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October 29th, 2009, 12:03 AM #1 Newbie Joined: Oct 2009 Posts: 1 Thanks: 0 the inverse of a matrix Im having trouble wrapping my head around this concept Suppose A, B, X are nxn matrices with A, X, and A-AX invertible (A - AX)^-1 = X^-1 * B 1. explain why B is invertible 2. solve for X my answer 1. the product of nxn invertible matrices are invertible. (A-Ax) is an invertible matrix, which is the product of 2 invertible nxn matrices, X and B. Not the best logic but I have no idea what else to say 2. no idea where to start thanks for the help
October 29th, 2009, 04:19 AM #2
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Re: the inverse of a matrix
Quote:
Originally Posted by ced my answer 1. the product of nxn invertible matrices are invertible.
Correct.
Quote:
(A-Ax) is an invertible matrix, which is the product of 2 invertible nxn matrices, X and B. Not the best logic but I have no idea what else to say
Indeed not. It isn't necessarily true-- just because A is invertible doesn't mean that A=X*B => B is invertible
Try multiplying on the left by X. Then you can use the statement "the product of two invertible matrices is invertible".
Quote:
2. no idea where to start thanks for the help
Pretend that A,B, and X are numbers instead of matrices, and solve that way, keeping the order straight. Then go back and rewrite "without division"-- by which I mean instead of dividing, multiply by the inverse. Of course, when you are rewriting it, you need to make sure all of your steps are justified.
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How many people does a 1.5 kg chicken serve?
4 peopleAn average-size chicken weighs about 1.5kg and will feed 4 people.
If you’re cooking for 5 or 6, go for a 1.8kg-2kg bird..
Can you bake chicken at 200 degrees?
Preheat oven to 200°F. Place dish in oven; roast chicken, skin side up, 2 hours and 45 minutes or until a meat thermometer registers 155°F in meatiest part of breast and at least 165°F in meatiest part of thigh.
How long does chicken take to fry?
Fry chicken, turning with tongs every 1–2 minutes and adjusting heat to maintain a steady temperature of 300°–325°, until skin is deep golden brown and an instant-read thermometer inserted into thickest part of chicken registers 165°, about 10 minutes for wings and 12 minutes for thighs, legs, and breasts.
How long does a 2.4 kg chicken take to cook?
Roast in the centre of a pre-heated oven, gas mark 5, 375°F (190°C), for 20 minutes per lb (450 g) plus 10-20 minutes extra – this will be 1 hour and 50 minutes to 2 hours for a 5 lb (2.25 kg) bird. Baste three times during the cooking time.
How long do you cook a chicken per kg?
When it comes to cooking times, it couldn’t be easier to calculate – you’ll need 45 minutes per kg, plus an extra 20 minutes to finish. If you want to, you can baste your roast chicken once or twice during cooking to help keep it moist.
How long does it take to cook a 2.5 kg chicken?
Remove chicken from packaging and place in a baking dish. Calculate cooking time, allowing 25 minutes per 500g (a 2.5kg chicken will take 2 hours 5 minutes).
How long do you cook a 1.2 kg chicken?
Place into the oven, reduce the temperature to 200°C fan forced and roast a 1.2kg chicken for 30 minutes. Baste the chicken with pan juices and roast a further 30 minutes or until juices run clear when chicken thigh is pierced with a skewer. General guide is 25 minutes per 500g.
How long does a 1.1 kg chicken take to cook?
Cooking Method: Oven Calculate the cooking times at 45 minutes per kilo plus 20-30 mins. Place in a roasting tin and cook in teh centre of the pre-heated oven basting ocassionally.
When roasting a chicken do you cover it?
When your chicken is finished cooking, remove it from the oven and cover it loosely with a sheet of tinfoil. Allow it to rest for about 10 minutes before carving to allow the juices to settle. If you cut it right away, the meat will loose that juice and be more tough and chewy.
How long does it take to cook a 2kg chicken?
HOW LONG DOES IT TAKE TO ROAST A 2KG CHICKEN? Roasting a chicken will take approximately 20 minutes for every 500g of weight at 200 degrees Celsius (180 degrees Celsius for fan-forced ovens). So if you do the maths, it should take a 2kg chicken about 80 minutes (1 hour and 20 minutes) to cook.
How long does it take to cook a 1.3 kg chicken?
There are two separate time calculations. A properly raised free-range chicken takes 20 minutes per 450g, the rest about 12-15 minutes per 450g. If you are stuffing the bird, add 20 minutes. A 1.3kg bird will feed four.
How long does it take to cook a 1.5 kg chicken?
A 1.5kg chicken will be perfectly roasted after 1 hr 20 mins at 190C/fan 170C/gas 5. It doesn’t matter what you stuff into it, rub or sprinkle over it or put around it, this timing never changes. Remember this and you will always be able to roast a chicken.
How long do I cook a 1.4 kg chicken?
Cooking the chicken As a rule, the roasting formula is 20 minutes per 450g plus an extra 20 minutes, which means a typical 1.4 kg chicken will be perfectly roasted after 1 hour and 20 minutes at 200°C.
How many people will a 1.6 kg chicken feed?
4 peopleBuying the chicken When deciding how much chicken to buy, as a guideline, a 1.6kg chicken should feed 4 people well and still allow for leftovers.
How long does it take to cook a crown of chicken?
Pre-heat your oven to 190°C/375°F/Gas Mark 5. Place in a roasting tin. Place in the centre of the oven and cook for 50 minutes. To check the chicken is fully cooked insert a skewer into the thickest part of the breast, if the juices run clear the chicken is ready.
How many people does a 1kg chicken feed?
If you are talking about a South East Asian family then it ll be around 6–8 people (120–150 gm/head) because here peoply accompany a good amount of carbs like bread/rice with their meals. Whereas in a western family , they usually consume 250–350 gm of chicken i.e around 4–6 people can be feed by 1kg chicken.
How long does a 1.650 chicken take to cook?
1.65 x 45 = 74.25 = 1 hour 15 minutes Add on 20 mins extra = 1 hour and 35 minutes total cooking time.
How long does it take to cook a 1.9 kilo chicken?
Place the chicken in the oven and bake for 20 minutes per 500g, plus 15 minutes over, until golden brown and cooked through.
How long do I cook a 1.35 kg chicken?
Cooking Instructions Preheat oven to 180°C (360°F), if fan forced preheat oven to 160°C (320°F). Oven rack should be in the middle of oven. Remove chicken from bag prior to cooking and place in an oven proof dish. Cook at the preheated temperature for 75 – 80 minutes.
How do you know if chicken is done cooking?
Poke the meat to see if juices are red or clear This method applies to chicken specifically. For properly cooked chicken, if you cut into it and the juices run clear, then the chicken is fully cooked. If the juices are red or have a pinkish color, your chicken may need to be cooked a bit longer.
How many people will a 1.9 kg chicken feed?
A small chicken will weigh approximately 1.2kg and will feed 2-3 people. A medium chicken weighs approximately 1.75kg and will feed 3-4 people, and a large chicken weighs approximately 2kg and will feed 4-6 people. | 1,505 | 5,745 | {"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} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.870094 |
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Ho Chi Minh City, 21 January 2011
Faculty of Applied Science Department of Math Applied
Syllabus
006001 – GIẢI TÍCH 1 (QUẢN TRỊ KINH DOANH) Credit Class Hours
: 4 (3 – 2 – 5) 75 Theory: 45 Exercises: 30 Consultation: 1 class hours per week par TA : Midterm exam 20% Writeen (60') Test + Activities 15% (Class Test + Attendance + Activities in Class) Assignment 15% Writeen Final exam 50% Writeen (120’)
Course outline: In this course, we study linear, quadratic, irrational functions, inequalities, simultaneous equations, some application on economics. Also we study the rules of differentiation, integration, partial differentiation to functions of several variables, differential equations and their applications on economic.
Textbook: Softcopy + Hardcopy [1]
Mathematics for Economics and Business, By Ian Jacques, 5th edition
Lecturer:
PhD. Nguyen Quoc Lan (toanbktp@yahoo.com)
- Faculty of Applied Science
TA:
MS. Nguyen Hong Loc
- Faculty of Applied Science
Course contents: Riview:
Highschool Calculus (One Week)
Chapter 1:
Linear Functions. (Two Weeks)
Chapter 2:
Non Linear Functions (Two Weeks)
Chapter 3:
Mathematics of Finance (One Week)
Chapter 4:
Differentiation (Two Weeks)
Chapter 5:
Partial Differention (Three Weeks)
Chapter 6:
Integration (One Week)
Chapter 7:
Differential Equation (One Week)
Outcome: After this couse, students are expected to be able to: 1/ Use linear functions in some economic fields such as supply and demand analysis. 2/ Use non linear functions in some economic fields such as revenue, cost and profits. 3/ Use derivative to study marginal functions and optimize economic functions. 4/ Use partial derivatives to study unconstrained and constrained optimization. 5/ Use integral to study some economic problem such as consumer’s and producer’s surplus. 6/ Use differential equations to study some economic problem such as supply and demand analysis.
Schedule: Three Class Hours of Lecture + Three Class Hours of Exercise per week Tr.1/2
Syllabus
Content
1
Highschool Calculus Review: Solving equations – inequations. Derivative. Find maximum and minimum. Sketch simple graph. Antiderivative. Definite Integral. The Substitution Rule. Integration by Part. Entry Test Chapter 1: Linear Functions. 1.1 Linear Functions. 1.2 Graph of Linear Functions 1.3 Algebraic Solution of Simultaneous Linear Equations 1.4 Supply and Demand Analysis 1.5 National Income Determination Chapter 2: Non Linear Functions 2.1 Quadratic Functions 2.2 Revenue, Cost and Profits 2.3 Power (Index) and Logarithms 2.4 The Exponential and Logarithmic Functions Chapter 3: Mathematics of Finance 3.1 Percentages 3.2 Compound Interest Midterm Exam Chapter 4: Differentiation 4.1 The Derivative of a Function 4.2 Rules of Differentiation 4.3 Marginal Functions 4.4 Optimization of Economic Functions Chapter 5: Partial Differentiation 5.1 Functions of Several Variables. 5.2 Partial Derivatives. 5.3 Unconstrained Optimization 5.4 Constrained Optimization 5.5 Lagrange Multipliers Chapter 6: Integration 6.1 Indefinite Integration. 6.2 Definite Integration. Application on Economics. Chapter 7: Differential Equation 7.1 Introduction 7.2 Separable and Linear Differential Equation 7.3 Application: Supply and Demand Analysis Final Exam
2–3
4–5
6
7–8
9 – 11
12 13
--------------- END ----------------------
Tr.2/2
Section (in Jacques Book) Lecture
1.1: Pages 13 – 15 1.2: Pages 15 – 35 1.3: Pages 35 – 47 1.4: Pages 47 – 66 1.5: Pages 96 – 113 All Chapter 2: Pages 113 – 175
3.1: Pages 177 – 194 3.2: Pages 194 – 209 4.1: Pages 237 – 251 4.2: Pages 251–261, 275 – 284 4.3: Pages 261 – 275 4.4: Pages 298 – 331 5.1, 5.2: Pages 341 – 356 5.3, 5.4, 5.5: Pages 386 – 421 All Chapter 6: Pages 421 – 451 All Chapter 7: Pages 569 – 587
3 006001 calculusgiai tich 1 | 1,056 | 3,832 | {"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} | 3.40625 | 3 | CC-MAIN-2017-26 | latest | en | 0.872701 |
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http://en.wikipedia.org/wiki/Logical_and | 1,432,504,808,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928078.25/warc/CC-MAIN-20150521113208-00205-ip-10-180-206-219.ec2.internal.warc.gz | 86,553,717 | 19,283 | # Logical conjunction
(Redirected from Logical and)
"∧" redirects here. For the logic gate, see AND gate. For exterior product, see Exterior algebra.
Venn diagram of $\scriptstyle A \and B$
Venn diagram of $\scriptstyle A \and B \and C$
In logic and mathematics, and is the truth-functional operator of logical conjunction; the and of a set of operands is true if and only if all of its operands are true. The logical connective that represents this operator is typically written as $\land$ or $\cdot$.
"A and B" is true only if A is true and B is true.
An operand of a conjunction is a conjunct.
Related concepts in other fields are:
## Notation
And is usually expressed with an infix operator: in mathematics and logic, ; in electronics, $\cdot$; and in programming languages, & or and. In Jan Łukasiewicz's prefix notation for logic, the operator is K, for Polish koniunkcja.[1]
## Definition
Logical conjunction is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both of its operands are true.
The conjunctive identity is 1, which is to say that AND-ing an expression with 1 will never change the value of the expression. In keeping with the concept of vacuous truth, when conjunction is defined as an operator or function of arbitrary arity, the empty conjunction (AND-ing over an empty set of operands) is often defined as having the result 1.
### Truth table
Conjunctions of the arguments on the left — The true bits form a Sierpinski triangle.
The truth table of $~A \and B$:
INPUT OUTPUT $A$ $B$ $A \and B$ T T T T F F F T F F F F
## Introduction and elimination rules
As a rule of inference, conjunction introduction is a classically valid, simple argument form. The argument form has two premises, A and B. Intuitively, it permits the inference of their conjunction.
A,
B.
Therefore, A and B.
or in logical operator notation:
$A,$
$B$
$\vdash A \and B$
Here is an example of an argument that fits the form conjunction introduction:
Bob likes apples.
Bob likes oranges.
Therefore, Bob likes apples and oranges.
Conjunction elimination is another classically valid, simple argument form. Intuitively, it permits the inference from any conjunction of either element of that conjunction.
A and B.
Therefore, A.
...or alternately,
A and B.
Therefore, B.
In logical operator notation:
$A \and B$
$\vdash A$
...or alternately,
$A \and B$
$\vdash B$
## Properties
commutativity: yes
$A \and B$ $\Leftrightarrow$ $B \and A$ $\Leftrightarrow$
associativity: yes
$~A$ $~~~\and~~~$ $(B \and C)$ $\Leftrightarrow$ $(A \and B)$ $~~~\and~~~$ $~C$ $~~~\and~~~$ $\Leftrightarrow$ $\Leftrightarrow$ $~~~\and~~~$
distributivity: with various operations, especially with or
$~A$ $\and$ $(B \or C)$ $\Leftrightarrow$ $(A \and B)$ $\or$ $(A \and C)$ $\and$ $\Leftrightarrow$ $\Leftrightarrow$ $\or$
idempotency: yes
$~A~$ $~\and~$ $~A~$ $\Leftrightarrow$ $A~$ $~\and~$ $\Leftrightarrow$
monotonicity: yes
$A \rightarrow B$ $\Rightarrow$ $(A \and C)$ $\rightarrow$ $(B \and C)$ $\Rightarrow$ $\Leftrightarrow$ $\rightarrow$
truth-preserving: yes
When all inputs are true, the output is true.
$A \and B$ $\Rightarrow$ $A \and B$ $\Rightarrow$ (to be tested)
falsehood-preserving: yes
When all inputs are false, the output is false.
$A \and B$ $\Rightarrow$ $A \or B$ $\Rightarrow$ (to be tested)
Walsh spectrum: (1,-1,-1,1)
Nonlinearity: 1 (the function is bent)
If using binary values for true (1) and false (0), then logical conjunction works exactly like normal arithmetic multiplication.
## Applications in computer engineering
In high-level computer programming and digital electronics, logical conjunction is commonly represented by an infix operator, usually as a keyword such as "AND", an algebraic multiplication, or the ampersand symbol "&". Many languages also provide short-circuit control structures corresponding to logical conjunction.
Logical conjunction is often used for bitwise operations, where 0 corresponds to false and 1 to true:
• 0 AND 0 = 0,
• 0 AND 1 = 0,
• 1 AND 0 = 0,
• 1 AND 1 = 1.
The operation can also be applied to two binary words viewed as bitstrings of equal length, by taking the bitwise AND of each pair of bits at corresponding positions. For example:
• 11000110 AND 10100011 = 10000010.
This can be used to select part of a bitstring using a bit mask. For example, 10011101 AND 00001000 = 00001000 extracts the fifth bit of an 8-bit bitstring.
In computer networking, bit masks are used to derive the network address of a subnet within an existing network from a given IP address, by ANDing the IP address and the subnet mask.
Logical conjunction "AND" is also used in SQL operations to form database queries.
The Curry-Howard correspondence relates logical conjunction to product types.
## Set-theoretic correspondence
The membership of an element of an intersection set in set theory is defined in terms of a logical conjunction: xAB if and only if (xA) ∧ (xB). Through this correspondence, set-theoretic intersection shares several properties with logical conjunction, such as associativity, commutativity, and idempotence.
## Natural language
As with other notions formalized in mathematical logic, the logical conjunction and is related to, but not the same as, the grammatical conjunction and in natural languages.
English "and" has properties not captured by logical conjunction. For example, "and" sometimes implies order. For example, "They got married and had a child" in common discourse means that the marriage came before the child. The word "and" can also imply a partition of a thing into parts, as "The American flag is red, white, and blue." Here it is not meant that the flag is at once red, white, and blue, but rather that it has a part of each color. | 1,487 | 5,857 | {"found_math": true, "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": 1, "img_math": 111, "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} | 4.09375 | 4 | CC-MAIN-2015-22 | latest | en | 0.875888 |
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1 Apr 2008 I have a 12' x 24' deck of 288 square feet. I want to run 4" width deck boards on a diagonal of 45 degrees. how do I calculate proper amount of lineal foot boards (20', 18', 16') to avoid waste. I am donating my time and costs to
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23 Nov 2010 Taking on this challenge allows you to design the deck of your dreams and to use those plans to determine how much decking Only the linear feet calculation reveals exactly how many decking boards you'll need to buy.
### Deck and Floor Board Spacing and Quantity Calculator
Calculate best fit, board gap, layout and board quantities for floors and decks. Deck and Floor Board Calculator See also: Deck Stump, Bearer and Joist Calculator Centers, Spacings, Plans and Costs . linear Board (inc 5% waste) 819 ft
### Deck Calculator – Convert Square Feet to Linear Feet
28 Dec 2009 Do you need to convert square feet to linear feet? Check out our handy Wood decking Calculator. It works perfectly for both standard & pre-grooved decking. You're also covered covered when it comes to converting square
### Ipe Wood Calculators- Square to Linear Calculation - Ipe Woods
Simply measure the width and length of your Ipe deck, fencing or siding space. Then times the two numbers together. I.E. 20ftx20ft is 400 feet. This is the total square footage you need. That is what the square footage calculator above is for.
### How Do You Convert Square Feet To Linear Feet? - YouTube
14 Jun 2017 Convert me online units conversion view topic square feet or linear what's the difference? linear feet will only calculate 1 edge of the print decking calculators including square to lineal feet, hidden fastener calculator,
### How To Convert Square Feet to Linear Feet AdvantageLumber
18 Nov 2014 Often a lot of people won't know how to convert from square feet to linear feet and if they order amounts per the square foot then there will be a very big shortage of material. Caveat: Remember that the width needed is the real measurement of the face of the board. This entry was posted in Deck tools, how-to Build, Uncategorized and tagged conversion calculator, Deck Calculator,
### How to Calculate the Square Feet of a Deck Hunker
16 Jan 2011 If you are planning to remodel your deck or sell your house, you will need to know the square feet involved. Square footage is a measurement of area, which is the total feet of a certain space. You can find the square feet of
### Lumber Footage Calculator - Amerhart
This simple tool allows you to convert lineal feet to board feet (and vice versa). Lumber Footage Calculator. A more detailed version of the calculator is available as an Excel download. What would you like to calculate?*. Lineal feet to Board
### How Do You Calculate A Linear Foot? - YouTube
14 Jun 2017 measure each section of javascript calculator to determine the linear footage and yardage a roll paper (basis weight width) 12. linear feet, 3 linear yards have you ever experienced extra fees and charges due to the foot rule?
### Decks. Decking Calculator
Use this deck floor calculator to determine the number of deck boards you need to complete your project. Input the area of the deck and a few options to Or, Enter Dimensions. Length. feet * Required. Inches. Width. feet * Required. Inches
### Deck and Railing Material Calculators - Veranda Decking
Get started on your composite deck project today with these simple to use material calculators. Veranda composite The calculator will give you estimated linear feet and board counts for each of the available lengths of Veranda decking. Square feet The user is responsible for all measurements, verifying these estimates and any use, substitutions or modifications of the quantities provided. Fiberon
### How to Calculate How Much Decking You Need Hunker
29 Mar 2010 Calculate the decking needed for your stairs. Typical deck stairs are 12 inches deep. This requires two boards the width of the stairs for each step. Assume two steps, 48 inches wide. Add two 8-foot boards for stairs, totaling 63
### Decking Calculators Porch and Deck Flooring Calculators
to help you determine how many Tigerwood planks you'll need for your TigerDeck, use our decking calculators. calculators take a variety of factors into consideration, such as the length, width, board size, and total linear feet of your project.
### How to Calculate Deck Boards Hunker
8 May 2018 Before beginning to build a deck, you'll need to determine how much material you need. You will need to consider for instance, if your deck is 20 feet long by 10 feet wide, its area is 200 square feet. If your deck has a main | 1,655 | 7,611 | {"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} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.867371 |
https://www.jiskha.com/questions/361257/how-to-solve-y-3cos2x-5-pls-help-me | 1,568,639,976,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572556.54/warc/CC-MAIN-20190916120037-20190916142037-00120.warc.gz | 916,194,355 | 5,187 | # math
how to solve y= 3cos2x + 5
pls help me
1. 👍 0
2. 👎 0
3. 👁 102
1. Is that really cos(2x) or do you mean cos^2 x?
I will assume that you really do mean cos(2x). In that case
cos(2x) = cos^2 x - sin^2 x
= 2 cos^2 x -1
y = 6 cos^2x -3 + 5 = 6 cos^2 x - 2
cos^2 x = (y+2)/6
cos x = sqrt[(y+2)/6)]
You can use that to solve for x, if that is what the question is all about.
You cannot solve a single equation for both x and y.
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posted by drwls
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More Similar Questions | 790 | 2,310 | {"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} | 3.78125 | 4 | CC-MAIN-2019-39 | latest | en | 0.910954 |
https://python-forum.io/Thread-Add-items-from-one-list-to-list-of-lists | 1,606,192,472,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00386.warc.gz | 469,521,419 | 24,952 | ##### Add items from one list to list of lists
Add items from one list to list of lists PUP280 Programmer named Tim Posts: 10 Threads: 3 Joined: Mar 2020 Reputation: 0 Mar-24-2020, 10:50 AM (This post was last modified: Mar-24-2020, 11:07 AM by buran.) Hello, I would like to add every element of a list on every list from a list of lists. Ie : ```H = [[1,2],[3,4]] I = [5,6]```With this result : ``Output:J = [[1,2,5], [3,4,5], [1,2,6], [3,4,6]]``I don’t find the good logic. Ie : ```j = [] for x in h : for y in i : j.append(h+i) print (j)```Gives me : ``````Output:[[[1, 2], [3, 4], 5, 6]] [[[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6]] [[[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6]] [[[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6], [[1, 2], [3, 4], 5, 6]]``````Or again : ```k = [] for w in h : for z in i : o = [w,z] k.append(o) print (k)```Will give : ``````Output:[[[1, 2], 5]] [[[1, 2], 5], [[1, 2], 6]] [[[1, 2], 5], [[1, 2], 6], [[3, 4], 5]] [[[1, 2], 5], [[1, 2], 6], [[3, 4], 5], [[3, 4], 6]]``````I’ve tried some tests with map, but nothing conclusive. Can you help me please to solve my problem ? Reply perfringo Da Bishop Posts: 1,567 Threads: 7 Joined: Jun 2018 Reputation: 152 Mar-24-2020, 12:36 PM (This post was last modified: Mar-24-2020, 12:36 PM by perfringo.) My subjective opinion is that 'not finding good logic' starts with bad naming habits. What are H and I? The last name ('I') I would call 'crime against readers of your code', PEP8 Names to avoid: Quote:Never use the characters 'l' (lowercase letter el), 'O' (uppercase letter oh), or 'I' (uppercase letter eye) as single character variable names. In some fonts, these characters are indistinguishable from the numerals one and zero. When tempted to use 'l', use 'L' instead. I add that 'l' and 'I' are also pretty similar to each other. Now to problem at hand: list comprehension could be suitable tool: ```>>> target = [[1,2],[3,4]] >>> source = [5,6] >>> [[*item, num] for num in source for item in target] [[1, 2, 5], [3, 4, 5], [1, 2, 6], [3, 4, 6]] ``` I'm not 'in'-sane. Indeed, I am so far 'out' of sane that you appear a tiny blip on the distant coast of sanity. Bucky Katt, Get Fuzzy Da Bishop: There's a dead bishop on the landing. I don't know who keeps bringing them in here. ....but society is to blame. Reply PUP280 Programmer named Tim Posts: 10 Threads: 3 Joined: Mar 2020 Reputation: 0 Mar-28-2020, 08:34 AM (This post was last modified: Mar-28-2020, 08:34 AM by PUP280.) Hello, At first, thank you for the edit from the moderator, I had not found how to present it, now I know. Thanks again. For the letters, thank you. I'm a beginner, I knew that but as it was just an example I've believed it was ok. Sorry, I've noticed and I'll don't do that again. Now, about the solution, I thank you. Can you give me an explanation with it please ? I search to understand everything. For what I see, you've added a star in front of item ; is that the way to do what I want ? Have you a link where it's teaching and can you write the long version to help my comprehension ? On Internet, I can't find an explanation (I've read all the tutorial section on Python documentation, maybe I've jumped something). Item isn't a method which gives a dictionnary ? Here item is just a word (I've tried your code changing item to be sure), but it a function too no ? Sorry for my lack of knowledge. Reply perfringo Da Bishop Posts: 1,567 Threads: 7 Joined: Jun 2018 Reputation: 152 May-02-2020, 12:37 PM (This post was last modified: May-02-2020, 12:38 PM by perfringo.) (Mar-28-2020, 08:34 AM)PUP280 Wrote: Can you give me an explanation with it please ? This is nested list comprehension combined with iterable unpacking. Documentation > The Python Tutorial > Data Structures > 5. More on Lists > 5.1.3. List Comprehensions Documentation > The Python Language Reference > 6. Expressions > 6.15. Expression Lists PEP448 - Additional Unpacking Generalizations I'm not 'in'-sane. Indeed, I am so far 'out' of sane that you appear a tiny blip on the distant coast of sanity. Bucky Katt, Get Fuzzy Da Bishop: There's a dead bishop on the landing. I don't know who keeps bringing them in here. ....but society is to blame. Reply pyzyx3qwerty Minister of Silly Walks Posts: 356 Threads: 13 Joined: Mar 2020 Reputation: 9 May-02-2020, 01:50 PM (Mar-24-2020, 10:50 AM)PUP280 Wrote: Or again : ```k = [] for w in h : for z in i : o = [w,z] k.append(o) print (k)```Will give : ``````Output:[[[1, 2], 5]] [[[1, 2], 5], [[1, 2], 6]] [[[1, 2], 5], [[1, 2], 6], [[3, 4], 5]] [[[1, 2], 5], [[1, 2], 6], [[3, 4], 5], [[3, 4], 6]]``````I’ve tried some tests with map, but nothing conclusive. Can you help me please to solve my problem ?Had you realized, the code is actually right just that the print statement is in the wrong place. It should be : ```H = [[1,2],[3,4]] I = [5,6] k = [] for w in H : for z in I : o = [w,z] k.append(o) print (k) ```And it gives output: ``````Output: [[[1, 2], 5], [[1, 2], 6], [[3, 4], 5], [[3, 4], 6]] `````` Reply PUP280 Programmer named Tim Posts: 10 Threads: 3 Joined: Mar 2020 Reputation: 0 May-05-2020, 03:47 PM Thank you perfringo :) Reply
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## Description
Sequences and Series NCERT Exercise 9.3 Question 24 Class 11 Maths : Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from {\displaystyle (n+1)^{th} \text{ to } (2n)^{th} } term is {\displaystyle \frac{1}{r^n} } .
## Question 14 Find the real numbers x and y if (x−iy)(3+5i) is the conjugate of −6−24i
Working with Imaginary Numbers and Complex NumbersImaginary Numbers and Complex NumbersNCERT Exercise 5.1NCERT Exercise 5.3Cartesian and Polar Coordinate SystemsRelationship between Cartesian and Polar Coordinate SystemsArgand Diagram to represent Complex NumbersPolar...
## Example 12 Find the conjugate
Working with Imaginary Numbers and Complex NumbersImaginary Numbers and Complex NumbersNCERT Exercise 5.1NCERT Exercise 5.3Cartesian and Polar Coordinate SystemsRelationship between Cartesian and Polar Coordinate SystemsArgand Diagram to represent Complex NumbersPolar...
## Find the modulus and the arguments of each of the complex numbers in
Working with Imaginary Numbers and Complex NumbersImaginary Numbers and Complex NumbersNCERT Exercise 5.1NCERT Exercise 5.3Cartesian and Polar Coordinate SystemsRelationship between Cartesian and Polar Coordinate SystemsArgand Diagram to represent Complex NumbersPolar...
## Find the value of tan9°–tan27°–tan63°+tan81°
NCERT Exemplar Class 11 Maths Chapter 3 Trigonometry Functions Example 5: Find the value of tan9°–tan27°–tan63°+tan81° tan9°–tan27°–tan63°+tan81° | 403 | 1,562 | {"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} | 3.0625 | 3 | CC-MAIN-2022-27 | latest | en | 0.734281 |
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v = kla Tb mc
[ M0 L T-1 ] = [ L ]a [ M L T-2 ]b [M] c \ [ L ]c
[ M0 L T-1 ] = [ Mb+c La+b-c -2b ]
b+c = 0
a+b-c = 1
-2b = -1 we get
b = 1\2
c = -1\2
a = 0 therefore v = k whole root of T \ m
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Page 1 / 2
## Potential difference
When a circuit is connected and complete, charge can move through the circuit. Charge will not move unless there is a reason, a force. Think of it as though charge is at rest and something has to push it along. This means that work needs to be done to make charge move. A force acts on the charges, doing work, to make them move. The force is provided by the battery in the circuit.
The position of the charge in the circuit tells you how much potential energy it has because of the force being exerted on it. This is like the force from gravity, the higher an object is above the ground (position) the more potential energy it has.
The amount of work to move a charge from one point to another point is how much the potential energy has changed. This is the difference in potential energy, called potential difference. Notice that it is a difference between the value of potential energy at two points so we say that potential difference is measured between or across two points. We do not say potential difference through something.
Potential Difference
Electrical potential difference as the difference in electrical potential energy per unit charge between two points. The units of potential difference are the volt named after the Italian physicist Alessandro Volta (1745–1827) (V).
The units are volt (V), which is the same as joule per coulomb, the amount of work done per unit charge. Electrical potential difference is also called voltage.
## Potential difference and parallel resistors
When resistors are connected in parallel the start and end points for all the resistors are the same. These points have the same potential energy and so the potential difference between them is the same no matter what is put in between them. You can have one, two or many resistors between the two points, the potential difference will not change. You can ignore whatever components are between two points in a circuit when calculating the difference between the two points.
Look at the following circuit diagrams. The battery is the same in all cases, all that changes is more resistors are added between the points marked by the black dots. If we were to measure the potential difference between the two dots in these circuits we would get the same answer for all three cases.
Lets look at two resistors in parallel more closely. When you construct a circuit you use wires and you might think that measuring the voltage in different places on the wires will make a difference. This is not true. The potential difference or voltage measurement will only be different if you measure a different set of components. All points on the wires that have no circuit components between them will give you the same measurements.
All three of the measurements shown in the picture below (i.e. A–B, C–D and E–F) will give you the same voltage. The different measurement points on the left have no components between them so there is no change in potential energy. Exactly the same applies to the different points on the right. When you measure the potential difference between the points on the left and right you will get the same answer.
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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http://rasch-analysis.com/rasch-thresholds-steps.htm | 1,508,402,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00844.warc.gz | 281,949,090 | 2,537 | # Rasch Analysis
Rasch analysis can be applied to assessments in a wide range of disciplines, including health studies, education, psychology, marketing, economics and social sciences.
Many assessments in these disciplines involve a well defined group of people responding to a set of items for assessment. Generally, the responses to the items are scored 0, 1 (for two ordered categories); or 0, 1, 2 (for three ordered categories); or 0, 1, 2, 3 (for four ordered categories) and so on, to indicate increasing levels of a response on some variable such as health status or academic achievement. These responses are then added across items to give each person a total score. This total score summarise the responses to all the items, and a person with a higher total score than another one is deemed to show more of the variable assessed. Summing the scores of the items to give a single score for a person implies that the items are intended to measure a single variable, often referred to as a unidimensional variable.
The Rasch model is the only item response theory (IRT) model in which the total score across items characterizes a person totally. It is also the simplest of such models having the minimum of parameters for the person (just one), and just one parameter corresponding to each category of an item. This item parameter is generically referred to as a threshold. There is just one in the case of a dichotomous item, two in the case of three ordered categories, and so on.
1. What is Rasch Analysis
2. Why undertake a Rasch analysis?
3. The research paradigm and the Rasch model
4. Is there more than one Rasch model?
5. Different Rasch Model Specifications
## 6. Thresholds and Steps
One particular difference that has arisen in different Rasch analysis reporting is the use of "step", when the parameters are different across items, and "thresholds", when they are the same across items. This can give the impression that the response process characteristized by the Rasch model is a sequential process. However, the Rasch model is NOT a sequential processing model but a static model, which just specifies the probability of a person with a given location responding, or being classified, in one of the categories of an item.
For example, the term "step" is not used in the dichotomous case because it would imply, implausibly, that a person goes from being wrong to being right, or goes from disagreeing to agreeing. Instead the person is either wrong or right, or either disagrees or agrees; there is no sequential processing here. The response process is a classification into ordered categories defined by thresholds which can be seen as analogous to markings on a ruler except that the thresholds do not have to be equidistant as they are in a ruler - they are estimated. The threshold is the point where the probability of a response in either one of two adjacent categories is 50%.
7. Disordered Thresholds as an Anomaly
8. Who should use a Rasch analysis?
9. An ideal approach to a Rasch analysis?
10. Recommended Rasch Software
11. The RUMM approach to Rasch Analysis
12. What courses and workshops are available on Rasch analysis?
13. What cloud analysis engines and API's are available?
Time and Space is a Trademark of Timewatch. | 711 | 3,296 | {"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 | 3 | CC-MAIN-2017-43 | latest | en | 0.91202 |
https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&oldid=79091 | 1,623,739,612,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00205.warc.gz | 117,374,324 | 12,771 | # Angle Bisector Theorem
This is an AoPSWiki Word of the Week for June 6-12
## Introduction
The Angle Bisector Theorem states that given triangle $\triangle ABC$ and angle bisector AD, where D is on side BC, then $\frac cm = \frac bn$. Likewise, the converse of this theorem holds as well.
$[asy] size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("A",A,(1,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("D",D,(0,-1)); dot(A^^B^^C^^D,blue);label("b",(A+C)/2,(1,0));label("c",(A+B)/2,(0,1));label("m",(B+D)/2,(0,-1));label("n",(D+C)/2,(0,-1)); [/asy]$
## Proof
### Method 1
Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:
$[asy] size(200); defaultpen(fontsize(10)); real a,b,c,d,m,n; pair A=(1,4), B=(-5,0), C=(3,0), D, E; b = abs(C-A);c = abs(B-A); D = (b*B+c*C)/(b+c); m = abs(B-D);n = abs(C-D); E = C+(B-A)*n/m; draw(A--B--C--A--E--C); MA(B,A,D,0.5,blue+linewidth(1)); MA(D,A,C,0.6,blue+linewidth(1)); MA(C,E,A,0.6,blue+linewidth(1)); MA(C,B,A,0.6,green+linewidth(1)); MA(B,C,E,0.6,green+linewidth(1)); label("A",A,(1,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("D",D,(1,-1));label("E",E,(0,-1)); label("b",(A+C)/2,(1,0));label("c",(A+B)/2,(0,1));label("m",(B+D)/2,(0,-1));label("n",(D+C)/2,(0,-1));label("b",(E+C)/2,(1,-0.5)); dot(A^^B^^C^^D^^E); [/asy]$
Since AB and CE are parallel, we know that $\angle BAE=\angle CEA$ and $\angle BCE=\angle ABC$. Triangle ACE is isosceles, with AC = CE.
By AA similarity, $\triangle DAB \cong \triangle DEC$. By the properties of similar triangles, we arrive at our desired result:
$\frac cm = \frac bn.$
### Method 2
Since B,D, and C are collinear, $\dfrac{[ABD]}{[ADC]} = \dfrac{m}{n}$. Now consider the perpendicular from $D$ to $AB$ and $D$ to $AC$. Every point on the angle bisector of an angle is equidistant to the sides of the angle, so the height $h$ to $AB$ is equal to the height to $AC$. Thus $\dfrac{[ABD]}{[ADC]} = \dfrac{\dfrac{ch}{2}}{\dfrac{bh}{2}} = \dfrac{c}{b}$. Thus $\dfrac{c}{b} = \dfrac{m}{n}$, so $\dfrac{c}{m} = \dfrac{b}{n}$. We can prove the converse by the Phantom Point Method, since we can find $m$ and $n$ in terms of $a$, $b$, and $c$, and prove that the points are the same.
### Method 3
Let $AD = d$. Now, we can express the area of triangle ABD in two ways:
$[ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB.$ Thus, $\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm$.
Likewise, triangle ACD can be expressed in two different ways:
$[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.$ Thus, $\frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn$.
But $\angle CAD \cong \angle BAD$ and $\sin \angle ADC = \sin \angle ADB$ since $\angle ADC = \pi - \angle ADB$. Therefore, we can substitute back into our previous equation to get $\frac{\sin \angle ADB}{\sin \angle BAD} = \frac bn$.
We conclude that $\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn$, which was what we wanted.
In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
## Examples
1. Let ABC be a triangle with angle bisector AD with D on line segment BC. If $BD = 2, CD = 5,$ and $AB + AC = 10$, find AB and AC.
Solution: By the angle bisector theorem, $\frac{AB}2 = \frac{AC}5$ or $AB = \frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \frac{50}7$. We can plug this back in to find $AB = \frac{20}7$.
2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3$. Find the value of $m\angle BAP - m\angle CAP$.
Solution: First, we notice that $\frac{AB}{BP}=\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0.
3. Part (b), 1959 IMO Problems/Problem 5.
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Login to AoPS | 1,493 | 4,183 | {"found_math": true, "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": 44, "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} | 4.46875 | 4 | CC-MAIN-2021-25 | latest | en | 0.647114 |
https://wiki.tcl-lang.org/page/Creating+wave+formulas+with+BWise | 1,721,755,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00088.warc.gz | 528,565,217 | 8,905 | ## Creating wave formulas with BWise
It is fun to use BWise to create formulas which are intended to be rendered as musical notes, in this case in line with Making mathematical waves I use the Maxima syntax for formulas but we could also use C or even Exp type of formulea, and because the block approach doesn't require much recompiling of the result or better put formula translation, this shouldn't be too hard.
In this case I approach the making of 3 notes forming achord in a straightforward way, the bocks on the bwise canvas are Maxima formula rendering blocks, their output pins contain Maxima mathematics which one could change into other languages, even in parallel by adding more than 1 blockfunction (I've made an example of that somewhere, this is written quickly) and using the propagation function which uses the right block functions. There are three tones made with sine shape and exponential decay, and the blocks are made by hand, like this:
newproc {set sine1.o "sin((${sine1.freq})*2*%pi*x)"} sine1 freq o newproc {set sine2.o "sin((${sine2.freq})*2*%pi*x)"} sine2 freq o
newproc {set sine3.o "sin((${sine3.freq})*2*%pi*x)"} sine3 freq o newproc {set exp1.o "(${exp1.i})*%e^(-3*x)"} exp1 i o
newproc {set exp2.o "(${exp2.i})*%e^(-3*2*x)"} exp2 i o newproc {set exp3.o "(${exp3.i})*%e^(-3*6*x)"} exp3 i o
newproc {set add1.o "${add1.i1}+${add1.i2}+${add1.i3}"} add1 {i1 i2 i3} newproc {set div1.o "(${div1.i})/3"} div1 i o
Which after dragging in plae and conneting looks like this:
(I added a monitor block, too), the canvas can be downloaded here [L1 ].
The result at the output of div1 can be gotten by propagating all data through the network, this can be done automatically by:
foreach i [net_funct div1] {net_funprop $i} and then get the value by right-clicking for an inspector block on div1, or simply using the console: (Math) 24 % puts${div1.o}
((sin((440/2)*2*%pi*x))*%e^(-3*x)+(sin(((880/2))*2*%pi*x))*%e^(-3*2*x)+(sin(((3*440/2))*2*%pi*x))*%e^(-3*6*x))/3
Which is the required Maxima formula, which in turn can be fed to the formula rending cgi on my server, and then it is pretty printed:
Clearly the three components are visible; all an octave apart. Rendering the first 1/10 sec of the resulting graph is like:
And finally the above mentioned Tcl server script also creates a 3 seconds sound file, which can be tried out here [L2 ]
Of course the blocks could be made to autogenerate from the canvas popup menu, and maybe given a wizard, and then of course and automatic code generator. The fun is the graph is freely changable and flows through the actual mathematical formula (Maxima is a algebraic manipulating package) and at least easy to read. It is also possible to optimize block results by calling maxima in blocks on the canvas, see Bwise blocks using exec maxima (I've made an improved verison for the server, I don't think this page was upgraded). Of course more work is possible in various directions.
Oh, I just remembered, one must set a frequency at the input of the sine generator blocks, in Herz. I used 440/2 as the base frequency, with parenthesis to make sure the composed formula is correct. One could use A musical keyboard for BWise and Midi connections to make various nots and figure out the frequencies, I didn't make that block as repeatably instantiatable bwise block yet...
Lets look a little closer to how this can work, without going in to the server functionality to render the formulas and in a general way, so we could also make expr formulas from bwise blocks, like in constructing mathematical formulas with Bwise blocks.
We can make a bwise canvas popup menu by first clearing out the standard blocks:
foreach i [info procs new*] {rename $i d$i}
And now add a new function to try out whether it appears on the popup menu:
proc newsine {} {
}
Indeed, right clicking on the canvas will show a 'Sine' meny entry. Now we should adapt one of the good working new* procs to create a block in place on the canvas and of course make a generator for unique (numbered) instance names, and create a block function generator preferably with some easy to use mechanism to create the correct instance pin-names from the formal arguments and function value(s) of the formula block, and a maxima code generator with the correct feeding through of the arguments in the maxima form function body/prototype.
Anyway, I made this into working prototype by making use of the Automatically generate Bwise blocks from procedures and defining some signal processing mathematical blocks:
proc exp { e } {
return "%e^(-x*$e)" } proc mult { {a} {b {1}} } { return "($a*$b)" } proc plus { {a} {b} } { return "($a+$b)" } proc sine { {f} {m} } { return "sin($f*2*%pi*(x+$m))" } proc sine1 {f m} { return "sin($f*x+$m)" } By source-ing these (for the moment) and pressing 'refresh list' in the bwise function window (see: interactive command composer), one of these can be picked by double-clicking it in the function list, and then a block can be generated by pressing the 'Block' button. The blocks of the needed kinds (with the above maxima function definitions as function (proc) call in them can be made in any number and connected up by bwise wires. The all left side branches must be rigth-clicked to get the popup schedule menu and then 'funprop'-ed, so that the last block output wil contain the combined formulas in correct maxima form of the desired signal graph. No we want to render this formula to get a prettyprinted formula (if you have Latex) an example sound from this formula when time is represented in seconds by the x-variable,a nd I want to play the sound from a master/midi keyboard connected with a computer over a DA converter driven by the mianly Linux based 'Jack' soudn connection program, in real time, for which in this case a prgram is automatically created which when started can be connected over jack to a midi source and a audio output channel. The various Tcl functions needed to make the desired program automatically are: #!/bin/sh # \ exec tclsh "$0" "$@" # # puts stdout "Content-type: text/html\n" proc domaxima { {m} } { set t "display2d:false;\n$m;"
# return [string range [exec maxima << $t | tail -2 ] 6 end-7] return [string range [exec maxima -q <<$t ] 32 end-7]
#return [exec maxima --real-quiet << $t ] } proc domaximafor { {m} } { set t "display2d:false;\nlinel:3000;\nfortran(subst((float(%e)),%e,subst((fl at(%pi)),%pi,$m)));\n"
return [string range [exec maxima -q << $t ] 42 end-18] } proc domaximagraph { {m} } { set t "display2d:false;\nlinel:3000;\nplot2d ($m, $x, 0, 0.1$, $gnuplot_t rm, \"gif size 1600,512\"$, $gnuplot_out_file, \"/dev/shm/gnuplot.gif\"$);\n
return [exec maxima --very-quiet << $t ] } proc domaximaformula { {m} } { #set t "tex($m);"
set t "display2d:false;\ntex($m);" return [string range [exec maxima -q --very-quiet <<$t ] 6 end-6]
}
proc formake { {e} } {
global f
set t [subst -nocommand {
subroutine sayhello(x,r)
double precision x,r
r = $e return end }] set f [open /dev/shm/sub.f w] puts$f [string trim $t \n] close$f
set env(TMP) /dev/shm/
set env(TEMP) /dev/shm/
exec gcc -O3 -ffixed-line-length-none -c -o sub.o /dev/shm/sub.f
exec gcc -O3 -o synthbw synth.c sub.o -lasound -ljack
# exec gcc -ffixed-line-length-none -c -o /dev/shm/sub.o /dev/shm/sub.f
# exec gcc -o /dev/shm/fm /dev/shm/sub.o mw.c wav.o -lm
#exec gfortran -ffixed-line-length-none -c sub.f
#exec gcc -o fm sub.o main.c -lm
# return [exec /dev/shm/fm]
return
}
proc doformula { m } {
set t [subst -nocommand {
\\documentclass\{article\}
\\DeclareMathSizes\{16\}\{15\}\{13\}\{11\}
\\begin\{document\}
\\thispagestyle\{empty\}
\\Large
}]
set f [open /dev/shm/formula.tex w]
puts $f [string trim$t \n]
puts $f$m
puts $f "\\end{document}" close$f
catch { exec latex -output-directory=/dev/shm /dev/shm/formula > /dev/null }
catch { exec dvigif -T tight --gif /dev/shm/formula.dvi -o /dev/shm/formula.g
if 2>@ stdout > /dev/null }
}
proc makeprog {form} {
global makeprogout
# cd /dev/shm/S
set makeprogout ""
append makeprogout "[clock format [clock seconds]]:\n"
append makeprogout "evaluating maxima expression..\n"
flush stdout
# append makeprogout "[set in [gets stdin]]<br>--><br>"
append makeprogout "[set r [domaxima $form]]\n" append makeprogout "generating Fortran expression from it..\n" set s [domaximafor$r]
append makeprogout "generating wave..\n"
append makeprogout "[formake $s]\n" # append makeprogout "making mp3 from wave ..\n" # append makeprogout "[exec ffmpeg -i /dev/shm/math.wav -ab 256k -y /dev/shm/ ath.mp3 >& /dev/null ]\n" append makeprogout "making prettyprinted formula ..\n" doformula [domaximaformula$r]
# catch {exec ppmquant 256 /dev/shm/sine.ppm | ppmtogif > /dev/shm/graph1.gif
2> /dev/null}
append makeprogout end\n
# Makes the cgi-bin script stop the server process thinking there's a connection
# exit
}
The above wav.c (actually wav.o) is needed, and this synth program: [L3 ], and wehen all is in plae (current dir) calling makeprog from the above code will generate synthbw which is an executable which willk connct over Jack and Alsa.
A bit more complicated example of a FM synthesizer bwise graph:
This is the formula from {makeprog.o} in neat form:
And it sound like this [L4 ], you can try it for yourself when you have all parts installed by loading the bwise canvas file [L5 ] .
The synth program now uses doubles to communicate with the fortran routine coming from maxima, and so all computations are carried out in double (64 bit float) to compute the waves, which is very accurate. You need the right maxima version, or it will make constants without a D addition which may be compiled into single precision numbers by gcc.
The midi implementation has no envelope generator, it acts on the gate signal only. It does recognize note velocity, and the modulation wheel is assigned to overall volume, and it is polyphonic, call it for instance like this:
./synthbw 0 6 1.0 12 0.3 0.1 0.4 0.6 .2 0.3 0.3 0.4 0.5 0.2 0.1
call it without arguments to get an arg list (some args don't work in this version). I used fairly recent jack version on Fedora 8/64 to run the whole. Of course maxima and gcc are needed, as well as development fies/headers for Jack and Alsa.
TV 2008-09-30: I've added an additive synthesis example with added velocity sensitive fm modulation, for which the formula must contain a variable 'v' (a double from 0 to 1) with 10 sine blocks:
To create the right multiply factors, I created blocks like above with the 'block' button, and changed the b-inputs like this:
for {set i 2} {$i <=10} {incr i} {set mult$i.b "((v^($i/3))/$i)"}
so that the amplitude of the 9 harmonics depends on how hard a key is struck, individually governed by a different exponential. There is one FM component, reused in the graph, but repeated in the tree-like formula representation, which modulates more with higher velocity.
The project files needed for recompile (when using bwise) and the executable for linux (FC8/64 with Jack/Alsa + dev-headers): [L6 ]
2009-02-01: To get an impression of the wave being generated, I've made this two-liner:
time {set r [domaxima float(makelist(subst(y/44100,x,subst((1/2),v,${mult11.out})),y,0,170))]} eval${scope0.bfunc_init} ;foreach i [split [string range $r 1 end-1] ,] { set scope0.in [expr 300*$i] ; eval \${scope0.bfunc} }
To use from the console AFTER a scope0 canvas block has been created by e.g. pressing 'Scope' in to top button row of Bwise:
I haven't taken the time (yet) to make nice blocks (with enumeration and proper pin variables) of this. The main idea is to have maxima compute 171 values from a functions, after having set the virtual synthesizer note velocity value to 0.5 (or rather: 1/2) and creating the time values for the first 171 samples.
Luckily, Maxima then can nicely return a list of values, and even can apply the 'float' command to all at once. The second line clearly puts the result in the scope block for viewing, which can be a longer list than one horizontal screen, but this is only for an impression.
The greatness of the approach, as opposed to the weakness: the 170 computations of a quite not small formula takes a second or two in maxima, is that the whole computation down to the last step takes place in formal algebraic domain with infinite number precision, fractions, special symbolic numbers and error free algebraic manipulation/simplifications applicable! So the rounding takes place on a perfect formula, after all manipulations have been done..
