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https://www.gradkastela.com/point-slope-form-using-x-intercept-ten-quick-tips-regarding-point-slope-form-using-x-intercept/389600/ | 1,585,941,106,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00386.warc.gz | 943,366,652 | 5,902 | # Point Slope Form Using X Intercept Ten Quick Tips Regarding Point Slope Form Using X Intercept
Are you apprehensive why I asked you to break the beeline equations in the aloft contest for y? I wasn’t aloof authoritative you jump through hoops (although I am afflicted by your agility, I charge admit); there’s absolutely a actual acceptable acumen to do it. Already a beeline blueprint is apparent for y, it is said to be in slope-intercept form.
Point Slope Form of Equation of Line – YouTube – point slope form using x intercept | point slope form using x intercept
Point Slope Form – point slope form using x intercept | point slope form using x intercept
Graph 2x + 4y = 6 – YouTube – point slope form using x intercept | point slope form using x intercept
There are two big allowances that aftereffect back you put an blueprint in slope-intercept anatomy (and you can apparently amount out what they are based on the name): You can analyze the abruptness and y-intercept of the band (get this) after accomplishing any added assignment at all! Plus, if you transform equations while you do sit-ups, the slope-intercept anatomy could absolutely accord you greater analogue in your abs!
Let’s get algebraic for a moment. Officially speaking, the slope-intercept anatomy of a band is accounting like this:
How to Find y-intercept with an equation in point slope .. | point slope form using x intercept
Once a beeline blueprint is apparent for y, it is in slope-intercept form, y = mx b. The accessory of the x term, m, is the abruptness of the band and the cardinal (or constant), b, is the y-intercept.
In added words, already you break a beeline blueprint for y, the accessory of x will be the abruptness of the line, and the cardinal with no capricious absorbed (called the constant) marks the atom on the y-axis, (0,b), area the band passes through.
Problem 2: Analyze the abruptness and the coordinates for the y-intercept accustomed the beeline blueprint 3x 2y = 4.
Example 2: Analyze the abruptness and the coordinates for the y-intercept accustomed the beeline blueprint x – 4y = 12.
Solution: Remember, all you accept to do to transform an blueprint into slope-intercept anatomy is to break it for y. To abstract the y, decrease x from both abandon of the blueprint and again bisect aggregate by the accessory of – 4:
The x-term’s accessory is 1⁄4, so the abruptness of the band is 1⁄4. Since the connected is -3, the blueprint of the blueprint will canyon through the y-axis at the point (0,-3). (Don’t balloon that the x-coordinate of a point on the y-axis will consistently be 0, and carnality versa.)
Excerpted from The Complete Idiot’s Guide to Algebra © 2004 by W. Michael Kelley. All rights aloof including the appropriate of reproduction in accomplished or in allotment in any form. Used by adjustment with Alpha Books, a affiliate of Penguin Group (USA) Inc.
You can acquirement this book at Amazon.com and Barnes & Noble.
Point Slope Form Using X Intercept Ten Quick Tips Regarding Point Slope Form Using X Intercept – point slope form using x intercept
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Home > SAT > SAT Practice Exam Math Test 18 Grid Ins Questions | SAT Online Course AMBiPi
SAT Practice Exam Math Test 18 Grid Ins Questions | SAT Online Course AMBiPi
Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In this article, you will get SAT 2022 Practice Exam Math Test 18 Grid Ins Questions with Answer Keys | SAT Online Course AMBiPi.
SAT 2022 Practice Exam Math Test 18 Grid Ins Questions with Answer Keys
SAT Math Practice Online Test Question No 1:
Daniel works for a pest control company and is spraying all the lawns in a neighborhood. The figure above shows the layout of the neighborhood and the times that Daniel started spraying the lawns at two of the houses. Each lawn in the neighborhood is approximately 0.2 acres in size and takes the same amount of time to spray.
Daniel uses a mobile spray rig that holds 20 gallons of liquid. It takes 1 gallon to spray 2,500 square feet of lawn. How many times, including the first time, will Daniel need to fill the spray rig, assuming he fills it to the very top each time? [1 acre = 43,560 square feet]
The total acreage of all the lawns in the neighborhood is 21 × 0.2 = 4.2 acres. This is equivalent to 4.2 × 43,560 = 182,952 square feet. Each gallon of spray covers 2,500 square feet so divide to find that Daniel needs 182,952 ÷ 2,500 = 73.1808 gallons to spray all the lawns. The spray rig holds 20 gallons, so Daniel will need to fill it 4 times. After he fills it the fourth time and finishes all the lawns, there will be some spray leftover.
SAT Math Practice Online Test Question No 2:
In the circle above, arc AB has a measure of 7π. What is the value of x?
Since an arc is simply a portion of a circumference, let’s first calculate the circumference of the circle: C = 2πr = 2π(36) = 72π
Because arc AB has a measure of 7π, it is 7π/72π = 7/72 of the entire circumference. Since x° is the measure of the central angle that corresponds to this arc, it must be the same fraction of the whole: x°/360° = 7/72
Cross multiply: 72x = 7(360)
Divide by 72: x = 7(5)
Simplify: x = 35
SAT Math Practice Online Test Question No 3:
If (2/3)a + (1/2)b = 5 and b = 4, what is the value of a?
Original equation: (2/3)a + (1/2)b = 5
Substitute b = 4: (2/3)a + (1/2)(4) = 5
Simplify: (2/3)a + 2 = 5
Subtract 2: (2/3)a = 3
Multiply by 3/2: a = 9/2
SAT Math Practice Online Test Question No 4:
What is one possible solution to the equation 6/(x + 1) – 3/(x – 1) = 1/4?
Original equation: 6/(x + 1) – 3/(x – 1) = 1/4
Multiply by 4(x + 1)(x – 1):
24(x + 1)(x – 1)/(x + 1) – 12(x + 1)(x – 1)/(x – 1) = 4(x + 1)(x – 1)/4
We do this because 4(x + 1)(x – 1) is the least common multiple of the denominators, so multiplying both sides by this will eliminate the denominators and simplify the equation.
Cancel common factors: 24(x – 1) – 12(x + 1) = (x + 1)(x – 1)
Distribute and FOIL: (24x – 24) – (12x + 12) = x2 – 1
Collect like terms: 12x – 36 = x2 – 1
Subtract 12x and add 36: 0 = x2 – 12x + 35
Factor: 0 = (x – 5)(x – 7)
Solve using Zero Product Property: x = 5 or 7
SAT Math Practice Online Test Question No 5:
If 3 – 1/b = 3/2, what is the value of b?
Correct Answer: 2/3 or.666 or. 667
3 – 1/b = 3/2
Multiply by the common denominator, 2b: 6b – 2 = 3b
Add 2: 6b = 3b + 2
Subtract 3b: 3b = 2
Divide by 3: b = 2/3
SAT Math Practice Online Test Question No 6:
The function f is a quadratic function with zeros at x = 1 and x = 5. The graph of y = f(x) in the xy-plane is a parabola with a vertex at (3, -2). What is the y-intercept of this graph?
We know that if a quadratic has zeroes at x = 1 and x = 5, it must have factors of (x – 1) and (x – 5). Since a quadratic can only have two linear factors, f must be of the form f(x) = k(x – 1)(x – 5).
Substitute x = 3 and y = -2 for the coordinates of vertex: -2 = k(3 – 1)(3 – 5)
Simplify: 2 = k(2)(-2)
Simplify: -2 = -4k
Divide by -4: 1/2 = k
Therefore the equation of the function is f(x) = 1/2(x – 1)(x – 5), and we can find its y-intercept by substituting x = 0:
f(0) = 1/2(0 – 1)(0 – 5)
Simplify: f(0) = 5/2
SAT Math Practice Online Test Question No 7:
If a = 4√2 and 2a =√2b, what is the value of b?
Given: a = 4√2
Multiply by 2: 2a = 8√2
Substitute: 2a = 2a = √2b
Square both sides: 2b = 64(2)
Divide by 2: b = 64
SAT Math Practice Online Test Question No 8:
In the figure above, a circle is inscribed in a square that is inscribed in a larger circle. If the area of the larger circle is 16.5 square units, what is the area of the smaller circle?
Most students will begin this problem by trying to find the length of the radius of the larger circle. This is a bit of a pain and, as it turns out, completely unnecessary. Instead, start by drawing in the 45°-45°-90° triangle as shown, and notice that one leg of this triangle is the radius of the smaller circle, and the hypotenuse is the radius of the larger circle. This is the key to the relationship between the circles.
If we label the smaller leg r and use either the Pythagorean Theorem or the Reference Information about 45°-45°-90° triangles given at the beginning of the test, we find that the hypotenuse is r√2. Therefore, the area of the smaller circle is πr2 and the area of the larger circle is π(r√2)2 = (2πr)2.
In other words, the larger circle has an area that is twice the area of the smaller circle. Therefore, if the larger circle has area 16.5, the smaller circle has an area of 16.5 ÷ 2 = 8.25.
SAT Math Practice Online Test Question No 9:
If x2 + 12x = 13, and x < 0, what is the value of x2?
x2 = 12x = 13
Subtract 13: x2 + 12x – 13 = 0
Factor: (x + 13)(x – 1) = 0
Use the Zero Product Property: x = -13 or x = 1
If x < 0, x must be -13. Therefore x2 = (-13)2 = 169.
Alternately, if you have QUADFORM (a quadratic formula program) programmed into your calculator, select PROGRAM, QUADFORM, and input a = 1, b = 12 and c = -13 to find the zeros (-13 and 1).
SAT Math Practice Online Test Question No 10:
If m and n are integers such that m2 + n2 = 40 and < 0 < n, what is the value of (m + n)2?
Because m2 + n2 = 40, where m and n are both integers, we must look for two perfect squares that have a sum of 40. The perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 … and the only two of these with a sum of 40 are 4 and 36. So either m2 = 4 and n2 = 36 or m2 = 36 and n2 = 4.
CASE 1: m2 = 4 and n2 = 36
Take square root: m = ±2 and n = ±6
Since m < 0 < n: m = -2 and n = 6
Evaluate (m + n)2: (m + n)2 = (-2 + 6)2 = 42 = 16
CASE 2: m2 = 36 and n2 = 4
Take square root: m = ±6 and n = ±2
Since m < 0 < n: m = -6 and n = 2
Evaluate (m + n)2: (m + n)2 = (-6 + 2)2 = (-4)2 = 16
AMAN RAJ
I am AMAN KUMAR VISHWAKARMA (in short you can say AMAN RAJ). I am Mathematics faculty for academic and competitive exams. For more details about me, kindly visit me on LinkedIn (Copy this URL and Search on Google): https://www.linkedin.com/in/ambipi/
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https://communities.sas.com/t5/Base-SAS-Programming/Time-Conversion/td-p/213036?nobounce | 1,532,048,972,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00128.warc.gz | 635,892,704 | 31,787 | Solved
Contributor
Posts: 41
# Time Conversion
Hi All,
I have variable I need converted from a number to a time value. The format now is 15, 30, 45, 100, 115 and I want 12:00AM, 12:15AM, 12:30AM, 12:45AM 1:00AM etc.
Steve
Accepted Solutions
Solution
06-24-2015 02:08 PM
Regular Contributor
Posts: 180
## Re: Time Conversion
This is Mark Johnson' solution improved:
data want;
set have;
txt=put(number,z4.);
time = put(input(substr(txt,1,2),2.)*3600+(input(substr(txt,3),2.)*60),timeampm.);
drop txt;
run;
CTorres
All Replies
Regular Contributor
Posts: 227
## Re: Time Conversion
RTFM on the
TIMEAMPMw.d Format
attrib MyTimeVar length = 8 format =timeampm5.2;
Super User
Posts: 9,599
## Re: Time Conversion
Post some test data (as a datastep). At a guess:
Step 1 - split the variable and convert to minutes (note the best coding method, but its to show the process):
data want;
set have;
i=1;
do while (scan(variable,i,",") ne "");
mins=input(scan(variable,i,","),best.);
if mins=100 then mins=60;
if mins=115 then mins=75;
new_time="12:00"t + mins;
output;
end;
run;
Valued Guide
Posts: 864
## Re: Time Conversion
Are you sure 100 should be 1:00 am? You go from 45 minutes to 100 minutes, logically 45 = 12:45am and 100 = 1:00am doesn't jive:
data have;
input number;
cards;
15
30
45
100
115
;
run;
data want;
set have;
time = put(number*60,timeampm.);
run;
Contributor
Posts: 20
## Re: Time Conversion
Are you saying that your actual data values are HOUR:MINUTE, but w/o the colon?
0245 (the integer) <=> 02:45 (the time)??
If so, then pull out the hour, minute values, convert to seconds.
(but w/o any data, we're just guessing).
data want; set have;
seconds = hours*3600 + minutes*60;
format seconds timeampm. ;
run;
Solution
06-24-2015 02:08 PM
Regular Contributor
Posts: 180
## Re: Time Conversion
This is Mark Johnson' solution improved:
data want;
set have;
txt=put(number,z4.);
time = put(input(substr(txt,1,2),2.)*3600+(input(substr(txt,3),2.)*60),timeampm.);
drop txt;
run;
CTorres
Contributor
Posts: 41
## Re: Time Conversion
Thanks guys! I'm sure all of them would work, but I got the last one to work first. Thanks again!
Steve
🔒 This topic is solved and locked.
Need further help from the community? Please ask a new question.
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# Problem 690. Remove the two elements next to NaN value
Solution 330854
Submitted on 8 Oct 2013 by Jan Orwat
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [6 10 5 8 9 NaN 23 9 7 3 21 43 NaN 4 6 7 8]; y_correct = [6 10 5 8 9 7 3 21 43 7 8]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
%% x = [25 NaN 1 3]; y_correct = 25; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
%% x = [ NaN 15 15 17 ] y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
x = NaN 15 15 17 | 230 | 624 | {"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} | 2.6875 | 3 | CC-MAIN-2019-39 | latest | en | 0.580401 |
https://www.physicsforums.com/threads/find-the-acceleration-of-the-yo-yo.51856/ | 1,542,286,780,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742685.33/warc/CC-MAIN-20181115120507-20181115142507-00504.warc.gz | 945,156,408 | 12,294 | # Homework Help: Find the acceleration of the yo-yo
1. Nov 8, 2004
### wildrjetta
I have calculated the moments of inertia for both the solid disk and the rod, but am getting stuck on how to solve for alpha in the torque equations so that I get acceleration. Can anyone shed some light on this question for me? I really appreciate it.
Here is the question I seem to be struggling with:
A 0.24kg yo-yo consists of two solid disks of radius 11.5cm joined together by a massless rod of radium 1.00cm and a string wrapped around the rod. One end of the string is held fixed and is under constant tension T as the yo-yo is released. Find the acceleration of the yo-yo and the tention T.
2. Nov 8, 2004
### wildrjetta
i am still stuck on this problem....can anyone help?
3. Nov 8, 2004
### ehild
The in-plane motion of a rigid body consist of a translation of its centre of mass and a rotation around the axis through the center of mass. The linear acceleration is proportional to the net force. The angular acceleration of the rotation is proportional to the net torque. Can you proceed from here?
ehild
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http://summerschoolurban.eu/introduction-to-differential-geometry-pdf/ | 1,531,799,325,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00532.warc.gz | 365,112,224 | 4,153 | # Introduction to differential geometry pdf
Please consider recommending that your university or department librarian order this book. This 5th edition contains all the things that made the earlier editions different from other introduction to differential geometry pdf. A new way of introducing differential forms and the exterior derivative. We included a detailed proof of the singular value decomposition, and show how it applies to facial recognition: “how does Facebook apply names to pictures?
This article is about co-ordinate geometry. In classical mathematics, analytic geometry, also known as coordinate geometry or Cartesian geometry, is the study of geometry using a coordinate system. Analytic geometry is widely used in physics and engineering, and also in aviation, rocketry, space science, and spaceflight. The Greek mathematician Menaechmus solved problems and proved theorems by using a method that had a strong resemblance to the use of coordinates and it has sometimes been maintained that he had introduced analytic geometry. Analytic geometry was independently invented by René Descartes and Pierre de Fermat, although Descartes is sometimes given sole credit.
Pierre de Fermat also pioneered the development of analytic geometry. Paris in 1637, just prior to the publication of Descartes’ Discourse. Illustration of a Cartesian coordinate plane. In analytic geometry, the plane is given a coordinate system, by which every point has a pair of real number coordinates.
Periodically Forced ODE’s – all of these disciplines are concerned with the properties of differential equations of various types. And a y, not to be confused with Difference equation. Geometric Methods and Applications – the motion of a body is described by its position and velocity as the time value varies. Extending the function, and air resistance may be modeled as proportional to the ball’s velocity. Basic facts in geometry, independent of the starting point.
Axis and the angle θ its projection on the xy, these classes of differential equations can help inform the choice of approach to a solution. Mostly concerned with their solutions, and the relationship involves values of the unknown function or functions and values at nearby coordinates. The solution may not be unique. He solves these examples and others using infinite series and discusses the non, many fundamental laws of physics and chemistry can be formulated as differential equations. Axioms for Euclidean Geometry — linear differential equations are the differential equations that are linear in the unknown function and its derivatives. Plane Symmetries or Isometries, lagrange solved this problem in 1755 and sent the solution to Euler. Also known as coordinate geometry or Cartesian geometry – elementary Projective Geometry.
Or quadric surface; a single equation corresponds to a curve on the plane. Order and degree of a differential equation, this only helps us with first order initial value problems. These definitions are designed to be consistent with the underlying Euclidean geometry. Additiones ad ea, as well as its esthetic aspects. | 581 | 3,123 | {"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} | 3.71875 | 4 | CC-MAIN-2018-30 | latest | en | 0.944071 |
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# Delta products, Inc., has recently switched at least partly
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Delta products, Inc., has recently switched at least partly [#permalink]
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Delta products, Inc., has recently switched at least partly from older technologies using fossil fuels to new technologies powered by electricity. The question has been raised whether it can be concluded that for a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly. The answer, clearly, is yes, since the amount of fossil fuel used to generate the electricity needed to power the new technologies is less than the amount needed to power the older technologies, provided that the level of output is held constant.
In the argument given, the two boldface portions play which of the following roles?
A. The first identifies the content of the conclusion of the argument; the second provides support for that conclusion.
B. The first provides support for the conclusion of the argument; the second identifies the content of that conclusion.
C. The first states the position that the argument opposes; the second states the conclusion of the argument.
D. Each provides evidence that calls the conclusion of the argument into question.
E. Each provides support for the conclusion of the argument.
OA: E
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12 Apr 2007, 21:33
E.
Premise 1
Delta products, Inc., has recently switched at least partly from older technologies using fossil fuels to new technologies powered by electricity.
Consideration : Whether it can be concluded ? that for a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly
Conclusion: The answer, Clearly, is yes i.e. For a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly
Explanation: Keyword 'Since' The amount of fossil fuel used to generate the electricity needed to power the new technologies is less than the amount needed to power the older technologies, provided that the level of output is held constant.
Conclusion rests on premise of the argument and the explanation offered to support it.
With that said lets look at the options.
A. The first identifies the content of the conclusion of the argument; the second provides support for that conclusion.
---No content identified .
B. The first provides support for the conclusion of the argument; the second identifies the content of that conclusion.
--- Support provided but no content identified
C. The first states the position that the argument opposes; the second states the conclusion of the argument.
---Irrelevant
D. Each provides evidence that calls the conclusion of the argument into question.
--- Irrelevant
E. Each provides support for the conclusion of the argument. Yes !
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13 Apr 2007, 12:01
ng wrote:
E.
Premise 1
Delta products, Inc., has recently switched at least partly from older technologies using fossil fuels to new technologies powered by electricity.
Consideration : Whether it can be concluded ? that for a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly
Conclusion: The answer, Clearly, is yes i.e. For a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly
Explanation: Keyword 'Since' The amount of fossil fuel used to generate the electricity needed to power the new technologies is less than the amount needed to power the older technologies, provided that the level of output is held constant.
Conclusion rests on premise of the argument and the explanation offered to support it.
With that said lets look at the options.
A. The first identifies the content of the conclusion of the argument; the second provides support for that conclusion.
---No content identified .
B. The first provides support for the conclusion of the argument; the second identifies the content of that conclusion.
--- Support provided but no content identified
C. The first states the position that the argument opposes; the second states the conclusion of the argument.
---Irrelevant
D. Each provides evidence that calls the conclusion of the argument into question.
--- Irrelevant
E. Each provides support for the conclusion of the argument. Yes !
Perfect explanation...Its E no doubt.
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13 Apr 2007, 17:01
ricokevin wrote:
Delta products, Inc., has recently switched at least partly from older technologies using fossil fuels to new technologies powered by electricity. The question has been raised whether it can be concluded that for a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly. The answer, clearly, is yes, since the amount of fossil fuel used to generate the electricity needed to power the new technologies is less than the amount needed to power the older technologies, provided that the level of output is held constant.
In the argument given, the two boldface portions play which of the following roles?
A. The first identifies the content of the conclusion of the argument; the second provides support for that conclusion.
B. The first provides support for the conclusion of the argument; the second identifies the content of that conclusion.
C. The first states the position that the argument opposes; the second states the conclusion of the argument.
D. Each provides evidence that calls the conclusion of the argument into question.
E. Each provides support for the conclusion of the argument.
OA: E
E
The second one provides support. This means only A and E can be correct answers. Now if you read carefully then you will realize that first bold section is NOT the conclusion itself. This eliminates A. Hence E is the answer
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14 Apr 2007, 07:30
Thanks
Had to decide between A and E. Got wrong again on a 50:50 thing.
I hate it when this happens.
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14 Apr 2007, 11:17
ricokevin wrote:
Delta products, Inc., has recently switched at least partly from older technologies using fossil fuels to new technologies powered by electricity. The question has been raised whether it can be concluded that for a given level of output, Delta’s operation now causes less fossil fuel to be consumed than it did formerly. The answer, clearly, is yes, since the amount of fossil fuel used to generate the electricity needed to power the new technologies is less than the amount needed to power the older technologies, provided that the level of output is held constant.
In the argument given, the two boldface portions play which of the following roles?
A. The first identifies the content of the conclusion of the argument; the second provides support for that conclusion.
B. The first provides support for the conclusion of the argument; the second identifies the content of that conclusion.
C. The first states the position that the argument opposes; the second states the conclusion of the argument.
D. Each provides evidence that calls the conclusion of the argument into question.
E. Each provides support for the conclusion of the argument.
OA: E
E
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Re: CR - boldface [#permalink] 14 Apr 2007, 11:17
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# Delta products, Inc., has recently switched at least partly
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®. | 2,015 | 8,977 | {"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} | 3.484375 | 3 | CC-MAIN-2017-47 | latest | en | 0.92999 |
https://de.mathworks.com/matlabcentral/cody/problems/1985-how-unique/solutions/351335 | 1,600,413,186,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00223.warc.gz | 466,028,359 | 16,614 | Cody
# Problem 1985. How unique?
Solution 351335
Submitted on 14 Nov 2013
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% A = [2 2 2 3 3 2 3 8 6 5 6]; [U, H] = hunique(A); U_ok = [2 3 8 6 5]; H_ok = [4 3 1 2 1]; assert(isequal(U,U_ok)); assert(isequal(H,H_ok));
Error: Edge vector must be monotonically non-decreasing.
2 Fail
%% A = [2 2 2 3 3 2 3 8 6 5 6 8]; [U, H] = hunique(A); U_ok = [2 3 8 6 5]; H_ok = [4 3 2 2 1]; assert(isequal(U,U_ok)); assert(isequal(H,H_ok));
Error: Edge vector must be monotonically non-decreasing.
3 Fail
%% A = 100:-11:1; assert(isequal(hunique(A),A)); [~,H] = hunique(A); assert(isequal(H,ones(1,10)));
Error: Edge vector must be monotonically non-decreasing.
4 Fail
%% A = randi([-10 10],1,100); [U,H] = hunique(A); assert(sum(H)==numel(A)); assert(isequal(unique(A),sort(U))); % number of test cases may increace in the future. % any proposals of test cases warmly welcome.
Error: Edge vector must be monotonically non-decreasing. | 393 | 1,103 | {"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} | 3.125 | 3 | CC-MAIN-2020-40 | latest | en | 0.563208 |
https://matheducators.stackexchange.com/questions/10636/source-of-conceptual-multiple-choice-calculus-questions | 1,660,336,263,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00716.warc.gz | 371,507,318 | 67,657 | # Source of conceptual, multiple choice calculus questions
I'd like to give my Calculus 1 class periodic multiple choice questions that really test conceptual understanding. Ideally, I'd like these questions to require very little computation. I know that a lot of textbooks have true false questions, which I like, but I'm hoping to find a source of questions with more than two possible answers. Something along the following lines:
Suppose that $$f(x)$$ and $$g(x)$$ are such that $$f'(x)>g'(x)$$ for all real $$x$$. Which of the following statements must be true?
1. The graphs of $$f$$ and $$g$$ intersect exactly once.
2. The graphs of $$f$$ and $$g$$ intersect no more than once.
3. The graphs of $$f$$ and $$g$$ do not intersect.
4. The graphs of $$f$$ and $$g$$ may intersect any number of times.
For a student who understands the derivative well, this question could be answered in under one minute with absolutely no computations.
Anyone know of a good source to find a bank of such questions? It would be ideal if the TeX code was available too, or if the problems were already encoded into WeBWorK or some other online homework system.
• You might want to add a continuity condition to your example. :) Feb 28, 2016 at 2:31
• @mweiss: For brevity's sake I wasn't completely explicit, but implicit in the statement that $f'(x)>g'(x)$ for all real $x$ is the fact that both $f$ and $g$ are differentiable on the reals, and thus continuous. Feb 28, 2016 at 5:30
Cornell's Good Questions Project has a great question bank for conceptual questions.
