBLEM
Hence, the expressions for area
in Equations 2, 3, and 4 can be written
as follows: n
A lim f ( xi )x
n
i 1
n
A lim f ( xi 1 )x
n
i 1
n
A lim f ( xi *) x
n
i 1
66. AREA PROBLEM
We can also rewrite Formula 1 in
the following way:
n
2 n(n 1)(2n 1)
i
i 1 6
67. AREA PROBLEM Example 3
Let A be the area of the region that lies under
the graph of f(x) = e-x between x = 0 and x = 2.
a. Using right endpoints, find an expression for A
as a limit. Do not evaluate the limit.
b. Estimate the area by taking the sample points
to be midpoints and using four subintervals
and then ten subintervals.
68. AREA PROBLEM Example 3 a
Since a = 0 and b = 2, the width of
a subinterval is:
2 0 2
x
n n
So, x1 = 2/n, x2 = 4/n, x3 = 6/n, xi = 2i/n,
xn = 2n/n.
69. AREA PROBLEM Example 3 a
The sum of the areas of the approximating
rectangles is:
Rn f ( x1 )x f ( x2 )x ... f ( xn )x
x1 x2 xn
e x e x ... e x
2/ n 2 4/ n 2 2n / n 2
e e ... e
n n n
70. AREA PROBLEM Example 3 a
According to Definition 2, the area is:
A lim Rn
n
2 2/ n 4/ n 6/ n 2n / n
lim (e e e ... e )
n n
Using sigma notation, we could write:
2 n 2i / n
A lim e
n n
i 1
71. AREA PROBLEM Example 3 a
It is difficult to evaluate this limit directly
by hand.
However, with the aid of a computer algebra
system (CAS), it isn’t hard.
In Section 5.3, we will be able to find A
more easily using a different method.
72. AREA PROBLEM Example 3 b
With n = 4, the subintervals of equal width
∆x = 0.5 are:
[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]
The midpoints of these subintervals are:
x1* = 0.25, x2* = 0.75, x3* = 1.25, x4* = 1.75
73. AREA PROBLEM Example 3 b
The sum of the areas
of the four rectangles
4
M 4 f ( xi *)x
i 1
f (0.25)x f (0.75)x f (1.25)x f (1.75)x
0.25 0.75 1.25 1.75
e (0.5) e (0.5) e (0.5) e (0.5)
0.25 0.75 1.25 1.75
(e
1
2 e e e ) 0.8557
74. AREA PROBLEM Example 3 b
With n = 10, the subintervals are:
[0, 0.2], [0.2, 0.4], . . . , [1.8, 2]
The midpoints are:
x1* = 0.1, x2* = 0.3, x3* = 0.5, …, x10* = 1.9
75. AREA PROBLEM Example 3 b
A M 10
f (0.1)x f (0.3)x f (0.5)x ... f (1.9) x
0.2(e 0.1 e 0.3 e 0.5 ... e 1.9 )
0.8632
76. AREA PROBLEM Example 3 b
From the figure, it
appears that
this estimate is better
than the estimate with n
= 4.
77. DISTANCE PROBLEM
Now, let’s consider the distance problem:
Find the distance traveled by an object during
a certain time period if the velocity of the
object is known at all times.
In a sense, this is the inverse problem of the velocity
problem that we discussed in Section 2.1
78. CONSTANT VELOCITY
If the velocity remains constant, then
the distance problem is easy to solve
by means of the formula
distance = velocity x time
79. VARYING VELOCITY
However, if the velocity varies,
it’s not so easy to find the distance
We investigate the problem in the following
example.
80. DISTANCE PROBLEM Example 4
Suppose the odometer on our car is
broken and we want to estimate the
distance driven over a 30-second time
81. DISTANCE PROBLEM Example 4
We take speedometer
every five seconds
and record them
in this table.
82. DISTANCE PROBLEM Example 4
In order to have the time and the velocity
in consistent units, let’s convert the velocity
(1 mi/h = 5280/3600 ft/s)
83. DISTANCE PROBLEM Example 4
During the first five seconds, the velocity
doesn’t change very much.
So, we can estimate the distance traveled during that
time by assuming that the velocity is constant.
84. DISTANCE PROBLEM Example 4
If we take the velocity during that time interval
to be the initial velocity (25 ft/s), then we
obtain the approximate distance traveled
during the first five seconds:
25 ft/s x 5 s = 125 ft
85. DISTANCE PROBLEM Example 4
Similarly, during the second time interval,
the velocity is approximately constant, and
we take it to be the velocity when t = 5 s.
So, our estimate for the distance traveled
from t = 5 s to t = 10 s is:
31 ft/s x 5 s = 155 ft
86. DISTANCE PROBLEM Example 4
If we add similar estimates for the other time
intervals, we obtain an estimate for the total
distance traveled:
(25 x 5) + (31 x 5) + (35 x 5)
+ (43 x 5) + (47 x 5) + (46 x 5)
= 1135 ft
87. DISTANCE PROBLEM Example 4
We could just as well have used the velocity
at the end of each time period instead of
the velocity at the beginning as our assumed
constant velocity.
Then, our estimate becomes:
(31 x 5) + (35 x 5) + (43 x 5)
+ (47 x 5) + (46 x 5) + (41 x 5)
= 1215 ft
88. DISTANCE PROBLEM Example 4
If we had wanted a more accurate
estimate, we could have taken velocity
readings every two seconds, or even
every second.
89. DISTANCE PROBLEM
Perhaps the calculations in Example 4
remind you of the sums we used earlier
to estimate areas.
90. DISTANCE PROBLEM
The similarity is
explained when we
a graph of the velocity
function of the car
and draw rectangles
whose heights are
the initial velocities for
each time interval.
91. DISTANCE PROBLEM
The area of the first
rectangle is 25 x 5 = 125,
which is also our
estimate for the distance
traveled in the first five
In fact, the area of each
rectangle can be
interpreted as a distance,
because the height
represents velocity and
the width represents time.
92. DISTANCE PROBLEM
The sum of the areas of
the rectangles is L6 =
1135, which is our initial
estimate for
the total distance
93. DISTANCE PROBLEM
In general, suppose an object moves
with velocity
v = f(t)
where a ≤ t ≤ b and f(t) ≥ 0.
So, the object always moves in the positive
direction.
94. DISTANCE PROBLEM
We take velocity readings at times
t0(= a), t1, t2, …., tn(= b)
so that the velocity is approximately constant
on each subinterval.
If these times are equally spaced, then
the time between consecutive readings is:
∆t = (b – a)/n
95. DISTANCE PROBLEM
During the first time interval, the velocity
is approximately f(t0).
Hence, the distance traveled is
approximately f(t0)∆t.
96. DISTANCE PROBLEM
Similarly, the distance traveled during
the second time interval is about f(t1)∆t
and the total distance traveled during
the time interval [a, b] is approximately
f (t0 )t f (t1 )t ... f (tn 1 ) t
n
f (ti 1 )t
i 1
97. DISTANCE PROBLEM
If we use the velocity at right endpoints
instead of left endpoints, our estimate for
the total distance becomes:
f (t1 ) t f (t2 )t ... f (tn )t
n
f (ti )t
i 1
98. DISTANCE PROBLEM
The more frequently we measure
the velocity, the more accurate our
estimates become.
99. DISTANCE PROBLEM Equation 5
So, it seems plausible that the exact distance
d traveled is the limit of such expressions:
n n
d lim f (ti 1 )t lim f (ti ) t
n n
i 1 i 1
We will see in Section 5.4 that this is indeed true.
100. Equation 5 has the same form as our
expressions for area in Equations 2 and 3.
So, it follows that the distance traveled
is equal to the area under the graph of
the velocity function.
101. In Chapters 6 and 8, we will see that other
quantities of interest in the natural and social
sciences can also be interpreted as the area
under a curve.
Examples include:
Work done by a variable force
Cardiac output of the heart
102. So, when we compute areas in this
chapter, bear in mind that they can
be interpreted in a variety of practical
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## Friday, January 14, 2011
### Open Problems
There are many problems in math that no one has been able to answer. A surprising number of these problems are easy to state, and even easy to play around with.Thinking about these problems can give a student more of an idea of what math really is.
I'd like to collect problems here that would be accessible to high school and community college students. James Tanton works with students on open problems, and they've gotten results. I think that requires finding little known problems. The problems I'm listing below have been worked on long enough that any new results from us amateurs are unlikely. But we can still have fun playing.
The word conjecture means something that people believe is likely to be true, but which has not been proved.
1. Goldbach's Conjecture: Every even number greater than 2 can be expressed as the sum of two primes.
2. Twin Prime Conjecture: There are an infinite number of twin primes (n and n+2).
3. Collatz Conjecture: Play this game: Start with a positive whole number, n1. If n1 is even divide by 2, if n1 is odd multiply by 3 and add 1. You now have n2. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1, which gives you a loop of 1, 4, 2, 1, ...
4. The Integer Brick Problem, to find a brick size in which the 3 lengths along the sides, the 3 diagonals along the faces, and the interior diagonal are all integers, or to prove that this cannot be done.
5. Is every prime larger than 3 the average of two primes? (from James Tanton)
6. A perfect number is a number that equals the sum of all its factors (including 1, but not including itself). Are there any odd perfect numbers? (from Wikipedia's list of open problems in mathematics)
7. The Lonely runner conjecture: if k + 1 runners with pairwise distinct speeds run round a track of unit length, will every runner be "lonely" (that is, be more than a distance 1 / (k + 1) from each other runner) at some time? (same source as #6)
8. Towers of Hanoi with 4 pegs: There is a stack of n disks on a peg, with the disks decreasing in size as you move up the stack. If you are allowed to move one disk at a time from one peg to another, and cannot put a bigger disk over a smaller one, what is the least number of moves it takes to move the stack. (seen twice recently, once here)
9. The Goormaghtigh conjecture: The only numbers that can be represented as 11...1 in more than one base are 31 and 8191. 31 can be represented in base 2 as 11111 and in base 5 as 111. 8191 can be represented in base 2 as 1111111111111 and in base 90 as 111. (We don't look 11 (two digits) in this problem, as any number n will be represented as 11 in the base n-1, making it easy to find numbers with two different representations that are all 1's. For example, 13 is 11 in base 12 and is 111 in base 3.) (thanks to Mary O'Keeffe)
10. Legendre's conjecture: There is a prime number between n2 and (n + 1)2 for every positive integer n. (mentioned in Bob and Ellen Kaplan's book, Hidden Harmonies)
Here's a good question: Which two problems on this list go together? If one were solved the other would be too. Are the two equivalent, or could one of them be solved and still leave the other open? (thanks to Christopher Sears)
What other open problems (simple to state and to play around with) can you add to this list?
[Edit date: 1/15]
1. Trisecting an angle (but I did see an interesting "hatchet proof" last summer)
2. I'm not sure what you mean by 'hatchet proof'.
This is not an open question. Trisecting the angle by standard Euclidean methods (straightedge and compass) has long been known to be impossible. (The Wikipedia article says it was proved in 1837.)
Trisecting the angle given a different set of operating instructions is possible. For instance, the origami axioms will do it. (I did this as a participant in a math circle, and loved it.)
3. Problem #5 is a special case of Goldbach's. I think it may be equivalent. I have a weekend's worth of induction ahead of me.
4. The Goormaghtigh conjecture is neat and simple to state in a way that students can understand it.
A repunit is a number composed entirely of ones.
The decimal number 31 can be expressed as a repunit in base 2 as 11111 and as a repunit in base 5 as 111.
Similarly, the decimal number 8,191 can be expressed as a repunit 1111111111111 in base 2 or as a repunit in base 90 as 111.
Strangely, enough, mathematicians have so far been unable to find any other numbers besides 31 and 8,191 that can be expressed as nontrivial repunits in more than one base!
The Goormaghtigh conjecture can be stated in multiple ways, but the simplest and most accessible statement is that 31 and 8,191 are the only such numbers.
(By non-trivial, I mean that the repunit representation must be at least three or more digits long. All numbers n can be stated as a two-digit repunit in base n-1, so that doesn't count.)
5. @Christopher, Thanks! I hadn't noticed. #5 Just deals with primes (or primes times two if you multiply both sides of the equation by 2), and Goldbach's deals with all even numbers, so I can imagine #5 being true and Goldbach's being false. Which to me means they're not exactly equivalent. Perhaps I'm missing something, though.
@Mary, Thanks! What a weird one... I see that both numbers are one less than a power of two (2^5-1 and 2^13-1). I wonder if anyone has proved that, if there are any others, they must also be 2^n-1.
6. Sorry, that was a silly comment. That would amount to proving that any answer has to have one of its bases be 2, not particularly deep most likely.
I'm including something about the relationship Christopher pointed out and the problem Mary pointed out in the post now. I'll keep editing the list as I learn of more problems.
7. I thought there was a way to start with problem #5 and fiddle with prime factorizations to show that Goldbach's Conjecture is equivalent. I played around with it this weekend, and I'm pretty sure that my line of reasoning won't work out.
If somebody was able to prove #5, that would lay a foundation for a proof of Goldbach's Conjecture.
8. What is the minimum number of colors needed to color the real plane such that no points 1 unit away from each other have the same color?
:) I'm at that conference we went to last year together, and a link to your blog was up on the screen for a good 30 minutes!
9. Wow, did they say why they had the link up? Was it one among many? I wish I were posting more lately, but after teaching all week, I'm all talked out, I guess.
So is that an open problem?
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GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video
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When quoting this document, please refer to the following
DOI: 10.4230/LIPIcs.ESA.2019.62
URN: urn:nbn:de:0030-drops-111831
URL: https://drops.dagstuhl.de/opus/volltexte/2019/11183/
Go to the corresponding LIPIcs Volume Portal
### Closing the Gap for Pseudo-Polynomial Strip Packing
pdf-format:
### Abstract
Two-dimensional packing problems are a fundamental class of optimization problems and Strip Packing is one of the most natural and famous among them. Indeed it can be defined in just one sentence: Given a set of rectangular axis parallel items and a strip with bounded width and infinite height, the objective is to find a packing of the items into the strip minimizing the packing height. We speak of pseudo-polynomial Strip Packing if we consider algorithms with pseudo-polynomial running time with respect to the width of the strip. It is known that there is no pseudo-polynomial time algorithm for Strip Packing with a ratio better than 5/4 unless P = NP. The best algorithm so far has a ratio of 4/3 + epsilon. In this paper, we close the gap between inapproximability result and currently known algorithms by presenting an algorithm with approximation ratio 5/4 + epsilon. The algorithm relies on a new structural result which is the main accomplishment of this paper. It states that each optimal solution can be transformed with bounded loss in the objective such that it has one of a polynomial number of different forms thus making the problem tractable by standard techniques, i.e., dynamic programming. To show the conceptual strength of the approach, we extend our result to other problems as well, e.g., Strip Packing with 90 degree rotations and Contiguous Moldable Task Scheduling, and present algorithms with approximation ratio 5/4 + epsilon for these problems as well.
### BibTeX - Entry
```@InProceedings{jansen_et_al:LIPIcs:2019:11183,
author = {Klaus Jansen and Malin Rau},
title = {{Closing the Gap for Pseudo-Polynomial Strip Packing}},
booktitle = {27th Annual European Symposium on Algorithms (ESA 2019)},
pages = {62:1--62:14},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-124-5},
ISSN = {1868-8969},
year = {2019},
volume = {144},
editor = {Michael A. Bender and Ola Svensson and Grzegorz Herman},
publisher = {Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik},
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# Could the end cap of the Pascal B (1) survive its trip through the atmosphere?
In 1957, the US had a nuclear test where a shaft was dug 152 meters into the ground. A 100 mm thick, 900 kg steel plug was installed and welded at the top of the shaft. Under it was 2 feet of concrete.
The bomb was detonated created a blast of less than 1 kt. The concrete was vaporized and the steel plug shot into the air at approximately 66 km / second (125,000 - 150,000 mph).
Some believe it became the first man made item to reach orbit, after all, all you need is approximately 7 km/sec to reach orbit, and this greatly exceeded that.
Others believe it was vaporized by the intense pressure of traveling though the atmosphere at this speed.
No one ever reported it coming down anywhere...
Is there a physics / mathematical way to determine what happened to this steel plug?
• It certainly didn't go into a closed orbit around the Earth as escape speed is only about 11 km/s. – dmckee Jun 25 at 21:13
• would you mean then that if it did make it into space, it would be traveling more or less in a straight line? I guess I am looking to see if it could have made it through the atmosphere or would the pressure have destroyed it. – Rick Jun 25 at 21:19
• The Test Site is big, and there is lots of nothing all around. If it came back down it could easily not be found. – Jon Custer Jun 25 at 21:41
• @JonCuster - at the speeds it went up, it would not have come down in the test zone. But my question still stands, could it have survived a trip through the atmosphere at the speeds it went up? – Rick Jun 25 at 23:30
• Let’s take the inverse - which meteoroids make it to the surface of the earth? science.howstuffworks.com/question486.htm - looks like it may have made it! – Paul Young Jul 3 at 2:31
I used the formula for meteoroid mass loss rate from https://www.spaceacademy.net.au/watch/debris/metflite.htm : $$dm/dt = (\Lambda A \rho_a v^3 m^{2/3} ) / ( 2 \zeta \rho_m^{2/3} )$$.
See below the legend and the values I chose (SI units, compare with the values at the link):
$$dm/dt$$ - mass loss rate
$$\Lambda$$=2 - heat transfer coefficient
$$A$$=1 - meteoroid shape factor
$$\rho_a$$=1.25 - atmospheric density
$$v$$=66000 - meteoroid speed
$$m$$=900 - meteoroid mass
$$\zeta=3\cdot 10^6$$ - heat of ablation of the meteoroid
$$\rho_m$$=7800 - meteoroid density.
I obtained $$dm/dt\approx 2.8\cdot 10^7$$. Thus, the meteoroid will lose mass comparable to its initial mass in $$\frac{m}{dm/dt}\approx 3\cdot 10^{-5} s$$. The meteoroid will travel $$v\frac{m}{dm/dt}\approx 2 m$$ within this time. Thus, the end cap probably did not survive in the atmosphere.
The results of the calculation seem strange. It is possible that the formula in the link is incorrect, or maybe I made some mistake, so take this just as an initial attempt to make an estimate.
• The mass loss rate depends on m (and v), so you can't just divide m by it to get the lifetime. Also, $\rho_a$ varies significantly with height and so also t. – Ryan Thorngren Jul 3 at 6:42
• @RyanThorngren : I was just making an estimate, so I estimated the time for the meteoroid to lose mass comparable to its initial mass, I did not mean that the meteoroid would lose all its mass over this time. Same for the atmospheric density - it does not change significantly over 2 meters. At this point I don't think numerically solving the differential equation is warranted, as we don't really know the precise values of the parameters. – akhmeteli Jul 3 at 6:51
• It took me a second to figure out that $2m$ meant "two meters" and not "twice the mass". (This is why it's better to write units in upright font and variables in italics.) – Michael Seifert Jul 3 at 7:52
• @MichaelSeifert : Thank you for the advice. – akhmeteli Jul 3 at 8:27
Building off of akhmeteli's excellent answer, I implemented the differential equations from the asteroid webpage in Mathematica. I then tried to tweak the numbers, within realistic bounds, to get the thing into space. In no realistic case was I able to get the thing more than a few hundred meters up before it completely burned away.
To maximize the distance travelled, we want $$\Lambda$$ and $$A$$ to be as small as possible; respectively, these correspond to the rate at which heat is transferred to the "asteroid" and the effective cross-sectional area of the object (taking into account turbulence). In addition, we want the heat of ablation $$\zeta$$ (the amount of heat require to vaporize a certain mass of the substance) to be relatively high, since this will reduce the rate at which mass is lost.
The parameter $$\Gamma$$ also has an effect; it describes the amount of drag experienced with the atmosphere. Interestingly, one can actually get the projectile higher by increasing the drag: a higher drag means the projectile slows down faster, but that means that the projectile can slow down enough enough that it doesn't burn up immediately.
My optimistic estimates are $$\Lambda \approx 0.15$$ (note that this number is used in the code example on the page) and $$A = 1$$ (which would be more streamlined than a sphere). I also used $$\zeta = 10\times 10^6$$ J/kg, since it was the highest "typical" value in the table.1 Finally, I used $$\Gamma = 0.5$$, an estimate given on that webpage for the lower atmosphere.
Here's the result of the simulation, with the parameters given above. The vaporization of the plug is complete at a height of 312 meters.
And here is the simulation for akhmeteli's parameters, with $$\Gamma = 0.5$$. The plug does not significantly change its velocity before it burns up; the final height is a little over 6 meters. As would be expected, this is within an order of magnitude of akhmeteli's back-of-the-envelope estimate.
If you tweak the unknown parameters of my "optimistic" case above, you can attain a height of 1 km if:
• $$\Lambda \approx 0.064$$ (more than twice as small)
• $$A \approx 0.031$$ (more than three times smaller)
• $$\zeta \approx 23.5 \times 10^6$$ J/kg (over twice as large.)
• $$\Gamma \approx 1.6$$ (much more drag—this slows it down sufficiently before too much of it burns away)
All in all, it seems unlikely that the plug got anywhere near space.
### Mathematica Code:
Feel free to tweak this code as you see fit. The code stops integration when either the mass of the steel falls below 1 gram, or the speed falls below 1 m/s. The code does implement a height-dependent atmospheric density via a simple exponential model, though it turns out not to be all that relevant for realistic parameters. The acceleration due to gravity is assumed to be constant.
Needs["DifferentialEquationsInterpolatingFunctionAnatomy"];
Λ = 0.15;(*heat transfer *)
A = 1;(*shape factor *)
Γ = 0.5; (*drag coefficient*)
ρa0 = 1.25 ;(*atmo. density *)
v0 = 66000; (*initial velocity*)
m0 = 900 ;(*initial mass*)
ζ = 7*10^6;(*heat of ablation*)
ρm = 7800;(*steel density*)
h = 7000; (*atmospheric "height"*)
a = Γ A ρa0 / ρm^(2/3);
b = Λ A ρa0/(2 ζ ρm^(2/3));
soln = NDSolve[{x''[t] == - a Exp[-x[t]/h] x'[t]^2/m[t]^(1/3) - 9.8 m[t],
m'[t] == -b Exp[-x[t]/h] x'[t]^3 m[t]^(2/3), x[0] == 0,
x'[0] == v0, m[0] == m0,
WhenEvent[{m[t] < 0.001, x'[t] < 1}, "StopIntegration"]},
{x, m}, {t, 0, 1000}]
{ti, tf} = First[InterpolatingFunctionDomain[x /. First[soln]]]
Plot[x'[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 66000},
AxesLabel -> {"Time (s)", "Velocity (m/s)"}]
Plot[m[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 900},
AxesLabel -> {"Time (s)", "Mass (kg)"}]
x[tf] /. First[soln]
1 It is not clear to me whether these are the appropriate units for $$\zeta$$; the page is unclear. They're dimensionally correct, though.