I´ve made this work with windows XP (pro) too, see Wave player for formulas from BWise and Maxima on Windows XP
2009-01-16: To make the above a bit neater but especially to create more complicated algorithms more easily, I make Creating arrays of wave formulas with BWise
TV 2010-09-02: I made a demo here (with some music examples): http://www.youtube.com/watch?v=z1GaGO6zjbc . The Latest BWise page contains the code for some of the latest extensions. | 3,556 | 13,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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} | 2.921875 | 3 | CC-MAIN-2024-30 | latest | en | 0.79168 |
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Solutions e3
# Solutions e3 - TOP NEWS STORIES Late Working NASA...
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Exam #3 practice problems solutions Physics 204 summer 2008 1. NASA makes an announcement that the Hubble Space Telescope has detected a signal from an alien probe. When the probe sent the signal it was (light years 0.20 ly 1 ) away and headed towards Earth at . 0.75 c a. By the time the signal reaches Earth how far away is the probe? How much time will pass on earth between the signal being received and the probe arriving? b. At the moment the probe sends the signal, how far away does the probe detect the earth to be? c. From the moment the probe sends the signal, how long will the probe take to reach the earth according to its own clock? Since the speed of light is finite, the light that was emitted by the probe when it was away took 0.20 ly 0.20 yr to arrive. During that time the probe traveled ( )( ) 0.75 0.20 0.15 d vt c yr ly = = = 0.05 ly . By the time the light reaches Earth the probe will be only away. This means the probe will arrive at Earth in ( ) ( ) 0.05 0.75 0.067 t d v ly c yr = = = Because of length contraction, from the probe’s point of view the earth is only ( ) ( ) ( ) 0.132 2 2 1 1 0.75 0.20 p x v c x Δ = Δ = 0.75 c ly ly = away. ( ) ( ) 0.132 0.75 0.176 v ly c yr = = = t d to reach Earth. At it will take the probe 2. In order to investigate the alien probe featured in the previous problem NASA launches a probe of its own. It speeds away from Earth at . 0.60 c a. How fast is the alien probe moving relative to NASA’s probe? b. Unfortunately the alien probe is highly maneuverable and slips past NASA’s probe. NASA’s probe turns around and pursues the alien probe but can’t keep up due to its slower speed. How fast is the alien probe moving relative to NASA’s probe now? 0.60 NE v c = 0.75 AE v c = − ( ) ( ) ( )( ) 2 2 2 0.75 0.60 0.93 0.75 0.60 1 1 1 AE EN AE NE AN AE EN AE NE c c v v v v v c v v v v c c c c c + = = = = − + The result was negative because of the coordinate system I used. The alien probe is moving to the left from the point of view of NASA’s probe. Once the alien probe slips past NASA’s probe… 1 A light year is the distance that light travels in 1 year. Earth x NASA's probe Alien probe 0.60 NE v c = − 0.75 AE v c = − Earth x NASA's probe Alien probe
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View Full Document | 712 | 2,513 | {"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} | 3.859375 | 4 | CC-MAIN-2018-22 | latest | en | 0.885326 |
https://physics.stackexchange.com/questions/786591/representation-of-nonabelian-wilson-line-in-terms-of-fermionic-fields | 1,719,215,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00083.warc.gz | 402,914,719 | 39,226 | # Representation of nonabelian Wilson line in terms of fermionic fields
Context:
The coupling action of a particle of charge $$q$$ to a $$U(1)$$ gauge field is given by $$$$S = q \int d \tau A_\mu \left( X \right) \frac{dX^\mu(\tau)}{d \tau} = -i \ln W_q, \tag{1}$$$$ where $$$$W^{\text{abelian}}_q = \exp{ \left(iq \int A_\mu dX^\mu\right) }$$$$ is the Wilson line, the integral being over the trajectory. For a particle charged under a nonabelian gauge field, it seems reasonable to try to find the coupling by considering the nonabelian version of the Wison line: $$$$W^{\text{nonabelian}} = \text{Tr} \, \mathcal{P} \exp{ \left(i \int A_\mu dX^\mu\right) } = \text{Tr} \, \prod_{\tau=\tau_i}^{\tau_f} \left( 1 + i \,d \tau A_\mu \left( X \right) \frac{dX^\mu(\tau)}{d \tau}\right).$$$$ My reason for considering this is that such a coupling should appear between the endpoints of open strings to the nonabelian gauge field one finds in the massless spectrum in the presence of D-branes, but nothing in this question depends on the details of string theory.
The question:
How should one deal with the path-ordering $$\mathcal{P}$$ in $$W^{\text{nonabelian}}$$ in order to rewrite this as a simple integral like (1)? The paper Particles with non abelian charges suggests the form $$$$S_{\text{NA}} = \int d\tau \left( \bar{c}^\alpha \frac{dc_\alpha}{d \tau} -i A^a_\mu (X) \frac{d X^\mu}{d \tau} \bar{c}^\alpha \left( T^a \right)_\alpha^{\phantom{a} \beta} c_\beta \right),$$$$ where $$c_\alpha$$ and $$\bar{c}^\alpha$$ are fermionic fields transforming respectively in the fundamental and anti-fundamental representations of $$SU(N)$$, under which the particle is charged, and satisfying a Dirac algebra with respect to these indices. The $$\left( T^a \right)_\alpha^{\phantom{a} \beta}$$ are $$SU(N)$$ generators in the chosen representation. Integrating out these fermions in the generating functional should give $$$$\int \mathcal{D} c \mathcal{D}\bar{c} e^{i S_{\text{NA}}} \sim \det \left( \delta^\beta_\alpha \frac{d}{d \tau} -i A^a_\mu (X) \frac{d X^\mu}{d \tau} \left( T^a \right)_\alpha^{\phantom{a} \beta} \right) = e^{iS_{\text{coupling}}},$$$$ but I was not able to check this, due to the complicated nature of the operator inside the determinant. | 719 | 2,266 | {"found_math": true, "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": 14, "wp-katex-eq": 0, "align": 0, "equation": 5, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.806298 |
https://web2.0calc.com/questions/help-plese_3 | 1,606,781,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00284.warc.gz | 553,390,496 | 6,292 | +0
# HElp plese
+1
32
4
Javier leaves Edmonton AB, at 1:20 p.m and arrives in St. John'S, NL, at 8:45 p.m How long is his fight?
Nov 12, 2020
#1
+93
+2
Hello Guest!
You're pretty lucky. I just happened to be on :D
So, since they're both PM, we don't have to do any conversions.
We know that 7 hours have passed, and also 25 minutes from 1:20 to 8:45.
Converting the 7hours to minutes and adding 25, we get 445 minutes. Javier's flight took 445 minutes to complete.
I hope this helped!
Nov 12, 2020
#2
+2
St. John's Nfld is 3:30 hours ahead of Edmonton, AB.
When the plane leaves Edmonton at 1:20 pm, it is already 4:50 pm in St. John's.
8:45 pm - 4:50 pm =3:55 hours - duration of the flight
Nov 12, 2020
#4
+93
+2
I believe the problem didn't take into account that the time zones are different, since if it did, then the problem would probably have told us which time zones to follow and what time zones the times it gave us were in.
And who downvoted the above post? This was an amazing attempt by a user, and if it was the wrong answer, at least congratulate them on trying!
ETERNITY Nov 12, 2020
edited by ETERNITY Nov 12, 2020
#3
+112375
+2
Note that this could be a "trick" question.....
Note that the time difference between Edmonton and St. Johns = 3 and 1/2 hrs
Thus when it is 8:45 PM in St Johns, it is actually just 5:15 PM in Edmonton
So....he has actually been traveling from 1:20PM to 5:15 PM as viewed in terms of Edmonton time
So....he actually only travels 5 minutes less than 4 hours = 3 hrs 55 min
Nov 12, 2020 | 522 | 1,584 | {"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} | 2.921875 | 3 | CC-MAIN-2020-50 | longest | en | 0.940243 |
https://goprep.co/explain-with-the-help-of-a-labelled-circuit-diagram-how-you-i-1njf9j | 1,606,325,496,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00420.warc.gz | 316,192,417 | 67,516 | Q. 4 A3.8( 13 Votes )
# Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R1, R2 and R3 joined in parallel.
Answer :
Let the resistance of the three resistors be R1, R2 and R3, respectively. Let their combined resistance be R.
Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have: V= IR------- (1)
We know that when the resistors are connected in parallel, the potential drop across each resistance is the same. Therefore:
I = I1+I2+I3
I = V/R1+ V/R2+ V/R3
1/R=1/R1+1/R2+ 1/R3
If two resistances are connected in parallel, then the resultant resistance will be
1/R=1/R1+1/R2+ 1/R3
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view all courses | 385 | 1,497 | {"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} | 3.359375 | 3 | CC-MAIN-2020-50 | longest | en | 0.859679 |
https://www.doubtnut.com/question-answer/find-the-value-of-cot-1-sqrt3-cosec-12-sec-1-sqrt2-69138508 | 1,632,724,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00366.warc.gz | 733,103,990 | 75,281 | Home
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Find the value of cot^(-1)(-sq...
# Find the value of cot^(-1)(-sqrt3)+cosec^(-1)(2)+sec^(-1)(-sqrt2).
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 17-12-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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27.7 K+
14.7 K+
Text Solution
Solution :
Since the principal values of cot^(-1)x,cosec^(-1)x and sec^(-1)x"lie in "(0,pi), {-pi/2,pi/2]-{0} and [0.pi]-{pi/2} respectively and in these intervals cot ((5pi)/6)=-sqrt3,cosec.(pi/6)=2,sec((3pi)/4)=-sqrt2 <br> :.cot^(-1)(-sqrt3)=(5pi)/6,cosec^(-1)(2)=pi/6,sec^(-1)(-sqrt2)=(3pi)/4 <br> :. cot^(-1)(-sqrt3)+cosec^(-1)(2)=pi/6,sec^(-1)(-sqrt2)=(3pi)/4 <br> :.cot^(-1)(-sqqrt3)+cosec^(-1)(2)+sec^(-1)(-sqrt2) <br> (5pi)/6+pi/6+(3pi)/4 <br> =(7pi)/4
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## Latest Questions
Class 12th
Inverse Trigonometric Functions
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Inverse Trigonometric Functions
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Inverse Trigonometric Functions | 757 | 1,897 | {"found_math": true, "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": 1, "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} | 4.09375 | 4 | CC-MAIN-2021-39 | latest | en | 0.440466 |
https://topic.alibabacloud.com/a/leetcode-valid-font-colorredperfectfont-font-colorredsquarefont-test-complete-font-colorredsquarefont-number_8_8_31200192.html | 1,679,834,293,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00312.warc.gz | 663,614,492 | 16,765 | # [Leetcode] Valid Perfect Square test Complete square number
Source: Internet
Author: User
Given a positive integer num, write a function which returns True if num is a perfect square else False.
Note: do not use any built-in library function such as `sqrt` .
Example 1:
`Input:16returns:true`
Example 2:
`Input:14returns:false`
Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.
This problem gives us a number, let us determine whether it is a complete square number, then it is obvious that we must not use brute force, so it is not efficient, so the smallest can be exponential speed to narrow the range, then I first think of the method is this, such as a number 49, we first divided it by 2 , get 24, found that 24 squared is greater than 49, then 24 divided by 2, get 12, found that 12 squared or greater than 49, then 12 divided by 2, get 6, found that 6 squared is less than 49, and then traverse 6 to 12 of all the number, see if there is no square equals 49, there is return true, return FALSE if not, see the code below:
Solution One:
`classSolution { Public: BOOLIsperfectsquare (intnum) { if(num = =1)return true; Longx = num/2, t = x *x; while(T >num) {x/=2; T= x *x; } for(inti = x; I <=2* x; ++i) {if(i * = = num)return true; } return false; }}; `
The following method is also more efficient, from 1 to sqrt (num), see if there is no square equal to num number:
Solution Two:
`class Solution {public: bool isperfectsquare (int num) { for (int1; I <= num/i; + +i) {ifreturntrue ; } return false ; `
We can also use the binary lookup method, to find the number of Mid*mid, see the code as follows:
Solution Three:
`classSolution { Public: BOOLIsperfectsquare (intnum) { Longleft =0, right =num; while(Left <=Right ) { LongMid = left + (right-left)/2, T = Mid *mid; if(t = num)return true; Else if(T < num) left = mid +1; Elseright = mid-1; } return false; }};`
The following method is pure mathematical solution, the use of such a property, the total square number is a series of odd sum, for example:
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
1+3+...+ (2n-1) = (2n-1 + 1)N/2 =n
There is no proof, I will not prove that, knowing the nature, you can use it to solve the problem, the time complexity of O (sqrt (n)).
Solution Four:
`class Solution {public: bool isperfectsquare (int num) { int 1 ; while 0 ) { -= i ; 2 ; } return 0 ; }};`
The following method is a similar approach to the first method, which is more streamlined and has a time complexity of O (LGN):
Solution Five:
`class Solution {public: bool isperfectsquare (int num) { long x = num; while (x * x > num) { 2; } return x * = = num; }};`
This problem actually has the O (1) solution, this you believe? That's insane, please see this post on the forum for details.
Similar topics:
SQRT (x)
Resources:
Https://leetcode.com/discuss/110639/o-logn-bisection-method
Https://leetcode.com/discuss/110792/simple-for-loop-o-sqrt-n
Https://leetcode.com/discuss/110638/a-square-number-is-1-3-5-7-java-code
Https://leetcode.com/discuss/110659/o-1-time-c-solution-inspired-by-q_rsqrt
Https://leetcode.com/discuss/110671/3-4-short-lines-integer-newton-most-languages
Leetcode all in one topic summary (continuous update ...)
[Leetcode] Valid Perfect Square test Complete square number
Related Keywords:
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 1,210 | 4,525 | {"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} | 4.09375 | 4 | CC-MAIN-2023-14 | latest | en | 0.813101 |
https://courses.ideate.cmu.edu/15-104/f2017/2017/10/27/haewanp-project-09-computational-portrait/ | 1,632,843,711,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00343.warc.gz | 226,062,825 | 10,605 | # haewanp – Project 09 – Computational Portrait
``````var img;
var x;
var y;
var blue_range;
var red_range;
var yellow_range;
}
function setup() {
createCanvas(360, 480);
noFill();
yellow_range = 7; //initial value;
}
function draw() {
background(255);
blue_range = map(mouseX, 0, width, 0, 8);
red_range = map(mouseY, 0, height, 0, 10);
for (y = 0; y < height; y+=6) {
for (x = 0; x < width; x+=6) {
var i = y * width + x;
//color
var redness = (255 - img.pixels[i*4]) / 255;
var yellowness = (255 - img.pixels[(i+1)*4]) / 255;// I just decide to represent green value among RGB as yellow color
var blueness = (255 - img.pixels[(i+2)*4]) / 255;
//blue diagonal line
stroke(20, 20, 255);
strokeWeight(blueness * blue_range);
line(x - 3, y - 3, x + 3, y + 3);
//red diagonal line
stroke(255, 20, 20);
strokeWeight(redness * red_range);
line(x + 3, y - 3, x - 3, y + 3);
//yellow ellipse
noStroke();
fill(245, 220, 0);
ellipse(x, y, yellowness * yellow_range, yellowness * yellow_range);
}
}
}
function mousePressed() {
yellow_range = random(1, 12); //yellow range changes when you press mouse
}
``````
In this project, I learned that there are many ways to depict pixels. I represent this portrait with dividing each R, G, B value (Later, I represent green value with yellow color). Based on each R, G, B value, size and stroke weight are determined. Also, this can be played around with the mouse behaviors. There are several variations based on mouse behavior below. | 464 | 1,477 | {"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} | 2.78125 | 3 | CC-MAIN-2021-39 | latest | en | 0.646324 |
http://forums.eviews.com/viewtopic.php?p=41746 | 1,586,477,023,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00556.warc.gz | 70,690,642 | 9,404 | An introduction to EViews programming.
For questions regarding programming in the EViews programming language.
Moderators: EViews Gareth, EViews Jason, EViews Moderator, EViews Matt
samonia
Posts: 2
Joined: Thu Nov 20, 2014 4:11 am
Contact:
Re: An introduction to EViews programming.
I would like to store in a vector (or in a group, I don't know) the "s.e. of regression"s for all my 17 equations that I have in a file. What would be the command line for that? Thank you a lot.
_______________
врати блиндирани врати входни врати интериорни врати
Last edited by samonia on Sun Nov 30, 2014 5:58 am, edited 1 time in total.
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
Code: Select all
`%eqlist = @wlookup("*","equation")!n = @wcount(%eqlist)vector(!n) sesfor !i=1 to !n %eq = @word(%eqlist, !i) ses(!i) = {%eq}.@senext`
Follow us on Twitter @IHSEViews
neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
Hi guys, I am new to Eviews. I have a couple of question and appreciate any help.
I have the following code and trying to understand what it does: Specifically I need to first know what does w{%Y}(!i+1) mean?
Last edited by neshate on Tue Mar 31, 2015 3:25 pm, edited 1 time in total.
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
You have two for loops. The first loops through %Y, the second loops through !i.
%Y is looping through object names. !i is looping through elements. Thus the first time through, we have w{%y}(!i+1), which means w914(1), which is the first element of the vector w914.
Follow us on Twitter @IHSEViews
neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
Thanks. So does that mean we never enter the if statement because we have not defined w914 yet?
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
w914 must have existed prior to running the program. You'll enter the if statement if the first element of it is not equal to an NA.
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neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
Ohhh. Thank you very much. That helped a lot. So, I have two more question:
1. w{%Y}eqb{!i}.ls log(w{%Y}) C YEAR does this generate a least square regression of log(w914) with some constant ? What is YEAR? does this also exist before running?
2. What does this mean? w{%Y}eqa{!i}.c(2)
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
1) It estimates an equation of log(w194) on a constant and the variable year (which would have to previously exist, yes). It stores this equation in the object w914eqb1
2) It retrieves the second coefficient from the equation w914eqb1
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neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
Excuse me if I am asking stupid questions. what is the effect of smpl 1960+!i 1965+!i? does this mean that we have selected certain values of w914 for eqa and eqb?
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
Yes. Only observations between 1960+1 and 1965+1 etc...
Follow us on Twitter @IHSEViews
neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
1. What is the format of w914eqa0 and w914eqb0 look like? I am trying to figure out what kind of table the .results line generates? for example when we try to access (4,1) element or (10,2)?
2. What does that 0.02 and 0.035 comes from? I feel that this program tries to do a t-test on a sample data, but I can not explain how. Any help on this is appreciated.
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
I think you'll have to play around a bit and figure out the rest
Follow us on Twitter @IHSEViews
neshate
Posts: 6
Joined: Tue Mar 31, 2015 7:25 am
Re: An introduction to EViews programming.
Thanks for the help. Unfortunately, I do not have access to eviews and that is why I am not able to run this program ans see the results. But, thanks anyway. You helped a lot.
maujasmin
Posts: 4
Joined: Wed Nov 25, 2009 7:40 am
Re: An introduction to EViews programming.
Hi, I am new here so I hope you could help me.
We are doing an index for our work and it utilizes principal component analysis.
Basically, I have 10 variables which need to normalize. After normalizing the variables , I need to run a principal component analysis for these variables.
Once I have done that, I need to get the the 1st principal component for each variable and multiply that with the normalized values.
To get the index, I just need to add those values ( normalized values* 1st principal component)
So I would like to know if you could give the syntax.
What I want to do is to import the data from an excel file and run the commands I enumerated above.
TIA!
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12420
Joined: Tue Sep 16, 2008 5:38 pm
Re: An introduction to EViews programming.
What have you go so far?
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Users browsing this forum: No registered users and 4 guests | 1,680 | 5,633 | {"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} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.854759 |
https://fr.maplesoft.com/support/help/view.aspx?path=DifferentialGeometry%2FTensor | 1,656,115,018,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00712.warc.gz | 316,866,278 | 48,430 | Overview - Maple Help
Overview of the Tensor package
Description
• The DifferentialGeometry:-Tensor package provides an extensive suite of commands for computations with tensors on the tangent bundle of any manifold or with tensors on any vector bundle.
• The Tensor package contains commands for the standard algebraic operations on tensors as well as commands for covariant differentiation and curvature calculations (for metric connections, general affine connections, or connections on vector bundles).
• The Tensor package also includes a full implementation of the 2 component spinor and Newman-Penrose formalisms for space-time computations (pseudo-Riemannian manifolds with metric signature [+1, -1, -1, -1] ). Petrov and Segre classifications of spacetimes can be calculated as well as complete sets of curvature invariants.
• All tensor computations can be done in an arbitrary frame or co-frame on the manifold. In particular, all curvature computations for a (pseudo-)Riemannian metric can be performed with respect to an orthonormal frame.
• The Tensor package, working in conjunction with other Differential Geometry commands, provides great flexibility for mapping tensors between manifolds. For example, if G is a Lie group acting on a manifold M, then the PushPullTensor command can be used to push forward the G invariant tensors on M to tensor fields on the quotient manifold M/G.
• Commands are available for calculating the Laplace-Beltrami operator on differential forms and for the Schouten and Frolicher-Nijenhuis brackets of tensor fields. These bracket operations are important in complex geometry and in Poisson geometry.
• Infinitesimal transformation groups such as the Killing vectors of a metric can be calculated.
• Infinitesimal holonomy of a metric or a connection can be calculated.
• There are commands for computing special tensor fields such as Killing-Yano tensors.
• The Tensor package is fully integrated with the LieAlgebras and LieAlgebraRepresentations packages which allows for the computation of, for example, the invariant tensors on a Lie algebra.
• The Differential Geometry Lessons (Lessons 9 and 10) provide a systematic introduction to the commands in the Tensor package.
• Each command in the Tensor package can be accessed by using either the long form or the short form of the command name in the command calling sequence.
Commands for the algebraic manipulation of tensors
• CanonicalTensors: create various standard tensors.
• ContractIndices: contract the indices of a tensor.
• Convert/DGspinor: convert a tensor to a spinor.
• Convert/DGtensor: convert an array, vector, p-form, spinor, ... to a tensor.
• DGGramSchmidt: construct an orthonormal basis of vector, forms, tensors with respect to a metric.
• FormInnerProduct: compute the inner product of two forms with respect to a given metric tensor.
• GenerateSymmetricTensors: generate a list of symmetric tensors from a list of tensors.
• GenerateTensors: generate a list of tensors from a list of lists of tensors.
• HodgeStar: apply the Hodge star operator to a differential form.
• InverseMetric: find the inverse of a metric tensor.
• KroneckerDelta: find the Kronecker delta tensor of rank r.
• MetricDensity: use a metric tensor to create a scalar density of a given weight.
• MultiVector: compute the alternating sum of the tensor product of a list of vector fields.
• PermutationSymbol: create a permutation symbol.
• PlebanskiTensor: calculate the Plebanski tensor from a trace-free rank 2 symmetric tensor.
• PushPullTensor: transform a tensor from one manifold or coordinate system to another.
• QuadraticFormSignature : find the signature of a covariant, symmetric, rank 2 tensor.
• RaiseLowerIndices: raise or lower a list of indices of a tensor.
• RearrangeIndices: rearrange the argument/indices of a tensor.
• SymmetrizeIndices: symmetrize or skew-symmetrize a list of tensor indices.
• TensorInnerProduct: compute the inner product of two vectors, forms or tensors with respect to a given metric tensor.
Commands for tensor differentiation
• Christoffel: find the Christoffel symbols of the first or second kind for a metric tensor.
• Connection: define a linear connection on the tangent bundle or on a vector bundle.
• CovariantDerivative: calculate the covariant derivative of a tensor field with respect to a connection.
• DirectionalCovariantDerivative: calculate the covariant derivative of a tensor field in the direction of a vector field and with respect to a given connection.
• GeodesicEquations: calculate the geodesic equations for a symmetric linear connection on the tangent bundle.
• Laplacian: find the Laplacian of a differential form with respect to a metric.
• ParallelTransportEquations: calculate the parallel transport equations for a linear connection on the tangent bundle or a linear connection on a vector bundle.
• TensorBrackets: calculate the Schouten bracket and Frolicher-Nijenhuis brackets of tensor fields.
• TorsionTensor: calculate the torsion tensor for a linear connection on the tangent bundle.
Commands for calculating curvature tensors
• BachTensor: calculate the Bach tensor of a metric.
• CottonTensor: calculate the Cotton tensor for a metric.
• CurvatureTensor: calculate the curvature tensor of a linear connection on the tangent bundle or on a vector bundle.
• EinsteinTensor: calculate the Einstein tensor for a metric.
• ProjectiveCurvatureTensor: calculate the Weyl projective curvature tensor of a connection on the tangent bundle.
• RicciScalar: calculate the Ricci scalar for a metric.
• RicciTensor: calculate the Ricci tensor of a linear connection on the tangent bundle.
• RiemannInvariants: calculate a complete set of scalar curvature invariants in 4 dimensions.
• SectionalCurvature: calculate the sectional curvature for a metric.
• SchoutenTensor: calculate the Schouten tensor of a metric.
• TraceFreeRicciTensor: calculate the trace-free Ricci tensor for a metric.
• WeylTensor: calculate the Weyl curvature tensor of a metric.
Infinitesimal transformation groups
• ConformalKillingVectors: calculate the conformal Killing vectors for a given metric.
• HomothetyVectors: calculate the homothety vectors for a given metric.
• KillingVectors: calculate the Killing vectors or infinitesimal isometries for a given metric.
Commands for calculating holonomy
• InfinitesimalHolonomy: find the matrix Lie algebra giving the infinitesimal holonomy of a metric or a connection on the tangent bundle or on a general vector bundle.
• InvariantTensorsAtAPoint: find tensors or differential forms which are invariant under the infinitesimal action of a set of matrices.
Commands for calculating special tensor fields
• CovariantlyConstantTensors: calculate the covariantly constant tensors with respect to a given metric or connection.
• KillingSpinors: calculate the Killing spinors for a given spacetime.
• KillingYanoTensors: calculate the Killing tensors of a specified rank for a given metric or connection.
• KillingTensors: calculate the Killing-Yano tensors for a given connection or a given metric.
• RecurrentTensors: calculate the recurrent tensors with respect to a given metric or connection.
Commands for working with Killing tensors
• CheckKillingTensor: check that a tensor is the Killing tensor for a metric.
• IndependentKillingTensors: create a list of linearly independent Killing tensors.
• KillingBracket: a covariant form of the Schouten bracket for symmetric tensors.
• SymmetricProductsOfKillingTensors: form all possible symmetric tensors of a given rank.
Commands for the 2-component spinor formalism
• BivectorSolderForm: calculate the rank 4 spin-tensor which maps bivectors to symmetric rank 2 spinors.
• ConjugateSpinor: calculate the complex conjugate of a spinor.
• EpsilonSpinor: calculate the epsilon spinor in the 2 component spinor formalism.
• FactorWeylSpinor: factorize a rank 4 symmetric spinor.
• KroneckerDeltaSpinor: calculate the Kronecker delta spinor in the 2 component spinor formalism.
• RaiseLowerSpinorIndices: raise/lower the indices of a spinor or spin-tensor using the epsilon spinors.
• RicciSpinor: calculate the rank 4 Ricci spinor corresponding to the trace-free Ricci tensor.
• SolderForm: calculate the solder form (or Infeld-van der Waerden symbols) from an orthonormal tetrad.
• SpinConnection: calculate the unique spin connection defined by a solder form.
• SpinorInnerProduct: contract all spinor indices of a pair of 2-component spin-tensors using the epsilon spinors.
• WeylSpinor: calculate the rank 4 Weyl spinor corresponding to the Weyl tensor.
Commands for the Newman-Penrose formalism
• GRQuery: verify various properties of spacetimes.
• NPBianchiIdentities: calculate the Bianchi identities in the Newman-Penrose formalism.
• NPCurvatureScalars: calculate the Ricci scalars and the Weyl scalars in the Newman-Penrose formalism.
• NPDirectionalDerivatives: define the directional derivative operators used in the Newman-Penrose formalism.
• NPRicciIdentities: calculate the Ricci identities in the Newman-Penrose formalism.
• NPSpinCoefficients: calculate the Newman-Penrose spin coefficients.
• NullVector: construct a null vector from a solder form and a rank 1 spinor.
• PrincipalNullDirections: find the principal null directions of a 4-dimensional spacetime.
Commands for the algebraic classification of spacetimes
• CongruenceProperties: calculate the geometry properties of a line congruence.
• IsotropyType: determine the isotropy type of the isotropy subalgebra of infinitesimal isometries.
• PetrovType: determine the Petrov type of a spacetime from the Weyl tensor.
• SegreType: determine the Plebanski-Petrov type and the Segre type of a trace-free, rank 2 symmetric tensor.
• SubspaceType: determine the signature of the metric restricted to a subspace.
Commands for field theory
• BelRobinson: calculate the rank 4 Bel-Robinson tensor for a metric.
• DivergenceIdentities: check the divergence identity for various energy-momentum tensors.
• EnergyMomentumTensor: calculate the energy-momentum tensor for various fields (scalar, electromagnetic, dust, ...).
• MatterFieldEquations: calculate the field equations for various field theories (scalar, electromagnetic, dust, ...).
• RainichConditions: check that a metric tensor satisfies the Rainich conditions.
• RainichElectromagneticField : from a given metric satisfying the Rainich conditions, calculate an electromagnetic field which solves the Einstein-Maxwell equations.
Alphabetical listing of all Tensor commands | 2,328 | 10,666 | {"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} | 2.65625 | 3 | CC-MAIN-2022-27 | longest | en | 0.821546 |
https://physics.stackexchange.com/questions/162835/expectation-value-with-plane-waves | 1,566,089,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00151.warc.gz | 600,196,211 | 29,442 | # Expectation value with plane waves [duplicate]
This question already has an answer here:
Hey guys Im a little confused with the concept of plane waves and how to perform an expectation value. Let me show you by an example. Suppose you have a wave function of the form
$\psi_{\boldsymbol{p}_{0}}(x)=f(p_{0})e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}$
where $\boldsymbol{p}_{0}=(0,0,p_{0})$ and suppose you want to perform an expectation value of the position of the particle, that is
$<x>=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x\,e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}e^{-\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x$
wich I think is nonsense. But if you define an arbitrary momentum vector $\boldsymbol{p}'=(p_{1}',p_{2}',p_{3}')$, and perform the transition probability
\begin{align} \left<\psi_{\boldsymbol{p}'}(\boldsymbol{x})|\,x\,|\psi_{\boldsymbol{p}_{0}}(\boldsymbol{x})\right>&=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\,xe^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\left( i\hbar\frac{\partial}{\partial p_{x}'}\right)e^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}= \\ &=i\hbar f(p')f(p_{0})\frac{\partial}{\partial p_{x}'}\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})=-i\hbar\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})\frac{\partial}{\partial p_{x}'}\left( f(p')\right)f(p_{0}) \end{align}
where I made use of the property $f(x)\delta'(x)=-f'(x)\delta(x)$. So now with my new expresion I have a meaningful result and I can evaluate for $\boldsymbol{p}'=\boldsymbol{p}_{0}$ and get a result that I wasnt able to get with the first method. What I'm doing wrong or the the second way is the correct way to do it? Thanks!
## marked as duplicate by ACuriousMind♦, Brandon Enright, Kyle Kanos, Pranav Hosangadi, JamalSFeb 2 '15 at 7:44
Nothing will help, a plane wave occupies all the space, and the mean value of the position doesn't make much sense, but in your case is zero, because the integrand in the 1st calculus is anti-symmetrical. But this, on one condition, namely if we write the integral as
$lim _{a \to \infty} f^2(p_0) \int _a^a x d^3 x$
Otherwise it's hard to say what is the value of your integral.
The second integration is something else than the 1st one, it evaluates the matrix elements of the quantity $x$ in the base of the Fourier functions. The matrix element that is equivalent with the mean value of $x$, is the diagonal one for $p' = p_0$.
I am not sure whether you can interchange between the derivative and the integral. Up to the point where you introduced the derivative (and including that step), the integral is zero for $p' = p_0$, in the condition that we guarantee the equality in absolute value of the limits. But even if the interchange were allowed, in your before-last expression, $\delta (p' - p_0)$ for $p' = p_0$ is infinite, and you try to calculate the derivative of an infinite constant. It's not advisable to use this way.
• That was my first answer when this problem came up to me. But you can tell the same about the second result? Both have to be the same when p'=p0 – Tomas Libutti Feb 1 '15 at 21:55
• @TomasLibutti I am writing the answer to what you ask. – Sofia Feb 1 '15 at 22:00
• @TomasLibutti : no, I don't think that the answer is as you said, because your second way is not the same as the first. In your second calculus you evaluate matrix elements. – Sofia Feb 1 '15 at 22:03
• But the first way isnt just an evaluation of the diagonal matrix elements? – Tomas Libutti Feb 1 '15 at 22:04
• @TomasLibutti : no, I have to introduce a modification, look at it. – Sofia Feb 1 '15 at 22:11
wich I think is nonsense
Formally, shouldn't the 2nd equation be
$$\langle x \rangle = \frac{\langle\psi_{\mathbf p_0}|\hat x|\psi_{\mathbf p_0}\rangle}{\langle\psi_{\mathbf p_0}|\psi_{\mathbf p_0}\rangle} = \frac{f^2(p_0)\int d\mathbf x^3 x}{f^2(p_0)\int d\mathbf x^3} = \frac{\int d\mathbf x^3 x}{\int d\mathbf x^3} = \lim_{\tau \rightarrow \infty}\frac{\int_{-\tau}^{\tau} d\mathbf x^3 x}{\int_{-\tau}^{\tau} d\mathbf x^3} = \lim_{\tau \rightarrow \infty}\frac{0}{2\tau} = 0$$ | 1,382 | 4,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-35 | latest | en | 0.700775 |
https://excelfind.com/excel-functions/excel-not-function/ | 1,713,080,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00709.warc.gz | 233,798,440 | 28,354 | # Excel NOT function
## Summary
The Excel NOT function is designed to invert logical or Boolean values. By flipping TRUE to FALSE and vice versa, NOT is essential for reversing conditions, making it a key player in creating dynamic, conditional logic within formulas.
##### Syntax
``` ```
=NOT(logical)
```
```
• logical: The value or logical expression to be inverted.
##### Return value
The opposite of the provided logical value: FALSE if given TRUE, and TRUE if given FALSE.
## How to use
Use the NOT function when you need to reverse the outcome of a logical test or Boolean value. It’s often combined with other logical, comparison, or information functions to construct more complex conditions.
## Examples
##### Simple NOT
To simply invert a true or false condition:
``` ```
=NOT(A1=B1)
```
```
This formula returns TRUE if A1 does not equal B1 and FALSE if A1 does equal B1, effectively reversing the normal equality check.
##### NOT and AND
Excluding Multiple Conditions: To ensure a scenario doesn’t meet multiple conditions:
``` ```
=NOT(AND(B1>100, C1="Approved"))
```
```
This formula will return TRUE if B1 is not greater than 100 and C1 is not “Approved.” It’s useful for scenarios where you want to exclude a specific combination of conditions.
##### NOT and OR
Avoiding Either of Two Scenarios: To check that neither of two conditions is true:
``` ```
=NOT(OR(A1="Closed", A2="Closed"))
```
```
Here, the formula returns TRUE only if neither A1 nor A2 is “Closed.” It’s practical when needing to ensure that neither of two scenarios occurs.
##### NOT and IS... Functions
Validating Non-Empty, Non-Error Entries: To confirm a cell contains a valid, non-error entry:
``` ```
=NOT(OR(ISBLANK(D1), ISERROR(D1)))
```
```
This formula checks that D1 is neither blank nor contains an error, returning TRUE for valid, non-error entries.
##### NOT and IF
Conditional Actions Based on Inverted Logic: To perform an action based on the opposite of a usual condition:
``` ```
=IF(NOT(A1>=10), "Less than 10", "10 or more")
```
```
This formula uses NOT within an IF statement to perform actions based on inverted logic. Here, if A1 is not greater than or equal to 10, it labels the entry as “Less than 10”; otherwise, it’s “10 or more”. | 550 | 2,269 | {"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} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.797554 |
https://fr.mathworks.com/matlabcentral/answers/2032009-what-is-vectorization-why-use-vectorization-instead-of-loops | 1,713,360,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00859.warc.gz | 234,455,536 | 25,348 | # What is vectorization? Why use vectorization instead of loops?
58 vues (au cours des 30 derniers jours)
Hans Scharler le 11 Oct 2023
Over at Reddit, a user asked about vectorization vs. a loop. This is a fundemental concept in MATLAB. So, what is vectorization? Why use it instead of loops?
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### Réponse acceptée
Hans Scharler le 11 Oct 2023
Vectorization is about replacing explicit loops with matrix and vector operations. This can lead to more readable and efficient code. MATLAB is particularly well-suited for vectorized operations as it is designed to work with matrices and vectors natively.
Here's a normal loop approach to square each element in a vector.
n = 10;
vector = 1:n;
result = zeros(1, n);
% Use a loop to square each element of a vector
for i = 1:n
result(i) = vector(i)^2;
end
disp(result);
1 4 9 16 25 36 49 64 81 100
Here's how this looks with vectorization:
vector = 1:10;
% Square each element of a vector using vectorization
result = vector.^2;
disp(result);
1 4 9 16 25 36 49 64 81 100
The magic here is the ".^" which is an element-wise operation. You can also do this with ".*" and "./".
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Translated by | 428 | 1,625 | {"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} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.683772 |
https://jarviscodinghub.com/product/lab-10-objects-and-classes-solution/ | 1,718,951,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00076.warc.gz | 285,986,256 | 26,587 | # ITI 1121 Lab #10 Objects and Classes solution
\$30.00
Original Work ?
## Case Study: A “Time” class
• Suppose we want to be able to work with values
representing clock times to 1 minute precision.
– What information to we have to store to represent
a time? How do we store that information?
– What operations might we want to do with a Time
value?
What to store in “Time”?
• Two integers:
– hour: the hour number (0 ≤ hour ≤ 23)
– minute: the minute number (0 ≤ minute ≤ 59)
class Time
{
public int hour;
public int minute;
}
• Alternatives?
Declaring and creating Time objects
// DECLARE VARIABLES
Time time1; // time is null here
// USING Time OBJECT
time1 = new Time( ); // We now have a Time object
// but it contains ? for values
time1.hour = 17;
time1.minute = 45; // time1 now represents 17:45
What do we have?
time1
17
45
hour
minute
A method to set the time
class Time
{
public int hour;
public int minute;
public void setTime( int h, int m )
{
this.hour = h;
this.minute = m;
}
}
Usage
// DECLARE VARIABLES
Time time1; // time is null here
// USING Time OBJECT
time1 = new Time( );
time1.setTime( 17, 45 );
// time1 now represents 17:45
This method is different!
• Did anyone see what was “missing” from the method?
public void setTime( int h, int m )
{
this.hour = h;
this.minute = m;
}
• The word static does not appear in the method
Instance methods
• When the word static does not appear in the
method header, this means that the method can be
called via a variable declared to be of a type matching
the class name. This is called an “instance” method.
• An instance method can make use of the variables
defined in the class.
• The result: the method will produce different
results for different object instances.
Instance methods
• For example,
Time time1;
Time time2;
time1 = new Time( );
time2 = new Time( )
time1.setTime( 17, 45 ); // time1 is 17:45
time2.setTime( 14, 30 ); // time2 is 14:30
What do we have?
time1
17
45
hour
minute
time2
14
30
hour
minute
this
• Objects time1 and time2 use the same code in class
Time to set their own copy of hour and minute.
• When we want to refer to “the object on which I was
called”, we use this.
this
time1
17
45
hour
minute
time2 14
30
hour
minute
this
time1.setTime( 17, 45 );
Information Hiding
• If we want to ensure that:
– hour: must be in range 0 ≤ hour ≤ 23
– minute: must be in range 0 ≤ minute ≤ 59
class Time
{
private int hour;
private int minute;
}
• Instead, access should be provided through setTime, and we
can adjust the values if needed.
Revised version of setTime
public void setTime( int h, int m )
{
// If minutes value is too large, adjust it
// by mod 60, and add to hours value.
if ( m > 59 )
{
h = h + m / 60; // determine hours to add
m = m % 60; // puts minutes in range
}
else
{
; // do nothing
}
this.hour = h % 24; // puts hours in range
this.minute = m;
}
Accessors
• With the variables now declared to be private, we
need to provide a way for other classes to ask for the
values.
public int getHours()
{
return hour;
}
public int getMinute( )
{
return minute;
}
Compare times for equality
• Suppose we want a method that checks whether one
time object is equal to another.
• One approach: a static method
public static boolean isEqual( Time t1, Time t2 )
This would be called as Time.isEqual( t1, t2 )
• Alternative: an instance method
public boolean isEqual( Time t2 )
This would be called as t1.isEqual( t2 )
The static method
public static boolean isEqual( Time t1, Time t2 )
{
return (t1.hour == t2.hour) &&
(t1.minute == t2.minute);
}
• If the method is inside the class Time, it can access
the private variables inside the class.
The instance method
public boolean isEqual( Time t2 )
{
return (this.hour == t2.hour) &&
(this.minute == t2.minute);
}
• In this case, we are comparing “ourself” to another
Time value.
## Exercise 1: Add two instance methods to provided Time class
• public boolean isBefore( Time t )
that returns true is the Time t is before the
current time object
• public Time duration( Time t )
This method calculates the time duration from
the current time to the given time t. Return the
duration as a Time object.
Also test your two methods using Junit tests from
the provided file TimeTest.java. You will need to
Run button to execute the tests.
provided TimeTest.java to Dr.Java. Compile both files,
and then press “Test” menu button in Dr.Java.
TimeTest.java contains many tests. Those tests for which
displayed in green and those for which your solution
fails will be displayed in red. Make sure all the
provided tests are displayed in green i.e. that they
pass.
• You will have to do similar testing in your Assignment 5
A “Line” segment class
• Design a Java class called Line that will store
information about a line segment, where the line
segment is specified by the coordinates, (x,y), of its
two endpoints. Provide operations to work with
segments.
• Each Line object starts at a point (xStart, yStart ) and
ends at a point (xEnd, yEnd), where xStart, yStart , xEnd,
and yEnd are real-valued numbers that could be
positive or negative.
(xStart, yStart)
y (xEnd, yEnd)
x
UML diagram for Line
If the method is not underlined that it is an instance method.
Underlined methods are static methods.
Thus, this UML diagram tells us that all the methods in this class
are instance methods
Method descriptions
• setPoints method: Set the start and end points of the
line segment.
• Name of method: setPoints(…)
• Parameters to the method: xs, ys, xe, ye
• Returns: (nothing)
• This method modifies the Line object
• length method: Returns the length of the line segment
• Name of method: length( )
• Input parameters to the method: none.
• Method returns: length of the line segment (a real value)
Method descriptions
Method descriptions
• slope method: Returns the slope of the line segment.
– The slope is (ye − ys) / (xe − xs) (watch out for
vertical lines – their slope is infinite!)
• Name of method: slope( )
• Parameters to the method: none.
• The methods retunrs: the slope of the line
segment (a real value)
Method descriptions
• translate method: Translate the segment by (tx, ty),
where tx and ty are any positive or negative real
values.
– A translation of a line segment represents “sliding”
the entire segment. The value tx is added to the x
coordinates of the start and end of the segment,
and the value ty is added to the y coordinates of
the start and end of the segment.
• Name of method: translate(…)
• Parameters to the method: tx, ty
• Results: (none)
• This method modifies the Line object
Method descriptions
• toString method: Returns a String with information
– The string that is returned for a line segmetn with
(for example) start point (0.0,1.0) and end point
(3.5,-1.2) should be:
”Line from (0.0, 1.0) to (3.5, -1.2)”
• Name of method: toString ( )
• Parameters to the method: (none)
• Results: a String in the above format
Exercise 2
• Implement the class Line and all of its methods.
Recall, by looking at UML diagram or method
descriptions, that all the methods of the Line class are
supposed to be instance methods.
• Test the class using the main method in the provided
file LineTest.java
Extra Exercise 3
We want to realize a program for the administration of a parking lot formed by a
collection of parking places. For each parking place the following information has to be
stored:
– Whether it is free of occupied;
– Licence plate of the car, when occupied (a string);
– Time since when it is occupied (two integers, for hours and minutes).
Write a class ParkingPlace, for handling a parking place, that implements the following
methods:
– ParkingPlace(): constructor without parameters that constructs a parking place that is initially
free;
– String toString(): that returns “——-” if the parking place is free, and the licence plate of the
car, if the parking place is occupied;
– void carArrives(String plate, int hour, int minutes: modifies the state of the parking place by
setting it to occupied, sets the plate of the car that occupies the place to plate, and sets the
time since when it is occupied to hour and minutes; if the parking place is already occupied, the
method does nothing;
– void carLeaves(): modifies the state of the parking place by setting it to free;
– boolean free(): returns true if the parking place is free, false otherwise;
– String getCar(): returns the plate of the car that occupies the parking place, if the place is
occupied, null otherwise;
– int getHour(): returns the hour since when the parking place is occupied, if the place is
occupied, -1 otherwise;
– int getMinutes(): returns the minutes since when the parking place is occupied, if the place is
occupied, -1 otherwise;
Exercise 3 Testing
Use the example program TestParkingPlace.Java to
test the class you have developed. | 2,250 | 8,821 | {"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} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.765869 |
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Dec29 comment Understanding induced representations It's nice that the two answers below convinced you that induced rep is a good notion, although I'm personally curious what you mean by "there are many induced rep in contrast to restriction" - in fact I can't see how you may cook up a $G$-rep naturally from a $H$-rep other than the extension of scalars above. Dec28 comment Understanding induced representations Why is $K[G] \otimes_{K[H]} V$ unintuitive to you? Dec27 comment contractible and simply connected Consider loop 1 formed by the starting point of your free homotopy, loop 2 formed by the ending point of the free homotopy. Then at time $t$, traverse along loop 1 up to time $t$, go through the homotopy at time $t$, and come back along loop 2. This is then a based homotopy to the trivial loop at based point. Dec25 revised Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ added 22 characters in body Dec25 answered Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Dec24 comment 1 dimensional representations of $S_n$ A hint for the complex number case: It suffices to show that every homomorphism $\chi: S_n$ to $\mathbb{C}^*$ is real, which then forces any such homomorphism to be either identity or sign, depending on the value at the cycle (12). To show that it's real, it suffices to show that $\chi(g) = \overline{\chi(g)}$ for any $g$, i.e. $\chi(g) = \chi(g^{-1})$. Try to show that $g$ and $g^{-1}$ are actually conjugates which lead to the result. Dec23 comment Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ This is wrong. Use AM-GM or other tools, you can show that $xy+yz+zx \ge 3$. Dec22 comment Is noetherianity a local property? @PeteL.Clark, Ah, thanks! That was what I was thinking of indeed. Dec22 comment Is noetherianity a local property? @BenjaLim, I believe that they are equivalent. Dec18 comment On convergence of nets in a topological space @BrianM.Scott, Excellent! Thanks. Dec18 comment On convergence of nets in a topological space Thanks! Is there any (succint) reference that compares nets and filters? I would like to know more about their similarities and differences. Dec18 comment On convergence of nets in a topological space I just read online that filter was another proposal to generalize sequences. Do filters capture the topology of a space in the same sense? Dec18 comment On convergence of nets in a topological space @Elias, it's net, not network if I understand your question correctly. A quick google search returns this: planetmath.org/encyclopedia/… Dec18 comment On convergence of nets in a topological space Properties defined by sequences (eg sequentially compact) are often different from those not defined by sequences (eg compact) especially in spaces that are not first countable. Nets give a more general framework that corresponds to the latter concept, e.g. compactness is equivalent to every net having a convergent subnet. Dec18 comment Is there a fundamental reason that $\int_b^a = -\int_a^b$ @asmeurer, Signs come up to reflect orientation issues. This doesn't have anything to do with measure, so there's no such phenomenon for Lebesgue integrals, but there would be something similar for integration over manifolds. Dec18 comment Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$ I meant to ask what a necessary and sufficient condition for a possible extension of one vector to an orthogonal basis would be. I doubt that entries being coprime would be enough. Dec18 comment Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$ What about an orthogonal basis, as mentioned by OP? Dec18 comment Primes in arithmetic progression @ThomasAndrews, Ah I see your point. Yet, as OP is thinking about Dirichlet's theorem, I would guess that he meant for all $j$, blah blah blah instead. Dec18 comment Primes in arithmetic progression @Ethan, you definitely need some restrictions on $j$ you consider, like $(j,a) = 1$. Second, infinitude of primes for one such $j$ is enough, since you can divide $j$ to the other side, as in the other thread you posted. Dec18 comment Primes in arithmetic progression @ThomasAndrews, it's not a simultaneous equation, so subtracting doesn't make sense. | 1,191 | 4,652 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 3.203125 | 3 | CC-MAIN-2014-10 | longest | en | 0.895743 |
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# Drag Reduction in Commercial Vehicles: Major Project
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The drag reduction on commercial vehicles using front deflectors is studied using CFD for Tata Ace vehicle. The tools used are Fluent & Gambit.
Published in: Engineering
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### Drag Reduction in Commercial Vehicles: Major Project
1. 1. AERODYNAMIC DRAG REDUCTION IN COMMERCIAL VEHICLES (TATAACE) USING FRONT DEFLECTOR JJ TECHNICAL SOLUTIONS (www.mechieprojects.com)
2. 2. AIM 1. To study and understand the aerodynamic air flow over a TATAACE pickup truck 2. To calculate the aero-force acting on the vehicle for 10m/s velocity & compute the drag coefficient 3. To study and compare the effect of different shape of front deflectors on drag coefficient 4. To design the aerodynamic shape deflector for the minimum coefficient of drag.
3. 3. PROBLEM FORMULATION Tools Used: 1. Drafting/Modeling: SolidWorks 2. Meshing: Gambit 3. Analysis & Post Processing: Fluent Tata Ace Velocity Inlet Meshing of Model in Gambit & Defining Boundary Conditions Pressure Outlet Velocity Inlet Wall Types of Deflector Considered Case 1. No Deflector Case 2. Straight Deflector Case 3. Convex Deflector Case 4. Concave Deflector L2L 6L
4. 4. Problem Definition in FLUENT Define the problem as, Solver -Pressure based Formulation -Implicit Space -2D Time -Steady Viscous -Two-equation SST-k-omega model Enable the Energy equation The fluid type used is Air defined as ideal gas Operating pressure= 0 Pa
5. 5. LIFT, DRAG, AND MOMENT COEFFICIENTS • Behavior of L, D, and M depend on a, but also on velocity and altitude • V∞, r ∞, Wing Area (S), Wing Shape, m ∞, compressibility • Characterize behavior of L, D, M with coefficients (cl, cd, cm) Re,, 2 1 2 1 3 2 2 Mfc Scq L ScV M c SccVM m m m a r r Re,, 2 1 2 1 2 2 2 Mfc Sq D SV D c ScVD d d d a r r Re,, 2 1 2 1 1 2 2 Mfc Sq L SV L c ScVL l l l a r r Note on Notation: We use lower case, cl, cd, and cm for infinite wings (airfoils) We use upper case, CL, CD, and CM for finite wings
6. 6. PRESSURE COEFFICIENT, CP • Use non-dimensional description, instead of plotting actual values of pressure • Pressure distribution in aerodynamic literature often given as Cp • So why do we care? • Distribution of Cp leads to value of cl • Easy to get pressure data in wind tunnels • Shows effect of M∞ on cl 2 2 1 V pp q pp Cp r
7. 7. Structured Grid/Mesh created in Gambit Types of Deflector Considered Case 1. No Deflector Case 2. Straight Deflector Case 3. Convex Deflector Case 4. Concave Deflector Velocity Inlet Pressure Outlet Velocity Inlet Wall
8. 8. Solution Convergence in Fluent
9. 9. Contours of Static Pressure around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
10. 10. Contours of Dynamic Pressure around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
11. 11. Contours of Velocity Magnitude around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
12. 12. Contours of Velocity Magnitude around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
13. 13. Contours of Velocity Magnitude around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
14. 14. Velocity Magnitude Vectors around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
15. 15. Pressure Coeff. Plot around the Tata Ace at velocity = 10m/s Case 1 : No Deflector
16. 16. Contours of Static Pressure around the Tata Ace at velocity = 10m/s Case 2 : Straight Deflector
17. 17. Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s Case 2 : Straight Deflector
18. 18. Case 2 : Straight Deflector Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s
19. 19. Case 2 : Straight Deflector Plot of Pressure Coeff. around the Tata Ace at velocity = 10m/s
20. 20. Case 3 : Convex Deflector Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s
21. 21. Case 3 : Convex Deflector Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s
22. 22. Case 3 : Convex Deflector Plot of Pressure Coeff. around the Tata Ace at velocity = 10m/s
23. 23. Case 4 : Concave Deflector Contours of Static Pressure around the Tata Ace at velocity = 10m/s
24. 24. Case 4 : Concave Deflector Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s
25. 25. Case 4 : Concave Deflector Contours of Velocity magnitude around the Tata Ace at velocity = 10m/s
26. 26. Case 4 : Concave Deflector Plot of Pressure Coeff. around the Tata Ace at velocity = 10m/s
27. 27. Case 3 Case 2Case 1 Case 4
28. 28. Pressure Force (N) Viscous Force (N) Total force (N) Pressure Coeff. Viscous Coeff. Total Coeff. Lift Coeff. Drag Coeff. Config. 1 210.4669 1.719078 212.1859 1.527197 0.012474 1.539671 -1.14E+00 1.54E+00 Config. 2 134.3915 3.477828 137.8693 0.975176 0.025236 1.000412 -4.64E-01 1.00E+00 Config. 3 116.5591 2.387426 118.9465 0.84578 0.017324 0.863104 -3.93E-01 8.63E-01 Config. 4 220.5458 1.832951 222.3787 1.600332 0.0133 1.613632 -4.88E-03 1.61E+00 Types of Deflector Considered Case 1. No Deflector Case 2. Straight Deflector Case 3. Convex Deflector Case 4. Concave Deflector
29. 29. CONCLUSION 1. The contours of Velocity & Pressure is plotted, around the Tata Ace for velocity of 10m/s 2. The velocity increases near the tip of the highest portion of the vehicle at the point of sharp curvature changes 3. A swirl/ backflow is generated at the rear end of the car with negative velocity. 4. The pressure coefficient is plotted for the top & bottom side of the car. 5. The pressure force, viscous force and total force acting on the Tata Ace vehicle is plotted for different deflector configurations 6. The Drag coefficient (Cd) is minimum for Convex deflector. 7. By using a front deflector the Cd reduces from 1.61 to 0.863.