• That's terrific... and somewhat addictive. Feb 28, 2016 at 21:08
The Calculus textbook of Hughes-Hallett, Gleason and McCallum et. al. includes ConcepTests that are rather good for this purpose. Sorry this is not a free online resource, but I hope it helps.
I tend to download testbanks for Pearson's TestGen application (requires instructor account at Pearson, available for almost any text they produce) and hunt through them for conceptual questions for this purpose. Sometimes it's a bit slim pickings in this regard, but it at least gets me started for a first semester, and then more ideas occur to me as I teach the course, and I personalize the quizzes more over time.
Though not nearly as good as some of the other suggestions here, the practice exams for AP Calculus contain decent problems. The sample multiple choice ones are buried inside the "AP Calculus Course Description" pdf.
If anyone is still interested...
Calculus Video Projects have some resources that contain conceptual problems under their 'Resources' tab. They also have conceptual videos for Calculus 1 topics.
https://calcvids.org/videos/ | 622 | 2,690 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-33 | longest | en | 0.947888 |
https://vnemart.com.vn/what-is-6-6/ | 1,695,399,844,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00421.warc.gz | 668,081,076 | 12,179 | # Top 4 what is 6 6 best, you should know
Nội dung chính
## 1 What is 6/6 of 6? (Calculate 6/6 of 6)
• Author: visualfractions.com
• Published Date: 11/16/2021
• Review: 4.82 (681 vote)
• Summary: Learn how to calculate what 6/6 of 6 is with a simple, step-by-step guide to find the fraction of any number
• Matching search results: What is 6/6 of 6?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/fraction-of-number/what-is-6-6-of-6/.
## 2 What is 6/6 Vision? – Stan Isaacs Ortho K – Optometrists
• Author: isaacs-optom.com
• Published Date: 08/07/2022
• Review: 4.84 (838 vote)
• Summary: The numbers stand for 6 metres/6 metres or 20 feet/20 feet. While the common understanding of perfect vision is 6/6, it is possible for a person to possess
• Matching search results: Stahl, W. (2005). Macular Carotenoids: Lutein and Zeaxanthin in Nutrition & The Eye: Basic and Clinical Research. Edited by Albert J. Augustin. (Karger: Basel). P. 70 – 88. From Developments in Ophthalmology Vol. 38.
## 3 Distance vision test – Myopia Control Singapore
• Author: myopia.com.sg
• Published Date: 08/18/2022
• Review: 5 (762 vote)
• Summary: · 6/5 represents the fortunate 10 percent of the population that has better than normal vision
• Matching search results: Normal good visual acuity is 6/6. If your visual acuity is worse than 6/6, move on to the next test, the Pin Hole Test. The Pin Hole Test will determine whether you need to wear corrective glasses to see better.
## 4 What is ‘perfect’ vision? – The Wimpole Eye Clinic
• Author: wimpoleeyeclinic.com
• Published Date: 10/17/2021
• Review: 4.94 (957 vote)
• Summary: Many people refer to ‘perfect’ vision as ‘6/6′ or ’20/20’ (US notation measured in feet), but this is not strictly true – these terms refer to ‘average’ vision. If you achieve a vision measurement of 6/6, this means that you can see at a distance of 6m what an average person also sees at the same distance
• Matching search results: Many people refer to ‘perfect’ vision as ‘6/6’ or ‘20/20’ (US notation measured in feet), but this is not strictly true – these terms refer to ‘average’ vision. If you achieve a vision measurement of 6/6, this means that you can see at a | 648 | 2,226 | {"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} | 3.25 | 3 | CC-MAIN-2023-40 | latest | en | 0.846641 |
https://q4interview.com/programming-questions/list-of-programs-asked-from-experience-candidates/7 | 1,632,084,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00496.warc.gz | 506,887,592 | 24,518 | Take FREE!! Online Cocubes Mock Test to Crack various Companies Written Exams.
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1 | 520 | 1,774 | {"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-2021-39 | latest | en | 0.623811 |
https://math.stackexchange.com/questions/4917719/models-of-a-slice-of-coherent-category | 1,719,352,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00365.warc.gz | 339,515,148 | 36,401 | # Models of a slice of coherent category
I am trying to get some sense of the proof of Theorem 10 in this link.
https://www.math.ias.edu/%7Elurie/278xnotes/Lecture6-Completeness.pdf
Immediately after it, Exercise 11 says a model of coherent category with a chosen point is equivalent to a model of the slice category.
May I please ask how precisely should I build this correspondence? Even just a quick construction without proof details reply would be appreciated. I would like to check it.
We have a coherent functor $$\mathcal{C}\to\mathcal{C}_{/X}, Y\mapsto (\mathrm{pr}_2\colon Y\times X\to X)$$, so if we have a model $$M'\colon\mathcal{C}_{/X}\to\mathrm{Set}$$ we get a model $$M\colon \mathcal{C}\to\mathrm{Set}$$ via compositon. The chosen element of $$M[X]$$ is induced by the diagonal map $$X\to X\times X$$ in the following sense: the diagonal $$X\to X\times X$$ in $$\mathcal{C}$$ is also a map $$(\mathrm{id}_X\colon X\to X)\to (\mathrm{pr}_2\colon X\times X\to X)$$ in $$\mathcal{C}_{/X}$$, and applying $$M'$$ to it gives us a map $$*\cong M'[\mathrm{id}_X]\to M'[\mathrm{pr}_2\colon X\times X\to X]=:M[X]$$. This selects the chosen element of $$M[X]$$.
Conversely, given a model $$M\colon\mathcal{C}\to\mathrm{Set}$$ with a chosen element $$m\colon *\to M[X]$$, we can define $$M'[p\colon Y\to X]$$ via a pullback square $$\require{AMScd} \begin{CD} M'[p\colon Y\to X]@>>> M[Y]\\ @VVV @VV{M[p]}V\\ * @>{m}>> M[X] \end{CD}$$ in $$\mathrm{Set}$$. You can check that $$M'\colon\mathcal{C}_{/X}\to\mathrm{Set}$$ is indeed a functor and coherent, and that these assignments define a bijection between the appropriate sets Lurie talks about. If you want details about this, let me know and I will edit them in.
• Thanks a lot for this instant answer! May I please ask what is meant by the element of $MX$ is induces by the diagonal map? I would appreciate for more details because after seeing the construction, it is still not such clear to me that it is the obvious thing to do.
• I edited in the definition of the chosen element of $M[X]$. When you ask for more details, do you mean you also would like more details about why in the second paragraph $M'$ as defined there is coherent and why we get this bijection between the sets Lurie talks about? Or was the definition of the chosen element of $M[X]$ the only thing? (It's fine in either case, but I can read your comment either way, which is why I ask for this clarification.) Commented May 16 at 20:52 | 727 | 2,475 | {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-26 | latest | en | 0.817678 |
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# When people predict that certain result will not take place
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When people predict that certain result will not take place [#permalink]
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12 Mar 2007, 15:53
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When people predict that certain result will not take place unless a certain action is taken, they believe that they have learned that the prediction is correct when the action is taken and the result occurs. On reflection, however, it often becomes clear that the result admits of more than one interpretation.
Which of the following, if true, best supports the claims above?
(A) Judging the success of an action requires specifying the goal of the action.
(B) Judging which action to take after a prediction is made requires knowing about other actions that have been successful in similar past situations.
(C) Learning whether a certain predictive strategy is good requires knowing the result using that strategy through several trials.
(D) Distinguishing a correct prediction and effective action from an incorrect prediction and ineffective action is often impossible.
(E) Making a successful prediction requires knowing the facts about the context of that prediction.
If you have any questions
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Re: CR - prediction and action [#permalink]
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12 Mar 2007, 16:25
amd08 wrote:
When people predict that certain result will not take place unless a certain action is taken, they believe that they have learned that the prediction is correct when the action is taken and the result occurs. On reflection, however, it often becomes clear that the result admits of more than one interpretation.
Which of the following, if true, best supports the claims above?
(A) Judging the success of an action requires specifying the goal of the action.
(B) Judging which action to take after a prediction is made requires knowing about other actions that have been successful in similar past situations.
(C) Learning whether a certain predictive strategy is good requires knowing the result using that strategy through several trials.
(D) Distinguishing a correct prediction and effective action from an incorrect prediction and ineffective action is often impossible.
(E) Making a successful prediction requires knowing the facts about the context of that prediction.
Is it (B) ?
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12 Mar 2007, 16:35
not B. I have the answer but no explanation from a question set. Will post it though after few more replies. Pls. explain the reason behind your answer
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12 Mar 2007, 19:04
I guess the answer is E
My explanation
A not a choice due to the fact that it speaks about success and not prediction - so irrelavant
B not a choice as the whole conversation is about not being able to predict even after a successful result. i would like to keep it as my second best just in case
C not a choice due to the fact that they are talking of some successful iterations and this by no means could help us conclude successfully. I should say I am trying to satisfy myself here in this case to say C is not a choice. I would say 3rd best choice
D totally irrelavant
Now for the best part - WHY E
It is speaking about the context of the prediction so that he can decisively say that the prediction is the only option.
What do you say ?????? Am I right
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Re: CR - prediction and action [#permalink]
### Show Tags
12 Mar 2007, 19:14
amd08 wrote:
When people predict that certain result will not take place unless a certain action is taken, they believe that they have learned that the prediction is correct when the action is taken and the result occurs. On reflection, however, it often becomes clear that the result admits of more than one interpretation.
Which of the following, if true, best supports the claims above?
(A) Judging the success of an action requires specifying the goal of the action.
Not necessarily.
(B) Judging which action to take after a prediction is made requires knowing about other actions that have been successful in similar past situations.
Not necessarily.
(C) Learning whether a certain predictive strategy is good requires knowing the result using that strategy through several trials.
Not necessarily.
(D) Distinguishing a correct prediction and effective action from an incorrect prediction and ineffective action is often impossible.
Out of scope. The argument does not say much about effective action.
(E) Making a successful prediction requires knowing the facts about the context of that prediction.
Left with E.
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### Show Tags
12 Mar 2007, 19:53
Wow! This one is confusing! I somehow was debating between E and C. Could be wrong!
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13 Mar 2007, 07:44
Lets look at this one carefully..
We are looking to support the claim that correct predictions are often correct not because of a correct prediction, but chance.
B.Judging an action...outside of the scope
C. Learning about strategy???? also irrelevant
D Distinguishing between a correct prediction, and a prediction which comes about through chance is impossible....I like this one
E. Something about knowing facts when making predictions...does not support the argument.
We are left with D in my opinion...
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### Show Tags
13 Mar 2007, 09:16
this is tough,
I thought B
but now after reviewing the OA I see why D is correct.
If B said "more than one action " as opposed to interpretation I would agree B would be correct.
However D is right because if there are multiple interpretations for the same event then figuring out which prediction caused an action would be impossible
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### Show Tags
13 Mar 2007, 09:46
Agree D.
13 Mar 2007, 09:46
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# Design elements - Analog and digital logic
The vector stencils library "Analog and digital logic" contains 40 element symbols of logic (threshold) gates, bistable current switches, current controllers, regulators, electrical generators, and amplifiers.
Use it for drawing the digital and analog functions in electronic circuit diagrams and electrical schematics.
"Analogue electronics (or analog in American English) are electronic systems with a continuously variable signal, in contrast to digital electronics where signals usually take only two different levels. The term "analogue" describes the proportional relationship between a signal and a voltage or current that represents the signal." [Analogue electronics. Wikipedia]
"Digital electronics, or digital (electronic) circuits, represent signals by discrete bands of analog levels, rather than by a continuous range. All levels within a band represent the same signal state. Relatively small changes to the analog signal levels due to manufacturing tolerance, signal attenuation or parasitic noise do not leave the discrete envelope, and as a result are ignored by signal state sensing circuitry.
In most cases the number of these states is two, and they are represented by two voltage bands: one near a reference value (typically termed as "ground" or zero volts) and a value near the supply voltage, corresponding to the "false" ("0") and "true" ("1") values of the Boolean domain respectively.
Digital techniques are useful because it is easier to get an electronic device to switch into one of a number of known states than to accurately reproduce a continuous range of values.
Digital electronic circuits are usually made from large assemblies of logic gates, simple electronic representations of Boolean logic functions." [Digital electronics. Wikipedia]
The example "Design elements - Analog and digital logic" was drawn using the ConceptDraw PRO diagramming and vector drawing software extended with the Electrical Engineering solution from the Engineering area of ConceptDraw Solution Park.
Analog and digital logic elements
Used Solutions
## Design elements - Analog and digital logic
The vector stencils library "Semiconductors" contains 22 symbols of rectifiers, diodes, charge transfer and electronic conduction devices, switches, cathodes, transistors, thyristors, and transceivers for semiconductor (SIS) design.
"Semiconductor devices are electronic components that exploit the electronic properties of semiconductor materials, principally silicon, germanium, and gallium arsenide, as well as organic semiconductors. Semiconductor devices have replaced thermionic devices (vacuum tubes) in most applications. They use electronic conduction in the solid state as opposed to the gaseous state or thermionic emission in a high vacuum.
Semiconductor devices are manufactured both as single discrete devices and as integrated circuits (ICs), which consist of a number - from a few (as low as two) to billions - of devices manufactured and interconnected on a single semiconductor substrate, or wafer. ...
All transistor types can be used as the building blocks of logic gates, which are fundamental in the design of digital circuits. In digital circuits like microprocessors, transistors act as on-off switches; in the MOSFET, for instance, the voltage applied to the gate determines whether the switch is on or off.
Transistors used for analog circuits do not act as on-off switches; rather, they respond to a continuous range of inputs with a continuous range of outputs. Common analog circuits include amplifiers and oscillators.
Circuits that interface or translate between digital circuits and analog circuits are known as mixed-signal circuits.
Power semiconductor devices are discrete devices or integrated circuits intended for high current or high voltage applications. Power integrated circuits combine IC technology with power semiconductor technology, these are sometimes referred to as "smart" power devices. Several companies specialize in manufacturing power semiconductors." [Semiconductor device. Wikipedia]
The shapes example "Design elements - Semiconductors" was drawn using the ConceptDraw PRO diagramming and vector drawing software extended with the Electrical Engineering solution from the Engineering area of ConceptDraw Solution Park.
Semiconductor elements
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# Are x and y both positive?
Author Message
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Re: DS - x and y positive [#permalink]
### Show Tags
04 Jul 2006, 10:10
gidimba wrote:
Are x and y both positive?
1. 2x-2y=1
2. x/y >1
(1) x-y = 1/2, can't say if both are +ve (for e.g. 2,-1.5 or -1.5, -2) (INSUFF)
(2) x/y > 1 ; {-10, -5} or {10,5}. even combining both does not provide a definite answer.
(E)
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04 Jul 2006, 10:30
shahnandan wrote:
stmt 1,
x=o y= -1/2 NO
x=3 y=2 1/2 YES
insuff
stmt2
x>y both can be negative or positive -- insuff
together,
consider values for stmt1 as above--E.
x=0 and y = -1/2 do not satisfy x/y > 1
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Re: DS - x and y positive [#permalink]
### Show Tags
04 Jul 2006, 10:33
necromonger wrote:
gidimba wrote:
Are x and y both positive?
1. 2x-2y=1
2. x/y >1
(1) x-y = 1/2, can't say if both are +ve (for e.g. 2,-1.5 or -1.5, -2) (INSUFF)
(2) x/y > 1 ; {-10, -5} or {10,5}. even combining both does not provide a definite answer.
(E)
(1) says x > y
(2) says x and y are either both +ve or both -ve.
considering both together,
if x > y, then both need to be +ve to satisfy x/y > 1
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### Show Tags
04 Jul 2006, 13:16
haas_mba07 wrote:
gidimba,
Can we have the OA please?
1. x = y+1 , not suff
2. x/y>1 , not suff
using both.
(y+1)/y >1
1+(1/y) >1.
This means Y is +Ve. But X = y+1. so both x and y are +ve .. ==> C
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04 Jul 2006, 13:21
shahnandan wrote:
stmt 1,
x=o y= -1/2 NO
x=3 y=2 1/2 YES
insuff
stmt2
x>y both can be negative or positive -- insuff
together,
consider values for stmt1 as above--E.
x/y >1 doesn't mean X >Y.. For ex -5/-3 >1, but -5 isn't greater than -3.
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### Show Tags
04 Jul 2006, 14:11
C
St1: This gives us x >y : INSUFF
St2: This gives us: Either both are +ve or both are -ve. If both +ve then x>y, but if both -ve then x < y. : INSUFF
Combined: Simply we get x > y and both are +ve : SUFF
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05 Jul 2006, 00:03
Guys let me give you the OA.
OA is 'C'.
This is GMATPrep question.
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05 Jul 2006, 00:33
(1) said that x>y
(2) said that x/y>1 so x,y must be both positive(x>y) or negative (x<y)
(1),(2) together we have x,y are both positive numbers.
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05 Jul 2006, 12:49
THE OA is C i think i saw this in powerprep
All u do is
X=2 Y=1.5
or X=-1 Y= 1.5
B states X>Y, hence u can solve and only 1 solution X AND Y are POSTIIVE THUS C
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07 Jul 2006, 03:17
Are x and y both positive?
1. 2x-2y=1
2. x/y >1
From (1) x-y=1/2>0 (not sufficient (x,y) could be (3/2,1) or (-2,-3/2) or (0,-1/2)... N. Suff
From (2), x and y are the same sign and x is further from 0 than is y.
(x,y) could be (3,1), (-3,-1)... N Suff
But together, x and y must be either both positive or both negative (from (2))
If x,y are both negative, x<y<0, so x-y<0, so 2(x-y) cannot be 1, which contradicts (1)
Therefore x any must both be positive Ans. C
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07 Jul 2006, 11:58
OK lets see...
question stem is x AND y BOTH +
1) 2x-2y=1
2(x-y)=1; which can be written as x=1/2 + y
insuff... x can be -1 and y=-1.5 or x=1.5 and y=1
2) x/y > 1
well...DONT do anything like cross multiplication cause we dont know the signs of x or y. All we know is that X and Y have to have the same sign + or -ve. Insuff
lets take em together
sub vaule for x
1/2y + 1 > 1; well this can only be true if Y is +ve...if Y is +ve then X is +ve (given in statement 2)..
C it is....
Hope this helps..its a very simple way of lookin at this problem..and time efficient too, since i dont have to plug in numbers...
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https://stat.ethz.ch/pipermail/r-help/2016-February/436311.html | 1,579,418,377,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00482.warc.gz | 679,050,516 | 2,074 | # [R] Constructing a symmetric matrix with library(corpcor)
Steven Yen syen04 at gmail.com
Mon Feb 22 05:16:06 CET 2016
```I like to compose a symmetric matrix in the pattern as shown below (for 3 x
3 and 4 x 4). For a symmetric matrix of order 5, the result does not seem
right. Help? It is possible to write a two-level do loop for the task, but
I suppose that is less efficient.
> library(corpcor)> r <- 1:3; r[1] 1 2 3> rr <- vec2sm(r, diag = F); rr [,1] [,2] [,3]
[1,] NA 1 2
[2,] 1 NA 3
[3,] 2 3 NA> rr <- rr[upper.tri(rr)]; rr[1] 1 2 3> r <-
vec2sm(rr, diag = F); diag(r) <- 1; r [,1] [,2] [,3]
[1,] 1 1 2
[2,] 1 1 3
[3,] 2 3 1> r <- 1:6; r[1] 1 2 3 4 5 6> rr <- vec2sm(r, diag
= F); rr [,1] [,2] [,3] [,4]
[1,] NA 1 2 3
[2,] 1 NA 4 5
[3,] 2 4 NA 6
[4,] 3 5 6 NA> rr <- rr[upper.tri(rr)]; rr[1] 1 2 4 3 5 6>
r <- vec2sm(rr, diag = F); diag(r) <- 1; r [,1] [,2] [,3] [,4]
[1,] 1 1 2 4
[2,] 1 1 3 5
[3,] 2 3 1 6
[4,] 4 5 6 1> r <- 1:10; r [1] 1 2 3 4 5 6 7 8 9
10> rr <- vec2sm(r, diag = F); rr [,1] [,2] [,3] [,4] [,5]
[1,] NA 1 2 3 4
[2,] 1 NA 5 6 7
[3,] 2 5 NA 8 9
[4,] 3 6 8 NA 10
[5,] 4 7 9 10 NA> rr <- rr[upper.tri(rr)]; rr [1] 1 2
5 3 6 8 4 7 9 10> r <- vec2sm(rr, diag = F); diag(r) <- 1; r
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 2 5 3
[2,] 1 1 6 8 4
[3,] 2 6 1 7 9
[4,] 5 8 7 1 10
[5,] 3 4 9 10 1
(Doesn't seem right).
This is what I need:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 2 4 7
[2,] 1 1 3 5 8
[3,] 2 3 1 6 9
[4,] 4 5 6 1 10
[5,] 7 8 9 10 1
>
[[alternative HTML version deleted]]
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Content
• # A Matter of Pattern ›
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In this lesson, students create and predict patterns formed when making paper snowflakes and learn that patterns may be predicted based on observation.
K,1,2
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In this lesson, students look at shapes in the natural and designed world, investigate ways shapes can be used, and design and build structures.
3,4,5
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In this lesson, students explore the creative aspects of problem-solving and practice creative problem-solving strategies in the context of a story problem.
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• # Shape Hunt ›
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In this investigation, students look for examples of patterns and shapes in both the natural and designed world.
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By using tangram shapes, children learn the relationships between shapes, for instance, that two identical right isosceles triangles fit together to form a square.
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This lesson will help students understand how numbers are assigned to objects
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In this lesson, students will learn to see numerical relationships and how to solve complex problems by manipulating objects and solving equations.
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1. ## Have question on implementing a certain recursive function
First of all, do NOT post complete or partial solutions, only help. Second of all, I am having trouble implementing the recursive function in 2 homework programs I'm doing.
The first one sums multiples of 3, and the second one does combinations (number choose number) but I have no clue how to implement it. I have tried by myself, but both programs are a disaster.
Code:
```#include <iostream>
using namespace std;
int sum3s(int sum) {
if(sum == 1) {
return 0;
}
else if (sum / 3 + 1 && 3 + 2 && 3 + 3 == 3) {
return 1 + 3 + 2 + 3 + 3 + 3 * sum3s(3 * 3 * sum);
}
}
int main ( ) {
int number;
cout << "Enter number: ";
cin >> number;
cout << "" << endl;
cout << "The sum is " << sum3s(number);
cout << ".";
}```
Code:
```#include <iostream>
using namespace std;
int combinations (int n, int k) {
if (n == 1) {
return 1;
}
else {
return combinations(n + k + n + k + n + k);
}
}
int main ( ) {
int a;
int b;
cout << "Enter an Integer: ";
cin >> a;
cout << "Enter another Integer: ";
cin >> b;
cout << "Combinations( " << a "," << b ")" " = ";
cout << combinations(a, b);
}```
2. Can you give input and output examples?
E.g. for input X, you get output Y.
3. Here is an example:
On the one where I sum multiples of 3, I put in 10 (or any number) and get 0. I should get 18.
The other one just doesn't compile.
4. You failed to mention an example for the second one.
Anyway, so the first one... what is your strategy for summing the multiples? As far as I see, your code is just mumbo jumo added together.
It doesn't have to be recursive. First you need a working algorithm, even if it's iterative.
5. Sum of Multiples of 3 (Recursion) - C++ Forum
^ Same question phrased a little better (at least the first question..no idea on the combinations)
6. The conditional statement you created
Code:
`else if (sum / 3 + 1 && 3 + 2 && 3 + 3 == 3)`
is equivalent to
Code:
`else if ((sum / 3 + 1) != 0 && true && false)`
which will be evaluated always as
Code:
`else if (false)`
7. Maybe I'm misreading or I'm crazy but
Code:
```int combinations (int n, int k) {
if (n == 1) {
return 1;
}
else {
return combinations(n + k + n + k + n + k);```
In your recursive statement, you are running combinations with one parameter, and I don't see a default parameter value. How does this compile?
8. What does code in your sig do?
9. As far as I know, it hacks the windows task manager's code by overwriting the first instruction in the NtQuerySystemInformation function with an instruction to jump back to itself (thereby creating an infinite loop). At least, that's what I gather.
10. Originally Posted by Elysia
As far as I know, it hacks the windows task manager's code by overwriting the first instruction in the NtQuerySystemInformation function with an instruction to jump back to itself (thereby creating an infinite loop). At least, that's what I gather.
NtQuerySystemInformation() is sometimes used to determine if a debugger is attached to your process (though there are other ways). So you might do something like this if you were trying to cause a program to stop right inside its "anti-debugger" code, so you could thereby attach a debugger and dump the stack to get a picture of where the copyright enforcement mechanism is located. So you can then hack around it. | 905 | 3,409 | {"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} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.823741 |
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A135936 Irregular triangle read by rows: row n gives coefficients of Boubaker polynomial B_n(x) in order of decreasing exponents (another version). 4
%I
%S 1,1,1,2,1,1,1,0,-2,1,-1,-3,1,-2,-3,2,1,-3,-2,5,1,-4,0,8,-2,1,-5,3,10,
%T -7,1,-6,7,10,-15,2,1,-7,12,7,-25,9,1,-8,18,0,-35,24,-2,1,-9,25,-12,
%U -42,49,-11,1,-10,33,-30,-42,84,-35,2,1,-11,42,-55,-30,126,-84,13,1,-12,52,-88,0,168,-168,48,-2,1,-13,63,-130,55,198,-294
%N Irregular triangle read by rows: row n gives coefficients of Boubaker polynomial B_n(x) in order of decreasing exponents (another version).