• So a 900kg steel plug, traveling at 125,000 mph only made it 6 meters? Would the blast pressure wave from below not assisted the plug by distorting / overcoming the resistance of the air pressure impacting the plug from above? – Rick Jul 4 at 12:48
The drag equation stands that the an object travelling through the air receives a force of d*v^2*a*q, if the aerodynamic coefficient (q) was 1 and the air density 1.2 then the cap would have received a pressure of 4961 Megapascals, steel can't stand that pressure.
Also the energy dissipated in 0.0001 seconds given by the same equation would be 31 gigajoules and it is required 0.61 gigajoules to melt 900 kg of steel so the only way the cap will survive the first 6 meters is if the 98% of the energy was dissipated to the air, which is hard because of the black body radiation at about 150000 kelvin (1000 J/(kg*k) * 6^3 kg (air around the cap) /31e+9 J).
Black body heat transfer in 0.0001 s would be 2 GJ .
• Does the fact that the steel will be surrounded by a shock layer change the pressure it will experience? If so, how? – Michael Seifert Jul 4 at 10:20
• the drag equation is based on dynamic pressure, which works for incompressible fluids, for making it work in air, the dynamic pressure needs to be multiplied, it is observed that at high mach numbers the drag coefficient is more stable and approaches a number. researchgate.net/figure/… – Sartem Cacartem Jul 4 at 15:24
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# Resistance, Capacitance, Voltage And Time Calculator
This calculator requires the use of Javascript enabled and capable browsers. This calculator is designed to calculate any one value in the group of Voltage, Capacitance, Resistance, Time of charge, and Instant Voltage. In the schematic rendering, the time required for the capacitor to charge to 63.2% of the battery voltage after the switch is closed is the product of the resistance and capacitance T=(R*C). For example, a rather common circuit is in which a 100 uF capacitor and a 100K resistor would require 10 seconds to charge to 7.6 volts using a 12 volt battery. Our calculator can be used to determine the capacitor voltage at other times after the switch in the circuit is closed (turned on). Any of the five variables can be determined by supplying the other four. For example, in the calculator defaults, a 1 Farad capacitor charging through a 1 ohm resistor and 1 volt battery will reach about 777 millivolts in 1.5 seconds. From left to right, the entries would be 1, .001, 1000000, 1500 with the Instantaneous Voltage field left blank, or zero. The result should be about 0.777. Only one field should be blank or a 0 value before each calculation. The general formula for these types of calculations is:
Vc = V * (1- e^(-t / R*C))
Where Vc is the capacitor voltage, V is the supply voltage from battery or other source, R is the resistance, C is the Capacitance, t is the time, and e = 2.71828 which is a constant in our calculator.
Designation Value Required Data - Any Four Values Supply Voltage Volts Resistance KOhms Capacitance Microfarads Time Milliseconds Instantaneous Voltage Volts
Version 1.4.3
Leave us a question or comment on Facebook Search or Browse Our Site
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Find answers, ask questions, and connect with our
community around the world.
Activity Discussion Math Mathematics
• # Mathematics
Posted by on December 7, 2023 at 4:35 pm
How do you think Mathematics play a role in our education other than academics ?
replied 3 months, 2 weeks ago 2 Members · 1 Reply
• ### Ayushi
Member
January 5, 2024 at 12:43 pm
0
Mathematics plays a significant role in our education beyond academics. Here are a few ways in which mathematics is relevant in various aspects of our lives:
1. Problem-solving skills: Mathematics teaches us how to approach problems logically, analyze information, and find solutions. These problem-solving skills are valuable in many areas of life, such as decision-making, critical thinking, and everyday challenges.
2. Quantitative reasoning: Mathematics develops our ability to reason quantitatively, enabling us to understand and interpret numerical information. This skill is essential in fields like finance, economics, statistics, and data analysis.
3. Financial literacy: Basic mathematical knowledge is crucial for managing personal finances, budgeting, understanding interest rates, calculating discounts, and making informed financial decisions. It helps individuals develop financial literacy and become more responsible with money.
4. Logical reasoning and deduction: Mathematics trains our minds to think logically, make deductions, and draw conclusions based on evidence and patterns. These skills are transferable to various domains, including problem-solving, scientific research, computer programming, and legal reasoning.
5. Analytical skills: Mathematics fosters analytical thinking by encouraging us to break down complex problems into smaller, more manageable components. Analytical skills are valuable in areas like engineering, computer science, business analysis, and research.
6. Pattern recognition: Mathematics helps us recognize patterns and relationships in data, which is relevant in fields such as data analysis, market research, and scientific discovery. Identifying patterns can lead to insights and advancements across various disciplines.
7. Technology and innovation: Mathematics is the foundation of many technological advancements and innovations. It underpins fields like computer science, artificial intelligence, cryptography, and data science. A strong mathematical background is often necessary to contribute to these rapidly evolving areas.
8. Everyday life and decision-making: Mathematics is present in our everyday lives, from calculating expenses, measuring ingredients for a recipe, and understanding probabilities and statistics, to interpreting charts and graphs in news articles. Mathematical literacy helps us make informed decisions and navigate the quantitative aspects of our world.
Start of Discussion
0 of 0 replies June 2018
Now
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Browse Prior Art Database
# Method of Generating Thermometer Code Output for Compare Operation Using Special Decoder
IP.com Disclosure Number: IPCOM000249539D
Publication Date: 2017-Mar-02
Document File: 2 page(s) / 18K
## Publishing Venue
The IP.com Prior Art Database
## Abstract
Described is a method of generating thermometer code output for compare operation using special decoder.
This text was extracted from a PDF file.
This is the abbreviated version, containing approximately 74% of the total text.
1
Method of Generating Thermometer Code Output for Compare Operation Using Special Decoder
This invention performs a 2^n compares between a n-bit binary input data value against 2^n
possible binary indexes and produces a 2^n-bit thermometer code that has values of 1 for all
indexes that are less than the n-bit binary input data value. The current invention uses a special
n-bit decoder that uses the n-bit binary input data value to produce a 2^n-bit output such that the
decoded value is 1 if the n-bit binary input data value is greater then the index of the
corresponding 2^n-bit decoder value.
Instead of implementing 2^n comparators, a n-bit special decoder circuit is implemented.
For each of the 2^n outputs of the decoder, the logic is such that the value of the output is
dependent upon the current index is strictly less than the n-bit input. So this is a circuit that is
kind of a decoder and comparator combined together to perform a special function. The circuit is
faster, smaller, and uses less power than prior art. The following logic truth table summarizes
the circuit for n=5-bits
Input = In<4:0>
Output=LT<31:0>
In<4> In<3> In<2> In<1> In<0> LT<0> LT<1> LT<2> LT<3> LT<4> ... LT<31>
0 0 0 0 0 0 0 0 0 0 ... 0
0 0 0 0 1 1 0 0 0 0 ... 0
0 0 0 1 0 1 1 0 0 0 ... 0
0 0 0 1 1 1 1 1 0 0 ... 0
0 0 1 0 0 1 1 1 1 0 ... 0
0 0 1 0 1 1 1 1 1 1 ... 0
0 0 1 1 0 1 1 1 1 1 ... 0
0 0 1 1 1 1 1 1 1 1 ... 0
0 1 0 0 0 1 1 1 1 1 ... 0
0 1 0 0 1 1 1 1 1 1 ... 0
0 1 0 1 0 1 1 1 1 1 ... 0
0 1 0 1...
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# Riddle #3284
Question: Use the numbers 1, 2, 4, 5, 6, and 8 and the symbols (x) and (=) to form a true equation. Note: Each number and symbol is to only be used once and no extra number or symbol is included.
Riddle Discussion
### Similar Riddles
##### When is 99 more than 100 (hard)
Question: When is 99 more than 100?
##### Riddle #134 (hard)
Question: You have a glass of water that looks about half full. How can you tell, only using the glass of water itself, if the glass is half full or not?
The glass is a right cylinder.
##### Riddle #270 (easy)
Question: Mr. Smith has 4 daughters. Each of his daughters has a brother.
How many children does Mr. Smith have?
##### Riddle #826 (medium)
Question: Jasmine has a toaster with two slots that toasts one side of each piece of bread at a time, and it takes one minute to do so.
If she wants to make 3 pieces of toast, what is the least amount of time she needs to toast them on both sides?
##### Riddle #170 (medium)
Question: How can you add 8 8's to make 1000?
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# Displaying Percentage Values Over 100%
Feb 4, 2008
I'm trying to make my spreadsheet display over 100% while calculating a long column of entries. Each entry is showing percentage cost per hour of a benchmark of \$65.00 per hour. When an entry for example is \$51.10 the percentage displays 0.79%, but if the entry is for example \$73.89 then the displayed value is 1.14%.
How do I make this display the percentage over the benchmark of 65 as 110, 115 or whatever it calculates out?
## Excel 2007 :: Displaying Zero Percentage In Single Cell
Jun 25, 2012
My Excel 2007 worksheet contains a cell where a percentage is manually input. A freight cost is calculated based on the input percentage. Typically, the percentage is 3-7% but once in a while freight is excluded and the percentage is zero.
My issue is that when a 0 is input the cell appears blank and I would like it to display 0.00%. The remainder of the worksheet needs to have the zero display turned off.
## 0 Values Are Not Displaying
May 31, 2012
I have two columns in my spreadsheet, and both are dates. Using a calculation (one date minus the other) I get the duration (in days) between the two.
I also apply color coding to say if the duration is > X, then color code as red, etc.
All the color coding is working fine for values that are either > or < 0, but anything where the two dates are the same (date 1 - date 2 = 0), the value shows as blank.
When I click on the cell, the value "0" is actually IN the cell, but it's now showing, as if the font was colored white or something, but it hasn't.
I don't have anything in the code telling the worksheet not to display "0" values, so I don't know where to go...
## Values Of Column Displaying In Rows
Mar 30, 2009
I have some values in the column and i want those values to be displayed in rows now. Example:
In the Excel sheet the data is in the form,
Column1 Column2 Column 3
I want in the format,
Column1
NE_TYPE
## Displaying Values In An Entire Column In Another Worksheet
Jun 26, 2008
Is there a formula/function in excel that can take all the cells in a column, lets say Column A, and paste/display it in another file without the spaces between the cells with values and without duplicates?
Worksheet1
Column A
1 Name
2
3 Mike
4 Rob
5 Ryan
6
7 Mindy
8 Paul
9
10 Rob
11
12 Mindy
13 Chris
Worksheet2
Column A
1 Name
2 Mike
3 Rob
4 Ryan
5 Mindy
6 Paul
7 Chris
8
9
10
11
12
13
## Displaying Multiple Values Without Using Macros Or ROWS Function
Mar 13, 2009
is to display a set of data based on filtered information.
My Data base:
Company Department Name
A X John
A Y Joe
A X Jane
B Y Bob
C Z Kate
A X Kerri
Based on user selection of Company and Department, I want to be able to display the relavent names.
If user chose Company A, and Department X, I want to be able to display
John
Jane
Kerri
I've used the ROWS, Index, Small combination that works perfectly (Please see sample below). However, since the this software doesn't support the ROWS function, and doesn't support Macros.
## Displaying Cell Values With Worksheet_Change By Value If Target Is In Certain Range
Nov 15, 2008
I have a spreadhseet where columns I and J (range from I6 to J300) serve as input cells, off to the right, 23 columns over in AF and AG respectively I have a hidden array formula (Index, match) calculating values based on input in either column I or J and several factors embedded in reference table in the same sheet. That works fine. I want cells in columns I and J to be interdependent, in other words, input in column I drives calculations in a hidden formula and I want the value of that calculaton to display in column J (in a adjacent cell input in I6 results in display in J6), but if I input value in J then this value will drive calculation in a hidden formula and display in I (let's say I is centimeters and J is inches). I have a code that works (I set it up as a try just for few rows) but only one code section at a time, not together. If I choose column I (#9) to go first in code, values update in J, but not the other way around, if I choose column J (#10) to go first in code, values update in I, but not the other way around. What am I doing wrong, I tried Target.address case, I tried Intersect ... is nothing then etc. They all work one at a time but not together. Here is the code as it stands now
## Excel 2003 :: Pivot Chart Displaying Added Values?
May 31, 2012
I'm using Windows XP with MS Excel 2003. I have a pivot table representing a survey. Let's say I've built the survey outside of excel and I've imported the response data into Excel. One of the questions in the survey is "ratings" and the possible valid responses for it is: "Excellent", "Good", or "Poor". In my data set in excel let's say I have 10 responses or rows and all the responses for the question on ratings are either "Excellent" or "Good". (There are no rows with a "Poor" value in the ratings column).
For example, let's say out of the 10 responses, 6 are "Excellent" and 4 are "Good". As such my Pivot chart shows two bars: one for the number of respones with "Excellent" (10) and another bar for the number of responses with "Good" (4). My delima is how to show a third bar showing "Poor" with a zero as the number of responses.
## Coloring Percentage Values
Oct 7, 2008
I need to coloring percentage values in excel 2000. What I would like to do is have percentages greater than zero shown in green, while negative values are red, and zero values are black. I currently have the custom formatting so negative values are red and all others are black, but I cant' seem to find a way to change it so greater than zero numbers are green and zeros are black. All I can do is chang positive and zero at the same time. Here is what I currently have:
## Percentage Difference Between Two Values
Sep 15, 2009
I am trying to create a formula that can work out the % difference between two values. The formula I am using is:
=(B1-A1)/ABS(A1)
So, where
A1=1, B1=2, = 100%
A1=-1, B1=2, = 300%
This seems to work perfectly, apart from where there are zero numbers involved. The answers I want to have are as follows:
A1=0, B1=0, = 0%
A1=0, B1=-1, = -100%
A1=0, B1=1, = 100%
A1=-1, B1=0, = 100%
A1=1, B1=0, = -100%
## Exclusion Of Values In SUM/Percentage
Jun 19, 2009
I am trying to exclude a certain value from an array of numbers to get a percentage. I have data that populates on its own and in the same columns regularly. The data contains a country name in column A, a year in columnB and a value in column C. I am trying to exclude the value for the U.S. in the totals so that I can get a percentage of values outside of the U.S.
So that the sum of all of the rows less the value corresponding to the U.S., divided by the sum of all of the values.
The problem is that the location of the U.S. value and U.S. designation varies with different inputs (this cannot be helped), so the formula must be able to follow it.
I have been able to do this but not without a lot of extra cells and rows, and I would like to have a single formula.
## Increase Values By X Percentage
Jan 16, 2008
How do I change numerous numbers in several different columns by a constant amount? example reduce each number in each of these columns by 50%.
## Changing Percentage Values Automatically
Dec 2, 2013
I'm designing a spreadsheet currently. I want to input a number into cell A1 - for example 13. Now the number 13 has a percentage value of 67.3%. When I enter the value into A1 I want excel to automatically fill cell A5 with the percentage value of 67.3%.
I need to have the numbers from 1-20 all with varying percentage values.
1 = 100%
2 = 97.8%
3 = 94.3% etc..
So essentially I would like to enter 13 into cell A1 and for cell A5 to automatically fill in the value of 67.3%, if I was to change the value in A1 to 10 it would also automatically change the percentage from 67.3% to 74.7%..
## Conditional Formatting For Percentage Values
Jan 26, 2014
Using conditional formatting to make all the cells that contain numbers and are formatted to percentage format make display in red color? let's say I want to set this conditional formatting rule initially after just opening the excel book and afterwards when I type in values/text etc. into the cells I want the cells where I have input numbers and chose the percentage format to come out in red color? and all the rest stay in black color?
## Deleting Rows With Non-percentage Values
Jun 9, 2009
What I would like to do is to delete the rows in the attached Excel file highlighted in yellow (whole numbers) and to keep the rows with percentage values. Ideally I would like to have only the rows with the string "Group:" remaining along with the rows that have percentage values in them.
What I'm going to be doing is then transferring these groupings of data to another excel sheet in order to create a time series of performance by "team leads". If anybody would be willing to help me out with the first part of this I would really appreciate it. I have some idea of how to go about it using conditional logic, but I'm just not comfortable enough in VBA to do this efficiently. It would probably take me hours just to get a simple version working....
## Finding Percentage Of A Range Of Values
Feb 13, 2009
I have a little table:
Columns A:B (separated by commas)
Row 1- Year One, 20
Row 2- Year One, 20
Row 3- Year One, 10
Row 4- Year One, 30
Row 5- Year One, 20
Row 6- Year One, 20
Row 7- Year Two, 10
Row 8- Year Two, 20
Row 9- Year Two, 20
Row 10- Year Two, 10
Row 11- Year Two, 20
Row 12- Year Two, 40
In the adjacent cells in columns C (C1:C12) I need a formula that would turn the value into a percentage of the total of the values for that year (e.g. in C1: 20 is 16.7% of 120, so the value displayed would be 16.7% (or 0.167)).
## One Value With Multiple Values Within - Need To Calculate As Percentage
Jan 16, 2014
I have a list of group ID codes, which contain 1 or more product codes within them.
Some product codes contain an "alternative" value (the actual value is irrelevant) and others do not. See example below
Group ID
Product Code
Alternative
56381
240027
160380
[Code] .....
Ultimately what I need to achieve is a percentage of how many product codes, within each group ID, contain an alternative.
So for the example above, group ID 56381 would have 50% codes with alternatives.
By counting the occurrences of a group ID and whether 1 particular code has an alternative I have calculated individual percentages against the size of the group e.g. product code 240027 (given a value of 1) divided by the total number of codes in the group (4), returns 0.25. Obviously doing this across the whole group would give my result (0.5 or 50%)
However some groups contain over 100 codes and the spreadsheet is 40K rows!
## Displaying Values Used In Criteria Selection (Date Ranges) As Part Of Query Output
Dec 20, 2011
Using MS Query in Excel, I've created a simple query that pulls its records from an SQL dbase. Here's the statement:
SELECT uvVisit.FacilityListName, uvVisit.DoctorListName, uvVisit.Date, uvVisit.PatientVisitId, uvVisit.PatientLast, uvVisit.PatientFirst
FROM CPS.dbo.uvVisit uvVisit
WHERE (uvVisit.Date Between ? And ?)
ORDER BY uvVisit.FacilityListName
The query runs fine and prompts the user to enter beginning & ending date ranges for the visit date when executed. So far...so good...but, this requires me to manually insert a line in Excel above the 1st record and type in something like: "For Date Range: MM/DD/YYY - MM/DD/YY" to denote the date range that the qualified records fall into (something the user wants to see).
However, I'd like to find a way automatically preface and display in the report's output (perhaps as the 1st line of the report in Excel??) something similar to what I'm already typing, and have it pull the beginning and ending MM/DD/YY values from those supplied by the user in the parameter.
## Add 2 Cells Show As Percentage Of 3rd Cell But Still Allow For 0 Values
May 23, 2013
I am trying to add 2 cell values together then show the total as a % of a value in a 3rd cell, however I also need it to allow for 0 values in the chosen cells without displaying an error message or it messes up the average formula elsewhere on the sheet?
## Adjusting Percentage Values On Cell Change
Jan 8, 2009
Good afternoon Gentlemen, I have a column of data, with a varying number of percentage values that add up to 100%, separated by "NA", i.e.
33%
33%
33%
NA
25%
25%
25%
25%
NA
100%
NA
20%
20%
20%
20%
20%
NA
Now... when I change one of the values I would like the others to even up, i.e. in the last example if I change a 20% to 50% I would like the others to change to 10%... any ideas?
## Take Cell Values From Six Worksheets And Adds Them Up As A Percentage
Jan 27, 2010
How would I write a formula that takes cell values from six worksheets and adds them up as a percentage (quarterly updates)? I realize this is wrong, but here's what I'd tried (for three sheets).
=+SUM(IF(EVH!F7="Y",25,0)+IF(FUN!F7="Y",25,0)+IF(HES!F7="Y",25,0))/3
## Formula To Determine Required Values To Achieve Target Percentage?
Aug 24, 2014
Data is;
83300 - hypothetical number of times I have fired my gun at target.
43209 - hypothetical number of times I have scored bullseye.
So, my bullseye percentage = 43209/83300, or ~51.87%.
Need formula to determine how many more consecutive bullseyes I need to shoot, in order to achieve 70% ratio.
Since each shot from now on will be a bullseye, both values (hits & shots) will increment together.
## Show Values As Percentage Difference - Quickly Change From Month To Week
Dec 18, 2013
I have a Pivot Table with fields for months and weeks. I also have a "Show Values as % Difference Field" that shows monthly or weekly % change. When I collapse the fields so that it goes from weekly to monthly (or vice versa), I have to manually change each Show Values As % Difference column. Is there a way to do this automatically or quickly?
## Calculating Percentage In Cell And Omitting Values When Specific Words Present
Apr 5, 2014
I have a daily report that gives a percentage based on what is typed into the "Supplier" and "Total Sales" columns but I'm having trouble writing the following formulas since my excel knowledge is relatively low.
1. Where a cell in the "Supplier" column says anything but "Den", "Burrowed" or "Studio Nyx" I need the "60%" and "40%" columns filled with the relevant formula (=B11*60% for example) and the "100%" column blank.
2. Additionally, if the cell contains "Den", "Burrowed" or "Studio Nyx" I need the "100%" column filled and the "60%" and "40%" blank.
## Return Range Of Numerical Values In Single Column Based On Frequency Percentage
Oct 25, 2005
I have a single column of numerical values that may repeat many times within
the column.
I require a flexible Formula:
Use an Input Cell for the specified and changeable Percentage(s) %.
Column may be filtered – so only take into account Visible Filtered Cells.
The Formula will calculate and Return the numerical range of values that fall
between the specified percentage % (using the Input Cell) e.g.; 70%.