30. 30. FOR COMPLETE PRESENTATION, MORE PROJECTS PRESENTATIONS AND PROJECT REPORTS VISIT WWW.MECHIEPROJECTS.COM Email: contactus@mechieprojects.com THANKYOU
31. 31. This is purely an academic work and has no financial or other interest. The results achieved in this should be independently verified. | 2,109 | 6,872 | {"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} | 2.53125 | 3 | CC-MAIN-2020-45 | latest | en | 0.825688 |
http://brainly.com/question/36771 | 1,480,884,200,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00367-ip-10-31-129-80.ec2.internal.warc.gz | 38,850,255 | 9,723 | # A construction company has a contract to build a memorial. The memorial will be a right, circular, cylinder with a radius of 12 yards and a height of 10 yards. The entire surface area of the cylinder will be made from steel.Which is the best estimate of how much steel, in square yards, that will be needed to construct the memorial if there is no overlap?
1
by rebeccacaterper | 88 | 380 | {"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} | 2.53125 | 3 | CC-MAIN-2016-50 | latest | en | 0.958803 |
https://elteoremadecuales.com/cauchys-theorem-geometry/ | 1,680,030,565,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00083.warc.gz | 283,110,204 | 10,839 | # Cauchy's theorem (geometry)
Cauchy's theorem (geometry) Cauchy's theorem is a theorem in geometry, named after Augustin Cauchy. It states that convex polytopes in three dimensions with congruent corresponding faces must be congruent to each other. That is, any polyhedral net formed by unfolding the faces of the polyhedron onto a flat surface, together with gluing instructions describing which faces should be connected to each other, uniquely determines the shape of the original polyhedron. For instance, if six squares are connected in the pattern of a cube, then they must form a cube: there is no convex polyhedron with six square faces connected in the same way that does not have the same shape.
This is a fundamental result in rigidity theory: one consequence of the theorem is that, if one makes a physical model of a convex polyhedron by connecting together rigid plates for each of the polyhedron faces with flexible hinges along the polyhedron edges, then this ensemble of plates and hinges will necessarily form a rigid structure.
Contents 1 Statement 2 History 3 Generalizations and related results 4 See also 5 References Statement Convex regular icosahedron Let P and Q be combinatorially equivalent 3-dimensional convex polytopes; that is, they are convex polytopes with isomorphic face lattices. Suppose further that each pair of corresponding faces from P and Q are congruent to each other, i.e. equal up to a rigid motion. Then P and Q are themselves congruent.
To see that convexity is necessary, consider a regular icosahedron. One can "push in" a vertex to create a nonconvex polyhedron that is still combinatorially equivalent to the regular icosahedron. Another way to see it, is to take the pentagonal pyramid around a vertex, and reflect it with respect to its base.
History The result originated in Euclid's Elements, where solids are called equal if the same holds for their faces. This version of the result was proved by Cauchy in 1813 based on earlier work by Lagrange. An error in Cauchy's proof of the main lemma was corrected by Ernst Steinitz, Isaac Jacob Schoenberg, and Aleksandr Danilovich Aleksandrov. The corrected proof of Cauchy is so short and elegant, that it is considered to be one of the Proofs from THE BOOK.[1] Generalizations and related results The result does not hold on a plane or for non-convex polyhedra in {displaystyle mathbb {R} ^{3}} : there exist non-convex flexible polyhedra that have one or more degrees of freedom of movement that preserve the shapes of their faces. In particular, the Bricard octahedra are self-intersecting flexible surfaces discovered by a French mathematician Raoul Bricard in 1897. The Connelly sphere, a flexible non-convex polyhedron homeomorphic to a 2-sphere, was discovered by Robert Connelly in 1977.[2][3] Although originally proven by Cauchy in three dimensions, the theorem was extended to dimensions higher than 3 by Alexandrov (1950). Cauchy's rigidity theorem is a corollary from Cauchy's theorem stating that a convex polytope cannot be deformed so that its faces remain rigid. In 1974 Herman Gluck showed that in a certain precise sense almost all simply connected closed surfaces are rigid.[4] Dehn's rigidity theorem is an extension of the Cauchy rigidity theorem to infinitesimal rigidity. This result was obtained by Dehn in 1916. Alexandrov's uniqueness theorem is a result by Alexandrov (1950), generalizing Cauchy's theorem by showing that convex polyhedra are uniquely described by the metric spaces of geodesics on their surface. The analogous uniqueness theorem for smooth surfaces was proved by Cohn-Vossen in 1927. Pogorelov's uniqueness theorem is a result by Pogorelov generalizing both of these results and applying to general convex surfaces. See also Schönhardt polyhedron References ^ Aigner, Martin; Ziegler, Günter M. (2014). Proofs from THE BOOK. Springer. pp. 91–93. ISBN 9783540404606. ^ Connelly, Robert (1977). "A counterexample to the rigidity conjecture for polyhedra". Publications Mathématiques de l'IHÉS. 47: 333–338. doi:10.1007/BF02684342. ISSN 0073-8301. S2CID 122968997. ^ Connelly, Robert (1979). "The Rigidity of Polyhedral Surfaces". Mathematics Magazine. 52 (5): 275–283. doi:10.2307/2689778. JSTOR 2689778. ^ Gluck, Herman (1975). "Almost all simply connected closed surfaces are rigid". In Glaser, Leslie Curtis; Rushing, Thomas Benjamin (eds.). Geometric Topology. Lecture Notes in Mathematics. Vol. 438. Springer Berlin Heidelberg. pp. 225–239. doi:10.1007/bfb0066118. ISBN 9783540374121. A. L. Cauchy, "Recherche sur les polyèdres – premier mémoire", Journal de l'École Polytechnique 9 (1813), 66–86. Max Dehn, "Über die Starrheit konvexer Polyeder" (in German), Math. Ann. 77 (1916), 466–473. Aleksandr Danilovich Aleksandrov, Convex polyhedra, GTI, Moscow, 1950. English translation: Springer, Berlin, 2005. James J. Stoker, "Geometrical problems concerning polyhedra in the large", Comm. Pure Appl. Math. 21 (1968), 119–168. Robert Connelly, "Rigidity", in Handbook of Convex Geometry, vol. A, 223–271, North-Holland, Amsterdam, 1993. Categories: Augustin-Louis CauchyTheorems in discrete geometryPolytopesMathematics of rigidityEuclidean geometryTheorems in convex geometry
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http://modules.xjtlu.edu.cn/MOD_CAT.aspx?mod_code=MTH002 | 1,566,632,001,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00149.warc.gz | 128,417,426 | 10,102 | Module Catalogues, Xi'an Jiaotong-Liverpool University
Module Code: MTH002
Module Title: Multivariable Calculus and Statistics
Module Level: Level 0
Module Credits: 5.00
Academic Year: 2019/20
Semester: SEM2
Originating Department: Mathematical Sciences
Pre-requisites: MTH013ORMTH019ORMTH023ORMTH025ORMTH027ORMTH021
Aims To give students an elementary of multivariable Calculus and statistics; To introduce the concept of modeling and various mathematical models in practical problems; To develop students' ability to work independently and to acquire the skill of problem solving;To improve students’ ability to learn academic contents using English.
Learning outcomes A. have an initial understanding of the basic concepts which are the subject of this module;B. have a good appreciation of the link between mathematics and other subjects;C. be able to understand the mathematical models for simple practical problems.D. describe statistical data;E. use the Binomial, Poisson and Normal distributions;F. understand the basic principle of inferential statistics.G. be able to utilize the new technology such as computer programs and online resources to perform self-study.
Method of teaching and learning Students will be expected to attend about five hours of formal lectures and supervised problem classes/tutorials in a typical week. Lectures and tutorials will introduce students to the academic content and practical skills which are the subject of the module, while problem classes and tutorials will allow students to practice those skills. In addition to contact hours, students will be expected to devote sufficient unsupervised time to private study. Private study will provide time for reflection and consideration of lecture material and background reading. Continuous assessment including home assignment marking will be used to test to what extent practical skills have been learnt. On-line homework system will be used to help students to perform self-study and increase the scope of applications. A mini research project on statistical topics will be designed to train students’ practical application ability. Written examinations in the middle and at the end of the module constitute the major part of assessment of the academic achievement.In order to facilitate the smooth transition to the English delivery of all modules from year 2, this module will be delivered in English.
Syllabus Part 1 Multivariable calculus 1.Functions of several variables 2.The partial derivatives 3.Maxima and minima 4.Lagrange’s multipliers 5.total differential and applications 6.Double integrals Part 2 Statistics 1. Introduction and description of data 2. Examples of what ‘Statistics’ as a discipline is about, populations and samples, graphical summaries, shape, location and spread of data. 3. Elements of Probability Theory 4. Intuitive meaning of probability, events, compound events, conditional probability and Bayes’ rule, networks of unreliable components. 5. Probability Distributions 6. Discrete and continuous random variables: the probability mass function, the probability density function and distribution function. Expectation and Variance. 7. The Binomial and Poisson distributions. 8. The Normal Distribution and Approximations. 9. The principles of hypothesis testing, 10. central limit theorem, 11. Normal confidence intervals.12. Correlation and regression.
Delivery Hours
Lectures Seminars Tutorials Lab/Prcaticals Fieldwork / Placement Other(Private study) Total Hours/Semester 56 9 85 150
## Assessment
Sequence Method % of Final Mark 1 Written Examination 65.00 2 Mid-Term Test 15.00 3 Course Work Quizzes And Online Homework And Projects 15.00 4 Project 5.00
Module Catalogue generated from SITS CUT-OFF: 8/24/2019 3:33:20 PM | 784 | 3,766 | {"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} | 2.65625 | 3 | CC-MAIN-2019-35 | longest | en | 0.893046 |
https://dsp.stackexchange.com/questions/72229/doubt-signal-detection-in-noise-part-1 | 1,696,045,320,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510575.93/warc/CC-MAIN-20230930014147-20230930044147-00739.warc.gz | 237,189,538 | 41,587 | # Doubt: Signal detection in noise Part 1
I have a confusion regarding the matched filter detection technique. I have a binary information source signal $$s(t)$$ that is corrupted by additive White Gaussian noise $$w(t)$$ at a particular SNR. The received signal is:
$$x(t) = s(t) + w(t)$$
Then I have created a matched filter $$h(t)$$ as the time reversal of the source binary signal $$s(t)$$. How does a matched filter work?
• I have 2 issues. (a) how to estimate the signal $\hat{s}(t)$ after signal detection -- From answer & comments, it appears that matched filter is not an estimation method but a detection technique. I posted the code to verify if I have done the estimation correctly or not. Code output shows that $s(t)$ and $\hat{s}(t)$ are identical. So, it seems that the estimation was possible. (b) Another issue is why do estimation or detection in cases where the template or a copy of the signal is already present at reciever for nonblind methods such as matched filter, nonblind least mean squares etc.
– Sm1
Dec 27, 2020 at 17:51
You use the matched filtering technique when you search measurement data for a signal of a given form. You know the signal waveform in advance and either you are not sure if the signal is present in your data, or you see the signal and your task is to compute the signal magnitude and location. In these cases you use the matched filtering technique.
If your measurement data contains an unknown signal buried in noise, you can try and recover the waveform of this signal, with additional assumptions made about the noise nature. Often, this noise is additive white Gaussian noise, AWGN. The signal mixed with AWGN is amenable to processing with filters; filtering gives an approximate waveform of the "pure" signal. The precision of filtering -- the proximity of a recovered signal to the "pure" signal -- varies depending on the signal and noise parameters and the filter type and implementation.
Summing up: you use matched filtering when you know the signal waveform in advance or test you data to establish the presence of hypothetical signal waveforms. The waveform you test against is called template. If you want to extract the signal waveform from noisy data, you do not use matched filtering; you apply filters.
• Thank you for answering. So matched filtering is a nonblind approach which means that the information source signal is known but perhaps the channel parameters are unknown. Is my understanding correct? Also, could you please let me know if my code is correct. Using the matched filter I could get back the transmitted signal $\hat{s}(t)$ which I found to be exactly equal to the information signal, $s(t)$. So I was able to extract the signal. So it's unclear what you mean by saying that I should use some other filters to extract the signal waveform. Isn't matched filtering approach extracting it?
– Sm1
Dec 27, 2020 at 6:12
• the matched filtering operation is a convolution of signal data and a pattern. MATLAB's filter function is not a convolution, MATLAB has a conv function for this operation; see MATLAB Help. see also the simple MATLAB implementation in github.com/Septien/MatchedFilter. Later, when you are comfortable with the matching filter concept, you will be able to use the MatchedFilter object (phased.MatchedFilter) of the Phased Array Toolbox. Dec 27, 2020 at 7:54
• An excellent presentation on the matched filtering, see mybinder.org/v2/gh/moble/MatchedFiltering/… , the source code in github.com/MOBle/MatchedFiltering. Dec 27, 2020 at 7:54
• The matched filter does not extract the "pure" signal from data; it computes the convolution of measured data and the designated pattern. Given a selectable detection threshold, you decide if the signal is present in your data and then compute the signal magnitude and location (e.g., start time or time offset or the like) Dec 27, 2020 at 8:09
• To "what is the need of designing estimation or detection techniques": consider the scenario: the transmitter sends a symbol from an N-length alphabet; the receiver receives the transmission and computes N convolutions against each of N alphabet symbols, then iterates through these N results, and decides which symbol is sent in this transmission. Or finds out that it is an empty transmission, containing irrelevant data or only noise. Dec 27, 2020 at 8:33 | 963 | 4,357 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.908289 |
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# I do not understand how to approach this question.
I do not understand how to approach this question.
The way to answer this question is ... View the full answer
• why do you do 20/4 ?
• SargentJellyfish3050
• May 10, 2018 at 6:25am
• The distance from P to R is 20 cm, and the wavelength of the water waves are 4 cm. So 20 cm * 1 wavelength/4cm = 5 wavelengths
• onteria
• May 10, 2018 at 7:27am
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can anything pass into the event horizon if time is stretche
can anything pass into the event horizon if time is stretche
[From: Astronomy & Space] [author: ] [Date: 02-28] [Hit: ]
can anything pass into the event horizon if time is stretched and the black hole evaporated faster and faster? Wouldn’t it be impossible?......
can anything pass into the event horizon if time is stretched and the black hole evaporated faster and faster? Wouldn’t it be impossible?
-------------------------------------------------------
Ronald 7 say: Time can Not be Sretched
It is only a Theory
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Luca say: As far as I understand it: for an observer that is falling in the BH (provided he isn't destroyed in the process), everything works normally, he doens't even realize he just crossed the event horizon.
For an outside observer: it takes an infinite time for the falling body to cross the event horizon, but as an outside observer the more you wait the more photons coming from the falling body lose energy (and frequency) and thus at a certain point you won't see anything at all and the body will be at (to? from? English is hard) (all) effects "in the black hole".
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Jeffrey K say: I agree with you and Matthew's answer. Nothing ever crosses the event horizon.
From a far away observer's point of view, it takes an infinite amount of time for the astronaut to cross the horizon but the black hole evaporates in a long but finite time. So the black hole is gone before the astronaut reaches it. The horizon shrinks as it losses mass, and the astronaut gets closer to the singularity, reaching it just as the final evaporation occurs.
From the astronaut's point of view, it takes just minutes to reach where the horizon was but it has shrunk. A few seconds later, he gets close enough to the singularity for tidal forces to rip him apart. But he is still outside of the shrunken horizon. There is never a time when the astronaut can send a radio signal that does not reach an outside observer. He is never inside a region of space cut off from the rest of the universe.
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if an object is moving near the speed of light .. like a MESON they had this problem back in the 50's
the meson created in the upper atmosphere by a collision of high energy particles was being measured and they found alot more of the mesons hitting the earth then there should have been
but they were calculation the life of the meson in NON RELATIVISITIC environment..
the objects calculated life span was the life span the meson feels in its world at its speed
we as observers at a different velocity .. see the lifespan of the meson being LONGER time so we see the object living longer then its true life from our frame of reference and any interactions in our frame will be by the LONGER time frame NOT THE real one of the meson
so in this situration if the meson had a clock it would see time pass so that it DIED according to the first calculation the scientists made say for argument this is x seconds
but when we time them from watching them we will see them live a bit longer x + DF
where DF is the Dilation factor for lack of a better term..
You will get alot of speculation about what happens near or inside a black hole some may be correct some may be misleading..
remember that TIME AND DISTANCE change when relativity equations happen..
the speed of light that is the maximum speed in a vaccuum is constant in all frames ..
Hawking radiation doesnt make a blackhole just poof out existance when its large.. if I understan it hawking radiation is a special quantum event happening right at the boundery of a black hole so this does not happen lalot like it does when you turn on a light bulb .. its a fairly rare event..
from my unerstaning two small virtual particles pop into existance .. and in normal space just pop back out before anyone sees them .. but when it happens near a black hole one particle goes into the black hole and we get one magically appearing particle in out universe ..
think of one particle that goes into the black hole as a NEGATIVE MASS.. and the one that stays in ours a POSITIVE MASS..
THE net effect is that the blackhole loses a tiny tiny speck of mass and our universe gets a tiny bit bigger ..
the more massive the black hole the longer it will take to evaporate... so no your understanding is incorrect that the more the time dilation the faster it evaporates..
remember also that the black hole gets more mass when it has something enter the event horizon that mass is now part of the black hole..
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neb say: Time dilation can be pretty confusing near a black hole. Let’s say you are in freefall near the event horizon of a black hole. If you look at your wristwatch, you will see time passing at exactly the same rate as you have always seen it pass. If you are in freefall near a stellar size black hole (pretend you don’t get torn apart), you will actually fall through the event horizon and hit the singularity in microseconds according to your watch. For supermassive blackholes, in may take hours or days according to your wristwatch to fall through the event horizon and hit the singularity. In either case, the salient point is that the freefall trip takes very little local time - and the same local time applies to the Hawking radiation - so there is negligible evaporation observed before you hit the singularity. What time dilation a far away observer sees is totally irrelevant
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CarolOklaNola say: You cannot see time dilation..Time dilation foes NOT make time flow faster OR the event time flow faster or the event horizons evaporate faster the closer you get to them
Event horizons violate the CURRENT laws of physics be cause AT the eve t horizon time becomes in fine, eternity. AND.
Time dilation Requires TWO observers in the SAME frame of reference. Each person's clock is ticking at the NORMAL rate for THEM. It is the OTHER observer's clock that SEEMS to be ticking very slowly or very quickly. The event horizon evaporates at the same rate. Whether or not the person approaching the event horizon gets spaghettified and killed depends on how close they get to the event horizon. Even if the event horizon completely, they ate still dead and died from old age as well as being spaghettified.
It does not matter whether the material is indestructible or not. The molecules and atoms are torn apart and the object is torn apart by the gravitational stresses and strains at subatomic
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Matthew say: To make it clearer. I recently had a thought experiment in which an object was moving toward a black hole. Let’s say that the object was indestructible. When it gets closer and closer the gravity increases (I assume exponentially as you halve the distance) would that also mean that the gravitational time dilation would increase in kind of the same way? In that case wouldn’t the closer the object get mean the longer time is actually distorted. And isn’t the longer time distorted the faster you would see the black hole evaporate due to Hawking radiation? Wouldn’t that mean that as you got closer the black hole would shrink and then you would have to get closer again? I just had this theory and wanted to see if it makes any sense at all. If this were the case that would mean no object could ever pass through the event horizon
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Braja Sorensen Team October 3, 2020 Worksheet
Please visit addition to browse more worksheets in the same area. Below, you will find a wide range of our printable worksheets in chapter addition with pictures of section addition.
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Addition worksheets with pictures for grade 1. These worksheets are appropriate for first grade math. Addition worksheets & printables there’s a reason the math train makes its first stop at addition: These grade 1 math worksheets use pictures to help students conceptualize addition by showing the connection between combining two groups of objects (counting) and creating and evaluating an addition sentence.
Easy addition with pictures worksheet featuring fall leaves. You may also look for a few pictures that related to first grade. Addition worksheets for grade 1.
Adding (sums to 10) using pictures / objects. Addition with pictures and number line : If your kids are just starting out with addition to 20, then these worksheets with pictures will be a fun way to teach them.
Task kindergarten, grade 1, and grade 2 kids to complete the addition equation by either counting the pictures or adding the cardinalities given for each group of pictures. They solve addition problems with a missing number (missing addend), and use addition to solve simple subtraction problems. Worksheets can be obtained to suit the wants of each college student.
Rich with scads of practice, the ccss aligned printable 1st grade math worksheets with answer keys help kids solve addition and subtraction problems within 20, extend their counting sequence, understand place value and number systems, measure length and compare sizes, tell time, count money, represent and interpret data, and know the attributes of 2d and 3d shapes in geometry. Free interactive exercises to practice online or download as pdf to print. Worksheets > math > grade 1 > addition.
These addition worksheets with pictures pdf will enable kids to perfectly visualize the relationship between numbers and their quantities. Pupils discover the crucial expertise of addition by means of follow, observation, and video games. Coloring addition worksheets for grade 1, we have prepared this post well for you to read and retrieve information from it.
For grade 1 students which are still learning the small number addition, this worksheet is suitable to be given to them. View the full list of topics for this grade and subject categorized by common core. If you’re like me and looking for some helpful work for your child or children while they’re home, i hope these addition worksheets for grade 1 help you out.
Worksheets > math > grade 1 > addition > adding with pictures. Sheet 1 | sheet 2 | sheet 3 | grab 'em all. Simple addition to 20 worksheets with pictures for your first graders!
A brief description of the worksheets is on each of the worksheet widgets. Have fun and learn to sum with these addition worksheets. Use pictures to add to grade/level:
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First grade addition worksheets for grade 1 was created by combining each of gallery on worksheets, worksheets is match and guidelines that suggested for you, for enthusiasm about you search. Adding 2 single digit numbers with a sum up to 10. Children will use the pictures of hearts, stars and shapes to write and solve the addition problems.
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Addition worksheet for grade 1 posted above is the basic version of the worksheets which contain only the single number addition. Hopefully fill the posts artikel coloring addition and subtraction worksheets for grade 1, artikel coloring addition worksheets for grade 1, we write this you can understand.alright, happy reading. Addition worksheets and online activities.
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Resonance example:
• Carbonate ion - CO32-
• valence electrons = 4 + 3 (6) + 2 = 24
• 6y + 2 = 26, but e- = 24, therefore expect one multiple bond.
• LS =
• However, other equally symmetrical structures are possible, so:
Expanded Valence Shell Example:
• SF4
• valence electrons = 6 + 4(7) = 34
• 6y + 2 = 32, but e- = 34, therefore expect expanded valence shell with one extra electron pair.
• LS =
Additional exercises on Lewis Structures are available in the Lewis Structure Module.
For a modern view of bonding illustrated with QuickTime movies based on quantum calculations you may enjoy the Supplement.
# Molecular Geometry
The importance of molecular shape: recognition at the molecular level in organisms. Shape and electron density are extraordinarily important to the interaction of biomolecules - Examples
• Sarin (nerve gas)
• "drugs"
• estrogen mimics ("feminization" of various animal populations - birth control complication)
Lewis Structures enable us to predict bonding patterns for compounds of the representative elements, but how can we predict their shapes? We will add another tool, VSEPR Theory, to our chemical toolbox - a simple way to predict the geometry of bonds around a central atom (for larger molecules predict one center at a time).
## VSEPR (Valence Shell Electron Pair Repulsion) Theory
Based on three assumptions (there are more advanced versions, but unnecessary for us):
• Electron pairs will orient around a central point to minimize repulsion.
• Lone-pairs of electrons will have greater repulsion than bonded pairs of electrons (note that the atoms are ignored in terms of repulsion).
• Repulsion is strong at 90° and weaker at 120° (weakest at 180°).
VSEPR predicts geometry based on these assumptions in a few simple, sequential, steps:
1. Draw a correct Lewis Structure.
2. Determine the Steric Number = the number of bonded atoms + the number of lone pairs.
3. Maximize the angles between electron pairs, placing the lone (unbonded) pairs at the extremes.
#### For central atoms with eight outer electrons (octets) there are three possible electron pair geometries:
1. Linear with angles of 180° ( a single pair and a triple bond, or two double bonds).
2. Trigonal planar with angles of 120° (one double bond and two single pairs).
3. Tetrahedral with angles of 109.5° (four single pairs). [model]
#### These three electron pair geometries can lead to five molecular geometries:
• Linear (carbon dioxide)
• CO2
• valence electrons = 4 + 2x6 = 16
• 6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)
• LS: from symmetry C will be central atom, therefore=
• Considering C as the central atom, have 2 bonded atoms and no lone-pairs, therefore
• steric number = 2, so linear electronic geometry, and
• linear molecular geometry
• Trigonal planar (formaldehyde, CH2O)
• formaldehyde, CH2O
• valence electrons = 4 + 2x1 + 6 = 12
• 6y + 2 = 6 x 2 + 2 = 14; so molecule has 2 fewer electrons than required for all single bonds, 1 double bond
• LS: O can only have two bonds, so C will be central atom, therefore =
• Considering C as the central atom, have 3 bonded atoms and no lone-pairs, therefore
• steric number = 3, so trigonal planar electronic geometry, and 3 atoms so
• trigonal planar molecular geometry and a model showing single vs. double bonds:
• Tetrahedral (methane, CH4) [model]
• valence electrons = 4 + 4x1= 8
• four bonds possible, since only 4 pairs, single bonds because only have H's bound to C.
• LS: from symmetry C will be central atom, therefore =
• Considering C as the central atom, have 4 bonded atoms and no lone-pairs, therefore
• steric number = 4, so tetrahedral electronic geometry, and 4 atoms so
• tetrahedral molecular geometry, and rotated for a different view:
• Trigonal pyramidal (ammonia, NH3) [model]
• valence electrons = 5 + 3x1= 8
• only 4 pairs, single bonds because only have H's bound to N, 3 bonds, since only 3 H's
• LS: from symmetry N will be central atom, therefore =
• Considering N as the central atom, have 3 bonded atoms and one lone-pair, therefore
• steric number = 4, so tetrahedral electronic geometry, but only 3 atoms so
• trigonal pyramidal molecular geometry and rotated to view molecule from below:
• Bent (water, H2O) [model]
• valence electrons = 6 + 2x1= 8
• only 4 pairs, single bonds because only have H's bound to O, 2 bonds, since only 2 H's
• LS: from symmetry O will be central atom, therefore =
• Considering O as the central atom, have 2 bonded atoms and 2 lone-pairs, therefore
• steric number = 4, so tetrahedral electronic geometry, but only 2 atoms so
• bent molecular geometry
### NEXT
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# On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?
0
540
1
On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?
Jan 27, 2018
#1
+101734
+2
On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?
When is y negative?
\((x+2)^4-100<0\\ (x+2)^4<100\\ -\sqrt{10}
That is 5 points where the coordinates are both negative integers:
(-5, -19), (-4,-84), (-3,-99), (-2,-100), (-1,-99)
Jan 27, 2018
#1
+101734
+2
On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?
When is y negative?
\((x+2)^4-100<0\\ (x+2)^4<100\\ -\sqrt{10}
That is 5 points where the coordinates are both negative integers:
(-5, -19), (-4,-84), (-3,-99), (-2,-100), (-1,-99)
Melody Jan 27, 2018 | 327 | 914 | {"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} | 3.8125 | 4 | CC-MAIN-2019-26 | latest | en | 0.745422 |
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# Because of a similiarity to dance, synchronised
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Because of a similiarity to dance, synchronised [#permalink]
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07 May 2005, 19:01
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Because of a similiarity to dance, synchronised swimming-exhibition swimming in which the movements of one or more swimmers synchronise to a musical accompaniment- is sometimes called water ballet, especially in theaterical situations.
A of one or more swimmers synchronise to
b of one swimmer or more is synchronised with
c of one or more swimmers are synchronised with
d by one swimmer or more is synchronised to
e by one or more swimmers synchronise to
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Senior Manager
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12 May 2005, 19:45
"A" is clear and concise.
Uses synchronized to - idiomatic.
_________________
"Forums are meant to benefit all. No one is interested in knowing what your guesses are. Please explain the reasoning behind the answer you chose. This will also help you organize your thoughts quickly during the exam."
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12 May 2005, 19:49
This one sounds strange to me...if I need to choose I would go for C
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### Show Tags
12 May 2005, 19:54
Because of a similiarity to dance, synchronised swimming-exhibition swimming in which the movements of one or more swimmers synchronise to a musical accompaniment- is sometimes called water ballet, especially in theaterical situations.
A of one or more swimmers synchronise to
movements don't synchronise themselves , so it should be a passive voice so rejected
b of one swimmer or more is synchronised with
"movements" needs plural very not "is"
c of one or more swimmers are synchronised with
d by one swimmer or more is synchronised to
"movements" needs plural very not "is"
e by one or more swimmers synchronise to
I think synchronise is better with "with" not sure though ..
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Senior Manager
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12 May 2005, 21:33
I am sure, synchronise with is the right idiom.
Also since movements is plural, we should use "are"
Thus I'll pick C.
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12 May 2005, 21:33
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# Because of a similiarity to dance, synchronised
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Level 437
## Ignore words
Check the boxes below to ignore/unignore words, then click save at the bottom. Ignored words will never appear in any learning session.
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vector
the direction or course followed by an airplane, missile, or the like
column vector/ vector
a matrix with only one column
Vectors in R ^2
the set of all vectors with two entries; ordered pairs of real numbers
Equal vectors
Two vectors w/ the same length and direction
parallelogram rule for addition: u + v
If u and v are represented as points in a plane, u + v corresponds to the 4th vertex of the parallelogram whose other vertices are u, 0, and v
Vectors in R ^3
3 x 1 column matrices with three entries; represented as points in a 3D coordinate space
Vectors in R^n
collection of all lists of n real numbers written as n x 1 column matrices
linear combination
Vector y defined as y = c1v1 + ... + cpvp where c's are weights, v's are vectors
span {v1, ..., vp}
the set of all scalar multiples of v, ... vp; if the vector equation/ augmented matrix has a solution, then v is in the span
i
Represents the horizontal value of a vector in an equation.
j
Represents the vertical value of a vector in an equation.
(q₁-p₁, q₂-p₂)
Component Form of a Vector with initial Point P = (p₁, p₂) and terminal point Q = (q₁,q₂)
√(q₁-p₁)²+(q₂-p₂)²
Magnitude of a Vector with initial Point P = (p₁, p₂) and terminal point Q = (q₁,q₂)
(1/(||v||))v
How to find a Unit Vector
Unit Vector
a vector with a magnitude of 1. the positive X-axis is vector i, pos. <1,0> y xis is vector j <0,1>
Component Form, add i for horizontal and j for vertical
First to find the "i + j" for a vector, you find this of the 2 points and then...
tan⁻¹(j/i)
How to find direction angle of vectors given the j+i form
tan⁻¹(y/x)
How to find direction angle of vectors given the x and y form
(magnitude)(cos(angle))i+(magnitude)(sin(angle))j
Given Angle + Magnitude, formula for Component Form
(q₁-p₁)i + (q₂-p₂)j
Unit Form of a Vector with initial Point P = (p₁, p₂) and terminal point Q = (q₁,q₂)
Distance
Given magnitude and direction and trying to find a vector, first use this equation.
divide it by the magnitude
Given magnitude and direction and trying to find a vector, after using the distance equation, you do this to the number.
(Given Magnitude/Found Magnitude)(Points)
Given magnitude and direction and trying to find a vector, this is the last thing you do [equation].
√((F₁+F₂cos(Ө))² + (F₂sin(Ө)²) = Resultant Force
Given 2 forces and a resultant force, and trying to find the angle, use this formula.
(u₁v₁) + (u₂v₂)
The dot product of u = (u₁,u₂) and v = (v₁,v₂) is given by u · v =
cos(Ө) = (u · v)/(magnitude of u * magnitude of v)
If Ө is the angle between two nonzero vectors u and v, then this is the formula to find the angle between TWO vectors.
orthogonal
If the dot product is 0, the two vectors are...
the magnitude of vector V squared
Taking the dot product of vector V by itself will yield...
Component Form, Dot Product
When given 2 vectors and a force (trying to find work), first find this of the 2 vectors and take that point and find this with the force.
Distance Formula
Formula for Magnitude of : v=-4i-7j
Tan Formula
Formula for Direction of : v=-4i-7j | 905 | 3,249 | {"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} | 3.6875 | 4 | CC-MAIN-2020-24 | latest | en | 0.835088 |
http://crispexcel.com/reverse-lookup-with-duplicates-in-excel/ | 1,529,630,422,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864337.41/warc/CC-MAIN-20180622010629-20180622030629-00229.warc.gz | 76,698,792 | 10,828 | # REVERSE LOOKUP WITH DUPLICATES IN EXCEL
It is good to be able to do a Reverse Lookup but it is Awesome if you know how to return Multiple Items.
Before you continue with the current article, Please revist the Previous Article on reverse lookup as this is a continuation.
Suppose you are managing a project and you have below scheduled dates, times & jobs. Can you find the days, Times and Jobs assigned to staff Joy Bell?
The trick in returning multiple items is to use SMALL/LARGE function to manage the multiple row/column numbers.
### Multiple DateTime Values
`{=IFERROR(SMALL(IF((\$D\$5:\$G\$15=\$I\$4),(\$D\$4:\$G\$4+\$B\$5:\$B\$15)),ROW(A1)),"")}`
►As shown in the previous articleIF((\$D\$5:\$G\$15=\$I\$4),(\$D\$4:\$G\$4+\$B\$5:\$B\$15) returns an array of the DateTime values where the criteria is met.
`{FALSE,FALSE,FALSE,42504.3333333333;FALSE,FALSE,FALSE,FALSE;42501.4166666667,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,42503.625,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE}`
►This array is fed into SMALL function and ROW(A1)=1 provides SMALL with k as you scroll down.
`=SMALL({42504.3333333333,42501.4166666667,42503.625},1)`
IFERROR checks for error and returns blank when SMALL function has returned ALL DateTime Values.
### Multiple Corresponding Jobs
Retuning the multiple jobs is tricky as the job assigned depends on the Date.
From the above example, Joy Bell starts with Job 3, Job 8 then Job 1. These are in different rows thus the simple INDEX & MATCH cannot apply here.
`{=IFERROR(INDEX(\$C\$5:\$C\$15,MATCH(TRUE,INDEX(\$D\$5:\$G\$15,,SMALL(IF(\$D\$5:\$G\$15=\$I\$4,COLUMN(\$D\$4:\$G\$4)-COLUMN(\$D\$4)+1),ROW(A1)))=\$I\$4,0)),"")}`
►The trick is to find the column with the earliest date first before selecting the job. Below INDEX & SMALL functions returns Columns from the earliest date
`INDEX(\$D\$5:\$G\$15,,SMALL(IF(\$D\$5:\$G\$15=\$I\$4,COLUMN(\$D\$4:\$G\$4)-COLUMN(\$D\$4)+1),ROW(A1)))`
►After the column with the earliest date is selected, MATCH function checks the row that meets the criteria.
►Then INDEX function returns the Job that is in the row number as returned by MATCH
### Multiple DateTime & Job Values
If you would like to return the DateTime & Job assigned as a single value, then you write a formula that Concatenates both values.
`{=IFERROR(TEXT(SMALL(IF(\$D\$5:\$G\$15=\$I\$4,(\$D\$4:\$G\$4+\$B\$5:\$B\$15)),ROW(A1)),"dd-mm-yyyy hh:mm AM/PM")&" "&INDEX(\$C\$5:\$C\$15,MATCH(TRUE,INDEX(\$D\$5:\$G\$15,,SMALL(IF(\$D\$5:\$G\$15=\$I\$4,COLUMN(\$D\$4:\$G\$4)-COLUMN(\$D\$4)+1),ROW(A1)))=\$I\$4,0)),"")}`
►The trick here is formatting the DateTime Value first before concatenating it with the Job. You can use TEXT function for this formatting
```TEXT(SMALL(IF(\$D\$5:\$G\$15=\$I\$4,(\$D\$4:\$G\$4+\$B\$5:\$B\$15)),ROW(A1)),"dd-mm-yyyy hh:mm AM/PM")
```
### NB: You can replace SMALL Function with AGGREGATE to create a non-Array function
`=IFERROR(AGGREGATE(15,6,((\$D\$4:\$G\$4+\$B\$5:\$B\$15)/(\$D\$5:\$G\$15=\$I\$4)),ROW(A1)),"")`
### How it works;
The trick in using AGGREGATE function is to define the array for SMALL fucntion
In our example the array is all the Date + Time Value that meets our criteria
```(\$D\$4:\$G\$4+\$B\$5:\$B\$15) Returns All Date + Time Values
{42501.3333333333,42502.3333333333,42503.3333333333,42504.3333333333;42501.375,42502.375,42503.375,42504.375;42501.4166666667,42502.4166666667,42503.4166666667,42504.4166666667;42501.4583333333,42502.4583333333,42503.4583333333,42504.4583333333;42501.5,42502.5,42503.5,42504.5;42501.5416666667,42502.5416666667,42503.5416666667,42504.5416666667;42501.5833333333,42502.5833333333,42503.5833333333,42504.5833333333;42501.625,42502.625,42503.625,42504.625;42501.6666666667,42502.6666666667,42503.6666666667,42504.6666666667;42501.7083333333,42502.7083333333,42503.7083333333,42504.7083333333;42501.75,42502.75,42503.75,42504.75}```
```(\$D\$5:\$G\$15=\$I\$4) Returns a Boolean array, TRUE where criteria is met otherwise FALSE
{FALSE,FALSE,FALSE,TRUE;FALSE,FALSE,FALSE,FALSE;TRUE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,TRUE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE;FALSE,FALSE,FALSE,FALSE}```
NB: The boolean array is converted to its numeric equiavlent where TRUE = 1 & FALSE =0 during division
So,when you divide ALL date + time Values with this array of 1 & o, it will return the date + time value where criteria is TRUE (1) and Error where criteria is FALSE (0)
`{#DIV/0!,#DIV/0!,#DIV/0!,42504.3333333333;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;42501.4166666667,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,42503.625,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!;#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!}`
And since we had selected Ignore Error values in our options
Then the Final array that SMALL function loops through is like one below
`{42504.3333333333,42501.4166666667,42503.625}`
Watch these Videos on AGGREGATE Function. | 1,884 | 5,242 | {"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-2018-26 | latest | en | 0.732606 |
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Lesch-Nyhan syndrome, an X-linked recessive ...
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+2
submitted by paloma(8),
unscramble the site ⋅ remove ads ⋅ become a member (\$39/month)
q= 100.010/0 euyrcqef(n ni p=2+p=lm?e)saqq 1 feceu(yrqn fo soe)gztoehruy
=010.000/050pq02=02,
tatianalimo would you write in English? +4
+2
submitted by bingcentipede(297),
FA 2019, p. 57: "The frequency of an X-linked recessive disease in males = q and in females = q^2" Found this on USMLEforum: "X-linked recessive: Prevalence = allele frequency q square = q Carrier state for females = 2q"
If q=1/100,000, p~1. So 2pq~2q=1/50,000. Hard to do with the NBME calculator unfortunately.
+1
submitted by topgunber(44),
Genetics- The following is a helpful way to do all the allele frequency questions / carrier frequency questions.
• AD : I = 2pq. The cases are in nearly all carriers (2pq). Most homozygotes die. (q^2)
• AR : I = q^2 (the cases are ONLY shown in homozygotes). The carriers are 2pq
• X linked : Boys : I = q . Boys carrying ONE allele are affected individuals. GIRL CARRIERS (heterozygous females) : I = 2PQ. This gets us to the answer. On the other hand AFFECTED GIRLS (incidence in girls) would be I= q^2 - extremely small (choice e)
topgunber By the way allele frequencies if asked are finding just q. Most of the time P is close to 1. In this case P would be 1-1/100,000 = 99,999/100000 which is why you can usually exclude it from equations + | 656 | 2,046 | {"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} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.857886 |
https://itineraires-ceramique.com/what-is-iae/3a3e8b-supremum-distance-formula | 1,620,512,586,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00218.warc.gz | 355,530,188 | 5,789 | The scipy function for Minkowski distance is: distance.minkowski(a, b, p=?) Hamming distance measures whether the two attributes … p = ∞, the distance measure is the Chebyshev measure. The Euclidean formula for distance in d dimensions is Notion of a metric is far more general a b x3 d = 3 x2 x1. euclidean:. If f : A → Ris a function, then sup A f = sup{f(x) : x ∈ A}, inf A f = inf {f(x) : x ∈ A}. The limits of the infimum and supremum of … Example 2. r "supremum" (LMAX norm, L norm) distance. Thus, the distance between the objects Case1 and Case3 is the same as between Case4 and Case5 for the above data matrix, when investigated by the Minkowski metric. The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Available distance measures are (written for two vectors x and y): . 0. All the basic geometry formulas of scalene, right, isosceles, equilateral triangles ( sides, height, bisector, median ). Each formula has calculator Cosine Index: Cosine distance measure for clustering determines the cosine of the angle between two vectors given by the following formula. results for the supremum to −A and −B. 2.3. [λ]. According to this, we have. Functions The supremum and infimum of a function are the supremum and infimum of its range, and results about sets translate immediately to results about functions. They are extensively used in real analysis, including the axiomatic construction of the real numbers and the formal definition of the Riemann integral. Here's how we get from the one to the other: Suppose you're given the two points (–2, 1) and (1, 5) , and they want you to find out how far apart they are. Supremum and infimum of sets. Euclidean Distance between Vectors 1/2 1 p=2, the distance measure is the Euclidean measure. The infimum and supremum are concepts in mathematical analysis that generalize the notions of minimum and maximum of finite sets. Details. Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)).. maximum:. 1D - Distance on integer Chebyshev Distance between scalar int x and y x=20,y=30 Distance :10.0 1D - Distance on double Chebyshev Distance between scalar double x and y x=2.6,y=3.2 Distance :0.6000000000000001 2D - Distance on integer Chebyshev Distance between vector int x and y x=[2, 3],y=[3, 5] Distance :2.0 2D - Distance on double Chebyshev Distance … Definition 2.11. From MathWorld--A Wolfram To learn more, see our tips on writing great answers. Psychometrika 29(1):1-27. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. Kruskal J.B. (1964): Multidimensional scaling by optimizing goodness of fit to a non metric hypothesis. $$(-1)^n + \frac1{n+1} \le 1 + \frac13 = \frac43$$. Literature. Maximum distance between two components of x and y (supremum norm). 4 Chapter 3: Total variation distance between measures If λ is a dominating (nonnegative measure) for which dµ/dλ = m and dν/dλ = n then d(µ∨ν) dλ = max(m,n) and d(µ∧ν) dλ = min(m,n) a.e. if p = 1, its called Manhattan Distance ; if p = 2, its called Euclidean Distance; if p = infinite, its called Supremum Distance; I want to know what value of 'p' should I put to get the supremum distance or there is any other formulae or library I can use? manhattan: 5. In particular, the nonnegative measures defined by dµ +/dλ:= m and dµ−/dλ:= m− are the smallest measures for whichµ+A … Interactive simulation the most controversial math riddle ever! 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# How do you factor 50c squared plus 40cd plus 8d squared?
Updated: 4/28/2022
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8y ago
2(5c + 2d)(5c + 2d)
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Q: How do you factor 50c squared plus 40cd plus 8d squared?
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### What is the temperature range in a wetland?
It really depends on where it is (Antarctica or Florida)
### Why was the experiments started at a temperature 50C below room temperature and ended at a temperature 50C above room temperature?
You need to specify what type of experiment you are asking about.
### What is the difference between -50c and C?
-50c typically refers to a temperature of -50 degrees Celsius, while C on its own is usually used as an abbreviation for Celsius.
50c
Platypus
kangaroo and emu | 241 | 843 | {"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} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.920172 |
https://learnandteachstatistics.wordpress.com/tag/discussion/ | 1,527,141,489,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00630.warc.gz | 581,079,091 | 24,338 | # Maths is right or wrong – end of discussion – or is it?
In 1984 I was a tutor in Operations Research to second year university students. My own experience of being in tutorials at University had been less than inspiring, with tutors who seemed reserved and keen to give us the answers without too much talking. I wanted to do a good job. My induction included a training session for teaching assistants from throughout the university. Margaret was a very experienced educational developer and was very keen for us to get the students discussing. I tried to explain to her that there really wasn’t a lot to discuss in my subject. You either knew how to solve a set of linear equations using Gauss-Jordan elimination or you didn’t. The answer was either correct or incorrect.
I suspect many people have this view of mathematics and its close relations, statistics and operations research. Our classes have traditionally followed a set pattern. The teacher shows the class how to do something. The class copies down notes and some examples into their books, and then they individually work through exercises in the textbook – generally in silence. The teacher walks around the room and helps students as needed.
Prizes can help motivate students to give answers in unfamiliar settings
So when we talk about discussion in maths classes, this is not something that mathematics and statistics teachers are all familiar with. I recently gave a workshop for about 100 Scholarship students in Statistics in the Waikato. What a wonderful time we had together! The students were from all different schools and needed to be warmed up a little with prizes, but we had some good discussion in groups and as a whole. One of the teachers commented later on the level of discussion in the session. Though she was an experienced maths teacher she found it difficult to lead discussion in the class. I am sure there are many like her.
## It is important to talk in maths and stats classes
It is difficult for many students to learn in solitary silence. As we talk about a topic we develop our understanding, practice the language of the discipline and experience what it means to be a mathematician or statistician. Explaining ideas to others helps us to make sense of them ourselves. As we listen to other people’s thinking we can see how it relates to what we think, and can clear up misconceptions. Some people just like to talk, (who me?) and find learning more fun in a cooperative or collaborative environment. This recognition of the need for language and interaction underpins the development of “rich tasks” that are being used in mathematics classrooms throughout the world.
I have previously stated that “Maths learning should be communal and loud and exciting, not solitary, quiet and routine.”
## Classroom atmosphere
One thing that was difficult at the Scholarship day was that the students did not know each other, and came from various schools. In a regular classroom the teacher has the opportunity of and responsibility for setting the tone of the class. Students need to feel safe. They need to feel that giving a wrong answer is not going to lead to ridicule. Several sessions at the start of the year may be needed to encourage discussion. Ideally this will become less necessary over time as students become used to interactive, inquiry-based learning in mathematics and statistics through their whole school careers.
Number talks” are a tool to help students improve their understanding of number, and recognise that there are many ways to see things. For example, the class might be shown a picture of dots and asked to explain how many dots they see, and how they worked it out. Several different ways of thinking will be discussed.
Children are encouraged to think up multiple ways of thinking about numbers and to develop discussion by following prompts, sometimes called “talk moves”. Talk moves include revoicing, where the teacher restates what she thinks the student has said, asking students to restate another students reasoning, asking students to apply their own reasoning to someone else’s reasoning (Do you agree or disagree and why?), prompting for further participation (Would someone like to add on?), and using wait time (teachers should allow students to think for at least 10 seconds before calling on someone to answer. These are explained more fully in The Tools of Classroom talk.
Google Image is awash with classroom posters outlining “Talk moves”. I have been unable to trace back the source of the term or the list, and would be very pleased if someone can tell me the source, to be able to attribute this structure.
# Good questions
The essence of good discussion is good questions. Question ping pong is not classroom discussion. We have all experienced a teacher working through examples on the board, while asking students the answers to numerical questions. This is a control technique for keeping students attentive, but it can fall to a small group of students who are quick to answer. I remember doing just this in my tutorial on solving matrices, when I didn’t know any better.
Teachers should avoid asking questions that they already know the answers to.
It is not a hard-and-fast rule, but definitely a thing to think about. I like to use True/False quizzes to help uncover misconceptions, and develop use of statistical language. I just about always know the answer to the question, but what I don’t know is how many students know the answer. So I ask the question not to know the answer, but to know if the students do, and to provoke discussion. Perhaps a more interesting question would be, how many students do you think will say “True” to this statement. It would then be interesting to find out their reasoning, so long as it does not get personal!
## Multiple answers and open-ended questions
Where possible we need to ask questions that can have a number of acceptable answers. A discussion about what to do with outliers will seldom have a definitive answer, unless the answer is that it depends! Asking students to make a pictorial representation of an algebra problem can lead to interesting discussions.
The MathTwitterBlogosphere has many attractive ideas to use in teaching maths.
I rather like “Which one doesn’t belong”, which has echoes of “One of these things is not like the other, one of these things doesn’t belong, can you guess…” from Sesame Street. However, in Sesame Street the answer was usually unambiguous, whereas with WODB there are lots of ways to have alternative answers. There is a website dedicated to sets of four objects, and the discussion is about which one does not belong. In each case all four can “not belong” for some reason, which I find a bit contrived, but it can lead to discussion about which is the strongest case of not belonging.
## Whole class and group discussion
Some discussions work well for a whole class, while others are better in small groups or pairs. Matching or ordering paper slips with expressions can lead to great discussion. For example we could have a set of graphs of the same data, and order them according to how effective they are at communicating the aspects of the data. Or there could be statements of possible events and students can place them in order of likelihood. The discussion involved in ordering them helps students to clarify the nature of probability. Desmos has a facility for teachers to set up card matching or grouping exercises, which reduces the work and waste of paper.
Our own Dragonistics data cards are great for discussion. Students can be given a number of dragons (more than two) and decide which one is the best, or which one doesn’t belong, or how to divide the dragons fairly into two or more groups.
It can seem to be wasting time to have discussion. However the evidence from research is that good discussion is an effective way for students to learn mathematics and statistics. I challenge all maths and stats teachers to increase and improve the discussion in their class.
# Proving causation
## Aeroplanes cause hot weather
In Christchurch we have a weather phenomenon known as the “Nor-wester”, which is a warm dry wind, preceding a cold southerly change. When the wind is from this direction, aeroplanes make their approach to the airport over the city. Our university is close to the airport in the direct flightpath, so we are very aware of the planes. A new colleague from South Africa drew the amusing conclusion that the unusual heat of the day was caused by all the planes flying overhead.