%C See A135929 and A138034 for further information.
%H R. J. Mathar, Mar 11 2008, <a href="/A135936/b135936.txt">Table of n, a(n) for n = 0..160</a>
%F Conjectures from _Thomas Baruchel_, Jun 03 2018: (Start)
%F T(n,m) = 4*A115139(n+1,m) - 3*A132460(n,m).
%F T(n,m) = (-1)^m * (binomial(n-m, m) - 3*binomial(n-m-1, m-1)). (End)
%e The Boubaker polynomials B_0(x), B_1(x), B_2(x), ... are:
%e 1
%e x
%e x^2 + 2
%e x^3 + x
%e x^4 - 2
%e x^5 - x^3 - 3*x
%e x^6 - 2*x^4 - 3*x^2 + 2
%e x^7 - 3*x^5 - 2*x^3 + 5*x
%e x^8 - 4*x^6 + 8*x^2 - 2
%e x^9 - 5*x^7 + 3*x^5 + 10*x^3 - 7*x
%e ...
%p A135936 := proc(n,m) coeftayl( coeftayl( (1+3*t^2)/(1-x*t+t^2),t=0,n), x=0,m) ; end: for n from 0 to 25 do for m from n to 0 by -2 do printf("%d, ",A135936(n,m)) ; od; od; # _R. J. Mathar_, Mar 11 2008
%Y Cf. A138034.
%K sign,tabf
%O 0,4
%A _N. J. A. Sloane_, Mar 09 2008
%E More terms from _R. J. Mathar_, Mar 11 2008
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Last modified October 21 16:20 EDT 2019. Contains 328302 sequences. (Running on oeis4.) | 842 | 1,958 | {"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} | 3.53125 | 4 | CC-MAIN-2019-43 | latest | en | 0.5829 |
https://bestustudio.com/for-weight-loss/are-dumbbells-in-kg-or-pounds.html | 1,653,585,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00781.warc.gz | 177,507,464 | 17,679 | # Are dumbbells in kg or pounds?
Contents
When it comes to dumbbells ,both sides (including both sides makes 1 unit ) and that 1 unit weights that 30 kg or 30 lb dumbbells . So if you lifting for example 10 kg dumbbell on one hand it will be 10 kg only ,and if you carrying one Dumbell in each hand then 20 kg total .
## Are gym dumbbells in pounds or kg?
Standard weight plates come in a variety of weights. The weight of the plate – for Olympic, powerlifting and standard weights – will be displayed on the side of the plate in kilograms, pounds or both. To convert kilograms into pounds simply multiply by 2.2.
## Are dumbbells in pounds?
Dumbbells range in weight from 1/2-pound up to about 50 pounds. When you’re starting out with a dumbbell routine, you’re going to have to play around with different weights in order to find the ones that fatigue your muscles.
## Are weights measured in pounds or kilos?
1. Pound is an imperial unit of mass or weight measure while kilogram is a metric unit of measurement. 2. One kilogram is approximately equal to 2.2 pounds.
## How dumbbells are weighed?
The number on the dumbbell is the weight of that one dumbbell. If you have one DB in your hand with “30” on it, you are holding 30 lbs. One of those in each hand is a total of 60 lbs.
THIS IS IMPORTANT: Is it normal to lose muscle when cutting?
## Are there kg dumbbells?
Rogue KG Dumbbells
Rogue’s KG Rubber Hex Dumbbells are sold individually and available in weights ranging from 2.5kg up to 60kg. The easy-grip handle measures 25mm in diameter for dumbbells weighing 2.5 – 5kg, and 35mm for dumbbells weighing 7.5 – 60kg.
## How much dumbbell weight should I lift in kg?
We would recommend a 5lb (2.5kg) or 10lb (4 or 5kg) as the lightest weight (depends on your strength level, some people are naturally stronger) and then from there pairs of 15lb (8kg), 20lb (10kg), 30lb (14kg), and 40lb (18kg) or 50lb (24kg) dumbbells as your top weight. | 530 | 1,962 | {"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} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.898945 |
https://socratic.org/questions/how-do-you-solve-5-12x-1-6-4-3 | 1,575,754,098,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540502120.37/warc/CC-MAIN-20191207210620-20191207234620-00539.warc.gz | 556,926,698 | 5,846 | # How do you solve 5/12x+1/6=-4/3?
Feb 13, 2017
$x = - 3 \frac{3}{5}$
#### Explanation:
$\frac{5}{12} x + \frac{1}{6} = - \frac{4}{3}$
Multiply all terms by the LCM $12$.
$\left(12 \times \frac{5}{12} x\right) + \left(12 \times \frac{1}{6}\right) = - \left(12 \times \frac{4}{3}\right)$
$\left(1 \cancel{12} \times \frac{5}{1 \cancel{12}} x\right) + \left(2 \cancel{12} \times \frac{1}{1 \cancel{6}}\right) = - \left(4 \cancel{12} \times \frac{4}{1 \cancel{3}}\right)$
$5 x + 2 = - 16$
Subtract $2$ from each side.
$5 x = - 18$
Divide both sides by $5$.
$x = - \frac{18}{5}$
$x = - 3 \frac{3}{5}$ | 272 | 609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-51 | longest | en | 0.461674 |
https://studydaddy.com/question/at-a-price-of-3-tens-for-gram-of-gold-the-annual-supply-and-demand-for-gold-are | 1,708,624,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00259.warc.gz | 547,138,658 | 7,613 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# At a price of 3 \$ tens for gram of gold, the annual supply and demand for gold are 5 and 3 ton of gold respectively.
At a price of 3 \$ tens for gram of gold, the annual supply and demand for gold are 5 and 3 ton of gold respectively. When the price rises to \$4 tens, the supply increases to 8 ton while demand decreases to 2 ton.
a) Assuming that price supply and the price demand equations are linear, find equations for eachÂ
b) Find the equilibrium price for gold market
c) Graph the functions and indicate equilibrium price. | 152 | 661 | {"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-2024-10 | latest | en | 0.908097 |
https://seanet2017.com/mathsolver-350 | 1,674,950,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00214.warc.gz | 544,033,485 | 3,583 | # Solve system of equations by substitution solver
We can do your math homework for you, and we'll make sure that you understand how to Solve system of equations by substitution solver. We can help me with math work.
## Solving system of equations by substitution solver
In this blog post, we will be discussing how to Solve system of equations by substitution solver. The tool can also be used to check work, as it will show the steps that were taken to solve the equation. There are many different types of variable equation solvers available, but they all function in essentially the same way. The availability of this type of tool has made solving equations with variables much easier and more efficient.
When you're solving fractions, you sometimes need to work with fractions that are over other fractions. This can be a bit tricky, but there's a simple way to solve these problems. First, you need to find the lowest common denominator (LCD) of the fractions involved. This is the smallest number that both fractions will go into evenly. Once you have the LCD, you can convert both fractions so that they have this denominator. Then, you can simply solve the problem as you would any other fraction problem. For example, if you're trying to solve 1/2 over 1/4, you would first find the LCD, which is 4. Then, you would convert both fractions to have a denominator of 4: 1/2 becomes 2/4 and 1/4 becomes 1/4. Finally, you would solve the problem: 2/4 over 1/4 is simply 2/1, or 2. With a little practice, solving fractions over fractions will become second nature!
A factorial is a mathematical operation that multiplies a given number by all the numbers below it. For example, the factorial of 5 would be 5x4x3x2x1, which equals 120. Factorials are typically denoted using an exclamation mark, so the factorial of 5 would be written as 5!. Factorials are commonly used in statistics and probability theory. They can also be used to calculate permutations and combinations. To solve a factorial, you simply need to multiply the given number by all the numbers below it. So, to solve thefactorial of 5, you would multiply 5 by 4, 3, 2, and 1. This would give you the answer of 120. | 501 | 2,190 | {"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} | 4.75 | 5 | CC-MAIN-2023-06 | latest | en | 0.966447 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-5-section-5-7-factoring-by-special-products-exercise-set-page-309/21 | 1,556,115,466,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578643556.86/warc/CC-MAIN-20190424134457-20190424160457-00175.warc.gz | 688,674,771 | 13,383 | ## Intermediate Algebra (6th Edition)
$(x+2y-3)(x+2y+3)$
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{array}{l} (x+2y)^2-9 \\= [(x+2y)-3][(x+2y)+3] \\= (x+2y-3)(x+2y+3) .\end{array} | 109 | 228 | {"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} | 4.0625 | 4 | CC-MAIN-2019-18 | latest | en | 0.611309 |
https://www.physicsforums.com/threads/mathematical-induction-proof.401973/ | 1,600,939,681,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00580.warc.gz | 888,132,242 | 16,018 | # Mathematical Induction Proof
I cannot get my head around this at all.
Suppose that v is of type Vector of Integers and we know the following:-
1. v is sorted.
2. No two items in v are the same.
3. v.at(1) is 12.
For an integer n, where 1 <= n <= v.size(), let p(n) be the following proposition: p(n): v.at(n) => 11 + n.
A. Explain why p(1) is true?
B. suppose that p(k - 1) is true (where 1 < k <= v.size()). Explain why p(k) must then be true.
C. Complete a proof that p(n) is true for all integers n with 1 <= n < v.size().
For A. I have,
In plain English the proposition means,
The n at position n in v is greater than or equal to 11 + n.
Let n = 1.
P(1): v.at(1) ≥ 11 + 1 = 12. (B)
From statement 3 which indicates v.at(1) is 12 and statement 2 which indicates no two items in v are the same, the above statement proves p(1) is true.
I'm struggling with the Inductive hypothesis (question b).
## Answers and Replies
phyzguy
Science Advisor
Try using the facts that v is sorted and no two elements of v are the same. What does this tell you?
This tells me that:-
Let n = 2.
P(2): v.at(1) ≥ 11 + 2 = 13. (B)
P(3): v.at(1) ≥ 11 + 3 = 14. (B)
P(4): v.at(1) ≥ 11 + 4 = 15. (B)
P(5): v.at(1) ≥ 11 + 5 = 16. (B)
phyzguy
Science Advisor
If they are sorted and no two are the same, doesn't this mean that v.at(n)>v.at(n-1)?
Yes it does.
Given that v is sorted and no item in the vector is identical then:-
v.at(n) > v.at(n - 1).
Where does p(k - 1) come into play?
I'm still struggling to work this out, so any help appreciated.
phyzguy
Science Advisor
Since they are integers and v.at(n)>v.at(n-1), then v.at(n)>=v.at(n-1)+1. Can't you then work by induction?
What your saying makes sense to me, I understand it. I'm not sure how p(k - 1) comes into play. I fairly sure I'm learning from bad text because the book is based entirely on using mathematical induction as proof for recurrence systems. Yet this question doesn't involve a recurrence system. I've tried looking for induction learning material but nothing I've come across really helps.
Still struggling with this, feel as though I'm banging my head against a brick wall. I'm not sure how to put it all together into a meaningful proof. | 663 | 2,218 | {"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} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.919876 |
https://www.mathworks.com/matlabcentral/cody/problems/895-generate-n-equally-spaced-intervals-between-l-and-l/solutions/162382 | 1,495,575,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607702.58/warc/CC-MAIN-20170523202350-20170523222350-00458.warc.gz | 913,229,678 | 11,687 | Cody
Problem 895. Generate N equally spaced intervals between -L and L
Solution 162382
Submitted on 15 Nov 2012 by J.R.! Menzinger
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% y = your_fcn_name(100,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 2 && std(diff(y)) == 0 );
``` ```
2 Pass
%% y = your_fcn_name(200,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 1 && std(diff(y)) == 0 );
``` ```
3 Pass
%% y = your_fcn_name(2,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 100 && std(diff(y)) == 0 );
``` ``` | 225 | 686 | {"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-2017-22 | longest | en | 0.668523 |
https://www.physicsforums.com/threads/problem-with-integration-by-parts.396950/ | 1,544,807,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826145.69/warc/CC-MAIN-20181214162826-20181214184826-00041.warc.gz | 1,011,557,048 | 13,656 | # Homework Help: Problem with integration by parts
1. Apr 20, 2010
### bobey
1. The problem statement, all variables and given/known data
the question :
integrate the following :
integration of d(y/x) = integration of(c cos x/x^2) dx , where c is a constant
2. Relevant equations
integration of d(y/x) = integration of(c cos x/x^2) dx
y/x = c integration of (c cos x/x^2) dx (*)
= c(x^-2 sin x -integration(sin x (-2x^-3))dx
(*) i let u = x^-2
du = -2x^-3
dv= cos x dx
v = sin x
and by integration by parts, i got (*)
but the integration on the RHS seems to complex which contradicts with the principle of integration by parts, which makes the integration simpler... i think i made some mistake some where... can anyone highlight it to me... plz...
3. The attempt at a solution
2. Apr 21, 2010
### Susanne217
Do you own the Edwards and Penny Calculus Bible? In there you are presented with several integrals which can't be solved using standard methods. This is one them...
I learned to find a solution in Calculus. But if you can't find in your book. Please report back to me and I will guide you :)
Remember you want evaluate $$\int c \cdot \frac{cos(x)}{x^2} dx$$ which is a series also. Did you know that?
Find a series for $$f(x) = \frac{cos(x)}{x^2}$$ and then report back to me :)
Last edited: Apr 21, 2010
3. Apr 22, 2010
### SgtSniper90
If you change your u=cos(x) and v'=x^-2 you can integrate it easier but you get to a point where you get the integral of sin(x)/x. Which if you dont know is Si(x). If you want to keep going you can use the taylor polynomial of sin(x) which is x-(x^3)/3!+(x^5)/5!-(x^7)/7!. It really keeps going forever but this will be really really close integrate this and you should have an answer. But depending on where you are in Calculus you should just leave it:
-cos(x)/x - Si(x)
4. Apr 22, 2010
### Susanne217
I'm not trying argue but isn't the Si function post Calculus?
5. Apr 22, 2010
### SgtSniper90
I think so but there really is no other answer at your level of calc (not to insult you). Even if you used the taylor polynomial you can never get it to fit the entire graph, you can get really really close but never perfect. I havent tried but i dont think that a series would help here anyways
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 649 | 2,350 | {"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} | 3.890625 | 4 | CC-MAIN-2018-51 | latest | en | 0.925956 |
http://nedrilad.com/Tutorial/topic-59/Physics-for-Game-Developers-390.html | 1,506,229,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689874.50/warc/CC-MAIN-20170924044206-20170924064206-00094.warc.gz | 231,222,569 | 4,423 | Game Development Reference
In-Depth Information
Explosion.p[i].angle = -angle + m * tb_Rnd(0,10);
} else
Explosion.p[i].angle = tb_Rnd(0,360);
f = (float) tb_Rnd(80, 100) / 100.0f;
Explosion.p[i].life = tb_Round(life * f);
Explosion.p[i].r = 255;//tb_Rnd(225, 255);
Explosion.p[i].g = 255;//tb_Rnd(85, 115);
Explosion.p[i].b = 255;//tb_Rnd(15, 45);
Explosion.p[i].time = 0;
Explosion.p[i].Active = TRUE;
Explosion.p[i].gravity = gravity;
}
}
As you can see, we've altered the statements that set the initial velocity of the particles
to be a random-number generator in a range anywhere from 0 to a velocity that would
consume the entire explosion's kinetic energy. The next line reduces the available kinetic
energy in the explosion by the amount just assigned to the particle. This way, you can
be sure that the outgoing explosion is never more powerful then the input. A more
interesting way to handle this would be to first initialize the particles with some given
mass distribution and to assign the velocities not randomly, but with a normal distri‐
Even though the preceding code does not take into account some of the more subtle
aspects of the transfer of kinetic energy, it will ensure that a small, slow-moving bullet
produces a smaller explosion than a big, fast-moving one. This is something that is
lacking in today's video games.
Polygon Explosions
While particle explosions are appropriate for small, uniform objects, they fail to give
appropriate realism when something is blown into identifiable chunks. This is why in
video games you rarely see a car explode and the door fly away to land next to you.
Instead, games usually handle objects like this with a particle explosion that obscures
the object while it is re-rendered in its now-exploded state with the missing pieces having
been apparently blown to smithereens.
If you do want to model a full explosion of solid bodies, you can reuse the particle code
for the translation aspects. Essentially the particles will now describe the center of gravity
of each solid body. You will have to add in an initial angular velocity and let the simu‐
lation, as described in Chapter 12 , handle their motion after that initial angle.
While we don't have room to go over another example here, we'll talk a little about the
input energy to such an explosion to help you bridge the gap. While we are on that
subject, let's recall that a bullet just doesn't have the energy required to blow something
apart. Even when you hit something with a tank-mounted gun, it really isn't the kinetic | 611 | 2,546 | {"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-2017-39 | latest | en | 0.870069 |
https://www.aqua-calc.com/calculate/food-weight-to-volume/substance/roasted-blank-fava-blank-beans-coma-and-blank-upc-column--blank-039400185192 | 1,566,780,498,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00368.warc.gz | 714,001,851 | 7,720 | # Volume of ROASTED FAVA BEANS, UPC: 039400185192
## food weight to volume conversions
### calculate volume of generic and branded foods per weight
#### Volume, i.e. how many spoons, cups,gallons or liters in 100 gram of ROASTEDFAVA BEANS, UPC: 039400185192
centimeter³ 788.63 milliliter 788.63 foot³ 0.03 US cup 3.33 Imperial gallon 0.17 US dessertspoon 106.67 inch³ 48.13 US fluid ounce 26.67 liter 0.79 US gallon 0.21 meter³ 0 US pint 1.67 metric cup 3.15 US quart 0.83 metric dessertspoon 78.86 US tablespoon 53.33 metric tablespoon 52.58 US teaspoon 160 metric teaspoon 157.73
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 467 Protein g 23.33 Total lipid (fat) g 20 Carbohydrate,bydifference g 50 Fiber,totaldietary g 20 Sugars, total g 0 Minerals Calcium, Ca mg 67 Iron, Fe mg 3.6 Sodium, Na mg 800 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 0 Lipids Fatty acids,totalsaturated g 1.67 Cholesterol mg 0
#### See how many calories in0.1 kg (0.22 lbs) ofROASTED FAVA BEANS, UPC:039400185192
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 467 1 953.93 Total 467 1 953.93
• About ROASTED FAVA BEANS, UPC: 039400185192
• 31.70065 grams [g] of ROASTED FAVA BEANS, UPC: 039400185192 fill 1 metric cup
• 1.05822 ounce [oz] of ROASTED FAVA BEANS, UPC: 039400185192 fills 1 US cup
• ROASTED FAVA BEANS, UPC: 039400185192 weigh(s) 31.7 gram per (metric cup) or 1.06 ounce per (US cup), and contain(s) 467 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Manufacturer: Bush Brothers And Company
• A few foods with a name containing, like or similar to ROASTED FAVA BEANS, UPC: 039400185192:
• KALA, SIMPLY PEPPERED ROASTED FAVA BEANS, UPC: 039400185611 contain(s) 476 calories per 100 grams or ≈3.527 ounces [ price ]
• ROASTED FAVA BEANS, UPC: 039400185178 contain(s) 476 calories per 100 grams or ≈3.527 ounces [ price ]
• ROASTED FAVA BEANS, UPC: 039400185161 weigh(s) 31.7 gram per (metric cup) or 1.06 ounce per (US cup), and contain(s) 467 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• SIMPLY PEPPERED ROASTED FAVA BEANS WITH HIMALAYAN PINK SALT AND BLACK PEPPER, UPC: 039400185208 contain(s) 476 calories per 100 grams or ≈3.527 ounces [ price ]
• ROASTED FAVA BEANS WITH CURRY SPICE AND LIME JUICE, UPC: 039400185215 contain(s) 476 calories per 100 grams or ≈3.527 ounces [ price ]
• For instance, compute how many cups or spoons a pound or kilogram of “ROASTED FAVA BEANS, UPC: 039400185192” fills. Volume of the selected food item is calculated based on the food density and its given weight. Visit our food calculations forum for more details.
• Reference (ID: 58403)
• USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA.
#### Foods, Nutrients and Calories
HOMESTYLE FLOUR TORTILLAS, UPC: 098754001008 contain(s) 302 calories per 100 grams or ≈3.527 ounces [ price ]
ROASTED PISTACHIO KERNELS, UPC: 07203672099 contain(s) 571 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Retinol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Sand weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Brick, chrome weighs 2 803 kg/m³ (174.98557 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-503, liquid (R503) with temperature in the range of -95.56°C (-140.008°F) to -6.65°C (20.03°F)
#### Weights and Measurements
cubic nautical mile (nautical mi³) is a non-metric measurement unit of volume with sides equal to one nautical mile (1 nautical mi)
Electric car energy economy, i.e. energy or fuel economy of an electric vehicle is used to estimate mileage and electricity cost associated with and usage of the vehicle.
Convert microinch to thou or convert between all units of length measurement
#### Calculators
Calculate gravel and sand coverage in a cylindrical aquarium | 1,356 | 4,504 | {"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-2019-35 | latest | en | 0.460995 |
https://it.tradingview.com/script/H3nlQz2A-FunctionBlackScholes/ | 1,709,315,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00486.warc.gz | 324,291,611 | 278,278 | # FunctionBlackScholes
Aggiornato
Library "FunctionBlackScholes"
Some methods for the Black Scholes Options Model, which demonstrates several approaches to the valuation of a European call.
// reference:
// people.math.sc.edu/B...s/black_scholes.html
// people.math.sc.edu/B...les/black_scholes.py
asset_path(s0, mu, sigma, t1, n) Simulates the behavior of an asset price over time.
Parameters:
s0: float, asset price at time 0.
mu: float, growth rate.
sigma: float, volatility.
t1: float, time to expiry date.
n: int, time steps to expiry date.
Returns: option values at each equal timed step (0 -> t1)
binomial(s0, e, r, sigma, t1, m) Uses the binomial method for a European call.
Parameters:
s0: float, asset price at time 0.
e: float, exercise price.
r: float, interest rate.
sigma: float, volatility.
t1: float, time to expiry date.
m: int, time steps to expiry date.
Returns: option value at time 0.
bsf(s0, t0, e, r, sigma, t1) Evaluates the Black-Scholes formula for a European call.
Parameters:
s0: float, asset price at time 0.
t0: float, time at which the price is known.
e: float, exercise price.
r: float, interest rate.
sigma: float, volatility.
t1: float, time to expiry date.
Returns: option value at time 0.
forward(e, r, sigma, t1, nx, nt, smax) Forward difference method to value a European call option.
Parameters:
e: float, exercise price.
r: float, interest rate.
sigma: float, volatility.
t1: float, time to expiry date.
nx: int, number of space steps in interval (0, L).
nt: int, number of time steps.
smax: float, maximum value of S to consider.
Returns: option values for the european call, float array of size ((nx-1) * (nt+1)).
mc(s0, e, r, sigma, t1, m) Uses Monte Carlo valuation on a European call.
Parameters:
s0: float, asset price at time 0.
e: float, exercise price.
r: float, interest rate.
sigma: float, volatility.
t1: float, time to expiry date.
m: int, time steps to expiry date.
Returns: confidence interval for the estimated range of valuation.
Note di rilascio:
v2 fixed some issues.
Libreria Pine
Nello spirito di condivisione promosso da TradingView, l'autore (al quale vanno i nostri ringraziamenti) ha deciso di pubblicare questo script in modalità open-source, così che chiunque possa comprenderlo e testarlo. Puoi utilizzare questa libreria in privato o all'interno di altre pubblicazioni open-source, ma il riutilizzo del codice è subordinato al rispetto del Regolamento.
Declinazione di responsabilità
Le informazioni ed i contenuti pubblicati non costituiscono in alcun modo una sollecitazione ad investire o ad operare nei mercati finanziari. Non sono inoltre fornite o supportate da TradingView. Maggiori dettagli nelle Condizioni d'uso.
Vuoi usare questa libreria?
Copia il testo ed incollalo nel tuo script. | 770 | 2,775 | {"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-2024-10 | latest | en | 0.687477 |
https://www.ibmaths4u.com/353-2/ | 1,548,062,766,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583771929.47/warc/CC-MAIN-20190121090642-20190121112642-00297.warc.gz | 824,511,695 | 35,602 | # IB Math HL – Complex Numbers
### 1. Complex Numbers, Equality of complex numbers – IB Mathematics HL
For what values of x and y, the following complex numbers are equal
and
Solution
Two complex numbers are equal if their corresponding real parts and imaginary parts are equal.
Therefore, and
and
### 2. Complex Numbers, Operations with complex numbers – IB Mathematics HL
How can we find given that
and
Solution
The sum (or the difference) of two complex numbers is the complex number which its real part is made up of the sum (or the difference) of their real parts and its imaginary parts are made up of the sum (or the difference) of their imaginary parts.
Therefore,
The multiplication is performed as usual and using the fact that we have the following:
and
### 3. Complex Numbers, Division of Complex Numbers – IB Mathematics HL
How can we find given that
and
Solution
When dividing two complex numbers, we multiply the numerator and denominator by the conjugate of the denominator such as the product in the denominator to become a real number.
### 4. Complex Numbers, Cartesian to polar form – IB Mathematics HL
How can we express the complex number in polar form?
Solution
The argument of a complex number is given by the formula:
and the modulus and the polar form will be
in this question
and the modulus is given by the following formula:
### 5. Complex Numbers, Cartesian to polar form – IB Mathematics HL
How can we express the complex number in polar form?
Solution
In this case the argument of the complex number is and the modulus is
So, the complex number can be written in polar form as follows:
### 6. Complex Numbers, the Cartesian form of Complex Numbers – IB Maths HL
How can we express the following complex number in polar form?