The Formula should Return two numeric values: a Start Value and an End Value
– NOT necessarily the minimum and maximum per se BUT the MIN and MAX of the
values that appear 70% of the time in the column; therefore, taking into
account Repeat / Duplicate values.
The calculated Results: the two numeric values will be returned to separate
cells on a new Sheet.
## VBA - Displaying Values In Multiple Lines If Multiple Checkbox's Are Selected
Sep 10, 2012
In my userform I have a list of check box's that can be selected. Currently if more than one is selected, they will appear in the spread sheet in the same line one after another with a space between them. How do I make it so they either appear with a comma appearing after each, so the next value appears on the next line below or most perferably the cell turning to a drop down list with the values?
Current code
VB:
If CheckBox1.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox1.Caption
If CheckBox2.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox2.Caption
If CheckBox3.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox3.Caption
If CheckBox4.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox4.Caption
If CheckBox5.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox5.Caption
If CheckBox6.Value = True Then ws.Cells(iRow, 2).Value = ws.Cells(iRow, 2).Value & " " & CheckBox6.Caption
Spreadsheet current cell appearence if all 6 are selected ' Chinnook EH101 Lynx Puma Sea King Fixed Wing'
Required appearence:
Chinnook
EH101
Lynx
Puma
Sea King
Fixed Wing
## Displaying And Un-Displaying Objects Using VBA..
Apr 21, 2007
Is It Possible to use VBA Coding to Display An Image (Object),
And Un-Display or Remove that Same Image (Object) using a VBA Code?
Example:
I have a Mailbox Picture that I would Like to Show, but only when a cell's
value = 5. If the Cell's Value is not 5, then don't display the mailbox.
## Conditional Formatting Based On Percentage And Insert Cell Based On Percentage
Apr 3, 2013
I have a workbook that I've built for a project. I've attached a sample workbook. What I'm trying to do, for the entire sheet or workbook if possible, is turn any Cell with a percentage of 30% to 49% yellow and any cell with a percentage of 50% or more Red. I would also like to move the ID's of the variable cells, for example Id number 9922, to the cells beside the description of the rows, Affected would be an example, if the information contained in the same row as the ID meets with a set of variables.
For example I only want the ID's moved if they correspond with IDsub 1-25. One more thing, the people who will be using this spreadsheet will be copying data from a website when it is imported it does not insert the values as numbers. I would like to format the cells, in example workbook they would be any of the cells labeled ID IDsub Variable or Number, so that anything put in that cell will automatically be converted to a number.
I should also probably add that the formatting will be done on approximately 80 rows a sheet with 47 sheets.
## Calculate Percentage Of A Percentage?
Apr 26, 2013
calculating the percentage of a percentage and writing the formula for excel.
There are 295 people in a room, of the 295, 75 or 25% are mothers. (I know how to calculate 25% - 75/295 = 25.42) of the 75 mothers 35 have 3 children, 32 have 2 children and 10 have 1 child.
35 is what percent of 25%
32 is what percent of 25%
10 is what percent of 25%
## Displaying Zero
Nov 24, 2009
I have tried different formula involving with decimal places.
If A1 has a whole number like 5141234 then the result would come out as a whole number =(A1-5148000)*1.000440935+48000
If A1 has three decimal places like 5142356.654 then the result would out as three decimal places
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https://www.dataunitconverter.com/gibibyte-per-hour-to-zebibit-per-hour
| 1,716,048,121,000,000,000
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crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00418.warc.gz
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# GiB/Hr to Zibit/Hr → CONVERT Gibibytes per Hour to Zebibits per Hour
expand_more
info 1 GiB/Hr is equal to 0.0000000000072759576141834259033203125 Zibit/Hr
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
Input Gibibytes per Hour (GiB/Hr) - and press Enter.
## Gibibytes per Hour (GiB/Hr) Versus Zebibits per Hour (Zibit/Hr) - Comparison
Gibibytes per Hour and Zebibits per Hour are units of digital information used to measure storage capacity and data transfer rate.
Both Gibibytes per Hour and Zebibits per Hour are the "binary" units. One Gibibyte is equal to 1024^3 bytes. One Zebibit is equal to 1024^7 bits. There are 137,438,953,472 Gibibyte in one Zebibit. Find more details on below table.
Gibibytes per Hour (GiB/Hr) Zebibits per Hour (Zibit/Hr)
Gibibytes per Hour (GiB/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Gibibytes that can be transferred in one Hour. Zebibits per Hour (Zibit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Zebibits that can be transferred in one Hour.
## Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr) Conversion - Formula & Steps
The GiB/Hr to Zibit/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Gibibyte) and target (Zebibit) data units.
Source Data Unit Target Data Unit
Equal to 1024^3 bytes
(Binary Unit)
Equal to 1024^7 bits
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Gibibyte to Zebibit in a simplified manner.
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
x 1024
x 1024
x 1024
x 1024
x 1024
Based on the provided diagram and steps outlined earlier, the formula for converting the Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr) can be expressed as follows:
diamond CONVERSION FORMULA Zibit/Hr = GiB/Hr x 8 ÷ 10244
Now, let's apply the aforementioned formula and explore the manual conversion process from Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Zebibits per Hour = Gibibytes per Hour x 8 ÷ 10244
STEP 1
Zebibits per Hour = Gibibytes per Hour x 8 ÷ (1024x1024x1024x1024)
STEP 2
Zebibits per Hour = Gibibytes per Hour x 8 ÷ 1099511627776
STEP 3
Zebibits per Hour = Gibibytes per Hour x 0.0000000000072759576141834259033203125
Example : By applying the previously mentioned formula and steps, the conversion from 1 Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr) can be processed as outlined below.
1. = 1 x 8 ÷ 10244
2. = 1 x 8 ÷ (1024x1024x1024x1024)
3. = 1 x 8 ÷ 1099511627776
4. = 1 x 0.0000000000072759576141834259033203125
5. = 0.0000000000072759576141834259033203125
6. i.e. 1 GiB/Hr is equal to 0.0000000000072759576141834259033203125 Zibit/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Gibibytes per Hour to Zebibits per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Gibibyte ?
A Gibibyte (GiB) is a binary unit of digital information that is equal to 1,073,741,824 bytes (or 8,589,934,592 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'gibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'gigabyte' (GB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### What is Zebibit ?
A Zebibit (Zib or Zibit) is a binary unit of digital information that is equal to 1,180,591,620,717,411,303,424 bits and is defined by the International Electro technical Commission(IEC). The prefix 'zebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'zettabit' (Zb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr)
Apply the formula as shown below to convert from 1 Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr).
A B C
1 Gibibytes per Hour (GiB/Hr) Zebibits per Hour (Zibit/Hr)
2 1 =A2 * 0.0000000000072759576141834259033203125
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Gibibytes per Hour (GiB/Hr) to Zebibits per Hour (Zibit/Hr) Conversion
You can use below code to convert any value in Gibibytes per Hour (GiB/Hr) to Gibibytes per Hour (GiB/Hr) in Python.
gibibytesperHour = int(input("Enter Gibibytes per Hour: "))
zebibitsperHour = gibibytesperHour * 8 / (1024*1024*1024*1024)
print("{} Gibibytes per Hour = {} Zebibits per Hour".format(gibibytesperHour,zebibitsperHour))
The first line of code will prompt the user to enter the Gibibytes per Hour (GiB/Hr) as an input. The value of Zebibits per Hour (Zibit/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Zebibits(Zibit) are there in a Gibibyte(GiB)?expand_more
There are 0.0000000000072759576141834259033203125 Zebibits in a Gibibyte.
#### What is the formula to convert Gibibyte(GiB) to Zebibit(Zibit)?expand_more
Use the formula Zibit = GiB x 8 / 10244 to convert Gibibyte to Zebibit.
#### How many Gibibytes(GiB) are there in a Zebibit(Zibit)?expand_more
There are 137438953472 Gibibytes in a Zebibit.
#### What is the formula to convert Zebibit(Zibit) to Gibibyte(GiB)?expand_more
Use the formula GiB = Zibit x 10244 / 8 to convert Zebibit to Gibibyte.
#### Which is bigger, Zebibit(Zibit) or Gibibyte(GiB)?expand_more
Zebibit is bigger than Gibibyte. One Zebibit contains 137438953472 Gibibytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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Two Plus Two Poker Forums Elaborating On My Multiplayer Toy Game Article
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12-20-2017, 09:44 AM #26 iamallin journeyman Join Date: Feb 2016 Location: Lonely planet Posts: 242 Re: Elaborating On My Multiplayer Toy Game Article Thank you whosnext. Very good effort indeed. I revisited some of those equations and have doubts about equation for c3a It doesn't include conditional probability that player 2 folded his hand. Should it be like this? C3a * (c3a-c1)/(1-c1) * 1/c2 = 1/3 Sent from my SM-N910W8 using Tapatalk
12-20-2017, 11:24 AM #27 iamallin journeyman Join Date: Feb 2016 Location: Lonely planet Posts: 242 Re: Elaborating On My Multiplayer Toy Game Article Edit : it seems the first set of equations were correct. P (p3 wins against p2 | p2 folded) = 1 In that case, solutions provided by whosnext are correct and we have the answers for a 4 player game. The size of the solution for n players ( including the blind player) is 2 ^ (n-2)
12-20-2017, 02:34 PM #28
whosnext
Pooh-Bah
Join Date: Mar 2009
Location: California
Posts: 4,413
Re: Elaborating On My Multiplayer Toy Game Article
Quote:
Originally Posted by iamallin Thank you whosnext. Very good effort indeed. I revisited some of those equations and have doubts about equation for c3a It doesn't include conditional probability that player 2 folded his hand. Should it be like this? C3a * (c3a-c1)/(1-c1) * 1/c2 = 1/3
I verified all the equations before I did my solution journey. (Not to diminish the derivation, but it is quite straightforward.)
The 4-player solution with antes is yet to be derived, though, as in the 3-player case, the underlying equations are unchanged except for the constant (the pot odds part).
The 5-player no-ante solution is probably straightforward to derive (mix of analytical and numerical).
To be honest, I am not sure what these GTO solutions of this highly stylized toy game reveal to us.
12-20-2017, 09:44 PM #29 whosnext Pooh-Bah Join Date: Mar 2009 Location: California Posts: 4,413 Re: Elaborating On My Multiplayer Toy Game Article Here is the solution for the 5-player game without antes. Following the lead from above, label the players P0, P1, P2, P3, and P4 (P0 bets blind 100% of the time). The other players can call or fold. The GTO solution consists of the minimum hand each player calls with on each possible branch of the game tree. As above, the GTO solution for many cases in the 5-player game tree devolve into the lower-order GTO solutions for the smaller games (derived and presented previously in the thread). The "new" GTO branches are as follows: C1 = .76996520 (P1's min call hand) C2 = .87267090 (P2's min call hand when P1 calls) C3A = .86606407 (P3's min call hand when P1 calls and P2 folds) C3B = .92579529 (P3's min call hand when P1 calls and P2 calls) C4A = .85920815 (P4's min call hand when P1 calls, P2 folds, P3 folds) C4B = .92130617 (P4's min call hand when P1 calls, P2 folds, P3 calls) C4C = .92408574 (P4's min call hand when P1 calls, P2 calls, P3 folds) C4D = .95543466 (P4's min call hand when P1 calls, P2 calls, P3 calls)
12-20-2017, 11:04 PM #30 iamallin journeyman Join Date: Feb 2016 Location: Lonely planet Posts: 242 Re: Elaborating On My Multiplayer Toy Game Article Incredible work whosnext. Thank you. I tried solving this in python and it was taking forever. Here are some of my observations 1. The effect of position is less pronounced than I was expecting it to be. When facing the same shoving range and getting same pot odds, players in later position only defend slightly more often than players in earlier positions ( 0.006 seems to be gap in the 5 player solution) 2. The effect of an extra player putting money in the pot is huge. Even though the next to act players are now getting better odds, they play much tighter than they were playing looser in the above observation. 3. Player 1's open shoving ( in this case calling) range is getting tighter the more players are in the pot. Very expected. But the jumps are smaller for each increment in game size. ( 0.63 for 3 player game, 0.71 for 4 player game and 0.77 for 5 player game) I wonder if there is a natural limit to it. That is no matter how many players are in the pot, player 1 can always shove top 1% of hands or something. Probably not true. Last edited by iamallin; 12-20-2017 at 11:11 PM.
12-21-2017, 11:16 AM #31 robert_utk old hand Join Date: Jan 2005 Location: ValueTown Posts: 1,686 Re: Elaborating On My Multiplayer Toy Game Article There should be a natural convergence for player 1 given infinite extra participants. Also, if player 1 could calculate this, and demonstrate the strategy to the rest of the world, then if player 1 actually limped then the game would be over, because any limp thereafter would be losing EV versus the first limper. This is what interests me in the infinite extension of these types of toy games. A first move that is “a look to the sky” that counters all future moves in infinite extension.
12-22-2017, 03:43 AM #32
David Sklansky
Administrator
Join Date: Aug 2002
Posts: 14,258
Re: Elaborating On My Multiplayer Toy Game Article
Quote:
Originally Posted by iamallin Incredible work whosnext. Thank you. I tried solving this in python and it was taking forever. Here are some of my observations 1. The effect of position is less pronounced than I was expecting it to be. When facing the same shoving range and getting same pot odds, players in later position only defend slightly more often than players in earlier positions ( 0.006 seems to be gap in the 5 player solution) 2. The effect of an extra player putting money in the pot is huge. Even though the next to act players are now getting better odds, they play much tighter than they were playing looser in the above observation. 3. Player 1's open shoving ( in this case calling) range is getting tighter the more players are in the pot. Very expected. But the jumps are smaller for each increment in game size. ( 0.63 for 3 player game, 0.71 for 4 player game and 0.77 for 5 player game) I wonder if there is a natural limit to it. That is no matter how many players are in the pot, player 1 can always shove top 1% of hands or something. Probably not true.
Cmon. Suppose there were a billion players and every opponent played the top 100 millionth of their hands
12-22-2017, 10:42 AM #33
robert_utk
old hand
Join Date: Jan 2005
Location: ValueTown
Posts: 1,686
Re: Elaborating On My Multiplayer Toy Game Article
Quote:
Originally Posted by David Sklansky Cmon. Suppose there were a billion players and every opponent played the top 100 millionth of their hands
Assuming antes, if first limper is optimal, then there will be no EV left for anyone else.
12-22-2017, 12:29 PM #34 iamallin journeyman Join Date: Feb 2016 Location: Lonely planet Posts: 242 Re: Elaborating On My Multiplayer Toy Game Article Thinking about infinite number of players may be counter productive to this thread. I apologise if it derailed the thread. I will post some of my final thoughts related to this topic and it may be better to start a new thread for the purely theoretical infinite player game. Please let me know if that's better. Instead of one player playing blind, let's say you have n players playing blind. ( or 1 player playing blind and the other n-1 players using a 100% shoving range) Strategy for the last player in n+1 player pool then is (1/1+n) ^(1/n) -> 1 as n goes to inf. So if you had everyone going all in blind, the last player can't call. If you flip the script, and say you knew everyone left to act after you is a degenerate gambler and will always call, as the first to act player you can't call anything. If you knew everyone left to act behind you is always folding, you can call with 0.5 So what should happen when you know everyone left to act behind you is neither a degenerate gambler nor a folding machine but an optimal player. If the answer is still to fold everything, then we are saying there is no difference between facing infinite optimal opponents and facing infinite degenerate gambling opponents. ( which may be true but is definitely insightful) Last edited by iamallin; 12-22-2017 at 12:40 PM.
12-22-2017, 08:44 PM #35
David Sklansky
Administrator
Join Date: Aug 2002
Posts: 14,258
Re: Elaborating On My Multiplayer Toy Game Article
Quote:
Originally Posted by whosnext Here is the solution for the 5-player game without antes. Following the lead from above, label the players P0, P1, P2, P3, and P4 (P0 bets blind 100% of the time). The other players can call or fold. The GTO solution consists of the minimum hand each player calls with on each possible branch of the game tree. As above, the GTO solution for many cases in the 5-player game tree devolve into the lower-order GTO solutions for the smaller games (derived and presented previously in the thread). The "new" GTO branches are as follows: C1 = .76996520 (P1's min call hand) C2 = .87267090 (P2's min call hand when P1 calls) C3A = .86606407 (P3's min call hand when P1 calls and P2 folds) C3B = .92579529 (P3's min call hand when P1 calls and P2 calls) C4A = .85920815 (P4's min call hand when P1 calls, P2 folds, P3 folds) C4B = .92130617 (P4's min call hand when P1 calls, P2 folds, P3 calls) C4C = .92408574 (P4's min call hand when P1 calls, P2 calls, P3 folds) C4D = .95543466 (P4's min call hand when P1 calls, P2 calls, P3 calls)
Very nice. So what is each of the players EVS before the game starts?
12-23-2017, 12:13 AM #36 whosnext Pooh-Bah Join Date: Mar 2009 Location: California Posts: 4,413 Re: Elaborating On My Multiplayer Toy Game Article Here is the solution for the 6-player game without antes. Following the lead from above, label the players P0, P1, P2, P3, P4, and P5 (P0 bets blind 100% of the time). The other players can call or fold. The GTO solution consists of the minimum hand each player calls with on each possible branch of the game tree. As above, the GTO solution for many cases in the 6-player game tree devolve into the lower-order GTO solutions for the smaller games (derived and presented previously in the thread). The "new" GTO branches are as follows: --- C1 = .80694545 (P1's min call hand) --- C2 = .89426354 (P2's min call hand when P1 calls) --- C3A = .88967823 (P3's min call hand when P1 calls and P2 folds) C3B = .93873171 (P3's min call hand when P1 calls and P2 calls) --- C4A = .88493582 (P4's min call hand when P1 calls, P2 folds, P3 folds) C4B = .93566347 (P4's min call hand when P1 calls, P2 folds, P3 calls) C4C = .93755983 (P4's min call hand when P1 calls, P2 calls, P3 folds) C4D = .96332643 (P4's min call hand when P1 calls, P2 calls, P3 calls) --- C5A = .88006664 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 folds) C5B = .93240807 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 calls) C5C = .93439073 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 folds) C5D = .96139927 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 calls) C5E = .93637226 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 folds) C5F = .96250749 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 calls) C5G = .96296510 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 folds) C5H = .97752185 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 calls)
12-23-2017, 01:33 AM #37 browni3141 old hand Join Date: Aug 2015 Location: South Florida Posts: 1,852 Re: Elaborating On My Multiplayer Toy Game Article How did you calculate theee values? Did you write a program to solve/estimate systems of equations? Did you solve them yourself? Did you use some recursive equilibrium search algorithm?
12-23-2017, 03:55 PM #38 whosnext Pooh-Bah Join Date: Mar 2009 Location: California Posts: 4,413 Re: Elaborating On My Multiplayer Toy Game Article I do not have any special software. I wrote some pretty basic functions to numerically find the solutions. It helps that some of the equations are solvable analytically (one variable in terms of one or more others). Also, even when equations are not analytically solvable, due to the nature of the game tree the variables "cascade" (sometimes in pairs) to a significant degree. So it is essentially just a matter of intelligently and efficiently searching at the highest levels and letting the possible solutions cascade down until all of them are satisfied.
12-24-2017, 08:08 PM #39
whosnext
Pooh-Bah
Join Date: Mar 2009
Location: California
Posts: 4,413
Re: Elaborating On My Multiplayer Toy Game Article
Here are the EV's for the GTO players in the respective toy games (without antes) solved so far.
PlayersP0P1P2P3P4P5
2
-0.250000
0.250000
3
-0.399056
0.167931
0.231125
4
-0.496559
0.125292
0.161640
0.209627
5
-0.565348
0.099747
0.122850
0.152260
0.190491
6
-0.616474
0.082576
0.098702
0.118191
0.142604
0.174401
12-25-2017, 04:19 PM #40 David Sklansky Administrator Join Date: Aug 2002 Posts: 14,258 Re: Elaborating On My Multiplayer Toy Game Article Very nice again. The answers seem logical and refute the illogical notion that the first guy after the blind is in some way not in the second worst spot. Since you are so good at this here three more questions. (I hope you are simulating rather than calculating.) With various number of players how often does the best hand win? (Obviously with two players the answer is 7/8.) For each player, what percentage of the time that it does indeed have the best hand does it still not win? For each player what percentage of the time does it win with the not best hand?
12-30-2017, 03:37 PM #41 David Sklansky Administrator Join Date: Aug 2002 Posts: 14,258 Re: Elaborating On My Multiplayer Toy Game Article Hopefully people realize that I simply wanted a simulation for the three questions above using the assumption that whosnext's results in post #36 are correct. Just tell your computer to deal out a billion hands using those rules and see who has the best hand and who actually wins. I don't want to dig out my Timex Sinclair and reread the three pages of instructions on basic Basic.
12-30-2017, 06:23 PM #42 whosnext Pooh-Bah Join Date: Mar 2009 Location: California Posts: 4,413 Re: Elaborating On My Multiplayer Toy Game Article Here is the solution for the 7-player game without antes. Following the lead from above, label the players P0, P1, P2, P3, P4, P5, and P6 (P0 bets blind 100% of the time). The other players can call or fold. The GTO solution consists of the minimum hand each player calls with on each possible branch of the game tree. As above, the GTO solution for many cases in the 7-player game tree devolve into the lower-order GTO solutions for the smaller games (derived and presented previously in the thread). The "new" GTO branches are as follows: --- C1 = .83386680 (P1's min call hand) --- C2 = .90968767 (P2's min call hand when P1 calls) --- C3A = .90632926 (P3's min call hand when P1 calls and P2 folds) C3B = .94788043 (P3's min call hand when P1 calls and P2 calls) --- C4A = .90286674 (P4's min call hand when P1 calls, P2 folds, P3 folds) C4B = .94565715 (P4's min call hand when P1 calls, P2 folds, P3 calls) C4C = .94702925 (P4's min call hand when P1 calls, P2 calls, P3 folds) C4D = .96887551 (P4's min call hand when P1 calls, P2 calls, P3 calls) --- C5A = .89931525 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 folds) C5B = .94331611 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 calls) C5C = .94474327 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 folds) C5D = .96748541 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 calls) C5E = .94616803 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 folds) C5F = .96828423 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 calls) C5G = .96861424 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 folds) C5H = .98094976 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 calls) --- C6A = .89569346 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 folds) C6B = .94085752 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 calls) C6C = .94233556 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 folds) C6D = .96601165 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 calls) C6E = .94381871 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 folds) C6F = .96684643 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 calls) C6G = .96720214 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 folds) C6H = .98008153 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 calls) C6I = .94529751 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 folds) C6J = .96767975 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 calls) C6K = .96801519 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 folds) C6L = .98056919 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 calls) C6M = .96835204 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 folds) C6N = .98076290 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 calls) C6O = .98086128 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 folds) C6P = .98811101 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 calls) --- In my next post I will report the GTO EV's for the players in this game.