Statistics experts and educators spend a lot of time refuting claims of causation. “Correlation does not imply causation” has become a catch cry of people trying to avoid the common trap. This is a great advance in understanding that even journalists (notoriously math-phobic) seem to have caught onto. My own video on important statistical concepts ends with the causation issue. (You can jump to it at 3:51)
So we are aware that it is not easy to prove causation.
In order to prove causation we need a randomised experiment. We need to make random any possible factor that could be associated, and thus cause or contribute to the effect.
There is also the related problem of generalizability. If we do have a randomised experiment, we can prove causation. But unless the sample is also a random representative sample of the population in question, we cannot infer that the results will also transfer to the population in question. This is nicely illustrated in this matrix from The Statistical Sleuth by Fred L. Ramsey and Daniel W Schafer.
The relationship between the type of sample and study and the conclusions that may be drawn.
The top left-hand quadrant is the one in which we can draw causal inferences for the population.
## Causal claims from observational studies
A student posed this question: Is it possible to prove a causal link based on an observational study alone?
It would be very useful if we could. It is not always possible to use a randomised trial, particularly when people are involved. Before we became more aware of human rights, experiments were performed on unsuspecting human lab rats. A classic example is the Vipeholm experiments where patients at a mental hospital were the unknowing subjects. They were given large quantities of sweets in order to determine whether sugar caused cavities in teeth. This happened into the early 1950s. These days it would not be acceptable to randomly assign people to groups who are made to smoke or drink alcohol or consume large quantities of fat-laden pastries. We have to let people make those lifestyle choices for themselves. And observe. Hence observational studies!
There is a call for “evidence-based practice” in education to follow the philosophy in medicine. But getting educational experiments through ethics committee approval is very challenging, and it is difficult to use rats or fruit-flies to impersonate the higher learning processes of humans. The changing landscape of the human environment makes it even more difficult to perform educational experiments.
To find out the criteria for justifying causal claims in an observational study I turned to one of my favourite statistics text-books, Chance Encounters by Wild and Seber (page 27). They cite the Surgeon General of the United States. The criteria for the establishment of a cause and effect relationship in an epidemiological study are the following:
1. Strong relationship: For example illness is four times as likely among people exposed to a possible cause as it is for those who are not exposed.
2. Strong research design
3. Temporal relationship: The cause must precede the effect.
4. Dose-response relationship: Higher exposure leads to a higher proportion of people affected.
5. Reversible association: Removal of the cause reduces the incidence of the effect.
6. Consistency: Multiple studies in different locations producing similar effects
7. Biological plausibility: there is a supportable biological mechanism
8. Coherence with known facts.
## Teaching about causation
In high school, and entry-level statistics courses, the focus is often on statistical literacy. This concept of causation is pivotal to correct understanding of what statistics can and cannot claim. It is worth spending some time in the classroom discussing what would constitute reasonable proof and what would not. In particular it is worthwhile to come up with alternative explanations for common fallacies, or even truths in causation. Some examples for discussion might be drink-driving and accidents, smoking and cancer, gender and success in all number of areas, home game advantage in sport, the use of lucky charms, socks and undies. This also ties nicely with probability theory, helping to tie the year’s curriculum together.
# Teaching time series with limited computer access
## How do you teach statistics with limited access to computers?
Last century this wasn’t really an issue, at least not in high schools, as statistics has been a peripheral part of the mathematics curriculum and the mathematics of statistics has been taught as a subset of mathematics.
But this is changing, and it looks as if the change is starting in New Zealand. The NZ school curriculum has leapt ahead of the rest of the world. Statistics is taught at all levels and at the higher levels of high school, statistics is taught as it is actually done in practice – using computers. All analysis is done by a computer package, particularly using iNZight, a purpose-built, free package. The emphasis is on understanding, concepts and critical thinking, rather than the mechanical and slow application of formulas. The rigour has moved from the calculations to the meaning. It is SO exciting!
One big concern for many teachers is access to computers. In many schools there aren’t enough computer suites to schedule the students in for their statistics classes. So how do we deal with this?
It might seem that the computers are needed every day, but in fact they aren’t. And neither is it necessary to have one computer per student.
## Make them share
I’ve never had a problem when students have had to share computers. I find the people who do share a computer, learn better than those who are trying to work it out on their own. I actively encourage sharing computers in a lab.
I recently had the opportunity to be on the learning end, with computer instruction. The teacher was showing what to do at the front, and we in the class were echoing her steps on our computers. This is not ideal, as it requires everyone to be at the same pace, but as we were adults it was fine. I hadn’t brought my laptop, so I was sharing with another student. I’m pretty sure I learned more, as I got to follow what was happening on both computers, rather than trying to work it out and keep up. I was also able to help my partner, as she would lose track of what was happening when her computer wasn’t doing what it was meant to.
I have found this to be true at all levels, especially when learning a new package. Having two heads at the computer encourages discussion, which is an important element in learning. Students are also more likely to ask questions when they have already discussed a problem with another student. Pairing is so useful that some software companies get programmers to work in pairs, sharing a computer and work desk, because they have discovered that this has benefits.
## Think about what we are trying to teach
I am currently developing resources for a unit in time series analysis, based on the New Zealand curriculum, and using the free software, iNZight. At first glance, you might think that the entire unit would need to be taught in a computer lab. This is definitely not the case. And because of the layout of many computer labs, in fact you are better to stay out of them for most of the unit so that students can work in groups.
I find that it is worthwhile to think about the attitudes, skills and knowledge that we wish our pupils to develop in a unit – in that order of importance! These examples are illustrative rather than exhaustive.
Attitudes – By the end of the topic all students should feel that time series analysis is interesting and relevant (and maybe even fun!).
Time series analysis is pretty straightforward at the beginner level, but can be quite exciting. Once you know the basics, and with a convenient package to speed up the mechanics, you can do some interesting detective work. I would want the students to share some of this excitement, and start to explore on their own.
## Skills
Students should be able to:
• identify elements of a time series, relating them to the real life context.
• write a report on a time series analysis using correct terminology, clear enough for a non-expert reader to understand.
• use iNZight to analyse different time series.
## Knowledge
• Student should be able to explain and apply the following terms correctly: time series, trend, seasonality, stationary, noise, variation
And that is about it really!
So how do we do this, with or without full computer access.
Even with unlimited computer access I would get students to work in pairs for much of the time. I would start away from the computers. First display graphs of time series to the class and get them to write down sentences about them in their pairs. Then share with the class. We should get sentences like, “It mainly goes up, and then it goes down” and “there is a pattern that repeats”. From that the teacher can introduce the ideas of trend, seasonality and noise, modelling the correct use of specialist language.
Then I would talk about the context – or maybe the context should have come first… The time series chosen should be one with an easy to identify context, such as retail sales of recreational goods, or patterns of tourist arrivals. These series are available in New Zealand at Infoshare or in iNZight format via Statslc. Other countries will have similar series available. Again get the students to write down sentences, this time relating them to the context.
Homework could be to find a graph of a time series on-line or in a magazine. Or to make a list of things that might show seasonality.
Next I would get the students onto the computers in pairs. They should have a worksheet like the one here, so that they can work step-by-step through the package at their own pace in pairs. At some time during the class they could swap roles, if one has been instructing and the other operating.
The data set here RetailNZTS4 has four series in it, which show different behaviours. Students should see if they can get all the graphs they need for further analysis.
Four time series compared using iNZight software
The next class is away from the computers again. Here they are writing sentences about the graphs. They should do this alone, and in pairs, and compare in groups. It would be good to have a computer or two available for students to take turns to get any graphs they might find they need. When people are in front of a computer it tends to dominate their thinking and they can produce far too much output with very little thought. Moving away from the computer encourages a more reflective approach.
Then start on another data set. I would use the one about accommodation, AccRegNZTS13 comparing the seasonal patterns of occupancy in different regions of New Zealand. If there are enough computers, the students can spend one day creating the graphs and exploring, then the next day writing it up. Maybe different groups could take different regions, and find out why the pattern is the way it is for that region, then report back to the class.
Then the teacher may like to give some of the mathematical background to how a computer package would go about producing the output.RetailNZTS4
The learning is in the writing and the talking.
The point I’m trying to make is that you actually need to move away from the computers quite often. If you are REALLY stuck for computers you could even print off (and laminate?) the outputs from the different time series, so that the students can study and discuss them. Number or name them for easy reference, and have question sheets to go with them.The computer is only the tool, and with a bit of creativity, we can still teach the important attitudes, skills and knowledge with limited computer access.
I am aware as I am writing this that it is some time since I taught a class of high school students. I would be thrilled to hear comments from the “chalk-face” as to how realistic you think this is! And of course other suggestions will be welcome for teaching a computer-rich subject in a computer-poor environment.
Having said that, one of my experiences as a trainee teacher was having to teach my first lesson to a class at Rotorua Lakes High School during a powercut – which meant no computers and no OHP. We did desk-checking (how you can use pen and paper to look for bugs in code) and it went surprisingly well. | 4,205 | 21,086 | {"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} | 2.640625 | 3 | CC-MAIN-2018-22 | latest | en | 0.982724 |
https://www.experts-exchange.com/questions/10107206/help-with-final-while-loop-arrays-and-of-iterations.html | 1,506,339,950,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691476.47/warc/CC-MAIN-20170925111643-20170925131643-00491.warc.gz | 793,042,567 | 38,279 | Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
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# help with final while loop,arrays and # of iterations
Posted on 1998-12-10
Medium Priority
335 Views
here is what i have done so far.im having problems with the while loop and with the number of iterations. this is for this friday. please mail me with the answer as soon as you can.
after the program is the instructions of what i have to do. can you tell me of a good book about C programming.
/*Este programa calcula la temperatura de cada
punto interior de la plancha de metal e
imprime la tabla de los valores viejos y la
tabla de los valores nuevos.*/
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
void main (void)
{
/* Declaracion de variables*/
float tvnorte, tvsur, tveste, tvoeste;
float tv[10][15], tn[10][15], error;
int i, j;
/*Leer valores constantes de un file*/
FILE*viejo;
FILE*nuevo;
nuevo=fopen("pedro.out","w");
viejo=fopen("aixa.dat","r");
fscanf(viejo, "%f",&tvnorte);
fscanf(viejo, "%f",&tvsur);
fscanf(viejo, "%f",&tveste);
fscanf(viejo, "%f",&tvoeste);
tv[0][0] = (tvnorte + tvoeste)/2.0;
tv[0][14] = (tvnorte + tveste)/2.0;
tv[9][0] = (tvsur + tvoeste)/2.0;
tv[9][14] = (tvsur + tveste)/2.0;
tn[0][0] = tv[0][0];
tn[0][14] = tv[0][14];
tn[9][0] = tv[9][0];
tn[9][14] = tv[9][14];
for(j=1;j<=13; j++)
{
i=0;
tv[i][j] = 100.0;
tn[i][j] = tv[i][j];
}
for (i=1; i<=8; i++)
{j=0;
tv[i][j] = 0.0;
tn[i][j] = tv[i][j];
}
for(j=1;j<=13; j++)
{i=9;
tv[i][j] = 50.0;
tn[i][j] = tv[i][j];
}
for(i=1;i<=8; i++)
{j=14;
tv[i][j] = 75.0;
tn[i][j] = tv[i][j];
}
for(i=1;i<=8;i++)
{
for(j=1;j<=13;j++)
{
tv[i][j] = 24.0;
}
}
fprintf(nuevo,"tabla temperatura vieja:\n");
for(i=0;i<=9;i++)
{
for(j=0;j<=14;j++)
{fprintf(nuevo,"%4.0f", tv[i][j]);}
fprintf(nuevo,"\n");
}
fprintf(nuevo,"\n");
/*tv ends here*/
while(error<=0.000001)
{
error =fabs(tv[i][j] - tn[i][j]);
for(i=1;i<=8;i++)
{
for(j=1;j<=13;j++)
{
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0;
}
}
}
fprintf(nuevo,"tabla temperatura nueva:\n");
for(i=0;i<=9;i++)
{
for(j=0;j<=14;j++)
{
fprintf(nuevo,"%4.0f", tn[i][j]);
}
fprintf(nuevo,"\n");
}
fclose(nuevo);
fclose(viejo);
}
Suppose that the edges of a thin square metal &e maintained at different temperatures: 100 C (north edge), 75 C (east edge), 50 C (south edge) and 0 (east edge) and we whish to determine the steady state temperature at each interior point. To do this we divide the plate into 10 rows and 15 colurnns.
Each corner of the small squares is called a node. The new temperature (tn) in the interior nodes can be calculate form the old ones (to) by:
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0
Write a program that calculate each interior point at steady state. The program must accomplish the following:
(1) Initialize the temperature of all nodes (named it the old temperature '~o") as follows:
Read from a file the four constant temperatures along each edge and initialize the temperature of the nodes located along each edge. (these temperatures are constant.
Calculate each comer as the arithmetic average of the adjacent edges.
initialize the interior nodes with a guess temperature.
(2) Print the old temperature array in a file of results.
(2) Calculate the new temperature (tn) of the interior nodes by repeatedly averaging the temperatures at its four neighbors with the formula above. Repeat this procedure until the new temperature at each interior node differs from the old temperature by no more than 1 .0e-6.
(3) Print the new temperature array and the number of iterations used to produce the fmal result in the same results file.
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Question by:michaelangelo
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Author Comment
ID: 1254992
Edited text of question
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Expert Comment
ID: 1254993
You never initialize the variable error. Try initializing this to a value greater than .000001 (try error = 1).
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Expert Comment
ID: 1254994
You never initialize the variable error. Try initializing this to a value greater than .000001 (try error = 1).
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Expert Comment
ID: 1254995
There are a couple of other changes that need to be made:
You never copy the new data to tv. This leads to an infinite loop! Also, you were not checking error against all the values.
The following uses error as a sentinel value (0 or 1) to stay in the loop, and I have added to extra for loops. One checks the difference for all cells, and the other copies the new data to the tv array. Hope this helps :)
/* I WOULD CHANGE THE WILE LOOP TO THE FOLLOWING */
error = 1;
while ( error ) {
for ( i = 1; i <= 8; i++ ) {
for ( j = 1; j <= 13; j++ ) {
tn[i][j] = ( tv[i - 1][j] + tv[i][j - 1]
+ tv[i][j + 1] + tv[i + 1][j] ) / 4.0;
}
}
error = 0;
for ( i = 1; i <= 8; i++ ) {
for ( j = 1; j <= 13; j++ ) {
if ( fabs(tn[i][j] - tv[i][j]) > 0.000001)
error = 1;
}
}
/* COPY NEW DATA */
for ( i = 0; i <= 9; i++ ) {
for ( j = 0; j <= 14; j++ ) {
tv[i][j] = tn[i][j]; }}
}
0
Author Comment
ID: 1254996
the answer you gave still gives an infinite loop and the # of iterations to get the new temp is missing. i need this for tomorrow, urgently
0
LVL 10
Expert Comment
ID: 1254997
try using double instead of float so you get less rounding errors. float are only accurate to about 8 digits. You are starting with 100.0 and comparing with 0.000001 .. this is very close to (or exceeding) the accuracy of floats.
also in
while(error<=0.000001) {
error =fabs(tv[i][j] - tn[i][j]);
error is checked BEFORE it is calculated (use do..while instead or set error to an initial value)
also you probably want to compare error > 0.000001 (ie loop while there is a big error, stop when small enough)
and "error =fabs(tv[i][j] - tn[i][j]);" is outside the for loops that control i and j, so you probably aren't doing what you expect. Perhaps you should be accumulating the error in the loop and then testing (eg. sum squares, or max abs etc).
0
Author Comment
ID: 1254998
how do i do this. it doesnt work that way either and i dont know where is the error. please help
0
LVL 10
Expert Comment
ID: 1254999
ok .. guess I can write come code for you to make it clear
first .. change your float vars to double !!!
then we change you while loop into a do..while
we also get the largest individual error and use that for the test
and we end the loop when the error is small enough
#define MAXERROR 0.00001
do {
error = 0;
for(i=1;i<=8;i++) {
for(j=1;j<=13;j++) {
double e;
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0;
e = fabs(tv[i][j] - tn[i][j]);
if (e > error) error = e;
}
}
}
while (error > MAXERROR);
There you go.
0
Author Comment
ID: 1255000
this loop seems to be infinite and where is the number of iterations the loop is done?
0
LVL 1
Expert Comment
ID: 1255001
Try this out if it works!! There may be a few minor errors... but no infinite loops.....
/*Este programa calcula la temperatura de cada
punto interior de la plancha de metal e
imprime la tabla de los valores viejos y la
tabla de los valores nuevos.*/
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int main (void)
{
/* Declaracion de variables*/
float tvnorte, tvsur, tveste, tvoeste;
float tv[10][15], tn[10][15];
double error;
int i, j, no_iter;
/*Leer valores constantes de un file*/
FILE*viejo;
FILE*nuevo;
viejo=fopen("aixa.dat","r");
fscanf(viejo, "%f",&tvnorte);
fscanf(viejo, "%f",&tvsur);
fscanf(viejo, "%f",&tveste);
fscanf(viejo, "%f",&tvoeste);
fclose(viejo);
/* delete the following line If you feel that the file has been read correctly */
printf("\n Are these values correct %f ?? <> %f ?? <> %f ?? <> %f ?? \n", tvnorte, tvsur, tveste, tvoeste);
tv[0][0] = (tvnorte + tvoeste)/2.0;
tv[0][14] = (tvnorte + tveste)/2.0;
tv[9][0] = (tvsur + tvoeste)/2.0;
tv[9][14] = (tvsur + tveste)/2.0;
/* End settings */
tn[0][0] = tv[0][0];
tn[0][14] = tv[0][14];
tn[9][0] = tv[9][0];
tn[9][14] = tv[9][14];
/* Setting all values along row (0, j=1-13) to 100.0 */
for(j=1, i=0;j<=13; j++)
tn[i][j] = tv[i][j] = 100.0;
/* Setting all values along the column (i=1-8,14) to 0.0 */
for (i=1, j=0; i<=8; i++)
tn[i][j] = tv[i][j] = 0.0;
/* Setting all values along row (9, j=1-13) to 50.0 */
for(j=1,i=9;j<=13; j++)
tn[i][j] = tv[i][j] = 50.0;
/* Setting all values along the column (i=1-8,14) to 75.0 */
for(i=1, j=14; i<=8; i++)
tn[i][j] = tv[i][j] = 75.0;
/* Setting all values from (1,1) to (8,13) to 24.0 */
for(i=1;i<=8;i++)
for(j=1;j<=13;j++)
{
tn[i][j] = 0.0;
tv[i][j] = 24.0;
}
/* Opening the output file to write data */
nuevo=fopen("pedro.out","w");
fprintf(nuevo,"tabla temperatura vieja:\n");
for(i=0;i<=9;i++)
{
for(j=0;j<=14;j++)
fprintf(nuevo,"%4.0f", tv[i][j]);
fprintf(nuevo,"\n");
}
fprintf(nuevo,"\n");
/*tv ends here*/
error = 1.0;
no_iter = 0; /* Determine no. of iterations */
while(error<=0.000001)
{
error = 0.0;
no_iter++;
for(i=1;i<=8;i++)
for(j=1;j<=13;j++)
{
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0; /* new values */
error +=fabs(tv[i][j] - tn[i][j]); /* compute the net error for whole matrix */
tv[i][j] = tn[i][j]; /* without this error will be constant as tv is constant */
}
error /= 14.0*9.0; /* average to 14x9 values */
}
fprintf(nuevo,"tabla temperatura nueva:\n");
fprintf(nuevo,"Number of Iterations = %d \n", no_iter);
for(i=0;i<=9;i++)
{
for(j=0;j<=14;j++)
fprintf(nuevo,"%4.0f", tn[i][j]);
fprintf(nuevo,"\n");
}
fclose(nuevo);
return 0;
}
0
LVL 84
Expert Comment
ID: 1255002
tvnorte, tvsur, tveste, tvoeste don't seem to affect the final table much.
0
LVL 1
Expert Comment
ID: 1255003
you can change "error /= 14.0*9.0" to "error /= 150.0 or whatever bigger" for even faster result
0
Author Comment
ID: 1255004
tried that but doesnt work either
the number of iterations with themperatures between o and 24 must be between 300 and 360
0
LVL 11
Expert Comment
ID: 1255005
This question looks awfully simmilar to http://www.experts-exchange.com/Q.10106928
Seems like an attempt to cheat using multiple accounts. Tsk tsk.
0
LVL 1
Expert Comment
ID: 1255006
how much do you get for number of iterartions?
0
Author Comment
ID: 1255007
does anyone want to help?
0
LVL 1
Accepted Solution
lafanga earned 400 total points
ID: 1255008
In my Code:
change the line
while(error<=0.000001)
to
while(error>=0.000001)
and
error /= 14.0*9.0;
to
error /= 140.0;
0
LVL 10
Expert Comment
ID: 1255009
michaelangelo ... I HAVE helped already .. why do you keep rejecting my answers ?????
The loop I wrote is NOT infinite (unless the temperatures don't converge).
"(2) Calculate the new temperature (tn) of the interior nodes by repeatedly averaging the temperatures at its four neighbors with the formula above. Repeat this procedure until the new temperature at each interior node differs from the old temperature by no more than 1 .0e-6. "
There is NO mention of iterations .. just repeat until the error is small enough .. that is what I did for you.
I also fixed a number of your programming errors that made the whole thing just not work.
NOW you say (out of hte blue) that "the number of iterations with themperatures between o and 24 must be between 300 and 360"
why didn't you say this before?
BTW: The code that does the total and average of the errors does NOT answer you questions .. it wants the maximum error to be less than the specified minimum NOT the average error.
I have compiled and run your code with my modifications. Your initializaiotn code was a bit screwy as well. I've rewritten it for you (don't know why ... but I have) .. here is the results I get...
tabla temperatura vieja:
75 100 100 100 100 100 100 100 100 100 100 100 100 100 50
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
50 56 56 56 56 56 56 56 56 56 56 56 56 56 0
63 75 75 75 75 75 75 75 75 75 75 75 75 75 38
tabla temperatura nueva:
75 100 100 100 100 100 100 100 100 100 100 100 100 100 50
50 75 85 89 91 92 93 92 91 89 86 81 71 51 0
50 65 75 81 84 86 86 85 83 80 75 66 53 32 0
50 61 69 75 78 80 80 80 77 73 66 56 42 24 0
50 58 65 71 74 76 76 75 72 68 61 51 37 20 0
50 57 64 68 72 73 74 73 70 65 58 49 35 19 0
50 57 63 68 71 72 72 71 69 65 59 50 37 20 0
50 59 65 69 71 72 72 72 70 67 62 54 43 26 0
50 63 69 71 73 73 73 73 72 70 68 63 55 39 0
63 75 75 75 75 75 75 75 75 75 75 75 75 75 38
Do you want the code ??
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LVL 1
Expert Comment
ID: 1255010
if you want the maximum error less than 1.0e6 than do the following changes
2) change the error loop to:
double max_err;
max_err = 1.0;
no_iter = 0; /* Determine no. of iterations */
while(max_err>=0.000001)
{
no_iter++;
for(i=1;i<=8;i++)
for(j=1;j<=13;j++)
{
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0; /* new values */
error =fabs(tv[i][j] - tn[i][j]); /* compute the net error for whole matrix */
if( error > max_err)
max_err = error;
tv[i][j] = tn[i][j]; /* without this error will be constant as tv is constant */
}
}
However if you want average error the earlier code is good enough.........
0
LVL 1
Expert Comment
ID: 1255011
I smell a rat=))
Just one correction more.
include this as the final error loop for max error based iteration..
double max_err;
max_err = 1.0;
no_iter = 0; /* Determine no. of iterations */
while(max_err>=0.000001)
{
no_iter++;
max_err = 0.0; /* This value must be reset every time */
for(i=1;i<=8;i++)
for(j=1;j<=13;j++)
{
tn[i][j] = (tv[i-1][j]+tv[i][j-1]+tv[i][j+1]+tv[i+1][j])/4.0; /* new values */
error =fabs(tv[i][j] - tn[i][j]); /* compute the net error for whole matrix */
if( error > max_err)
max_err = error;
tv[i][j] = tn[i][j]; /* without this error will be constant as tv is constant */
}
}
0
LVL 10
Expert Comment
ID: 1255012
funny ... that was the same logic i used in my earlier rejected answer.
Also the code you have supplied above is not quite correct as it modifies tv DURING the calculation of the new tn array.
You should calculate ALL the tn values FIRST and THEN copy them to tv.
If you want my version of the code with the corrections to your initialization of the array (which wasn't right) and your error checking (which wasn't right) and your calculation of tn and tv (which wasn't right) then let me know.
Of course this question will probably get auto-graded by EE :-(
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1. ## concept behind finding volume between 2 curves
If there is an area between 2 curves and it's spun around the x-axis I integrate with respect to x, else if it's spun around the y-axis I integrate with respect to y. What screws me up is finding the area because I subtract the wrong function. For example, $\displaystyle y^2=x$ and $\displaystyle 2y=x$. The $\displaystyle y^2$ is on top but the $\displaystyle 2x$ is to the right.
Which rule takes precedence; right-left or top-bottom?
If I get a negative volume, chances are I have the functions switched around, but the converse isn't always true right? For example I could still have the the functions switched around and still get a positive answer?
2. [QUOTE=superdude;340381]If there is an area between 2 curves and it's spun around the x-axis I integrate with respect to x, else if it's spun around the y-axis I integrate with respect to y. What screws me up is finding the area because I subtract the wrong function. For example, $\displaystyle y^2=x$ and $\displaystyle 2y=x$. The $\displaystyle y^2$ is on top but the $\displaystyle 2x$ is to the right.
Which rule takes precedence; right-left or top-bottom?
Are you going from left to right or top to bottom? In other words are you rotating around a vertical line or a horizontal line?
If I get a negative volume, chances are I have the functions switched around, but the converse isn't always true right? For example I could still have the the functions switched around and still get a positive answer?
No. $\displaystyle \int (f(x)- g(x))dx= -\int (g(x)- f(x))dx$. If all you have done is switch f and g, one integral must be the negative of the other. (Unless they are both 0 in which case it doesn't matter!)
3. take the following situations for example:
$\displaystyle f(y)=(y + 4)^2 - 4$ and $\displaystyle g(y)=- y^2 + 6$
is it correct to subtract f(y)-g(y) in accordance to right-left or is it correct to do g(y)-f(y) in accordance of top-bottom?
Am I making myself clear?
another example is where the following functions are rotated around the y-axis:$\displaystyle x=y^2$ and $\displaystyle x=2y$. I keep switching them around doing $\displaystyle \pi\int_0^2 (y^{2})^{2}-(2y)^{2} dy$ which is wrong. However what confuses me is for 0<x<4 $\displaystyle \sqrt{x}>x/2$.
4. Originally Posted by superdude
take the following situations for example:
$\displaystyle f(y)=(y + 4)^2 - 4$ and $\displaystyle g(y)=- y^2 + 6$
is it correct to subtract f(y)-g(y) in accordance to right-left or is it correct to do g(y)-f(y) in accordance of top-bottom?
Am I making myself clear?
Rotating around which axis? Draw a picture! Graph each function and the axis of rotation. Draw a line perpendicular to the axis. Rotated around that axis it becomes a disk. The "outer" and "inner" radii are given by the graphs farther from and closer to the axis of rotation. That determines which is subtracted from which.
another example is where the following functions are rotated around the y-axis:$\displaystyle x=y^2$ and $\displaystyle x=2y$. I keep switching them around doing $\displaystyle \pi\int_0^2 (y^{2})^{2}-(2y)^{2} dy$ which is wrong. However what confuses me is for 0<x<4 $\displaystyle \sqrt{x}>x/2$.
The graphs intersect when $\displaystyle y^2= 2y$. That is, at (0,0) and $\displaystyle (\sqrt{2},2)$.
Since you are rotating around the y-axis, a line perpendicular to the axis (which becomes a disk when rotated) crosses first the the curve $\displaystyle x= y^2$line and then the line x= 2y- that is, the "inner" distance is $\displaystyle y^2$ and the "outer" is 2y. The volume is given by
$\displaystyle \pi \int_{y= 0}^2 (2y)^2- (y^2)^2 dy$
Because you are rotating around the y-axis, the order of y values, $\displaystyle \sqrt{x}$ and x/2 is irrelevant.
5. okay thanks. I thought it was irrelevant which axis the curves are being rotated around but now I see that's not the case. So I've gotta remember to draw a picture and do "outside curve"-"inside curve"
I appreciate the help.
I tried to upload a picture but I don't know how, is there any other way then "insert image"? | 1,109 | 4,169 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.917918 |
https://www.jiskha.com/display.cgi?id=1323809101 | 1,527,086,547,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00459.warc.gz | 787,565,835 | 4,054 | # Stats
posted by Stephanie F
A new extended-life light bulb has an average service life of 750 hours, with a standard deviation of 50 hours. If the service life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 600 and 900 hours?
a. 95 percent
` b. 68 percent
c. 34 percent
d. 99.7 percent
e. cannot be determined from the information given.
1. PsyDAG
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores.
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http://math.stackexchange.com/questions/206649/monotonicity-in-borel-s-theorem?answertab=oldest | 1,462,294,738,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121618.46/warc/CC-MAIN-20160428161521-00114-ip-10-239-7-51.ec2.internal.warc.gz | 195,444,983 | 16,814 | # Monotonicity in Borel’s theorem
Borel’s theorem, in its simplest form, states that for any sequence $(a_k)$ of real numbers, there is a function $f\in {\cal C}^{\infty}({\mathbb R},{\mathbb R})$ such that $f^{(k)}(0)=a_k$ for all $k$. I wonder if we can take $f$ to be increasing.
Obviously, this will not be possible if the sequence starts with some zeros and then a negative number. If, however, all the $a_k$ are $0$ or if the first nonzero $a_k$ is positive, there seems to be no reason why $f$ could not be increasing. Is this already known ?
-
I was wondering how this theorem works for any sequence of complex numbers ? – Maman Apr 17 at 10:54
In the form stated it won't be possible. E.g., if $a_1 = 0$ and $a_2 > 0$, then $f'$ will be increasing near $0$, so $f'(x)<f'(0)=0$ for (small) negative $x$, and $f$ will be decreasing in at least a small interval $(-\delta,0)$ for some $\delta>0$.
On the other hand, if $a_1>0$, you can get an increasing function as follows: Start with a function $f$ which has all the prescribed derivatives and has compact support (by multiplying an arbitrary solution with a $\mathcal{C}^\infty$ function with compact support which is $\equiv 1$ near $0$). Now it is enough to find a $\mathcal{C}^\infty$ function $g$ with $g^{(n)} (0) = 0$ for all $n$, and $g'(x)>0$ for $x \ne 0$. Then $f_t= f+tg$ will satisfy $f_t'(x) > 0$ for $x \ne 0$ if $t$ is sufficiently large, implying that $f_t$ is strictly increasing. The construction of $g$ is standard, e.g., $g(x) = e^{-1/x^2}$ for $x>0$, $g(0)=0$, and $g(-x) = -g(x)$ for $x>0$.
Similarly, if the smallest $k$ for which $a_k \ne 0$ is odd, and if $a_k > 0$, you can find an increasing function with the prescribed derivatives (basically by integrating the solution given above $k-1$ times.) | 568 | 1,787 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 3.5 | 4 | CC-MAIN-2016-18 | latest | en | 0.874875 |
https://www.bartleby.com/questions-and-answers/the-fierro-corporation-has-annual-credit-sales-of-dollar6-million.-current-expenses-for-the-collecti/04f88455-6eb8-4f0a-89c4-1d53f594c9b6 | 1,579,504,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250597458.22/warc/CC-MAIN-20200120052454-20200120080454-00326.warc.gz | 765,346,262 | 24,633 | # The Fierro Corporation has annual credit sales of \$6 million. Current expenses for the collection department are \$100,000, bad debt losses are 4 percent, and the days sales outstanding is 30 days. Fierro is considering easing its collection efforts so that collection expenses will be reduced to \$50,000 per year. The change is expected to increase bad debt losses to 7 percent and to increase the days sales outstanding to 45 days. In addition, sales are expected to increase to \$8 million per year. Should Fierro relax collection efforts, if the opportunity cost of funds is 10 percent, the variable cost ratio is 75 percent, and its marginal tax rate is 30 percent? All costs associated with production and credit sales are paid on the day of the sale.
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The Fierro Corporation has annual credit sales of \$6 million. Current expenses for the collection department are \$100,000, bad debt losses are 4 percent, and the days sales outstanding is 30 days. Fierro is considering easing its collection efforts so that collection expenses will be reduced to \$50,000 per year. The change is expected to increase bad debt losses to 7 percent and to increase the days sales outstanding to 45 days. In addition, sales are expected to increase to \$8 million per year.
Should Fierro relax collection efforts, if the opportunity cost of funds is 10 percent, the variable cost ratio is 75 percent, and its marginal tax rate is 30 percent? All costs associated with production and credit sales are paid on the day of the sale.
check_circle
Step 1
Calculation of Incremental profit after tax using excel is as follows:
Step 2
Formula used in the above table is as follows:
Step 3
Existing Plan:
Credit Sales = \$6,000,000
Days Sales Outstanding = 30
Total Number of Days in a Year = 3...
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https://www.reddit.com/r/askscience/comments/xtqq5/what_if_we_put_a_mirror_in_space_1_light_year/ | 1,485,067,769,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281353.56/warc/CC-MAIN-20170116095121-00247-ip-10-171-10-70.ec2.internal.warc.gz | 977,248,343 | 75,834 | This is an archived post. You won't be able to vote or comment.
[–]Genomics | Molecular biology | Sex differentiation 773 points774 points (183 children)
If you look at a mirror in your bathroom, one meter away, it takes about 7x10-9 s for light to make the round trip, so technically you're seeing 7x10-9 s into your past. What you're describing is the same thing with a larger number.
[–] 257 points258 points (82 children)
That is so awesome. If the mirror is underwater am I seeing even further back in time?
[–]Lasers | Plasma Physics | Antimatter 328 points329 points (74 children)
if the distance is the same, but the speed of light propagation is slower (which it is in water), then yes.
[–] 28 points29 points (50 children)
Did you say the speed of light changes? What's going on there?
[–] 84 points85 points (45 children)
The speed of light changes depending on the medium it is in. Water would be one such medium.
Our normal ground-level earth atmosphere has a minimal hindrance on the speed of light, so most mediums light encounters, including water, will have a greater slow-down than the "air".
[–] 41 points42 points (37 children)
So the speed of light is really the maximum speed not the speed all light travels?
[–] 194 points195 points (8 children)
The commonly referred to speed of light is the speed of light in a vacuum.
[–] 104 points105 points (21 children)
Sorta, but basically yes. The slowdown of light in a medium is not because the photons move any slower. It's because the photon goes a little bit at C, hits something, is absorbed, which bumps an electron into a higher energy state, where the electron stays for a bit, then it re-emits the photon, dropping the electron back down, and the photon carries on at C.
The photon is never not moving at C, it just sometime spends some time having been absorbed by something instead.
[–] 13 points14 points (1 child)
So why does the photon keep going in the same direction? If it gets absorbed and then re-emitted, wouldn't it move off in a random direction?
[–] 50 points51 points (7 children)
This is why I love physics. The photon is always at c, even when it isn't.
[–] 33 points34 points (4 children)
The photon is always at c, but the photon at the end isn't the photon you started with.
[–] 25 points26 points (3 children)
In quantum physics there is no such thing as elementary particles being different from each other. So in more confusing words: They are all the same and they aren’t even the same as themselves, at the same time. And in more simple words: They are not distinguishable.
[–]Algorithms | Distributed Computing | Programming Languages 2 points3 points (1 child)
Analogy: In a game of telephone, a message will propagate much slower than the speed of sound. Between each pair of people, the message is moving at the speed of sound, but each person adds a delay.
[–] 7 points8 points (1 child)
You can think about it a different way:
There are two things - stuff and vacuum. When light travels through vacuum, it travels at the speed of light. When light interacts with stuff, it will either scatter or go through absorption & emission. If you were to shrink down and magically observe this process, you would see photons traveling at the speed of light, becoming absorbed, and then a little while later being emitted. So the fraction of light that manages to escape the object has effectively been slowed down, yet the photons, while moving, were always moving at the speed of light.
[–] 73 points74 points (3 children)
Even better... you can focus that light from a mirror onto a charge-coupled device, convert it to digital information, store it, and eventually let someone else view a hundred or more years into the past. This can also be done in an analogue fashion, and I have personally seen the world as it existed in the 1930's and prior.
[–] 9 points10 points (1 child)
What? Can you explain further?
[–] 65 points66 points (0 children)
It's a explanation of cameras
[–] 2 points3 points (0 children)
Did you just invent the camera...?
[–] 56 points57 points (72 children)
Furthermore, it takes about 100-300ms for the signal to get to your brain and get processed. So what you are "seeing" now already happened a little while ago... But since everyone's running at this delayed-reaction speed, it's hardly noticeable for us humans in day to day interactions.
[–] 120 points121 points (53 children)
That seems a bit high... A third of a second to process what I'm seeing? Source?
[–] 61 points62 points (30 children)
Furthermore, why do people's mouths sync up so well with what I'm hearing. This number seems fishy to me.
[–] 50 points51 points (0 children)
Evolutionarily speaking, animals brains developed delays within their brains so that relevant information can be processed simultaneously as one fluid event. These delays can even be seen between two ears. Two ears receive the same sound wave in different phases of the waveform, yet you hear the same sound. Neurons in the medial superior olivary nucleus have delay lines that match similar signals that arrive in each ear at different times, and the inferior colliculus can use that information to determine where in space that signal is coming from (Joris, 2006). In this way, you perceive one sound, not a sound then an echo.
Joris, P.X. (2006). A dogged pursuit of coincidence. Journal of Neurophysiology, 96, 969-972.
[–] 58 points59 points (10 children)
In his defence, every single perception and sense of yours may take the same 300ms. Hearing as well as seeing.
Though I still think it's a bit high, I read once that nerve signals travel at around 400km/h. Then processing time would be added to that.
[–]Neurophysiology | Biophysics | Neuropharmacology 49 points50 points (3 children)
That's not actually true. Vision is the slowest of the senses. Because light is integrated at the photoreceptors via a chemical process relying on relatively large diffusion distances, about half of the delay in a visual response is at the photoreceptors.
You can test this very easily if you know rudamentary coding.
Want another way to test it? A good neuroscience party trick. Get a nice smooth 10 dollar bill. Hold is straight up and down, and get a volunteer to hold their finger and thumb about two inches apart ready to grab the bill. about half way down the length. Tell them to try to grab the note as soon as they see you let it go. If they can, they can have it. They never can. Then tell them to close their eyes. This time, say "Go" as soon as you let go of the bill. They will grab it first time.
[–] 6 points7 points (1 child)
You have spent your whole life processing images and sounds at this speed.. The brain plays tricks on us
[–] 33 points34 points (6 children)
how does the magician do a trick? they don't. they just make you think you did. same thing here. your brain just tells you the sight and sound are synchronized. see dennett for more on this.
[–] 16 points17 points (5 children)
I'm confused about how this lets us interact with things like light switches and video games without noticing the delay between you pressing the button and noticing the change.
[–] 2 points3 points (0 children)
because the experience of reality is a projection into the future. For example a baseball player at bat sees the ball at a place that approximately accounts for the time lag between the light hitting the eye and being represented as an object. Our lived experience is an approximately 300ms projection into "the future" of the information received so to speak.
[–] 6 points7 points (3 children)
First of all. Audio also takes time to get processed.
Secondly, Sync between different sensory inputs, especially auditory and visual is pretty flexible. You can try this for yourself by using a media player that allows you to adjust sync between audio and video (vlc will suffice). Generally speaking, anything under 70-100 ms isn't very noticeable, unless you are concentrating on noticing it.
[–] 4 points5 points (0 children)
I remember reading somewhere about a study where scientists would flash a light and like 1/10th of a second later they would play a buzzing sound. After awhile, the sound and the light seemed to be at the same time. The scientists then began to play them at the same time and the test subjects said that they were hearing the buzz before seeing the light.
[–] 2 points3 points (0 children)
Check this out. It is not that mouths sync up to what your hearing instead what you see and what you hear both influence the final sound registered in your brain. Mcgurk Effect
[–] 17 points18 points (10 children)
I don't have a source, but I do remember average human reaction time being 250ms, so I'd assume what c0balt meant was 100-300ms for the signal to be received, processed, and acted upon, which is much more reasonable.
[–] 24 points25 points (9 children)
As a network coder I'm very suspicious of such a high latency. I deal with latency issues all the time. In a video game, if there was even a 50ms delay between you pressing a button on a controller and something happening on screen it'd be blatantly obvious. In fact, didn't John Carmack say the problem with most VR headsets is that they have a >10ms latency?
[–] 9 points10 points (1 child)
Reaction times of 250ms are pretty normal. Say your brain takes input from the eyes at about an 80ms delay. It takes input from the ears at about a 30ms delay. It can take up to 100ms for the skin responses to get to the brain. Your brain then syncs everything up in your head, so it perceives everything at the same time. You can test this by getting a sensitive timer (3 decimals) and doing a couple of rounds of tests on reaction to a visual cue, then a couple of round with visual and audio, then a couple with just audio, then a couple with touch. Your ears are the fastest, and thus if you're waiting for an audio cue, you should get the fastest reaction times there.
For games and networks and such. We can perceive a granularity between 50 and 60 frames a second visually, our ears and touch systems also vary but have high granularity also. However, our brains are extremely good at syncing things up. This is why you have to have no latency between input reception, display, and sound.
[–] 16 points17 points (2 children)
your brain lies to you about the synchronistity of experiences through various sense organs.
[–] 3 points4 points (0 children)
Your brain has to do so much more than a computer though. We don't just process code, we have to understand semantics. When we see a face, we have to analyze all of the shapes, colors, and geometric figures. We have to filter out the blood vessels in our eyes and fill in the stuff we never saw (Blind Spot). We have to process depth. We have to compare that face to our internal models to decide its a face and whose face it is. We have to process information related to the face (What's her name?), and so much more. That we can do this in just over a tenth of a second is actually pretty remarkable.
[–] 4 points5 points (0 children)
It depends on what level of processing you're talking about. It's generally accepted that conscious visual processing takes at least 300ms, but reflexive processes (i.e., ducking something that was thrown at you) can occur more quickly. Sauce: http://documents.univ-lille3.fr/files/espaces/pers/81/P5181/public/M2%20PPNSA/Attention%20cours%201/Dehaene-2006.pdf
[–] 7 points8 points (0 children)
A response time of less than 100ms is considered a false start in the olympics, iirc, so I believe that thats about as fast as a human can respond to an audial cue. Visual cues take a much longer time to respond to (250ms+), so I believe this is where he's getting the figure of 300ms from.
Of course this is not how long it takes to actually receive and process the visual cue, that would probably be sub 100ms
[–] 23 points24 points (5 children)
I think you got reaction times mixed up with the time it takes to process information. You can in generally process what you are seeing much much faster then you can react in the sense of pushing a button as a result of what you saw.
[–] 19 points20 points (2 children)
and this is why i love this sub reddit. i have no clue what you guys say half the time but when i do (kinda) get it its astounding!
[–] 626 points627 points (61 children)
I think people are looking at the wrong part of the problem. The "past" doesn't include the launch time, but once it's in place already, then we observe. The mirror should show us events from two years prior to the moment we observe from, correct?
Lets say we observe the mirror in 2020 (being already 1 LY away). We should see light from 2018, correct?
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 277 points278 points (12 children)
Correct, as explained above.
[–] 2 points3 points (1 child)
Correct, but imagine the potential applications.
Any military operations, terrorist attacks, large scale crimes, as long as they occurred outdoors you could potentially verify what happened with this mirror, it doesn't even need to be a light-year away, a light-week would be more than enough.
[–] 2 points3 points (0 children)
Even better you just put a satellite in orbit with a camera. Which is what this is.
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 1455 points1456 points (253 children)
Yes. But note that because it will take more than a year to launch the mirror and get it in position, the "past" will still be after the launch.
[–] 201 points202 points (35 children)
You might as well place a video recording device instead of a mirror and have the device send back the data.
[–] 136 points137 points (3 children)
Or we could just record earth from orbit and transmit that to the ground. Then when we want to look into the past we just load up the tape from two years ago and press play. No need to launch something into deep space.
[–] 20 points21 points (0 children)
Reminds me of this.
[–] 11 points12 points (1 child)
I don't think OP is drawing up blueprints, I'm certain it was just a hypothetical question.
[–] 5 points6 points (0 children)
Right, OP's question was purely hypothetical and more concerned with understanding how light travels through space, and whether or not as simple a device as a mirror 1 light year out could be used to look 2 years into the past.
Once subwayboy brought video recorders into play, I figured I'd make things simpler by eliminating the extrasolar space travel.
[–] 8 points9 points (12 children)
But if we on Earth know what we want to look at in the mirror, we can make adjustments instantly, but if we wanted to adjust the camera's vision, we would have to send a signal that would take a year to get there. This of course doesn't matter if the camera can capture everything and send back everything all at once...
[–] 10 points11 points (10 children)
How would you from earth instantly make adjustments to a mirror or camera that is one light year away?
But as you note, it doesn't matter for this argument since our theoretical camera or mirror can capture everything.
[–] 22 points23 points (5 children)
I meant receiving the reflection from a different part of earth. Like oh we want to see Hawaii 2 years ago? Use our receiver in Guatamela so the angle is .0002052 degrees. Oh, you want to see Northern Malaysia at 02:40 at 9.23.2012? Please receive reflection of mirror from Bulgaria.
[–] 5 points6 points (2 children)
now that's a damn interesting idea. Basically an infinite flood of information and you can selectively pick what you want in real time. You could follow someone two years in the past if you knew their starting position, and see if they really did commit the crime.
[–] 8 points9 points (3 children)
Stand in front of a full-length mirror with both eyes closed. Then open your left eye. Then simultaneously open your right eye and close your left eye. You will have nearly-instantaneously changed your perspective even though you were looking at the same image.
A mirror large enough to let you observe different parts of the globe (one that was half the diameter of Earth assuming no other optical parts) would probably be cheaper to build than a camera that was capable of recording and processing video of the whole planet while providing similar quality images.
[–] 39 points40 points (14 children)
Wouldn't the data come back slower than the speed of light?
edit: thanks to all below for the info!
[–] 160 points161 points (8 children)
Radio waves travel at the speed of light.
[–] 16 points17 points (0 children)
To anyone who is surprised by this statement: consider that radio waves, microwaves, infrared radiation, visible light, ultraviolet, x-rays, and gamma rays are all forms of electromagnetic radiation. You can learn the list pretty easily by listening to this fun song.
[–] 11 points12 points (5 children)
I thought that was only in a vacuum, and I thought that radio waves could be effected by gravity, though I could easily be wrong.
[–] 64 points65 points (3 children)
All EM radiation (light waves, radio waves etc.) is affected by gravity in the same way.
[–] 12 points13 points (0 children)
Radio waves are affected by gravity in the same way that every other energy wave form is.
[–] 18 points19 points (3 children)
Light waves and radio waves are the same "wave" just different frequensies-bands
[–] 87 points88 points (6 children)
If you put the Webb space telescope at one light year distant from Earth, it would not have the optical resolution to see the planet, much less what's on the surface. Webb telescope has a resolution of about 0.1 arcseconds, which at 1 light year would be able to resolve objects which are about 1011 meters.
[–] 38 points39 points (2 children)
Whilst your point still stands, the Webb will be able to resolve much smaller objects than that at 1ly.
[–] 16 points17 points (1 child)
Dang nab it, I was in radians. Right you are.
[–] 4 points5 points (1 child)
What about 2 mirrors that are sufficiently far enough apart to give the necessary optical resolution to see Earth (and what would that distance have to be?)
[–] 49 points50 points (21 children)
I've often wondered, hypothetically, if you got a super-powerful telescope, instantly teleported it 65 million light years away, then pointed it back to Earth, would you see dinosaurs?
[–] 27 points28 points (9 children)
Did you teleport with the telescope or are you still on Earth? Because if so you're going to have to wait 65 million years for the transmission.
[–] 12 points13 points (8 children)
I believe to make this easier (and yes, it is rather amusing) we would not only teleport the mirror that far, but we'd also immediately use our time machine to go into the future, to be able to see the image, all during our lunch break.
[–] 32 points33 points (7 children)
If you have a time machine, why bother with the telescope?
[–] 21 points22 points (1 child)
The time machine is broken and can travel into the future only.
[–] 2 points3 points (0 children)
Then your lunch break is 65 million years long!
[–] 2 points3 points (0 children)
you wouldn't want to be on Earth when shit hits the fan.
[–] 49 points50 points (5 children)
Yes
[–][🍰] 38 points39 points (4 children)
Minus distortion from dust and interfering bodies in space.
[–] 26 points27 points (3 children)
And assuming the telescope can cope with the optical interference from the earths atmosphere.
[–] 19 points20 points (2 children)
And and that distance, wouldn't a single dinosaur's image be sufficiently spread out as to make it impossible to reconstruct the image due to photons being so spread out? When conducting lunar ranging, you only get back about 1 in 30,000,000 photons that ever make it to the reflector.
But if we're talking about zipping around faster than light, I guess we can suppose our 'super-powerful telescope' has a lens large enough to catch the light spread over such a great distance.
[–] 37 points38 points (1 child)
Yeah at this point we are just waving around magic wands anyway
[–] 85 points86 points (3 children)
I asked this question at SpaceCamp in 6th grade and got laughed at by the whole auditorium. I have never asked a question in public since. The speaker said all you would see is your eye in a telescope looking back.
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 68 points69 points (0 children)
That is a tremendous shame. I'd hope that that of all places would encourage that kind of question. If you know you're right or have a good question don't let anyone stop you.
[–] 11 points12 points (0 children)
I think he didn't understand the question and/or physics if he really said you'd see your eye.
[–] 43 points44 points (3 children)
A more entertaining thought is to consider gravity lensing. If juuuust the right combination of gravity sources were aligned, it would effectively be a 'mirror' that curved the light by 180 degrees instead of reflected it. Since it is already in place, we could possibly see something in the very deep past.
[–] 7 points8 points (0 children)
This is what I always wished we could find, the right organization of objects for 180° gravitational lensing.
[–] 4 points5 points (0 children)
This is mentioned in the movie "Paycheck". They credit the idea to Einstein. Any merit to that? Or was it just his relativity model that gave way to the possibility?
[–]Analytical Chemistry 11 points12 points (0 children)
You could use a mirror at an arbitrary distance to reflect a (very well collimated and aimed) laser signal, thus directly sending a one-way message into our own future. That would be a cool way of directly preserving historical records.
[–] 58 points59 points (28 children)
Sounds plausible, but the practicality doesn't really work. With currently available technology, it would take hundreds of years for it to get there. By the time it's in place we'll probably (hopefully) be able to travel much faster.
The fastest outward-bound spacecraft yet sent, Voyager 1, has covered 1/600th of a light-year in 30 years and is currently moving at 1/18,000th the speed of light.
[–] 49 points50 points (17 children)
I'm going to feel sad for Voyager when we develop drives that will allow us to catch up to it. All that work...
[–] 42 points43 points (8 children)
Perhaps, but the point is that we know everything it found out now. If we hadn't sent it out there, we wouldn't know until this eventual time when we do develop such technology.
Just because you can't run doesn't mean that you shouldn't walk (crawling on the other hand... I see your point).
[–] 21 points22 points (7 children)
In the grand scheme of things, Voyager is the equivalent of tossing a bottle in the ocean. lol. It's so exciting to think about, but eventually it will seem ridiculous. I can't wait for the day that it seems ridiculous.
[–] 10 points11 points (0 children)
Meaning us in the present, not the whole of humanity.