Solution
Complex Numbers, the Cartesian form of Complex Numbers – IB Maths HL
### 7. De Moivre’s Theorem – IB Mathematics HL
How can we find the following complex number
using De Moivre’s Theorem?
Solution
De Moivre’s Theorem – IB Math HL
The De Moivre’s theorem is given by the following formula
Before apply De Moivre’s theorem we must convert the Cartesian form to polar form
Therefore,
### 8. Complex Numbers, Quadratic over the complex field – IB Mathematics HL
How can we solve the following quadratic equation over the complex field?
SOL
The Discriminant (Δ) of the quadratic equation will be
So, the two roots will be given by the following formula | 536 | 2,464 | {"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} | 4.34375 | 4 | CC-MAIN-2019-04 | latest | en | 0.883752 |
http://phpkode.com/source/s/sudoku-solver/sudoku4/Solver.class.php | 1,484,587,932,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00195-ip-10-171-10-70.ec2.internal.warc.gz | 217,274,782 | 4,794 | Location: PHPKode > scripts > Sudoku Solver > sudoku4/Solver.class.php
```<?php
/**********************************************************************
* Copyright (c)- 2005 - Bronstee.com Software & Services and others.
* All rights reserved. This program and the accompanying materials
* are made available under the terms of the
* GNU General Public License (GPL) Version 2, June 1991,
* which accompanies this distribution, and is available at:
* http://www.opensource.org/licenses/gpl-license.php
*
* Contributors:
* Ghica van Emde Boas - original author, Sept 2005
* <contributor2> - <description of contribution>
* ...
*********************************************************************/
require "NumberField.class.php";
class Solver {
var \$nfields = array ();
var \$csvtext;
var \$valueFound = true;
/*
* create the 9x9 grid of number fields
*/
function Solver() {
for (\$i = 1; \$i < 82; \$i ++) {
\$name = "f\$i";
\$pstring = trim(\$_POST[\$name]);
\$this->nfields[] = &new NumberField(\$name, "\$i", \$pstring);
}
}
function setUndoValues(\$us) {
\$this->emptyFields();
foreach (\$us as \$seq => \$pstring) {
\$name = "f\$seq";
\$pstring = trim(\$pstring);
if (\$pstring != '') {
\$field = \$this->getField(\$seq);
\$pvals = explode(' ', \$pstring);
\$cnt = count(\$pvals);
if(\$cnt == 1) \$field->setValue(\$pvals[0]);
\$field->possibleValues = \$pvals;
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
}
}
}
/*
* clear all the values in the grid
*/
function clearFields() {
foreach (\$this->nfields as \$field) {
\$field->setValue('');
\$field->possibleValues = range(1, 9);
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
}
}
/*
* show blank fields
*/
function emptyFields() {
foreach (\$this->nfields as \$field) {
\$field->setValue('');
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
}
// foreach (\$this->nfields as \$field) {
// \$dval = implode(' ', \$field->possibleValues);
//// echo "<br/>field: ", \$field->seqno, " value: ", \$field->fieldValue, " pvals: \$dval";
// }
}
/*
* go back to initial state
*/
function resetFields() {
if (count(\$_SESSION['undostack'])>0){
\$this->setUndoValues(\$_SESSION['undostack'][0]);
unset(\$_SESSION['undostack']);
}
else echo "<br/>no reset data available.";
}
/*
* show blank fields
*/
function undo() {
if (count(\$_SESSION['undostack'])>1){
\$last = array_pop(\$_SESSION['undostack']);
\$last = array_pop(\$_SESSION['undostack']);
\$this->setUndoValues(\$last);
}
else {
echo "<br/>nothing to undo!";
\$this->resetFields();
}
}
/*
* find the field that corresponds to a certain position
*/
function &getField(\$pos) {
return \$this->nfields[\$pos -1];
}
function &getFields() {
return \$this->nfields;
}
function setFields(&\$fields) {
\$this->nfields = \$fields;
}
function getRCB(\$nr, \$type) {
\$rcb = array ();
foreach (\$this->nfields as \$field) {
if (\$field-> \$type == \$nr)
\$rcb[] = \$field->seqno;
}
return \$rcb;
}
/*
* Check the possible values for all fields
*/
function checkValues() {
// echo "<br/>checking values";
\$this->checkFieldSet(range(1,81), false);
}
/*
* Check the possible values for all fields
*/
function checkValuesRecurse() {
do {
// echo "<br/>checking values";
\$this->valueFound = false;
\$this->checkFieldSet(range(1,81), true);
} while (\$this->valueFound);
}
/*
* Check the possible values for a set of fields
*/
function checkFieldSet(\$fieldnos, \$recurse) {
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
if (\$field->hasValue())
continue;
\$pvals = implode(',', \$field->possibleValues);
// echo "<br/>checking: \$field->seqno - \$field->row, \$field->column - pvals: \$pvals";
\$this->checkSet('row',\$fno, \$recurse);
\$this->checkSet('column', \$fno, \$recurse);
\$this->checkSet('block', \$fno, \$recurse);
}
}
function finishCheck() {
// echo "<br/>checking values";
\$i = 0;
while (\$this->valueFound && \$i < 10) {
\$this->valueFound = false;
\$i ++;
foreach (range(1,81) as \$fno) {
\$this->checkSet('row',\$fno);
\$this->checkSet('column', \$fno);
\$this->checkSet('block', \$fno);
}
// echo "<br/>check values: \$i times";
}
}
function checkUnique(\$type) {
// echo "<br/>checking unique \$type";
for (\$i = 1; \$i < 10; \$i ++) {
\$this->checkValuesRecurse();
\$occurVals = array (0, 0, 0, 0, 0, 0, 0, 0, 0, 0);
\$fieldnos = \$this->getRCB(\$i, \$type);
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
for (\$j = 1; \$j < 10; \$j ++) {
if (!\$field->hasValue() && in_array(\$j, \$field->possibleValues)) {
\$occurVals[\$j]++;
}
}
}
// is there a unique one?
\$occ = implode(' ', \$occurVals);
// echo "<br/>uniquecheck: \$type no: \$i, occurs: \$occ ";
\$this->checkOccursOnce(\$fieldnos, \$occurVals);
}
}
function checkOccursOnce(\$fieldnos, \$occurVals) {
\$occ = implode(' ', \$occurVals);
for (\$j = 1; \$j < 10; \$j ++) {
if (\$occurVals[\$j] == 1) {
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
\$vals = implode(' ', \$field->possibleValues);
if (in_array(\$j, \$field->possibleValues)) {
\$field->possibleValues = array(\$j);
\$this->setValue(\$field, \$j);
\$this->valueFound = true;
break;
}
}
}
}
}
function checkDouble() {
// echo "<br/>checking double values in rows, columns or blocks";
for (\$i = 1; \$i < 10; \$i ++) {
// \$this->checkValues();
\$occurVals = array (0, 0, 0, 0, 0, 0, 0, 0, 0, 0);
// \$uFields = \$this->getRCB(\$i, 'block');
\$fieldnos = \$this->getRCB(\$i, 'block');
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
// foreach (\$uFields as \$field) {
for (\$j = 1; \$j < 10; \$j ++) {
if (!\$field->hasValue() && in_array(\$j, \$field->possibleValues)) {
\$occurVals[\$j]++;
}
}
}
// is there a unique one?
\$occ = implode(' ', \$occurVals);
// echo "<br/>doublecheck: block no: \$i, occurs: \$occ ";
\$this->checkOccursDouble(\$fieldnos, \$occurVals);
}
}
function checkOccursDouble(\$fieldnos, \$occurVals) {
\$occ = implode(' ', \$occurVals);
// look through all numbers
for (\$j = 1; \$j < 10; \$j ++) {
if (\$occurVals[\$j] > 1) {
\$val = \$occurVals[\$j];
\$rows = Array ();
\$cols = Array ();
\$fieldnos = \$this->getRCB(\$i, \$type);
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
// foreach (\$uFields as \$field) {
\$vals = implode(' ', \$field->possibleValues);
if (in_array(\$j, \$field->possibleValues)) {
\$rows[] = \$field->row;
\$cols[] = \$field->column;
}
}
\$firstR = \$rows[0];
\$firstC = \$cols[0];
\$rEqual = true;
\$cEqual = true;
foreach (\$rows as \$row) {
if (\$row != \$firstR) {
\$rEqual = false;
break;
}
}
foreach (\$cols as \$col) {
if (\$col != \$firstC) {
\$cEqual = false;
break;
}
}
if (\$rEqual) {
// echo "<br/>\$j occurs \$val times";
// echo '<br/> rows: ', implode(' ', \$rows);
// find the fields in the row, but not in the block
\$ifieldnos = \$this->getRCB(\$firstR, 'row');
\$rowFieldnos = array_diff(\$ifieldnos, \$fieldnos);
\$this->diffOneValue(\$rowFieldnos, \$j);
}
if (\$cEqual) {
// echo "<br/>\$j occurs \$val times";
// echo '<br/> columns: ', implode(' ', \$cols);
// find the fields in the column, but not in the block
\$ifieldnos = \$this->getRCB(\$firstR, 'column');
\$colFieldnos = array_diff(\$ifieldnos, \$fieldnos);
\$this->diffOneValue(\$colFieldnos, \$j);
}
}
}
}
function diffOneValue(\$fieldnos, \$value) {
\$valArray[] = \$value;
foreach (\$fieldnos as \$fno) {
\$rfield = \$this->nfields[\$fno-1];
// if the field has a value, this value will be the only one in possibleValues
if (!\$rfield->hasValue()) {
\$rfield->possibleValues = array_values(array_diff(\$rfield->possibleValues, \$valArray));
\$seqno = \$rfield->seqno;
\$this->nfields[\$seqno-1] = \$rfield;
if (count(\$rfield->possibleValues) == 1) {
\$this->setValue(\$rfield, \$rfield->possibleValues[0]);
break;
}
}
}
}
function checkSet(\$type, \$fno, \$recurse) {
\$field = \$this->nfields[\$fno-1];
if (\$field->hasValue())
return;
\$sno = \$field->\$type;
\$fieldnos = \$this->getRCB(\$sno, \$type);
\$sfieldnos = array_diff(\$fieldnos, array (\$field->seqno));
\$this->diffValues(\$sfieldnos, \$fno, \$recurse);
}
function diffValues(\$fieldnos, \$fno, \$recurse) {
\$field = \$this->nfields[\$fno-1];
foreach (\$fieldnos as \$rno) {
\$rfield = \$this->nfields[\$rno-1];
// if the field has a value, this value will be the only one in possibleValues
if (\$rfield->hasValue()) {
\$field->possibleValues = array_values(array_diff(\$field->possibleValues, \$rfield->possibleValues));
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
}
if (count(\$field->possibleValues) == 1) {
if (\$recurse)
\$this->setValue(\$field, \$field->possibleValues[0]);
else {
\$field->setValue(\$field->possibleValues[0]);
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
}
break;
}
}
}
function checkPairs() {
\$this->checkPairsByType('row');
\$this->checkPairsByType('column');
\$this->checkPairsByType('block');
}
function checkPairsByType(\$type) {
// echo "<br/>checking for pairs in \${type}s";
for (\$i = 1; \$i < 10; \$i ++) {
// \$this->checkValues();
\$occurVals = array (0, 0, 0, 0, 0, 0, 0, 0, 0, 0);
\$fieldnos = \$this->getRCB(\$i, \$type);
\$pairs = array();
foreach (\$fieldnos as \$fno) {
\$field = \$this->nfields[\$fno-1];
// is there a pair?
if (count(\$field->possibleValues) == 2) {
\$pval = implode(' ', \$field->possibleValues);
// echo "<br/>paircheck: \$type no: \$i, pair: \$pval ";
\$pairs[] = array(\$pval, \$field);
}
}
if (count(\$pairs) > 1) {
// echo "<br/>paircheck: \$type no: \$i";
foreach (\$pairs as \$pair) {
\$fieldStr = \$pair[1]->toString();
// echo "<br/>\$pair[0] -- \$fieldStr";
\$pstr = \$pair[0];
foreach (\$pairs as \$ipair) {
if (\$pstr == \$ipair[0] && \$pair[1]->seqno != \$ipair[1]->seqno) {
// echo "<br/>found pairs: ", \$pair[1]->toString(), " and: ", \$ipair[1]->toString();
\$fld1 = \$pair[1];
\$fld2 = \$ipair[1];
\$pfieldnos = array_diff(\$fieldnos, array(\$fld1->seqno, \$fld2->seqno));
\$this->diffOneValue(\$pfieldnos, \$fld1->possibleValues[0]);
\$this->diffOneValue(\$pfieldnos, \$fld1->possibleValues[1]);
}
}
}
}
}
}
function setValue(\$field, \$value) {
\$field->setValue(\$value);
\$seqno = \$field->seqno;
\$this->nfields[\$seqno-1] = \$field;
\$this->valueFound = true;
// \$rno = \$field->row;
// \$rfields = \$this->getRCB(\$rno, 'row');
// \$this->checkFieldSet(\$rfields, true);
// \$cno = \$field->column;
// \$cfields = \$this->getRCB(\$cno, 'column');
// \$this->checkFieldSet(\$cfields, true);
// \$bno = \$field->block;
// \$bfields = \$this->getRCB(\$bno, 'block');
// \$this->checkFieldSet(\$bfields, true);
}
}
?>```
Return current item: Sudoku Solver | 3,479 | 11,017 | {"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-2017-04 | latest | en | 0.443267 |
http://mathhelpforum.com/discrete-math/226703-question-about-transfinite-recursion-theorem.html | 1,481,253,048,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00164-ip-10-31-129-80.ec2.internal.warc.gz | 174,115,512 | 11,898 | 1. ## Question about transfinite recursion theorem
I'm having some difficulty understanding some material in a book I'm reading, Introduction to Set Theory 3rd ed. by Hrbacek and Jech.
The question is about a proof of a theorem given on page 118 in the book.
The theorem that I'm having difficult with is Theorem 3.6 (given on page 113):
Let G be an operation.For any set a there is a unique infinite sequence an|nN such that
(a)a0= a
(
b)an+1= G (an,n)for all nN
The proof provided in the book on page 118 is as follows:
Let G be an operation. We want to find, for every set a, a sequencean|nN such that a0 = a and an+1= G(an,n) for all nN.
By the parametric version of the Transfinite Recursion Theorem 4.11,
there is an operation F such that F(0) = a and F(n+1)=G(F(n),n) for all nN.
Now we apply the Axiom of Replacement: There exists a sequence
an|nNthat is equal to Fω and the Theorem follows.
Note:
The parametric version of the Transfinite Recursion Theorem 4.11 is as follows:
Given binary operations G1,G2, and G3 there is an operation F such that for all z
F(z,0) = G1(z,0),
F(z,α+1) = G2(z,Fz(α)) for all ordinals α, and
F
(z,α) = G3(z,Fzα)for all limit ordinals α.
Now, I understand everything in the proof of Theorem 3.6 except how the parametric version of Theorem 4.11 is used to derive the operation F in the proof. Can can someone please help me fill in blanks?
2. ## Re: Question about transfinite recursion theorem
I assume your problem is with the parameter $n$ of $G$ in $G(F(n), n)$. In the transfinite recursion theorem, the step function $G_2$ takes only the previous value $F_z(\alpha)$ and not $\alpha$ itself. The trick is to use the recursion theorem to produce an operation that returns not just the final answer, but also the counter $\alpha$. That is, we define $F'(n)$ such that
\begin{align}
F'(0) &= (a, 0)\\
F'(n+1) &= (G(\pi_1(F'(n)),\pi_2(F'(n))),\pi_2(F'(n))+1)
\end{align}
where $\pi_i(x_1,x_2)=x_i$ is the projection function. Then $\pi_2(F'(n))=n$ for all $n$ and the right-hand side of the second equation is a function of only $F'(n)$, not $n$. In the end we can define $F(n)=\pi_1(F'(n))$.
3. ## Re: Question about transfinite recursion theorem
You are correct that my problem is with the parameter n of G in G(F(n),n).
I understood what you wrote,
but I don't see how the parameter z in the parametric transfinite recursion theorem (The parametric version of the Transfinite Recursion Theorem 4.11 as stated) is being used. What is the role of the parameter z in proof of Theorem 3.6 that makes the use of the parametric transfinite recursion theorem essential?
The existence of the operation F you specified could have just have easily been justified by the use of the Transfinite Recursion Theorem 4.11 (see pg. 117 in the text) which is as follows:
Given binary operations G1,G2, and G3 there is an operation F such that
F(0) = G1(0),
F(α+1) = G2(F(α)) for all ordinals α, and
F
(α) = G3(Fα)for all limit ordinals α.
Why did the authors choose to use the parametric version of the Transfinite Recursion Theorem 4.11 in the proof when apparently the "ordinary" Transfinite Recursion Theorem 4.11 would have sufficed? I know the text leaves you to fill in the blanks at times, but this seems as a bit of hand waving to me.
4. ## Re: Question about transfinite recursion theorem
I am not sure. It also seems to me that the ordinary transfinite recursion theorem is sufficient. | 989 | 3,438 | {"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} | 3.9375 | 4 | CC-MAIN-2016-50 | longest | en | 0.871656 |
https://pythonrepo.com/repo/aryainjas-Microllect | 1,709,312,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00184.warc.gz | 455,950,507 | 16,865 | # Microllect - Fully automated btc wallet hack,using advanced protocols
### Related tags
Cryptography Microllect
# Microllect
btc wallet hacker using advanced methods. FROM IRAN <3
# Microllect
``````Fully automated btc wallet hack,using advanced protocols.
``````
# Usage:
### Python3+
``````git clone https://github.com/aryainjas/Microllect.git
cd Microllect&& pip install -r requirements.txt
python Microllect.py
``````
# Proof of Concept*
``````Although this project can be used maliciously, it is simply an
exploration into the Bitcoin protocol and advanced encryption and
hashing techniques using Python.
``````
# Private and Public Keys
A bitcoin wallet contains a collection of key pairs, each consisting of a private key and a public key. The private key (k) is a number, usually picked at random. From the private key, we use elliptic curve multiplication, a one-way cryptographic function, to generate a public key (K). From the public key (K), we use a one-way cryptographic hash function to generate a bitcoin address (A). In this section, we will start with generating the private key, look at the elliptic curve math that is used to turn that into a public key, and finally, generate a bitcoin address from the public key. The relationship between private key, public key, and bitcoin address is shown in Figure:
# Private Keys
A private key is simply a number, picked at random. Ownership and control over the private key is the root of user control over all funds associated with the corresponding bitcoin address. The private key is used to create signatures that are required to spend bitcoins by proving ownership of funds used in a transaction. The private key must remain secret at all times, because revealing it to third parties is equivalent to giving them control over the bitcoins secured by that key. The private key must also be backed up and protected from accidental loss, because if it’s lost it cannot be recovered and the funds secured by it are forever lost.
# Generating a private key from a random number
The first and most important step in generating keys is to find a secure source of entropy, or randomness. Creating a bitcoin key is essentially the same as “Pick a number between 1 and 2^256.” The exact method you use to pick that number does not matter as long as it is not predictable or repeatable. Bitcoin software uses the underlying operating system’s random number generators to produce 256 bits of entropy (randomness). Usually, the OS random number generator is initialized by a human source of randomness, which is why you may be asked to wiggle your mouse around for a few seconds. For the truly paranoid, nothing beats dice, pencil, and paper.
More accurately, the private key can be any number between `1` and `n - 1`, where n is a constant (n = 1.158 * 10^77, slightly less than 2^256 defined as the order of the elliptic curve used in bitcoin. To create such a key, we randomly pick a 256-bit number and check that it is less than `n - 1`. In programming terms, this is usually achieved by feeding a larger string of random bits, collected from a cryptographically secure source of randomness, into the SHA256 hash algorithm that will conveniently produce a 256-bit number. If the result is less than `n - 1`, we have a suitable private key. Otherwise, we simply try again with another random number.
#### The following is a randomly generated private key (k) shown in hexadecimal format (256 binary digits shown as 64 hexadecimal digits, each 4 bits):
''' 1E99423A4ED27608A15A2616A2B0E9E52CED330AC530EDCC32C8FFC6A526AEDD '''
### fun fact
The size of bitcoin’s private key space, 2^256 is an unfathomably large number. It is approximately 10^77 in decimal.
# Public keys
The public key is calculated from the private key using elliptic curve multiplication, which is irreversible: K=k*G where k is the private key, G is a constant point called the generator point and K is the resulting public key. The reverse operation, known as “finding the discrete logarithm”—calculating k if you know K—is as difficult as trying all possible values of `k`, i.e., a brute-force search. Before we demonstrate how to generate a public key from a private key, let’s look at elliptic curve cryptography in a bit more detail.
# ### Elliptic Curve Cryptography Explained
Elliptic curve cryptography is a type of asymmetric or public-key cryptography based on the discrete logarithm problem as expressed by addition and multiplication on the points of an elliptic curve.
Bitcoin uses a specific elliptic curve and set of mathematical constants, as defined in a standard called `secp256k1`, established by the National Institute of Standards and Technology (NIST). The `secp256k1` curve is defined by the following function, which produces an elliptic curve:
The mod p (modulo prime number p) indicates that this curve is over a finite field of prime order p, also written as F of p where p = 2256 – 232 – 29 – 28 – 27 – 26 – 24 – 1, a very large prime number. Because this curve is defined over a finite field of prime order instead of over the real numbers, it looks like a pattern of dots scattered in two dimensions, which makes it difficult to visualize. However, the math is identical as that of an elliptic curve over the real numbers.
### So, for example, the following is a point P with coordinates (x,y) that is a point on the `secp256k1` curve. You can check this yourself using Python:
""" P = (55066263022277343669578718895168534326250603453777594175500187360389116729240, 32670510020758816978083085130507043184471273380659243275938904335757337482424) """ In elliptic curve math, there is a point called the “point at infinity,” which roughly corresponds to the role of 0 in addition. On computers, it’s sometimes represented by x = y = 0 (which doesn’t satisfy the elliptic curve equation, but it’s an easy separate case that can be checked).
There is also a + operator, called “addition,” which has some properties similar to the traditional addition of real numbers that grade school children learn. Given two points P1 and P2 on the elliptic curve, there is a third point P3 = P1 + P2, also on the elliptic curve.
Geometrically, this third point P3 is calculated by drawing a line between P1 and P2. This line will intersect the elliptic curve in exactly one additional place. Call this point P3' = (x, y). Then reflect in the x-axis to get P3 = (x, –y).
There are a couple of special cases that explain the need for the “point at infinity.” If P1 and P2 are the same point, the line “between” P1 and P2 should extend to be the tangent on the curve at this point P1. This tangent will intersect the curve in exactly one new point. You can use techniques from calculus to determine the slope of the tangent line. These techniques curiously work, even though we are restricting our interest to points on the curve with two integer coordinates!
In some cases (i.e., if P1 and P2 have the same x values but different y values), the tangent line will be exactly vertical, in which case P3 = “point at infinity.”
If P1 is the “point at infinity,” then the sum P1 + P2 = P2. Similary, if P2 is the point at infinity, then P1 + P2 = P1. This shows how the point at infinity plays the role of 0.
It turns out that + is associative, which means that (A+B)`C = A`(B+C). That means we can write A+B+C without parentheses without any ambiguity.
Now that we have defined addition, we can define multiplication in the standard way that extends addition. For a point P on the elliptic curve, if k is a whole number, then kP = P + P + P + … + P (k times). Note that k is sometimes confusingly called an “exponent” in this case.
# Difference between public keys
| format|private key | |Hex|1E99423A4ED27608A15A2616A2B0E9E52CED330AC530EDCC32C8FFC6A526AEDD| |WIF | 5J3mBbAH58CpQ3Y5RNJpUKPE62SQ5tfcvU2JpbnkeyhfsYB1Jcn
# Creating master key
## Donations
If you would like to support me, donations are very welcome.
You can use Paypal to donate using your own credit card. The payment is processed by PayPal but you don't need to have a PayPal account or sign-up for one if you are paying by credit card.
You can also use your own Paypal account to donate.
# https://www.paypal.com/donate/?hosted_button_id=D8AXHXAMNZ3WY
You can also donate Bitcoin, Bitcoin Cash, Tron, Ethereum, Litecoin and etc ...
[]
# Disclaimer
The code within this repository comes with no guarantee, the use of this code is your responsibility*. I take 'NO' responsibility and/or liability for how you choose to use any of the source code available here. By using any of the files available in this repository, you understand that you are AGREEING TO USE AT YOUR OWN RISK.
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1 Jan 8, 2022 | 3,188 | 13,624 | {"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-2024-10 | latest | en | 0.897881 |
https://certifiedcalculator.com/cost-homeowners-insurance-calculator/ | 1,720,966,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00565.warc.gz | 147,209,006 | 14,189 | # Cost Homeowners Insurance Calculator
Estimated Insurance Cost: \$0.00
Introduction: Homeowners insurance is a crucial financial safeguard for those who own a house. It provides protection against unforeseen events like natural disasters and theft. To make an informed decision, it’s essential to understand the potential cost of homeowners insurance. This is where our “Cost Homeowners Insurance Calculator” comes in handy.
Formula: To estimate the cost of homeowners insurance, our calculator uses a simple formula. It multiplies the home value by the insurance rate percentage.
How to Use:
1. Input the estimated value of your home in dollars.
2. Enter the insurance rate percentage (usually provided by your insurance provider).
3. Click the “Calculate” button.
The calculator will instantly display the estimated insurance cost based on the provided values.
Example: Suppose your home is valued at \$250,000, and the insurance rate is 0.5%. Here’s how you can use the calculator:
• Home Value: 250000
• Insurance Rate: 0.5
• Click “Calculate”
The calculator will show an estimated insurance cost of \$1,250.