12-30-2017, 06:34 PM #43
whosnext
Pooh-Bah
Join Date: Mar 2009
Location: California
Posts: 4,413
Re: Elaborating On My Multiplayer Toy Game Article
Here are the GTO EV's for the players in the respective toy games (without antes) solved so far. The values below are the exact EV's whereas previously I used simulations for a few of the cases. I have now programmed the complete game tree and derived each player's exact EV on each branch of the tree.
PlayersP0P1P2P3P4P5P6
2
-0.250000
0.250000
3
-0.399056
0.167931
0.231125
4
-0.496559
0.125292
0.161640
0.209627
5
-0.565278
0.099662
0.122836
0.152215
0.190565
6
-0.616430
0.082656
0.098620
0.118121
0.142637
0.174396
7
-0.656085
0.070577
0.082212
0.096025
0.112792
0.133744
0.160735
12-31-2017, 02:41 PM #44
whosnext
Pooh-Bah
Join Date: Mar 2009
Location: California
Posts: 4,413
Re: Elaborating On My Multiplayer Toy Game Article
Quote:
Originally Posted by David Sklansky Very nice again. The answers seem logical and refute the illogical notion that the first guy after the blind is in some way not in the second worst spot. Since you are so good at this here three more questions. (I hope you are simulating rather than calculating.) With various number of players how often does the best hand win? (Obviously with two players the answer is 7/8.) For each player, what percentage of the time that it does indeed have the best hand does it still not win? For each player what percentage of the time does it win with the not best hand?
Here are three tables which answer your questions. The tables are based upon the results of simulations. Each simulation consisted of 100 million trials using the GTO strategies for each of the toy games from 2 to 7 players.
PlayersPct of Time Overall Best Hand Wins Pot
2
87.50%
3
86.46%
4
86.25%
5
86.11%
6
85.97%
7
85.84%
PlayersPct of the times P0 does NOT Win Pot when he has Overall Best HandPct of the times P1 does NOT Win Pot when he has Overall Best HandPct of the times P2 does NOT Win Pot when he has Overall Best HandPct of the times P3 does NOT Win Pot when he has Overall Best HandPct of the times P4 does NOT Win Pot when he has Overall Best HandPct of the times P5 does NOT Win Pot when he has Overall Best HandPct of the times P6 does NOT Win Pot when he has Overall Best Hand
2
0.00%
25.00%
3
0.00%
25.48%
15.15%
4
0.00%
26.35%
18.12%
10.52%
5
0.00%
27.06%
20.43%
13.55%
8.40%
6
0.00%
27.61%
22.10%
16.23%
10.76%
7.47%
7
0.00%
28.05%
23.37%
18.31%
13.26%
9.07%
7.09%
PlayersPct of the times P0 does NOT have Overall Best Hand when he Wins PotPct of the times P1 does NOT have Overall Best Hand when he Wins PotPct of the times P2 does NOT have Overall Best Hand when he Wins PotPct of the times P3 does NOT have Overall Best Hand when he Wins PotPct of the times P4 does NOT have Overall Best Hand when he Wins PotPct of the times P5 does NOT have Overall Best Hand when he Wins PotPct of the times P6 does NOT have Overall Best Hand when he Wins Pot
2
20.00%
0.00%
3
26.84%
3.20%
1.70%
4
30.56%
4.46%
3.60%
4.76%
5
33.02%
5.09%
4.46%
4.91%
8.10%
6
34.82%
5.45%
4.95%
5.05%
6.72%
11.19%
7
36.20%
5.70%
5.29%
5.21%
6.06%
8.72%
13.89%
I have saved all the raw totals of these simulations so would be able to easily answer any other questions in case I misunderstood what you were asking.
01-20-2018, 12:57 AM #45 whosnext Pooh-Bah Join Date: Mar 2009 Location: California Posts: 4,413 Re: Elaborating On My Multiplayer Toy Game Article Here is the solution for the 8-player game without antes. Following the lead from above, label the players P0, P1, P2, P3, P4, P5, P6, and P7 (P0 bets blind 100% of the time). The other players can call or fold. The GTO solution consists of the minimum hand each player calls with on each possible branch of the game tree. As above, the GTO solution for many cases in the 8-player game tree devolve into the lower-order GTO solutions for the smaller games (derived and presented previously in the thread). The "new" GTO branches are as follows: --- C1 = .85428329 (P1's min call hand) --- C2 = .92122710 (P2's min call hand when P1 calls) --- C3A = .91866516 (P3's min call hand when P1 calls and P2 folds) C3B = .95467593 (P3's min call hand when P1 calls and P2 calls) --- C4A = .91603133 (P4's min call hand when P1 calls, P2 folds, P3 folds) C4B = .95299330 (P4's min call hand when P1 calls, P2 folds, P3 calls) C4C = .95403058 (P4's min call hand when P1 calls, P2 calls, P3 folds) C4D = .97298036 (P4's min call hand when P1 calls, P2 calls, P3 calls) --- C5A = .91333368 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 folds) C5B = .95123216 (P5's min call hand when P1 calls, P2 folds, P3 folds, P4 calls) C5C = .95230644 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 folds) C5D = .97193267 (P5's min call hand when P1 calls, P2 folds, P3 calls, P4 calls) C5E = .95337839 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 folds) C5F = .97253406 (P5's min call hand when P1 calls, P2 calls, P3 folds, P4 calls) C5G = .97278325 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 folds) C5H = .98347919 (P5's min call hand when P1 calls, P2 calls, P3 calls, P4 calls) --- C6A = .91058239 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 folds) C6B = .94939225 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 calls) C6C = .95050143 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 folds) C6D = .97082989 (P6's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 calls) C6E = .95161236 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 folds) C6F = .97145508 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 calls) C6G = .97172204 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 folds) C6H = .98282647 (P6's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 calls) C6I = .95271983 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 folds) C6J = .97207901 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 calls) C6K = .97233166 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 folds) C6L = .98319243 (P6's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 calls) C6M = .97258516 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 folds) C6N = .98333858 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 calls) C6O = .98341244 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 folds) C6P = .98969571 (P6's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 calls) --- C7A = .90778936 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 folds, P6 folds) C7B = .94747429 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 folds, P6 calls) C7C = .94861542 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 calls, P6 folds) C7D = .96967056 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 folds, P5 calls, P6 calls) C7E = .94976314 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 folds, P6 folds) C7F = .97031875 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 folds, P6 calls) C7G = .97060241 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 calls, P6 folds) C7H = .98213768 (P7's min call hand when P1 calls, P2 folds, P3 folds, P4 calls, P5 calls, P6 calls) C7I = .95091158 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 folds, P6 folds) C7J = .97096796 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 folds, P6 calls) C7K = .97123756 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 calls, P6 folds) C7L = .98251884 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 folds, P5 calls, P6 calls) C7M = .97150821 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 folds, P6 folds) C7N = .98267544 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 folds, P6 calls) C7O = .98275462 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 calls, P6 folds) C7P = .98928520 (P7's min call hand when P1 calls, P2 folds, P3 calls, P4 calls, P5 calls, P6 calls) C7Q = .95205537 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 folds, P6 folds) C7R = .97161524 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 folds, P6 calls) C7S = .97187121 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 calls, P6 folds) C7T = .98289914 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 folds, P5 calls, P6 calls) C7U = .97212826 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 folds, P6 folds) C7V = .98304762 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 folds, P6 calls) C7W = .98312370 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 calls, P6 folds) C7X = .98951344 (P7's min call hand when P1 calls, P2 calls, P3 folds, P4 calls, P5 calls, P6 calls) C7Y = .97238614 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 folds, P6 folds) C7Z = .98319659 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 folds, P6 calls) C7AA = .98327109 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 calls, P6 folds) C7AB = .98960364 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 folds, P5 calls, P6 calls) C7AC = .98334582 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 folds, P6 folds) C7AD = .98964743 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 folds, P6 calls) C7AE = .98967206 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 calls, P6 folds) C7AF = .99346786 (P7's min call hand when P1 calls, P2 calls, P3 calls, P4 calls, P5 calls, P6 calls) --- In my next post I will report the GTO EV's for the players in this game.
01-20-2018, 01:03 AM #46
whosnext
Pooh-Bah
Join Date: Mar 2009
Location: California
Posts: 4,413
Re: Elaborating On My Multiplayer Toy Game Article
Here are the GTO EV's for the players in the respective toy games (without antes) solved so far. The values below are the exact EV's whereas previously I used simulations for a few of the cases. I have now programmed the complete game tree and derived each player's exact EV on each branch of the tree.
PlayersP0P1P2P3P4P5P6P7
2
-0.250000
0.250000
3
-0.399056
0.167931
0.231125
4
-0.496559
0.125292
0.161640
0.209627
5
-0.565278
0.099662
0.122836
0.152215
0.190565
6
-0.616430
0.082656
0.098620
0.118121
0.142637
0.174396
7
-0.656085
0.070577
0.082212
0.096025
0.112792
0.133744
0.160735
8
-0.687795
0.061564
0.070409
0.080683
0.092819
0.107482
0.125720
0.149118
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CC-MAIN-2018-17
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latest
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en
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http://colin.barker.pagesperso-orange.fr/lpa/cry_mult.htm
| 1,656,696,911,000,000,000
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#### Multiplicative Cryptarithm Solver
Solves cryptarithms consisting of the product of two numbers. For example:
• `NO * WAR = PEACE`
• `CAT * DOG = FIGHT`
• `TWO * SIX = TWELVE`
• `CINQ * SIX = TRENTE`
• `CRAFT * OF = PROLOG`
• `IOWA * OHIO = FLORIDA`
```/* cry([[N,O], [W,A,R], [P,E,A,C,E]]). */
/* cry([[C,A,T], [D,O,G], [F,I,G,H,T]]). */
/* cry([[T,W,O], [S,I,X], [T,W,E,L,V,E]]). */
/* cry([[C,R,A,F,T], [O,F], [P,R,O,L,O,G]]). */
/* cry([[C,I,N,Q], [S,I,X], [T,R,E,N,T,E]]). */
/* cry([[I,O,W,A], [O,H,I,O], [F,L,O,R,I,D,A]]). */
cry([As,Bs,Cs]):-
sub_products(As, Bs, [], SubProducts),
columns(SubProducts, [], Columns),
reverse(Cs, Totals),
solve(Totals, Columns, [0,1,2,3,4,5,6,7,8,9], 0),
As \= [0|_], Bs \= [0|_], Cs \= [0|_],
write([As,Bs,Cs]), nl,
fail.
/* e.g. sub_products([n,o], [w,a,r], [], */
/* [[p(o,w),p(o,a),p(o,r)],[p(n,w),p(n,a),p(n,r)]]). */
sub_products([], _, Zss, Zss).
sub_products([X|Xs], Ys, Zss0, Zss):-
sub_products_1(Ys, X, Zs),
sub_products(Xs, Ys, [Zs|Zss0], Zss).
/* e.g. sub_products_1([w,a,r], n, [p(n,w),p(n,a),p(n,r)]). */
sub_products_1([], _, []).
sub_products_1([Y|Ys], X, [p(X,Y)|Zs]):-
sub_products_1(Ys, X, Zs).
/* e.g. columns([[p(o,w),p(o,a),p(o,r)],[p(n,w),p(n,a),p(n,r)]], [], */
/* [[p(o,r)],[p(n,r),p(o,a)],[p(n,a),p(o,w)],[p(n,w)]]). */
columns([], Yss, Zss):-
reverse(Yss, Zss).
columns([Xs|Xss], Yss0, Yss):-
columns_1(Xs, [[]|Yss0], Yss1),
columns(Xss, Yss1, Yss).
/* e.g. columns_1([p(n,w),p(n,a),p(n,r)], [[],[p(o,w)],[p(o,a)],[p(o,r)]], */
/* [[p(n,w)],[p(n,a),p(o,w)],[p(n,r),p(o,a)],[p(o,r)]]). */
columns_1([], Yss, Yss).
columns_1([X|Xs], [], [[X]|Ys]):-!,
columns_1(Xs, [], Ys).
columns_1([X|Xs], [Ys|Yss], [[X|Ys]|Zss]):-
columns_1(Xs, Yss, Zss).
/* This is where all the work is done. */
solve([], _, _, 0).
solve([Z], [], Digits, Carry):-!,
solve([Z], [[]], Digits, Carry).
solve([Z|Zs], [Xs|Xss], Digits, Carry):-
column_sum(Xs, Digits, Digits1, Carry, Sum),
Digit is Sum mod 10,
select(Z, Digits1, Digits2),
Digit = Z,
Carry1 is Sum // 10,
solve(Zs, Xss, Digits2, Carry1).
column_sum([], Digits, Digits, Sum, Sum).
column_sum([p(A,B)|Xs], Digits0, Digits, Sum0, Sum):-
select(A, Digits0, Digits1),
select(B, Digits1, Digits2),
Sum1 is Sum0 + A * B,
column_sum(Xs, Digits2, Digits, Sum1, Sum).
/* select(X, Ys, Zs) is true either if X is bound and the list Zs is equal */
/* to the list Ys, or if X is not bound and the list Zs is the result of */
/* removing one occurrence of the element X from the list Ys. */
select(X, Ys, Ys):-nonvar(X), !.
select(X, Ys, Zs):-remove(X, Ys, Zs).
```
| 1,095
| 2,763
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.109375
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.384892
|
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=DRD
| 1,500,685,166,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549423839.97/warc/CC-MAIN-20170722002507-20170722022507-00066.warc.gz
| 442,432,287
| 107,757
|
Back to list of Stocks See Also: Seasonal Analysis of DRDGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
# Fourier Analysis of DRD (DRDGOLD Ltd)
DRD (DRDGOLD Ltd) appears to have interesting cyclic behaviour every 92 weeks (1.7395*cosine), 70 weeks (.7472*cosine), and 46 weeks (.4958*cosine).
DRD (DRDGOLD Ltd) has an average price of 8.55 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 1/3/2000 to 7/10/2017 for DRD (DRDGOLD Ltd), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
08.55262 0
11.64751 7.0188 (1*2π)/915915 weeks
2-3.23363 3.39523 (2*2π)/915458 weeks
3-2.24496 -.22953 (3*2π)/915305 weeks
4.09022 -2.14921 (4*2π)/915229 weeks
51.18701 -1.78431 (5*2π)/915183 weeks
6.93505 .10493 (6*2π)/915153 weeks
7.61726 -.17761 (7*2π)/915131 weeks
81.06703 .54201 (8*2π)/915114 weeks
9-.01484 1.76879 (9*2π)/915102 weeks
10-1.7395 .4375 (10*2π)/91592 weeks
11-.34834 -.46896 (11*2π)/91583 weeks
12.00627 -.28547 (12*2π)/91576 weeks
13.74721 .03202 (13*2π)/91570 weeks
14.20411 .32414 (14*2π)/91565 weeks
15.43624 .47034 (15*2π)/91561 weeks
16.07032 .40434 (16*2π)/91557 weeks
17.23527 .22055 (17*2π)/91554 weeks
18-.23248 .52498 (18*2π)/91551 weeks
19-.30762 .22765 (19*2π)/91548 weeks
20-.49581 -.4195 (20*2π)/91546 weeks
21.18277 -.49715 (21*2π)/91544 weeks
22.26526 -.01912 (22*2π)/91542 weeks
23.35301 .14365 (23*2π)/91540 weeks
24.19021 .31495 (24*2π)/91538 weeks
25-.10725 .43635 (25*2π)/91537 weeks
26-.25511 .33659 (26*2π)/91535 weeks
27-.39415 -.03009 (27*2π)/91534 weeks
28.10864 -.32521 (28*2π)/91533 weeks
29.38007 .10386 (29*2π)/91532 weeks
30.2552 .43476 (30*2π)/91531 weeks
31.13065 .54443 (31*2π)/91530 weeks
32-.18469 .34201 (32*2π)/91529 weeks
33-.13448 .1465 (33*2π)/91528 weeks
34.00003 .00287 (34*2π)/91527 weeks
35.06745 .20513 (35*2π)/91526 weeks
36.07917 .18821 (36*2π)/91525 weeks
37.0055 .07349 (37*2π)/91525 weeks
38.10146 -.11604 (38*2π)/91524 weeks
39.18273 .22352 (39*2π)/91523 weeks
40.00953 .13642 (40*2π)/91523 weeks
41-.0755 .17677 (41*2π)/91522 weeks
42-.10751 -.00645 (42*2π)/91522 weeks
43-.02039 -.04563 (43*2π)/91521 weeks
44-.10568 .01044 (44*2π)/91521 weeks
45.04365 -.03759 (45*2π)/91520 weeks
46.1115 .05552 (46*2π)/91520 weeks
47.1238 .20935 (47*2π)/91519 weeks
48-.19922 .32079 (48*2π)/91519 weeks
49-.19841 -.02701 (49*2π)/91519 weeks
50.02577 -.15169 (50*2π)/91518 weeks
51.24989 -.10798 (51*2π)/91518 weeks
52.4624 .23813 (52*2π)/91518 weeks
53-.08611 .44522 (53*2π)/91517 weeks
54-.21997 .15406 (54*2π)/91517 weeks
55-.20998 .05976 (55*2π)/91517 weeks
56-.06381 -.06265 (56*2π)/91516 weeks
57-.04101 -.20017 (57*2π)/91516 weeks
58.29858 -.0375 (58*2π)/91516 weeks
59.12368 .21725 (59*2π)/91516 weeks
60-.1316 .15019 (60*2π)/91515 weeks
61-.05959 .07526 (61*2π)/91515 weeks
62-.09876 .12922 (62*2π)/91515 weeks
63-.13656 -.0262 (63*2π)/91515 weeks
64-.06091 -.02512 (64*2π)/91514 weeks
65.08719 -.01743 (65*2π)/91514 weeks
66.03837 .05378 (66*2π)/91514 weeks
67.03928 .26989 (67*2π)/91514 weeks
68-.23073 .1217 (68*2π)/91513 weeks
69-.03294 -.08883 (69*2π)/91513 weeks
70.09531 -.02491 (70*2π)/91513 weeks
71.10793 .04631 (71*2π)/91513 weeks
72.05735 .01703 (72*2π)/91513 weeks
73.03939 .03269 (73*2π)/91513 weeks
74.06107 .03201 (74*2π)/91512 weeks
75.01888 .05484 (75*2π)/91512 weeks
76-.056 -.0508 (76*2π)/91512 weeks
77.07604 -.04423 (77*2π)/91512 weeks
78.10956 .00543 (78*2π)/91512 weeks
79.01329 .14454 (79*2π)/91512 weeks
80-.08652 .09046 (80*2π)/91511 weeks
81-.12612 .05522 (81*2π)/91511 weeks
82-.07819 -.06417 (82*2π)/91511 weeks
83-.00926 -.05031 (83*2π)/91511 weeks
84.12905 .0169 (84*2π)/91511 weeks
85.00744 .1534 (85*2π)/91511 weeks
86.04775 .06903 (86*2π)/91511 weeks
87-.06319 .11371 (87*2π)/91511 weeks
88-.10404 .00534 (88*2π)/91510 weeks
89-.0857 -.03528 (89*2π)/91510 weeks
90.07604 -.09633 (90*2π)/91510 weeks
91.12254 -.04371 (91*2π)/91510 weeks
92.10002 .1736 (92*2π)/91510 weeks
93-.04185 .14568 (93*2π)/91510 weeks
94-.1187 .03465 (94*2π)/91510 weeks
95-.04191 -.10325 (95*2π)/91510 weeks
96.04802 .01264 (96*2π)/91510 weeks
97-.06645 .07322 (97*2π)/9159 weeks
98-.06439 -.04202 (98*2π)/9159 weeks
99.04059 -.09228 (99*2π)/9159 weeks
100.11596 .02648 (100*2π)/9159 weeks
101-.00691 .11049 (101*2π)/9159 weeks
102-.04798 .02558 (102*2π)/9159 weeks
103-.05286 -.00416 (103*2π)/9159 weeks
104.01444 -.02876 (104*2π)/9159 weeks
105.02322 .06465 (105*2π)/9159 weeks
106-.11097 -.00862 (106*2π)/9159 weeks
107.07784 -.09269 (107*2π)/9159 weeks
108.02247 .06279 (108*2π)/9158 weeks
109-.02539 .02713 (109*2π)/9158 weeks
110.04182 -.02302 (110*2π)/9158 weeks
111.04293 .05708 (111*2π)/9158 weeks
112-.02876 .03183 (112*2π)/9158 weeks
113.03816 .03334 (113*2π)/9158 weeks
114.03984 .07614 (114*2π)/9158 weeks
115-.11074 -.00796 (115*2π)/9158 weeks