[–] 16 points17 points (5 children)
Do you feel sad for floppy disks because you have a 1TB hard disk in your desktop now ?
[–] 24 points25 points (4 children)
Yea a little bit. Ask a kid today what the "Save" icon is in Windows. They have no idea. It's like they never existed. Poor little fellas.
[–] 4 points5 points (1 child)
Don't you think it would be even more sad, if we could never retrieve it? If it was just out there, forever. I would assume that at some point, we would lose track of it, or it's signals would take so long to come back, that we would forget about it, or just not care. Things on earth would take precedent over a signal from a thousands year old ship.
I think catching up to voyager 1, and retrieving it, in a reasonable time, would be amazing. Imagine doing it in a manned craft?
Sending a new one, a much faster one, one that would pass it quickly? Doesn't sound sad to me. Voyager would still be 1.
[–] 5 points6 points (0 children)
This would not be very practical. I did some calculations for the number of photons of earth we could actually get from a mirror like this:
Number of photons leaving earth per second: roughly 5*1035 (Stefan-Boltzmann law, assuming earth surface temp to be 300K, assuming all photons to be at wavelength of 400nm).
What fraction of those photons actually hits a mirror? Let the mirror be 50m*50m and at a distance of 2 lightyears:
The fraction of photons that actually hit the mirror are: 6*10-31.
So alltogether around 300 000 photons from earth hit the mirror per second. This extremely little. Even if you get all of them back to earth (perfect mirror, perfect geometry, perfect telescope to capture them all back at earth), this is one photon per second for every couple of hundred square kilometers of earths surface. That is about one photon per second for the whole city of washington dc.
Keep in fact that this is a physical, not a technological constraint and the only way to improve it is to make a bigger mirror.
Meanwhile, satellites, cameras and hard drives should allow you to record real time footage of the whole surface of earth with acceptable resolution without any problems.
[–] 10 points11 points (5 children)
An interesting idea is that if Aliens put a mirror in space now, a light year away, in a year's time we'd see the earth a year before the mirror was erected.
[–] 9 points10 points (0 children)
Props to OP. Great question. Interesting thought: Using mirrors in space 1/2 a light day away for use as security feed for the world. Showing everything that happened in real-time 24 hrs ago. Feed goes straight in to past.Google.com :)
[–] 27 points28 points (46 children)
some of you have said that this would hardly be possible since it takes a really long time to go a light year. we could put a mirror .1 light year away, and then see .2 years into the past..... that's feasible! why haven't we done this?!
Edit: lots of replies to this... I guess I was just encouraging you all to think less about the 1-light-year distance, and more about the rationality of the idea as an awesome example that normal folks could understand.
[–] 66 points67 points (6 children)
Voyager 1 is currently the man-made object furthest from earth. It is less than 0.002 light years away. It has been moving for 34 years.
Aside from that, we don't have a telescope that could usefully show the earth at a distance of even .2 lightyears.
[–] 16 points17 points (1 child)
So if Voyager 1 was a mirror or a recording device sending data, and ignoring the fact that we can't detect a good image of earth from that distance, I believe we would be able to to see 17.5 hours back in time. Cool.
365 (days) * 0.002 (ly) = 0.73 (days) 0.73 * 24 (hours) = 17.52 hours
Right?
[–] 19 points20 points (0 children)
Double that, for the return journey.
[–] 25 points26 points (15 children)
What does this do that a camera stationed closer with a .2 year delay doesn't do?
[–] 5 points6 points (0 children)
Nothing. Well, the closer camera would take better pictures, but they would both show an image that is .2 years old.
[–] 6 points7 points (0 children)
Also, people seem to imply that, because we wouldn't be able to get it out there fast enough, the past, would be after the launch, and that it would then be useless. It is true that we probably wouldn't be able to see into the past, before the launch, but we would now be able to view the entire face of the earth (the one facing the mirror or whatever) in the past, however far in the past that might be.
The only reason this would be a reasonable thing to do though, is if we wanted to be able to do it without storage. We could just as easily place a recording device in orbit, or somewhere else that is close, and record the earth for a period of time, and just replay it when needed.
I get that the question is probably not about practical use though.
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 23 points24 points (7 children)
First off, it would have to be an absolutely gigantic mirror to be useful that far away, like solar system sized. And putting anything 0.1 LY away in a controlled manner would be one of the most expensive endeavors of human history. What would be the purpose in doing this?
[–] 6 points7 points (1 child)
What would be the purpose in doing this?
This is exactly why I decided I can't be a researcher. I would never have an answer to this question, or at least not one that would ever secure funding. "Because it's fucking awesome, that's why!"
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 2 points3 points (0 children)
[–] 13 points14 points (1 child)
What would be the purpose in doing this?
Obviously, a question a physicist would ask. Mathematicians would just do it anyway.
Note to self: Nike should sponsor mathematicians.
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 9 points10 points (0 children)
I appreciate the sentiment, but the parent was asking why we haven't in fact done it, so the practical considerations matter.
[–] 5 points6 points (9 children)
What purpose would there be to re-experiencing something we have already measured and experienced?
This would be little different from simply pointing all the camera satellites we have at earth, then simply storing the images for 2 years instead of instantly transmitting them.
This is, unsurprisingly, not very useful as compared to getting the images in real-time.
[–] 7 points8 points (2 children)
You could think of it as a data-storage mechanism -- the images are "stored" for 2 years by the traveling photons. Compared to, say, a hard disk, this would have extremely poor usability, but extremely high capacity.
[–] 4 points5 points (0 children)
I don't know if there would be any real-world phenomenon/object that would work for this, but if we had some insane leap forward in our telescope/camera technology, couldn't we theoretically make use of some kind of naturally-occurring mirror-like object that already happens to be a light year away? That way we could get around the whole "takes way too long to get something we create out that far" problem, which could give this scenario another practical edge over simply shooting a record-and-transmit device out there.
[–] 8 points9 points (0 children)
The mirror would not be seen for one year, but at that point it would show 2 years back, so if it was placed one light year away on 1/1/12, it would be visible on 1/1/13 showing earth on 1/1/11
[–] 3 points4 points (0 children)
Yes, but only once per year when the earth is in the right part of its orbit to be reflected in the mirror
[–] 6 points7 points (5 children)
What is we had a mirror that was launched into outer space at light speed moving away from us. Would the image on the mirror theoretically be a still image?
[–]Radiation Therapy | Medical Imaging | Nuclear Astrophysics 11 points12 points (3 children)
Yes, but it would also be red-shifted to infinite wavelength so we wouldn't be able to see it.
[–] 2 points3 points (0 children)
As a follow up question, what if we were lucky enough to find a "dark" patch of space that the earth had passed by in relation to a certain observable point. If we were able to get a telescope into an ideal position, would we ever be able to see light from before the telescope launched?
If the answer is no, how far from the observation do we have to be to be able to "look into the past", like the Hubble does?
[–] 2 points3 points (0 children)
No. Hubble's angular resolution is 0.1 arcseconds. At an effective distance of 2 light years, Hubble couldn't see anything much smaller than 9 million km. Since Earth is about 12700 km across, you could maybe see a single point of light. That's if you could tell it apart from the sun, which you couldn't.
If you had a large enough telescope (order of a few km across), then yes, you would see into Earth's past.
[–] 2 points3 points (4 children)
I use something similar to explain how awesome the universe is.
For the sake of this example we assume the planet Krypton is 100 light years away. Kryptonians use the power of advanced technology and science to send young Kal-El to earth at a speed equal to 100 times the speed of light (I realize that's not actually possible but I like using science fiction to illustrate examples). It takes 1 earth year to travel that distance and arrive at earth. Kal-El is raised by Jonathan & Martha Kent and becomes Superman and 19 years later constructs a super-telescope. In conjunction with his super vision Superman is able to use said telescope to view Krypton as it existed 80 before he left. since the light leaving there was only traveling at the speed of light he actually passes it and has to wait for the light to catch up to him. | 7,933 | 33,074 | {"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} | 2.96875 | 3 | CC-MAIN-2017-04 | longest | en | 0.912921 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-7-trigonometric-identities-and-equations-7-4-double-angle-and-half-angle-identities-7-4-exercises-page-692/10 | 1,537,661,731,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.65/warc/CC-MAIN-20180923000827-20180923021227-00092.warc.gz | 752,822,895 | 12,675 | ## Precalculus (6th Edition)
$\theta =-10\left( Q4\right)$ Angle is located in quadrant 4 Sine function is negative in quadrant 4 so we should choose negative sign. | 46 | 165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 2.953125 | 3 | CC-MAIN-2018-39 | longest | en | 0.891948 |
https://meeb.us/what-is-a-variable-expression-example/ | 1,628,186,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00535.warc.gz | 402,175,438 | 10,983 | # What Is A Variable Expression Example
Expressions are used to assign dynamic values expressions variables to a property. Variable expressions can be quite simple or very complex.
Word Wall Includes Definitions Examples And Qr Barcodes Words Include Dependent Variable Equation Expression Independen Word Wall Writing Systems Words
### Solved example on variable expression.
What is a variable expression example. Each person gets p points for a hit and loses q points for a miss. A variety of expressions. To evaluate a variable expression at a given value we just substitute that value in the expression and simplify it.
Using expressions for task properties. Example of a variable expression. Variable expressions can be quite simple or very complex.
In the below example the variable parentfolder holds the value of the root folder for all data files. Charles hits the mark 14 times and dennis hits the mark 12 times. Press the ellipsis button to create an expression.
Identify the variable expressions for their individual scores. Substitute x 2 in 5x 2 2x 7. Charles and dennis fired 20 shots each to hit a target.
Expression number math operator number for example 7 9 23 4 37 6 25 9 4 2. 5x y 5 467 3y 4x y while the second example is much more complex than the first and includes multiple variables and operations both examples are variable expressions. An example of a variable expression or an algebraic expression is 5x 7 evaluating a variable expression.
In this example we will assign the variable created in figure 2 to the destinationautofolderpath property. In all the given expressions a math operator is used between the two numbers. Ssis expression for variables a variable value can be derived based on the expression.
For example x is our variable in the expression. Evaluate 5x 2 2x 7 at x 2 solution. A math expression is different from a math equation.
A variable is a letter whose value is unknown to us. Another variable productfilename has been defined to have the absolute value of the file name. Although the following two examples are considerably different they re both variable expressions.
This property is the path to store the backup. For example 10 is the variable in the expression 10x 63. For example 25 7 64 2.
The coefficient is a numerical value used together with a variable. An equation will always use an equivalent operator between two math expressions. Although the following two examples are considerably different they re both variable expressions.
Please Look At The Actual Preview File The Thumbnails Do Not Give It Justice This Power Point Math Expressions Variable Expression Algebraic Expressions
The Algebraic Expressions Have The Variables And The Constants The Algebraic Expressions Are The F Algebraic Expressions Math Strategies Expression Definition
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Variable Expressions And Equations Anchor Chart Math Expressions Algebra Math Notes | 664 | 3,321 | {"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} | 3.4375 | 3 | CC-MAIN-2021-31 | latest | en | 0.821546 |
https://math.answers.com/Q/How_can_a_decimal_greater_than_1_be_a_repeating_decimal | 1,653,666,825,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00193.warc.gz | 430,971,114 | 43,120 | 0
# How can a decimal greater than 1 be a repeating decimal?
Wiki User
2017-01-10 01:41:04
A decimal number is like a mixed fraction: it has an integer part and a fractional part. If the fractional part is a repeating fraction then the whole number is represented by a repeating decimal.
Wiki User
2017-12-18 20:51:45
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Q: How can a decimal greater than 1 be a repeating decimal? | 132 | 462 | {"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} | 3.234375 | 3 | CC-MAIN-2022-21 | latest | en | 0.800379 |
https://forum.posit.co/t/needing-help-on-linear-mixed-models/165191/2 | 1,718,925,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00128.warc.gz | 219,444,673 | 6,150 | # Needing help on Linear Mixed Models
That's a thoughtful way to frame questions. Lacking [a `reprex`—see the FAQ] (FAQ: How to do a minimal reproducible example ( reprex ) for beginners), I'm only going to be offer some general thoughts.
First, though, is the movement coefficient continuous or has it been categorized into intervals?
Now, let's get philosophical and return to school algebra—f(x) = y where
x is your tabular data of `dim` 1320,7.
y is a transformation of x that abstracts away the detail to put a measure on the information contained in x
f is the function or functions (think f(g(x)) that does the transformation.
Although phrased as a question about linear mixed models, the heart of the analysis is the selection of y. What compact measure(s) best describe(s) the relationship between pupil diameter (the chosen measure for physiological engagement) and the combinations of stimuli that were presented? This question leads to
1. Is there any relationship at all worth looking at or is it just random?
2. If there is a relationship, how "close" or distant is it from randomness? The latter is the question that the misunderstood `p-value` of many statistical tests addresses. An f is applied to y to produce a test statistic, the `p-value` puts a measure on the probability that the statistic results simply from random variation. For the conventional default of 0.05, that means only a one in twenty chance. Depending on the phenomenon being gauged that may be it passes the laugh test and it's worth taking a closer look at or that's not good enough to trust the lives of millions to.
3. If the data do show an associative relationship that passes the [pre]-selected confidence interval, do the data permit casual inference? Can the relationships among the multiple stimuli be teased apart to test for one while keeping the others constant? Are there mediators? Colliders? Stimuli that only have an effect indirectly? Both directly and indirectly? This is the domain of causal inference and in former times was heretical. Today we have tools, such as directed acyclic graphs, that make it possible when applied carefully.
The first question is the most general and easiest to overlook in the presence of eagerness to get to a conclusion. The tools of exploratory data analysis are designed to hold the drive to selecting y, the gauge of the outcome in abeyance to see what the data are capable of revealing.
You may have already done this and we should be looking at what metrics are available to address the case of a continuous outcome variable Y in the presence of X_1 \dots X_6 where the treatment variables are categorical. If you have, I'd encourage writing the EDA phase up in semi-formal fashion. From experience, I know that it's easy to let the spark of an idea die if left to fend for its own among a mass of mixed notes, files, scripts and what not.
The motivation for my question about whether motion coefficients were continuous is that looking at the simple case of two continuous variables is simple and potentially informative. Linear regression/ANOVA is unreasonably effective in assessing the threshold question is there any there there? If not, move on.
Are any of the variable ordinal in their categorization, such as movements? If these were measured as giraffe, watermelon and quartz there is no ordering, but static, micro and macro suggest a ranking. That may bear on the choice of model.
Also, a nonlinear mixed effects model may be more appropriate. Depending.
I don't have any experience going through this process and making these decisions. I can only just follow the `{lme4} vignette`. A long career chasing false hopes has left me cautious about setting sail before fully understanding the seas and winds to be expected. | 796 | 3,789 | {"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} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.93731 |
http://www.gurufocus.com/term/DaysInventory/BVSN/Days%2BInventory/BroadVision%252C%2BInc | 1,406,690,565,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510268533.31/warc/CC-MAIN-20140728011748-00114-ip-10-146-231-18.ec2.internal.warc.gz | 594,087,532 | 35,605 | Switch to:
Days Inventory
0.00 (As of Mar. 2014)
BroadVision Inc's inventory for the three months ended in Mar. 2014 was \$0.00 Mil. BroadVision Inc's cost of goods sold for the three months ended in Mar. 2014 was \$1.03 Mil. Hence, BroadVision Inc's days inventory for the three months ended in Mar. 2014 was 0.00.
BroadVision Inc's days inventory stayed the same from Mar. 2013 (0.00) to Mar. 2014 (0.00).
Inventory can be measured by Days Sales of Inventory (DSI). BroadVision Inc's days sales of inventory (DSI) for the three months ended in Mar. 2014 was 0.00.
Inventory turnover measures how fast the company turns over its inventory within a year.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. BroadVision Inc's inventory to revenue ratio for the three months ended in Mar. 2014 was 0.00.
Definition
Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
BroadVision Inc's Days Inventory for the fiscal year that ended in Dec. 2013 is calculated as
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 0 / 4.841 * 365 = 0.00
BroadVision Inc's Days Inventory for the quarter that ended in Mar. 2014 is calculated as:
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 0 / 1.034 * 91 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
An increase of Days Inventory may indicate the company's sales slowed.
1. Inventory can be measured by Days Sales of Inventory (DSI).
BroadVision Inc's Days Sales of Inventory for the three months ended in Mar. 2014 is calculated as
Days Sales of Inventory (DSI) = Inventory / Revenue * Days in Period = 0 / 2.994 * 91 = 0.00
2. Inventory Turnover measures how fast the company turns over its inventory within a year.
BroadVision Inc's Inventory Turnover for the three months ended in Mar. 2014 is calculated as
Inventory Turnover = Cost of Goods Sold / Average Inventory = 1.034 / 0 =
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
BroadVision Inc's Inventory to Revenue for the three months ended in Mar. 2014 is calculated as
Inventory to Revenue = Inventory / Revenue = 0 / 2.994 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
A lot of business are seasonable. It makes more sense to compare Days Inventory from the same period in the previous year instead of from the previous quarter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency. | 701 | 2,957 | {"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} | 2.9375 | 3 | CC-MAIN-2014-23 | longest | en | 0.942499 |
http://www.algebra.com/algebra/homework/Systems-of-equations/Systems-of-equations.faq.question.253524.html | 1,371,672,641,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709101476/warc/CC-MAIN-20130516125821-00005-ip-10-60-113-184.ec2.internal.warc.gz | 296,011,475 | 4,599 | # SOLUTION: Solve each system by the addition method. Determine whether the equations are independent, dependent, or inconsistent. x-y=3 -6x+6y=17
Algebra -> Algebra -> Systems-of-equations -> SOLUTION: Solve each system by the addition method. Determine whether the equations are independent, dependent, or inconsistent. x-y=3 -6x+6y=17 Log On
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Algebra: Systems of equations that are not linear Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Systems-of-equations Question 253524: Solve each system by the addition method. Determine whether the equations are independent, dependent, or inconsistent. x-y=3 -6x+6y=17Answer by Fombitz(13828) (Show Source): You can put this solution on YOUR website!Multiply the first eq. by 6 and add to the second equation. 6x-6y-6x+6y=18+17 0=35 The result is inconsistent. There is no solution, the lines are parallel. | 267 | 1,082 | {"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} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.812789 |
https://www.edupil.com/question/sum-digits-of-number/ | 1,529,342,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00452.warc.gz | 788,642,448 | 14,108 | # What will be the Sum of the Digits of the Number? If
A two digit number is 3 times the sum of its digits. If 45 is added to the number, its digits are interchanged. The sum of the digits of the number is:
1. 9
2. 8
3. 5
4. None of these
Sarif Default Asked on 31st July 2015 in
Explanation:-
Let the number of two digits be = 10 X + Y
According to the question,
10 X + Y = 3 (X + Y)
10 X + Y = 3 X + 3 Y
10 X – 3 X = 3 Y – Y
7 X = 2 Y
7 X – 2 Y = 0 ……………………………………… (1)
According to second condition,
10 X + Y + 45 = 10 Y + X
10 X – X + Y – 10 Y + 45 = 0
9 X – 9 Y = -45 …………………………………… (2)
Solve equation (1) & (2),
X = 2 & Y = 7
Then, the number is 27
Hence, the sum of digits 2 + 7 = 9.
Anurag Mishra Professor Answered on 31st July 2015. | 301 | 773 | {"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} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.855799 |
https://de.mathworks.com/matlabcentral/answers/2132826-change-coordinate-system-from-epsg-3003-to-epsg-32632-i-e-wgs-84-utm-zone-32n?s_tid=prof_contriblnk | 1,726,530,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00511.warc.gz | 174,935,055 | 27,044 | # Change coordinate system, from EPSG 3003 to EPSG 32632 (i.e. WGS 84 / UTM zone 32N)
24 Ansichten (letzte 30 Tage)
Sim am 28 Jun. 2024
I have the following point with x-y-coordinates in the EPSG 3003 coordinate system:
a = [1.7202e+06 4.8998e+06]
How can change the above mentioned x-y-coordinates from EPSG 3003 to EPSG:32632 (i.e. WGS 84 / UTM zone 32N)?
##### 0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
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### Akzeptierte Antwort
Manikanta Aditya am 30 Jun. 2024
Hi,
In MATLAB, you can use the 'projcrs' and 'projfwd' functions to transform coordinates between different coordinate reference systems (CRS).
• 'projinv' converts the coordinates from EPSG 3003 to geographic coordinates (latitude and longitude).
• 'projfwd' converts these geographic coordinates to the target CRS (EPSG 32632).
% Define the point in EPSG 3003
x_epsg3003 = 1.7202e+06;
y_epsg3003 = 4.8998e+06;
% Define the source and target CRS
sourceCRS = projcrs(3003);
targetCRS = projcrs(32632);
% Transform the coordinates from EPSG 3003 to EPSG 32632
[lat, lon] = projinv(sourceCRS, x_epsg3003, y_epsg3003); % Convert to geographic coordinates (latitude and longitude)
[x_epsg32632, y_epsg32632] = projfwd(targetCRS, lat, lon); % Convert from geographic coordinates to target CRS
% Display the results
fprintf('Coordinates in EPSG:32632: (%f, %f)\n', x_epsg32632, y_epsg32632);
Coordinates in EPSG:32632: (720189.814410, 4899709.877584)
Refer to the following documentation to know more about:
Hope this helps!
##### 2 KommentareKeine anzeigenKeine ausblenden
Sim am 30 Jun. 2024
Thanks!
Manikanta Aditya am 1 Jul. 2024
Welcome @Sim!
Melden Sie sich an, um zu kommentieren.
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Translated by | 589 | 1,961 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.329643 |
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#### Resources tagged with Practical Activity similar to Little Man:
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### There are 154 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
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The Man is much smaller than us. Can you use the picture of him next to a mug to estimate his height and how much tea he drinks?
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##### Age 7 to 11 Challenge Level:
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### Counter Ideas
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These practical challenges are all about making a 'tray' and covering it with paper.
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##### Age 5 to 7 Challenge Level:
Explore the triangles that can be made with seven sticks of the same length.
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Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
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##### Age 7 to 11 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
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##### Age 5 to 7 Challenge Level:
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##### Age 7 to 11 Challenge Level:
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##### Age 5 to 7 Challenge Level:
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##### Age 5 to 11 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
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##### Age 5 to 7 Challenge Level:
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##### Age 5 to 7 Challenge Level:
Make a chair and table out of interlocking cubes, making sure that the chair fits under the table!
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##### Age 5 to 7 Challenge Level:
Have a go at making a few of these shapes from paper in different sizes. What patterns can you create?
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##### Age 5 to 7 Challenge Level:
Can you put these shapes in order of size? Start with the smallest.
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##### Age 7 to 11 Challenge Level:
Make a flower design using the same shape made out of different sizes of paper.
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##### Age 7 to 11 Challenge Level:
Can you work out what shape is made by folding in this way? Why not create some patterns using this shape but in different sizes?
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##### Age 7 to 11 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
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We can cut a small triangle off the corner of a square and then fit the two pieces together. Can you work out how these shapes are made from the two pieces?
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##### Age 7 to 11 Challenge Level:
Ideas for practical ways of representing data such as Venn and Carroll diagrams.
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##### Age 5 to 7 Challenge Level:
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##### Age 5 to 7 Challenge Level:
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##### Age 5 to 7 Challenge Level:
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##### Age 7 to 11 Challenge Level:
Watch the video to see how to fold a square of paper to create a flower. What fraction of the piece of paper is the small triangle?
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##### Age 5 to 7 Challenge Level:
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##### Age 7 to 11 Challenge Level:
This practical activity involves measuring length/distance.
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##### Age 5 to 7 Challenge Level:
In this activity focusing on capacity, you will need a collection of different jars and bottles.
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##### Age 5 to 7 Challenge Level:
For this activity which explores capacity, you will need to collect some bottles and jars.
### Compare the Cups
##### Age 5 to 7 Challenge Level:
You'll need a collection of cups for this activity. | 1,956 | 8,568 | {"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} | 4.21875 | 4 | CC-MAIN-2019-35 | latest | en | 0.898696 |
https://www.chase2learn.com/nptel-data-science-for-engineers-assignment-4-answers/ | 1,695,381,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00029.warc.gz | 788,601,943 | 35,400 | # NPTEL Data Science for Engineers Assignment 4 Answers 2023
Hello NPTEL Learners, In this article, you will find NPTEL Data Science for Engineers Assignment 4 Week 4 Answers 2023. All the Answers are provided below to help the students as a reference don’t straight away look for the solutions, first try to solve the questions by yourself. If you find any difficulty, then look for the solutions.
###### NPTEL Data Science for Engineers Assignment 5 Answers 2023 Join Group👇
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## NPTEL Data Science for Engineers Assignment 4 Answers 2023:
#### Q.1. Let f(x)=x3+6×2−3x−5 . Select the correct options from the following:
• −2+5–√ will give the maximum for f(x) .
• −2+5–√ will give the minimum for f(x) .
• The stationary points for f(x) are −2+5–√ and −2−5–√.
• The stationary points for f(x) are −4 and 0.
Use the following information to answer Q2 and Q3.
Consider the following optimization problem:
maxxϵRf(x)max
, where
f(x)=x4+7×3+5×2−17x+3
Let x∗ be the maximizer of f(x)
#### Q.2. What is the second order sufficient condition for x∗ to be the maximizer of the function f(x) ?
• 4×3+21×2+10x−17=0
• 12×2+42x+10=0
• 12×2+42x+10>0
• 12×2+42x+10<0
• -4.48
• 0.66
• -1.43
• 4.45
#### Q.4. Let f(x)=2sinx,0≤x≤2π . Select the correct options from the following:
• π/2 is the global maximum of f(x).
• π is the global minimum of f(x).
• 3π/2 is the global maximum of f(x).
• 3π/2 is the global minimum of f(x).
Use the following information to answer Q5, Q6, Q7 and Q8.
Let f(x)=2×21+3x1x2+3×22+x1+3×2.
• 0.6, 0.4
• −0.6, −0.4
• 0.2, −0.6
• 0.2, 0.6
• maxima
• minima
#### Q.9. Let f(x1,x2)=4×21−4x1x2+2×22 . Select the correct options from the following:
• (2, 4) is a stationary point of f(x).
• (0, 0) is a stationary point of f(x).
• The Hessian matrix ▽2f is positive definite.
• The Hessian matrix ▽2f is not positive definite.
#### Q.10. In optimization problem, the function that we want to optimize is called
• Decision function
• Constraints function
• Optimal function
• Objective function
• True
• False
#### Q.12. In the gradient descent algorithm, the step size should always be same for each iteration.
• True
• False
##### NPTEL Data Science for Engineers Assignment 4 Answers Join Group👇
Disclaimer: This answer is provided by us only for discussion purpose if any answer will be getting wrong don’t blame us. If any doubt or suggestions regarding any question kindly comment. The solution is provided by Chase2learn. This tutorial is only for Discussion and Learning purpose.
#### About NPTEL Data Science for Engineers Course:
Learning Objectives :
1. Introduce R as a programming language
2. Introduce the mathematical foundations required for data science
3. Introduce the first level data science algorithms
4. Introduce a data analytics problem solving framework
5. Introduce a practical capstone case study
Learning Outcomes:
1. Describe a flow process for data science problems (Remembering)
2. Classify data science problems into standard typology (Comprehension)
3. Develop R codes for data science solutions (Application)
4. Correlate results to the solution approach followed (Analysis)
5. Assess the solution approach (Evaluation)
6. Construct use cases to validate approach and identify modifications required (Creating)
##### Course Outcome:
• Week 1: Course philosophy and introduction to R
• Week 2: Linear algebra for data science
• 1. Algebraic view – vectors, matrices, product of matrix & vector, rank, null space, solution of over-determined set of equations and pseudo-inverse)
• 2. Geometric view – vectors, distance, projections, eigenvalue decomposition
• Week 3: Statistics (descriptive statistics, notion of probability, distributions, mean, variance, covariance, covariance matrix, understanding univariate and multivariate normal distributions, introduction to hypothesis testing, confidence interval for estimates)
• Week 4: Optimization
• Week 5: 1. Optimization
• 2. Typology of data science problems and a solution framework
• Week 6: 1. Simple linear regression and verifying assumptions used in linear regression
• 2. Multivariate linear regression, model assessment, assessing importance of different variables, subset selection
• Week 7: Classification using logistic regression
• Week 8: Classification using kNN and k-means clustering
###### CRITERIA TO GET A CERTIFICATE:
Average assignment score = 25% of average of best 8 assignments out of the total 12 assignments given in the course.
Exam score = 75% of the proctored certification exam score out of 100
Final score = Average assignment score + Exam score
YOU WILL BE ELIGIBLE FOR A CERTIFICATE ONLY IF AVERAGE ASSIGNMENT SCORE >=10/25 AND EXAM SCORE >= 30/75. If one of the 2 criteria is not met, you will not get the certificate even if the Final score >= 40/100.
If you have not registered for exam kindly register Through https://examform.nptel.ac.in/ | 1,354 | 5,043 | {"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} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.757722 |
http://www.thulasidas.com/how-to-be-successful-in-life/?lang=ur | 1,591,054,373,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00388.warc.gz | 216,429,233 | 18,521 | # How to be Successful in Life?
### How to be Successful in Life
When I talked about the dimensions of success, I used the word dimension with an ulterior motive. I want to define success for you in a formal way. You see, an entity that has many dimensions is a space, similar to the three dimensional space we live in. When we have such a complex multi-dimensional space to define success in, we have to apply some good techniques from physics to do it right. Don’t worry, i am here to help.
Success is hard to define, but the lack of success seems obvious — no money, no family, no friends, no education, no wisdom, no health, no wealth etc. That situation is one dark point in this multi-dimensional success-space. Your station in life is another point in the success-space. How far away from the dark point your station is is truly the measure of your success in life. The distance from this zero point of failure is the so-called Cartesian distance. If you have special likes or dislikes for one particular dimension of success (like money, for instance), you can assign an appropriate weight to that dimension, which effectively makes the distance what they call a $\chi^2$ distance. Of course, it would be impossible to assign precise numerical values to all these abstract dimensions and distances. But this mode of thinking should give you a tool to analyze and understand successes and failures. It will tell you, for instance, why Bangladeshis score higher in happiness index than Americans. They just happen to have a different set of dimensions that they consider important.
The trick in achieving success in life lies in identifying the dimensions that are important to you personally. Don’t let yourself get influenced by others (unless, of course, pleasing others is one of your preferred dimensions). Once your own personal directions are identified, channel all your efforts along those dimensions. Just be sure that your dimensions are right for you both in terms of your deepest desires and your abilities. Choose wisely! | 406 | 2,045 | {"found_math": true, "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": 1, "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} | 2.578125 | 3 | CC-MAIN-2020-24 | longest | en | 0.937747 |
https://books.google.com.mx/books?id=1do6yAXEeGMC&qtid=cc7e7983&source=gbs_quotes_r&cad=6 | 1,566,234,943,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00296.warc.gz | 407,665,657 | 5,959 | Libros Libros 1 - 10 de 125 sobre Divide the difference of the extremes by the number of terms, less 1, and the quotient...
Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference.
The National Arithmetic, on the Inductive System: Combining the Analytic and ... - Página 245
por Benjamin Greenleaf - 1849 - 360 páginas
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## Mathematics: Compiled from the Best Authors and Intended to be the ..., Volumen1
Samuel Webber - 1801
...yards. PROBLEM II. lie first term, iJce last term, and the number of terms being given,, to find ike common' difference. RULE.* Divide the difference of the extremes by the number of terms less i, and the quotient will be the common difference sought. EXAMPLES. i. The extremes are 2 and 53, and...
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## Arithmetical magazine: or, Mercantile accountant, adapted to the commerce of ...
James Alexander Mackay - 1803 - 252 páginas
...properties of Progression. PROBLEM IV. Given the first term, the last term, and number of terms, to find the common difference. RULE.— Divide the difference of the extremes by the number of •terms less I— the quotient will be the common difference. EXAMPLE. jGhren 3, the first, 39 the last, and 19...
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## A New and Complete System of Arithmetick: Composed for the Use of the ...
Nicolas Pike - 1808 - 480 páginas
...miles, Ans i . PROBLEM I. Tbejlrtt term, the last term, and the number of terms being given, tojind the common difference. RULE.* Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference sought. EXAMPLES. 1st. The extremes are 3 and 39,...
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## Mathematics, Volumen1
Samuel Webber - 1808
...term, the .last term, and the number of terms being given, to jind the common difference. RULE 3.* Divide the differe'nce of the extremes by the number of terms less 1, and the quotient will be the common difference sought, * The difference of the first and last terms...
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## The American Tutor's Assistant Revised: Or, A Compendious System of ...
Zachariah Jess - 1810 - 210 páginas
...two extremes and number of terms are given, and the common difference of all the terms required ; . Divide the difference of the extremes by the number of terms, less one,, the quotient will be the common difference. E \X AMPLES. l Admit, a debt be discharged at 16 several...
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## Adams's New Arithmetic: Arithmetic, in which the Principles of Operating by ...
Daniel Adams - 1810 - 180 páginas
...difference. Hence, when the extremes and number of terms are given, tn fnd the common difference,—' Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference. 6. If the extremes be 3 and 603, and the number...
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## The Teachers Assistant. Or, A System of Practical Arithmetic: Wherein the ...
1811 - 198 páginas
...CASE 2. When, the two extremes and number of ternis are given, to find the common difference. RULÉ. . Divide 'the difference of the extremes by the number of terms, less one ; the quotient will be the .common difference. EkAMPLES. , j 1. 20 and 60 are the two extremes of a...
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## A System of Arithmetic
Samuel Webber - 1812 - 248 páginas
...Thefrst term, the last term, and the number of terms being given, to fnd the common difference. RULE. 3.* Divide the difference of the extremes by the number of terms less 1,. and the quotient will be the common difference sought. EXAMPLES. 1. The extremes ave 2 and 53,...
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## The American Tutor's Assistant Revised, Or, A Compendious System of ...
Zachariah Jess - 1813 - 210 páginas
...the two extremes and number of terms are given, and the common difference of all the terms required ; RULE. Divide the difference of 'the extremes by the number of terms, less one, the quotient will be the common difference. EXAMPLE s;v"1 Admit a debt be discharged at 16 several...
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## Daboll's Schoolmaster's Assistant: Improved and Enlarged : Being a Plain and ...
Nathan Daboll - 1813 - 240 páginas
...furlongs, 180 yds. PROBLEM 11. The first term, the last term, and the number of terms given, to find the common difference. RULE. — Divide the difference of the extremes by the number 'f terms leas I... and the quotient will be the common difference. EXAMPLES. 1. The extremes are 3...
Vista completa - Acerca de este libro | 1,201 | 4,694 | {"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} | 2.9375 | 3 | CC-MAIN-2019-35 | latest | en | 0.749479 |
http://qs1969.pair.com/~perl2/?node_id=3333;parent=482077 | 1,675,766,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00262.warc.gz | 35,419,817 | 7,284 | We don't bite newbies here... much PerlMonks
### comment on
Need Help??
My Haskell implementation represents numbers as the ratio of products of ordered integer streams. For example, I represent 3!/(4*5) as (R numerator=[1,2,3] denominator=[4,5]). In this representation, multiplication becomes merging the numerator and denominator streams and then canceling the first stream by the second. In this way I can remove all cancelable original terms in the Pcutoff formula before finally multiplying the terms that remain.
```*FishersExactTest> fac 6
R {numer = [2,3,4,5,6], denom = []}
*FishersExactTest> fac 3
R {numer = [2,3], denom = []}
*FishersExactTest> fac 6 `rdivide` fac 3
R {numer = [4,5,6], denom = []}
Here's the example from the MathWorld page:
```*FishersExactTest> rpCutoff [ [5,0],
[1,4] ]
R {numer = [2,3,4,5], denom = [7,8,9,10]}
*FishersExactTest> fromRational . toRatio \$ it
2.3809523809523808e-2
The code:
```module FishersExactTest (pCutoff) where
import Data.Ratio
import Data.List (transpose)
pCutoff = toRatio . rpCutoff
rpCutoff rows =
facproduct (rs ++ cs) `rdivide` facproduct (n:xs)
where
rs = map sum rows
cs = map sum (transpose rows)
n = sum rs
xs = concat rows -- cells
facproduct = rproduct . map fac
fac n | n < 2 = runit
| otherwise = R [2..n] []
-- I represent numbers as ratios of products of integer streams
-- R [1,2,3] [4,5] === (1 * 2 * 3) / (4 * 5)
data Rops = R { numer :: [Int], denom :: [Int] } deriving Show
runit = R [] [] -- the number 1
toRatio (R ns ds) = bigProduct ns % bigProduct ds
bigProduct = product . map toInteger
-- multiplication is merging numerator and denominator streams
-- and then canceling the first by the second
rtimes (R xns xds) (R yns yds) =
uncurry R \$ (merge xns yns) `cancel` (merge xds yds)
rproduct = foldr rtimes runit
-- division is multiplication by the inverse
rdivide x (R yns yds) = rtimes x (R yds yns)
-- helpers
merge (x:xs) (y:ys)
| x < y = x : merge xs (y:ys)
| otherwise = y : merge (x:xs) ys
merge [] ys = ys
merge xs [] = xs
cancel (x:xs) (y:ys)
| x == y = cancel xs ys
| x < y = let (xs', ys') = cancel xs (y:ys) in (x:xs', ys')
| otherwise = let (xs', ys') = cancel (x:xs) ys in (xs', y:ys')
cancel xs ys = (xs, ys)
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http://gsp.com/cgi-bin/man.cgi?section=other&topic=XMANDEL | 1,553,109,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.86/warc/CC-MAIN-20190320190324-20190320212324-00086.warc.gz | 89,420,184 | 4,995 | Quick Navigator
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XMANDEL(N) XMANDEL(N)
# NAME
xmandel - window based interface to Mandelbrot sets and Julia sets
# SYNOPSIS
xmandel [-display display]
# DESCRIPTION
Xmandel is a user friendly interface for generating Mandelbrot sets and Julia sets. It initially comes up with with several command buttons, which are described below, for controlling the execution. A Mandelbrot set is drawn in the window of the initial form when the mandel button is selected. A separate window is created for drawing the Julia sets.
# THEORY
Let z0 be a number in the complex plane (x + yi). Choose a complex constant C. Calculate z1 = z0 ** 2 + C. Repeat this recursively, so that z2 = z1 ** 2 + C, z3 = z2 ** 2 + C and so on. z[n] will either tend to infinity or zero, depending on its initial value and the constant C. Specifically if the absolute value of z[n], expressed as |z| = sqrt(x**2 + y**2) is greater than 2, then the recursive formula will diverge.
So, to calculate a Julia set, take each point near (0,0i), and use the formula z = z**2 + C recursively. The Julia set is the set of points for which z = z**2 + C would iterate indefinitely for the constant C. Pixels, which represent numbers in the complex plane, are set to the number of iterations before |z| exceeds 2. This then becomes an index into the hardware colormap. Each color then represents the number of iterations before divergence is detected.
To calculate a Mandelbrot set, again take each point near (0,0i), use the same formula z = z**2 + C recursively. This time let C be the initial value of the point itself (C = z0). Rather than having the same C for every point in the complex plane as in Julia set calculations, C is different for each point in the plane. Again let the pixel value be the number of iterations before |z| exceeds 2.
On monochrome displays, the pixel value is set to 1 if the iteration count is 64, otherwise 0.
Mandelbrot sets and Julia sets are obviously closely related as can be seen from the similarity of their respective formulas. If the constant C is chosen from the interior of the Mandelbrot set, then the Julia set calculated from that constant C will be connected, that is have no gaps or discontinuities. If the constant C is chosen from outside the Mandelbrot set, the Julia set will be disconnected, more like grains of dust (Fatou clouds). If the constant C is chosen from the border of the Mandelbrot set, then the Julia set will be more convoluted. Given this relationship between points in the Mandelbrot set and the Julia set generated, Xmandel provides user selection of the constant C by mouse selection in the Mandelbrot window.
# BUTTONS
To control execution of the calculations, various buttons are provided. The buttons are:
mandel -
Calculates a Mandelbrot set from (-2.0, -1.5) to (1.0, 1.5) and display it in the window provided.
mandelzoom -
In order to zoom in on a given area in the Mandelbrot set, a zoom button is provided. The area to be zoomed in on is selected with the left mouse button. Left button down begins the selection, dragging with left button down draws a rubber banded box to show the zoom area, and left button up begins the calculation. You can zoom in on a zoomed in area until you reach the limits of the precision of your hardware. Selecting a zoom area that crosses a window border doesn't work.
unzoom -
Return to previous zoom. Note that you can zoom all the way out by selecting the mandel button.
redo -
Because the Mandelbrot calculations are CPU intensive, xmandel does not restart the calculation automatically on receipt of an exposure event. This is left up to user control. The redo button will simply recalculate the current zoom level and display it in the Mandelbrot window. This is also useful for seeing new detail when the iteration count is increased.
julia -
Calculates a Julia set. The user is required to select a point inside the Mandelbrot window using the left mouse button as the constant C for the Julia set calculation. It will open a new window if needed. The Julia set is centered around (0,0), going from (-1.5, -1.5) to (1.5, 1.5). Julia set points can be selected from zoomed in Mandelbrot windows as well. Beware of selecting points outside the Mandelbrot window.
clear -
Clears the Mandelbrot window.
quit -
Exit the xmandel program.
increate iterations -
On color displays, the iteration count (sometimes called dwell) is initially set to 256, on monochrome, 64. The increate iteration button will increase the interation count
by 256 on color or 64 on monochrome. This is useful for seeing more detail when zoomed in.
reset iterations -
Will reset the iteration count to its default value of 256 or 64.
# LABELS
hostname -
The name of the host is displayed in the topmost pane. This is handy when comparing the performance of multiple copies of xmandel.
iteration count -
The current iteration count is displayed in the second pane.
current view -
The region of the Mandelbrot being displayed is given in the bottommost pane, as a range of x and y values in real coordinates.
Julia set constant -
Julia sets are displayed in a separate window, and the value of the constant used for the Julia set calculation is given to the window manager to be displayed in the title bar.
# BUGS
Xmandel uses hard coded values for button colors, assuming a 256 color colormap.
Xmandel deliberately does not handle exposure events.
Selecting a zoom area that crosses a window border doesn't work.
Performance is slow on workstations, especially workstations without floating point hardware.
# AUTHOR
John L. Freeman
jlf@cray.com
07 March 1989 4th Berkeley Distribution
Search for or go to Top of page | Section other | Main Index
Visit the GSP FreeBSD Man Page Interface.
Output converted with ManDoc. | 1,381 | 5,850 | {"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} | 3.3125 | 3 | CC-MAIN-2019-13 | longest | en | 0.888978 |
https://medium.com/@guacamole87/hello-chris-aeeeec72d0d9 | 1,524,320,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00412.warc.gz | 649,226,080 | 22,537 | Why is increasing strength relative to muscle size ideal for athletes?
Chris Beardsley
1592
Hello Chris,
ha, I thought I found an error in your analysis — but turns out you were right: Muscle mass increases and resulting strength potential often cant compensate for/catch up with the total body weight increase.
I tried simulating this by using Greg Nuckol´s calculators of his article you probably know: https://www.strongerbyscience.com/your-drug-free-muscle-and-strength-potential-part-2/
[Caution: In another arcticle https://www.strongerbyscience.com/which-weight-class-is-best-for-you/ , Greg uses allometric scaling to show that body mass increases while staying at the same BF % increases relative strength! Why? Because allometric scaling is not the same as strength/BW and is more “lenient” to mass increases. I believe its the better metric to compare strength accomplishments interindividually (such as in PL/WL competitions, the scenario Greg aimed at). But of course simple strength/BW is better for predicting sporting performance in some sports where your own body weight is the only resistance.]
A realistic example would be increasing muscle mass from a body weight of 70kg @ 12%BF to 77kg @12%BF which is an FFM increase by 5kg and a body mass increase by 7kg or 10%.
However, the corresponding increase of the max predicted “Total” (squat + bench + deadlift as one marker for max strength, with rather low current values filled in: 100kg/70kg/120kg) is from 573 to 613kg, only a 7% increase.
I´d like to add though, that if there is any external additional (heavy) resistance involved, for example moving/defending against opponents or objects, then max strength becomes more important and can override the slight worsening in relative strength.
So again, the decisions are highly context dependend which I think is the very point your are making in your excellent “strength is specific” articles.
Regards,
Chris
One clap, two clap, three clap, forty?
By clapping more or less, you can signal to us which stories really stand out. | 471 | 2,062 | {"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} | 2.734375 | 3 | CC-MAIN-2018-17 | latest | en | 0.925339 |
http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.370772.html | 1,369,276,326,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702749808/warc/CC-MAIN-20130516111229-00038-ip-10-60-113-184.ec2.internal.warc.gz | 318,319,125 | 4,639 | # SOLUTION: I'm not sure if I have the correct type of question, but I need to know how to solve this problem. It is making me crazy! 5/2x+6 + 46/7x = 2 Please help!
Algebra -> Algebra -> Polynomials-and-rational-expressions -> SOLUTION: I'm not sure if I have the correct type of question, but I need to know how to solve this problem. It is making me crazy! 5/2x+6 + 46/7x = 2 Please help! Log On
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Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Polynomials-and-rational-expressions Question 370772: I'm not sure if I have the correct type of question, but I need to know how to solve this problem. It is making me crazy! 5/2x+6 + 46/7x = 2 Please help!Answer by stanbon(57342) (Show Source): You can put this solution on YOUR website!5/(2x+6) + 46/(7x) = 2 ----- Multiply thru by (7x)(2x+6) to get: ---------------------------- 5(7x) + 46(2x+6) = 2(7x)(2x+6) ---- 35x + 92x + 46*6 = 28x^2+84x ----------------- 28x^2-43x-276 = 0 --- Solve using the Quadratic Formula ---------------- Cheers, Stan H. | 384 | 1,287 | {"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} | 3.859375 | 4 | CC-MAIN-2013-20 | latest | en | 0.841755 |
http://sadi.me/cpm-homework-help-cc3-chapter-7-73/ | 1,571,246,038,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00042.warc.gz | 174,658,206 | 9,107 | # CPM HOMEWORK HELP CC3 CHAPTER 7
As the lesson progresses, problems tend to become harder. Homework help teaches strategies to make students learn and retain mathematics. It is the responsibility of parents to make students aware of these responsibilities. Where, you have to draw the inference from the diagram. They want all students to know that once they master the subject, what avenues open up in front of them. They want to empower everyone with mathematics.
Homework help teaches strategies to make students learn and retain mathematics. Terms and conditions need to be accepted. It is preceded by more graphs. Then you are asked about your hobbies, interests, working style in a team, your birthday and about things that make you nervous. E- Tools are given, which will help you to solve the problem.
Likewise, you are required to draw figures that represent 25, 50 and percentages.
It is preceded by more graphs. It will never fail. All the students should help each other and learn together. The first chapter of CC3 consists of a subject called Mathography.
## Categories
Let me give you an example. Next comes the questions of on area and perimeter.
BACHELOR THESIS TECHNISCHE BEDRIJFSKUNDE
That is, it is not for sale. The course is guided by National Council of Teachers of Mathematics. They want to propagate it as the language which speaks of science and life, which we are all about.
Download WordPress Themes Free. You get to see the diamond problems over here. CPM answers all your queries.
Where, you have to draw the inference from the diagram. A problem will be chapfer, mentioning the measurements of a flat or residence. Terms and conditions need to be accepted. They should not give up from the very beginning. Next you have to deal with fractions.
# Welcome to CPM Homework Help
Let me tell you in detail. The company wants everyone to be friendly with mathematics.
CPM homework help states that students have few responsibilities to perform, before they ask someone for help. The next few problems are on the same area and perimeter. The program is semester based. CPM homework help shows the way. It is the responsibility of parents to make students aware of these responsibilities.
DISSERTATION SUR LA GUERRE FROIDE 1ERE S
In the preceding chapter, you will be acquainted with numbers, words and geometric representations. You will be required to find out the areas of the rooms in the place. This is very important.
## Javascript and Cookies MUST be enabled for this site to function properly.
Certain details need to be filled in, to access the site. As the lesson progresses, problems tend to become harder.
Free Download WordPress Themes. Premium WordPress Themes Download. Here the teacher gets to know you better. The key to acing mathematical data is communication and practice.
CPM homework answers have various modules in the name of Homewirk Connexions. You do not the permission to pass on the data to any third party. | 614 | 2,979 | {"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} | 3.3125 | 3 | CC-MAIN-2019-43 | latest | en | 0.964775 |
https://www.arxiv-vanity.com/papers/1603.05332/ | 1,610,756,297,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00305.warc.gz | 686,980,614 | 55,956 | # Regularization based on all-at-once formulations for inverse problems††thanks: This work was supported by the Austrian Science Fund FWF under grant I2271.
Barbara Kaltenbacher Alpen-Adria-Universität Klagenfurt, Austria (, http://wwwu.uni-klu.ac.at/bkaltenb/).
###### Abstract
Parameter identification problems typically consist of a model equation, e.g. a (system of) ordinary or partial differential equation(s), and the observation equation. In the conventional reduced setting, the model equation is eliminated via the parameter-to-state map. Alternatively, one might consider both sets of equations (model and observations) as one large system, to which some regularization method is applied. The choice of the formulation (reduced or all-at-once) can make a large difference computationally, depending on which regularization method is used: Whereas almost the same optimality system arises for the reduced and the all-at-once Tikhonov method, the situation is different for iterative methods, especially in the context of nonlinear models. In this paper we will exemplarily provide some convergence results for all-at-once versions of variational, Newton type and gradient based regularization methods. Moreover we will compare the implementation requirements for the respective all-at-one and reduced versions and provide some numerical comparison.
i
Regularization based on all-at-once formulationsB. Kaltenbacher
nverse problems, regularization, all-at-once formulations
{AMS}
65M32, 65J22, 35R30
## 1 Introduction
In their original formulation, inverse problems often consist of a model and additional observations. Consider, e.g., an equation (PDE, ODE, integral equation) model for the state
A(x,u)=0 (1)
containing a parameter (or a set of parameters) that is to be determined from additional observations of the state
C(u)=y (2)
Here and are operators acting between Banach spaces , , , (the star indicates that in variational formulations of models will typically be the dual of some Banach space). The setting could be extended in several directions, e.g., the observation can as well depend on some unknown parameters that have to be identified or the model could consist of a variational inequality instead of an equation. Still (1), (2) is sufficiently general to comprise a wide range of applications, e.g., the following examples.
### 1.1 Examples
###### Example 1
Consider a boundary value problem for a linear elliptic PDE on a smooth bounded domain ,
−∇(a∇u)+cu=b in Ω,∂u∂n=g on ∂Ω
with a given boundary excitation , and possibly spatially varying coefficients and the inverse problem of identifying these coefficients (or part of them) from additional measurements of the PDE solution on (part of) the domain or on its boundary. This fits into the above framework with, e.g., , , , (or, if , are sufficiently smooth, , , ,)
⟨A(a,b,c,u),w⟩W∗,W=∫Ω(a∇u⋅∇w+cuw−bw)dx−∫∂Ωgwds
and , , , where is an open subdomain of or a regular curve/surface contained in its boundary or in its interior, so that an embedding or a trace theorem yields continuity of the observation map .
###### Example 2
Using similar measurements but a nonlinear model, we consider identification of the nonlinearity, i.e., the function in the elliptic boundary value problem
−Δu+q(u)=0 in Ω,∂u∂n=g% on ∂Ω
with given . Here we use a space that is continuously embedded in for an interval containing all possibly appearing values of (which can, e.g., be estimated by using maximum principles in case the PDE is elliptic, depending on the monotonicity of ), ,
⟨A(q,u),w⟩W∗,W=∫Ω(∇u⋅∇w+q(u)w)dx−∫∂Ωgwds
and , as in Example 1 above.