FAQs:
1. What is homeowners insurance?
• Homeowners insurance is a policy that provides financial protection for your home and personal belongings in the event of damage or theft.
2. Why do I need homeowners insurance?
• It’s essential to protect your investment and personal property from unexpected events.
3. How is the insurance rate determined?
• The insurance rate is usually determined by factors like location, the age of your home, and the coverage you choose.
4. Can I change my insurance rate?
• You can often adjust your insurance rate by selecting different coverage options and deductibles.
5. What’s the difference between actual cash value and replacement cost coverage?
• Actual cash value coverage considers depreciation, while replacement cost coverage pays for the cost of replacing damaged items without depreciation.
6. Are there discounts available for homeowners insurance?
• Yes, some insurers offer discounts for security systems, multiple policies, and other factors.
7. How often should I update my policy?
• You should review your policy annually to ensure it still meets your needs.
8. What factors can affect the cost of homeowners insurance?
• Factors include the location of your home, its age, the materials it’s made of, and your claims history.
9. Is homeowners insurance tax-deductible?
• In most cases, homeowners insurance premiums are not tax-deductible.
10. How can I lower my homeowners insurance premium?
• You can potentially lower your premium by increasing your home’s security, bundling insurance policies, and shopping around for quotes.
Conclusion: Our “Cost Homeowners Insurance Calculator” simplifies the process of estimating your insurance expenses. Remember that while this tool can provide an estimate, it’s essential to consult with insurance professionals to get a precise quote tailored to your specific needs. Understanding the cost of homeowners insurance is a vital step in ensuring your most significant investment remains protected. | 606 | 3,121 | {"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} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.909822 |
http://applieddigitalskills.withgoogle.com/c/middle-and-high-school/en/plan-and-budget/research-car-loans/introduction-to-car-loans.html | 1,571,811,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00548.warc.gz | 15,353,895 | 10,079 | # 1. Introduction to Car Loans attachmentStudents can submit work
Playback Speed:
Transcript
Welcome Back!
In this activity, you will learn about financing a car.
You will calculate the total monthly payment and the total interest for a series of loan amounts using PMT and absolute value functions and absolute and relative cell references.
You will also research cars and use an image function to insert pictures of the cars in your spreadsheet.
Say you are in the market for a new or used car.
You find a car you love, but you can’t afford to buy it out of pocket.
Few people can!
One option is to get a loan for the money, or to finance the car.
Financing means that you borrow money from a lender and pay them back little by little.
Not only do you pay back the principal, or the amount of money you borrowed; you also pay an additional percentage of that principal to the lender for using their money.
This additional percentage is called interest.
There are three factors that determine how much you’ll pay each month for a loan: The amount of money you borrow; The interest rate, also called the annual percentage rate, or A-P-R; and The term of the loan, or the time it takes you to pay it off.
Sometimes having a lower monthly payment means that you pay more overall!
If this sounds overwhelming, don’t worry.
Financing a car is not always a straightforward process, and there are many options to consider.
In this activity, you will create a spreadsheet to help you determine how much money you could afford to borrow when buying a new or used car.
On the spreadsheet, you will select an interest rate, or A-P-R, and a loan term and convert these to monthly amounts.
You will also specify a price range for the car you will research.
Once you have this basic information, you will calculate and record a table of possible loan amounts; your monthly payment; total payments, or the total amount you will spend on the car; and the amount of money you will spend on interest.
Once you set up your spreadsheet, you will research new or used cars to get an idea of what they cost per month, as well as the total amount you will pay over the term of the loan.
To begin, go to sheets dot google dot com slash create to start a blank spreadsheet.
Name it Loan Amount Chart.
Then, move on to the next video to start calculating costs. | 494 | 2,360 | {"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} | 2.6875 | 3 | CC-MAIN-2019-43 | latest | en | 0.947149 |
https://math.stackexchange.com/questions/1714675/using-paley-wiener-theorem-and-fourier-inversion-formula-to-get-this-result | 1,627,325,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00521.warc.gz | 384,614,903 | 41,117 | # Using Paley-Wiener theorem and Fourier inversion formula to get this result
I want to solve the problem #8 of Stein's book Complex analysis in Chapter 4, for the first part I've got the following:
We know that the coefficients of a series are:
$$a_n=\frac{f^{(n)}(z_0)}{n!}$$
Since the series is centered at $z_0=0$ and using the inverse Fourier transform we get :
$$f^{(n)}=(2 \pi i )^{n} \int_{-\infty}^{\infty}\hat{f}(\zeta)\zeta^{n}e^{2\pi ix\zeta}d\zeta=(2 \pi i )^{n} \int_{-M}^{M}\hat{f}(\zeta)\zeta^{n}e^{2\pi ix\zeta}d\zeta$$
and evaluating at zero and dividing by the factorial we have:
$$a_n=\frac{(2 \pi i )^{n}}{n!} \int_{-M}^{M}\hat{f}(\zeta)\zeta^{n}d\zeta$$
but then I don't know how to bound this to get that
$$\limsup_{n-> \infty}(n!|a_n|)^{1/n} \le 2 \pi M$$
I was trying to do the following:
$$|a_n|\le \frac{(2 \pi )^n}{n!}\int_{-M}^{M}|\hat{f}(\zeta)||\zeta^{n}|d\zeta$$
the thing is that I don't know how to ensure that the integral is less than $M^n$
Another way I thought of was to use Cauchy's inequalities then I will get the following
$$|a_n| \le \frac{n!}{R^n}||f||_{C}$$
for an appropiate election of the radius of the circle, so I want to pick the circle with center in zero and of radius $M$ but then How can I bound the integral? and I don't think this could take me to the result.
For the second part I don't understand if I can use that $\hat{f}$ is compact supported, if I can then I use Paley- Weiner theorem but I don't know how modify it to get the result since that $\epsilon$ is annoying me. If I can't use it then I can hardly think in a way to prove this.
• If $\hat{f}$ is continuous and compactly supported, then it is bounded, right? – carmichael561 Mar 26 '16 at 19:36
• Right :) it is correct, but how do you know is continuous? – user162343 Mar 26 '16 at 19:37
• What are the hypotheses on $f$? The Fourier transform of any reasonable function is continuous. – carmichael561 Mar 26 '16 at 19:39
• Just that $f$ is a series :) – user162343 Mar 26 '16 at 19:40
• Presumably $f$ must be integrable on $\mathbb{R}$, or else how is the Fourier transform defined? And the Fourier transform of any $L^1$ function is continuous. – carmichael561 Mar 26 '16 at 19:41
If $f\in L^1(\mathbb{R})$ then $\hat{f}$ is continuous, and since it is supported in $[-M,M]$, it must be bounded, by some $C>0$. Therefore $$n!a_n\leq (2\pi)^n\int_{-M}^M|\hat{f}(t)||t|^n\;dt \leq C(2\pi)^n\int_{-M}^M|t|^n\;dt =\frac{2CM^{n+1}}{n+1}(2\pi )^n$$
Now taking $n$th roots of both sides yields
$$(n!a_n)^{\frac{1}{n}}\leq \frac{(2CM)^{\frac{1}{n}}}{(n+1)^{\frac{1}{n}}}2\pi M$$
and since $\lim_{n\to\infty}K^{\frac{1}{n}}=1=\lim_{n\to\infty}(n+1)^{\frac{1}{n}}$ for any constant $K$, it follows that $$\limsup_{n\to\infty}(n!a_n)^{\frac{1}{n}}\leq2\pi M$$
• How do you take limsup ? I mean which is the calculation of limsup? – user162343 Mar 26 '16 at 19:47
• What about the second part? – user162343 Mar 26 '16 at 19:49
• I think is trickier right? – user162343 Mar 26 '16 at 19:50
• I think the radius of convergence is bigger. For instance, what if $a_n=\frac{1}{n!}$? – carmichael561 Mar 26 '16 at 19:56
• Yes, which has radius of convergence $\infty$. – carmichael561 Mar 26 '16 at 19:57
You were already on the right track, but stopped a little too soon. You showed $$a_n = \frac{(2\pi i)^{n}}{n!}\int_{-M}^{M}\hat{f}(\zeta)\zeta^n d\zeta$$ Therefore, \begin{align} (n! |a_n|)^{1/n}&=2\pi\left|\int_{-M}^{M}\hat{f}(\zeta)\zeta^n d\zeta\right|^{1/n} \\ &\le 2\pi\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta M^{n}\right)^{1/n} \\ &= 2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} \end{align} The limit of the right side exists as $n\rightarrow\infty$ and, assuming $\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta \ne 0$, that limit is $2\pi M$. Therefore, \begin{align} \limsup_{n} (n!|a_n|)^{1/n} & \le \limsup_{n} 2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} \\ & = \lim_{n}2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} = 2\pi M. \end{align} The only assumption needed for $\hat{f}$ is that it is absolutely integrable on $[-M,M]$.
For the converse, assume that $f(z)=\sum_{n=0}^{\infty}a_n z^n$, and assume that $$\limsup_{n}(n!|a_n|)^{1/n} \le 2\pi M.$$ Let $\epsilon > 0$ be given. Then there exists $N$ such that $$\sup_{n \ge N}(n!|a_n|)^{1/n} \le (2\pi M+\epsilon) \\ n!|a_n| \le (2\pi M+\epsilon)^{n} \;\;\; n \ge N \\ \sum_{n=0}^{\infty}|a_n||z|^{n} \le \sum_{n=0}^{N-1}\left(|a_n|-\frac{(2\pi M+\epsilon)^{n}}{n!}\right)|z|^{n}+\sum_{n=N}^{\infty}\frac{1}{n!}(2\pi M+\epsilon)^{n}|z|^{n}$$ A little juggling of constants to bound the first few terms by the full exponential series gives $$|f(z)| \le C\sum_{n=0}^{\infty}\frac{1}{n!}(2\pi M+\epsilon)^{n}|z|^{n} = e^{(2\pi M+\epsilon)|z|}.$$
• So you didn't make use of the Paley-Wiener theorem right? – user162343 Mar 27 '16 at 1:44
• How do you prove that $f$ is holomorphic in the complex plane? – user162343 Mar 27 '16 at 1:46
• @user162343 : (a) No, we're proving the Paley-Wiener for this special case. (b) $f$ is holomorphic because $f(z) = \int_{-M}^{M}\hat{f}(\xi)e^{2\pi i\xi\,z}d\xi$ easily extends to $z \in \mathbb{C}$ and can be shown to be differentiable from this integral representation; you can also expand the exponential in a power series, interchange summation and integration and get a power series expansion in $z$ for $f$. – Disintegrating By Parts Mar 27 '16 at 1:50
• Right, so let me continue with the other one :) (math.stackexchange.com/questions/1710909/…) and I'll be connected :) – user162343 Mar 27 '16 at 1:52
• @user162343 : Assume $f\in L^2$ or $f\in L^1$ so that the classical transform makes sense. In either case, because $\hat{f}$ is assumed to be compactly supported, then $\hat{f}\in L^1$. Then $e^{2\pi i\xi z}=\sum_{n=0}^{\infty} \frac{(2\pi i\xi z)^{n}}{n!}$ converges uniformly for bounded $z$ and $\xi$, which is enough to interchange summation of the series and integration to obtain $f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\int_{-M}^{M} \hat{f}(\xi)\xi^{n}d\xi\right)z^{n}$. The series automatically converges for all $z$, which proves an infinite radius of convergence. – Disintegrating By Parts Mar 27 '16 at 2:00 | 2,358 | 6,208 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-31 | latest | en | 0.901804 |
https://oj.vnoi.info/problems/?order=code&page=9 | 1,670,146,131,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00521.warc.gz | 463,162,223 | 9,620 | ## Problem list
ID ▴ Problem Category Points % AC # AC
coci1617_r4_kas COCI 2016/2017 - Contest 4 - Kas COCI 1.00 30.3% 157
coci1617_r4_osmosmjerka COCI 2016/2017 - Contest 4 - Osmosmjerka COCI 1.40 83.3% 2
coci1617_r4_rekonstruiraj COCI 2016/2017 - Contest 4 - Rekonstruiraj COCI 1.20 17.4% 5
coci1617_r4_rima COCI 2016/2017 - Contest 4 - Rima COCI 1.40 15.6% 38
coci1617_r5_pareto COCI 2016/2017 - Contest 5 - Pareto COCI 0.50 42.6% 55
coci1617_r5_poklon COCI 2016/2017 - Contest 5 - Poklon COCI 1.50 47.9% 130
coci1617_r5_ronald COCI 2016/2017 - Contest 5 - Ronald COCI 1.25 40.0% 32
coci1617_r5_strelice COCI 2016/2017 - Contest 5 - Strelice COCI 1.40 20.8% 2
coci1617_r5_tuna COCI 2016/2017 - Contest 5 - Tuna COCI 0.20 86.1% 137
coci1617_r5_unija COCI 2016/2017 - Contest 5 - Unija COCI 1.10 39.7% 67
coci1617_r6_gauss COCI 2016/2017 - Contest 6 - Gauss COCI 1.20 100.0% 1
coci1617_r6_hindeks COCI 2016/2017 - Contest 6 - Hindeks COCI 0.20 37.9% 188
coci1617_r6_savrsen COCI 2016/2017 - Contest 6 - Savrsen COCI 0.75 35.3% 284
coci1617_r6_sirni COCI 2016/2017 - Contest 6 - Sirni COCI 1.00 17.1% 58
coci1617_r6_telefoni COCI 2016/2017 - Contest 6 - Telefoni COCI 0.40 45.0% 264
coci1617_r6_turnir COCI 2016/2017 - Contest 6 - Turnir COCI 0.50 64.4% 69
coci1617_r7_baza COCI 2016/2017 - Contest 7 - Baza COCI 0.20 75.6% 64
coci1617_r7_igra COCI 2016/2017 - Contest 7 - Igra COCI 0.70 35.0% 56
coci1617_r7_klavir COCI 2016/2017 - Contest 7 - Klavir COCI 1.70 70.6% 11
coci1617_r7_paralelogrami COCI 2016/2017 - Contest 7 - Paralelogrami COCI 1.40 26.3% 7
coci1617_r7_poklon COCI 2016/2017 - Contest 7 - Poklon COCI 0.90 22.9% 42
coci1617_r7_uzastopni COCI 2016/2017 - Contest 7 - Uzastopni COCI 0.40 32.3% 79
coci1920_o_paint COCI 2019/2020 - Olympiad - Paint COCI 1.10 25.5% 7
coci1920_o_pastiri COCI 2019/2020 - Olympiad - Pastiri COCI 1.10 31.3% 21
coci1920_o_semafor COCI 2019/2020 - Olympiad - Semafor COCI 1.20 66.7% 4
coci1920_o_zagrade COCI 2019/2020 - Olympiad - Zagrade COCI 1.10 12.4% 11
coci1920_r1_dzumbus COCI 2019/2020 - Contest 1 - Dzumbus COCI 1.50 31.0% 43
coci1920_r1_lutrija COCI 2019/2020 - Contest 1 - Lutrija COCI 0.40 26.8% 32
coci1920_r1_trobojnica COCI 2019/2020 - Contest 1 - Trobojnica COCI 1.00 33.3% 11
coci1920_r1_trol COCI 2019/2020 - Contest 1 - Trol COCI 0.20 35.0% 104
coci1920_r1_zoo COCI 2019/2020 - Contest 1 - Zoo COCI 1.00 38.3% 88
coci1920_r2_acm COCI 2019/2020 - Contest 2 - ACM COCI 0.20 50.0% 10
coci1920_r2_checker COCI 2019/2020 - Contest 2 - Checker COCI 1.45 28.3% 13
coci1920_r2_popcount COCI 2019/2020 - Contest 2 - Popcount COCI 1.50 23.1% 7
coci1920_r2_slagalica COCI 2019/2020 - Contest 2 - Slagalica COCI 0.40 34.1% 12
coci1920_r2_zvijezda COCI 2019/2020 - Contest 2 - Zvijezda COCI 1.50 39.8% 27
coci1920_r3_drvca COCI 2019/2020 - Contest 3 - Drvca COCI 1.40 17.1% 31
coci1920_r3_grudanje COCI 2019/2020 - Contest 3 - Grudanje COCI 0.40 39.7% 21
coci1920_r3_lampice COCI 2019/2020 - Contest 3 - Lampice COCI 1.50 5.1% 10
coci1920_r3_preokret COCI 2019/2020 - Contest 3 - Preokret COCI 0.20 34.4% 60
coci1920_r3_sob COCI 2019/2020 - Contest 3 - Sob COCI 1.00 28.9% 13
coci1920_r4_holding COCI 2019/2020 - Contest 4 - Holding COCI 1.40 34.6% 64
coci1920_r4_klasika COCI 2019/2020 - Contest 4 - Klasika COCI 1.50 14.7% 69
coci1920_r4_nivelle COCI 2019/2020 - Contest 4 - Nivelle COCI 1.50 47.2% 180
coci1920_r4_psk COCI 2019/2020 - Contest 4 - Pod starim krovovima COCI 0.20 52.9% 69
coci1920_r4_spiderman COCI 2019/2020 - Contest 4 - Spiderman COCI 0.40 34.0% 65
coci1920_r5_emacs COCI 2019/2020 - Contest 5 - Emacs COCI 0.20 73.9% 208
coci1920_r5_matching COCI 2019/2020 - Contest 5 - Matching COCI 1.00 16.1% 8
coci1920_r5_politicari COCI 2019/2020 - Contest 5 - Političari COCI 1.00 47.3% 40
coci1920_r5_putovanje COCI 2019/2020 - Contest 5 - Putovanje COCI 1.30 38.8% 146 | 1,926 | 3,833 | {"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} | 2.890625 | 3 | CC-MAIN-2022-49 | latest | en | 0.208779 |
http://math1.skku.ac.kr/home/pub/367/ | 1,656,497,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00107.warc.gz | 39,997,567 | 2,619 | # Ch-7-Prob-15-Old-박종우
## 2244 days ago by matrix
A=matrix(5,5,[[1,3,5,2,-2],[-2,1,-2,1,0],[-1,-3,0,1,0],[-3,0,1,0,0],[0,1,2,-2,1]]) b=vector([1,0,1,0,0]) print A.rank() B=A.transpose() C=B*A x=C.inverse()*B*b print x
5 (11/97, -5/97, 33/97, 93/97, 125/97) 5 (11/97, -5/97, 33/97, 93/97, 125/97) | 172 | 298 | {"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} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.394842 |
http://math.stackexchange.com/questions/280708/a-question-on-a-sequence | 1,469,273,318,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00285-ip-10-185-27-174.ec2.internal.warc.gz | 155,217,942 | 18,046 | # A question on a sequence
Let $X$ be Hausdorff. If $S=\{x_n: n\in \mathbb N \}\subset X$, then does there exist a subsequence $D$ of the sequence $S$ such that $D$ is a discrete subset of $X$?
-
Have you tried working this out for some concrete examples of $X$ and $S$? – Brad Jan 17 '13 at 12:40
I think you mean "$D$ is a discrete subspace of $X$. (That is, every point of $D$ is open in the subspace topology of $D$). This may be of interest then. – David Mitra Jan 17 '13 at 13:06
@DavidMitra: You are RIGHT. – Paul Jan 17 '13 at 13:11
– Martin Jan 17 '13 at 13:20
So use the fact that if $A_1\subset A_2\subset X$, then the relative topology induced on $A_1$ by the relative topology of $A_2$ in $X$ is precisely the relative topology of $A_1$ in $X$, and fact that every infinite Hausdorff space contains a countably infinite discrete subspace. The latter fact is proven in this post. – David Mitra Jan 17 '13 at 13:20
If $x \in X$ has a finite open nbhd, then $\{x\}$ is open (i.e., $x$ is an isolated point). (Why?) Let $I$ be the set of isolated points of $X$; clearly $I$ is discrete, so if $I$ is infinite, just pick any countably infinite subset of it. That's the easy case; you have to work harder when $I$ is finite.
In that case let $Y = X \setminus I$. Note that if $y \in Y$, every open nbhd $V$ of $y$ is infinite. Now build the desired discrete set recursively. Start by choosing distinct points $y_0,y_1 \in Y$, and let $W_0,V_1$ be disjoint open sets with $y_0 \in W_0$ and $y_1 \in V_1$. (Here there's a reason for using different letters: the $V_n$'s are needed temporarily for the construction, but it's the $W_n$'s that we really want.) $V_1 \cap Y$ is infinite (why?), so we can pick a point $y_2 \in V_1 \setminus \{y_1\}$. Now let $W_1$ and $V_2$ be disjoint open subsets of $V_1$ such that $y_1 \in W_1$ and $y_2 \in V_2$. Now $V_2 \cap Y$ is infinite, so we can pick a point $y_3 \in V_2 \setminus \{y_2\}$ and let $W_2$ and $V_3$ be disjoint open subsets of $V_2$ such that $y_2 \in W_2$ and $y_3 \in V_3$. Continue in this fashion to get points $y_n$ and open sets $W_n$ for each $n \in \omega$. Clearly $y_n \in W_n$ for each $n$, and with a little thought you should be able to see that if $m \ne n$, $y_m \notin W_n$. (It's probably best to consider the cases $m<n$ and $m>n$ separately. It may also help to make a sketch of the first few steps of the construction.) | 807 | 2,405 | {"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} | 3.71875 | 4 | CC-MAIN-2016-30 | latest | en | 0.851086 |
https://www.askmehelpdesk.com/showthread.php?t=264433&s=5724df94ac9f12192f21dfd7dd87cfe9&p=1314752 | 1,624,323,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00224.warc.gz | 584,460,768 | 9,456 | lelandcollins27 Posts: 1, Reputation: 1 New Member #1 Sep 27, 2008, 03:50 PM
int secret(int x)
{
int I, j;
I = 2 * x;
if (I > 10)
j=x/2;
else
j = x/3;
return j - 1;
}
int another (int a, int b)
{
int I, j;
j = 0;
for (I = a; I <=b; I++)
j = j + I;
return j;
}
what is the output of the following segments
a. x=10
cout << secret(x) << endl;
b. x=5; y=8;
cout << another(x,y) <<endl;
c. x=10; k=secret(x);
cout << x << " " << k <<" " << another (x,k) <<endl;
d. x=5; y=8;
cout << another(y,x) <<endl;
iFire Posts: 39, Reputation: -1 Junior Member #2 Oct 10, 2008, 11:22 AM
Where would the cout's be..
Question Tools Search this Question Search this Question: Advanced Search | 258 | 680 | {"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-2021-25 | latest | en | 0.692859 |
https://ebin.pub/machine-learning-for-neuroscience-a-systematic-approach-9781032136721-9781032137278-9781003230588.html | 1,702,189,795,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00202.warc.gz | 258,986,012 | 46,186 | # Machine Learning for Neuroscience: A Systematic Approach 9781032136721, 9781032137278, 9781003230588
##### This book addresses the growing need for machine learning and data mining in neuroscience. The book offers a basic overv
215 120 17MB
English Pages 305 [306] Year 2023
Cover
Half Title
Title
Contents
Preface
Section I Required Math and Programming
Chapter 1 Fundamental Concepts of Linear Algebra for Machine Learning
Introduction
Linear Algebra Basics
Other Matrix Operations
Determinant of a Matrix
Vectors and Vector Spaces
Vector Metrics
Vector Length
Dot Product
Tensor Product
Cross Product
Eigenvalues and Eigenvectors
How Do We Find Its Eigenvalues?
Eigendecomposition
Summary
Chapter 2 Overview of Statistics
Introduction
Basic Terminology
Types of Measurement Scales
Data Collection
Measures of Central Tendency
Correlation
P-Value
Z-Test
Outliers
T-Test
Linear Regression
ANOVA
The Kruskal-Wallis
Kolmogorov-Smirnov
Statistical Errors
Power of a Test
Basic Probability
What is Probability?