116.05432 -.0527 (116*2π)/9158 weeks
117.08058 .00427 (117*2π)/9158 weeks
118.089 .08148 (118*2π)/9158 weeks
119.03709 .10547 (119*2π)/9158 weeks
120-.05597 .11711 (120*2π)/9158 weeks
121-.09823 -.01834 (121*2π)/9158 weeks
122-.0655 -.09409 (122*2π)/9158 weeks
123.09784 -.08918 (123*2π)/9157 weeks
124.0881 .0736 (124*2π)/9157 weeks
125.00267 .07539 (125*2π)/9157 weeks
126-.00948 .05325 (126*2π)/9157 weeks
127-.03339 .01011 (127*2π)/9157 weeks
128-.00937 .03179 (128*2π)/9157 weeks
129-.02651 -.0314 (129*2π)/9157 weeks
130.00197 -.02811 (130*2π)/9157 weeks
131.07818 -.04078 (131*2π)/9157 weeks
132.07475 .00185 (132*2π)/9157 weeks
133.02577 .02636 (133*2π)/9157 weeks
134.0234 .02155 (134*2π)/9157 weeks
135.01598 .01473 (135*2π)/9157 weeks
136-.0112 .0253 (136*2π)/9157 weeks
137.0015 .04817 (137*2π)/9157 weeks
138-.05538 .02322 (138*2π)/9157 weeks
139-.08165 -.08028 (139*2π)/9157 weeks
140.0549 -.0605 (140*2π)/9157 weeks
141.0208 .038 (141*2π)/9156 weeks
142-.01082 .00373 (142*2π)/9156 weeks
143.02364 -.00192 (143*2π)/9156 weeks
144.0164 .0252 (144*2π)/9156 weeks
145-.0008 .05389 (145*2π)/9156 weeks
146-.04176 .02603 (146*2π)/9156 weeks
147-.01741 -.03129 (147*2π)/9156 weeks
148.02846 -.04377 (148*2π)/9156 weeks
149.04904 .02905 (149*2π)/9156 weeks
150.02667 -.04499 (150*2π)/9156 weeks
151.05564 .02625 (151*2π)/9156 weeks
152.02044 .02855 (152*2π)/9156 weeks
153-.00866 .04172 (153*2π)/9156 weeks
154.04207 -.0518 (154*2π)/9156 weeks
155.05823 .01227 (155*2π)/9156 weeks
156.01413 .04429 (156*2π)/9156 weeks
157.01254 .05598 (157*2π)/9156 weeks
158-.01641 .05291 (158*2π)/9156 weeks
159-.03362 -.01396 (159*2π)/9156 weeks
160-.0123 -.03996 (160*2π)/9156 weeks
161.05809 .0016 (161*2π)/9156 weeks
162.07641 .06688 (162*2π)/9156 weeks
163.00531 .08372 (163*2π)/9156 weeks
164-.09131 .01701 (164*2π)/9156 weeks
165-.01577 -.06528 (165*2π)/9156 weeks
166.11873 -.02429 (166*2π)/9156 weeks
167.09852 .04718 (167*2π)/9155 weeks
168.07925 .10591 (168*2π)/9155 weeks
169-.03109 .07358 (169*2π)/9155 weeks
170-.04159 -.00433 (170*2π)/9155 weeks
171.01398 -.06473 (171*2π)/9155 weeks
172.0555 .0031 (172*2π)/9155 weeks
173-.02135 -.00622 (173*2π)/9155 weeks
174.02029 -.03915 (174*2π)/9155 weeks
175.07899 -.0422 (175*2π)/9155 weeks
176.07249 .06233 (176*2π)/9155 weeks
177-.03246 .08787 (177*2π)/9155 weeks
178-.07673 .02483 (178*2π)/9155 weeks
179-.03234 -.04409 (179*2π)/9155 weeks
180-.00883 -.01952 (180*2π)/9155 weeks
181.03215 -.01669 (181*2π)/9155 weeks
182.00758 .00365 (182*2π)/9155 weeks
183.06842 -.00842 (183*2π)/9155 weeks
184.07514 .05177 (184*2π)/9155 weeks
185.05114 .11794 (185*2π)/9155 weeks
186-.0881 .07254 (186*2π)/9155 weeks
187-.0603 -.02446 (187*2π)/9155 weeks
188.04756 -.10362 (188*2π)/9155 weeks
189.10344 .00482 (189*2π)/9155 weeks
190.02134 .02687 (190*2π)/9155 weeks
191.04917 .02111 (191*2π)/9155 weeks
192.04009 .02277 (192*2π)/9155 weeks
193-.0125 .02885 (193*2π)/9155 weeks
194-.00008 -.03279 (194*2π)/9155 weeks
195.02754 -.02198 (195*2π)/9155 weeks
196.05339 -.00687 (196*2π)/9155 weeks
197.05304 .03905 (197*2π)/9155 weeks
198.02529 .03144 (198*2π)/9155 weeks
199.02271 .0384 (199*2π)/9155 weeks
200.00599 .0667 (200*2π)/9155 weeks
201-.05381 .00854 (201*2π)/9155 weeks
202.01729 -.05572 (202*2π)/9155 weeks
203.08611 -.02151 (203*2π)/9155 weeks
204.07204 .04569 (204*2π)/9154 weeks
205.04446 .06025 (205*2π)/9154 weeks
206.01007 .02826 (206*2π)/9154 weeks
207-.01002 -.00392 (207*2π)/9154 weeks
208.02215 -.02247 (208*2π)/9154 weeks
209.06076 -.00512 (209*2π)/9154 weeks
210.054 .03984 (210*2π)/9154 weeks
211.02094 .04246 (211*2π)/9154 weeks
212-.00854 .04251 (212*2π)/9154 weeks
213-.03047 -.00269 (213*2π)/9154 weeks
214-.00223 .0046 (214*2π)/9154 weeks
215.05529 -.0201 (215*2π)/9154 weeks
216.05089 .06708 (216*2π)/9154 weeks
217-.00174 .06359 (217*2π)/9154 weeks
218-.00845 -.01801 (218*2π)/9154 weeks
219.05976 .01762 (219*2π)/9154 weeks
220-.02663 .0351 (220*2π)/9154 weeks
221.01248 -.05285 (221*2π)/9154 weeks
222.05908 -.02506 (222*2π)/9154 weeks
223.05466 .02888 (223*2π)/9154 weeks
224.02005 .07461 (224*2π)/9154 weeks
225-.05131 .02528 (225*2π)/9154 weeks
226-.02327 -.05615 (226*2π)/9154 weeks
227.0034 -.02294 (227*2π)/9154 weeks
228.05958 .00273 (228*2π)/9154 weeks
229.03129 .04536 (229*2π)/9154 weeks
230.00104 .04536 (230*2π)/9154 weeks
231.02315 .03795 (231*2π)/9154 weeks
232-.01546 .05936 (232*2π)/9154 weeks
233-.03149 .00913 (233*2π)/9154 weeks
234-.00679 -.00973 (234*2π)/9154 weeks
235.03683 -.00593 (235*2π)/9154 weeks
236.06988 .01958 (236*2π)/9154 weeks
237.07299 .04996 (237*2π)/9154 weeks
238-.01149 .10937 (238*2π)/9154 weeks
239-.03138 .01792 (239*2π)/9154 weeks
240-.02654 .01376 (240*2π)/9154 weeks
241-.00964 -.02789 (241*2π)/9154 weeks
242.01813 -.01372 (242*2π)/9154 weeks
243.04398 -.02722 (243*2π)/9154 weeks
244.08342 .00897 (244*2π)/9154 weeks
245.06126 .09274 (245*2π)/9154 weeks
246-.0201 .09382 (246*2π)/9154 weeks
247-.08447 .03947 (247*2π)/9154 weeks
248-.033 -.01399 (248*2π)/9154 weeks
249-.02987 -.03063 (249*2π)/9154 weeks
250.08087 -.04581 (250*2π)/9154 weeks
251.05567 .02112 (251*2π)/9154 weeks
252.04832 .06933 (252*2π)/9154 weeks
253-.02984 .06861 (253*2π)/9154 weeks
254-.03056 .03418 (254*2π)/9154 weeks
255-.05537 -.0372 (255*2π)/9154 weeks
256.02563 -.07503 (256*2π)/9154 weeks
257.07365 -.03 (257*2π)/9154 weeks
258.06262 .0485 (258*2π)/9154 weeks
259-.01208 .07773 (259*2π)/9154 weeks
260-.03266 .0273 (260*2π)/9154 weeks
261-.05363 .01064 (261*2π)/9154 weeks
262-.00972 -.06511 (262*2π)/9153 weeks
263.03475 -.0094 (263*2π)/9153 weeks
264.05205 .01555 (264*2π)/9153 weeks
265.0351 .04734 (265*2π)/9153 weeks
266.00826 .08945 (266*2π)/9153 weeks
267-.03268 .03154 (267*2π)/9153 weeks
268-.02971 .00064 (268*2π)/9153 weeks
269-.02072 -.04378 (269*2π)/9153 weeks
270.05413 -.04657 (270*2π)/9153 weeks
271.10945 .02048 (271*2π)/9153 weeks
272.04838 .09655 (272*2π)/9153 weeks
273-.03811 .05916 (273*2π)/9153 weeks
274-.06134 -.00075 (274*2π)/9153 weeks
275-.03372 -.02795 (275*2π)/9153 weeks
276.01059 -.0648 (276*2π)/9153 weeks
277.07303 -.01593 (277*2π)/9153 weeks
278.05179 .04903 (278*2π)/9153 weeks
279-.00216 .03265 (279*2π)/9153 weeks
280.01469 .04996 (280*2π)/9153 weeks
281-.03197 .04829 (281*2π)/9153 weeks
282-.02255 .00741 (282*2π)/9153 weeks
283-.05186 -.01869 (283*2π)/9153 weeks
284.02719 -.05317 (284*2π)/9153 weeks
285.08125 .00482 (285*2π)/9153 weeks
286.03605 .04206 (286*2π)/9153 weeks
287.02727 .03816 (287*2π)/9153 weeks
288-.01563 .058 (288*2π)/9153 weeks
289-.05306 .02134 (289*2π)/9153 weeks
290-.03065 -.03158 (290*2π)/9153 weeks
291-.01496 -.05149 (291*2π)/9153 weeks
292.05098 -.04166 (292*2π)/9153 weeks
293.04918 .04242 (293*2π)/9153 weeks
294-.0122 .04161 (294*2π)/9153 weeks
295-.01055 .01912 (295*2π)/9153 weeks
296-.01089 .04042 (296*2π)/9153 weeks
297-.07295 -.00926 (297*2π)/9153 weeks
298.01229 -.05532 (298*2π)/9153 weeks
299.04145 -.00914 (299*2π)/9153 weeks
300.05467 .02023 (300*2π)/9153 weeks
301.00869 .04292 (301*2π)/9153 weeks
302.00388 .04145 (302*2π)/9153 weeks
303-.03264 .01503 (303*2π)/9153 weeks
304-.00555 -.0453 (304*2π)/9153 weeks
305.00395 -.04558 (305*2π)/9153 weeks
306.06458 -.04016 (306*2π)/9153 weeks
307.06944 .04964 (307*2π)/9153 weeks
308-.0043 .04443 (308*2π)/9153 weeks
309-.04017 .01397 (309*2π)/9153 weeks
310.01399 -.02159 (310*2π)/9153 weeks
311-.00653 -.00832 (311*2π)/9153 weeks
312.0371 -.00844 (312*2π)/9153 weeks
313.01577 .03299 (313*2π)/9153 weeks
314-.02646 .01413 (314*2π)/9153 weeks
315.00296 -.00086 (315*2π)/9153 weeks
316-.00973 .03392 (316*2π)/9153 weeks
317-.01748 -.0325 (317*2π)/9153 weeks
318.03573 -.01991 (318*2π)/9153 weeks
319.0447 .00086 (319*2π)/9153 weeks
320.02067 .03364 (320*2π)/9153 weeks
321.03692 .02582 (321*2π)/9153 weeks
322.01835 .04891 (322*2π)/9153 weeks
323-.00393 .03865 (323*2π)/9153 weeks
324-.04225 .01446 (324*2π)/9153 weeks
325-.00554 -.07572 (325*2π)/9153 weeks
326.06883 -.01037 (326*2π)/9153 weeks
327.0255 .04165 (327*2π)/9153 weeks
328.00355 .01384 (328*2π)/9153 weeks
329.01958 .0217 (329*2π)/9153 weeks
330-.02488 .0286 (330*2π)/9153 weeks
331-.04281 -.01265 (331*2π)/9153 weeks
332-.01879 -.0395 (332*2π)/9153 weeks
333.0295 -.02451 (333*2π)/9153 weeks
334.04673 .0359 (334*2π)/9153 weeks
335-.02625 .0433 (335*2π)/9153 weeks
336.00116 -.00381 (336*2π)/9153 weeks
337.0046 .02467 (337*2π)/9153 weeks
338-.03023 .01715 (338*2π)/9153 weeks
339-.01436 -.01633 (339*2π)/9153 weeks
340.02712 -.02968 (340*2π)/9153 weeks
341.03111 -.00545 (341*2π)/9153 weeks
342.03254 .02997 (342*2π)/9153 weeks
343-.00933 .02875 (343*2π)/9153 weeks
344-.02451 -.03956 (344*2π)/9153 weeks
345.02107 -.02691 (345*2π)/9153 weeks
346.00981 -.00895 (346*2π)/9153 weeks
347.04606 -.01566 (347*2π)/9153 weeks
348.00268 .0217 (348*2π)/9153 weeks
349-.00612 -.00307 (349*2π)/9153 weeks
350-.00218 -.00368 (350*2π)/9153 weeks
351.0272 .0143 (351*2π)/9153 weeks
352-.01152 .04252 (352*2π)/9153 weeks
353-.0247 -.01277 (353*2π)/9153 weeks
354.01516 -.01069 (354*2π)/9153 weeks
355.02182 .00419 (355*2π)/9153 weeks
356.00872 .01773 (356*2π)/9153 weeks
357-.00655 .00979 (357*2π)/9153 weeks
358-.01784 -.0344 (358*2π)/9153 weeks
359.04188 -.0435 (359*2π)/9153 weeks
360.05571 .00816 (360*2π)/9153 weeks
361.01268 .03249 (361*2π)/9153 weeks
362-.00901 -.00566 (362*2π)/9153 weeks
363-.00772 .00839 (363*2π)/9153 weeks
364.01404 -.01729 (364*2π)/9153 weeks
365.00832 .03356 (365*2π)/9153 weeks
366-.01038 .00563 (366*2π)/9153 weeks
367-.01024 -.01718 (367*2π)/9152 weeks
368.02453 .00247 (368*2π)/9152 weeks
369.00473 .00482 (369*2π)/9152 weeks
370.01448 -.01597 (370*2π)/9152 weeks
371.04628 .01234 (371*2π)/9152 weeks
372.00558 .02567 (372*2π)/9152 weeks
373-.00323 .00244 (373*2π)/9152 weeks
374-.01707 -.01774 (374*2π)/9152 weeks
375.01208 -.0284 (375*2π)/9152 weeks
376.04006 -.0208 (376*2π)/9152 weeks
377.02159 .03236 (377*2π)/9152 weeks
378-.03691 -.00904 (378*2π)/9152 weeks
379-.00052 -.02404 (379*2π)/9152 weeks
380.00894 -.02447 (380*2π)/9152 weeks
381.04254 .00068 (381*2π)/9152 weeks
382.01001 .03087 (382*2π)/9152 weeks
383-.01291 .02167 (383*2π)/9152 weeks
384.00435 -.01117 (384*2π)/9152 weeks
385.00496 .02789 (385*2π)/9152 weeks
386-.01101 -.0051 (386*2π)/9152 weeks
387-.0023 -.003 (387*2π)/9152 weeks
388-.00427 -.01564 (388*2π)/9152 weeks
389.01748 -.01926 (389*2π)/9152 weeks
390.0457 -.01957 (390*2π)/9152 weeks
391.04606 .02183 (391*2π)/9152 weeks
392.00026 .03834 (392*2π)/9152 weeks
393-.0196 -.00754 (393*2π)/9152 weeks
394.02198 -.0368 (394*2π)/9152 weeks
395.02198 -.00434 (395*2π)/9152 weeks
396.03089 -.00201 (396*2π)/9152 weeks
397.01167 .00915 (397*2π)/9152 weeks
398.01221 .01339 (398*2π)/9152 weeks
399-.0202 .00924 (399*2π)/9152 weeks
400.00295 -.01804 (400*2π)/9152 weeks
401-.00474 -.03036 (401*2π)/9152 weeks
402.04398 -.03012 (402*2π)/9152 weeks
403.06495 .00914 (403*2π)/9152 weeks
404.02179 .0407 (404*2π)/9152 weeks
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• Status: Solved
• Priority: Medium
• Security: Public
• Views: 1114
# Nested loop in C# ---- Console Application
Hi Experts!
How would I change this code to display the balances for three people A,B,C? ask the for their deposits, but it needs to loop until the N is entered by the user, calculate the interest rate of 10%, and finally total the balance plus the interest rate. for A,B, or C.... then exit
``````using System;
public class LoopingBankBal
{
public static void Main()
{
double bankBal = 1000;
const double INT_RATE = 0.04;
string inputString;
char response;
Console.Write("Do you want to see your balance? Y or N ...");
response = Convert.ToChar(inputString);
while(response == 'Y')
{
Console.WriteLine("Bank balance is {0}", bankBal.ToString("C"));
bankBal = bankBal + bankBal * INT_RATE;
Console.Write("Do you want to see next year’s balance? Y or N ...");
response = Convert.ToChar(inputString);
}
Console.WriteLine("Have a nice day!");
}
}
``````
0
December2000
• 6
• 5
• 3
2 Solutions
Commented:
I'm a bit confused with what's required.
BTW, is it an assignment? :)
The current code is wrong. It will show the same value every time you enter "Y". The current code never asks to enter any deposit amounts...
0
Commented:
So I don't want to just give you the code in case this is a homework assignment. But I can try to answer your questions:
The first thing I would do is create a method that collected the deposit information for a single person and I would call that method from Main 3 times, once for each person. For discussion purposes I'll call it CollectDeposits and it could return a Decimal type value. The loop until N is entered would be inside the CollectDeposits method.
So after we called CollectDeposits 3 times, we would have a 3 decimal values called a, b, and c.
Then you multiply each value by .1 to calculate the interest, lets call those 3 values interestA,interestB,interestC. Then you add interestA to a, interestB to b, etc. Then you WriteLine the values for each user.
0
Author Commented:
Thanks @ anarki_jimbel I worked on it and got this so far but, it is not elegant and I want to use a string N instead of the 0.... So, I figured it out pretty much.... just thought it was a better way to do it.. usually asking the question helps me go in the right direction
`````` using System;
public class TotalDeposits
{
public static void Main()
{
double purchase;
double total = 0;
string inputString;
string inputName;
const double QUIT = 0;
Console.WriteLine( "Enter Accont Name a, b, c");
Console.WriteLine("Enter deposit amount ");
purchase = Convert.ToDouble(inputString);
while (purchase != QUIT)
{
total += purchase * (1.1); // INTEREST
Console.WriteLine("Enter next deposi amount, or " +
QUIT + " to quit ");
purchase = Convert.ToDouble(inputString);
}
Console.WriteLine("{1} balance is {0}", total.ToString("C"), inputName);
}
}
``````
0
Commented:
Here is how to call a method (please forgive me if you already know).
inside Main: Decimal a = CollectDeposit();
private static Decimal CollectDeposit()
{
// get the string, convert it to a decimal and return it.
}
One (of many) ways to convert a string to a decimal is with Decimal.TryParse() which returns a false if the conversion fails.
For example:
Decimal deposit;
if (Decimal.TryParse(enteredStringValue, out deposit))
{
// deposit now contains a valid decimal
}
else
{
}
Note that I didn't actually compile this but it should be very close.
0
Commented:
Also, it might look like extra work to create a separate method for a small piece of the problem, but if you do it you will find that it will make it much easier for you to get the program working correctly.
0
Commented:
I agree to MajorBigDeal that moving some code to separate methods makes code cleaner.
But in whole your code is pretty OK. However, you may run your loop as many times as you want. And basically, for any name, not just A,B and C. Is that what you want?
If you need to restrict account names, you may have an array or List with account names (hard-coded) and every time user enters an account name you check if it is valid (i.e., it exists in the array.
0
Author Commented:
Now that I think of it... when the program QUIT's it should print the total for A, B, and C... the way it is it will only return results for one at a time. I need to reconstruct the loop or should I use a case statement for the A,B, and C?
0
Commented:
Hmmm, I'd probably use Dictionary class, something like Dictionary<string, double>,
where key is an account name (a,b or c) and the value is the account amount. Initially the dictionary have all 3 users with 0 value.
Are you allowed to use this class or any other collections?
We run a loop like for each user in dictionary keys (pseudo code):
for each string user in usersdictionary.keys
{
print user name
save amount to the dictionary:
usersdictionary[user]=amount
//do not display anything her
}
Because we have three users - the above loop runs 3 times.
After it is done - run a similar loop again and calculate and print all required data
0
Author Commented:
Never used dictionary looks cool! How do you code to save usersdictionary[user]=amount?
`````` Console.WriteLine("Enter deposit amount ");
purchase = Convert.ToDouble(inputString);
while (purchase != QUIT)
{
total += purchase * (1.1); // INTEREST
Console.WriteLine("Enter next deposi amount, or " +
QUIT + " to quit ");
purchase = Convert.ToDouble(inputString);
}
``````
As a method?
0
Author Commented:
Use it as a class?
``````Dictionary<string, string> KeyValue = new Dictionary<string, string>();
foreach (KeyValuePair<string, string> item in KeyValue) {
Console.WriteLine("{0} = {1}", item.Key, item.Value);
}
``````
This is totally cool but, I think dictionary is above my head .... Oh well back to the drawing board.
0
Commented:
Dictionary is simple. Let me a bit time to give you an example :)
0
Commented:
OK, look at this example. I show here how to use a Dictionary. By the way, Lists and Dictionaries are very extensively used
`````` class Program
{
static void Main(string[] args)
{
// cretae a dictionary where key is a account name and value - account amount
// amout is initially zero
Dictionary<string, double> accounts = new Dictionary<string, double>();
// checked what we have
foreach (KeyValuePair<string, double> item in accounts)
{
Console.WriteLine("Account {0}, amount = {1}", item.Key, item.Value);
}
// enter new amounts:
// We cannot modify the dictionary in foreach loop
// Therefore copy keys to a list:
List<string> accountNames = new List<string>(accounts.Keys);
foreach(string accountName in accountNames)
{
Console.WriteLine(" Please enter new amount for the Account {0}", accountName);
}
// check again
foreach (KeyValuePair<string, double> item in accounts)
{
Console.WriteLine("Account {0}, amount = {1}", item.Key, item.Value);
}
}
}
``````
0
Author Commented:
Thank you! Sort of lost as to where I place the loop with a dictionary... Meaning every account should be able to loop until all the deposits are complete and the accounts are totaled at the end. This whole concept is cool though....
0
Author Commented:
Thank you!
0
## Featured Post
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• 3
Tackle projects and never again get stuck behind a technical roadblock.