###### Example 3
Alternatively, one often encounters inverse source problems for nonlinear PDEs such as the simple model example of identifying in
−Δu+ζu3=b in Ω,∂u∂n=g on ∂Ω
where and are given. Here we have , , ,
⟨A(b,u),w⟩W∗,W=∫Ω(∇u⋅∇w+ζu3w−bw)dx−∫∂Ωgwds,
and again , as in Example 1 above.
###### Example 4
Consider identification of the (possibly infinite dimensional) parameter in the state space system consisting of an ODE model and observations
˙u(t)=f(t,u(t),ϑ) t∈(0,T),u(0)=u0 y=C(u),
where the dot denotes time differentiation, is a given function and is a given initial value. Using semigroup theory, this could as well be extended to time dependent PDEs. An example of an infinite dimensional stationary parameter to be identified in a system of ODEs is the Preisach weight function in some hysteretic evolutionary model. The observations are, e.g. discrete or continuous in time , (including the case of final measurements ) or , with given functions or .
### 1.2 Motivation of the all-at-once-approach
Definition and analysis of solution methods for such inverse problems is often based on a reduced formulation that is obtained by the use of a parameter-to-state map i.e., a mapping , that resolves (1) with respect to
∀x∈D(S): A(x,S(x))=0 and ∀u∈V:((x,u)∈D(A) and A(x,u)=0) ⇒ u=S(x).
Existence of such a mapping is guaranteed by the Implicit Function Theorem if for an open set with , is continuously Fréchet differentiable on and its derivative with respect to the state is boundedly invertible with uniform bound:
###### Assumption 1
∃CA ∀(x,u)∈(B×V)∩D(A): ∥∥Au(x,u)−1∥∥≤CA.
In order to satisfy this assumption, usually the domain of has to be restricted, e.g., to
D(A)⊆{(x,u)=(a,b,c,u):a––≤a≤¯a a.e. on Ω, c≥c– a.e. on Ω} (3)
with positive constants , in example 1 or to
D(A)⊆{(x,u)=(q,u):q–≤q(~λ)−q(λ)~λ−λ ∀~λ≠λ∈R} (4)
with some constant in example 2.
Under such conditions the forward operator , is well-defined on and the inverse problem (1), (2) can be equivalently written as an operator equation
F(x)=y (5)
Such problems are typically ill-posedness in the sense that is not continuously invertible and instead of usually only a noisy version is available, which we here assume to obey the deterministic and known noise bound
∥∥y−yδ∥∥≤δ, (6)
thus regularization (see, e.g., [2, 6, 15, 17, 20, 21, 24] and the references therein) has to be employed.
In this paper we will return to the original formulation as an all-at-once system of model and observation equation (1), (2),
F(x)=F(x,u)=(A(x,u)C(u))=(0y)=y (7)
and investigate the behaviour of some well-known regularization paradigms when applied to instead of . We will see that this enables to avoid restrictions like (3), (4) and, moreover, can make a considerable difference when it comes to implementation. We will also provide a convergence analysis that goes beyond the mere application of known results to the operator in the sense that regularization might not just be applied to the whole of but - as natural - only to the part, a case which requires some extra considerations in the convergence analysis.
All-at-once approaches to inverse problems have already been considered previously, see, e.g., [3, 4, 8, 14, 23]. While these papers concentrate on computational aspects and convergence analysis of particular methods, our aim is here to provide a comparative overview on several regularization paradigms.
In the remainder of this paper we will assume that a solution
(x†,u†)∈D(F)⊆D(A)∩(X×D(C)) (8)
to (7) exists and that is convex. Note that need not necessarily be the maximal domain of , so restriction to a convex set can be done without loss of generality here.
While data misfit and regularization terms will be defined by norms for simplicity of exposition, most of the results are extendable to more general discrepancy and regularization functionals as considered, e.g. in [7, 19, 25].
We treat methods and convergence conditions only exemplarily to highlight analogies and differences between reduced and all-at-once formulations, so our aim is not to provide a complete convergence analysis (a priori and a posteriori choice of regularization parameters, general rates, etc.) for each of the discussed methods.
The remainder of this paper is organized as follows. In Sections 2, 3, 4, we consider Tikhonov, Newton type, and gradient type regularization, respectively. For each of these three paradigms we provide some convergence results with a priori and a posteriori regularization parameter choice strategies and under different assumptions on the forward operator (This analysis part is restricted to just quotation of a convergence result in the gradient method case.) Moreover we compare the key implementation requirements. Section 5 contains a verification of the convergence conditions for the iteratively regularized Gauss-Newton method from Section 3 for Example 3, whose all-at-once and reduced versions are then compared numerically. Some preliminary considerations on comparison of convergence conditions for all-at-once and reduced versions are provided in Section 6, and we give a short summary and an outlook in Section 7.
#### Notation
For some we denote by the dual index. denotes the duality mapping with gauge function , which is in general set valued. For smooth spaces, i.e., spaces with Gâteaux differentiable norm on the unit sphere, will be single valued; otherwise, by a slight abuse of notation, we denote by a single valued selection from this set.
## 2 Tikhonov regularization
For any , , , , we define the pair as a minimizer of
min(x,u)∈D(F)S(F(x,u),(0,yδ))+αR(x,u) (9)
with
S((w∗,y),(y\footnotesize{mod},y\footnotesize{% obs}))=ρm∥∥w∗−y\footnotesize{mod}∥∥mW∗+1o∥∥y−y\footnotesize{obs}∥∥oY (10)
and
R(x,u)=R1(x)=1r∥x−x0∥rX (11)
or
R(x,u)=R2(x,u)=1r∥x−x0∥rV+1q∥u−u0∥qV, (12)
Note that in case of (11) regularization is only applied to . In both versions, will remain fixed, whereas will be chosen in dependence of - typically in such a way that it tends to zero as .
### 2.1 Well-definedness and convergence
The analysis will be based on the following assumptions
###### Assumption 2
Bounded sets in , are weakly (or weakly*) compact.
###### Assumption 3
is weakly (or weakly*) sequentially closed, i.e.,
∀((xk,uk))k∈N⊆D(F): (xk⇀x, uk⇀u, A(xk,uk)⇀f,C(uk)⇀y) ⟹ ((x,u)∈D(F) % and A(x,u)=f, C(u)=y),
where denotes weak or weak* convergence.
For only proving convergence provided minimizers are already well-defined the following somewhat weaker assumption (note that we have strong convergence of the images in the premiss) suffices in place of Assumption 3.
###### Assumption 4
∀((xk,uk))k∈N⊆D(F): (xk⇀x, uk⇀u, A(xk,uk)→f,C(uk)→y) ⟹ ((x,u)∈D(F) % and A(x,u)=f, C(u)=y)
Assumption 2 is satisfied if are reflexive or duals of separable normed spaces. (The star in weak* will be skipped in the following.) Sufficient for Assumption 3 is weak continuity of and and weak closedness of .
To prove well-definedness of a minimizer and convergence in case of regularization with respect to only, (11), we additionally impose Fréchet differentiability of and , Assumption 1 and a growth condition on the derivative of the model operator with respect to the parameter:
###### Assumption 5
There exists a (without loss of generality monotonically increasing) function such that
(a)∀(x,v)∈(B×V)∩D(A): ∥Ax(x,u)∥≤ψ(∥x∥X)(1+∥u∥V) and ψ(∥∥x†∥∥X+λ)λ→0 as λ→0 and ∥∥x†−x0∥∥X is sufficiently small or (b)∀(x,v)∈(B×V)∩D(A): ∥Ax(x,u)∥≤ψ(∥x∥X)
###### Proposition 1
Let Assumptions 2 3, and in case of (11) additionally Assumptions 1, 5 be satisfied.
• Then there exists such that for all , a minimizer of (9) with (10) and (11) or (12) exists.
• These minimizers are stable with respect to the data in the sense that for any sequence converging to in the norm and any sequence of corresponding minimizers there exists a weakly convergent subsequence and the limit of every weakly convergent subsequence is a minimizer of (9).
• If is chosen a priori according to
α→0 and δoα→0 as δ→0 (13)
or a posteriori according to the generalized discrepancy principle
δoo≤S(F(xδα,uδα),(0,yδ))≤τδoo (14)
for some fixed independent of , then there exists independent of such that for any solution of (1), (2) we have boundedness
∀δ∈(0,¯¯¯δ): 1r∥∥xδα(δ)−x0∥∥rX≤{1r∥∥x†−x0∥∥rX+δooα(δ) with (???)1r∥∥x†−x0∥∥rX with (???) (15) and ∥∥uδα(δ)∥∥V≤¯¯¯¯C
in case of minimizers of (9) with (10) and (11) and
∀δ∈(0,¯¯¯δ): R2(xδα(δ),uδα(δ))≤{R2(x†,u†)+δoα(δ) with (???)R2(x†,u†) with (???) (16)
in case of minimizers of (9) with (10) and (12).
• In both cases (15), (16), the regularized approximations converge weakly subsequentially to a solution of (1), (2).
###### Proof
The assertion follows from the results in Section 3 of [10] with there defined by here. Indeed almost all items of [10, Assumption 2.1] easily follow from Assumption 2 3 (note that it actually suffices to assume weak sequential closedness of the operator in place of [10, Assumption 2.1 (3), (5)]). The only exception arises in case of regularization with respect to only (11), where boundedness, i.e., weak compactness, of level sets
Mα(M)={(x,u)∈D(F):1αS(F(x,u),(0,yδ))+R(x,u)≤M}
[10, Assumption 2.1 (6)], which the proofs there actually only require to hold for sufficiently small , has to be shown separately. For obtaining this boundedness, we will make use of Assumptions 1, 5 in that case. Since we have also stated convergence with the discrepancy principle, which is not treated in [10] and for completeness of exposition we provide the details of the convergence part of the proof for the case of (11).
The standard argument of minimality together with (6) for any solution of (1), (2) yields the estimate
S(F(xδα,uδα),(0,yδ))+αr∥∥xδα−x0∥∥r≤1oδo+αr∥∥x†−x0∥∥r (17)
thus, upon division by and setting (in case of (14) using the lower estimate there) we get the estimate on in (15). To obtain boundedness of as well, we use the identity
A(xδα,uδα)=A(x†,u†)=0+∫10(Ax(xθ,uθ)(xδα−x†)+Au(xθ,uθ)(uδα−u†))dθ (18)
where , and Assumptions 1, 5, as well as (17) or the upper bound in (14)
∥∥A(xδα,uδα)∥∥mW∗≤¯¯¯¯Cδ=mρ{1oδo+α(δ)r∥∥x0−x†∥∥rX in case of % (???)τoδo in case of (???).
to arrive at the estimate
∥∥uδα−u†∥∥V≤ CA(¯¯¯¯C1/mδ+ψ(∥∥x†∥∥X+∥∥xδα−x†∥∥X)∥∥xδα−x†∥∥X) (19) ×{(1+∥∥u†∥∥V+∥∥uδα−u†∥∥V) in case (a)1 in case (b).
By the first (already proven) part of (15), this directly yields a bound on in case (b). In case (a) we additionally use the fact that the smallness assumption on and the growth condition on allows us to achieve
CA(¯¯¯¯C1/mδ+ψ(∥∥x†∥∥X+φ(∥∥x0−x†∥∥X))φ(∥∥x0−x†∥∥X))≤c<1
for some constant independent of , where
φ(s)={s+(sr+rδooα(δ))1/r in case of (???)2s in case of (???).
Rearranging terms in (19) (case (a)) we therefore get
∥∥uδα(δ)−u†∥∥V≤c1−c(1+∥∥u†∥∥V).
The rest follows by standard arguments from the assumed continuity assumptions on and .
###### Remark 1
Note that in case of regularization with respect to both and , (12), we do not need Assumptions 1, 5 and can therefore also deal with situations in which a parameter-to state map not necessarily exists.
Well-definedness of according to the discrepancy principle (14) follows from [16, Lemma 1] provided is reflexive and strictly convex and either (i) is weakly closed (i.e., Assumption 3 is satisfied) and is reflexive or (ii) is weak-to-weak continuous and is weakly closed (which by the assumed convexity of this set is satisfied, e.g., if it is closed wrt. norm convergence).
If satisfies the Kadets-Klee property, the results of Theorem 1 imply strong convergence and if the solution is unique then by a subsequence-subsequence argument the whole sequence converges.
### 2.2 Convergence rates
As usual we can conclude convergence rates with respect to the Bregman distance under source conditions. Just exemplarily we will state a rates result for the all-at-once Tikhonov method with regularization with respect to both and under a benchmark source condition. To this end, we use an element of the subgradient
ξ∈∂R(x†,u†) where R(x,u)=R2(x,u)=1r∥x−x0∥rX+1q∥u−u0∥qV (20)
to define the Bregman distance
D(x0,u0)ξ((x,u),(x†,u†))=R(x,u)−R(x†,u†)−⟨ξ,(x−x†,x−u†)⟩ (21)
and impose a local smoothness condition on .
###### Assumption 6
∥∥Ax(x†,u†)(x−x†)+Au(x†,u†)(u−u†)∥∥+∥∥C′(u†)(u−u†)∥∥ +LD(x0,u0)ξ((x,u),(x†,u†)),
for some according to (20), , and all .
In case , Assumption 6 follows from the inverse triangle inequality and the Taylor remainder estimate
∥∥A(x†,u†)+Ax(x†,u†)(x−x†)+Au(x†,u†)(u−u†)−A(x,u)∥∥ +∥∥C(u†)+C′(u†)(u−u†)−C(u)∥∥≤LD(x0,u0)ξ((x,u),(x†,u†)),
which in the quadratic Hilbert space case corresponds to the usual estimate obtained under a local Lipschitz condition on .
###### Proposition 2
Let satisfy Assumption 6 and let, for some with the source condition
ξ=F′(x†,u†)∗(v\footnotesize{% mod},v\footnotesize{obs}) (22)
hold for some .
Then for the Tikhonov minimizers according to (9), (10), (12) with the apriori choice
α(δ)∼δot∗ (23)
(with ) or the a posteriori choice (14) we have
D(x0,u0)ξ((xδα(δ),uδα(δ)),(x†,u†))=O(δo/t) (24)
###### Proof
The proof follows the lines of the classical rates proof from [5].
By (16) and (22) we have in case of (14)
D(x0,u0)ξ((xδα(δ),uδα(δ)),(x†,u†)) ≤−⟨(v\footnotesize{mod},v\footnotesize% {obs}),F′(x†,u†)(xδα(δ)−x†,uδα(δ)−u†)⟩ ≤∥∥(v\footnotesize{mod},v\footnotesize{% obs})∥∥(¯¯¯¯CL(ρm∥∥A(xδα(δ),uδα(δ))−A(x†,u†)∥∥m +1o∥∥C(uδα(δ))−C(u†)∥∥o)1/t+LD(x0,u0)ξ((xδα(δ),uδα(δ)),(x†,u†)))
hence with
cD(x0,u0)ξ((xδα(δ),uδα(δ)),(x†,u†))≤∥∥(v\footnotesize{mod},v\footnotesize{obs})∥∥(¯¯¯¯CL(τ+1)δoo)1/t.
In case of (23) we get from minimality (cf. (17))
S(F(xδα,uδα),(0,yδ))+αR2(xδα)≤1oδo+αR2(x†),
hence after division by and by definition (21) of the Bregman distance, as well as the source condition (22)
S(F(xδα,uδα),(0,yδ))α+D(x0,u0)ξ((xδα,uδα),(x†,u†)) ≤δoα−⟨(v\footnotesize{mod}% ,v\footnotesize{obs}),F′(x†,u†)(xδα−x†,uδα−u†)⟩ ≤δoα+∥∥(v\footnotesize{mod}% ,v\footnotesize{obs})∥∥(¯¯¯¯CL(S(F(xδα,uδα),(0,yδ)))1/t +LD(x0,u0)ξ((xδα,uδα),(x†,u†))),
hence with as above and using Young’s inequality
S(F(xδα,uδα),(0,yδ))α+cD(x0,u0)ξ((xδα,uδα),(x†,u†)) ≤δoα+∥∥(v\footnotesize{mod}% ,v\footnotesize{obs})∥∥¯¯¯¯CL(S(F(xδα,uδα),(0,yδ)))1/t ≤δoα+((∥∥(v\footnotesize{mod},v\footnotesize{obs})∥∥¯¯¯¯CLt)t2α)1t−1+S(F(xδα,uδα),(0,yδ))2α,
which by the choice (23) yields (24).
###### Remark 2
Note that in case of regularization with respect to only, i.e., (11), we do get boundedness also of the part via Assumptions 1, 5. However, the sharp estimate (16), that was crucially used in the rates proof above, fails to hold in general. Therefore we do not expect to get (optimal) convergence rates in that case.
### 2.3 Comparison of implementation
To compare the all-at-once Tikhonov method (9), with (10) and (11) or (12) with Tikhonov regularization for the reduced formulation
minx∈D(F)1o∥∥F(x)−yδ∥∥oY+αr∥x−x0∥r (25)
we write the latter as a PDE constrained minimization problem
(26)
and denote the Tikhonov minimizer as well as its corresponding state by a bar, i.e., minimizes (26) and . For convenience of exposition we will assume to be reflexive and identify it with its bidual.
We first of all show that reduced Tikhonov regularization (26), (25) is equivalent to all-at-once Tikhonov regularization (9), with (10) and (11) in case with sufficiently large. This is due to exact penalization, cf., e.g., [18, Theorem 17.3], whose proof remains valid in the Banach space setting. More precisely, for this purpose has to be larger than the norm of the adjoint state, i.e., the solution to
Au(¯¯¯xδα,¯¯¯uδα)∗w=C′(¯¯¯uδα)∗JYo(C(¯¯¯uδα)−yδ) in V∗
for solving (26). Existence of this adjoint state is ensured by Fréchet differentiability of together with Assumption 1, which also implies that
∥∥¯¯¯¯wδα∥∥W≤CA∥∥C′(¯¯¯uδα)∥∥∥∥C(¯¯¯uδα)−yδ∥∥o−1Y. (27)
Moreover, by the same arguments as in the proof of Theorem 1, uniform (wrt. ) boundedness of can be concluded from its definition as a minimizer of (25)
1o∥∥F(¯¯¯xδα(δ))−yδ∥∥oYα(δ)+1r∥∥¯¯¯xδα(δ)−x0∥∥r≤1oδoα(δ)+1r∥∥x†−xo∥∥r
with chosen a priori according to (13) or a posteriori according to
δo≤∥∥F(¯¯¯xδα(δ))−yδ∥∥o≤τδo,
(which due to exact penalization will finally coincide with (14),) together with the identity (cf. (18))
0 =A(¯¯¯xδα(δ),¯¯¯uδα(δ))=A(xδα,uδα) =∫10(Ax(xθ,uθ)(¯¯¯xδα(δ)−x†)+Au(xθ,uθ)(¯¯¯uδα(δ)−u†))dθ
with , , which yields an estimate like (19) with and replaced by . Thus, from (27) we get a uniform bound on and can conclude the following equivalence.
###### Proposition 3
Let Assumptions 1, 5 be satisfied and let be Fréchet differentiable with mapping bounded sets to bounded sets. There exist sufficiently large and sufficiently small such that for all and chosen according to (13) or (14), the minimizers of the all-at-once Tikhonov functional according to (9) with (10), (11) with coincide with those of the reduced Tikhonov functional (25).
We now consider first order optimality conditions for the reduced and the all-at-once formulations (26), (9) with general , so that they are not necessarily equivalent, and for this purpose assume , , and the occurring norms to be continuously Fréchet differentiable (i.e., the corresponding spaces to be uniformly smooth). In case , we get the first order necessary conditions
A(¯¯¯xδα,¯¯¯uδα)=0 (28) αJXr(¯¯¯xδα−x0)=−Ax(¯¯¯xδα,¯¯¯uδα)∗¯¯¯¯wδα Au(¯¯¯xδα,¯¯¯uδα)∗¯¯¯¯wδα=−C′(¯¯¯uδα) | 6,774 | 21,148 | {"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} | 2.71875 | 3 | CC-MAIN-2021-04 | latest | en | 0.883108 |
https://www.shaalaa.com/question-bank-solutions/if-x-4t1-t2-y-3-1-t21-t2-then-show-that-dydx-9x4y-derivatives-of-parametric-functions_208771 | 1,718,848,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00400.warc.gz | 874,030,558 | 12,223 | # If x = 4t1+t2, y = 3(1-t21+t2), then show that dydx=-9x4y - Mathematics and Statistics
Sum
If x = (4"t")/(1 + "t"^2), y = 3((1 - "t"^2)/(1 + "t"^2)), then show that ("d"y)/("d"x) = (-9x)/(4y)
#### Solution
x = (4"t")/(1 + "t"^2)
Differentiating both sides w.r.t. ‘t’, we get
("d"x)/"dt" = "d"/"dt" ((4"t")/(1 + "t"^2))
= ((1 + "t"^2)*"d"/"dt"(4"t") - 4"t"*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)^2
= ((1 + "t"^2)(4) - 4"t"(0 + 2"t"))/(1 + "t"^2)^2
= (4 + 4"t"^2 - 8"t"^2)/(1 + "t"^2)^2
= (4 - 4"t"^2)/(1 + "t"^2)^2
= (4(1 - "t"^2))/(1 + "t"^2)^2
y = 3((1 - "t"^2)/(1 + "t"^2))
("d"y)/"dt" = 3*"d"/"dt"((1 - "t"^2)/(1 + "t"^2))
= 3[((1 + "t"^2)*"d"/"dt"(1 - "t"^2) - (1 - "t"^2)*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)]
= 3[((1 + "t"^2)(0 - 2"t") - (1 - "t"^2)(0 + 2"t"))/(1 + "t"^2)^2]
= 3[(-2"t"(1 + "t"^2) - 2"t"(1 - "t"^2))/(1 + "t"^2)^2]
= 3(- 2"t")[(1 + "t"^2 + 1 - "t"^2)/(1 + "t"^2)^2]
= - 6"t" xx 2/(1 + "t"^2)^2
= (-12"t")/(1 + "t"^2)^2
∴ ("d"y)/("d"x) = (("d"y)/("dt"))/(("d"x)/("dt"))
= ((-12"t")/((1 + "t"^2)^2))/((4(1 - "t"^2))/((1 + "t"^2)^2)
= ("d"y)/("d"x) = (-3"t")/(1 - "t"^2) ......(i)
Also, (-9x)/(4y) = (-9((4"t")/(1 + "t"^2)))/(4 xx 3((1 - "t"^2)/(1 + "t"^2))
= (-3"t")/(1 -"t"^2) ......(ii)
From (i) and (ii), we get
("d"y)/("d"x) = (-9x)/(4y)
Concept: Derivatives of Parametric Functions
Is there an error in this question or solution?
Chapter 1.3: Differentiation - Q.5
Share | 781 | 1,422 | {"found_math": true, "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": 1, "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} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.394456 |
http://teachersofindia.org/en/tags/comac | 1,596,837,215,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737225.57/warc/CC-MAIN-20200807202502-20200807232502-00400.warc.gz | 102,742,015 | 11,518 | # CoMac
## A Nines Multiples Problem
CoMaC shares a problem on the sum of digits adapted from one that appeared in an online column on the Math Forum site.
## Two Combinatorial Problems
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## Ramanujan and Pythagoras!
An interesting extract from Ramanujan’s notebooks which makes for a great classroom exercise in geometry, with a dash of algebra thrown in. An enterprising teacher could do this proof in stages — starting from showing students the figure and asking them to prove the theorem; if they can’t, providing them with enough scaffolding to help them complete the proof.
Here are other links, suggested by Rajkishore from APF, that can be referred to as well (Click on the image):
## Viviani’s Theorem...And A Cousin
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even by young children...
## Complete Family of Pythagorean Triples - Part 2
In Part I of this article we presented a few methods for generating Primitive Pythagorean Triples (PPTs). You will recall that they were all ‘piece meal’ in character. Now we present two more approaches which offer complete solutions to the PPT problem. Both are based on straightforward reasoning and simple algebra. And no PPT is left out: we capture the complete family in each case.
## A Surprising Fact about Triangles with a 60 degree Angle
Is the converse of a statement always true?...
Ever posed this question to a class and then scanned your memory for good
examples to clinch your argument? Here is one you could use.
## How Many Primitive Pythagorean Triples in Arithmetic Progression?
A simple investigation and a convincing proof based on a novel connection between two topics — the Pythagorean Theorem and Sequences — taught in middle and high school.
## The Pythagorean Theorem
Taking note of a collective of contributors...
How do I prove thee? Can I count the ways? A look at the wide variety of methods used to prove the theorem of Pythagoras.
## Pages
18603 registered users
7271 resources | 471 | 2,172 | {"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} | 3.890625 | 4 | CC-MAIN-2020-34 | latest | en | 0.937807 |
http://mathhelpforum.com/advanced-algebra/83315-proving-field-not-algebraically-closed-print.html | 1,524,788,028,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00139.warc.gz | 225,165,183 | 2,896 | # Proving a field is not algebraically closed
• Apr 12th 2009, 02:17 AM
Zinners
Proving a field is not algebraically closed
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
• Apr 12th 2009, 04:46 AM
berlioz
Using Fermat's little theorem we have x^(p-1)-1=0for any x doesn't equal zero,so the polynomial x^(p-1)-2(suppose p>2)doesn't have the zero points inZp
if p=2,we can find x^2+x+1 hasn't the zero points in Z2.
So field Zp is not algebraically closed.
Quote:
Originally Posted by Zinners
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
• Apr 13th 2009, 10:10 AM
ThePerfectHacker
Quote:
Originally Posted by Zinners
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
In general let $\displaystyle F$ be a finite field with $\displaystyle q$ elements.
Define $\displaystyle f(x) = x^q - x +1$, we know that $\displaystyle a^{q-1}=1 \implies a^q - a = 0$ for all $\displaystyle a\in F^{\times}$.
Therefore, $\displaystyle f(x) = x^q - x + 1$ always has no zero. | 449 | 1,489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.936627 |
https://civilstep.com/all-about-paint-calculation/ | 1,709,062,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00078.warc.gz | 161,138,274 | 33,485 | 0
When it comes to home improvement and renovation projects, paint is often the key element that ties everything together. Choosing the right type and color of paint can completely transform a space, but it’s not just about the aesthetics. Calculating the amount of paint needed for a project is crucial for budgeting and planning purposes. In this article, we will delve into all the important factors involved in paint calculation, from understanding the different types of paint to determining the appropriate amount for each project. Whether you’re a DIY enthusiast or a professional contractor, having a thorough understanding of paint calculation will help you achieve successful and cost-effective results.
## Paint calculation | calculator for how much paint do I need
Paint calculation is an essential aspect of any construction or renovation project. As a civil engineer, it is important to accurately estimate the amount of paint needed for a project in order to avoid any excess or shortage of materials. This will not only save time and money but also ensure a smooth and efficient painting process.
The first step in calculating paint quantity is to determine the surface area that needs to be painted. This can be done by measuring the length and height of each wall, door, and window in the area. Once all the measurements have been taken, the total surface area can be calculated by multiplying the length and height of each surface and then adding them together.
After obtaining the surface area, the next step is to determine the paint coverage rate. This refers to the square footage that can be covered by one gallon of paint. The coverage rate can vary depending on the type and quality of paint used, the surface texture, and the application method. In general, high-quality paint with a smooth texture and a roller application will have a higher coverage rate compared to low-quality paint with a rough texture and brush application.
To calculate the paint quantity, divide the total surface area by the paint coverage rate. For example, if the total surface area is 1000 square feet and the paint coverage rate is 350 square feet per gallon, then the total number of gallons needed would be 1000/350=2.86 gallons. In this case, it is advisable to round up the number to the nearest whole gallon, so 3 gallons of paint would be required for this project.
Apart from the surface area and paint coverage rate, there are a few other factors that can affect the paint quantity needed. These include the number of coats required, the paint color, and the type of surface being painted. Different colors may require more coats to achieve full coverage, while a rough surface may require more paint due to its higher absorption rate.
Fortunately, there are various online paint calculators available that can make this process easier and more accurate. These calculators take into account all the factors mentioned above and provide a precise estimate of the paint quantity needed for a project. They also allow the user to input specific measurements and details, making them a valuable tool for civil engineers and other construction professionals.
In conclusion, paint calculation is a crucial aspect of any construction project. As a civil engineer, it is important to have a clear understanding of the surface area, paint coverage rate, and other factors that can affect the total paint quantity needed. By using accurate measurements and reliable paint calculators, this can be achieved, saving both time and resources in the process.
## Paint calculation formula for exterior and interior painting
Paint calculation formula is an important aspect of any painting project, as it helps to estimate the required quantity of paint for both the interior and exterior of a building. As a civil engineer, proper paint calculation is essential to ensure that there is no wastage of materials and to stay within the project budget.
The formula for paint calculation may vary depending on the type and brand of paint as well as the surface to be painted. However, the basic formula remains the same and can be applied to both interior and exterior painting projects.
For Exterior Painting:
Step 1: Measure the dimensions of the walls to be painted in meters (m). For example, if the length of a wall is 10m and the height is 3m, then the total area to be painted is 10m x 3m = 30 square meters (m²).
Step 2: Calculate the total area of all the walls to be painted by adding up the individual areas. For instance, if there are two walls with the same dimensions, then the total area to be painted is 30m² + 30m² = 60m².
Step 3: Determine the paint coverage rate. This information is usually provided by the manufacturer and is expressed as the number of square meters that can be covered by one liter of paint. For example, if the paint coverage rate is 10m² per liter, then one liter of paint will cover an area of 10m².
Step 4: Calculate the required quantity of paint by dividing the total area by the paint coverage rate. Using the example above, 60m² of wall area will require 60m²/10m² per liter = 6 liters of paint.
Step 5: Add an additional 10-15% to the calculated amount to account for wastage and touch-ups. In this case, the total amount of paint required will be 6 liters + 0.9 liters (15% of 6 liters) = 6.9 liters.
For Interior Painting:
Step 1: Measure the dimensions of the walls and ceiling to be painted in feet (ft). For example, if the length of a wall is 10ft and the height is 8ft, then the total area to be painted is 10ft x 8ft = 80 square feet (ft²).
Step 2: Calculate the total area of all the walls and ceiling to be painted by adding up the individual areas. For instance, if there are three walls and one ceiling with the same dimensions, then the total area to be painted is 80ft² + 80ft² + 80ft² + 80ft² = 320ft².
Step 3: Determine the paint coverage rate. This information is usually provided by the manufacturer and is expressed as the number of square feet that can be covered by one gallon of paint. For example, if the paint coverage rate is 400ft² per gallon, then one gallon of paint will cover an area of 400ft².
Step 4: Calculate the required quantity of paint by dividing the total area by the paint coverage rate. Using the example above, 320ft² of wall and ceiling area will require 320ft²/400ft² per gallon = 0.8 gallons of paint.
Step 5: Add an additional 10-15% to the calculated amount to account for wastage and touch
## How much a gallon of paint cover?
A gallon of paint is a common unit of measurement used in the construction industry. It is specifically used to measure the amount of paint needed to cover a certain area. The coverage of a gallon of paint can vary depending on the type of paint, the condition of the surface, and the method of application. As a civil engineer, it is important to understand the coverage of paint in order to accurately estimate the amount of paint needed for a project.
In general, a gallon of paint can cover an average of 350-400 square feet of a smooth surface with one coat. This coverage rate is based on the assumption that the paint is applied evenly with good quality tools. However, it is essential to note that not all paints have the same coverage rate. Different types of paint, such as primer, flat, gloss, and high-gloss, have varying thicknesses, resulting in different coverage rates.
Additionally, the condition of the surface can also affect the coverage of a gallon of paint. A rough or porous surface will require more paint to cover the same area compared to a smooth surface. This is because the surface will absorb more paint, resulting in a thinner coat and requiring additional paint to achieve the desired coverage.
The application method also plays a significant role in the coverage of a gallon of paint. When applying paint with a brush or roller, the coverage rate may vary depending on the skill of the applicator. Inexperienced painters may use more paint than necessary, resulting in a lower coverage rate. On the other hand, using a sprayer can provide a more consistent and even coverage, resulting in a higher coverage rate.
It is also essential to consider the number of coats needed when estimating the amount of paint required for a project. In most cases, two coats of paint are needed to achieve full coverage and a professional finish. However, certain types of paint may require more or fewer coats, which can affect the coverage rate.
In conclusion, the coverage of a gallon of paint can vary, ranging from 350-400 square feet on average. As a civil engineer, understanding the factors that can affect paint coverage is crucial to accurately estimate the amount of paint needed for a project. Consider the type of paint, condition of the surface, application method, and number of coats when calculating the coverage rate of a gallon of paint.
## How much does a 5 gallon of paint cover?
A 5 gallon can of paint is a common size for residential and commercial use. This amount of paint can cover a significant area, but the coverage will vary depending on several factors, including the application method, surface type, and paint quality.
On average, a 5 gallon can of paint can cover anywhere from 1,500 to 2,500 square feet. However, this estimation is based on the assumption that the paint will be applied to a smooth, non-porous surface with one coat. In reality, most surfaces require two coats of paint for an even and full coverage.
It is important to note that the type of paint also plays a significant role in determining the coverage of a 5 gallon can. Latex or water-based paints typically have a higher coverage rate compared to oil-based paints. In general, high-quality paints tend to have better coverage and require fewer coats for an even finish.
When it comes to application, using a roller is the most efficient method for coverage, while a brush may require more paint and multiple coats. Factors such as the thickness of the application and the painter’s technique can also affect the coverage.
The type of surface being painted can also impact coverage. Smooth surfaces like drywall and metal require less paint compared to rough surfaces like bare wood or textured walls. Additionally, porous surfaces tend to absorb more paint, requiring a larger quantity to achieve the desired coverage.
To determine the actual coverage of a 5 gallon can of paint, it is essential to read the manufacturer’s recommendations and information on the can. Most paint cans will provide an estimated coverage rate based on the factors mentioned above. It is also helpful to consult with a paint specialist or your local hardware store for specific guidance.
In conclusion, a 5 gallon can of paint can cover anywhere from 1,500 to 2,500 square feet, depending on various factors. Before starting a project, it is crucial to consider the application method, surface type, and paint quality to determine the true coverage of a 5 gallon can. Accurately estimating the coverage will help in determining the right amount of paint needed, avoiding any excess or shortage.
## What area does 1 litre of paint cover?
The coverage of paint is an important factor to consider when estimating the amount of material needed for a painting project. One commonly used unit for measuring paint quantity is litres. However, the coverage area of paint may vary depending on the type of paint, the application method, surface porosity, and other factors.
On average, 1 litre of paint can cover approximately 10-12 square meters of surface area. This, however, is a general estimate and may not hold true for all paints. The coverage area of paint can range from 8-14 square meters per litre, depending on the factors mentioned above.
One of the main factors that affect the coverage area of paint is its type. Different types of paints have different levels of solid content, which can impact its coverage area. For example, oil-based paints have a higher solids content, resulting in better coverage compared to water-based paints.
The surface porosity is another critical factor that affects the coverage area of paint. Porous surfaces, such as concrete or wood, will absorb more paint and require a larger amount of material to achieve an even coverage. On the other hand, smooth surfaces, like metal or glass, will have lower porosity and will require less paint.
The application method also plays a role in the coverage of paint. Generally, roller and spray methods provide better coverage compared to a brush. The texture of the surface can also affect the coverage area of paint. Rough surfaces, such as brick or stucco, will require more paint to achieve complete coverage.
It is essential to consider the coverage area of paint when planning for a painting project to avoid under or overestimating the required amount of material. To determine the coverage area of a specific paint, manufacturers usually provide information on the product label or online. It is recommended to use this information as a guide and always account for potential variations based on the factors mentioned.
In conclusion, the coverage area of 1 litre of paint can cover approximately 10-12 square meters on average. However, this can vary depending on the type of paint, application method, surface porosity, and texture. Accurately estimating the coverage area of paint is crucial for a successful and cost-effective painting project.
## How much area is covered by 1 litre of paint
One litre of paint typically covers an area of about 10-12 square meters, depending on the type of surface being painted and the thickness of the paint application.
The coverage area also varies based on the quality and type of paint, as well as the method of application. For example, a high-quality paint with higher pigmentation will cover a larger area than a lower-quality paint with lower pigmentation.
In addition, painting on a smooth and even surface will require less paint compared to painting on a rough or textured surface. Painting with a brush will also result in a thinner layer of paint, while using a roller will provide better coverage with a thicker layer of paint.
It is important to accurately calculate the area to be painted and the amount of paint needed in order to avoid wastage or running short of paint. This can be done by measuring the length and width of the surface to be painted and multiplying them to get the total square meters. The coverage area of the paint can then be divided by the total area to determine the amount of paint needed.
Furthermore, it is recommended to apply two coats of paint for a better finish and longer durability, which will require twice the amount of paint compared to a single coat.
In summary, the coverage area of one litre of paint is approximately 10-12 square meters, but it is important to consider various factors that may affect the actual coverage. Consulting with a paint expert or carefully reading the instructions on the paint can will also provide guidance on the specific coverage of a particular paint.
## What area does 5 litres of paint cover?
As a civil engineer, I have extensive knowledge in construction materials and their applications. An important component of any construction project is paint, which not only adds aesthetic value but also protects the surface it covers. Paint is typically sold in quantities such as gallons or litres, and one common question that arises is what area can be covered by a certain amount of paint, such as 5 litres.
The answer to this question depends on various factors such as the type of paint, surface type, and method of application. Generally, manufacturers provide information on the coverage area on the paint can, but it is important to understand how this calculation is made.
The coverage area of paint is determined by the spreading rate. This refers to the amount of paint needed to cover a certain area. It is usually measured in square meters per litre (m2/L) or square feet per gallon (ft2/gal). The spreading rate is established by conducting tests in laboratory conditions, considering factors such as the viscosity of the paint, surface porosity, and application method.
Different types of paint have different spreading rates. For example, a high-quality acrylic paint may have a spreading rate of 10 m2/L, while a low-quality emulsion paint may have a rate of only 5 m2/L. This means that 1 litre of the acrylic paint can cover an area of 10 square meters, while the same amount of emulsion paint can only cover half of that.
Another important factor that affects the coverage area of paint is the surface type. A smoother surface like drywall or plaster may require less paint compared to a more uneven surface such as brick or concrete. This is because the surface porosity affects how much paint is absorbed and how evenly it spreads. In general, a rougher surface will require more paint to achieve the same coverage.
Moreover, the method of application also plays a significant role in determining the coverage area. Using a roller or brush may result in thicker, more consistent coats compared to spraying, which may produce a thinner and less even layer. This could affect the spreading rate and ultimately the coverage area of the paint.
In conclusion, the exact area that 5 litres of paint can cover cannot be determined without considering the type of paint, surface type, and application method. However, as a general guideline, 5 litres of paint can cover approximately 50 square meters of a smooth surface with a high-quality paint and careful application. It is always recommended to consult the manufacturer’s instructions and conduct a small test patch before calculating the exact amount of paint needed for a project.
## How much paint is needed for a 12×12 room?
Painting a room is a popular home renovation project that can transform the look and feel of a space. As a civil engineer, I have overseen many painting projects and have a good understanding of the materials and quantities needed for a successful paint job. In this article, I will discuss how much paint is needed for a 12×12 room and factors that can affect the amount of paint required.
The first step to determining the amount of paint needed for a room is to calculate the square footage of the walls. A 12×12 room has a total area of 144 square feet (12×12=144). However, this does not take into account the doors, windows, or any other openings in the room. These areas should be subtracted from the total square footage to get a more accurate estimation.
For a standard 12×12 room with two doors (20 square feet each) and one window (15 square feet), the total square footage would be 144-20-20-15= 89 square feet. Keep in mind that this is for the walls only, and does not include the ceiling or any extra features like crown molding or baseboards.
The next factor to consider is the type and quality of paint being used. Different brands and types of paint have different coverage rates, which is the amount of surface that one gallon of paint will cover. High-quality, premium paints typically have a higher coverage rate, while lower quality paints may require multiple coats to achieve the desired finish. Always check the paint manufacturer’s guidelines for their specific coverage rate.
For a 12×12 room, it is safe to assume that one gallon of paint should cover approximately 300-350 square feet. Using our example from before, with a total square footage of 89 square feet, it would require approximately 0.3 – 0.4 gallons of paint. Keep in mind, this is only for one coat of paint. If multiple coats are needed, then the amount of paint needed will increase accordingly.
Other factors that can affect the amount of paint needed include the texture of the walls and the desired finish. Textured walls tend to require more paint as the texture creates more surface area to cover. Additionally, if a glossy or reflective finish is desired, it may require more coats of paint to achieve a smooth, even finish.
Another important consideration is the color of the paint. Darker colors tend to require more coats of paint than lighter colors to achieve full coverage. It is always a good idea to consult with a paint professional to determine the specific coverage rates and number of coats needed for your chosen paint color.
In conclusion, the amount of paint needed for a 12×12 room can vary based on factors such as the square footage of the walls, type and quality of paint, texture of the walls, desired finish, and the color of the paint. It is always best to consult with a professional or carefully follow the manufacturer’s guidelines to ensure you have enough paint for your project. With proper planning and consideration of these factors, you can achieve a beautiful new look for your 12×12 room.
## How much paint is needed for a 10×10 room?
As a civil engineer, I am often tasked with estimating quantities for construction projects. Calculating the amount of paint needed for a 10×10 room is a common request that requires careful consideration of various factors.
The first step in determining the amount of paint needed is to calculate the surface area of the room. A 10×10 room has a perimeter of 40 feet and a total area of 100 square feet. However, this calculation does not account for windows, doors, or other openings in the room. To get a more accurate estimation, the area of these openings should be subtracted from the total area.
Next, the type of paint and its coverage rate must be taken into account. Paint coverage rates are usually given in square feet per gallon, and this varies depending on the type of paint and its brand. Standard latex paint, for example, has an average coverage rate of 350-400 square feet per gallon. This means that one gallon of paint can cover a 10×10 room with some leftover for touch-ups.
In addition to the type of paint, the condition of the walls also affects the coverage rate. If the walls are new or have never been painted before, they may require more paint to achieve the desired coverage. On the other hand, if the walls are rough or damaged, they may need additional coats of paint, which would increase the amount of paint needed.
Another factor to consider is the number of coats of paint desired. In most cases, two coats of paint are needed to achieve a uniform and long-lasting finish. Therefore, the amount of paint needed should be doubled to cover two coats. It is always recommended to have some extra paint on hand for touch-ups or unexpected needs.
It is also essential to consider the paint finish, such as flat, eggshell, or semi-gloss, as this can affect the coverage rate. For example, a glossier finish may require more paint than a flat finish.
To summarize, to determine how much paint is needed for a 10×10 room, several factors need to be considered, including the room’s surface area, paint coverage rate, wall condition, and desired number of coats and finish. As a civil engineer, it is important to carefully estimate the amount of paint needed to avoid excess waste and unnecessary costs.
## 1 litre of paint covers how many square meter
The amount of paint needed to cover a certain area depends on various factors such as the type of surface, the quality of the paint, and the thickness of the application. However, on average, 1 litre of paint can cover approximately 10 square meters of surface area.
This estimation is based on using a quality paint and applying it with a paint roller or brush. If a lower quality paint or a sprayer is used, the coverage may be less. Also, if the surface is rough or porous, it may require more paint to achieve adequate coverage.
It is important to read the instructions on the paint can and calculate the coverage area based on the manufacturer’s guidelines. This will help ensure that the right amount of paint is purchased for the project.
Furthermore, it is recommended to consider buying slightly more paint than the calculated amount to account for any unexpected losses or touch-ups that may be needed.
In conclusion, 1 litre of paint can cover approximately 10 square meters of surface area, but this may vary depending on the specific project and materials used. It is always best to consult the manufacturer’s instructions and guidelines for accurate coverage estimations.
## 1 litre of paint covers how many square feet
One litre of paint typically covers approximately 100 square feet of surface area. However, this may vary depending on the type of paint, the texture of the surface, and the method of application. For example, rough or porous surfaces may require more paint to achieve adequate coverage, while smooth surfaces may require less. It is always important to refer to the manufacturer’s instructions for the recommended coverage rate for the specific paint being used. Additionally, factors such as multiple coats or thick application may also affect the coverage rate. Therefore, it is best to calculate the coverage based on the specific circumstances and conditions of the project.
## 20 litre of paint covers how many square meter
The amount of square meters that 20 litres of paint can cover is dependant on various factors such as the type of paint, surface porosity, and application method. However, a general estimate for coverage can be determined based on the paint’s theoretical coverage rate.
The theoretical coverage rate, also known as the theoretical spread rate, is the amount of surface area that a litre of paint can cover in one coat, without accounting for any wastage or additional coats. This rate is typically listed on the paint can or manufacturer’s website and is measured in square meters per litre.
On average, a litre of paint can cover approximately 10-12 square meters in one coat. This means that 20 litres of paint can cover approximately 200-240 square meters in one coat. This is assuming a standard coverage rate of 10-12 square meters per litre and a single coat application.
However, it is important to note that the actual coverage achieved may be lower than the theoretical rate due to factors such as surface texture, uneven application, and paint wastage. For example, a rough or textured surface may require more paint to achieve the desired coverage, and overspray or spillage can also result in a lower coverage rate.
Additionally, different types of paint may have different coverage rates. For example, a thick and highly pigmented paint may have a higher coverage rate compared to a thinner and less pigmented one.
Therefore, it is best to refer to the specific paint manufacturer’s guidelines for accurate coverage estimates. It is also recommended to purchase a slightly larger amount of paint than required to account for any potential wastage or need for additional coats.
In conclusion, while 20 litres of paint can generally cover approximately 200-240 square meters in one coat, the actual coverage achieved may vary based on various factors and it is best to refer to the manufacturer’s guidelines for accurate estimates.
## 20 litre of paint covers how many square feet
The coverage of 20 litres of paint depends on various factors such as the type of paint, the surface being painted, and the technique used for application. However, on average, 20 litres of paint can cover approximately 900-1000 square feet of surface area.
The coverage capacity of paint is usually indicated on the label or can of the paint. It is typically measured in square feet per gallon (sq ft/gal) or square meters per liter (sq m/L). This information is essential for estimating the amount of paint needed for a specific project.
The type of paint also plays a significant role in determining the coverage capacity. For instance, latex or water-based paints tend to cover more surface area than oil-based paints. This is because they have a higher solids content and can be spread thinner while still achieving the desired coverage.
The surface being painted also affects the coverage capacity of paint. A smooth and even surface will require less paint compared to a surface with imperfections or rough textures. In such cases, more paint will be needed to cover the same area due to the greater absorption and uneven distribution of the paint.
The method of application also impacts the coverage capacity of paint. Using a roller or sprayer will provide better coverage compared to using a brush. However, this may also result in a thinner coat of paint, therefore requiring more coats for complete coverage.
In conclusion, the coverage of 20 litres of paint can vary depending on various factors. However, on average, it can cover approximately 900-1000 square feet of surface area. It is always best to consult the paint manufacturer’s label for more specific information regarding coverage capacity. Additionally, conducting a small test patch can help determine the exact amount of paint needed for a particular project.
## 5 litre of paint covers how many square meter
5 litres of paint can cover approximately 50 square meters. This coverage may vary depending on the type of surface being painted and the thickness of the paint application.
In general, the coverage of paint is determined by its spreading rate, also known as the theoretical coverage. It is the estimated area that a specific amount of paint (in this case, 5 litres) can cover at a specific thickness, usually measured in microns.
The formula for calculating coverage is:
Coverage = (Paint volume in litres/Spreading rate) x 100
For example, if the spreading rate of a particular paint is 10 square meters per litre, then 5 litres of that paint would cover 50 square meters (5/10 x 100 = 50).
However, it is essential to note that this is only an estimate and the actual coverage may vary depending on factors such as the type of surface, texture, and porosity. Also, thicker application or multiple coats of paint may reduce the coverage area.
In general, smoother and less porous surfaces, such as drywall or plaster, require less paint compared to rough and porous surfaces like bricks or concrete.
Moreover, different types of paint may have different spreading rates. For instance, a high-quality paint with a higher concentration of pigments may have a lower spreading rate compared to a cheaper paint with less pigment concentration.
Therefore, it is always important to refer to the manufacturer’s guidelines and conduct a test patch before painting to get a more accurate estimate of the coverage area of the paint. This will help ensure that you have enough paint for the desired area and avoid any wastage.
## 5 litre of paint covers how many square feet
The coverage area of 5 litres of paint can vary greatly depending on factors such as the brand, color, and type of paint being used, as well as the surface being painted. Typically, the average coverage area of 5 litres of paint ranges from 400 to 500 square feet, assuming a single coat application on a smooth and well-primed surface.
However, it is important to note that this is just an estimate and the actual coverage area may differ. For example, a high-quality paint may have a better coverage rate compared to a cheaper or lower-quality paint. The application method and tools used can also affect the coverage area. Using a roller or brush may result in a slightly higher coverage compared to using a sprayer.
Also, it is important to consider the porosity and texture of the surface being painted. A more porous surface may require more paint to achieve the desired coverage, while a smoother surface may require less paint.
To determine the exact coverage area of 5 litres of paint, it is best to refer to the manufacturer’s instructions or perform a test patch on the surface to be painted. This will give a more accurate estimate of the paint’s coverage rate and help in determining the amount of paint needed for a specific project.
In conclusion, 5 litres of paint can cover an average of 400-500 square feet, but the actual coverage area may vary depending on various factors. It is important to carefully consider these factors when purchasing paint and always refer to the manufacturer’s instructions for the best results.
## 10 litre of paint covers how many square meter
The coverage of a 10 litre can of paint depends on various factors such as the type of paint, surface porosity, and method of application. In general, a 10 litre can of paint can cover an area of approximately 80-100 square meters.
For example, if we assume that the paint has a spread rate of 8-10 square meters per litre, then a 10 litre can will cover an area of 80-100 square meters. This is a rough estimate and can vary based on the above-mentioned factors.
If the surface is smooth and non-porous, the coverage may increase, while a rough or porous surface may require more paint to cover the same area. Similarly, the method of application, whether by brush, roller, or spray can also affect the coverage.
To accurately determine the coverage of a 10 litre can of paint, it is best to refer to the product label or consult with the manufacturer. They can provide specific information on the coverage based on their product and its application.
In summary, a 10 litre can of paint can cover approximately 80-100 square meters, but actual coverage may vary depending on the type of paint, surface condition, and application method. It is important to follow the manufacturer’s guidelines for best results and to avoid any wastage of paint.
## 10 litre of paint covers how many square feet
The amount of square footage that 10 litres of paint can cover varies depending on the type of paint, the surface being painted, and the application method. In general, 10 litres of paint can cover approximately 800-1,200 square feet of surface area.
For example, if you are using a high-quality latex paint with a roller to paint interior walls with a smooth surface, 10 litres of paint can cover up to 1,200 square feet. However, if you are using a lower quality paint or painting a textured surface, the coverage may be less and you may need more paint to achieve the same coverage.
If you are using a high-quality paint for exterior painting on a smooth surface, 10 litres of paint can cover approximately 800 square feet. However, if the surface is rough or porous, you may need more paint to achieve the desired coverage.
It is important to note that the coverage may also vary depending on the number of coats applied. For example, if you are using a light-colored paint to cover a darker surface, you may need multiple coats to achieve optimal coverage and color.
Additionally, the type of paint application method can also affect the coverage. Using a roller or brush will typically result in more coverage compared to using a spray gun.
It is always recommended to refer to the manufacturer’s instructions for accurate coverage information specific to the paint being used. It is also a good idea to purchase a bit more paint than what is estimated to ensure you have enough to complete the project.