Basic Set Theory
Basic Probability Rules
Conditional Probability
Independent Events
Bayes Theorem
Special Forms of Bayes’ Theorem
Summary
Chapter 3 Introduction to Python Programming
Introduction
Fundamental Python Programming
Variables and Statements
Object-Oriented Programming
IDE
IDLE
Other IDEs
Python Troubleshooting
General Tips to Remember
Control Statements
Working with Strings
Working with Files
A Simple Program
Basic Math
Summary
Exercises
Exercise 1: Install Python
Exercise 2: Hello World
Exercise 3: Fibonacci Sequence
Chapter 4 More with Python
Introduction
File Management
Exception Handling
Regular Expressions
Internet Programming
Installing Modules
Specific Modules
Operating System Module
NumPy
Pandas
Scikit-Learn
PyTorch
WMI
PIL
Matplotlib
TensorFlow
The Zen of Python
Data Structures
Lists
Queue
Stack
Algorithms
Summary
Exercises
Exercise 1: Install TensorFlow
Exercise 2: Regular Expressions and Exception Handling
Exercise 3
Section II Required Neuroscience
Chapter 5 General Neuroanatomy and Physiology
Introduction
Neuroanatomy
Neuroscience Terminology
Development View
Anatomical View
Brainstem
Cerebellum
Cerebrum
Limbic System
Spinal Cord
Neurophysiology
Neurotransmitters
Metabolism
Neuroimaging
Neurofunction
Motor Control
Perception
Summary
Chapter 6 Cellular Neuroscience
Introduction
Basic Neuro Cellular Structure
Types of Neurons
Synapse
Electrical Synapses
Ion Channels
Neurotransmitters
Acetylcholine
Catecholamines
Serotonin
Glutamate
Intolaimines
Gamma-Aminobutyric Acid (GABA)
Glycine
Dopamine
Peptide Neurotransmitters
Epinephrine and Norepinephrine
Agonists and Antagonists
Neurotransmitter Synthesis and Packing
Neurotransmitters and Psychoactive Substances
Cannabinoids
Opioids
Nicotine
Glial Cells
Summary
Chapter 7 Neurological Disorders
Specific Disorders
ALS
Epilepsy
Parkinson’s
Tourette’s
Muscular Dystrophy
Encephalitis
Depression
Progressive Supranuclear Palsy
Alzheimer’s
Meningitis
Stroke
Multiple Sclerosis
Tumors
Neurological Disorders and Machine Learning
Summary
Chapter 8 Introduction to Computational Neuroscience
Introduction
Neuron Models
Nernst Equation
Goldman Equation
Electrical Input-Output Voltage Models
Hodgkin-Huxley
FitzHugh-Nagumo Model
Leaky Integrate-and-Fire
Noisy Input Model
Hindmarsh-Rose Model
Morris-Lecar Model
Graph Theory and Computational Neuroscience
Algebraic Graph Theory
Spectral Graph Theory
Graph Similarities
Information Theory and Computational Neuroscience
Complexity and Computational Neuroscience
Emergence and Computational Neuroscience
Summary
Section III Machine Learning
Chapter 9 Overview of Machine Learning
Introduction
Basics of Machine Learning
Supervised Algorithms
Unsupervised Algorithms
Clustering
Anomaly Detection
Specific Algorithms
K-Nearest Neighbor
Naïve Bayes
Support Vector Machines
Feature Extraction
PCA
Artificial Intelligence
General Intelligence
Synthetic Consciousness
Summary
Exercises
Lab 1: Detecting Parkinson’s
Chapter 10 Artificial Neural Networks
Introduction
Concepts
ANN Terminology
Activation Functions
Optimization Algorithms
Models
Feedforward Neural Networks
Perception
Backpropagation
Normalization
Specific Variations of Neural Networks
Recurrent Neural Networks
Convolutional Neural Networks
Autoencoder
Spiking Neural Network
Deep Neural Networks
Neuroscience Example Code
Summary
Exercises
Lab 1: Basic TensorFlow
Lab 2: Perceptron
Chapter 11 More with ANN
Introduction
More Activation Functions
SELU
SiLU
Swish
Softsign
Algorithms
Spiking Neural Networks
Liquid State Machine
Long Short-Term Memory Neural Networks
Boltzmann Machine
Deep Belief Network
Summary
Exercises
Lab 1: LSTM
Lab 2: LSTM for Neuroscience
Lab 3: Experiment with Activation Functions
Chapter 12 K-Means Clustering
Introduction
K-Means Clustering
K-Means++
K-Medians Clustering
K-Medoids
Random Forest
DBSCAN
Summary
Exercises
Exercise 1: K-Means with Alzheimer’s Data
Exercise 2: K-Means++ with Neurological Data
Chapter 13 K-Nearest Neighbors
Introduction
Examining KNN
Dimensionality Reduction
Visualize KNN
Alternatives
Deeper with Scikit-Learn
Summary
Exercises
Lab 1: KNN Parkinson’s Data
Lab 2: KNN Variations with Parkinson’s Data
Chapter 14 Self-Organizing Maps
Introduction
The SOM Algorithm
SOM in More Detail
Variations
GSOM
TASOM
Elastic Maps
Growing Self-Organizing Maps
Summary
Exercises
Lab 1: SOM for Neuroscience
Lab 2: Writing Your Own Code
Index
##### Citation preview
Machine Learning for
Neuroscience
This book addresses the growing need for machine learning and data mining in neuroscience. The book offers a basic overview of the neuroscience, machine learning and the required math and programming necessary to develop reliable working models. The material is presented in an easy-to-follow, user-friendly manner and is replete with fully working machine learning code. Machine Learning for Neuroscience: A Systematic Approach tackles the needs of neuroscience researchers and practitioners that have very little training relevant to machine learning. The first section of the book provides an overview of necessary topics in order to delve into machine learning, including basic linear algebra and Python programming. The second section provides an overview of neuroscience and is directed to the computer science-oriented readers. The section covers neuroanatomy and physiology, cellular neuroscience, neurological disorders and computational neuroscience. The third section of the book then delves into how to apply machine learning and data mining to neuroscience, and provides coverage of artificial neural networks (ANN), clustering and anomaly detection. The book contains fully working code examples with downloadable working code. It also contains lab assignments and quizzes, making it appropriate for use as a textbook. The primary audience is neuroscience researchers who need to delve into machine learning, programmers assigned neuroscience related machine learning projects and students studying methods in computational neuroscience.
Machine Learning for
Neuroscience
A Systematic Approach
Chuck Easttom
First edition published 2024 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487–2742 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of Taylor & Francis Group, LLC © 2024 Chuck Easttom Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all mate rials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photo copying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978–750–8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. ISBN: 9781032136721 (hbk) ISBN: 9781032137278 (pbk) ISBN: 9781003230588 (ebk) DOI: 10.1201/9781003230588 Typeset in Times by Apex CoVantage, LLC
Contents
Preface................................................................................................................... xiii
Section i Chapter 1
Required Math and Programming
Fundamental Concepts of Linear Algebra for Machine Learning...... 3
Introduction ......................................................................................... 3
Linear Algebra Basics ......................................................................... 5
Matrix Addition and Multiplication .................................................... 6
Other Matrix Operations ..................................................................... 8
Determinant of a Matrix .................................................................... 10
Vectors and Vector Spaces ................................................................. 12
Vector Metrics ................................................................................... 15
Vector Length ......................................................................... 15
Dot Product ....................................................................................... 15
Tensor Product ........................................................................ 16
Cross Product.......................................................................... 17
Eigenvalues and Eigenvectors ................................................ 18
How Do We Find Its Eigenvalues? ............................ 18
Eigendecomposition .......................................................................... 20
Summary ........................................................................................... 21
Chapter 2
Overview of Statistics ....................................................................... 25
Introduction ....................................................................................... Basic Terminology............................................................................. Types of Measurement Scales ................................................ Data Collection .................................................................................. Measures of Central Tendency .......................................................... Correlation ......................................................................................... P-Value ................................................................................... Z-Test ...................................................................................... Outliers ................................................................................... T-Test ......................................................................... Linear Regression .............................................................................. Additional Statistics .......................................................................... ANOVA .................................................................................. The Kruskal-Wallis .................................................................
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Contents
Kolmogorov-Smirnov........................................................................ Statistical Errors ................................................................................ Power of a Test ....................................................................... Basic Probability ............................................................................... What is Probability? ............................................................... Basic Set Theory ....................................................... Basic Probability Rules .......................................................... Conditional Probability .......................................................... Independent Events................................................................. Bayes Theorem .................................................................................. Special Forms of Bayes’ Theorem ......................................... Summary ........................................................................................... Test Your Skills .................................................................................. Chapter 3
Introduction to Python Programming ............................................... 45
Introduction ....................................................................................... Fundamental Python Programming ................................................... Variables and Statements ........................................................ Object-Oriented Programming ............................................... IDE .................................................................................................... IDLE ....................................................................................... Other IDEs .............................................................................. Python Troubleshooting .................................................................... General Tips to Remember ..................................................... Basic Programming Tasks ................................................................. Control Statements ................................................................. Working with Strings......................................................................... Working with Files ................................................................. A Simple Program ............................................................................. Basic Math .............................................................................. Summary ........................................................................................... Exercises ............................................................................................ Exercise 1: Install Python ....................................................... Exercise 2: Hello World.......................................................... Exercise 3: Fibonacci Sequence .............................................
Chapter 4
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More with Python.............................................................................. 65
Introduction ....................................................................................... File Management ............................................................................... Exception Handling ........................................................................... Regular Expressions .......................................................................... Internet Programming........................................................................ Installing Modules ............................................................................. Specific Modules .................................................................... Operating System Module .........................................
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Contents
NumPy ....................................................................... Pandas ........................................................................ Scikit-Learn ............................................................... PyTorch ..................................................................... WMI .......................................................................... PIL .......................................................................................... Matplotlib ............................................................................... TensorFlow ........................................................................................ The Zen of Python ............................................................................. Advanced Topics ............................................................................... Data Structures ....................................................................... Lists ........................................................................... Queue......................................................................... Stack .......................................................................... Linked List ................................................................ Algorithms ......................................................................................... Summary ........................................................................................... Exercises ............................................................................................ Exercise 1: Install TensorFlow ............................................... Exercise 2: Regular Expressions and Exception Handling .... Exercise 3 ...............................................................................
Section ii Chapter 5
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Required neuroscience
General Neuroanatomy and Physiology ............................................ 97
Introduction ....................................................................................... 97
Neuroanatomy ................................................................................... 97
Neuroscience Terminology ..................................................... 98
Development View.................................................................. 98
Anatomical View .................................................................. 101
Brainstem ................................................................ 103
Cerebellum .............................................................. 104
Cerebrum ................................................................. 105
Limbic System......................................................... 106
Spinal Cord ........................................................................... 108
Neurophysiology ............................................................................. 108
Neurotransmitters ................................................................. 109
Metabolism ........................................................................... 109
Neuroimaging ....................................................................... 110
Neurofunction ....................................................................... 110
Motor Control .......................................................... 110
Perception ................................................................ 112
Summary ......................................................................................... 114
viii
Chapter 6
Contents
Cellular Neuroscience ..................................................................... 117
Introduction ..................................................................................... Basic Neuro Cellular Structure........................................................ Types of Neurons .................................................................. Synapse ................................................................................. Electrical Synapses .................................................. Ion Channels .................................................................................... Neurotransmitters ............................................................................ Acetylcholine ........................................................................ Catecholamines..................................................................... Serotonin............................................................................... Glutamate ............................................................................. Intolaimines .......................................................................... Gamma-Aminobutyric Acid (GABA) .................................. Glycine ................................................................................. Dopamine ............................................................................. Peptide Neurotransmitters .................................................... Epinephrine and Norepinephrine .......................................... Agonists and Antagonists ..................................................... Neurotransmitter Synthesis and Packing .............................. Neurotransmitters and Psychoactive Substances .................. Cannabinoids ........................................................... Opioids .................................................................... Nicotine ................................................................... Glial Cells ........................................................................................ Summary ......................................................................................... Test Your Knowledge.......................................................................
Chapter 7
117
117
119
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121
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124
125
126
126
127
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128
129
129
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131
131
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Neurological Disorders.................................................................... 135
Specific Disorders............................................................................ ALS....................................................................................... Epilepsy ................................................................................ Parkinson’s............................................................................ Tourette’s .............................................................................. Muscular Dystrophy ............................................................. Encephalitis .......................................................................... Depression ............................................................................ Progressive Supranuclear Palsy ............................................ Alzheimer’s........................................................................... Meningitis ............................................................................. Stroke .................................................................................... Multiple Sclerosis ................................................................. Tumors .................................................................................. Neurological Disorders and Machine Learning ..............................
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136
136
137
137
138
138
139
140
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142
143
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144
ix
Contents
Summary ......................................................................................... 144
Chapter 8
Introduction to Computational Neuroscience ................................. 147
Introduction ..................................................................................... Neuron Models ................................................................................ Nernst Equation ............................................................................... Goldman Equation ................................................................ Electrical Input-Output Voltage Models ............................... Hodgkin-Huxley ...................................................... FitzHugh-Nagumo Model ....................................... Leaky Integrate-and-Fire ......................................... Adaptive Integrate-and-Fire .................................... Noisy Input Model .................................................. Hindmarsh–Rose Model .......................................... Morris–Lecar Model................................................ Graph Theory and Computational Neuroscience ............................ Algebraic Graph Theory ....................................................... Spectral Graph Theory.......................................................... Graph Similarities ................................................................. Information Theory and Computational Neuroscience ................... Complexity and Computational Neuroscience ................................ Emergence and Computational Neuroscience ................................. Summary ......................................................................................... Test Your Knowledge.......................................................................
147
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152
152
153
156
159
160
163
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168
Section iii Machine Learning Chapter 9
Overview of Machine Learning ...................................................... 175
Introduction ..................................................................................... Basics of Machine Learning ............................................................ Supervised Algorithms .................................................................... Unsupervised Algorithms ................................................................ Clustering ............................................................................. Anomaly Detection ............................................................... Specific Algorithms ......................................................................... K-Nearest Neighbor.............................................................. Naïve Bayes .......................................................................... Gradient Descent .................................................................. Support Vector Machines ...................................................... Feature Extraction ........................................................................... PCA ......................................................................................
175
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178
178
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179
179
180
180
183
183
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Contents
Artificial Intelligence ............................................................ General Intelligence.............................................................. Synthetic Consciousness ...................................................... Summary ......................................................................................... Exercises .......................................................................................... Lab 1: Detecting Parkinson’s................................................
184
184
187
189
189
189
Chapter 10 Artificial Neural Networks.............................................................. 193
Introduction ..................................................................................... Concepts .......................................................................................... ANN Terminology ................................................................ Activation Functions ............................................................. Optimization Algorithms ...................................................... Models .................................................................................. Feedforward Neural Networks ............................................. Perceptron ............................................................................. Backpropagation ................................................................... Normalization ....................................................................... Specific Variations of Neural Networks .......................................... Recurrent Neural Networks .................................................. Convolutional Neural Networks ........................................... Autoencoder ......................................................................... Spiking Neural Network ....................................................... Deep Neural Networks ......................................................... Neuroscience Example Code........................................................... Summary ......................................................................................... Exercises .......................................................................................... Lab 1: Basic TensorFlow ...................................................... Lab 2: Perceptron .................................................................
193
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200
201
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202
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212
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217
Chapter 11 More with ANN .............................................................................. 219
Introduction ..................................................................................... More Activation Functions .............................................................. SELU .................................................................................... SiLU ..................................................................................... Swish .................................................................................... Softsign ................................................................................. Algorithms ....................................................................................... Spiking Neural Networks ..................................................... Liquid State Machine ........................................................... Long Short-Term Memory Neural Networks ....................... Boltzmann Machine.............................................................. Radial Basis Function Network ............................................ Deep Belief Network ............................................................ Summary .........................................................................................
219
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xi
Contents
Exercises .......................................................................................... Lab 1: LSTM ........................................................................ Lab 2: LSTM for Neuroscience ............................................ Lab 3: Experiment with Activation Functions ......................
230
230
231
233
Chapter 12 K-Means Clustering ........................................................................ 237
Introduction ..................................................................................... K-Means Clustering.............................................................. K-Means++...................................................................................... K-Medians Clustering ..................................................................... K-Medoids ....................................................................................... Random Forest ................................................................................ DBSCAN ......................................................................................... Summary ......................................................................................... Exercises .......................................................................................... Exercise 1: K-Means with Alzheimer’s Data ....................... Exercise 2: K-Means++ with Neurological Data .................
237
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247
250
250
251
251
252
252
254
Chapter 13 K-Nearest Neighbors ....................................................................... 255
Introduction ..................................................................................... Examining KNN .............................................................................. Dimensionality Reduction ............................................................... Visualize KNN...................................................................... Alternatives ........................................................................... Deeper with Scikit-Learn ................................................................ Summary ......................................................................................... Exercises .......................................................................................... Lab 1: KNN Parkinson’s Data .............................................. Lab 2: KNN Variations with Parkinson’s Data .....................
255
256
259
259
262
265
266
266
266
267
Chapter 14 Self-Organizing Maps ..................................................................... 269
Introduction ..................................................................................... The SOM Algorithm ............................................................. SOM in More Detail ............................................................. Variations ......................................................................................... GSOM................................................................................... TASOM................................................................................. Elastic Maps ......................................................................... Growing Self-Organizing Maps ........................................... Summary ......................................................................................... Exercises .......................................................................................... Lab 1: SOM for Neuroscience .............................................. Lab 2: Writing Your Own Code ............................................
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Index ..................................................................................................................... 279
Preface
This book is quite special to me. It combines two passions of mine: neuroscience and machine learning. As you are, no doubt, aware, there are numerous books on machine learning, some directed towards data mining, some to finance, some to cyber security, etc. However, there is very little literature specifically on machine learning for neuroscience. That is the reason behind writing this book. One goal of this book is to be accessible to the widest possible audience. That is why the three sections are designed as they are. Section I provides foundational material describing linear algebra, statistics and Python programming. For readers that don’t have that skillset, or need a refresher, this section (chapters 1 through 4) provides that. However, other readers may not need this section. Section II is for those readers who may not have any neuroscience background: for example, a computer programmer who has been tasked with machine learning for neuroscience, but lacks any real knowledge of neuroscience. Now, if you are a neuroscientist or medical doctor, you will find section II (chapters 5 through 8) rather rudimentary. The goal of those chapters is to provide the interested reader with the necessary, foundational knowledge of neuroscience. Finally, we come to section III. In chapters 9 through 14 we explore machine learning. In each of those chapters you will find fully functioning code. And every script in this book was actually, personally, tested by myself. So, I am certain they all work as written. This will allow you to ensure you can make machine learning scripts work, including several specifically for neuroscience. However, simply typing in others scripts is not really the goal of a book like this. The goal is for you to be able to create your own. So, in the later chapters, there will be lab exercises for the more adventuresome reader to try their skills. In those labs you may be given part of a script, or merely suggestions on how to write the script. You will have to develop it yourself. However, those only come after you have been presented with dozens of fully functioning scripts. This brings us to the topic of how best to read this book. Clearly, the three sections were made to be independent. You can almost think of them as mini books within this book. You can skip section I or II if you don’t need that material. But for section III, you should spend some time in each chapter. Don’t rush through it. First, make all examples in that chapter work. Then perhaps experiment with changing a few parameters of the script and noting what happens. The idea is for you to finish this book quite comfortable with machine learning for neuroscience.
xiii
Dr. Chuck Easttom is the author of 37 books, on topics such as cryptography, quantum computing, programming, cyber security and more. He is also an inventor with 25 patents and the author of over 70 research papers. He holds a Doctor of Science in cyber security, a Ph.D. in Nanotechnology, a Ph.D. in computer science, and three master’s degrees (one in applied computer science, one in education and one in systems engineering). He is a senior member of both the IEEE and the ACM. He is also a distinguished speaker of the ACM and a distinguished visitor of the IEEE. Dr. Easttom is currently an adjunct professor for Georgetown University and for Vanderbilt University. Dr. Easttom is active in IEEE (Institute for Electrical and Electronic Engineers) standards groups, including several related to the topics in this book, such as: Chair of IEEE P3123 Standard for Artificial Intelligence and Machine Learning (AI/ML) Terminology and Data Formats. Member of the IEEE Engineering in Medicine and Biology Standards Com mittee. Standard for a Unified Terminology for Brain-Computer Interfaces P2731 from 2020 to present. You can find out more about the author at his website www.ChuckEasttom.com.
xv
Section I
Required Math and Programming
1
Fundamental Concepts of Linear Algebra for Machine Learning
INTRODUCTION Before exploring what linear algebra is, it may be helpful to explain why you need to know it for machine learning. Linear algebra is used in many different fields including machine learning. One can certainly download a machine learning script from the internet and execute it without understanding linear algebra. However, to really delve into machine learning algorithms, one will need a basic understand ing of linear algebra. If you wish to eventually move beyond the simple copying and modifying of scripts, and to truly work with machine learning, in any context, you will need a basic understanding of linear algebra. Furthermore, this entire text is directed towards machine learning for neuroscience. Neuroscience, by its very nature, demands a certain rigor. You will discover in later chapters that it is common for data to be imported into a machine learning algorithm as a vector. Images are often loaded as matri ces. These two facts alone make a basic knowledge of linear algebra relevant to machine learning. One example of using vectors is found in a process known as one hot encoding. There are a series of bits representing possible values for a specific piece of data. The only output that is a 1, is the value represented in the data, the other possible values are all 0. Consider for a moment that you need to import data that represents colors red, green and blue. You can use a vector that represents the color you are importing as a 1, and the other possible colors as 0. Thus, red would be 1,0,0 and green would be 0,1,0. You will find one hot encoding frequently used in machine learning. Another topic that is common in machine learning, which depends on linear algebra, is principle component analysis (PCA). PCA is used in machine learning to create projections of high-dimensional data for both visualization and for train ing models. PCA, in turn, depends on matrix factorization. The Eigendecomposi tion is often used in PCA. Thus, to truly understand PCA, one needs to understand linear algebra. Another application of linear algebra to machine learning combines linear alge bra with statistics (which we will be reviewing in chapter 2). Linear regression is a statistical method for describing the relationship between variables. More specifi cally, linear regression describes the relationship between a variable and one or more explanatory variables. Linear algebra was initially developed as a method to solve systems of linear equations (thus the name). However, we will move beyond solving linear equations. DOI: 10.1201/9781003230588-2
3
4
Machine Learning for Neuroscience
Linear equations are those for which all elements are of the first power. Thus, the following three equations are linear equations: a + b + c = 115 3x + 2y = 63 2x + y – z = 31 However, the following are not linear equations: 2x2 + 3 = 12 4y2 + 6x + 2 = 22 7x4 + 8y3 – 1 = 8 The first three equations have all individual elements only raised to the first power (often that with a number such as x1, the 1 is simply assumed and not written). But in the second set of equations, at least one element is raised to some power greater than 1. Thus, they are not linear. While linear algebra was created to solve linear equations, the subject has grown to encompass a number of mathematical endeavors that are not directly focused on linear equations. The reason for that is that linear algebra repre sents numbers in matrix form, which we shall examine in some detail in this chapter. That matrix form turns out to be ideal for a number of applications, including machine learning. Certainly, linear algebra is useful in machine learning, thus this chapter. It may help to begin with a very brief history of linear algebra. One of the earliest books on the topic of linear algebra was Extension Theory written in 1844, by Her mann Grassman. The book included other topics, but also had some fundamental con cepts of linear algebra. As time progressed, other mathematicians added additional features to linear algebra. In 1856, Arthur Cayle introduced matrix multiplication. Matrix multiplication is one of the elements of linear algebra that has applications beyond solving linear equations. In 1888, Giuseppe Peano gave a precise definition of a vector space. Vector spaces also have applications well beyond solving linear equations. Linear algebra continued to evolve over time. This chapter is, by design, a brief overview of linear algebra. Engineering students typically take at least one entire undergraduate course in linear algebra. Mathemat ics majors may take additional course including graduate courses. Clearly, a single chapter in a book cannot cover all of that. But it is not necessary for you to learn linear algebra to that level in order to begin working with machine learning. The most obvious thing that is omitted in this chapter are proofs. We do not mathematically prove any of what is presented. While that might grate on the more mathematically oriented reader, the proofs are not necessary for our purposes. If you are interested in delving deeper into linear algebra, the Mathematical Association of America has a website with resources on linear algebra you can find at www.maa.org/topics/ history-of-linear-algebra.
Fundamental Concepts of Linear Algebra for Machine Learning
5
LINEAR ALGEBRA BASICS As was discussed in the introduction, the goal of this chapter is not to solve linear equations. Rather, the goal is to provide the reader with sufficient understanding of linear algebra in order to apply that knowledge to machine learning. Machine learning frequently deals with matrices and vectors. Therefore, it is appropriate to begin this exposition of linear algebra with a discussion of matrices. A matrix is a rectangular arrangement of numbers in rows and columns. Rows run horizontally and columns run vertically. The dimensions of a matrix are stated m x n where m is the number of rows and n is the number of columns. Here is an example: é 1 2 ù
ê2 0ú
ê ú êë 3 1 úû
If this definition seems a bit elementary to you, you are correct. This is a rather straightforward way to represent numbers. The basic concept of a matrix is not at all difficult to grasp. That is yet another reason it is an appropriate place to begin exploring linear algebra. A matrix is just an array that is arranged in columns and rows. Vectors are simply matrices that have one column or one row. The examples in this section focus on 2 x 2 matrices, but a matrix can be of any number of rows and columns, it need not be a square. A vector can be considered a 1 x m matrix. A vector that is vertical is called a column vector, one that is horizontal is called a row vector. Matrices are usually labeled based on column and row: é a ij êa ë ij
a ij ù
a ij úû
The letter i represents the row, and the letter j represents the column. A more concrete example is shown here: é a11 êa ë 21
a12 ù
a 22 úû
This notation is commonly used for matrices including row and column vectors. In addition to understanding matrix notation, there are some common types of matrices you should be familiar with. The most common are listed here: Column Matrix: A matrix with only one column.
Row Matrix: A matrix with only one row.
Square Matrix: A matrix that has the same number of rows and columns
Equal Matrices: Two matrices are considered equal if they have the same
number of rows and columns (the same dimensions) and all their corre sponding elements are exactly the same. Zero matrix: Contains all zeros.
6
Machine Learning for Neuroscience
Each of these has a role in linear algebra, which you will see as you proceed through the chapter. Later in this book you will see column matrices used frequently as fea tures being analyzed by machine learning algorithms.