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| 2.6875
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| 84,430,486
| 12,445
|
# FHSST Physics/Waves/Definition
Waves and Wavelike Motion The Free High School Science Texts: A Textbook for High School Students Studying Physics. Main Page - << Previous Chapter (Units) - Next Chapter (Vectors) >> Definition - Types of Waves - Properties of Waves - Practical Applications: Sound Waves - Practical Applications: Electromagnetic Waves - Equations and Quantities
# What are waves?
Waves are phenomena that everyone experiences constantly; water waves, sound waves, light waves, human waves when the home team scores... the list goes on. When asked what makes a wave a wave, the most common responses would probably be that a wave is something that moves, or propagates, or perhaps that it is something that repeats over and over again. These properties do capture the essential qualities of waves. Now we must determine these properties quantitatively, and discover what governs their behavior.
Generally, a wave is defined as any phenomenon which can be modeled by a function of the form $f(\overline{k} \overline{r} - \omega t)$ where the r-vector represents a position in space, and t represents a time, and the k-vector and omega are both constants. Don't be intimidated by the vectors in the argument - most of our time at first will be spent on one-dimensional waves. If the wave is in only one spatial dimension 'x', for instance a wave travelling on a taut string, it can be written simply as $f(k x - w t)$.
Any function of this form "propagates" along the $\overline{k}$ direction over time. As time increases, the argument of the function increases; over time the form of the function effectively advances through space. Try coming up with functions of this form, and plot them at time t = 0, then plot them again at a later time. This progressive property will become obvious. Try to figure out the velocity with which your function advances! (we will study this later) The negative sign in front of the time term causes the wave to propagate in the direction defined as positive (if that seems confusing, try plotting more functions over time, and examine the results). If you replace the negative with a positive (or instead consider a negative value of omega), the wave will propagate in the negative direction.
A very special and important case of a wave is the mathematical function $f(\overline{k} \overline{r} - \omega t) = sin(\overline{k} \overline{r} - \omega t)$, or in one dimension, $f(k x - \omega t) = sin(k x - \omega t)$. This is a sinusoidal wave - it oscillates up and down infinitely in both directions, and moves as time progresses. I mentioned that waves have the quality of repeating over and over, the quality of periodicity. However, many functions of the form mentioned above do not seem to repeat. However, you will find that ALL waves can be decomposed into a sum of many of these simple, infinitely repeating waves when you learn about Fourier transformations.
More than any other concept, physicists are finding that waves characterize the structure of the universe at every scale imaginable. As you learn about the physics of waves in everyday life, keep an open mind towards finding waves and wave behavior everywhere you turn.
Let's consider a very well-known case of a wave phenomenon: water waves. Waves in water consist of moving peaks and troughs. A peak is a place where the water rises higher than when the water is still and a trough is a place where the water sinks lower than when the water is still.
So waves have peaks and troughs. This could be our first property for waves. The following diagram shows the peaks and troughs on a wave.
In physics we try to be as quantitative as possible. If we look very carefully we notice that the height of the peaks above the level of the still water is the same as the depth of the troughs below the level of the still water.
Waves are repetitions of physical quantity in a periodic manner, carrying energy in the process. The water waves, for example, can be visualized to repeat any of the physical quantities like "peaks", "troughs", "potential energy" or "kinetic energy". Even, we can visualize water waves as the motion of disturbance (energy). It is the energy aspect of waves that is central to the understanding of different types of waves, many of which are not visible.
Looking closely at the water wave, we can recognize that crests and troughs basically represent of extreme potential and kinetic energies in addition to representing rise and fall of water from the still level. At the peak, energy is only potential, whereas energy is only kinetic at the trough. Similarly, propagation of electromagnetic wave is associated with repetitions of magnetic and electric field in space with certain periodicity. As existence of electrical or magnetic fields does not require any medium, electromagnetic waves can move even in the absence of any medium.
## Characteristics of Waves : Amplitude
We use symbols agreed upon by convention to label the characteristic quantities of the waves. The characteristic height of a peak and depth of a trough is called the amplitude of the wave. The vertical distance between the bottom of the trough and the top of the peak is twice the amplitude. To put it simply, the amplitude is the distance of the wave from the medium, to the crest or trough
Worked Example 1
Question: (NOTE TO SELF: Make this a more exciting question) The height of the wave from the medium is 2m. What is the distance from the peak to the trough. What
The amplitude is 2m. (Read above paragraph to know why). The distance from the peak to trough is 4m.
## Characteristics of Waves : Wavelength
Look a little closer at the peaks and the troughs. The distance between two adjacent (next to each other) peaks is the same no matter which two adjacent peaks you choose. So there is a fixed distance between the peaks.
Similarly, you'll notice that the distance between two adjacent troughs is the same no matter which two troughs you look at. But, more importantly, its is the same as the distance between the peaks. This distance which is a characteristic of the wave is called the wavelength.
Waves have a characteristic wavelength. The symbol for the wavelength is the Greek letter lambda, $\lambda$.
The wavelength is the distance between any two adjacent points which are in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They don't have to be peaks or trough but they must be separated by a complete number of waves.
## Characteristics of Waves : Period
Now imagine you are sitting next to a pond and you watch the waves going past you. First one peak, then a trough and then another peak. If you measure the time between two adjacent peaks you'll find that it is the same. Now if you measure the time between two adjacent troughs you'll find that its always the same, no matter which two adjacent troughs you pick. The time you have been measuring is the time for one wavelength to pass by. We call this time the period and it is a characteristic of the wave.
The period of the wave is denoted with the symbol $T$.
## Characteristics of Waves : Frequency
There is another way of characterising the time interval of a wave. We timed how long it takes for one wavelength to go past. We could also turn this around and say how many waves go by in 1 second.
We can easily determine this number, which we call the frequency and denote f. To determine the frequency, how many waves go by in 1s, we work out what fraction of a waves goes by in 1 second by dividing 1 second by the time it takes T. If a wave takes 1/2 a second to go by then in 1 second two waves must go by. $\frac{1}{\frac{1}{2}} =2$. The unit of frequency is the Hz or s-1.
Waves have a characteristic frequency.
$f=\frac{1}{T}$
f : frequency (Hz or s-1) T : period (s)
generally, the frequency of a wave is the number of crests that pass by per unit time.
## Characteristics of Waves : Speed
Now if you are watching a wave go by you will notice that they move at a constant velocity. Thinking back to rectilinear motion you will be able to remember that we know how to work out how fast something moves. The speed is the distance you travel divided by the time you take to travel that distance. This is excellent because we know that the waves travel a distance $\lambda$ in a time T. This means that we can determine the speed.
$v = \frac{\lambda}{T}$
v : speed (m.s-1) $\lambda$ : wavelength (m) T : period (s)
There are a number of relationships involving the various characteristic quantities of waves. A simple example of how this would be useful is how to determine the velocity when you have the frequency and the wavelength. We can take the above equation and substitute the relationship between frequency and period to produce an equation for speed of the form
$v = f\lambda$
v : speed (m.s-1) $\lambda$ : wavelength (m) f : frequency (Hz or s-1)
Is this correct? Remember a simple first check is to check the units! On the right hand side we have speed which has units ms-1. On the left hand side we have frequency which is measured in s-1 multiplied by wavelength which is measure in m. On the left hand side we have ms-1 which is exactly what we want.
### Speed of a wave through strings
The speed of a wave traveling along a vibrating string (v) is directly proportional to the square root of the tension (T) over the linear density (μ):
$v=\sqrt{\frac{T}{\mu}} \,$
μ is equal to the mass of the string divided by the length of the string.
$\mu={\frac{M}{L}}$
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22. ### I Self balancing stick based on flywheels
ah yes I understand now. thanks.
23. ### I Self balancing stick based on flywheels
Sorry, I was confusing it with the cubli, which minimizes the flywheel velocities so that gyroscopic effects are negligible. I was wondering, if you stopped just short of the motor reaching max rotational speed, would there be a way to "reset" the default rotational speed back to zero, or are...
24. ### I Self balancing stick based on flywheels
that was not what I asked. my question is about how a restoring torque is limited by how long the motor can speed up for. what I mean is, the flywheel can only general a restoring torque through ACCELERATION, what this implies, is if you need to generate a constant restoring force, you need to...
26. ### I Self balancing stick based on flywheels
so I recently saw a video where a stick had two flywheels attached on the top, they would accelerate or decelerate based on the pitch of the stick. my question= if you delivered enough impulse do the stick, would the stick stop being able to balance? since the restoring torque is based on...
27. ### Interactive Holograms in Iron Man
I assume most people know what I'm talking about. I've done some mild googling regarding hologram technology, but currently technology isn't nearly as robust as what Stark has in the movies, and I'm not so interested in that aspect anyways. How would you set up a room so that you can interact...
28. ### Programs Tell me (EE major) some reasons to take more physics classes
My situation: I am an EE major, first year, 6 quarters ahead in math (meaning i have 6 "extra" slots for electives). I will probably take many CS courses, or maybe switch into a combined CS and EE major (like ECE or CSE), partially due to interest and necessity of coding skills nowadays. I also...
29. ### Studying Necessity of formal math education for Electrical Engineer?
So, I am majoring in Electrical engineering, and probably doing a decent amount of Physics and Computer science courses until I can no longer squeeze it into my schedule. In the far future, I definitely plan to work in industry, and hopefully in R&D type engineering work. Will I ever in my...
30. ### B How to track the center of mass of a human in a video
So a few months or years ago I learned that most of the energy from swinging comes from raising the center of mass at the lowest point of a swing. I also recently found a method of swinging that maximizes this energy gain. You squat on the swing, and stand up every time it gets to the lowest...
31. ### Calculus Reviewing Multivariable calculus to skip in college
Hi, I'm asking for a friend who will be majoring in chemical engineering. We have already taken Calculus I, II, and III under a course offered by a local community college. Admittedly, it was taught from stewart's series of calculus books, and we did exactly zero proofs in the class, and all...
32. ### B Viability of log-log transformation for some data
I have taken AP Statistics, and this is for the final project. What we have learned consists of some simple significant tests (t test, z test for proportions, two sample tests, chi squared, and logarithmic transforms). My partner and I are considering creating a scatter plot of distance between...
34. ### B Dimension of subset containing two circles
Clarification: I meant just the two curves. Not the space in between, but I see from your explanation that it is one dimensional. Thanks.
35. ### B Dimension of subset containing two circles
So I am reading a calculus book, and went online to find explanations for why a circle is 1D. Theres the explanations that say something about zooming in very close and seeing that it's indistinguishable from a Real line. Or you can specify any point on it with only one variable Or if there was...
36. ### How to measure the output of a generator
Some clarifications: We are using a box fan as the wind source. The blades will only sweep out a circle with a 15 inch diameter. (it was even smaller in science olympiad, so this feels rather liberating) Yes, we won't be trying to use this to produce usable electricity. We are varying the...
37. ### How to measure the output of a generator
I am doing a math project. Our goal is to use Mathematica to design some small wind turbine blades, 3D print them, and then attaching it to a small dc motor. I did something very similar for science Olympiad, where we had been measuring the "power" by measuring the voltage between the two wires...
38. ### Can Vibrating Shoes Help Us Walk on Water?
Would shoes that vibrate very quickly enable us to walk on water? Why or why not?
39. ### I Two sheets of glass stick together. Why?
but going back to the original question, the two most relevant forces are then just adhesion (much smaller), and air pressure, right?
40. ### I Two sheets of glass stick together. Why?
I think i recall approximately that the force of an infinite charged plane on a charge L away from the plane is the same as the force of a sphere of radius L/2 on the point, if the point is right on the surface of the sphere. It should be trivial to extend this to mass, since the gravity and...
41. ### I Two sheets of glass stick together. Why?
When the sheets are very smooth and are pressed together strongly, I guess they could be viewed somewhat as a unit, so the force to pull them apart would be approximately equal to or less than the atmospheric pressure times the area of the sheets, but definitely not more than, is this correct...
42. ### Constructing a graph of WiFi strength
Wow thanks for all the input, once my college applications are done I'll probably dabble a bit in Java and Android app developer and get started on the project.
43. ### Constructing a graph of WiFi strength
I have way too much time to spare, so I came up with this fun little project. What i want to do is use my phone or laptop to incrementally collect data on WiFi strength at my location at that time, using GPS (longitude and latitude), and once I have enough data, (I'm fine if it takes like...
44. ### Soda bubble formation rate on plastic straw and glass cup
Something i found while doing research, they state that PET has a specific property to absorb bubbles, but why? Plastic straws have a specific property to adsorb bubbles. This property is more significant in PET (poly ethylene terephthalate) and hence these bottles serve as good containers to...
45. ### Soda bubble formation rate on plastic straw and glass cup
I observed that if you pour soda into a glass cup and leave the straw inside, bubbles seem to form on the plastic cup at a much much quicker rate. Why is this? Why do bubbles form on the straw at all? Not sure if this should be in general physics or chemistry
46. ### B Trash can fire tornado
but also, the fire tends to get much bigger once it starts spinning, why is that?
47. ### B Trash can fire tornado
Why does this happen? When the trash can is not moving, the air gets drawn in through the mesh and flows more or less straight to the fire. I'm thinking that a rotating trashcan "pushes" the wind in the direction that it's rotating, so that the wind no longer blows towards the fire, but...
48. ### Programs Physics minor or physics major
Perhaps you might consider that I'm not taking it to get better at ee, but rather for interest or knowledge? The lead researcher at my dad's company that makes ultrasound oceanography devices took a lot of physics classes and says it's been useful very frequently
49. ### Why the temperature of a moving jar of gas doesn't increase
Ah thanks, makes much more sense than my friends explanation haha
50. ### Why the temperature of a moving jar of gas doesn't increase
The frame in which the average velocity is zero, so the jar.
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## What is VSWR?
VSWR is defined as the ratio of the maximum voltage to the minimum voltage in standing wave pattern along the length of a transmission line structure. It varies from 1 to (plus) infinity and is always positive. Unless you have a piece of slotted line-test equipment this is a hard definition to use, especially since the concept of voltage in a microwave structure has many interpretations.
Sometimes VSWR is called SWR to avoid using the term voltage and to instead use the concept of power waves. This in turn leads to a mathematical definition of VSWR in terms of a reflection coefficient. A reflection coefficient is defined as the ratio of reflected wave to incident wave at a reference plane. This value varies from -1 (for a shorted load) to +1 (for an open load), and becomes 0 for matched impedance load. It is a complex number. This helps us because we can actually measure power.
The reflection coefficient, commonly denoted by the Greek letter gamma (Γ), can be calculated from the values of the complex load impedance and the transmission line characteristic impedance which in principle could also be a complex number.
Γ = (Zl – Z0)/(Zl + Z0)
The square of | Γ | is then the power of the reflected wave, the square hinting at a historical reference to voltage waves.
Now we can define VSWR (SWR) as a scalar value:
VSWR= (1 + | Γ |)/(1 – | Γ |) or in terms of s-parameters: VSWR= (1 + | S11 |)/(1 – | S11 |)
This is fine but what has it to do with common usage in ads and specifications. Generally, VSWR is sometimes used as a stand-in for a figure of merit for impedance matching. Sometimes this simplification of a scalar quantity and it’s restricted definition can lead to confusion in the matter of a source to load match. Most of the time there is no problem but, technically, VSWR derives from the ratio using the load impedance and the characteristic impedance of the transmission line in which the standing waves reside and not specifically to a source to load match. I prefer to think of VSWR as a figure of merit and to use the reflection coefficient whenever I am trying to solve problems.
By the way, if you think you have never experienced a standing wave personally, it’s very unlikely. Standing waves in a microwave oven are the reason that food is cooked unevenly (the turntable is a partial solution to that problem). The wavelength of the 2.45 GHz signal is about 12 centimeters, or about five inches. Nulls in the radiation (and heating) will be separated at a distance similar to wavelength.
### Standing waves in nature
What’s a standing wave? Luckily there are tons of examples in nature. Any stringed instrument such as a guitar or piano makes music using standing waves. But what about a traveling wave that reflects off of an object and creates a standing wave due to constructive interference?
Let’s go to the beach. Breakers roll in off the ocean, come up on the sand, and disappear; no standing wave occurs. What’s happening? The beach is absorbing all (or at least most) of the energy, in effect it is “matched” to the wave front.
Now let’s go next door to marina where all of those expensive yachts are moored… chances are there are vertical concrete seawalls inside the marina to allow owners to bring their boats close enough so that only a small walkway is needed to get to them.
Now notice the breakwater that extends around the marina, with only a narrow opening for boats to go in and out. That’s there because the vertical walls in the marina offer near perfect reflection to moving waves (an “open circuit”).
Without the breakwater wall (which absorbs energy) huge standing waves are possible due to constructive interference, and all those boats would bob up and down like crazy corks and eventually everything would get smashed to tiny bits.
### Other ways to express impedance mismatches
The reflection coefficient is what you’d read from a Smith chart. A reflection coefficient with a magnitude of zero is a perfect match, a value of one is perfect reflection.
The symbol for reflection coefficient is uppercase Greek letter gamma (). Note that the reflection coefficient is a complex value, so it includes an angle.
Unlike VSWR, the reflection coefficient can distinguish between short and open circuits. A short circuit has a value of -1 (1 at an angle of 180 degrees), while an open circuit is one at an angle of 0 degrees.
Quite often we refer to only the magnitude of the reflection coefficient. The symbol for this is the lower case Greek letter ρ.
The return loss of a load is merely the magnitude of the reflection coefficient expressed in decibels. The correct equation for return loss is:
Return loss = -20 x log [mag(Γ)]
Thus in its correct form, return loss will usually be a positive number. If it’s not, you can usually blame measurement error.
The exception to the rule is something with negative resistance, which implies that it is an active device (external DC power is converted to RF) and it is potentially unstable (it could oscillate).
Not something you have to worry about if you are just looking at coax cables!
However, many engineers often omit the minus sign and talk about “-9.5 dB return loss” for example. People that find it necessary to correct engineers who do this have underwear that is too tight.
Here are the equations that convert between VSWR, reflection coefficient (Γ) and return loss (RL) as well as mismatch loss (ML), which we will cover later). Note that |Γ| in these equations always stands for the magnitude of the complex reflection coefficient and is itself a scalar term in these equations:
Let’s end our discussion with a table of reflection VSWR, refection coefficient and return loss values (and remember that our VSWR calculator can provide any values you need).
If you want to impress your friends, memorize as much of this table as you can. Yes, rounding off is permitted.
### Calculating VSWR from impedance mismatches
The mismatch of a load ZL to a source Z0 results in a reflection coefficient of:
Γ=(ZL-Z0)/(ZL+Z0)
Note that the load can be a complex (real and imaginary) impedance. If you can’t remember in which order the numerator is subtracted (did we just say “ZL-Z0” or Z0-ZL“?), you can always figure it out by remembering that a short circuit (ZL=0) is on the left side of the Smith chart (angle = -180 degrees) which means Γ=-1 in this case, which means that the minus sign belongs in front of Z0.
The magnitude of the reflection coefficient is given by:
ρ=mag(Γ)
For cases where ZL is a real number,
ρ=abs((ZL-Z0)/(ZL+Z0))
Note that “abs” means “absolute value” here. VSWR can be calculated from the magnitude of the reflection coefficient:
VSWR=(1+ρ)/(1-ρ)
For cases where ZL is real, with a little algebra you’ll see there are two cases for VSWR, calculated from load impedance:
For ZL<Z0: VSWR=Z0/ZL
For ZL>Z0: VSWR=ZL/Z0
Just remember to divide the larger impedance by the smaller impedance, because VSWR is always greater than 1. Hey, this calculation is so easy you can do it in your head!!!
Let’s look at the special case where you mix up 50 ohm parts into a 75 ohm system (or vice-versa). In either case, the resulting VSWR is 1.5:1. Yes, we did that without a calculator. While we’re at it, the reflection coefficient is:
Γ=(75-50)/(75+50)=0.2
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Can I compute on right part of a fat arrow?
Let’s assume you have this code:
``````data HList :: [*] -> * where
Cons :: p -> HList a -> HList (p ': a)
End :: HList '[]
-- convinience to not write explicit signature
class CBottom
type CSat = () :: Constraint
type CNSat = CBottom
type ConstStr = '[Char]
data Pair a b where
MkPair :: a -> b -> Pair a b
-- Check if hlist have all elements of type Pair '[Char] Type
type family RecurseAndCheck a (b :: HList t) where
RecurseAndCheck a End = CSat
RecurseAndCheck a (Cons (b :: a) r) = RecurseAndCheck a r
RecurseAndCheck a (Cons (b :: m) r) = CNSat
-- Any hlist that is constructed from Pairs of (ConstStr, Type) is a TypeClass
class (forall t p (a :: HList t) . RecurseAndCheck (Pair ConstStr p) a)
=> TypeClass a where
retrieve :: a -> (s :: ConstStr) -> Pair a s
``````
Unfortunately, compiler shows me a horrible error:
`Occurs check: cannot construct the infinite kind: t ~ a : t • In the second argument of ‘RecurseAndCheck’, namely ‘(Cons (b :: a) r)’`
So two questions:
• What can I do to vanquish the error?
• Do you see anything awkward in this code?
I think both arguments of the type family need to be types. In your case the second argument (c :: HList t) is a value not a type.
You can do this, but it still feels awkward:
``````type family RecurseAndCheck a bs where
RecurseAndCheck a (HList '[]) = CSat
RecurseAndCheck a (HList (a : bs)) = RecurseAndCheck a (HList bs)
RecurseAndCheck _ _ = CNSat
-- Any hlist that is constructed from Pairs of (ConstStr, Type) is a TypeClass
class (forall t p . RecurseAndCheck (Pair ConstStr p) (HList t))
=> TypeClass a where
retrieve :: a -> (s :: ConstStr) -> Pair a s
``````
P.S. I would define `type CNSat = Int ~ Bool`, then you cannot accidentally make an instance.
1 Like
Thank you for response, I am still in a learning phase and your remarks are helpful.
Unfortunately, the stopper is now `Quantified predicate must have a class or type variable head` and it seems to be a major hindrance here (despite my ignorance, of course). Doing troubleshooting, I found at gitlab a discussion that mentioned this error and it looks dormant. So it seems like it is not possible to compute on the left side of the arrow, even with closed type families, is this statement correct?
Oh, yes! It is definitely better.
1 Like
You’re doing some quite advanced Haskell for someone in a learning phase . I feel like I am only a few steps ahead of you in the learning process, so I can’t give any definitive answers.