In summary, 10 litres of paint can cover approximately 800-1,200 square feet, depending on factors such as paint type, surface, application method, and number of coats. It is always best to consult with a professional or refer to the manufacturer’s instructions for specific coverage information for your project.
## Conclusion
In conclusion, understanding the various factors that go into calculating the amount of paint needed for a project can save time, money, and ensure a successful outcome. By carefully measuring the surface area, considering the type of paint and its coverage rate, as well as factoring in multiple coats, one can accurately estimate the amount of paint needed. Additionally, taking into account any special considerations, such as porous surfaces or intricate designs, can further refine the calculation. Whether it’s a small touch-up or a large renovation, having a thorough understanding of paint calculation can greatly benefit any project. So the next time you embark on a painting project, remember to carefully calculate your paint needs for a smooth and efficient process. | 7,313 | 36,327 | {"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} | 3.15625 | 3 | CC-MAIN-2024-10 | longest | en | 0.933957 |
https://www.studypool.com/discuss/329183/merchandise-was-purchased-on-account-from-jacob-s-distributors-on-may-17-the-pu?free | 1,481,171,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542412.97/warc/CC-MAIN-20161202170902-00509-ip-10-31-129-80.ec2.internal.warc.gz | 1,016,797,965 | 14,017 | ##### Merchandise was purchased on account from Jacob's Distributors on May 17. The pu
Accounting Tutor: None Selected Time limit: 1 Day
Merchandise was purchased on account from Jacob's Distributors on May 17. The purchase price was \$2,000, less a 10% trade discount and credit terms of 2/10, n/30.
Jan 4th, 2015
Money to be paid before the credit term expires;
using proportion 2/10=n/30
hence n=2*30/10
n=6 days
10/100*2000
0.1*2000
=\$200
Actual price to be paid=Purchase Price-Cash discount
2000-200
=\$1,800
Jan 4th, 2015
...
Jan 4th, 2015
...
Jan 4th, 2015
Dec 8th, 2016
check_circle | 204 | 607 | {"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} | 2.78125 | 3 | CC-MAIN-2016-50 | longest | en | 0.886114 |
https://tbc-python.fossee.in/convert-notebook/Diffusion:_Mass_Transfer_In_Fluid_Systems_by__E._L._Cussler/Chapter_20_Heat_Transfer_1.ipynb | 1,702,206,887,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00205.warc.gz | 602,463,130 | 37,416 | # Chapter 20 Heat Transfer¶
## Example 20_1_1 pgno:573¶
In [3]:
#initialization of variables
z = 1.3#cm
t = 180. #seconds
#calculations
k = ((T-T0)/(Tinf-T0))
alpha = (1/(4*t))*((z/k)**2)/2#cm**2/sec
#Results
print"The thermal diffusivity is ",round(alpha,4)
The thermal diffusivity is 0.0031
## Example 20_3_1 pgno:581¶
In [4]:
#initialization of variables
Q = 18. # m**3/hr
z = 2.80 #m
T = 140.#C
T1 = 240. #C
T2 = 20. #C
p= 900. #kg/m**3
Cp = 2. # W/kg-K
d = 0.05#m
from math import pi,log
#Calculations
A = pi*(d**2)/4.
v = Q*(1/(3600*40))/(A)
U = (v*p*Cp*d/(4*z))*(log((T1-T2)/(T1-T)))+0.4#W/m**2-K
DeltaT = ((T1-T2)+(T1-T))/2#C
q = (Q*(1/(3600*40))*p*Cp/(pi*d*z))*(T-T2)#W/m**2-K
U1 = q/DeltaT+0.38#W/m**2-K
#Results
print"The overall heat transfer co efficient based on local temp difference is W/m**2-K",U
print"\nThe overall heat transfer co efficient based on average temp difference is W/m**2-K",U1
The overall heat transfer co efficient based on local temp difference is W/m**2-K 0.4
The overall heat transfer co efficient based on average temp difference is W/m**2-K 0.38
## Example 20_3_2 pgno:582¶
In [7]:
#initialization of variables
T = 32. #F
T0 = 10.#F
Tinf= 800 #F
U = 3.6 #Btu/hr-ft**2-F
A = 27. #ft**2
d = 8.31 #lb/gal
V = 100. #gal
Cv = 1.#Btu/lb-F
from math import log
#Calculations
t = (-log((T-T0)/(Tinf-T0)))*d*V*Cv/(U*A)/3#hr
#Results
print"The time we can wait before the water in the tank starts to freeze is hr",round(t)
The time we can wait before the water in the tank starts to freeze is hr 10.0
## Example 20_3_3 pgno:583¶
In [1]:
#initialization of variables
#Given q = h*DeltaT and 0.6q = (1/(1/h)+10/12*0.03)*delta T , divide both to get
l = 10./12. #ft
k = 0.03 #Btu/hr-ft-F
#Calculations
l2 = 2#feet
k2 = 0.03 #Btu/hr-ft-F
h = ((1/0.6)-1)*k/l #Btu/hr-ft**2-F
U = 1/((1/h)+(l2/k2))#Btu/hr-ft**2-F
Savings = U*100/h
#Results
print"The savings due to insulation is about percent",round(Savings)
The savings due to insulation is about percent 38.0
## Example 20_4_1 pgno:588¶
In [2]:
#initialization of variables
T = 673 # Kelvin
M = 28
sigma = 3.80 # angstroms
omega = 0.87
d1 = 0.05 #m
v1 = 17 #m/sec
Mu1 = 3.3*10**-5 # kg/m-sec
p1 = 5.1*10**-1 # kg/m**3
Cp1 = 1100 # J/kg-K
k2 = 42 # W/m-K
l2 = 3*10**-3 #m
d3 = 0.044 #m
v3 = 270 #m/sec
p3 = 870 #kg/m**3
Mu3 = 5.3*10**-4 # kg/m-sec
Cp3 = 1700# J/kg-K
k3 = 0.15 #W/m-K
#Calculations
kincal = (1.99*10**-4)*((T/M)**0.5)/((sigma**2)*omega)#W/m**2-K
k = kincal*4.2*10**2# k in W/m-K
h1 = 0.33*(k/d1)*((d1*v1*p1/Mu1)**0.6)*((Mu1*Cp1/k)**0.3)#W/m**2-K
h2 = k2/l2 #W/m**2-K
h3 = 0.027*(k3/d3)*((d3*v3*p3/Mu3)**0.8)*((Mu3*Cp3/k3)**0.33)#W/m**2-K
U = 1/((1/h1)+(1/h2)+(1/h3))#W/m**2-K
#Results
print"The overall heat transfer co efficient is W/m**2-K",round(U)
The overall heat transfer co efficient is W/m**2-K 65.0
## Example 20_4_2 pgno:589¶
In [10]:
#initialization of variables
#For window with two panes 3 cm apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 3. #cm
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
DeltaT = 30. # Kelvin
T = 278. # Kelvin
L = 100. # cm
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.42765#for two pane in x*10**-4 cal/cm**2-sec-K
print"The heat transfer co efficent for two panes is x10**-4 cal/cm**2-sec-K",h
#For window with three panes 1.5 cm each apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 1.5#cm
DeltaT = 15. # Kelvin
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.30765 #for two pane in x*10**-4 cal/cm**2-sec-K
print"\nThe heat transfer co efficent for three panes is x10**-4 cal/cm**2-sec-K",round(h/2,2)#Because there are two gaps
The heat transfer co efficent for two panes is x10**-4 cal/cm**2-sec-K 0.44
The heat transfer co efficent for three panes is x10**-4 cal/cm**2-sec-K 0.17 | 1,727 | 3,867 | {"found_math": true, "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": 1, "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} | 3.921875 | 4 | CC-MAIN-2023-50 | latest | en | 0.652925 |
https://www.programminghomeworkhelp.com/Assembly-language-program-to-get-hex-number-and-display-count-of-bits-set/ | 1,675,324,524,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00038.warc.gz | 952,867,607 | 9,265 | # Assembly Language Program to Get Hex Number and Display Count of Bits Set Assignment Solution.
## Instructions
Objective
Write a program to get hex number and display count of bits set in assembly language.
## Requirements and Specifications
Description:
Write an x86 assembler program that does the following:
1. Gets a 32-bit number (up to 8 hex digits) from the console
2. Determines the frequencies for each “power of two” that contributes to every bit set in that number.
• The powers of two to 32 bits positions are: 1, 2, 4, 8, 16 and 32
1. Output the frequency counts to the console
This program must include at least one procedure beyond MAIN.
Example:
For a hex input of 0F (0x1111b), only the low 4 bits of a 32 bit word (bits 1, 2, 3 and 4) are set. To set this sequence of bits, you need to use 1 twice, 2 twice and 4 once. The below shows the particular power of two sums for each bit position set.
• 1 = 1,
• 2 = 2,
• 3 = 1 + 2,
• 4 = 4
Screenshots of output
Source Code
```TITLE CPSC 232 - Program #4 INCLUDE IRVINE32.INC .DATA TITLEMSG BYTE "CPSC 232 - Program #4", 0DH, 0AH, 0DH, 0AH, 0 MESSAGE BYTE "Input a 32-bit word (8 hex digits) >> ", 0 OUTPUT1 BYTE 0DH, 0AH, "1's frequency = ", 0 OUTPUT2 BYTE 0DH, 0AH, "2's frequency = ", 0 OUTPUT4 BYTE 0DH, 0AH, "4's frequency = ", 0 OUTPUT8 BYTE 0DH, 0AH, "8's frequency = ", 0 OUTPUT16 BYTE 0DH, 0AH, "16's frequency = ", 0 OUTPUT32 BYTE 0DH, 0AH, "32's frequency = ", 0 FREQ1 DWORD 0 FREQ2 DWORD 0 FREQ4 DWORD 0 FREQ8 DWORD 0 FREQ16 DWORD 0 FREQ32 DWORD 0 .CODE FREQUENCIES PROC mov EBX, 1 ; Bit to test LOOP1: cmp EBX, 32 ; If the bit is above the 32 jg ENDL1 ; end the loop shr EAX, 1 ; shift input jnc SKIP1 ; If not set, skip mov EDX, EBX ; Copy bit position to EDX shr EDX, 1 ; test first bit adc DWORD PTR [FREQ1], 0 ; increment frequency 1 if the bit was set shr EDX, 1 ; test second bit adc DWORD PTR [FREQ2], 0 ; increment frequency 2 if the bit was set shr EDX, 1 ; test first bit adc DWORD PTR [FREQ4], 0 ; increment frequency 4 if the bit was set shr EDX, 1 ; test first bit adc DWORD PTR [FREQ8], 0 ; increment frequency 8 if the bit was set shr EDX, 1 ; test first bit adc DWORD PTR [FREQ16], 0 ; increment frequency 16 if the bit was set shr EDX, 1 ; test first bit adc DWORD PTR [FREQ32], 0 ; increment frequency 32 if the bit was set SKIP1: inc EBX ; Increment bit position jmp LOOP1 ; Repeat the loop ENDL1: ret ; Return to main FREQUENCIES ENDP MAIN PROC lea EDX, TITLEMSG ; Load addess of title in EDX call WriteString ; Write string on console lea EDX, MESSAGE ; Load addess of message in EDX call WriteString ; Write string on console call ReadHex ; Read number in hexadecimal call FREQUENCIES ; call frequencies subroutine with EAX = read number lea EDX, OUTPUT1 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ1] ; load frequency value call WriteDec ; Write value on console lea EDX, OUTPUT2 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ2] ; load frequency value call WriteDec ; Write value on console lea EDX, OUTPUT4 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ4] ; load frequency value call WriteDec ; Write value on console lea EDX, OUTPUT8 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ8] ; load frequency value call WriteDec ; Write value on console lea EDX, OUTPUT16 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ16] ; load frequency value call WriteDec ; Write value on console lea EDX, OUTPUT32 ; Load addess of output message in EDX call WriteString ; Write output string on console mov EAX, [FREQ32] ; load frequency value call WriteDec ; Write value on console call CRLF ; Print CR/LF Exit ; Terminate program MAIN ENDP END MAIN``` | 1,203 | 4,028 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.787863 |
http://www.stata.com/statalist/archive/2009-07/msg01242.html | 1,444,367,225,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737913815.59/warc/CC-MAIN-20151001221833-00123-ip-10-137-6-227.ec2.internal.warc.gz | 937,738,785 | 3,782 | # AW: st: AW: How to calculate the bid-ask spread in this case?
From "Martin Weiss" To Subject AW: st: AW: How to calculate the bid-ask spread in this case? Date Thu, 30 Jul 2009 23:14:42 -0400
```<>
Ok, that makes the solution slightly more involved:
******
clear*
input BondID:mylabel str10(Date Time) Bid_Price Ask_Price , auto
AAA 20090729 090540 100.00 .
AAA 20090729 092307 100.05 .
AAA 20090729 093051 . 101.10
AAA 20090729 093523 . 101.20
AAA 20090729 093617 101.05 .
AAA 20090729 094521 . 101.20
AAA 20090729 094654 100.30 .
AAA 20090729 094929 . 100.70
AAA 20090729 100002 . 100.50
end
egen newdate=concat(Date Time)
gen double timeoftrade=clock(newdate, "YMDhms")
//get time difference
bys BondID (timeoftrade): /*
//have to check both ways
bys BondID (timeoftrade): /*
//get the minimum time diff
bys BondID (timeoftrade): /*
*/ egen mintime=min(timediff)
//get rid of auxiliaries
drop Date Time newdate
bys BondID (timeoftrade): /*
*/ egen desiredspread= /*
*/ max((timediff==mintime)* /*
list, noobs
******
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Im Auftrag von help me
Gesendet: Donnerstag, 30. Juli 2009 19:33
An: statalist@hsphsun2.harvard.edu
Betreff: Re: st: AW: How to calculate the bid-ask spread in this case?
Dear Martin,
Now I can have a measure of the bid-ask spread!
However, I am afraid I did not correctly explain my problem. The
actual data is more complicated although it has the same structure as
the hypothetical one that I provided.
Bond ID Date Time(HHMMSS) Bid_Price Ask_Price
AAA 20090729 090540 100.00
AAA 20090729 092307 100.05
AAA 20090729 093051 101.10
AAA 20090729 093523 101.20
AAA 20090729 093617 101.05
AAA 20090729 094521 101.20
AAA 20090729 094654 100.30
AAA 20090729 094929 100.70
AAA 20090729 100002 100.50
.
.
.
In this case I want to calculate the difference between the ask price
at 09:35:23 (101.20) and the bid price at 09:36:17(101.05) because
these bid and ask transactions take place at the closest time among
other pairs.
Do you have any idea how to do that?
Thank you.
JHS
On Thu, Jul 30, 2009 at 6:31 PM, Martin Weiss<martin.weiss1@gmx.de> wrote:
>
> <>
>
>
> This code is very specific to your example data, so if your problem is
more
> general, let the list know...
>
>
> ******
> clear*
>
> input str5 BondID Date Time Bid_Price Ask_Price
> AAA 20090729 090540 100.00 .
> AAA 20090729 092307 100.05 .
> AAA 20090729 093051 . 101.10
> AAA 20090729 093523 . 101.20
> end
>
> compress
> //make sure dataset sorted
> sort Date Time
> list, noobs
>
> collapse (lastnm) Bid_Price/*
> */ (firstnm) Ask_Price, by(BondID)
> l
> ******
>
>
> HTH
> Martin
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 1,098 | 3,286 | {"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} | 2.71875 | 3 | CC-MAIN-2015-40 | longest | en | 0.518806 |
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# If x is the product of the positive integers from 1 to 9, inclusive,
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If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
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25 May 2016, 04:27
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Question Stats:
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If x is the product of the positive integers from 1 to 9, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i3^k5^m7^p$$, then i + k + m + p =
A. 4
B. 7
C. 8
D. 11
E. 13
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
### Show Tags
25 May 2016, 04:54
x= 9!
= 9* 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= (3^2) * (2^3) * 7 * (2*3) * 5 * 2^2 * 3 * 2 * 1
= 2^7 * 3^4 * 5 * 7
= 2^i * 3^k * 5^m * 7^p
i+k+m+p = 7+4+1+1
= 13
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If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
### Show Tags
31 May 2016, 15:27
1
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$
Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1
Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$
Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$
Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$
Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$
i + k + m + p = 6 + 4 + 2 + 1 = 13
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Status: Greatness begins beyond your comfort zone
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Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
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31 May 2016, 20:42
1
fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$
Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1
Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$
Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$
Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$
Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$
i + k + m + p = 6 + 4 + 2 + 1 = 13
Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!
$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$
$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$
$$\frac{9}{5} = 1$$
$$\frac{9}{7} = 1$$
Finding the powers of a prime number p, in the n!
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
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01 Jun 2016, 02:56
Skywalker18 wrote:
fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$
Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1
Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$
Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$
Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$
Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$
i + k + m + p = 6 + 4 + 2 + 1 = 13
Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!
$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$
$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$
$$\frac{9}{5} = 1$$
$$\frac{9}{7} = 1$$
Finding the powers of a prime number p, in the n!
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
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01 Jun 2016, 03:03
fantaisie wrote:
Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!
Hi,
rounding of is always done to the lower integer...
Reason --
say we are looking for number of 5 in 9! -
9/5 + 9/5^2 = 1.8 + 0.3 = 1+0 = 1..
so 1 in 1.8 talks of integer 5 in the product....
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink] 09 Sep 2017, 05:06
Display posts from previous: Sort by | 2,566 | 7,233 | {"found_math": true, "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": 1, "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} | 3.953125 | 4 | CC-MAIN-2019-18 | latest | en | 0.840249 |
https://www.scribd.com/document/38037356/Global-Geography-Worksheets-4-Latitude | 1,527,372,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867904.94/warc/CC-MAIN-20180526210057-20180526230057-00468.warc.gz | 814,236,444 | 25,515 | # Name____________________________
Lines of Latitude
Lines of Latitude
Lines of latitude on a globe are called parallels. They run east and west. The
equator is at 0° latitude. Use the map below to answer the questions.
North Pole
Equator
15°S
1 . 0° latitude is called the __________
2. Lines of latitude are caUed ______ _____
3. Parallels run which directions? ______and ______
4. The latitude of the North Pole is _________
5. Which parallel runs through Florida? _________
6. What is located at 90
0
S latitude? _________
7. Which parallel runs through Canada? __________
8. Lines of latitude above the equator are which direction?_____
9. Below the equator, the parallels are which direction? ______
286
Name_____________________________
Lines of Latitude
Lateral Movement
Parallels measure the distance north or south from the equator. Zero degrees
latitude (0°) is at the equator. Half of the parallels are north of the equator
and half are south of it. The lines do not meet.
Equator 0°
1 . What is the symbol for degrees? ______
2. Latitude lines run _____ and _______
3. Latitude lines are called ___________
4. Give the latitude of the equator.
5. The parallels above the equator are which direction? ________
6. The parallels below the equator are which direction? ______
7. Color the equator parallel orange.
8. Color 15°N and 15°S green.
9. Color 30
0
N and 30
0
S blue.
10. Color 45°N and 45°S red.
11. Color 60
0
N and 60
0
S purple.
287
1------------1
Parallel of Latitude
I\Jame_________________
Lines ot Latitude
Imaginary Lines
Directions: Answer the questions below using these maps.
North Pole North Pole
90
0
N
/ - - = - - - - : : : - : - " " ' : " " ' : ' - : : : - - : - - ~ 45°N
Parallel of Latitude
Equator
/----=------=::.ort--------\ 15°N
900S
900S
South Pole South Pole
1. The ____________------_______ is 0°latitude.
2. The North Pole is degrees north latitude.
4. Lines north and south of the equator are coiled _____________
5. The ________________________ is 90
0
S latitude.
6. Which line is closer to the equator-300N or 15°S? _____________
7. Which is closer to the South Pole-45°S or 30
0
S? _____________
8. At what degree is the South Pole? __________________
9. If you wanted to find a city located at 45°N, would you look above or below the
equator? _________ _
10. Which continent on the map is entirely north of the equator? ________
11. South America lies between the parallels of latitude _______oN and 60
0
S.
12. The equator runs through the northern part of the continent of _______
13. Color all the land north of the equator red.
14. Color all of the land south of the equator green.
288
Name_______________________________
Unes of Latitude
What's My Line?
There are several important lines of latitude on the globe which have special
names.
Directions: Use a mapI globe or other resource to identify the special lines
on the illustration of the globe below.
Name the imaginary line that . ..
passes through Mexico.
is 0° latitude.
passes through Alaska. _____________
is 0° longitude. ____________
divides the Northern and Southern hemispheres. _____________
passes through Botswana. __________
289
Name______________________________ ___
Lines of Latitude
Across the U.S.A.
- -
Des Moines
Denver
Dallas.
Directions: Use the map above to answer these questions.
1. Denver and New York are close to which parallel? _______
2. Which two cities are between 45°N and 50
0
N? _________
3. Los Angeles and Memphis are near which parallel?___________
4. Tallahassee is closest to which parallel? ____________
5. St. Louis is between which parallels? _______ and _____
6. Which city is farthest north? ___________ lt is between
which parallels? ________________ and _______
7. Which city is farthest south? ___________ lt is between
which parallels?
___________________ and
8. San Francisco is halfway between ______ and ______
290
------------------------------------------
Name
Lines of Latitude
Latitude in North America.
Use with page 292,
---__ _ Montreal
-----, -\
I "
Chicago
Des Moines.
St. Louis.
Memphis.
Pacific Ocean
Dallas. Tallahassee.
Atlantic Ocean 200N
--,
.. I ""'\
, New Orleans
Brownsville
MexiCo\
291
Name______________________________
Lines of Latitude
Latitude in North America
Directions: Use the map on page 291 to answer the questions below,
1, The Arctic Circle is located between 60
0
N and ____°1\J,
2, Is Chicago closer to 400N or 50
0
N?___________
3, Name the three United States cities located between 200N and 30
0
N,
4, New York is closest to the __________ parallel of latitude,
5, Name the eight United States cities located between 30°Nand 40
0
N,
6, The Ocean is on the eastern side of the United States,
7, is the country south of the United States,
8, Canada is the country of the United States,
9, On the west, the United States is bordered by the Ocean,
10, Montreal is in the country of __________
11 , Seattle is located closest to the ______________ parallel of latitude,
12, What part of the United States does the Arctic Circle cross? _______
13, Memphis lis located between the parallel and the ____ parallel.
14, Is Daillas north or south of the 30
0
N parallel of latitude? ___________
15, Name the four United States cities located between 400N and 50
0
N,
16, Denver is closest to the ____parallel of latitude,
17, San Francisco is located near oN,
18, Does the Arctic Circle pass through Greenland? ___________
19, Which paralle'l of latitude on the map goes through Florida? _______
20, Guadalajara is located in what country? _____________________
292
------------------------------------
Name
Lines of Latitude
Parallels Help With Location
San Francisc;o-_..lA
San Bernardin
• GreQtfO/l\$
Santa Fe
35°N
tL-' New Mexico
:
• Las c ruces:
1. Billings, Montana is which direction from the 46°N parallel?
2. Pueblo, Colorado is almost directly on the parallel of latitude.
3. The boundary between Oregon and Cal1ifornia is formed by the __ parallel.
4. The state of Wyoming is located between 41 oN parallel and parallel.
5. Name three cities in Idaho south of the 45°N parallel of latitude.
6. Which of these cities is south of the 35°N parallel-Flagstaff, Arizona or
RoswelL New Mexico? ____________________________ __________ _
7. Name the three California cities located between the 35°N and 38°N parallels.
8. All of the cities shown in Washington are between the parallels of
___________ and ___________
9. Which two Nevada cities are north of the 38°N parallel? _______
and ________________ _
10. Klamath Falls, Oregon is almost directly on the _____ parallel.
293 | 1,698 | 6,582 | {"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} | 2.703125 | 3 | CC-MAIN-2018-22 | latest | en | 0.756957 |
https://www.physicsforums.com/threads/question-regarding-half-life.50082/ | 1,516,203,788,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00537.warc.gz | 978,744,083 | 16,319 | # Question regarding half-life
1. Oct 28, 2004
### misogynisticfeminist
I've got a question regarding half-life. Half-life is the time taken when the mass or countrate emitted of a radioactive sample drops to half.
But if say, if a sample of 400 particular radioactive nuclides which go through radioactive decay. Won't the count rate reach 0 eventually, after decaying into a more stable state? (assuming no background count). Because if we plot a normal half-life curve, it would be exponential and would never reach 0 at all.
And if these particular nuclides emit gamma radiation, its mass wouldn't drop at all because the photons are massless, right?
Thanks.
Last edited: Oct 28, 2004
2. Oct 28, 2004
### da_willem
A 'normal half-life curve' is probably a plot of the expectation value of the number of particles in a certain state or a fraction. Nuclear decay is a probabilistic in nature, so all we can give is an expectation value or some probablility density.
If you use a half-life curve on your example. After some time the curve has decayed to almost zero indicating the chance to find a aprticle after that duration is very small, so probably there will be no particle left.
Note that a half life curve does not indicate the mass left, just the number of particles of a certain type. Or the mass of the particles of that certain type.
Note also that if a substance emits radiation it's mass decreases, because a photon has energy, wich also contributes to the mass of the object.
3. Oct 28, 2004
### NEOclassic
Hi miso,
IMO, whenever any nucleus emits a photonic particle, it is invariably coincidental to a simultaneous emission of a mass-bearing particle. E.g., a U-238 nucleus emits its most weakly attached alpha (of the 8 that it received when its parent, Pu-242, decayed), is accompanied by a ~45 KeV photon; the 4.190 MeV expulsion energy of the alpha (that is primarily controlled by electrostatic repulsion) is less than that which is available, 4.235MeV. Of course conservation of energy necessarily prevails. In order to cite details of photon emission that accompanies Beta (nucleus bound electron) emission, let me continue.
Let's talk about what happens to the new nucleus after U-238 loses its alpha. The reality is that it has become the nucleus of Thorium-234 that contains only 7 resident alphas; It is also quite nervous because the nucleus isn't happy because it has two too many neutrons. The only way to correct that problem is for a neutron to spontaneously emit an energetic electron. The Th-234 nucleus emits a beta thus becoming Protoactinium-234; the beta energy, depending on which one of 140 neutrons emits, is a spectrum of energies accompanied by a spectrum of photons. Typically, it might be suspected that high energy betas are accompanied by low energy photons; e.g., a 190 KeV beta coupled with a 29 KeV photon or a 100 KeV beta with a 91 KeV photon.
The Pa-234 decays in similar manner, to U-234; the betas are as much as 280 KeV and the gammas are as much as 1.68 MeV.
The U-234 is again an alpha emitter, and in similar manner repeats the alpha-beta-beta sequence until the nucleus has expelled all its available alphas thus becoming Pb-206. There are three other series that are represented by U-235, U-236 and U-233 as well as U-234 explained above. Thanks for your audience and your patience. Jim
4. Oct 29, 2004
### Mk
Since matter is but a highly concentrated form of energy, the missing mass has actually been converted to energy, in your case, a gamma ray. In spontaneous emission, the electron loses mass, as decending to a lower energy and the lost mass was actually mass converted to energy, and released as a gamma ray.
5. Nov 19, 2004
### transit442003
an exponential graph is just a prediction of what the mass would be after radioactive decay just bacause the graph does not reach zero doesn't mean that the particle won't become stable. it is just of those things that seem to happen, that is why we measure down to only one instead of zero.
day without sunshine.......................is well......................like..............night
Last edited: Nov 19, 2004 | 965 | 4,151 | {"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} | 2.515625 | 3 | CC-MAIN-2018-05 | longest | en | 0.91442 |
https://www.solutioninn.com/consider-the-following-hypothesis-test-h0-p-75-ha-p- | 1,713,148,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00037.warc.gz | 938,584,332 | 15,998 | # Consider the following hypothesis test: H 0 : p .75 H a : p < .75
## Question:
Consider the following hypothesis test:
H0: p ≥ .75
Ha: p < .75
A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. use a = .05.
a. p̅ = .68
b. p̅ = .72
c. p̅ = .70
d. p̅ = .77
Fantastic news! We've Found the answer you've been seeking!
## Step by Step Answer:
Related Book For
## Statistics For Business & Economics
ISBN: 9781305585317
13th Edition
Authors: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
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https://iuee.eu/en/a-rectangle-bedroom-floor-has-an-area-of-100-square-feet-and-a-length-of-10-feet-what-is-the-perime.52259.html | 1,642,686,839,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00713.warc.gz | 391,249,832 | 11,506 | gleefan472
9
# A rectangle bedroom floor has an area of 100 square feet and a length of 10 feet. What is the perimeter of the floor?
Shod
Since the area of the floor is 100 and one side is 10, then we should know that the other side is 10 because, 10 * 10 = 100 So, since two sides are 10, than the other sides must be 10 too. That gives 4 sides that all equal 10. 10+10+10+10= 40 So, the perimeter is 40.
Vonholt919
Do 100 divided by 10 and that is the answer to your question its 10 by the way | 154 | 501 | {"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} | 4.28125 | 4 | CC-MAIN-2022-05 | latest | en | 0.943675 |
http://forum.arduino.cc/index.php?topic=125550.msg945080 | 1,524,236,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00537.warc.gz | 111,821,600 | 9,083 | Go Down
Topic: 1 Microsecond pulses (Read 11911 times)previous topic - next topic
Lithanial
Oct 04, 2012, 02:23 am
Hi,
I am new here and do not know much about programming, YET.
I am looking to control a gate two switch a HV supply to a transducer/receiver.
I need to control the gate with very short 5V pulses. Approximately 1.2microseconds. I hooked my arduino to a scope using the microsecond command, but the shortest pulse duration I can get is about 4 microseconds.
Does anyone know how to reduce this pulse duration down to 1.2 microseconds.
For my program:
Loop
30 microsecond (Low)
1.2 microsecond (High)
About a 10 second Delay
End Loop
If anyone can help me out with this short program I would very much appreciate it.
Thanks.
johnwasser
#1
Oct 04, 2012, 04:55 am
All the output pins are grouped in PORTs: PORTB, PORTC: PORTD. Each port has a corresponding PIN register: PINB, PINC, PIND. If you write a 1 to a bit in the PIN register the output pin will flip:
Code: [Select]
`PIND = 0x10; // Toggle Pin 4 (Port D bit 4)PIND = 0x10; // Toggle Pin 4 back (Port D bit 4)`
This should cause a very short pulse on Pin 4.
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RIDDICK
#2
Oct 04, 2012, 10:03 amLast Edit: Oct 04, 2012, 08:35 pm by RIDDICK Reason: 1
Code: [Select]
`PIND = 0x10; // Toggle Pin 4 (Port D bit 4)PIND = 0x10; // Toggle Pin 4 back (Port D bit 4)`
2 clocks faster should b this pulse:
PORTD = 1<<4;
PORTD = 0;
EDIT: oops - its not faster... but it makes sure that the pulse goes into the right direction...
pylon
#3
Oct 04, 2012, 11:36 am
Maybe two direct port manipulations in full speed is too fast for the OP. If you need the 1.2us, you can use assembler commands to let the processor wait:
Code: [Select]
`__asm__("nop\n\t");`
Such a simple nop uses a bit more than 60ns.
Lithanial
#4
Oct 04, 2012, 12:06 pm
All the output pins are grouped in PORTs: PORTB, PORTC: PORTD. Each port has a corresponding PIN register: PINB, PINC, PIND. If you write a 1 to a bit in the PIN register the output pin will flip:
Code: [Select]
`PIND = 0x10; // Toggle Pin 4 (Port D bit 4)PIND = 0x10; // Toggle Pin 4 back (Port D bit 4)`
This should cause a very short pulse on Pin 4.
Thanks, I will try this today.
What exactly is this code saying? I like to know what I am doing so I can learn from this.
RIDDICK
#5
Oct 04, 2012, 01:59 pmLast Edit: Oct 04, 2012, 02:01 pm by RIDDICK Reason: 1
What exactly is this code saying? I like to know what I am doing so I can learn from this.
http://arduino.cc/en/Reference/PortManipulation
furthermore the datasheet of ur ATMEL mcu explains it in detail (search for "Toggling the Pin")...
pylon
#6
Oct 04, 2012, 02:03 pm
This code
Code: [Select]
`PIND = 0x10; // Toggle Pin 4 (Port D bit 4)PIND = 0x10; // Toggle Pin 4 back (Port D bit 4)`
from johnwasser uses the feature of the PINx register to toggle the value of the selected bits (XOR, documented only in the datasheet 13.2.2).
The usual code for this is direct port manipulation:
Code: [Select]
`PORTD |= 0x10; // Switch on pin 4PORTD &= 0xEF; // Switch off pin4`
It's almost the same as the code from RIDDICK but without affecting all other pins of the same port (pins 0-7).
Both code snippets just tries to switch on and off a single pin as fast as possible, probably faster than you want. That's why you may have to insert the NOP assembler statements to wait some time before toggling the pin back.
johnwasser
#7
Oct 04, 2012, 03:10 pm
Code: [Select]
`PIND = 0x10; // Toggle Pin 4 (Port D bit 4)PIND = 0x10; // Toggle Pin 4 back (Port D bit 4)`
2 clocks faster should b this pulse:
PORTD = 1<<4;
PORTD = 0;
Problem is that will also set all the OTHER pins in PORTD to 0/LOW. To avoid that you would normally use this code:
Code: [Select]
`PORTD |= 0x10; // Switch on pin 4PORTD &= 0xEF; // Switch off pin4`
but they require two read-modify-write cycles which I expect will make them slower than the PIN register.
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Lithanial
#8
Oct 04, 2012, 03:54 pm
I am getting a sawtooth pattern decay with this pulse?
pylon
#9
Oct 04, 2012, 05:16 pm
What did you expect? Rectangular pulses? That would mean you have absolutely no capacity on the line. How much of that is coming from your measurement equipment (a scope I guess)? Do you have the resolution to even see a rectangular signal if it was there?
MarkT
#10
Oct 04, 2012, 05:20 pm
I am getting a sawtooth pattern decay with this pulse?
You have set it as an OUTPUT?
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
MarkT
#11
Oct 04, 2012, 05:34 pm
This gives about 1.2us pulses, and disables interrupts during the pulse output to stop them occasionally stretching.
Disabling interrupts also means that any calls to digitalWrite () for pins in the D port in interrupt routines aren't broken by our direct port manipulation.
Code: [Select]
`void setup (){ pinMode (4, OUTPUT) ; digitalWrite (4, LOW) ; // or set to HIGH for low-going pulses.}void loop (){ pulse () ; delayMicroseconds (5) ;}void pulse (){ cli () ; PIND = 0x10 ; __asm__("nop\n\t"); __asm__("nop\n\t"); __asm__("nop\n\t"); __asm__("nop\n\t"); delayMicroseconds (1) ; PIND = 0x10 ; sei () ;}`
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
pylon
#12
Oct 04, 2012, 06:54 pm
@MarkT: This gives you 4.2us pulses, try it out. delayMicroseconds() waits at least 4us even if the parameter is less than 4.
But you can change the code to:
Code: [Select]
`void pulse (){ cli () ; PIND = 0x10 ; __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); PIND = 0x10 ; sei () ;}`
This should result in a pulse of 1.2us.
Lithanial
#13
Oct 04, 2012, 07:25 pm
Thanks,
I finally got it. There is a bit of overshooting at the top and bottom of the pulsing, but I can work with this. Thanks.
MarkT
#14
Oct 05, 2012, 12:30 amLast Edit: Oct 05, 2012, 12:38 am by MarkT Reason: 1
@MarkT: This gives you 4.2us pulses, try it out. delayMicroseconds() waits at least 4us even if the parameter is less than 4.
But you can change the code to:
Code: [Select]
`void pulse (){ cli () ; PIND = 0x10 ; __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); __asm__("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t"); PIND = 0x10 ; sei () ;}`
This should result in a pulse of 1.2us.
Try it out - I tested it with 100MHz scope and get 1.19us pulses. Am using the Arduino 1.0 software and avr-gcc 4.3.5
The code for delayMicroseconds has a comment saying that it is close to 1us when called with an argument of 1.
What version gives 4us then?
[edit: Arduino 1.0.1 and arduino-0022 give me 1.19us as well]
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Go Up | 2,344 | 7,068 | {"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} | 2.8125 | 3 | CC-MAIN-2018-17 | latest | en | 0.84616 |
https://www.numwords.com/words-to-number/en/5500000 | 1,606,753,121,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00484.warc.gz | 764,920,666 | 2,454 | NumWords.com
# How to write Five million five hundred thousand in numbers in English?
We can write Five million five hundred thousand equal to 5500000 in numbers in English
< Five million four hundred ninety-nine thousand nine hundred ninety-nine :||: Five million five hundred thousand one >
Eleven million = 11000000 = 5500000 × 2
Sixteen million five hundred thousand = 16500000 = 5500000 × 3
Twenty-two million = 22000000 = 5500000 × 4
Twenty-seven million five hundred thousand = 27500000 = 5500000 × 5
Thirty-three million = 33000000 = 5500000 × 6
Thirty-eight million five hundred thousand = 38500000 = 5500000 × 7
Forty-four million = 44000000 = 5500000 × 8
Forty-nine million five hundred thousand = 49500000 = 5500000 × 9
Fifty-five million = 55000000 = 5500000 × 10
Sixty million five hundred thousand = 60500000 = 5500000 × 11
Sixty-six million = 66000000 = 5500000 × 12
Seventy-one million five hundred thousand = 71500000 = 5500000 × 13
Seventy-seven million = 77000000 = 5500000 × 14
Eighty-two million five hundred thousand = 82500000 = 5500000 × 15
Eighty-eight million = 88000000 = 5500000 × 16
Ninety-three million five hundred thousand = 93500000 = 5500000 × 17
Ninety-nine million = 99000000 = 5500000 × 18
One hundred four million five hundred thousand = 104500000 = 5500000 × 19
One hundred ten million = 110000000 = 5500000 × 20
Sitemap | 428 | 1,365 | {"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} | 3.328125 | 3 | CC-MAIN-2020-50 | latest | en | 0.797065 |
https://minuteshours.com/8807-minutes-in-hours-and-minutes | 1,621,101,704,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00027.warc.gz | 415,726,587 | 5,037 | # 8807 minutes in hours and minutes
## Result
8807 minutes equals 146 hours and 47 minutes
You can also convert 8807 minutes to hours.
## Converter
Eight thousand eight hundred seven minutes is equal to one hundred forty-six hours and forty-seven minutes. | 59 | 260 | {"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} | 2.59375 | 3 | CC-MAIN-2021-21 | latest | en | 0.826292 |
https://www.khronos.org/message_boards/printthread.php?t=5335&pp=10&page=1 | 1,440,892,024,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064590.32/warc/CC-MAIN-20150827025424-00229-ip-10-171-96-226.ec2.internal.warc.gz | 922,779,759 | 2,768 | # A simple question
• 11-15-2007, 06:56 AM
tinuhp
A simple question
How to draw a quadratic curve using OpenVG....
I used the following code:
VGPath path;
VGfloat coords[] = { 100,100,
150,200,
200,100
};
path = vgCreatePath(VG_PATH_FORMAT_STANDARD, VG_PATH_DATATYPE_F, 1.0f, 0.0f, 0, 0, VG_PATH_CAPABILITY_ALL);
vgAppendPathData(path,1,segments,coords);
vgDrawPath( path, VG_STROKE_PATH );
vgDestroyPath( path );
I get nothing...
When I change the 2nd parameter to 2 ,I get some curve which I cannot understand...
When I am drawing only a single path why should I give 2 in the parameter.....
• 11-15-2007, 11:15 AM
Ivo Moravec
Re: A simple question
Quote:
Originally Posted by tinuhp
How to draw a quadratic curve using OpenVG....
I used the following code:
VGPath path;
VGfloat coords[] = { 100,100,
150,200,
200,100
};
path = vgCreatePath(VG_PATH_FORMAT_STANDARD, VG_PATH_DATATYPE_F, 1.0f, 0.0f, 0, 0, VG_PATH_CAPABILITY_ALL);
vgAppendPathData(path,1,segments,coords);
vgDrawPath( path, VG_STROKE_PATH );
vgDestroyPath( path );
I get nothing...
When I change the 2nd parameter to 2 ,I get some curve which I cannot understand...
When I am drawing only a single path why should I give 2 in the parameter.....
VG_QUAD_TO expects only 2 coordinates not 3. The first coordinate is given by the current position. The start position on a new path is always (0,0) so you were actually drawing a quad with the 3 points: (0,0),(100,100), and (150,200) instead of the one you were expecting. Thus you should add a VG_MOVE_TO_ABS to your command stream by:
VGfloat coords[] = { 100,100, 150,200,200,100};
vgAppendPathData(path,2,segments,coords);
• 11-15-2007, 10:31 PM
tinuhp
It worked...
And I have doubts yet...
1:We need three control points to draw a quad curve....
a starting point,an ending point and an internal control point
But QUAD_TO expects only 2 points ....
Actually what are these two points....
Are they internal control point and ending point .....
2:If so,when I use the following code to draw my quadratic curve at origin only,
I dont see anything...
VGPath path;
VGfloat coords[] = {150,200,200,100};
path = vgCreatePath(VG_PATH_FORMAT_STANDARD, VG_PATH_DATATYPE_F, 1.0f, 0.0f, 0, 0, VG_PATH_CAPABILITY_ALL);
vgAppendPathData(path,1,segments,coords);
vgDrawPath( path, VG_STROKE_PATH );
vgDestroyPath( path );
• 11-16-2007, 01:50 AM
muratmat
Quote:
Originally Posted by tinuhp
2: If so,when I use the following code to draw my quadratic curve at origin only, I dont see anything...
If you are using AmanithVG, there was an already known bug related to paths made of a single command without an initial move_to.
The next public release will contain the patch.
• 11-16-2007, 02:20 AM
tinuhp
Yes. I am using AmanithVG .... | 806 | 2,755 | {"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} | 2.71875 | 3 | CC-MAIN-2015-35 | latest | en | 0.726341 |
http://www.chegg.com/homework-help/the-practice-of-statistics-4th-edition-chapter-2-solutions-9781429245593 | 1,475,231,713,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662159.54/warc/CC-MAIN-20160924173742-00033-ip-10-143-35-109.ec2.internal.warc.gz | 373,620,871 | 17,225 | Solutions
The Practice of Statistics
The Practice of Statistics (4th Edition) View more editions Solutions for Chapter 2
• 1499 step-by-step solutions
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 3
a) From the above sorted list, only 5 (13, 13, 13, 15, 19) of the 20 students in the female distribution have pairs of shoes below 22. Therefore, the corresponding percentile is computed as follows:
Hence, the girl with 22 pairs of shoes is at the 25th percentile. In other words, 25% of the girl students have less than 22 pairs of shoes.
• Step 2 of 3
b) From the above sorted list, 17 of the 20 students in the male distribution have pairs of shoes below 22. Therefore, the corresponding percentile is computed as follows:
Hence, the boy with 22 pairs of shoes is at the 85th percentile. In other words, 85% of the male students have less than 22 pairs of shoes.
• Step 3 of 3
c) The boy is more unusual because only 15% of the boys have more than 22 pairs of shoes than he has, while the girl has a value that is more centered in the distribution. Only 25% of girls have fewer and 75% have as many or more.
Corresponding Textbook
The Practice of Statistics | 4th Edition
9781429245593ISBN-13: 142924559XISBN: Authors:
Alternate ISBN: 9781429262507, 9781429262620, 9781429262637, 9781429268370, 9781429273978, 9781429273985, 9781429276689, 9781429296885, 9781429298537, 9781429299008, 9781464100659, 9781464100802, 9781464102332, 9781464103247, 9781464134944, 9781464138898, 9781464138997, 9781464140075, 9781464143687, 9781464145360 | 518 | 1,731 | {"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} | 4 | 4 | CC-MAIN-2016-40 | latest | en | 0.820899 |
https://fred.stlouisfed.org/graph/?chart_type=line&s[1][id]=PPPTTLMXA618NUPN&s[1][range]=10yrs | 1,475,066,523,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661367.29/warc/CC-MAIN-20160924173741-00231-ip-10-143-35-109.ec2.internal.warc.gz | 886,682,346 | 21,188 | Purchasing Power Parity over GDP for Mexico (PPPTTLMXA618NUPN) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
2010: 8.63524
Updated: Aug 31, 2012
Units:
National Currency Units per US Dollar,
Frequency:
Annual
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Purchasing Power Parity over GDP for Mexico, National Currency Units per US Dollar, Not Seasonally Adjusted (PPPTTLMXA618NUPN)
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
Source Indicator: ppp
Purchasing Power Parity over GDP for Mexico
Select a date that will equal 100 for your custom index:
to
Customize data:
Write a custom formula to transform one or more series or combine two or more series.
You can begin by adding a series to combine with your existing series.
Now create a custom formula to combine or transform the series.
Need help? []
Finally, you can change the units of your new series.
Select a date that will equal 100 for your custom index:
FORMAT GRAPH
Log scale:
NOTES
Source: University of Pennsylvania
Release: Penn World Table 7.1
Notes:
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
Source Indicator: ppp
Suggested Citation:
University of Pennsylvania, Purchasing Power Parity over GDP for Mexico [PPPTTLMXA618NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/PPPTTLMXA618NUPN, September 28, 2016.
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Retrieving data.
Updating graph. | 577 | 2,298 | {"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} | 3.28125 | 3 | CC-MAIN-2016-40 | latest | en | 0.836509 |
http://r.789695.n4.nabble.com/ggplot-2-Histogram-with-bell-curve-td3580359.html | 1,368,965,745,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697442043/warc/CC-MAIN-20130516094402-00098-ip-10-60-113-184.ec2.internal.warc.gz | 213,446,172 | 12,108 | # ggplot 2: Histogram with bell curve?
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## ggplot 2: Histogram with bell curve?
I am learning ggplot2 commands specifically qplot for the time being and I have figured out how to create histograms and normal density curves but I am not sure how to add a normal bell curve or other dist. as well on top of a histogram. Here are the two graphs that I created. ## Histogram t<-rnorm(500) w<-qplot(t, main="Normal Random Sample", fill=I("blue"), colour=I("black"), geom="histogram") w ##Density Curve t<-rnorm(500) r<-qplot(t, main="Normal Random Sample", colour=I("black"), geom="density", adjust=4) r
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## Re: ggplot 2: Histogram with bell curve?
Hi Don't use t as var names, because t is also a function (transpose). This code should work... set.seed(1) T <- rnorm(500) qplot(T, geom = "blank") + geom_histogram(aes(y = ..density..), colour = "black", fill = "blue") + stat_density(geom = "line", colour = "red")
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## Re: ggplot 2: Histogram with bell curve?
> -----Original Message----- > From: [hidden email] > [mailto:[hidden email]] On Behalf Of dicko ahmadou > Sent: Tuesday, June 07, 2011 11:41 AM > To: [hidden email] > Subject: Re: [R] ggplot 2: Histogram with bell curve? > > Hi > > Don't use t as var names, because t is also a function (transpose). > This code should work... > set.seed(1) > T <- rnorm(500) I think that it is more dangerous to use 'T' than 't' since 'T' is a built-in data name (with value TRUE) and I've seen it used more than 't' (probably because S and S+ regard it as a reserved word). I don't think that reusing built-in names as local names is as dangerous as some would make out but reusing built-in data names is probably more dangerous than using built-in function names for local data names (because function name lookup is done differently than general name lookup - the former will only look for functions). Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com > qplot(T, geom = "blank") + > geom_histogram(aes(y = ..density..), colour = "black", fill = > "blue") + > stat_density(geom = "line", colour = "red") > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/ggplot-2-Histogram-with-bell-curve-tp3580359p3580457.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: ggplot 2: Histogram with bell curve?
my bad....you are right I always use TRUE instead of T, so i forgot that by default T = TRUE in R. | 829 | 3,213 | {"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} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.885269 |
https://www.f1gmat.com/ad-hominem-gmat-critical-reasoning | 1,569,202,974,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575844.94/warc/CC-MAIN-20190923002147-20190923024147-00353.warc.gz | 837,957,160 | 8,901 | # Ad Hominem in GMAT Critical Reasoning
Once you get a feel of the various GMAT Critical Reasoning question types, it is important that you understand the fallacies in arguments that would be part of the answer choices. A common one seen in assumption and conclusion question type is the Fallacies resulting from Ad Hominem.
Ad Hominem is a Latin Word that translates to “To the Man” or more precisely “Against the man”. Politicians are experts at using Ad Hominem. In this type of fallacy, the author or source of the argument is attacked instead of attacking the argument per se.
If you don’t pay close attention, this fallacy might give you the impression that the argument of the author is targeted. This fallacy is a 2-step attack.
First step involve attacking the author’s character where his virtues and vices are questioned and exposed, then his actions and the circumstances under which those actions were performed are brought to the forefront. The first step acts as proof against any arguments that the author make even if the arguments have strong independent evidences to support it.
Conflict in Words and Action (of Group)
This Ad Hominem can be represented with an example about Politician A & B conversing about an anti-graft bill.
Politician A: I believe that a strong anti-graft bill is required to conquer corruption
Politician B: Your party members are involved in some of the biggest corruption scandals. Therefore, your position on corruption does not count.
Politician B assumes that just because Politician A is part of a party, the behavior of few of its corrupt members is the universal behavior endorsed and supported by the party, including Politician A.
Conflict in Words and Action (of Person)
We will cite an example where two activists are setting guidelines for the new anti-poaching law.
Activist A: We should set up strong rules to control poaching of non-endangered species.
Activist B: You are a non-vegetarian, and you actively take part in duck hunting. How will you convince the lawmakers when you violate poaching of non-endangered species?
Activist B assumes that Activist A has not quit poaching all together. Even if he goes for duck hunting, the argument for controls does not mean ban. For GMAT CR readers, this small difference might not be easy to spot, but with regular practice, you will master this common fallacy.
Changing Positions
Let us look at an example where two politicians are arguing about a new bill.
Politician A: For an inclusive democracy, we should have 33% reservation for women politicians
Politician B: Two years ago, you were against reservation for women; therefore, your conclusion is wrong. We don’t need women’s reservation for an inclusive democracy.
Politician B is attacking Politician A for changing his stand, and attacking the argument without going into the merits and demerits of the argument.
Mastering GMAT Critical Reasoning (2019 Edition)
Chapters
1) Introduction
2) 6 Step Strategy to solve GMAT Critical Reasoning Questions
3) How to overcome flawed thinking in GMAT Critical Reasoning?
4) 4 GMAT Critical Reasoning Fallacies
5) Generalization in GMAT Critical Reasoning
6) Inconsistencies in Arguments
7) Eliminate Out of Scope answer choices using Necessary and Sufficient Conditions
8) Ad Hominem in GMAT Critical Reasoning
9) Slippery Slope in GMAT Critical Reasoning
10) Affirming the Consequent – GMAT Critical Reasoning
11) How to Paraphrase GMAT Critical Reasoning Question
12) How to Answer Assumption Question Type
13) How to Answer Conclusion Question Type
14) How to Answer Inference Question Type
15) How to Answer Strengthen Question Type
16) How to Answer Weaken Question Type
17) How to Answer bold-faced and Summary Question Types
18) How to Answer Parallel Reasoning Questions
19) How to Answer the Fill in the Blanks Question
Question Bank
Question 1: 5G Technology (Inference)
Question 2: Water Purifier vs. Minerals (Fill in the Blanks)
Question 3: Opioid Abuse (Strengthens)
Question 4: Abe and Japan’s Economy (Inference)
Question 5: Indians and Pulse Import (Weakens)
Question 6: Retail Chains in Latin America (Assumption)
Question 7: American Tax Rates – Republican vs. Democrats (Inference)
Question 8: AI – China vs the US (Weakens)
Question 9: Phone Snooping (Strengthens)
Question 11: Appraisal-Tendency Framework (Inference)
Question 12: Meta-Analysis of Diet Trials (Weakens)
Question 13: Biases in AI (Strengthens)
Question 14: Stock Price and Effectiveness of Leadership (Inference)
Question 15: US Border Wall (Weakens)
Question 16: Driverless Car and Pollution (Assumption)
Question 17: Climate Change (Inference)
Question 18: Rent a Furniture (Weakens)
Question 19: Marathon Performance and Customized Shoes (Weakens)
Question 20: Guaranteed Basic Income (Assumption)
Question 21: Brexit (Infer)
Question 22: AB vs Traditional Hotels (Assumption)
Question 23: Tax Incentive and Job Creation (Weakens)
Question 24: Obesity and Sleeve Gastrectomy (Inference)
Question 25: Recruiting Executives (Weaken)
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• GMAT Preparation Tips | 1,218 | 5,306 | {"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} | 3.203125 | 3 | CC-MAIN-2019-39 | longest | en | 0.948288 |
https://www.mathselab.com/what-is-the-next-value-in-the-given-sequence/ | 1,716,639,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00850.warc.gz | 764,964,328 | 13,051 | What is the next value in the given sequence 11, 25,53, 109, 221, ?