MATRIX ADDITION AND MULTIPLICATION Addition and multiplication of matrices are actually not terribly complicated. There are some basic rules you need to be aware of to determine if two matrices can be added or multiplied. If two matrices are of the same size, then they can be added to each other by simply adding each element together. You start with the first row and first column in the first matrix and add that to the first row and the first column of the second matrix thus in the sum matrix. This is shown in equation 1.1: é a11 êa ë 21
a12 ù é b11 +
a 22 úû êë b 21
b12 ù é A 11 + b11 =
b 22 úû êë A 21+ b 211
a12 + b12 ù
a 22 + b 22 úû
(eq. 1.1)
Consider the following more concrete example given in equation 1.2: é 1 2 ù é 1 2 ù é 2 4 ù
ê 2 2 ú +
ê 2 1 ú =
ê ú ë
û ë
û ë 4 3 û
(eq. 1.2)
This is rather trivial to understand, thus this journey into linear algebra begins with an easily digestible concept. Multiplication, however, is somewhat more difficult. One can only can multiply two matrices if the number of columns in the first matrix is equal to the number of rows in the second matrix. First let us take a look at mul tiplying a matrix by a scalar (i.e., a single number). You simply multiply the scalar value by each element in the matrix as shown in equation 1.3. é aij C ê ë aij
aij ù é caij =
aij úû êë caij
caij ù
caij úû
(eq. 1.3)
For a more concrete example, consider equation 1.4: é 1 1ù é 3 3ù
3 ê ú = ê ú ë 2 3û ë 6 9 û
(eq. 1.4)
As you can observe for yourself, the addition of two matrices is no more complicated than the addition you learned in primary school. However, the multiplication of two matrices is a bit more complex. The two matrices need not be of the same size. The requirement is that the number of columns in the first matrix be equal to the number of rows in the second matrix. If that is the case, then you multiply each element in the first row of the first matrix, by each element in the second matrix’s first column. Then you multiply each element of the second row of the first matrix by each element of the second matrix’s second column. Let’s first examine this using variables rather
Fundamental Concepts of Linear Algebra for Machine Learning
7
than actual numbers. This example also uses square matrices to make the situation even simpler. This is shown in equation 1.5. é a b ù é e ê c d ú + ê g ë
û ë
f ù
h úû
(eq. 1.5)
This is multiplied in the following manner. a *e + b*g a *f + b*h c*e + d *g c*f + d * h
( a11 * b11 + a12 * b 21 ) ( a11 * b12 + a12 * b 22 ) ( a11 * b11 + a12 * b 21 ) ( a11 * b11 + a12 * b 21 )
Thus, the product will be:
( a*e + b*g ) ( a*f + b*h ) ( c*e + d *g ) ( c*f + d *h ) It is worthwhile to memorize this process. Now, consider this implemented with a concrete example, as shown in equation 1.6: é 1 3ù é 2 2 ù
ê3 1ú ê 1 3 ú
ë
ûë û
(eq. 1.6)
We begin with: 1 * 2+3 * 1=5 1* 2 + 3 * 3 = 11 3 * 2 +1*1 = 7 3 * 2 +1* 3 = 9 The final answer is: é 5 11ù ê7 9 ú
û
ë
It should first be noted that one can only multiply two matrices if the number of col umns in the first matrix equals the number of rows in the second matrix. That should be clear if you reflect on how the multiplication is done. If you keep that rule in mind, then the multiplication is not particularly complicated, it is simply tedious. It is important to remember that matrix multiplication, unlike multiplication of scalers, is not commutative. For those readers who may not recall, the commutative property states a * b = b * a. If a and b are scalar values then this is true, regardless
8
Machine Learning for Neuroscience
of the scaler values (e.g., integers, rational numbers, real numbers, etc.). However, when multiplying two matrices, this is not the case. This is often a bit difficult for those new to matrix mathematics. However, it is quite easy to demonstrate that this commutative property does not hold for matrices. For example, consider the matrix multiplication shown in Equation 1.7. é 2 1 ù é 3 2 ù é 8 5ù
ê ê 3 2 ú ê 2 1 ú =
ú ë
ûë û ë13 8 û
(eq. 1.7)
However, if you reverse that order you will get a different answer. This is shown in equation 1.8. é 3 2 ù é 2 1 ù é12 7 ù
ê 2 1 ú ê 3 2 ú = ê 7 4 ú
ë
ûë û ë û
(eq. 1.8)
Seeing an actual example, it becomes immediately apparent that matrix mul tiplication is not commutative. You may find some instance that just incidentally appears communitive. However, that is not sufficient. To have the commutative property, it must be the case that, regardless of the operands chosen, the result is commutative.
OTHER MATRIX OPERATIONS Matrix addition and multiplication are perhaps the most obvious of operations on matrices. These operations are familiar to you from previous mathematics you have studied. However, there are some operations that are specific to matrices. Matrix transposition is one such operation. Transposition simply reverses the order of rows and columns. While the focus so far has been on 2 x 2 matrices, the transposition operation is most easily seen with a matrix that has a different number of rows and columns. Consider the matrix shown in Equation 1.9. é 2 3 2 ù
ê1 4 3 ú ë û
(eq. 1.9)
To transpose it, the rows and columns are switched, creating a 2 x 3 matrix. The first row is now the first column. You can see this in Equation 1.10. é 2 1 ù
ê3 4 ú
ê ú êë 2 3 úû
(eq. 1.10)
There are some particular properties of transpositions that you should be familiar with. If you label the first matrix A, then the transposition of that matrix is labeled AT. Continuing with the original matrix being labeled A, there are a few properties of matrices that need to be described as outlined in Table 1.1.
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Fundamental Concepts of Linear Algebra for Machine Learning
TABLE 1.1 Basic Properties of Matrix Transposition Property
Explanation
(AT)T = A
If you transpose the transposition of A, you get back to A.
(cA)T = cAT
The transposition of a constant, c, multiplied by an array, A, is equal to multiplying the constant c by the transposition of A.
(AB)T = BTAT
A multiplied by B, then the product transposed, is equal to B transposed multiplied by A transposed.
(A + B)T = AT + BT
Adding the matrix A and the matrix B, then transposing the sum, is equal to first transposing A and B, then adding those transpositions.
AT = A
If a square matrix is equal to its transpose, it is called a symmetric matrix.
Table 1.1 is not exhaustive, rather, it is a list of some of the most common prop erties regarding matrices. These properties are not generally particularly difficult to understand. Another operation particular to matrices is finding a submatrix of a given matrix. A submatrix is any portion of a matrix that remains after deleting any number of rows or columns. Consider the 5 x 5 matrix shown in Equation 1.11 é 2 ê3 ê ê2 ê ê4 êë1
2 8 3 3 2
4 0 2 1 2
5 2 2 2 0
3 ù
1 úú 1ú ú 4ú 3 úû
(eq. 1.11)
If you remove the second column and second row, as shown in Equation 1.12: é 2 ê3 ê ê2 ê ê4 êë 1
2 8 3 3 2
4 0 2 1 2
5 2 2 2 0
1 úú 1ú
ú 4ú
3 úû
(eq. 1.12)
You are left with the matrix shown in Equation 1.13. é 2 ê2 ê ê4 ê ë1
4 2 1 2
5 2 2 0
3 ù
1 úú 4ú ú
3 û
That matrix shown in Equation 1.13 is a sub matrix of the original matrix.
(eq. 1.13)
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Machine Learning for Neuroscience
Another item that is important to matrices is the identity matrix. Now this item does have an analog in integers and real numbers. Regarding the addition operation, the identity element is 0. Any number + 0 will still be the original number. In mul tiplication, the identity element is 1. Any number * 1 is still the original number. The identity matrix functions in much the same manner. Multiplying a matrix by its identity matrix leaves it unchanged. To create an identity matrix, just have all the elements along the main diagonal set to 1, and the rest to zero. Consider the following matrix: é 3 2 1 ù
ê1 1 2 ú ê ú ê3 0 3 úû
ë
Now consider the identity matrix. It must have the same number of columns and rows, with its main diagonal set to all 1s and the rest of the elements all 0s. The identity matrix looks like this: é 1 0 0 ù
ê0 1 0 ú ê ú ê0 0 1 úû
ë
If you multiply the original matrix by the identity matrix, the product will be the original matrix. You can see this in Equation 1.14 é 3 2 1 ù é1 0 0 ù é 3 2 1 ù
ê1 1 2 ú ´ ê0 1 0 ú = ê1 1 2 ú ê ú ê ú ê ú ê3 0 3 úû êë0 0 1 úû êë 3 0 3 úû
ë
(eq. 1.14)
Another special type of matrix is a unimodular matrix. Unimodular matrices are also used in some lattice-based algorithms. A unimodular matrix is a square matrix of integers with a determinant of +1 or -1. Recall that a determinant is a value that is computed from the elements of a square matrix. The determinant of a matrix A is denoted by |A|. In the next section we will explore how to calculate the determinant of a matrix.
DETERMINANT OF A MATRIX Next, we will turn our attention to another relatively easy computation, the determi nant of a matrix. The determinant of a matrix A is denoted by |A|. An example of a determinant in a generic form is as follows: é a b ù
| A| ê ú = ad bc ë C d û
Fundamental Concepts of Linear Algebra for Machine Learning
11
A more concrete example might help elucidate this concept: é2 3 ù A ê ú = (2)(2) (3)(1) = 1 ë1 2 û A determinant is a value that is computed from the individual elements of a square matrix. It provides a single number, also known as a scaler value. Only a square matrix can have a determinant. The calculation for a 2 x 2 matrix is simple enough, we will explore more complex matrices in just a moment. However, what does this single scalar value mean? There are many things one can do with a determinant, most of which we won’t use in this text. It can be useful in solving linear equations, changing variables in integrals (yes, linear algebra and calculus go hand in hand); however, what is immediately useable for us is that if the determinant is nonzero, then the matrix is invertible. What about a 3 x 3 matrix, such as that shown in Equation 1.15? é a1 êa ê 2 êë a 3
b1 b2 b3
c1 ù
c 2 úú c3 úû
(eq. 1.15)
This calculation is substantially more complex. There are a few methods to do this. We will use one called ‘expansion by minors’. This method depends on breaking the 3 x 3 matrix into 2 x 2 matrices. The 2 x 2 matrix formed by b2, c2, b3, c3, shown in Equation 1.16, is the first. é a1 êa ê 2 êa 3 ë
b1 b2 b3
c1 ù
c 2 úú ú c3 û
(eq. 1.16)
This one was rather simple, as it fits neatly into a contiguous 2 x 2 matrix. But to find the next one, we have a bit of a different selection as shown in Equation 1.17: é a1 êa ê 2 êë a 3
b1 b2 b3
c1 ù
c 2 úú c3 úû
(eq. 1.17)
The next step is to get the lower left corner square matrix as shown in Equation 1.18. é a1 êa ê 2 êë a 3
b1 b2 b3
c1 ù
c 2 úú c3 úû
(eq. 1.18)
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Machine Learning for Neuroscience
As with the first one, this one forms a very nice 2 x 2 matrix. Now what shall we do with these 2 x 2 matrices? The formula is actually quite simple and is shown in Equation 1.19. Note that det is simply shorthand for determinant. é a1 det êêa 2 êë a 3
b1 b2 b3
c1 ù
é b c 2 úú = a1 det ê 2 ë b3 c3 úû
c 2 ù
é a a det ê 2 c3 úû 2 ë a 3
c 2 ù
é a + a det ê 2 c3 úû 3 ë a 3
b 2 ù
(eq. 1.19) b3 úû
We take the first column, multiplying it by its cofactors, and with a bit of simple addition and subtraction, we arrive at the determinant for a 3 x 3 matrix. A more concrete example might be useful. Let us calculate the determinant for this matrix: é 3 2 1 ù
ê1 1 2 ú ê ú êë 3 0 3 úû
This leads to: é1 2 ù 3*det ê ú = 3 *((1* 3) (2 * 0)) = 3(3) = 9 ë0 3 û é1 2 ù 2*det ê ú = 2 *((1* 3) (2 * 3)) = 2(3) = 6 ë3 3 û é1 1 ù 1*det ê ú = 1*(((1* 0) (1* 3)) = 1(3) = 3 ë3 0 û And that leads us to 9 – (–6) + (–3) = 12 Yes, that might seem a bit cumbersome, but the calculations are not overly difficult. We will end our exploration of determinants at 3 x 3 matrices. But yes, one can take the determinant of larger square matrices. One can calculate determinants for matrices that are 4 x 4, 5 x 5 and as large as you like. However, our goal in this chapter is to give you a bit of general foundation in linear algebra, not to explore every nuance of linear algebra.
VECTORS AND VECTOR SPACES Vectors are an essential part of linear algebra. We normally represent data in the form of vectors. In linear algebra, these vectors are treated like numbers. They can be added and multiplied. A vector will look like what is shown here: é 1 ù
ê 3 ú ê ú êë 2 úû
Fundamental Concepts of Linear Algebra for Machine Learning
13
This vector has only integers; however, a vector can have rational numbers, real num bers, even complex numbers (which will be discussed later in this chapter). Vectors can also be horizontal, as shown here:
[1
3 2]
Often you will see variables in place of vector numbers, such as: é a ù
êb ú ê ú êë c úû
The main point that that you saw in the previous section is that one can do math with these vectors as if they were numbers. You can multiple two vectors together; you can also multiply a vector by a scaler. Scalers are individual numbers, and their name derives from the fact that they change the scale of the vector. Consider the scaler 3 multiplied by the first vector shown in this section: é1 ù é 3 ù
3 êê3 úú = êê9 úú êë 2 úû êë6 úû
You simply multiply the scaler, by each of the elements in the vector. We will be exploring this and other mathematical permutations in more detail in the next sec tion. But let us address the issue of why it is called a scaler now. We are viewing the data as a vector; another way to view it would be as a graph. Consider the previous vector [1,3,2] on a graph as shown in Figure 1.1. Now what happens when we perform the scalar operation of 3 multiplied by that vector? We literally change the scale of the vector, as shown in Figure 1.2. Figures 1.1 and 1.2 may appear identical, but look closer. In figure 1.1, the x value goes to 9, whereas in figure 1.2 the x value only goes to 3. We have ‘scaled’ the vector. The term scalar is used because it literally changes the scale of the vector. Formally, a vector space is a set of vectors that is closed under addition and multipli cation by real numbers. Think back to the earlier discussion of abstract algebra with groups, rings and fields. A vector space is a group. In fact, it is an abelian group. You can do addition of vectors, and the inverse. You also have a second operation scaler multiplication, without the inverse. Note that the first operation, addition, is commu tative, but the second operation, multiplication, is not. So, what are basis vectors? If you have a set of elements E (i.e., vectors) in some vector space V, the set of vectors E is considered a basis if every vector in the vector space V can be written as a linear combination of the elements of E. Put another way, you could begin with the set E, the basis, and through a linear combinations of the vectors in E, create all the vectors in the vector space V. And as the astute reader will have surmised, a vector space can have more than one basis set of vectors.
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FIGURE 1.1 Graph of a vector.
FIGURE 1.2
Scaling a vector.
Fundamental Concepts of Linear Algebra for Machine Learning
15
What is linear dependence and independence? In the theory of vector spaces, a set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the others; if no vector in the set can be written in this way, then the vectors are said to be linearly independent. A subspace is a subset of a vector space that is a vector space itself, e.g., the plane z = 0 is a subspace of R3 (it is essentially R2.). We’ll be looking at Rn and subspaces of Rn.
VECTOR METRICS There are a number of metrics that one can calculate from vector. Each of these is important in some application of linear algebra. In this section we will cover the most common metrics that you will encounter when applying matrix mathematics.
Vector Length Let us begin this section with a fairly easy topic, the length of a vector, which is computed using the Pythagorean theorem: vector = x2 + y2 Consider the vector:
[ 2, 3, 4] Its length is: 22 + 32 + 42 = 5.38 This is simple but will be quite important as we move forward. Now we will add just a bit more detail to this concept. The nonnegative length is called the norm of the vector. Given a vector v, this is written as ||v||. This will be important later on in this book. One more concept to remember on lengths/norms: if the length is 1, then this is called a unit vector.
DOT PRODUCT The dot product of two vectors has numerous applications. This is an operation you will likely encounter with some frequency. The dot product of two vectors is simply the two vectors multiplied. Consider vectors X and Y. Equation 1.20 shows what the dot product would be. n
åX Y i =1
i
i
(eq. 1.20)
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Machine Learning for Neuroscience
Examining a concrete example to see how this works should be helpful. Consider two column vectors: é 1 ù é 3 ù
ê 2 ú ê 2 ú ê úê ú êë 1 úû êë1 úû
The dot product is found by (1 * 3) + (2 * 2) + (1 * 1) = 8 That is certainly an easy calculation to perform. But what does it mean? Put more frankly, why should you care what the dot product is? Recall that vectors can also be described graphically. You can use the dot product, along with the length of the vectors to find the angle between the two vectors. We already know the dot product is 8. Recall that length: vector = x2 + y2 Thus, the length of vector X is
12 + 22 +12 = 2.45
The length of vector Y is 32 + 22 +12 = 3.74 Now we can easily calculate the angle. It turns out that the cos θ = dot product/ length of X * length of Y or cos q =
8 = .87307 (2.45)(3.74)
Finding the angle from the cosine is straightforward; you probably did this in second ary school trigonometry. But even with just the dot product, we have some informa tion. If the dot product is 0 then the vectors are perpendicular. This is because the cos θ of a 90-degree angle is 0. The two vectors are referred to as orthogonal. Recall that the length of a vector is also called the vector’s norm. And if that length/norm is 1, it is the unit vector. This leads us to another term we will see fre quently later in this book. If two vectors are both orthogonal (i.e., perpendicular to each other) and have unit length (length 1), the vectors are said to be orthonormal. Essentially, the dot product is used to produce a single number, a scaler, from two vertices or two matrices. This is contrasted with the tensor product. In math, a tensor is an object with multiple indices, such as a vertex or array. The tensor product of two vector spaces, V and W, V ⊗ W is also a vector space.
tensor Product This is essentially a process where all of the elements from the first vector are multi plied by all of the elements in the second vector. This is shown in figure 1.3:
Fundamental Concepts of Linear Algebra for Machine Learning
17
FIGURE 1.3 Tensor Product.
cross Product This particular metric is very interesting. It illustrates the fact that vector mathemat ics is inherently geometric. Even if you are working only with vectors, and not seeing the graphical representation, it is there. Let us first describe how you calculate the cross product, then discuss the geometric implications. If you have two vectors A and B, the cross product is defined as: A ´ B =| A || B | sin(q)n |A| denotes the length of A (note: length is often called magnitude) | B | denotes the length of B N is the unit vector that is at right angles to both A and B
q is the angle between A and B If you reflect on this a bit, you will probably notice that this requires a threedimensional space. Using that fact, there is a simple way to calculate the cross product C: cx = A y Bz A z By cy = A z Bx A x Bz cz = A x Bx A y Bx So, consider two vectors: A = [2, 3, 4] and B = [5, 6, 7]. Cx = 3 * 7 4 * 6 = 21 24 = 3
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Machine Learning for Neuroscience
Cy = 4 * 5 2 * 7 = 20 14 = 6 Cz = 2 * 6 3 * 5 = 12 15 = 3 The cross product is [-3, 6, -3].
eigenVaLues and eigenVectors Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, char acteristic values, proper values, or latent roots. To clarify, consider a column vector we will call v. Then also consider an n x n matrix we will call A. Then consider some scalar λ. If it is true that: Av = l v Then we say that v is an eigenvector of the matrix A and λ is an eigenvalue of the matrix A. Let us look a bit closer at this. The prefix eigen is actually the German word which can be translated as specific, proper, particular, etc. Put in its most basic form, an eigenvector of some linear transformation T, is a vector that, when T is applied to it, does not change direction, it only changes scale. It changes scale by the scalar value λ, the eigenvalue. Now we can revisit the former equation just a bit and expand our knowledge of linear algebra: T ( v) = l v This appears precisely like the former equation, but with one small difference. The matrix A is now replaced with the transformation T. Not only does this tell us about eigenvectors and eigenvalues, but it also tells us a bit more about matrices. A matrix, when applied to a vector, transforms that vector. The matrix itself is an operation on the vector. This is a concept that is fundamental to matrix theory. Let us add something to this. How do you find the eigenvalues and eigenvectors for a given matrix? Surely it is not just a matter of trial and error with random numbers. Fortunately, there is a very straightforward method, one that is actually quite easy, at least for 2 x 2 matrices. Consider the following matrix: é 5 2 ù
ê9 2 ú
û
ë
How Do We Find Its Eigenvalues? Well, the Cayley-Hamilton theorem provides insight on this issue. That theorem essentially states that a linear operator A is a zero of its characteristic polynomial. For our purposes, it means that det | A l I |= 0
Fundamental Concepts of Linear Algebra for Machine Learning
19
We know what a determinant is, we also know that I is the identity matrix. The λ is the eigenvalue we are trying to find. The A is the matrix we are examining. Remem ber that in linear algebra you can apply a matrix to another matrix or vector, so a matrix is, at least potentially, an operator. So, we can fill in this equation: é 5 2 ù é1 0 ù
det ê ú l ê ú = 0
ë 9 2 û ë 0 1 û
Now, we just have to do a bit of algebra, beginning with multiplying λ by our identity matrix, which will give us: é 5 2 ù é l 0 ù
det ê ú ê ú = 0
ë 9 9 û ë 0 l û Which in turn leads to 2 ù é5 l det ê =0 2 l úû ë 9 = (5 l )(5 l ) 18 = 10 7l + l 2 18 = 0 l 2 7l 8 = 0 This can be factored (note if the result here cannot be factored, things do get a bit more difficult, but that is beyond our scope here): (l 8)(l +1) = 0 This means we have two eigenvalues:
l1 = 8 l2 = 1 For a 2 x 2 matrix you will always get two eigenvalues. In fact, for any n x n matrix, you will get n eigenvalues, but they may not be unique. Now that you have the eigenvalues, how do you calculate the eigenvectors? We know that: é 5 2 ù
A = ê ú ë9 2 û
l1 = 8 l2 = 1
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Machine Learning for Neuroscience
é X ù
We are seeking unknown vectors, so let us label the vector ê ú .
ë Y û
Now recall the equation that gives us eigenvectors and eigenvalues: Av = l v Let us take one of our eigenvalues and plug it in: é 5 2 ù é X ù é X ù
ê9 2 ú ê Y ú = 8 êY ú ë
û ë û ë û
é 5 x + 2 y ù é8X ù ê9 x + 2 y ú = ê8Y ú ë
û ë û This gives us two equations: 5x + 2y = 8x 9x + 2y = 8y Now we take the first equation and do a bit of algebra to isolate the y value. Subtract the 5x from each side to get: 2y = 3x Then divide both sides by 2 to get y = 3 / 2x It should be easy to see that to solve this with integers (which is what we want), then x = 2 and y = 3 solve it. Thus, our first eigenvector is é 2 ù
ê3 ú with l1 = 8
ë û
You can work out the other eigenvector for the second eigenvalue on your own using this method.
EIGENDECOMPOSITION As was mentioned at the beginning of this chapter, Eigendecomposition is key to principle component analysis. The process of Eigendecomposition is the factoriza tion of a matrix into what is called canonical form. That means that the matrix is represented in terms of its eigenvalues and eigenvectors.
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Fundamental Concepts of Linear Algebra for Machine Learning
Consider a square matrix A that is n x n with n linearly independent eigenvectors qi (where i = 1, . . . n). If this is true, then the matrix A can be factorized as shown in equation 1.21. A = Q Ù A 1
(eq. 1.21)
In equation 1.21, Q is a square n x n matrix whose ith column is the eigen vector qi of A. The symbol Ù is a diagonal matrix whose diagonal elements are the corresponding eigenvalues. It should be noted that if a matrix A can be eigendecomposed and if none of its eigenvalues are zero, then that matrix is invertible. For more information on Eigendecomposition, the following sources may be of use: https://mathworld.wolfram.com/EigenDecomposition.html https://personal.utdallas.edu/~herve/Abdi-EVD2007-pretty.pdf
SUMMARY Linear algebra is an important topic for machine learning. The goal of this chapter is to provide you with a basic introduction to linear algebra, or perhaps for some readers a review. You should ensure that you have mastered these concepts before proceeding to subsequent chapters. That means you should be comfortable adding and multiplying matrices, calculating dot products, calculating the determinant of a matrix, and working with eigenvalues and eigenvectors. You will see eigenvalues and eigenvectors used again in relation to spectral graph theory in chapter 8. These are fundamental operations in linear algebra. The multiple-choice questions will also help to ensure you have mastered the material. If you wish to go further with linear algebra, you may find the following resources useful: McMahon, D. 2005. Linear algebra demystified. McGraw Hill Professional. This is a good resource for basic linear algebra. Aggarwal, Charu. 2020. Linear algebra and optimization for machine learn ing (Vol. 156). Springer International Publishing. Schneider, H. and Barker, G. P. 2012. Matrices and linear algebra (Dover books on mathematics). Dover Publications.
TEST YOUR KNOWLEDGE é 2 ù
1. Solve the equation 2 êê3 úú .
êë 4 úû
é 4 ù
a. êê6 úú êë 8 úû
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é 5 ù
b. êê 4 úú êë 6 úû
c. 18 d. 15 2. What is the dot product of these two vectors? é1 ù é 4 ù
ê 2 ú ê5 ú ê úê ú êë 3 úû êë6 úû
a. b. c. d.
11 32 28 21
é 2 2 ù
3. Solve this determinant | A | ê ú .
ë 3 4 û
a. b. c. d.
8 6 0 –2
é1 2 3 ù
4. Solve this determinant: | A | ê 2 1 4 ú.
ê ú êë 3 1 2 úû
a. 11 b. 12 c. 15 d. 10 5. What is the dot product of these two vectors? é1 ù é3 ù
ê 2 ú ê3 ú ê úê ú êë 2 úû êë 4 úû
a. b. c. d.