I only really focused on fixing the error. Now that I’m reading it a bit better the quantification seems strange. The left side of the “fat arrow” is not connected to the right side at all. Now it is saying something like "If all heterogeneous lists consists of only pairs then you can make an instance for `TypeClass`". But clearly you can come up with some HLists that do not contain only those specific pairs. So you probably want something more like:
``````class (forall p . RecurseAndCheck (Pair ConstStr p) (HList t))
=> TypeClass t where
retrieve :: HList a -> (s :: ConstStr) -> Pair a s
``````
But this is still not really what you want (and it will still give the same error) because of that `forall p`. You probably want to define `RecurseAndCheck` slightly differently:
``````type family RecurseAndCheck bs where
RecurseAndCheck (HList '[]) = CSat
RecurseAndCheck (HList (Pair ConstStr _ : bs)) = RecurseAndCheck a (HList bs)
RecurseAndCheck _ _ = CNSat
``````
This is less general, but it allows you to completely remove the foralls from the left side of the “fat arrow”. But at this point I’m also running out of ideas because the type of `retrieve`: `a -> (s :: ConstStr) -> Pair a s` does not really make sense to me. Can you explain it a bit more?
1 Like
Sure. I thought that it’s how term level strings are represented at the type level. I just found out that mature haskellres call it Symbol, so I’ll google that. The idea was to construct a record with functions, basically.
To clarify a bit more, I made TypeClass type class to simplify the signature of a particular function.
My intention was to try to implement ‘nominal instances’ as ‘things that carry implementation’ so for example you could create multiple records with functions and later use them in context at will. Can’t make up convincing examples of how it would look like, sorry . Anyway, I was mostly trying to stretch it for fun.
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Created by zhengkevin
# Division Worksheets
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Christmas Division #5
For all you merry mathematicians, celebrate the Christmas season with a festive worksheet! Here are some simple division problems to solve.
Christmas Division #4
Enjoy a festive math sheet for the Christmas season! This one has simple division presents in store to keep the learning going over the winter break.
Christmas Division #3
Ho ho ho! Here are some colorful presents from Santa, filled with simple division problems for your third grader to solve.
Christmas Division #2
Keep the learning going over the holiday break!Santa has brought some colorful presents filled with simple division problems for your child to solve.
Christmas Division #1
Santa has brought a colorful page of simple division problems for your child to solve. Keep the learning going this holiday, but still fun!
Put your child's math skills to the test with a fun division worksheet. He'll practice some simple division problems, a great way to review times tables!
Baseball Division #4
Math super-stars, take the plate with some baseball division! Here are 12 simple division problems that will help introduce your child to a new math function.
Baseball Division #5
Take a swing at math practice with some baseball division! Here are 12 simple division problems that will help introduce your child to a new math function.
Baseball Division
Take a swing at math practice with some baseball division! Here are 12 simple division problems that will help your child review his times tables.
St. Patrick's Day Division #5
Can your little leprechaun collect all of these lucky clovers? Give him a fun challenge this St. Patrick's Day, with some division problems to solve.
Halloween Division #4
Don't be doomed by division! Help your child stay sharp this Halloween with a spooky division practice sheet.
St. Patrick's Day Division #2
Give your little lucky charm a fun way to practice math this St. Patrick's Day!
Halloween Division #2
Calling all division demons! Stay sharp this Halloween with a spooky division practice sheet for beginners.
Valentine's Day Division #2
Prepare your third grader for Valentine's Day with these festive division problems, a great way to help her review her multiplication tables.
Thanksgiving Division #5
Get into the spirit of the Thanksgiving season with a helpful math sheet full of festive division problems! This will help her review her times tables.
Thanksgiving Division #3
This festive division worksheet will keep learning alive for your math student as she calls on her memory of her multiplication tables to solve each problem.
Football Division #3
For all you sports fans, here's a great football math sheet that will challenge your child's basic division skills.
Football Division #2
Kick off the school season right with a fun, football math sheet. Your student can practice some basic division problems, a great way to review times tables!
Football Division #1
Is your child going pro with his times tables? Give him a new challenge with some football division problems, a perfect way to review multiplication tables.
Is your math whiz acing his times tables? Give him a great challenge with this basketball division sheet, where he'll practice division facts.
Take a break from boring textbook math, and practice division with a basketball math sheet! Your child will work on simple division.
Get your child pumped up for math practice with a fun basketball division sheet! He'll try to complete all the division equations as quick as possible.
Halloween Division #5
If your child has mastered his multiplication tables, help him stay sharp this Halloween with a spooky division practice sheet.
Halloween Division #3
For a monster math whiz who has mastered his multiplication tables, stay sharp this Halloween with a spooky division practice sheet.
Halloween Division
Division isn't scary ... unless it's surrounded by ghosts, pumpkins, and tombstones! Do some Halloween-themed division in this math worksheet.
Football Division #5
If your child is a times tables pro, toss him a challenge with some football division! He'll be kicking "field goals" as he solves each basic division problem.
Football Division #4
Division can be a tough concept! Give your child a kick-off with football division, where he'll practice basic division problems.
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0
# When you reverse two-digit prime numbers are they also prime?
Wiki User
2016-04-16 15:44:04
No. For example, 61 is a Prime number; 16 is not a prime number.
Wiki User
2016-11-19 18:16:55
Study guides
4 cards
## What is the prime factorization of 954
➡️
See all cards
4.22
32 Reviews
Wiki User
2016-08-06 22:45:31
Not necessarily. For example, 23 is prime but 32 not.
Wiki User
2016-07-03 10:06:45
No. 23 is a prime but 32 is not.
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# How do you solve the following system: 4x+y=-7, 2x + 3y = 8 ?
Apr 25, 2018
$\left(- 2.9 , 4.6\right)$
#### Explanation:
Rearrange the second equation to get:
$2 x = 8 - 3 y$
Also:
$2 \left(2 x\right) + y = - 7$
$2 \left(8 - 3 y\right) + y = - 7$
$16 - 6 y + y = - 7$
$- 5 y = - 23$
$y = \frac{23}{5} = 4.6$
Now we put this in:
$4 x + \frac{23}{5} = - 7$
$4 x = - 7 - \frac{23}{5} = \frac{- 35 - 23}{5} = - \frac{58}{5}$
$x = - \frac{58}{20} = - 2.9$
$\left(- 2.9 , 4.6\right)$
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# Homework Help: Particle at rest; v(t) = 0
1. Oct 28, 2013
### Sage Hopkins
(1.)
I have a "particle in motion" problem that is asking me when a particle is at rest, which I understand to be when velocity = v(t) = 0, so
v(t) = - (π/4) sin (πt/4) = 0.
The given answer is as follows:
- (π/4) sin (πt/4) = 0
sin (πt/4) = 0
πt/4 = πn.
t = 0,4,8 seconds.
(2.) Can someone please explain to me how 0 becomes πn, and/or what specific mathematical concept(s) I need to review?
2. Oct 28, 2013
### vanhees71
What's the full problem statement?
Of course $\sin(n \pi)=0$ for all $n \in \mathbb{Z}$.
3. Oct 28, 2013
### Sage Hopkins
The full problem statement goes:
A particle moves according to a law of motion s = cos(πt/4), t >= 0, where t is measured in seconds and s in feet.
There are several sub-questions from here about velocity, acceleration, graphs, etc.., but the one that I got stuck on is
(c.) When is the particle at rest for t <= 10.
After differentiating the given function for f', understanding the answer to this question was as simple as reviewing the unit circle and the graph of sin for me, as elementary as it may be.
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Third edition of Artificial Intelligence: foundations of computational agents, Cambridge University Press, 2023 is now available (including the full text).
### 7.2.1 Evaluating Predictions
If e is an example, a point estimate for target feature Y is a prediction of a particular value for Y on e. Let pval(e,Y) be the predicted value for target feature Y on example e. The error for this example on this feature is a measure of how close pval(e,Y) is to val(e,Y), where val(e,Y) is the actual value for feature Y in e.
For regression, when the target feature Y is real valued, both pval(e,Y) and val(e,Y) are real numbers that can be compared arithmetically.
For classification, when the target feature Y is a discrete variable, a number of alternatives exist:
• When Y is binary, one value can be associated with 0, the other value with 1, and a prediction can be some real number. The predicted and actual values can be compared numerically.
• When the domain of Y has more than two values, sometimes the values are totally ordered and can be scaled so a real number can be associated with each value of the domain of Y. In this case, the predicted and actual values can be compared on this scale. Often, this is not appropriate even when the values are totally ordered; for example, suppose the values are short, medium, and long. The prediction that the value is short ∨ long is very different from the prediction that the value is medium.
• When the domain of Y is {v1,...,vk}, where k>2, a separate prediction can be made for each vi. This can be modeled by having a binary indicator variable associated with each vi which, for each example, has value 1 when the example has value vi and the indicator variable has value 0 otherwise. For each training example, exactly one of the indicator variables associated with Y will be 1 and the others will be 0. A prediction gives k real numbers - one real number for each vi.
Example 7.2: Suppose the trading agent wants to learn a person's preference for the length of holidays. Suppose the holiday can be for 1, 2, 3, 4, 5, or 6 days.
One representation is to have a real-valued variable Y that is the number of days in the holiday.
Another representation is to have six real-valued variables, Y1,...,Y6, where Yi represents the proposition that the person would like to stay for i days. For each example, Yi=1 when there are i days in the holiday, and Yi=0 otherwise.
The following is a sample of five data points using the two representations:
Example Y e1 1 e2 6 e3 6 e4 2 e5 1
Example Y1 Y2 Y3 Y4 Y5 Y6 e1 1 0 0 0 0 0 e2 0 0 0 0 0 1 e3 0 0 0 0 0 1 e4 0 1 0 0 0 0 e5 1 0 0 0 0 0
A prediction for a new example in the first representation can be any real number, such as Y=3.2.
In the second representation, the learner would predict a value for each Yi for each example. One such prediction may be Y1=0.5, Y2=0.3, Y3=0.1, Y4=0.1, Y5=0.1, and Y6=0.5. This is a prediction that the person may like 1 day or 6 days, but will not like a stay of 3, 4, or 5 days.
In the following definitions, E is the set of all examples and T is the set of target features.
There are a number of prediction measures that can be defined:
• The absolute error on E is the sum of the absolute errors of the predictions on each example. That is,
e∈EY∈T|val(e,Y)-pval(e,Y)|.
This is always non-negative, and is only zero when the predictions exactly fit the observed values.
• The sum-of-squares error on E is
e∈EY∈T (val(e,Y)-pval(e,Y))2.
This measure treats large errors as worse than small errors. An error twice as big is four times as bad, and an error 10 times as big is 100 times worse.
• The worst-case error on E is the maximum absolute error:
maxe∈E maxY∈T|val(e,Y)-pval(e,Y)|.
In this case, the learner is evaluated by how bad it can be.
Example 7.3: Suppose there is a real-valued target feature, Y, that is based on a single real-valued input feature, X. Suppose the data contains the following (X,Y) points:
(0.7,1.7),(1.1,2.4),(1.3,2.5),(1.9,1.7),(2.6,2.1),(3.1,2.3),(3.9,7).
Figure 7.2: Linear predictions for a simple prediction example. Filled circles are the training examples. P1 is the prediction that minimizes the absolute error of the training examples. P2 is the prediction that minimizes the sum-of-squares error of the training examples. P3 is the prediction that minimizes the worst-case error of the training examples. See Example 7.3.
Figure 7.2 shows a plot of the training data (filled circles) and three lines, P1, P2, and P3, that predict the Y-value for all X points. P1 is the line that minimizes the absolute error, P2 is the line that minimizes the sum-of-squares error, and P3 minimizes the worst-case error of the training examples.
Lines P1 and P2 give similar predictions for X=1.1; namely, P1 predicts 1.805 and P2 predicts 1.709, whereas the data contain a data point (1.1,2.4). P3 predicts 0.7. They give predictions within 1.5 of each other when interpolating in the range [1,3]. Their predictions diverge when extrapolating from the data. P1 and P3 give very different predictions for X=10.
The difference between the lines that minimize the various error measures is most pronounced in how they handle the outlier examples, in this case the point (3.9,7). The other points are approximately in a line.
The prediction with the least worse-case error for this example, P3, only depends on three data points, (1.1,2.4), (3.1,2.3), and (3.9,7), each of which has the same worst-case error for prediction P3. The other data points could be at different locations, as long as they are not farther away from P3 than these three points.
In contrast, the prediction that minimizes the absolute error, P1, does not change as a function of the actual Y-value of the training examples, as long as the points above the line stay above the line, and those below the line stay below. For example, the prediction that minimizes the absolute error would be the same, even if the last data point was (3.9,107) instead of (3.9,7).
Prediction P2 is sensitive to all of the data points; if the Y-value for any point changes, the line that minimizes the sum-of-squares error will change.
There are a number of prediction measures that can be used for the special case where the domain of Y is {0,1}, and the prediction is in the range [0,1]. These measures can be used for Boolean domains where true is treated as 1, and false is treated as 0.
• The likelihood of the data is the probability of the data when the predicted value is interpreted as a probability:
e∈EY∈T pval(e,Y)val(e,Y)(1-pval(e,Y))(1-val(e,Y)).
One of val(e,Y) and (1-val(e,Y)) is 1, and the other is 0. Thus, this product uses pval(e,Y) when val(e,Y)=1 and (1-pval(e,Y)) when val(e,Y)=0. A better prediction is one with a higher likelihood. The model with the greatest likelihood is the maximum likelihood model.
• The entropy of the data is the number of bits it will take to encode the data given a code that is based on pval(e,Y) treated as a probability. The entropy is
-∑e∈EY∈T [val(e,Y) log pval(e,Y) + (1-val(e,Y)) log (1-pval(e,Y))].
A better prediction is one with a lower entropy.
A prediction that minimizes the entropy is a prediction that maximizes the likelihood. This is because the entropy is the negative of the logarithm of the likelihood.
• Suppose the predictions are also restricted to be {0,1}. A false-positive error is a positive prediction that is wrong (i.e., the predicted value is 1, and the actual value is 0). A false-negative error is a negative prediction that is wrong (i.e., the predicted value is 0, and the actual value is 1). Often different costs are associated with the different sorts of errors. For example, if there are data about whether a product is safe, there may be different costs for claiming it is safe when it is not safe, and for claiming it is not safe when it is safe.
We can separate the question of whether the agent has a good learning algorithm from whether it makes good predictions based on preferences that are outside of the learner. The predicting agent can at one extreme choose to only claim a positive prediction when it is sure the prediction is positive. At the other extreme, it can claim a positive prediction unless it is sure the prediction should be negative. It can often make predictions between these extremes.
One way to test the prediction independently of the decision is to consider the four cases between the predicted value and the actual value:
actual positive actual negative predict positive true positive (tp) false positive (fp) predict negative false negative (fn) true negative (tn)
Suppose tp is the number of true positives, fp is the number of false positives, fn is the number of false negatives, and tn is the number of true negatives. The precision is (tp)/(tp+fp), which is the proportion of positive predictions that are actual positives. The recall or true-positive rate is (tp)/(tp+fn), which is the proportion of actual positives that are predicted to be positive. The false-positive error rate is (fp)/(fp+tn), which is the proportion of actual negatives predicted to be positive.
An agent should try to maximize precision and recall and to minimize the false-positive rate; however, these goals are incompatible. An agent can maximize precision and minimize the false-positive rate by only making positive predictions it is sure about. However, this choice worsens recall. To maximize recall, an agent can be risky in making predictions, which makes precision smaller and the false-positive rate larger. The predicting agent often has parameters that can vary a threshold of when to make positive predictions. A precision-recall curve plots the precision against the recall as these parameters change. An ROC curve, or receiver operating characteristic curve, plots the false-positive rate against the false-negative rate as this parameter changes. Each of these approaches may be used to compare learning algorithms independently of the actual claim of the agent.
• The prediction can be seen as an action of the predicting agent. The agent should choose the action that maximizes a preference function that involves a trade-off among the costs associated with its actions. The actions may be more than true or false, but may be more complex, such as "proceed with caution" or "definitely true." What an agent should do when faced with uncertainty is discussed in Chapter 9.
Example 7.4: Consider the data of Example 7.2. Suppose there are no input features, so all of the examples get the same prediction.
In the first representation, the prediction that minimizes the sum of absolute errors on the training data presented in Example 7.2 is 2, with an error of 10. The prediction that minimizes the sum-of-squares error on the training data is 3.2. The prediction the minimizes the worst-case error is 3.5.
For the second representation, the prediction that minimizes the sum of absolute errors for the training examples is to predict 0 for each Yi. The prediction that minimizes the sum-of-squares error for the training examples is Y1=0.4, Y2=0.1, Y3=0, Y4=0, Y5=0, and Y6=0.4. This is also the prediction that minimizes the entropy and maximizes the likelihood of the training data. The prediction that minimizes the worst-case error for the training examples is to predict 0.5 for Y1, Y2, and Y6 and to predict 0 for the other features.
Thus, whichever prediction is preferred depends on how the prediction will be evaluated.
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# Help Clarifying Answers-Linear Acceleration & Conservation of Momentum
Hey all,
I’m brushing up on some physics in my spare time, and I was wondering if someone knowledgeable here could help me clarify the reasons behind a couple answers…much thanks in advance! =)
Question 1) The greatest linear acceleration of the center of mass of a baseball bat will be produced by pushing with a force F at…(picture attached)
A) position 1
B) position 2
C) position 3
D) all the same
E) not enough information
Relevant equation: Force = mass[itex]_{center of mass}[/itex] * acceleration[itex]_{center of mass}[/itex]
Attempt at solution: The solution tells me that it’s "D) all the same," and that using F = ma, you can see that linear acceleration is not related to the position where the force is applied. However, this doesn’t seem intuitive to me.
If I have a bat sitting on a table, and I push at position 1 or 3, I would have expected there to be some but less linear acceleration (and more rotational acceleration), so that position 2 would have the greatest linear acceleration. I remember reading that rotational acceleration is calculated independently of linear acceleration.
Am I overanalyzing and treating this as more than a statics problem? That as soon as it starts rotating if pushed in position 1 or 3, that the scenario is no longer valid because, realistically force F (if it continues in the same direction) is no longer pushing at the bat perpendicular to its motion? So in that case, I should think of this force more like a quick, impulsive force than a force over a prolonged period of time?
Question 2) Suppose you are on a cart, initially at rest on a track with very little friction. You throw balls at a partition that is rigidly mounted on the cart. If the balls bounce straight back as shown in the figure, is the cart put in motion? (image attached)
A) it moves to the right
B) it moves to the left
C) it remains in place
D) not enough information
Relevant equation:
Conservation of momentum: 0 = m_person+cart* v_person+cart + m_ball * v_ball
Attempt at solution: Taking the person, cart and balls as a system, there are no external forces in the direction of motion and we can say that momentum is conserved. The solution tells me that the answer is "A) it moves to the left." Again, if I use the equation to answer, I arrive at that conclusion. Or if I imagine putting a curtain covering the contraption and the only thing I see is a ball bouncing out, I can visualize that that must make the cart move to the left. HOWEVER…
If I analyze it from the following approach, I get stuck (probably overanalyzing it again). Let’s say the person starts out with the ball in his hand, primed to throw it (he’s like a gun). He throws it, and the ball goes to the left, so the cart + person goes to the right in response (like recoil). Then the ball hits the cart partition and bounces off elastically. How do you justify that after that action the cart + person move to the left (and doesn’t just stop the cart)? If anything, doesn’t the ball hit the wall with a force F (that it left the person’s hand in), and cancel it out?
In a variation of this problem, I was told that if the ball is inelastic and smashes/sticks to the wall instead of bouncing off, that in this case the cart would stop moving to the right once the ball hits the wall.
???????
Again, any clarification would be great and super appreciated! =)
Attached Images
bat.JPG (13.0 KB) cart.JPG (16.1 KB)
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# Newton rule (or method)
(0.001 seconds)
## 9 matching pages
##### 1: 3.8 Nonlinear Equations
###### §3.8(ii) Newton’s Rule
Newton’s rule is given by … Newton’s rule can also be used for complex zeros of $p(z)$. …
##### 2: 8.25 Methods of Computation
A numerical inversion procedure is also given for calculating the value of $x$ (with 10S accuracy), when $a$ and $P\left(a,x\right)$ are specified, based on Newton’s rule3.8(ii)). …
##### 3: 29.20 Methods of Computation
A second approach is to solve the continued-fraction equations typified by (29.3.10) by Newton’s rule or other iterative methods; see §3.8. …
##### 4: 9.17 Methods of Computation
Zeros of the Airy functions, and their derivatives, can be computed to high precision via Newton’s rule3.8(ii)) or Halley’s rule3.8(v)), using values supplied by the asymptotic expansions of §9.9(iv) as initial approximations. …
##### 5: 10.74 Methods of Computation
Newton’s rule3.8(i)) or Halley’s rule3.8(v)) can be used to compute to arbitrarily high accuracy the real or complex zeros of all the functions treated in this chapter. …Newton’s rule is quadratically convergent and Halley’s rule is cubically convergent. …
##### 6: 6.18 Methods of Computation
Zeros of $\operatorname{Ci}\left(x\right)$ and $\operatorname{si}\left(x\right)$ can be computed to high precision by Newton’s rule3.8(ii)), using values supplied by the asymptotic expansion (6.13.2) as initial approximations. …
##### 7: 4.45 Methods of Computation
For $x\in[-1/e,\infty)$ the principal branch $\operatorname{Wp}\left(x\right)$ can be computed by solving the defining equation $We^{W}=x$ numerically, for example, by Newton’s rule3.8(ii)). …
##### 8: 3.3 Interpolation
and compute an approximation to $a_{1}$ by using Newton’s rule3.8(ii)) with starting value $x=-2.5$. …Then by using $x_{3}$ in Newton’s interpolation formula, evaluating $\left[x_{0},x_{1},x_{2},x_{3}\right]f=-0.26608\;28233$ and recomputing $f^{\prime}(x)$, another application of Newton’s rule with starting value $x_{3}$ gives the approximation $x=2.33810\;7373$, with 8 correct digits. …
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Question: Is it possible to be 30 pounds overweight and still be healthy? My blood pressure and cholesterol are normal, and I have no health issues other than being overweight.
Answer: Weight is one important measure of health. But it's not the only thing to be considered when assessing how healthy you are overall. Other factors play a role as well, such as how active you are and the amount of muscle vs. fat you have in your body. Taken together, these variables can help give you a more comprehensive view of your health, now and into the future.