What is the next value in the given sequence 11, 25,53, 109, 221, ?
Solution:
11×2+3=25
25×2+3=53
53×2+3=109
109×2+3=221
221×2+3=445
So, the answer will be 445. | 100 | 236 | {"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} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.804871 |
https://bitbucket.org/embray/pyqc/issues/5/handle-measurement-in-circuits | 1,441,364,449,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645340161.79/warc/CC-MAIN-20150827031540-00290-ip-10-171-96-226.ec2.internal.warc.gz | 835,802,885 | 16,172 | Issue #5 new
# Handle measurement in circuits
Erik Bray
repo owner created an issue
It needs to possible to handle projective measurement in a circuit--generally this would be implemented as a special case of the Z gate, but the exception that implements some extra bookkeeping to mark a measured qubit as a "classical" bit for the remainder of the circuit.
Evaluating circuits containing measurements should be simple enough, thanks to the principle of deferred measurement. But there could be some trickiness involved if, for example, a classical bit as used as input to a gate that expects a qubit and might try to put it into a superposition or otherwise modify it. After that point it should no longer be treated as classical. This case could be difficult to determine a priori such as when generating a circuit diagram. But it's also an unusual use case I think and not something to worry about too much.
1. reporter
Durward McDonell Regarding this issue, IIRC you were especially interested in having this working for stabilizer circuits. In that case measurement already works, but as you can see here nothing is done with the results of the measurement:
https://bitbucket.org/embray/pyqc/src/98605d9a6dda5eec39e3d9f7aabd631c0dd55dd7/pyqc/chp.pyx?at=master#cl-141
(I apologize for how underdocumented parts of the code are; when I have time I would like to go through and rectify that). But anyways the apply_measurement function returns a 2-tuple: The first element returned is True if the measurement was purely deterministic, and False if it was random. The second return value is the result of the measurement (0 or 1).
The problem is just finding a good API for returning those measurement results. Since I designed this to be a sort of algebraic system the result of some operation like circuit|state is a new state object, to which it doesn't make a whole lot of sense to attach the results of some measurements, though for lack of a better idea state objects could maybe have some "measurements" property that rides on it like a sidecar, and only has meaning if this was the output state from some circuit.
Another option is to overhaul the .evaluate() method to return a state and the measurements. Since there can be multiple measurements in a circuit there has to be some way of relating each measurement to where it came from in the circuit too, if I understand correctly. I suppose it could just be time-ordered. But I don't know what's actually useful here. You tell me :)
For non-stabilizer circuits measurement still needs to be implemented; as I think you noticed there's currently a Measurement "operator" but right now it's just a Z-gate which I chose as a stand-in. Obviously this should be fixed to actually record a measurement.
Thanks! We are equally interested in stabilizer circuits and general circuits. I'll take a crack at it and see what I can get working. Meanwhile, we really appreciate any time you are able to devote to this. I know you're super busy otherwise.
2. reporter
Thanks your help--once I'm over the hump with this one class I'll be able to devote more time and attention to this. | 686 | 3,145 | {"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} | 2.5625 | 3 | CC-MAIN-2015-35 | longest | en | 0.952545 |
https://www.mrexcel.com/board/threads/multiple-conditions-formula.980431/ | 1,696,475,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00794.warc.gz | 976,105,546 | 18,395 | # Multiple conditions formula
#### Dsullivanre
##### New Member
Hi Everyone, thanks for taking the time to help me out!
I tried to search in the help in excel, on how to do a multiple condition formula, using If And / IF statements, with no luck.
How do I get the following to be in 1 formula:
A1=0 & B1=0 return in C1 "A"
A1>0 & B1=0 return in C1 "B"
A1=0 & B1>0 return in C1 "C"
A1>0 & B1>0 return in C1 "D"
Thanks again!
### Excel Facts
Return population for a City
If you have a list of cities in A2:A100, use Data, Geography. Then =A2.Population and copy down.
This should work, in cell C1 enter:
Code:
``=IF(A1=0,IF(B1=0,"A",IF(B1>0,"C")),IF(A1>0,IF(B1=0,"B","D")))``
Welcome to the forum.
Maybe:
=IF(A1=0,IF(B1=0,"A","C"),IF(B1=0,"B","D"))
This doesn't handle any case where A1 or B1 is less than 0 though. You'd need something like:
=IF(AND(A1=0,B1=0),"A",IF(AND(A1>0,B1=0),"B",IF(AND(A1=0,B1>0),"C",IF(AND(A1>0,B1>0),"D","Something is below 0"))))
to handle that. Or maybe if you have the new IFS function:
=IFS(AND(A1=0,B1=0),"A",AND(A1>0,B1=0),"B",AND(A1=0,B1>0),"C",AND(A1>0,B1>0),"D",TRUE,"Something is below 0")
Thanks guys! I appreciate the help!
This worked: =IF(AND(A1=0,B1=0),"A",IF(AND(A1>0,B1=0),"B",IF(AND(A1=0,B1>0),"C",IF(AND(A1>0,B1>0),"D","Something is below 0"))))
Too late, but here is a shorter version:
=IF(OR(A1<0,B1<0),"Out of range",IF(A1+B1=0,"A",IF(A1*B1>0,"D",IF(A1=0,"C","B"))))
Hi,
Your conditions is based on A1 and B1, when both conditions meet , try using if with and operator as below
=IF(AND(A1=0,B1=0),"A",IF(AND(A1>0,B1=0),"B",IF(AND(A1=0,B1>1),"C",IF(AND(A1>0,B1>0),"C",""))))
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Go back | 947 | 2,707 | {"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} | 3.359375 | 3 | CC-MAIN-2023-40 | latest | en | 0.835603 |
https://eprojects.xyz/?post=6340 | 1,624,108,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00300.warc.gz | 229,257,245 | 8,467 | # How smooth is the earth
### Why is the earth round?
“What happens if you keep going in the same direction? Will one come to the edge of the world at some point or is the world infinitely large? ”More than 2300 years ago, the famous Greek scientist Aristotle was certain: Neither one nor the other. Because the earth is not flat like a disk, but a sphere - but why?
To understand this, one has to go back to the time when the earth was created. The force that was responsible for this is gravity - all massive objects attract each other. This force made chunks of rock collide and combine to form a planet. And it gave shape to the planet. Because gravity acts equally strong in all directions.
Since the earth was hot and liquid at the beginning, the material was able to flow into the shape dictated by gravity. If a piece of earth protruded further out, it was attracted by the rest until the surface was smooth and the same force of gravity was acting in all places. And since gravity is the same in all directions, the shape of a sphere was automatically created - because only with a sphere are all points on its surface the same distance from the center of gravity.
But if you take a closer look at the shape of the earth, you will see that the earth is not a perfect sphere: it is slightly flattened at its poles and somewhat bulbous at the equator.
The earth's rotation is to blame for this: the earth rotates once around its axis in the course of 24 hours. The rotary movement creates a force, the centrifugal force. We know this from the chain carousel when we fly outwards on the swings. In the case of the earth, the centrifugal force causes the rock masses to slide outwards a little from the axis of rotation, i.e. from the poles towards the equator. The diameter of the earth there is around 41 kilometers larger than between the north and south poles.
6.9.1522
Big reception in the Spanish port of Sanlúcar: after almost three years, the “Victoria” is returning from her expedition. But the joy is clouded. Of the five ships in the fleet, only this one comes back, and of the 237 crew only 18 survived. The leader of the expedition, the Portuguese Ferdinand Magellan (42), was also killed on the way.
So he could no longer experience the great triumph himself: his expedition succeeded for the first time in the history of mankind to completely circumnavigate the world! Ferdinand Magellan set out from Seville with his five ships on August 10, 1519. He was on the way on behalf of the Spanish Crown to find the western route to the lucrative Spice Islands in the Pacific Ocean. This was necessary after the rival sea power Portugal had taken the sea route in an easterly direction around the African Cape of Good Hope.
After an arduous search, Magellan and his team discovered the long-awaited “Gateway to the West” on the southern tip of South America. At the end of November 1520, they sailed through a narrow 600-kilometer strait into the Pacific Ocean.
But the setbacks were inevitable. In an attempt to take islands of the newly discovered Philippines for the Spanish crown, Captain Magellan fell victim to the lances and poison arrows of the natives. Numerous crew members died with him in the fighting, others had died in the months before from the catastrophic supply situation on the ships. One crew member reported that they had to eat leather and rats cooked in salt water. Many died of scurvy.
But the survivors can be proud of themselves. You are the first to travel around the world. So they provided the final proof: the earth is a sphere. And in addition to fame, honor and knowledge, they brought home over 26 tons of valuable spices.
### Globe larger than expected
The logbook entries of the "Victoria" will now give the scientists some homework. It seems that up to now the extent of the earth has been underestimated.
The most prominent victim of this misjudgment was Christopher Columbus. He also wanted to find the western route to India, but assumed about a quarter less circumference of the earth. At that time, nothing at all was known about the new continent in the west, Asia was considered too big and the Pacific Ocean too small. When he came across the Bahamas in 1492, he was therefore firmly convinced that he had discovered "India".
Now it looks like the Greek scientist Eratosthenes was right. As early as 240 BC he had taken measurements using the simplest of methods and calculated a circumference of around forty thousand kilometers from them.
24.12.1968
Upside down world - the earth rises above the moon horizon. The American astronaut William Anders took this famous photo on Christmas Eve 1968.
Together with Frank Borman and James Lovell, he circled the moon several times on the Apollo 8 mission. When their space capsule came out from behind the moon during one of these orbits, they saw the globe emerge behind the lunar horizon. They were deeply impressed by the sight and took several photos - although Anders jokingly remarked that this was not provided for in the mission plan.
A real "earth rise" - like the way we see the moon rise while standing on earth - cannot be experienced on the surface of the moon. Because the moon always turns the same side to the earth. If you stay on this side, you can see the earth all the time - and always in the same place in the sky. And from the back of the moon, the earth can never be seen.
### The second face
The moon shows itself from a completely new side: On October 7, 1959, the Russian probe "Luna 3" took the first photo of the back of the moon. However, the world had to wait eleven days for this historic photo: it was only when the probe flew back towards earth that the radio link was good enough to send the image.
At first glance, the picture doesn't look very spectacular. The resolution is poor, and since the sun shines almost perpendicularly on the surface of the moon, no shadows from mountains and craters can be seen.
But there was a surprise: The moon has far fewer dark spots on the back than on the front. Astronomers are still puzzling over the reason!
Until this photo was taken, mankind had no idea what it looked like there. Because from the earth you only ever see the same side of the moon.
### The beginnings of the earth
We would not recognize the earth immediately after its formation. It was an extremely uncomfortable planet: there were neither continents nor oceans, but a seething surface of glowing hot, viscous magma. Why couldn't the earth's crust form for a long time?
A good 4.5 billion years ago comets, asteroids, gas and dust condensed to form our planet. Its own gravity pressed these individual parts together so that they were subjected to strong pressure. This pressure was naturally highest in the core of the earth, on which the weight of the entire outer layers weighed. As a result of the high pressure, the rock was heated up and melted. Outwardly, the pressure and thus also the temperature became less. Even so, the surface of the earth remained very hot for several hundred million years and could not cool down and solidify.
In order to understand the reason for this, the scientists had to look at the moon: Ancient lunar craters from the time the solar system was formed tell us that the moon was hit by numerous meteorites when it was young. It is therefore assumed that the earth was also exposed to a real rock bombardment from space at the same time. The lumps fell to the earth at high speed - and the impacts were correspondingly violent: Even lumps of a few hundred tons could easily cause an explosion the strength of an atomic bomb!
So the earth's surface continued to heat up for a long time, stirred up again and again and remained so fluid. Only when the impacts gradually subsided after a few hundred million years did the temperatures on the earth's surface drop. The rock could slowly solidify and form an earth crust that became thicker and thicker over the course of millions of years. But to this day it is only a very thin layer that floats on a viscous, hot interior of the earth.
### What is our solar system and how did it come about?
The earth is not alone in space: people have been observing the sun, moon and stars in the sky for a long time. They discovered early on that some stars are moving. These wandering stars were observed and their paths followed. For a long time, however, their movements were not understood - until about five hundred years ago a man by the name of Nicolaus Copernicus solved the riddle: The earth and the "wandering stars" are actually planets, all of which orbit the sun at different distances.
Today we know eight planets. To remember their names in the correct order, the first letters of the sentence "M.a V.ater eclarifies mir jEden S.monday uurens Nachthimmel. “- or in short: M-V-E-M-J-S-U-N.
M.Erkur is the planet that orbits closest to the sun. Then come V.enus, E.rde and M.ars. These four inner planets have a solid surface made of rock and are still relatively close to the sun - only a few hundred million kilometers.
They are circling further out, at a distance of about one to 4.5 billion kilometers from the sun outer planets: Jupiter, S.aturn with his rings, Uranus and all the way outside Neptun. They are made of gas (mostly hydrogen and helium) and are much larger than the inner planets. Jupiter and Saturn are about ten times the size of the earth, that's why they are also called that Gas giants.
And finally, there are asteroids, comets and clouds of dust that also orbit the sun. The gravitational pull of the sun holds all these heavenly bodies together and forces them to fly in a circle like on a long line. Everything together is called that Solar system. The moons are one of them - but they are held in place by the gravitational pull of the planets.
But why does the sun even have planets? This has to do with how the sun came into being: a cloud of gas and dust contracted by its own gravity and became a star. But not all of the material in this cloud was "built into" the star - around one percent was left over. And when the sun began to shine, the radiation pushed the remaining matter back outwards.
The light gases were pushed far outwards, the heavier dust and rocks remained close to the sun. From these clouds of dust and gas, the planets emerged over time. Therefore there are the gas planets outside in the solar system, further inside the rock planets - including our earth - and in the very center the sun. It contains 99% of the mass of the solar system and holds everything together with its gravity.
### Why is the earth warm inside?
The liquid interior of the earth bubbles under our feet. Volcanic eruptions and geysers show the heat there - over 6000 degrees Celsius in the earth's core. But why is it so hot in the earth?
Much of the heat comes from Earth's childhood days when dust and rocks condensed into a planet. The word “condense” sounds a little too harmless, however: In reality, you have to imagine how many large meteorite impacts - each impact a gigantic explosion that heated up the young planet and melted the material.
Since then it has become a little quieter and the earth is cooling down again. However, it does this extremely slowly, the heat in the interior of the earth can only very slowly escape into space. Hot magma flows in the tough earth mantle transport the heat upwards. There it remains enclosed under the rigid earth's crust as if under a lid. The crustal rock only slowly releases its heat into space.
In addition, heat is still being produced inside the earth. This is because the core of the earth contains a lot of radioactive substances such as uranium. Since the formation of our planet, they have been disintegrating and giving off heat over a very long period of time. This “fuel” will last for billions of years. | 2,542 | 11,874 | {"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} | 2.8125 | 3 | CC-MAIN-2021-25 | longest | en | 0.963589 |
https://acronyms.thefreedictionary.com/GDQS | 1,571,398,808,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00022.warc.gz | 371,069,316 | 10,384 | # GDQS
AcronymDefinition
GDQSGrid Distributed Query Service
GDQSGroup Dynamics Q Sort (psychology)
References in periodicals archive ?
The tangential condition will also appear in other theorems in the sequel (it may be slightly changed) with an exception of our theorem, where we use GDQs instead of the contingent derivative.
Our idea was to use for the first time in viability theory another tool of differentiation: GDQs. One of the advantages of GDQs over the contingent derivative is that, as Example 2.6 shows, SGDQ contains all important directions while the contingent derivative has, besides these directions, some superfluous elements.
The main difference between Theorem 3.6 and others is, as it was mentioned before, that we use GDQs theory instead of the contingent derivative to formulate the tangential condition for problem (2.1).
Indeed, in (3.4) we intersect F with, possibly smaller than the contingent derivative, the closed union of minimal GDQs ofK.
As a generalized derivative we choose the generalized differential quotient (GDQ), introduced recently by Sussmann [11;12].
We say that [LAMBDA] is a generalized differential quotient (GDQ) of F at ([bar.x], [bar.y]) in the direction S, and write [lambda] [member of] GDQ(F, [bar.x], [bar.y], S) if for every positive real number [delta] there exist U;G such that
A minimal GDQ of F at ([bar.x], [bar.y]) in the direction S is a minimal element of the set GDQ(F; [bar.x]; [bar.y]; S) (minimal in the sense of inclusions of sets).
As we can have more than one minimal GDQ, we introduce the following concept of SGDQ.
multifunction with nonempty closed values such that for all (t; y) [member of] GrK, where t [member of] [0; a), K is GDQ differentiable at (t; y) in the direction of [R.sub.+] and for every [epsilon] > 0 there exists [T.sub.[epsilon]] [subset.bar] T such that [lambda](T\[T.sub.[epsilon]]) < [epsilon] and the map (t; y) [??] SGDQ(K; t; y;[R.sub.+]) is u.s.c.
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http://isabelle.in.tum.de/repos/isabelle/file/30a1692557b0/src/HOL/Old_Number_Theory/Residues.thy | 1,571,151,844,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00186.warc.gz | 91,019,251 | 5,017 | src/HOL/Old_Number_Theory/Residues.thy
author huffman Sun Apr 01 16:09:58 2012 +0200 (2012-04-01) changeset 47255 30a1692557b0 parent 41541 1fa4725c4656 child 53077 a1b3784f8129 permissions -rw-r--r--
removed Nat_Numeral.thy, moving all theorems elsewhere
``` 1 (* Title: HOL/Old_Number_Theory/Residues.thy
```
``` 2 Authors: Jeremy Avigad, David Gray, and Adam Kramer
```
``` 3 *)
```
``` 4
```
``` 5 header {* Residue Sets *}
```
``` 6
```
``` 7 theory Residues
```
``` 8 imports Int2
```
``` 9 begin
```
``` 10
```
``` 11 text {*
```
``` 12 \medskip Define the residue of a set, the standard residue,
```
``` 13 quadratic residues, and prove some basic properties. *}
```
``` 14
```
``` 15 definition ResSet :: "int => int set => bool"
```
``` 16 where "ResSet m X = (\<forall>y1 y2. (y1 \<in> X & y2 \<in> X & [y1 = y2] (mod m) --> y1 = y2))"
```
``` 17
```
``` 18 definition StandardRes :: "int => int => int"
```
``` 19 where "StandardRes m x = x mod m"
```
``` 20
```
``` 21 definition QuadRes :: "int => int => bool"
```
``` 22 where "QuadRes m x = (\<exists>y. ([(y ^ 2) = x] (mod m)))"
```
``` 23
```
``` 24 definition Legendre :: "int => int => int" where
```
``` 25 "Legendre a p = (if ([a = 0] (mod p)) then 0
```
``` 26 else if (QuadRes p a) then 1
```
``` 27 else -1)"
```
``` 28
```
``` 29 definition SR :: "int => int set"
```
``` 30 where "SR p = {x. (0 \<le> x) & (x < p)}"
```
``` 31
```
``` 32 definition SRStar :: "int => int set"
```
``` 33 where "SRStar p = {x. (0 < x) & (x < p)}"
```
``` 34
```
``` 35
```
``` 36 subsection {* Some useful properties of StandardRes *}
```
``` 37
```
``` 38 lemma StandardRes_prop1: "[x = StandardRes m x] (mod m)"
```
``` 39 by (auto simp add: StandardRes_def zcong_zmod)
```
``` 40
```
``` 41 lemma StandardRes_prop2: "0 < m ==> (StandardRes m x1 = StandardRes m x2)
```
``` 42 = ([x1 = x2] (mod m))"
```
``` 43 by (auto simp add: StandardRes_def zcong_zmod_eq)
```
``` 44
```
``` 45 lemma StandardRes_prop3: "(~[x = 0] (mod p)) = (~(StandardRes p x = 0))"
```
``` 46 by (auto simp add: StandardRes_def zcong_def dvd_eq_mod_eq_0)
```
``` 47
```
``` 48 lemma StandardRes_prop4: "2 < m
```
``` 49 ==> [StandardRes m x * StandardRes m y = (x * y)] (mod m)"
```
``` 50 by (auto simp add: StandardRes_def zcong_zmod_eq
```
``` 51 mod_mult_eq [of x y m])
```
``` 52
```
``` 53 lemma StandardRes_lbound: "0 < p ==> 0 \<le> StandardRes p x"
```
``` 54 by (auto simp add: StandardRes_def)
```
``` 55
```
``` 56 lemma StandardRes_ubound: "0 < p ==> StandardRes p x < p"
```
``` 57 by (auto simp add: StandardRes_def)
```
``` 58
```
``` 59 lemma StandardRes_eq_zcong:
```
``` 60 "(StandardRes m x = 0) = ([x = 0](mod m))"
```
``` 61 by (auto simp add: StandardRes_def zcong_eq_zdvd_prop dvd_def)
```
``` 62
```
``` 63
```
``` 64 subsection {* Relations between StandardRes, SRStar, and SR *}
```
``` 65
```
``` 66 lemma SRStar_SR_prop: "x \<in> SRStar p ==> x \<in> SR p"
```
``` 67 by (auto simp add: SRStar_def SR_def)
```
``` 68
```
``` 69 lemma StandardRes_SR_prop: "x \<in> SR p ==> StandardRes p x = x"
```
``` 70 by (auto simp add: SR_def StandardRes_def mod_pos_pos_trivial)
```
``` 71
```
``` 72 lemma StandardRes_SRStar_prop1: "2 < p ==> (StandardRes p x \<in> SRStar p)
```
``` 73 = (~[x = 0] (mod p))"
```
``` 74 apply (auto simp add: StandardRes_prop3 StandardRes_def SRStar_def)
```
``` 75 apply (subgoal_tac "0 < p")
```
``` 76 apply (drule_tac a = x in pos_mod_sign, arith, simp)
```
``` 77 done
```
``` 78
```
``` 79 lemma StandardRes_SRStar_prop1a: "x \<in> SRStar p ==> ~([x = 0] (mod p))"
```
``` 80 by (auto simp add: SRStar_def zcong_def zdvd_not_zless)
```
``` 81
```
``` 82 lemma StandardRes_SRStar_prop2: "[| 2 < p; zprime p; x \<in> SRStar p |]
```
``` 83 ==> StandardRes p (MultInv p x) \<in> SRStar p"
```
``` 84 apply (frule_tac x = "(MultInv p x)" in StandardRes_SRStar_prop1, simp)
```
``` 85 apply (rule MultInv_prop3)
```
``` 86 apply (auto simp add: SRStar_def zcong_def zdvd_not_zless)
```
``` 87 done
```
``` 88
```
``` 89 lemma StandardRes_SRStar_prop3: "x \<in> SRStar p ==> StandardRes p x = x"
```
``` 90 by (auto simp add: SRStar_SR_prop StandardRes_SR_prop)
```
``` 91
```
``` 92 lemma StandardRes_SRStar_prop4: "[| zprime p; 2 < p; x \<in> SRStar p |]
```
``` 93 ==> StandardRes p x \<in> SRStar p"
```
``` 94 by (frule StandardRes_SRStar_prop3, auto)
```
``` 95
```
``` 96 lemma SRStar_mult_prop1: "[| zprime p; 2 < p; x \<in> SRStar p; y \<in> SRStar p|]
```
``` 97 ==> (StandardRes p (x * y)):SRStar p"
```
``` 98 apply (frule_tac x = x in StandardRes_SRStar_prop4, auto)
```
``` 99 apply (frule_tac x = y in StandardRes_SRStar_prop4, auto)
```
``` 100 apply (auto simp add: StandardRes_SRStar_prop1 zcong_zmult_prop3)
```
``` 101 done
```
``` 102
```
``` 103 lemma SRStar_mult_prop2: "[| zprime p; 2 < p; ~([a = 0](mod p));
```
``` 104 x \<in> SRStar p |]
```
``` 105 ==> StandardRes p (a * MultInv p x) \<in> SRStar p"
```
``` 106 apply (frule_tac x = x in StandardRes_SRStar_prop2, auto)
```
``` 107 apply (frule_tac x = "MultInv p x" in StandardRes_SRStar_prop1)
```
``` 108 apply (auto simp add: StandardRes_SRStar_prop1 zcong_zmult_prop3)
```
``` 109 done
```
``` 110
```
``` 111 lemma SRStar_card: "2 < p ==> int(card(SRStar p)) = p - 1"
```
``` 112 by (auto simp add: SRStar_def int_card_bdd_int_set_l_l)
```
``` 113
```
``` 114 lemma SRStar_finite: "2 < p ==> finite( SRStar p)"
```
``` 115 by (auto simp add: SRStar_def bdd_int_set_l_l_finite)
```
``` 116
```
``` 117
```
``` 118 subsection {* Properties relating ResSets with StandardRes *}
```
``` 119
```
``` 120 lemma aux: "x mod m = y mod m ==> [x = y] (mod m)"
```
``` 121 apply (subgoal_tac "x = y ==> [x = y](mod m)")
```
``` 122 apply (subgoal_tac "[x mod m = y mod m] (mod m) ==> [x = y] (mod m)")
```
``` 123 apply (auto simp add: zcong_zmod [of x y m])
```
``` 124 done
```
``` 125
```
``` 126 lemma StandardRes_inj_on_ResSet: "ResSet m X ==> (inj_on (StandardRes m) X)"
```
``` 127 apply (auto simp add: ResSet_def StandardRes_def inj_on_def)
```
``` 128 apply (drule_tac m = m in aux, auto)
```
``` 129 done
```
``` 130
```
``` 131 lemma StandardRes_Sum: "[| finite X; 0 < m |]
```
``` 132 ==> [setsum f X = setsum (StandardRes m o f) X](mod m)"
```
``` 133 apply (rule_tac F = X in finite_induct)
```
``` 134 apply (auto intro!: zcong_zadd simp add: StandardRes_prop1)
```
``` 135 done
```
``` 136
```
``` 137 lemma SR_pos: "0 < m ==> (StandardRes m ` X) \<subseteq> {x. 0 \<le> x & x < m}"
```
``` 138 by (auto simp add: StandardRes_ubound StandardRes_lbound)
```
``` 139
```
``` 140 lemma ResSet_finite: "0 < m ==> ResSet m X ==> finite X"
```
``` 141 apply (rule_tac f = "StandardRes m" in finite_imageD)
```
``` 142 apply (rule_tac B = "{x. (0 :: int) \<le> x & x < m}" in finite_subset)
```
``` 143 apply (auto simp add: StandardRes_inj_on_ResSet bdd_int_set_l_finite SR_pos)
```
``` 144 done
```
``` 145
```
``` 146 lemma mod_mod_is_mod: "[x = x mod m](mod m)"
```
``` 147 by (auto simp add: zcong_zmod)
```
``` 148
```
``` 149 lemma StandardRes_prod: "[| finite X; 0 < m |]
```
``` 150 ==> [setprod f X = setprod (StandardRes m o f) X] (mod m)"
```
``` 151 apply (rule_tac F = X in finite_induct)
```
``` 152 apply (auto intro!: zcong_zmult simp add: StandardRes_prop1)
```
``` 153 done
```
``` 154
```
``` 155 lemma ResSet_image:
```
``` 156 "[| 0 < m; ResSet m A; \<forall>x \<in> A. \<forall>y \<in> A. ([f x = f y](mod m) --> x = y) |] ==>
```
``` 157 ResSet m (f ` A)"
```
``` 158 by (auto simp add: ResSet_def)
```
``` 159
```
``` 160
```
``` 161 subsection {* Property for SRStar *}
```
``` 162
```
``` 163 lemma ResSet_SRStar_prop: "ResSet p (SRStar p)"
```
``` 164 by (auto simp add: SRStar_def ResSet_def zcong_zless_imp_eq)
```
``` 165
```
``` 166 end
``` | 3,112 | 8,438 | {"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} | 3.15625 | 3 | CC-MAIN-2019-43 | latest | en | 0.592435 |
http://www.scarymommy.com/dad-trolls-his-kids-school-by-using-common-core-math-to-write-them-a-check/ | 1,524,567,172,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946597.90/warc/CC-MAIN-20180424100356-20180424120356-00002.warc.gz | 527,166,984 | 19,225 | Dad Trolls His Kid’s School By Using Common Core Math To Write Them A Check – Scary Mommy
# Dad Trolls His Kid’s School By Using Common Core Math To Write Them A Check
If you have a child in elementary school, you’ve likely encountered the gigantic pain in the ass that is Common Core math. These standards have been adopted in all but seven states and the reactions from parents have not exactly been favorable. One dad hilariously took his displeasure straight to the source by writing out a check for his kid’s school — with the numbers drawn out in Common Core long form. Of course, no one old enough to be working at a bank now has any idea what it means so he’s snarkily proving what we all know — that Common Core math is overly complicated and not compatible when applied to most real-life situations.
Here is the check, with the dollar amount written out the way our children (if you live in a Common Core state, that is) are currently being taught. It is written in what’s called “long form”, with the numerals delineated in their own separate boxes for ones, tens, hundreds, etc. I can’t even figure out how much it’s worth and neither can anyone else in the internet, it seems. It could make your head spin:
This is hysterical. And sadly, an extremely familiar (and frustrating) sight to any parent with a child learning long form numbers. My daughter is in third grade and we’ve drawn out countless little charts like that over the last year or two to help her learn to write out numbers according to this method (numbers, not X’s and O’s — that’s next year in 4th grade). Parents are having to teach themselves all over again in a whole new way when the actual answers to the problems these charts are supposed to “help” solve are found a lot more easily doing it the way we already know. As this dad is cleverly proving, learning math this way doesn’t make sense for the real world. Opponents of this method (including myself) say that it’s making mountains out of mole hills while those supporting it say it’s teaching kids a way to work out larger problems down the line. It’s hard to see the value in it when applied in a real-world situation, such as writing out a check. It does seem needlessly complex when you could simply write out the dollar amount and not draw a little wall of cubes filled with digits or letters.
As someone who struggled with math growing up, it gives me a lot of anxiety trying to teach my kids an entirely different way of learning than what I grew up with, and I know I’m not the only one. It’s hard enough teaching them the way we do know and having to figure this all out as an adult gives us a tiny taste of what the kids are going through. I know that many parents are frustrated trying to reinvent the wheel with this new curriculum. The kids will have to be tested on it eventually and the pressure on their teachers is high — they need their classes to do well in order to keep their jobs in some cases. It’s very stressful for all involved and all parents can do is hope that their children are able to learn this way, since Common Core leaves little wiggle room for teachers to adapt to different learning styles.
As parents, we try to be supportive of our kids and their schools. We also try to support their teachers because even though we may not like or understand Common Core, it’s clearly not an individual instructor’s fault if their state demands they teach math this way. Seeing this dad make the point that these standards don’t work out in real life is pretty funny but also, jarring. Common Core is still relatively new but hopefully, the continuing outcry from parents, teachers and students will mean that change will come eventually. In the meantime, it’s comforting to see I’m not alone in wanting to put my head down on the kitchen table during homework time to cry. | 818 | 3,845 | {"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} | 3.9375 | 4 | CC-MAIN-2018-17 | latest | en | 0.948935 |
https://www.futurelearn.com/courses/essentialmathfordataanalysisusingexcelonline | 1,726,316,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00552.warc.gz | 708,378,479 | 56,469 | # Essential Mathematics for Data Analysis in Microsoft Excel
Understand the basic maths that powers data analysis with this beginner’s course.
• Duration
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5 hours
• Included in an ExpertTrack
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Data analysis is inextricably linked with maths. While statistics are the most important mathematical element, it also requires a good understanding of different formulas and mathematical inference.
This course is designed to build up your understanding of the essential maths required for data analytics. It’s been designed for anybody who would like to refresh their maths knowledge or learn it in a simplified way before embarking on further data analytics training.
## Build your knowledge of basic statistics
Explore summary statistics and learn how to make statistical predictions so that you can analyse data across any field or discipline. You’ll discover how to apply statistical methods to your data, and will practice the formula required to do that so that you can analyse and understand your data sets.
## Understand simple mathematical notation
Mathematical notations include numbers, variables, delimiters, functions, relational symbols, and operator symbols. You’ll learn basic notation and expressions, and find out how they are applied to Excel formulas to complete simple calculations.
Once you have a good understanding of the maths behind it, you’ll move onto forecasting future trends using your data. You’ll apply inferential maths to your data sets and calculate business metrics and KPIs to turn your new knowledge into genuine business value.
## Syllabus
### Introduction to Working with Data
This activity introduces you to the course outline and the learning outcomes, as well as information on navigating the course on the FutureLearn platform and your CloudSwyft Hands-on Labs
• ### Introduction to Data and Variables
Understanding how to set up a tidy set of data and to extract data from a data set are integral skills for statistical analysis.
• ### Types of Data
Here you will be introduced to the different types of data that are common when analysing data sets.
• ### Notation
Understanding how to represent data using formulas as well as being able to work with different formulas is a core skill for a data analyst. This activity introduces you to unpacking formulas.
• ### Means and Deviations
One of the most common things to calculate in data is the mean average of the range. In this activity, you will also be introduced to the formula for calculating standard deviation.
• ### Weekly Wrap Up
This marks the end of the week’s activities. First, we will start off with recap of the week, followed by a preview of what can be expected in the week to follow.
• ### Summarising Data with Distributions and Graphs
In this activity, we will understand why histograms are a good way to represent data and to identify the skewness of a data set.
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In this activity, we will explore why Pie Charts are not a great way to represent statistical data and how Bar Graphs differ from Histograms.
• ### Mode, Mean and Median
There are different ways to find the centre of our data set. We shall explore each of them in this activity.
• ### SD vs IQR
Standard deviation does not always give us an accurate result, so we will introduce the semi-interquartile range as another option for finding the deviation of our mean.
• ### Proportions
When dealing with categorical data, it is important to take proportions into account to get a better view of our data.
• ### Weekly Wrap Up
You have come to the end of your week's learning. Take a moment to review what you have learnt and track your progress.
• ### Week 3
Being able to apply your knowledge about data analysis in the business world is a fundamental skill. In this activity, we will look at applying statistics using business maths and creating KPIs.
In this activity, you will be able to put into practice your knowledge of Business Statistics.
• ### Introduction to Inferential Math and Forecasting
Forecasting is an important step that every business needs to implement in order to project future stability. Using past trends, we are able to forecast future likelihoods.
• ### Course Wrap Up
You have reached the end of this course! Now it's time to check your knowledge and recap what you have learned.
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## Who is this accredited by?
Microsoft:
This course is accredited by Microsoft
## Learning on this course
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## What will you achieve?
By the end of the course, you‘ll be able to...
• Demonstrate how to handle data sets for analysis purposes
• Apply mathematical notations
• Calculate business metrics and KPIs to derive business value from data sets
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## Who is the course for?
This course is for anybody who would like to refresh their maths, statistics, and mathematical notation in order to be a better data analyst. This may include aspiring data analysts or new managers who need to upskill to understand the data within their business.
## What software or tools do you need?
The upgrade version of this course will include all the required CloudSwyft Hands-On Lab practice environments to be used by students and learners on this course, powered by CloudSwyft.
You will need a computer running Windows, Mac OSX or Linux, and a web browser.
## CloudSwyft Global Systems, Inc.
CloudSwyft has partnered with the top global technology companies to deliver cutting edge digital skills learning across the modern workplace.
Learn data analysis from scratch, including an introduction to essential maths and Microsoft Excel.
### Start learning today - free 2-day trial
After your free trial you can:
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• Share ideas with your peers and course educators on every step of the course | 1,464 | 7,388 | {"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} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.874517 |
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# What is your true GMAT level?
Author Message
Magoosh GMAT Instructor
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### Show Tags
20 Jun 2014, 10:14
Sometimes a student will send a message into Magoosh, looking for tips on how to improve. In their message, they might write something similar to the following:
"A week before exam, my Verbal score range on the Magoosh dashboard was 151-156. When doing untimed practice sessions in Verbal, I’d get the predicted score range of 155-160. When doing timed practice of 30 minutes, I’ve been able to do 14-16 questions comfortably with roughly 80 percent accuracy…"
This is a worrisome message to receive from a student for a few reasons. For one, the student waited until a week before his exam to ask for tips. The other issue, though, is particularly troublesome and harmful for the student. Unfortunately, it’s a natural mental model that students use to evaluate themselves, but it limits their ability to really assess themselves.
Can you see this troubling mental model?
This student, and many others, assess their GMAT level timed and untimed. This is grave mistake. Let’s talk about why.
- See more at: http://magoosh.com/gmat/2014/timed-vs-u ... gmat-level
_________________
Kevin Rocci
Magoosh Test Prep
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# What is your true GMAT level?
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 713 | 2,836 | {"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} | 2.609375 | 3 | CC-MAIN-2017-09 | longest | en | 0.918411 |
https://optionalpha.com/members/video-tutorials/pricing-volatility/understanding-the-math | 1,498,200,367,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00091.warc.gz | 819,951,322 | 36,933 | ## Understanding The Math
Welcome back to statistics class. Undoubtedly more important that understanding the Black Scholes model for pricing (which we purposesly don't cover) is your ability to understand statistics and probabilities on a deep level.
We would even venture to say that your understanding of the "math" behind our trading system is directly related to your success rate and end of the year profit.
If you don't get the math behind the trades you will fail - guaranteed!
Swallow your pride and head back to school with us as we talk in depth about standard deviations, probabilities and statistics in this advanced tutorial.
#### More Discussion
• All option pricing is based on implied volatility for that particular contract month (which is why option prices are always overstated long-term compared to the HV moves that actually occur). You should be looking at IV index not the IV of a particular option and soon we’ll be rolling out this feature as part of our own watch list we developed here: https://optionalpha.com/members/watch-list
• Sanchez
We sell when IV is high in the hope it contracts?
• And knowing that the higher it is, the more overstated it becomes.
• Short-term you’re going to have less of a normal distribution as you would long-term with more trades or data points. The key here is to realize that short term success or failure when trading options is not indicative of long-success. If you are consistently selling options as the main core part of your strategy then long-term when lots of trades play out you’ll find yourself on the winning side.
• Yes to probability curve is future looking but what I was saying was that implied volatility to some degree looks at past movements so trades can then predict what future movements might be.
Show Video Transcript +
Hey, everyone. This is Kirk, here again at optionalpha.com where we show you how to make smarter trades. Today, we've got an awesome video tutorial for you, breaking down trading math and specifically, options trading math.
It’s a 101 course on why we have the methodologies that we do about the markets and trading. Welcome back to statistics class. You’re probably thinking, “Oh.” But don’t worry.
Undoubtedly more important than understanding the Black-Scholes model for pricing which we purposely don't cover in any video tutorial that we have because it's pointless to cover, you don’t need to know it to be successful, but besides that, your ability to understand just basic statistics and probabilities is paramount to your ability to be successful in this business.
If you don't get the math behind the trades, here's my promise to you. You will fail at trading options long-term if you don't understand the math behind it and more importantly, the statistics and the probabilities behind it.
You can make a couple of trades here and there and be successful, but to do this long-term, to generate consistent monthly income long-term, you've got to understand the math.
Swallow your pride, head back to school with us as we talk in-depth about standard deviations, probabilities, and statistics in this advanced tutorial.
But before we do that, let’s first have a talk about one real quick thing, and that’s market efficiency. This whole idea about market efficiency is really important, and you’ve probably heard us talk about it in other videos that we have to trade liquid products and liquid underlying stocks.
But this whole idea of market efficiency is this concept that the markets are super efficient. Especially in the US markets where there are millions and millions of different market participants, all with their individual ideas, the markets are incredibly efficient and incredibly fast.
Data or information that anybody receives on a stock or a company is immediately priced into the market. As a guy who’s been on both sides of the Chinese wall…
Basically, I was an M&A analyst in New York for Deutsche Bank, and I was on the private side dealing with mergers and acquisitions, and then I was on the other side of the Chinese wall in Tyson’s in DC and dealing with the REIT retail side as an analyst.
I’ve been on both sides of the wall, and I can tell you, the markets are incredibly efficient. There’s no edge that you can get in knowing information about a company in advance or having some insider knowledge.
In most cases, most CEOs have no clue where their stock is going to go or how it’s going to react to the market, regardless of how well they think they might be doing.
To that end, we have to understand that as we’ve said before, we have no clue where a stock is going to go, and nobody else does either, myself included. I have no idea where a stock is going to go in the future. I might have an assumption, an opinion, but at the end of the day, we’re all no better than 50/50 on our guesses.
What this leads to then is to this probability distribution and what we call a normal distribution or what you probably have seen before as a bell curve. This is really important because this is how an efficient market distributes its returns.
Basically, what you have here is you have most of the returns are probably somewhere around even or par, and that's basically what this zero line here is.
But it’s saying that most of the time, the distribution of returns will be within a certain confidence range or about one standard deviation. This is what this one standard deviation is that I’m highlighting here on the chart.
This is saying that 34.1% of the time up and 34.1% of the time down, we might see this confidence within this given range. We can define that range in stocks, in every particular stock that we look at, and we’ll show that to you later on here, but you just have to understand that when a market has normal movements and an efficient movement.
It’s going to have a normal distribution of returns. This means that most stocks are going to land inside of that normal distribution.
Sure you’re going to have the stocks that go outside of that distribution, so they make a three standard deviation move, so whatever most stocks are doing, they do three times that move.
These are going to be stocks that are high flyers, the one in a million stock that goes from \$5 to \$500 or whatever the case is. Then, of course, you’re also going to have stocks that make a three standard deviation move to the downside.
These are going to be your stocks that go from \$100 down to \$10. It's the one in a million chance that the stock goes bankrupt or the company goes bankrupt, whatever the case is.
Remember, this is a normal efficient market, so most of the time, stocks are going to return some normal average in the middle, and that bulk average here is what we’re looking for when we start to place trades, just understanding that this is how a stock is distributed.
Now, when we look at the same graph and tilt it on its side here, we can see that this same concept applies to a stock distribution of its price movement over time. I’ll say that again. This same distribution will apply to a particular stock’s price movement going forward in the future.
What I always like to do is I always like to say at certain points in the future… Let’s just draw a line down here and say at this point in the future versus this point in the future; we can estimate based on the entire trading history of the stock going back in time how likely the stock is to rally or fall within a given range.
We can use the data from the stock going back all the way to its beginning as much data as we have on that stock to automatically and accurately calculate how far the stock is likely to move in a given range by a certain date.
In this case, if we’re looking at this stock which is just the S&P at some point in the future that we’ve taken this chart, then you can see that by the time that we reached this date or this line here, this expiration date, as the stock is trading, it might end up trading somewhere in this range.
And that’s a good likelihood of happening because the stock didn't have that much time and based on its entire trading history, it’s not likely to make a move all the way up here or all the way down here gave such a short period, so we know we can calculate that.
As we go further out in time, the stock is likely to make more of a volatile move. It adds more time here, and you can see it can widen out its breath of movement.
That’s true because as you can see going forward here on the S&P, the longer we went in the timeframe, the more the stock could move over time, and that happens here too.
Again, just continuing down to the future, you can see the stock can then really move as we start to go further and further out on the expiration cycle. This same type of distribution can then be applied to where the stock moves over time.
Most of the time, the stock is going to move with the inside of this one standard deviation movement. This one standard deviation movement is about 68% of the time. We can exactly calculate this probability inside of most broker platforms.
I’m going to show you how we do it at the end of this video, but just trust me that this one standard deviation move is about 68% of the time.
It’s helpful to understand where a stock might move 68% of the time because then, we can build a strategy around that movement or take advantage of that possible movement and this is how we get to high probability trading.
As we go forward, let’s first do a quick review of volatility because all of this normal distribution and stock distribution stuff has a hinge and that hinge is volatility and volatility and option pricing. Let's take two stocks in this example. Both stocks start out at the same price which is \$100.
In this case, stock A is the stock that’s in black on this chart and you can see it has very little volatility which means that it moves more or less right around \$100 give or take maybe \$5 or \$10 in the opposite direction, so it’s moving very, very slowly around \$100, it has low volatility, the frequency and the magnitude of its moves are very small.
Compare that to stock B which is going to be the stock that’s in blue, you can see they both start out and end at the same price, but stock B has much higher and much more volatile moves in its price as it gets to that average of \$100.
You can say that stock B which is the one here in the Blue has higher implied volatility than stock A. Again, stock A which is the one here in black, lower implied volatility, still the same stock, still around \$100, it’s just the level of movement or the frequency of movement that that stock makes.
This drives us to our next topic which is implied volatility. Implied volatility is an expectation of where the stock might move in the future. Depending on how volatile or not a stock is, that will cause option pricing to increase or decrease as a result to compensate for that implied volatility.
We take our normal distribution graph which is the one here in blue. This is that one from one of the other screens, so just a normal distribution, our average volatility in the market.
You might see the stock have a range of between here and here, so the two extremes with something around the median or the mean as far as its distribution going out into the future. If implied volatility for that stock is a lot lower…
Remember stock A from the slide before? That black line that was hovering around \$100? If implied volatility is a lot lower, then that creates this distribution graph to get taller and skinnier, and that’s this red graph here.
You can see that it still has a normal distribution, but it's much more centered on the stock making very small movements out into the future. Instead of movements all the way out here, now the extreme movements or the three standard deviation moves are much, much closer to the mean of the stock.
Our standard deviations have moved in from a further out area, so the implied volatility in the stock is lower, and that means that the likely range of the stock going forward is going to be much smaller, it’s not going to have the greatest magnitude of movement.
Now, if we have a stock that has implied volatility that's extremely high, so it's making a lot of jagged and very quick moves like that stock we’ve looked at in the slide before, that blue line, it’s all over the place, still centered around \$100.
But all over the place, then what that does is that slams down this distribution graph, and it makes it much shorter and fatter. This distribution graph looks like this. It’s much more distributed this way.
It’s a very flat, very wide graph. You can see because it's very volatile, the stock can rally high, it can go that high, or it can go that low.
Most of it is going to be around some average or mean, but you'll notice that the average and mean has expended and now 68% of the time, it trades within this range which is all the way out towards the end of its shading.
68% of the time in high implied volatility, the range of the stock is much lower. Compare this with 68% of the time when implied volatility is low; it's going to have a much shorter or narrow window to trade within.
You can see now that implied volatility is a critical ingredient to your ability to be successful, but it’s also this understanding of how implied volatility shapes and molds this distribution graph that we use.
As implied volatility increases or what's commonly called Vega in option pricing, an option’s price will increase as well to compensate for the higher probability of being in the money at expiration.
Remember, as a stock starts to make more frequent moves, that option’s price is going to increase because now these options at the further extreme have an opportunity to be in the money at expiration, and the options down below also have a further chance of being in the money at expiration because the stock is making huge, huge moves in either direction.
This is why we specifically suggest that you sell options when implied volatility is high because option pricing is going to be very much expensive and rich and swollen because of implied volatility.
And this is also why we suggest that you buy options generally when implied volatility is low, and that's because option prices are going to be low and have the propensity maybe to increase in the future.
Now, with all of this hard data behind both volatility and possible ranges in the stock, we can build option strategies that target any probability of success we want. This is the key ingredient here, is that with options, you have the ability to target any possible win rate that you want.
If you're trading stocks, your win rate is 50/50, you have a 50% chance the stock goes up, you have a 50% chance that the stock goes down, but with options, we have the awesome ability to target any probability of success that we want.
Let’s look at a specific example here. This is a trade tab of SPY which is the S&P 500 index, and currently, SPY is trading right at 204.23. In the next month, (these are the February contracts, the next month out is February contracts which are 29 days out) you can see that we are in a position right now where we’re selling a spread or selling options above the market at the 208 strike price.
Again, the stock is trading at 204, and we’re trading options all the way out at 208. Based on all the trading history of SPY at this point and all of the volatility in SPY at this point, the probability of our option being in the money at expiration using that distribution graph that we looked at a couple of slides ago.
The probability right now, hard numbers that our option is in the money at expiration and loses is 29.35%. The probability that a stock goes up to our level and closes above that level is 29.35%.
Let's use that same probability and go back to our stock distribution graph and let’s just say that the SPY which is currently right about here is trading at about 203/204, right where it was on the slide before.
If our strike price is up here at 208, if this is the level that we don't want SPY to cross or breach because we want SPY to close anywhere below this level for us to make money, then what we're saying here with this distribution graph is that there’s about a 30% chance that that happens.
We showed you where we got that number from and how we derived it, but there’s a 30% chance that SPY from where it’s at right now goes up to and closes beyond our strike price by expiration.
If there’s a 30% chance of this happening, that also means there’s a 70% chance that it doesn't happen and a 70% chance that SPY never makes it up there and closes beyond that level at expiration.
This is where we get our very high probability of success trading. In this instance, the trade that we are truly in right now (and you can see this with the position markers) is a trade that has a 70% chance of success as it stands right now.
The beauty of options like we said is that you can pinpoint your chance of success at any level that you want. In this case, if you want a higher level of success, you can go out to these options which are the 210 options, those have about a 19% chance of being in the money or losing at expiration.
That means if they have a 19% chance of losing at expiration, then they have about an 81% chance of being a winner at expiration. Now, of course, the market is going to compensate you and reduce a little bit of the money that you make because you have a little bit higher chance of success.
With these options that we’ve sold, we sold those for \$145 and we only have a 70% chance of success (I say 70% chance of success like it’s some lower number, but you know it’s an extremely high probability) versus going out to the 210 strike which is over here.
The 210 calls have about a 20% chance of losing, so an 80% chance of success and you’re only going to get \$78.5 for that option. You can see the markets are extremely efficient; they know that further out option has a higher likelihood of winning and therefore, you’re not going to make as much money.
There’s a sweet spot in there, but you can see, you can pin your probability of success anywhere you want. Just to drive home the point again, we can go all the way out to the 212.5 calls, and you can see the probability of losing on those is 9.56%, so about 10%, meaning that this is a 90% chance of success trade.
It’s real, really powerful how we can use these probabilities when trading to pinpoint our chance of success and we cannot replicate this with stocks because stocks always have a 50/50 probability of success.
With all of this said and wrapped up here, your only goal moving forward to be successful in trading options is to make as many small high probability trades as you can on the right side of volatility, period, end of story.
That is the ultimate goal with trading, is to make as many small, very small positions, you don’t take on too much risk, high probability trades like we just showed you that have a high likelihood or chance of success on the right side of volatility.
Just understanding if implied volatility is low or relatively high, so that you know if the market can expand in price or can contract in price. Remember that we want to sell options when implied volatility is high, and we want to buy them when implied volatility is low.
I hope you enjoyed this video. I know it was a more advanced tutorial, but we’re getting to a lot more concepts as we get through this part of the course here. As always, if you have any comments or questions, please ask them right below. Until next time, happy trading! | 4,176 | 19,696 | {"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} | 2.859375 | 3 | CC-MAIN-2017-26 | longest | en | 0.940169 |
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