Health care providers often assess the impact of a person's weight on their health using a calculation called the body mass index, or BMI. BMI is calculated by dividing your weight in kilograms by your height in meters squared. To find your BMI quickly, you can go to Mayo Clinic's website at mayoclinic.org, and enter your height and weight into the site's online BMI calculator.
#### Latest Video
BMI values between 18.5 and 24.9 are considered normal. Values between 25 and 30 are considered overweight, and values greater than 30 are considered obese. Generally, a BMI that's more than 30 is associated with higher risks to health. These risks include a higher likelihood of developing diseases and health problems such as heart disease, diabetes and high blood pressure.
However, BMI doesn't always provide the full story regarding health risks for some people. That's because it does not take into consideration body composition, particularly the percentage of your body that is fat vs. muscle.
For example, if you lead a very active lifestyle, regularly participating in both aerobic exercise and weight training activities, you may have a healthy percentage of body fat despite having a BMI above the normal range. So, in that situation, a higher BMI does not necessarily translate to higher health risks. It's important to note, though, this situation is less likely when BMI values are higher than 35. Beyond that point, additional weight is much more likely to be distributed as fat and not muscle.
It's also possible to have a normal BMI while your body fat percentage is high enough to increase health risks. People with this condition, known as normal weight obesity, may have the same serious health risks as does someone who is obese. This is especially true for individuals who have a high percentage of body fat around the waist. Research has shown people who carry a high proportion of body fat at the waist have increased health risks.
To get the most accurate assessment of your health, first find your BMI. Then, take a look at your lifestyle. If your BMI is less than 35 and you exercise regularly - participating in at least 150 minutes of physical activity per week - you may not be at an increased health risk. If your BMI is 30 or higher, and you don't have an active lifestyle, your health may be at risk. If so, talk to your health care provider about changes you may be able to make to improve your health now, as well as lower your risk for health problems in the future.
To submit a question, write to: medicaledge@mayo.edu.
### Welcome to the discussion.
Keep it Clean. Please avoid obscene, vulgar, lewd, racist or sexually-oriented language.
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Education.com
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The Washer Method for Volumes of Solids for AP Calculus
based on 3 ratings
By — McGraw-Hill Professional
Updated on Oct 24, 2011
Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus
The volume of a solid (with a hole in the middle) generated by revolving a region bounded by 2 curves:
; where f (x) = outer radius & g (x) = inner radius.
; where p(y ) = outer radius & q(y) = inner radius.
About a line x = h:
About a line y = k:
Example 1
Using the Washer Method, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x in the first quadrant about the x-axis.
Step 1. Draw a sketch. See Figure 12.4-12.
To find the points of intersection, set x = x3 x 3x = 0 or x (x2 – 1)=0, or x = – 1, 0, 1. In the first quadrant x = 0, 1.
Step 2. Determine the outer and inner radii of a washer, whose outer radius = x; and inner radius = x3.
Step 3. Set up an integral.
Step 4. Evaluate the integral.
Verify your result with a calculator.
Example 2
Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region in Example 1 about the line y = 2.
Step 1. Draw a sketch. See Figure 12.4-13.
Step 2. Determine the outer and inner radii of a washer. The outer radius = (2 – x3) and inner radius = (2 – x ).
Step 3. Set up an integral.
Step 4. Evaluate the integral.
and obtain .
The volume of the solid is .
Example 3
Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x2 and x = y2 about the y-axis.
Step 1. Draw a sketch. See Figure 12.4-14.
Intersection points: y = x2; x = y2 y =
Set x2 = x4 = x4x = 0 x = 0 or x = 1
x = 0, y = 0 (0, 0)
x = 1, y = 1 (1, 1).
Step 2. Determine the outer and inner radii of a washer, with outer radius:
x = and inner radius: x = y2.
Step 3. Set up an integral.
.
Step 4. Evaluate the integral
Enter and obtain .
The volume of the solid is .
Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus
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# Embedding images in QR code
Posted 2 years ago
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I guess a story-telling type post would attract more upvotes and probably give some insight about how to 'solve problems' using Mathematica, so I would go into details and try to explain not only the code but also how I figured out how to write them.
To begin with, here's three QR code generated with the code, check it by yourself, they are actually scan-able~ It's also amazing that even very fine details of the image can be shown in the QR code (Note that it would better if you view these QR with your glass off XD)
# How this works?
In fact, this form of QR code is not my original idea, I came across such type of QR code on internet but failed to find its origin. So I tried to figure out the principle by myself.
Carefully observe the image, one can find out that there's something odd about this QR:
The marker on the corner are three times coarser than the majority of the QR. So I initially hypothesize that the QR recognition algorithm would first average the brightness of a segment, turning it into a normal QR and then recognize it. The code I used is as follows:
Block[{img = Import["http://community.wolfram.com//c/portal/getImageAttachment?filename=mathematica1.png&userId=1340903"], dat, partitioned},
dat = ImageData@ImagePad[Binarize[ImageResize[img, Scaled[1/3]]], -4];
partitioned = Partition[dat, {3, 3}];
Grid[{ImageResize[#, Dimensions@dat], BarcodeRecognize@#} & /@
{Image@dat,Binarize@Image@Map[Mean@*Flatten, partitioned, {2}]}]
]
The result proved me wrong as the averaged version cannot be properly recognized. Then further observer the QR code, I found that there are mysterious dots even in the places which should be purely white, also the dots are a bit too structural. So I suspect that normal QR code recognition algorithm only takes the color of the center dot, so I added this to the previous code:
Map[#[[2,2]]&,partitioned,{2}]
then it worked out properly!
As we've already cracked the theory, we can now generate some of our own.
# How to generate?
#### QR code generation
First we can use BarcodeImage to generate a QR code, for example here I would use: "This is a sample QR generated by Mathematica!" as the content of the QR code:
text = "This is a sample QR generated by Mathematica!";
qrraw = BarcodeImage[text, {"QR", "H"}, 1]
BarcodeRecognize@qrraw
#### Image processing
Then we create a black and white image to use as background, for example here we use the wolfram wolf icon:
Import, convert to grayscale and adjust the grayscale a bit:
img=ColorConvert[Rasterize[Graphics[{
Inset[Import["http://community.wolfram.com//c/portal/getImageAttachment?filename=wolframwolf.png&userId=1340903"],{.6,.4},Automatic,.8],
Text[Style["WOLFRAM",Bold,14],{.5,.92}]
},PlotRange->{{0,1},{0,1}},ImageSize->3ImageDimensions@qrraw]],"Grayscale"]^.45
which returns:
Note that in order to get enough resolution while keeping the QR code easy to scan, the dimension of the QR code is best in the range of [25,50], one can test that using ImageDimensions@img and adjust it by changing the error correction level by setting {"QR",lev} where lev can be "L", "M", "Q", or "H".
#### Merging
Then we should merge this two image together. Here we use the technique of dither to display grayscale image using only white and black pixels. In the process of dithering, we should notice that at center of each 9*9 pixel the value should correspond to the value in the QR image, or the QR code would be invalid. The code could be easily written out as follows:
dithering[imgdat_, qrdat_] :=
Block[{imgdat1 = imgdat, dimx, dimy, tmp1, tmp2, f = UnitStep[# - .5] &},
{dimx, dimy} = Dimensions@imgdat;
Quiet@Do[
(*Rounding*)
tmp1 = If[Mod[{i, j}, 3] == {2, 2}, qrdat[[(i + 1)/3, (j + 1)/3]], f[imgdat1[[i, j]]]];
tmp2 = Clip[imgdat1[[i, j]] - tmp1, {-.5, .5}];
(*Diffuse Error*)
imgdat1[[i, j]] = tmp1;
imgdat1[[i, j + 1]] += 0.4375 tmp2;
If[j != 1, imgdat1[[i + 1, j - 1]] += 0.1875 tmp2];
imgdat1[[i + 1, j]] += 0.3125 tmp2;
imgdat1[[i + 1, j + 1]] += 0.0625 tmp2
, {i, dimx}, {j, dimy}];
imgdat1
]
Special attention should be paid to the handling key pixels of the QR code, the error created by introducing it should not be ignored, but its influence should be limited in a range, so here a Clip in error is required, while in a traditional dithering process it would be redundant.
Apply dithering to the image and we have:
Image[ditherdat=dithering[ImageData@img, ImageData@qrraw]]
#### Refinement
One can see that the shape of the original image is quite well preserved and key points of the QR code are properly dealt with. Then the final step is to process the key features on the corner and edge of the QR code, which is quite trivial:
replicate = (Flatten[ConstantArray[#, {3, 3}], {{3, 1}, {4, 2}}] &);
refineqr[qrdat_] :=
Block[{qrd = qrdat, d = Length[qrdat]},
(*Corner*)
(qrd[[#1 ;; 24 #1 ;; #1, #2 ;; 24 #2 ;; #2]] = replicate[qrd[[2 #1 ;; 23 #1 ;; 3 #1, 2 #2 ;; 23 #2 ;; 3 #2]]]) & @@@ {{1, 1}, {1, -1}, {-1, 1}};
(*Edge*)
qrd[[22 ;; d - 21, 19 ;; 21]] = Transpose[qrd[[19 ;; 21, 22 ;; d - 21]] = replicate[{Mod[Range[(d + 1)/3 - 14], 2]}]];
qrd]
Then apply this to previously get result, we get the final result, which is scan-able:
Image[final = refineqr@ditherdat]
BarcodeRecognize@%
It's usually favourable to have a 3x zoom to the image:
Image@replicate@final
A fully packed version is shown in the attachment notebook file, where:
createqr[text,img]
would generate the same result.
Further optimizations could include using machine learning to further refine the display effect. Sharper lines, Less interfering key points and more could be expected.
ENJOY~
# Update
@Henrik Schachner kindly remind me that the previous QR is not that easy to scan with average QR scanning software. So I made some tiny updates to make the QR more standardized and much more easier to scan:
refineqr[qrdat_] :=
Block[{qrd = qrdat, d = Length[qrdat], temp = Fold[ArrayPad[#1, 1, #2] &, {{{0}}, 1, 0}], p},
p = Position[Round@ListCorrelate[temp, qrdat[[2 ;; ;; 3, 2 ;; ;; 3]], {{1, 1}, {-1, -1}}, 0, Abs@*Subtract], 0, 2];
(*Corner*)
(qrd[[#1 ;; 24 #1 ;; #1, #2 ;; 24 #2 ;; #2]] = replicate[qrd[[2 #1 ;; 23 #1 ;; 3 #1, 2 #2 ;; 23 #2 ;; 3 #2]]]) & @@@ {{1, 1}, {1, -1}, {-1, 1}};
(*Edge*)
qrd[[22 ;; d - 21, 19 ;; 21]] = Transpose[ qrd[[19 ;; 21, 22 ;; d - 21]] = replicate[{Mod[Range[(d + 1)/3 - 14], 2]}]];
(*Special*)
(qrd[[3 #1 - 2 ;; 3 #1 + 12, 3 #2 - 2 ;; 3 #2 + 12]] = replicate@temp) & @@@ p;
qrd]
after this update, the QR code would look like this:
After minor manual edition, it could be like:
Maybe this would be easier to scan due to the newly add correction block on the right-down corner.
Also, I think I found a better realization using same basic design principle here.
Attachments:
6 Replies
Sort By:
Posted 2 years ago
Hi Jingxian Wang,thank you for sharing this nice idea together with the code! Meanwhile I have been playing around with it quite a bit. As far as I can say yet your idea is working well under "laboratory conditions", i.e. creating those QR images and applying BarcodeRecognize - all within the Mathematica environment. But when printed on paper and trying the recognition using a QR scanner of e.g. some Android device it obviously does not work. Any idea?Regards -- Henrik
Posted 2 years ago
That's weird... I've actually tested all these QR using my phone before uploading and just now and all of them worked, and one of these codes, the one with "POA" on, is used in a real event in my school and most people can scan them properly...I suspect that you're using a not-that-strong QR scanner, in China we mainly use the QR code scanner in Wechat, which is quite powerful. I usually get a correct result when I scan the QR with Wechat, but my phone's built-in scanner and Facebook app's scanner would fail to recognize it sometimes and also numerous online recognizer failed to recognize it at all... A small trick is to vibrate your phone a bit when scanning, sometimes it would help the phone scanner to locate the QR. (Have a try~ maybe it would work this time~)I'm aware that this form of QR is more fragile than normal QR so I suggested that "the dimension of the QR code is best in the range of [25,50]", but it seems that it is just a bit too hard for average QR scanners even if the size is in this range :(
Posted 2 years ago
Hi~ I've updated my post a bit to improve success rate in scanning, now Wechat/Facebook/my phone's built-in scanner can all easily recognize the QR code~ Can you successfully scan the QR now? ;)
Posted 2 years ago
Dear Jingxian Wang,many thanks for your effort! I am glad to tell you that now the QRs can be recognized successfully! The fourth little square obviously made the difference. Best regards from Germany -- Henrik
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# Aptitude Percentage Test Paper 1
1) 40 % of 280 =?
1. 112
2. 116
3. 115
4. 120
Explanation:
x % of a given number 'n' =
x = 40 and n = 280
∴ 40 % of 280 =* 280 = 112
2) Whose 35% is 280?
1. 700
2. 750
3. 800
4. 850
Explanation:
Let the required value is x.
As per question; 35% of x = 280
3) 45 *? = 35% of 900
1. 6
2. 7
3. 9
4. 4
Explanation:
Let the missing value is x.
∴ 45 * x =*900
45 x = 315
4) If 40% of an amount is 250, what will be 60% of that amount?
1. 300
2. 320
3. 375
4. 400
Explanation:
Let the amount be x.
As per question: 40% of x = 250
Now, 60% of 625 =* 625 = 375
5) If 30 % of 1520 + 40 % of 800 = x % of 5000, find the value of x.
1. 14.42%
2. 15.52%
3. 12.22%
4. 18.82%
Explanation:
As per question:
Percentage Aptitude Test Paper 1
Percentage Aptitude Test Paper 2
Percentage Aptitude Test Paper 3
Percentage Aptitude Test Paper 4
Percentage Aptitude Test Paper 5
Percentage Aptitude Test Paper 6
Percentage Aptitude Test Paper 7
Percentage Aptitude Test Paper 8
Percentage Concepts
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# Special case of nonlinear first order ordinary differential equation.
#### sola maths
Hi there,
I've having problems solving a particular nonlinear ODE. Any help/suggestions will be highly appreciated.
The nonlinear ODE is:
v[t]*v'[t] + (4*v[t])/(t^2 - 1) = t/(t^2 - 1)
Thank you.
Related Differential Equations News on Phys.org
#### jackmell
If:
$$z=\int f(x)dx$$
then:
$$\frac{dz}{dx}=f(x)$$
or:
$$\frac{dx}{dz}=\frac{1}{f}$$
now just substitute back into the original DE
#### sola maths
Thanks for your reply, I understand your point but if you look at the example in the link I posted, you will see that the substitution is not as straightforward as you put it.
Seems to me that there's some sort of parametrization to be done.
#### jackmell
. . . what did I get myself into. But that's ok, need to be willing to try things in math to succeed and not let the risk of failure stop you but I'd rather not make two mistakes in one day. So I think it's this below but I'd have to work on it more with real problems to verify it ok.
Suppose you have:
$$vv'=f(t)v+g(t)$$
and you let:
$$z=\int f dt$$
to obtain:
$$v\frac{dv}{dz}=v+\Phi(t)$$
where:
$$\Phi(t)=\frac{g(t)}{f(t)}$$
and suppose we're able to solve:
$$v\frac{dv}{dz}=v+\Phi(z)=v+\frac{g(z)}{f(z)}$$
say for example the solution is:
$$v(z)=z^3-3\sin(z)$$
but z is paramaterized by t so that we can write the solution for the original problem as:
$$v(t)=z^3-3\sin(z);\quad z=\int f(t)$$
I'm not sure at this point but it's what I'm going with for now until I can verify that or someone can correct me.
Also, I should mention this holds for the simple case:
$$vv'=tv+t$$
which can be solved exactly but that's still no guarantee my explanation above is correct for the general case. Never worked on this type of problem before
Last edited:
#### jackmell
Ok, after reviewing I don't think that's correct. If we have:
$$y\frac{dy}{dz}=y+\Phi(t)=y+\frac{g(t)}{f(t)}$$
and:
$$z=\int f(t)=h(t)$$
then I believe we'd have to compute the inverse:
$$t=h^{-1}(z)$$
then solve:
$$y\frac{dy}{dz}=y+\Phi\big(h^{-1}(z)\big)$$
I'm probably making a mess out of this but I'd at least like to demonstrate how sometimes real math is not so neat and pretty and quick and often fought with many wrong turns and maybe I'm still going wrong.
#### sola maths
Pheeww! Mathematics indeed could be rattling, been on the problem for sometime now.
The problem I foresee with your last review is computing:
$$t=h^{-1}(z)$$
Can't see how that will be done.
#### jackmell
Well I probably don't have it exactly right yet. However if this was my problem, I would go to the library and find one of those Russian texts cited in the Eqnworld reference and get some simple test cases with the answers and study them and maybe the references would explain how the parameterization is used. I'll look at it more later.
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mersenneforum.org > Math Linear algebra proof
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2007-02-06, 02:35 #1 Damian May 2005 Argentina 101110102 Posts Linear algebra proof Can someone point me to the proof that if the geometric multiplicity of each eigenvalue is equal to the corresponding algebraic multiplicity, then the matrix is diagonalizable. Thanks in advance.
2007-02-06, 17:03 #2 ewmayer ∂2ω=0 Sep 2002 Repรบblica de California 5×2,351 Posts Could you please define "geometric multiplicity?" (It sounds like something relating to the eigenspace, but I want to be sure.)
2007-02-06, 22:04 #3
Damian
May 2005
Argentina
2×3×31 Posts
Quote:
Originally Posted by ewmayer Could you please define "geometric multiplicity?" (It sounds like something relating to the eigenspace, but I want to be sure.)
The geometric multiplicity is the dimenion of the eigenspace.
2007-02-06, 22:18 #4 ewmayer ∂2ω=0 Sep 2002 Repรบblica de California 5·2,351 Posts Then it's quite simple: for a repeated eigenvalue (the only case one need be concerned about w.r.to possible nondiagonalizability), if the geometric multiplicity of the corresponding eigenspace is equal to the algebraic multiplicity of the eigenvalue (call that K), that means that one can find K linearly independent eigenvectors, hence the matrix is diagonalizable. Put another way, one only winds up with a Jordan form (nondiagonalizability) if the eigenspace is rank-deficient. In that case the best one can do is to find a set of pseudo-eigenvectors (real eigenvectors plus some non-eigenvectors to "fill in" the rank-deficient elements of the eigenspace corresponding to the particular problematic repeated eigenvalues) which "nearly" diagonalize the matrix.
2007-02-08, 16:28 #5 Damian May 2005 Argentina BA16 Posts Thanks, and other question Thank you very much. I've got a new question: is there a proof that for distinct eigenvalues, there correspond linear independent eigenvectors, that does not use mathematical induction? Thanks in advance, Damian.
2007-02-08, 17:13 #6
ewmayer
2ω=0
Sep 2002
Repรบblica de California
267538 Posts
Quote:
Originally Posted by Damian Thank you very much. I've got a new question: is there a proof that for distinct eigenvalues, there correspond linear independent eigenvectors, that does not use mathematical induction?
This would appear to follow directly from the definition of an eigenvector. Try this: assuming that for 3 distinct eigenvalues l1,l2,l3 with corresponding eigenvectors x,yz, one of the eigenvectors is a linear combination of the other 2. e.g. z = a*x+b*y. Multiply by the matrix, and you should pretty quickly get a contradiction.
2007-02-12, 19:31 #7
ewmayer
2ω=0
Sep 2002
Repรบblica de California
2DEB16 Posts
Quote:
Originally Posted by ewmayer This would appear to follow directly from the definition of an eigenvector. Try this: assuming that for 3 distinct eigenvalues l1,l2,l3 with corresponding eigenvectors x,yz, one of the eigenvectors is a linear combination of the other 2. e.g. z = a*x+b*y. Multiply by the matrix, and you should pretty quickly get a contradiction.
OK, I verified that this does lead to a simple proof, but one still needs to also show that given a starting point of one {eigenvalue,eigenvector} pair, the eigenvector for the *second* distinct eigenvalue must be LI of the first, which in this case reduces to "not a multiple" of. Again easy to show, but in the end it does amount to proof by induction.
2007-02-12, 20:27 #8
xilman
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2DC916 Posts
Quote:
Way Out in Hilbert Space
An infinite-dimensional exit? If so, where does it lead to?
Paul
Last fiddled with by xilman on 2007-02-12 at 20:28 Reason: Fix tag
2007-02-12, 22:25 #9
ewmayer
2ω=0
Sep 2002
Repรบblica de California
5·2,351 Posts
Quote:
Originally Posted by xilman An infinite-dimensional exit? If so, where does it lead to?
Like the sign (you know, the one that pops up almost everywhere) says: In Hilbert Space, all roads converge (if they converge) to a li'l place called Norm's Functional Rest Stop. I'm hoping to get there at some point so I can start to unload some of my collection of old vinyl Lp's, but failing that, simply to achieve closure.
p.s.: Norm's is best-known for its "eat a burger, drink a beer and smoke adjoint" special, but interestingly, they also offer a nice lineup of Cauchy foods.
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ยฌ โจ โง โ โ โ โ โ โ โ โ โ โด โต โค โฅ โข โจ โซค โฃ โฆ โฏ โฎ โฐ โฑ
โซ โฌ โญ โฎ โฏ โฐ โ โ ฮด โ โฑ โ โ
๐ข๐ผ ๐ฃ๐ฝ ๐ค๐พ ๐ฅ๐ฟ ๐ฆ๐๐ ๐ง๐ ๐จ๐ ๐ฉ๐๐ ๐ช๐ ๐ซ๐
๐ฌ๐ ๐ญ๐ ๐ฎ๐ ๐ฏ๐ ๐ฐ๐ ๐ฑ๐ ๐ฒ๐ ๐ด๐๐ ๐ต๐ ๐ถ๐ ๐ท๐๐ ๐ธ๐ ๐น๐ ๐บ๐
| 2,107
| 5,999
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.21875
| 3
|
CC-MAIN-2023-14
|
latest
|
en
| 0.825208
|
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