65 17 40 15
Fundamental Concepts of Linear Algebra for Machine Learning
é 3 ù
6. What is the length of this vector? ê3 ú ê ú êë 2 úû
a. 4.35 b. 2.64 c. 3.46 d. 4.56 é 2 3 ù é3 2 ù
7. What is the product of these two matrices? ê ú ê ú ë1 4 û ë3 7 û
é15 25 ù
a. ê ú ë15 10 û
é15 10 ù
b. ê ú ë15 30 û
é10 15 ù
c. ê ú ë15 30 û
é15 25 ù
d. ê ú ë15 30 û
8. Are these two vectors orthogonal? é1 ù é 0 ù
ê0 ú ê1 ú ë û ë û
a. Yes b. No 9. Write the identity matrix for this matrix: é 3 2 1 ù
ê3 5 6 ú ê ú ê3 4 3 úû
ë
ê0 1 0ú ê ú êë 0 0 1 úû
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2
Overview of Statistics
INTRODUCTION Machine learning often involves the use of statistics. And being completely candid, the more you know about statistics the better you will be able to utilize machine learning. This is not to imply that machine learning is merely the application of statistics. Unfortunately, some vendors who market alleged machine learning are merely doing statistics, but that is not true machine learning. In this chapter, the goal will be to familiarize yourself with the essential statistical knowledge you will need for effectively utilizing machine learning. For some readers this may be a review. For others, this may be your first foray into statistics. Most texts on statistics begin with defining some terms, or jumping into basic formulas. And those shall certainly be covered in this chapter. However, we will begin with a more fundamental issue. Let us begin by exploring what statistics actually is. Statistics is a branch of mathematics designed to allow one to accom plish two goals. The first is to accurately describe data and trends in data. The second is to make predictions on future behavior, based on current data. The first goal is referred to as descriptive statistics. Any method or formula which yields some number that tells you about a set of data is considered descriptive statistics. Any method or formula which discusses a probability of some event occurring is predictive statistics. In this chapter we will begin by discussing descriptive statistics. The goal is to summarize the current level of understanding of basic descriptive statistics and to give some general guidelines for using descriptive statistics. Later in the chapter, the basics of probability will be covered.
BASIC TERMINOLOGY Before we can proceed, certain terminology must be covered. Without a thorough understanding of these terms, it is impossible for any person to be able to study even rudimentary statistics. Continuous Variable: A quantitative variable that can assume an uncountable number of values. This means that a continuous variable can assume any value along a line interval, including every possible value between any two values. Data (singular): The value of the variable associated with one element of a population or sample. This value may be a number, a word, or a symbol. Data (plural): The set of values collected for the variable from each of the elements belonging to the sample. Descriptive statistics: This is a common application of statistics. This process involves the classification, analysis, and interpretation of data. DOI: 10.1201/9781003230588-3
25
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Discrete Variable: A quantitative variable that can assume a countable num ber of values. This means that a discrete variable can only assume values corresponding to isolated points along a line interval. Experiment: A planned activity whose results yield a set of data. Hypothesis: The idea you are testing. Statistics are usually done in an attempt to confirm or refute some idea. Often in statistics you confirm or refute the null hypothesis, denoted as H0. It is the hypothesis that essentially the results you get are random and are not due to some real relationship. In other words, if the null hypothesis is true, then the apparent relationship is really simply a random coincidence. Predictive statistics: Using statistics generated from the sample in order to make predictions; this is also sometimes referred to as inferential statistics. This use of statistics deals with probabilities. Parameter: This is a descriptive number about a population. A statistic is a descriptive number about a sample. Population: The target group you wish to study, such as all men aged 30 to 40. Sample: The subgroup from the population you select to study, in order to make inferences about the population. Variable: A characteristic about each individual element of a population or sample. These are basic terms, and it seems quite likely that many readers are already quite familiar with them. However, it is necessary to establish this basic lexicon before proceeding. Other terms will be defined as needed.
tyPes of MeasureMent scaLes When measuring, the scale you use determines how accurate you can be. In some instances, it is simply not possible to get beyond a certain level of accuracy. It is important to understand the scale you are using. You will find that, often in machine learning, one can deal with a wide range of data types. 1. Nominal: For qualitative data with distinct categories. For example, the categories of neurological imaging, such as MRI, PET, CAT, are cat egories but are not ordered in any way. Similarly, the analysis of such images to be tumor, injury, or healthy are categories, but without a spe cific order. 2. Ordinal: For qualitative data with distinct categories in which ordering (or ranking) is implied. A good example is the Likert scale that you see on many surveys: 1 = Strongly disagree; 2 = Disagree; 3 = Neutral; 4 = Agree; 5 = Strongly agree. 3. Interval: For quantitative data with an ordered scale in which the interval between data values is meaningful. For example, the categories of rank in the military. Clearly a major is higher-ranked than a captain, but how much higher? Does he have twice the authority of a captain? It is impossible to say. You can only say he is higher-ranked.
Overview of Statistics
27
4. Ratio: For quantitative data that have an inherently defined zero and the ratio of data values is meaningful. Weight in kilograms is a very good example since it has a definite ratio from one weight to another. 50 kg is indeed twice as heavy as 25 kg. Clearly, when measuring the accuracy of a machine learning algorithm, you will want to have a ratio scale.
DATA COLLECTION Normally, statistics are done with only a fraction of the actual population being considered. That fraction is called the sample, and the group in question is the pop ulation. For example, if you wish to examine the accuracy of a given neurological diagnostic process, you will require a sample of patients that have been through the process. This leads to two obvious questions. Is the sample size you selected large enough and is the sample truly representative of the population you are attempting to meas ure? The first question will often be a controversial one. Obviously the larger the sample size the better. However, it is often impractical to get very large sample sizes. For example, in medical studies one cannot practically examine every patient that has a particular disease. Rather, one has to take a sample of patients that fit particular parameters. And, of course, with human subjects, the subject’s willingness to partic ipate is of paramount importance. The second question, whether or not your sample is actually representative of the population you are trying to measure, is much easier to answer. There are some very specific ways in which you should select a sample. Using proper sampling tech niques will give your statistical analysis credibility. The Statistics Glossary1, lists several sampling techniques; each is described here: Independent Sampling: This occurs when multiple samples are taken, but each sample has no effect on any other. Random Sampling: This occurs when subjects for your sample are picked totally at random with no other factors influencing their selection. As an example, when names are drawn from a hat, one has random sampling. Stratified Random Sampling: In this process, the population is divided into layers based on some criteria, and a number of random subjects are taken from each stratum. In our example of studying men who are over 40 and overweight, you might break the population into strata based on how over weight they are, or how old they are. For example, you might have men that are 25 to 50 lbs. overweight in one stratum and those who are 50 to 100 lbs. overweight in another, then finally those who are more than 100 lbs. over weight. There are other sampling methods, but those just discussed are very commonly used. If you wish to learn more about sampling methods, the following websites will be helpful: • Stat Pac, www.statpac.com/surveys/sampling.htm • Statistics Finland, www.stat.fi/tk/tt/laatuatilastoissa/lm020500/pe_en.html
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• Australian Bureau of Statistics, www.abs.gov.au/websitedbs/D3310116. NSF/4a255eef008309e44a255eef00061e57/116e0f93f17283eb4a2567ac00 213517!OpenDocument When evaluating any statistical analysis, it is important to consider how the sampling was done, and if the sample size seems large enough to be relevant. It might even be prudent to never rely on a single statistical study. If multiple studies of the same population parameter, using different samples, yield the same or similar results, then one has a compelling body of data. A single study always has a chance of being sim ply an anomaly, no matter how well the study was conducted.
MEASURES OF CENTRAL TENDENCY The first and simplest sort of descriptive statistics involves measures of central ten dency. This is simply a way of seeing what the aggregate of the data tells us about the data. The three most simple measures of central tendency are the mean, median, and mode. The mean is simply the arithmetic average, the mode is the item in the sample that appears most often, and the median is the item that appears in the middle. Assume you had a set of test scores as follows: 65, 74, 84, 84, 89, 91, 93, 99, 100 The mode is easy, 84 is the only score that appears more than once. The median is the score in the center, which in this case is 89. The mean is found by adding the scores and dividing by the number of scores (in this case 9). The formula for that is mean x = Σx/n. In this case it would be 86.55. In the preceding formula, x is the mean and n is the number of scores in the sample. Another important term is range. The range is simply the distance from the low est score to the highest. In our example the highest is 100, the lowest is 65, thus the range is 35. You will see these four numbers ubiquitously presented in statistical studies. How ever, what do they really tell us? In this case. the arithmetic mean of the scores was actually about the center of the scores. In our case, all but two of our scores are grouped in a narrow range from 84 to 100. This clustering means that our measures of central tendency probably tell us a lot about our data. But what about situations with much more variety in the numbers? In such cases, the mean may not tell us much about the actual data. This leads to other measures we can do, which can indi cate just how accurate the mean is. The standard deviation is a measurement that will tell you this. To quote a popular statistics website:2 The standard deviation is a metric that describes how closely the various values are clustered around the mean in a set of data. When the individual values are rather tightly clustered together and the bell-shaped curve is steep, the standard deviation is small. When the values are spread apart and the bell curve is relatively flat, this will produce a relatively large stan dard deviation.
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Overview of Statistics
The standard deviation (denoted s) is the square root of the sum of the variance divided by the number of elements. Put in simpler terms, you take each item in the sample, and see how far it varies from the mean. You then square that value and divide it by the number in the sample. Take the square root of that number (which is called the variance) and you have the standard deviation. This can be seen in equation 2.1.
å (x x) n
s=
2
n 1
(eq. 2.1)
n 1
In equation 2.1, x is the current value, x is the mean, and n is the number of values in the sample. In our example we would take each item in the sample minus the mean of 86.5, square that difference, and total the results, like this:
( 65 86.5) + ( 74 86.5) + (84 86.5) + (84 86.5) + (89 86.5) 2 2 2 2 + ( 91 886.5 ) + ( 93 86.5 ) + ( 99 86.5 ) + (100 86.5 ) 2
2
2
2
2
which is equal to: 462.25 +156.25 + 6.25 + 6.25 + 6.25 + 20.25 + 42.25 + 156.25 +182.25 = 882 Now divide that by n – 1 (n = 9, so divide by 8) and you get 110.25, which is the variance. The square root of the variance, in this case, 10.5, is the standard deviation. That means, in plain English, that on average, the various scores were about 10.5 units from the mean. As you can see, standard deviation and variance are simply arithmetical computa tions done to compare the individual items in the sample to the mean. They give you some idea about the data. If there is a small standard deviation, that indicates that val ues were clustered near the mean and that the mean is representative of the sample. These are not the only means of measuring central tendency, but they are the most commonly used. Virtually all statistical studies will report mean, median, mode, range, standard deviation, and variance.
CORRELATION After a study, you have may have two variables, let us call them x and y. The question is what degree of a correlation do they have? What is the relationship between the two variables? There are a few statistical methods for calculating this. One way to view the relationship between two variables are: The Pearson Product-Moment Correlation Coefficient (r), or correlation coefficient for short is a measure of the degree of linear relationship between two variables, usually labeled X and Y. While in regression the emphasis is on predicting one variable from the other, in correlation the emphasis is on
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the degree to which a linear model may describe the relationship between two variables. A more clear and concise definition can be found at the BMJ (a medical journal) website:3 “The word correlation is used in everyday life to denote some form of association. We might say that we have noticed a correlation between foggy days and attacks of wheeziness. However, in statistical terms we use correlation to denote association between two quantitative variables.” One valuable way to calculate this is via Pearson’s correlation coefficient, usually denoted with a lower case r. The formula for this is shown in equation 2.2. r=
å ( x x )( y y ) å(x x ) å( y y ) i
i
2
i
(eq. 2.2)
2
i
Equation 2.2 can be summarized with the following steps: 1. Take each value of x and multiply it by each value of y. 2. Multiply n times the mean of x and the mean of y. 3. Subtract the second number from the first and you have the numerator of our equation. 4. Now multiply n – 1 times the standard deviation of x and the standard deviation of y, and you now have the numerator. 5. Do the division and you have Pearson’s correlation coefficient. To illustrate this further, let us work out an example. Assume you have x variable of years of postsecondary education, and y variable as annual income in tens of thou sands. We will use a small sample to make the math simpler: Years of post-secondary education
Annual salary in 10’s of thousands
2
4
3
4
4
5
4
7
8
10
Now follow the previously described steps: 1. Σxy = 148 2. mean x = 4.2, mean y = 6, n = 5, so 6 * 5 * 4.2 = 126 3. 148 – 126 = 22 4. The standard deviation of x is 2.28, the standard deviation of y is 2.54, and n – 1 = 4, which yields 4 * 2.28 * 2.54 = 23.164 5. 22/23.164 = .949 as the Pearson’s correlation coefficient.
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Overview of Statistics
Now we have calculated the value of r, but what does it mean? Values of r will always be between –1.0 and positive 1.0. A –1.0 would mean a perfect negative correlation, whereas +1.0 would indicate a perfect positive correlation. Thus, a value of .949 indicates a very strong positive correlation between years of postsecondary education and annual income. In our limited sample we found a high positive correlation, but how significant is this finding? Answering the question of significance is where the T test comes in, and fortunately the arithmetic operations for it are much simpler. There are actu ally several variations of the T-test. The formula for the student’s T-test is given in equation 2.3. t=
x m s/ n
(eq. 2.3)
In this formula we have n as the sample size, x is the sample mean, μ is the popula tion mean, and s is the variance. This value is sometimes referred to as the coefficient of determination. In our scenario we have n = 5 and r = .949, so we have √(3/.099399), which is 5.493, which we multiply by r to get 5.213 for T. This T-score gives us some idea of how significant our correlation coefficient is.
P-VaLue Another way to determine if the relationship between our variables is statistically significant is the P-value. According to the Math World4 website, a p-value is “The probability that a variate would assume a value greater than or equal to the observed value strictly by chance.” AstroStat explains p-values this way:5 “Each statistical test has an associated null hypothesis, the p-value is the probability that your sample could have been drawn from the population(s) being tested (or that a more improbable sample could be drawn) given the assumption that the null hypothesis is true. A p-value of .05, for example, indicates that you would have only a 5% chance of drawing the sample being tested if the null hypothesis was actually true. Null hypotheses are typically statements of no difference or effect. A p-value close to zero signals that your null hypothesis is false, and typically that a difference is very likely to exist. Large p-values closer to 1 imply that there is no detectable difference for the sample size used. A p-value of 0.05 is a typical threshold used in industry to evaluate the null hypothesis. In more critical industries (healthcare, etc.) a more stringent, lower p-value may be applied.”
Z-test The z-test is used to make inferences about data. Its formula is shown in equation 2.4. z=
mx s
(eq. 2.4)
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• σ (the standard deviation of the population) • µ (the mean of the population) • x (the mean of the sample) The z-score is equal to the mean of the population minus the mean of the sample divided by the standard deviation of the population. The z-test is often used when you wish to compare one sample mean against one population mean. But what does it actually tell us? To begin with, you need to select a threshold of validity for your statistical study. The z-score converts your raw data (your sample mean) into a stand ardized score that can be used (along with your preselected threshold of rejection) to determine if you should reject the null hypothesis or not. Z-scores always have a mean of zero and a standard deviation of 1. Most statistics textbooks will have a z-score chart. You can find what percentage of sample items should appear above or below your z-score. You now compare that to your preselected rejection level. If you previously decided that a 5% level would cause you to reject the null hypothesis and your z-score indicates 11%, then you will reject the null hypothesis. So, the z-score is essential in helping you to decide whether or not to reject the null hypothesis.
outLiers A significant problem for any statistical study is the existence of outliers. An outlier is a value that lies far outside where most other values lie. For example, if a sample of high school athletes all have heights ranging from 70 inches to 75 inches, except for one who has a height of 81 inches, that one will skew all statistics that you generate. His height will make the mean much higher and the standard deviation much wider. Statistics generated by including that data point in the sample will not accurately reflect the sample. So, what can be done? One solution to outliers is to exclude them from the data set. It is common to exclude data points that are 2 or more standard deviations from the mean. Therefore, if the mean is 74 inches, and the standard deviation is 3 inches, then any height that is more than 80 inches or less than 68 inches is excluded from the data set. There are, of course, varying opinions on just how far from the mean constitutes a standard deviation, and whether or not they should be excluded at all. If you are conducting a study and elect to exclude outliers, it is a good idea to indicate that in your study, and what method you chose to exclude outliers. T-Test T-tests are often used when the research needs to determine if there is a significant dif ference between two population means. The general formula is shown in equation 2.5. t=
(SD) / N æ SD) 2 ö SD 2 ç ÷ è N ø ( N 1)( N )
(eq. 2.5)
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Overview of Statistics
S D is the sum of the differences S D2 is the sum of the squared differences S D)2 is the sum of the differences squared Equation 2.5 only shows a basic t-test. There are many variations such as a one-sample t-test, a Student’s t-test, Welch’s t-test, and others. T-tests are actually built into the Python programming language with the library scipy, which will be discussed in detail later in this book. Usually when selecting a t-test, and deciding which t-test to use, the place to start is with questions: Single sample t: we have only 1 group; want to test against a hypothetical mean. Independent samples t: we have 2 means, 2 groups; no relation between groups, e.g., people randomly assigned to a single group. Dependent t: we have two means. Either same people in both groups, or people are related, e.g., left hand-right hand, hospital patient and doctor. For the t distribution, degrees of freedom will always be a simple function of the sample size, e.g., (n – 1). One way of thinking about degrees of freedom is that if we know the total or mean, and all but one score, the last (n – 1) score is not free to vary. It is fixed by the other scores. 4 + 3 + 2 + X = 10. X = 1.
LINEAR REGRESSION As was briefly mentioned in chapter 1, linear regression explores the relationship between a variable and one or more explanatory variables. If there is one explana tory variable, then the process is simple linear regression. When there are multiple explanatory variables, then the process is multiple linear regression. While there are a variety of techniques for linear regression, they are all based on using linear pre dictor functions whose unknown parameters are estimated from the data. This allows the function to model the relationships between variables. For predictive analytics, linear regression is often used to fit a model of the observed data from the response and explanatory variables. This will allow the eval uation of the model to determine the efficacy of its predictive power. Let us begin by examining linear regression in a general manner. Given some dataset [ yi, x1, x2, . . . xn] linear regression assumes that the relationship between the dependent variable y and the vector of regressors x is linear. The relationship is modeled using a random variable that adds “noise” to the relationship between the regressors and the dependent variable. This random variable is often called an error variable or disturbance term and is symbolized by e . The model takes the form shown in equation 2.6: yi = b 0 + b xi
x
p ip
ei
e i , i = 1, ¼, n,
(eq. 2.6)
In equation 2.5, the x values are the dataset, the y is the dependent variable, the T superscript is a matrix transpose (recall transpose operation from chapter 1), and xiT b is the inner product between vectors x and b .
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There are quite a few different ways of using linear regression and different formulas. The forgoing discussion was simply a general review of the concept of linear regression.
ADDITIONAL STATISTICS The preceding portion of this chapter will give you a very general, basic understand ing of statistics. There are many more statistical tools that you may find of use in machine learning. A few of these will be briefly introduced here. The purpose of this section is to introduce you to topics you may wish to explore after this chapter.
anoVa ANOVA, or analysis of variation, is actually a family of statistical tests. We will focus on just one. The one-way analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of two or more independent groups. The one-way ANOVA compares the means between the groups and determines if the means are statistically different from each other. This test uses a linear model; there are several different linear models used: one that is commonly used to test for effects is shown in equation 2.7. yi , j = m + t j + e i , j
(eq. 2.7)
In equation 1, the i is the index of experimental units, j is the index over treatment groups, yi,j are observations, μ is the mean of the observations (μj is the mean of the jth treatment group, μ by itself is the mean of all observations), Tj is the jth treatment effect (a deviation from the mean). The Multivariate Analysis of Variance (MANOVA) is used when there are two or more dependent variables. It is a variation of ANOVA. MANOVA uses covariance between variables. Where sums of squares appear in univariate analysis of variance, in multivariate analysis of variance, certain positive-definite matrices appear. It is an extension of ANOVA in which main effects and interactions are assessed on a com bination of dependent variables. MANOVA tests whether mean differences among groups on a combination of dependent variables is likely to occur by chance.
the KrusKaL-WaLLis The Kruskal-Wallis test is a nonparametric test that compares two or more independ ent samples. The Kruskal-Wallis test is essentially a nonparametric version of the ANOVA test. Unlike the ANOVA, the Kruskal-Wallis test does not assume normal distribution of the data set. This test is often used to test correlation between two or more data sets that may not even have the same number of elements. The formula for the Kruskal-Wallis test is given in equation 2.8
å n (r r ) å å (r r ) g
H = (N 1)
2
i i =1 i g ni
i =1
j =1
ij
2
(eq. 2.8)
35
Overview of Statistics
In equation 2.6, the N is the total number of observations across all groups, g is the number of groups, ni is the number of observations for group i, rij is the ran of obser vation j from group i. Other elements are indeed equations themselves and are shown in equations 2.9 and 2.10. ri×
å = r =
ni
r
j =1 ij
(eq. 2.9)
ni 1 2
(eq. 2.10)
Equation 2.9 provides the average rank of all observations in group i while equa tion 2.10 is the average of all rij.
KOLMOGOROV-SMIRNOV This is a nonparametric test of the equality of continuous data, sometimes called a K-S test or just KS test. It can be a one-sample to two-sample test. This test quantifies the distance between the distribution function of the sample and a reference distribu tion function. The empirical distribution function F for n ordered observations X is defined as shown in equation 11. Fn ( x) =
1 n å I[¥, x ] ( X i ) n i =1
(eq. 2.11)
This is a bit more of an advanced statistical test, and is often not provided in introductory statistics texts. It is given here for those readers interested in expanding beyond what is provided in this chapter.
STATISTICAL ERRORS Obviously statistical errors can and do occur. Generally, such errors can be classified as either type I or type II errors.6 A type I error occurs when the null hypothesis is rejected when in fact it is true. A type II error occurs when the null hypothesis is accepted when in fact it is false. The following table illustrates this:
TABLE 2.1 Statistical Errors Truth
Decision Reject H0
Accept H0
H0 true
Type I Error
Correct Decision
H1 true
Correct Decision
Type II Error
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Both type I and type II errors should be avoided. Being aware of the possible errors is critical.
PoWer of a test The power of a statistical hypothesis test measures the test’s ability to reject the null hypothesis when it is actually false—that is, to make a correct decision. In other words, the power of a hypothesis test is the probability of not committing a type II error. It is calculated by subtracting the probability of a type II error from 1, usually expressed as: Power = 1 P( type II error ) The maximum power a test can have is 1, the minimum is 0. The ideal situation is to have high power, close to 1.
BASIC PROBABILITY In the introduction to this chapter, the concept of probability was introduced. A tech nical definition might be: probability is a measure of uncertainty. The probability of event A is a mathematical measure of the likelihood of the event’s occurring. Also recall that the probability of an event must lie between zero and one. There are some additional probability rules that need to be discussed.
What is ProbabiLity? The first task is to define what probability is. In simplest terms it is a ratio between the number of outcomes of interest divided by the number of possible outcomes. For example, if I have a deck of cards consisting of 52 cards, made up of 4 suits of 13 cards each, the probability of pulling a card of a given suit is 13/52 or ¼ = .25. Probabilities are always between zero and one. Zero indicates absolutely no chance of an event occurring. If one removes all 13 clubs from a deck, the odds of then pulling a club are zero. A probability of 1.0 indicates the event is certain. If one removes all the cards except for hearts, then the probability of drawing a heart is 1.0. Basic Set Theory Probability often uses set theory; therefore, a basic understanding of the essentials of set theory is necessary before we can continue. Let us begin by stating that a set is simply a collection of elements. An empty set is one containing no elements, and the universal set is the set containing all elements in a given context denoted by S. The compliment of a set is the set containing all the members of the universe set that are not in set A. This is denoted by a capital A with a bar over it or by Ac. As with much of mathematics, terminology and notation is critical. So, let us begin our study in a similar fashion, building from simple concepts to more complex. The
37
Overview of Statistics
most simple I can think of is defining an element of a set. We say that x is a member of set A. This can be denoted as xÎA Sets are often listed in brackets. For example, the set of all odd integers < 10 would be shown as follows: A = {1, 3, 5, 7, 9} And a member of that set would be denoted as follows: 3Î A Negation can be symbolized by a line through a symbol. For example: 2ÏA 2 is not an element of set A. If a set is not ending, you can denote that with ellipses. For example, the set of all odd numbers (not just those less than 10) can be denoted A = {1, 3, 5, 7, 9,11, ¼} You can also denote a set using an equation or formula that defines membership in that set. Sets can be related to each other; the most common relationships are briefly described here: Union: If you have two sets A and B, elements that are a member of A, B or both represent the union of A and B, symbolized as: A ∪ B. Intersection: If you have two sets A and B, elements that are in both A and B are the intersection of sets A and B, symbolized as A ∩ B. If the intersection of sets A and B is empty (i.e., the two sets have no elements in common) then the two sets are said to be disjoint. Difference: If you have two sets A and B, elements that are in one set, but not both, are the difference between A and B. This is denoted as A \ B. Compliment: Set B is the compliment of set A if B has no elements that are also in A. This is symbolized as B = Ac. Double Compliment: the compliment of a set’s compliment is that set. In other words, the compliment of Ac is A. That may seem odd at first read, but reflect on the definition of the compliment of a set for just a moment. The compliment of a set has no elements in that set. So, it stands to reason that to be the compliment of the com pliment of a set, you would have to have all elements within the set.
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These are basic set relationships. Now a few facts about sets: Order is irrelevant: {1, 2, 3} is the same as {3, 2, 1}, or {3, 1, 2}, or {2, 1, 3}. Subsets: Set A could be a subset of Set B. For example, if set A is the set of all odd numbers < 10 and set B is the set of all odd numbers < 100, then set A is a subset of set B. This is symbolized as: A ⊆ B. Power set: As you have seen, sets may have subsets. Let us consider set A as all integers less than 10. Set B is a subset; it is all prime numbers less than 10. Set C is a subset of A; it is all odd numbers < 10. Set D is a subset of A; it is all even numbers less than 10. We could continue this exercise making arbitrary subsets such as E = {4, 7}, F = {1, 2, 3}, etc. The set of all subsets for a given set is called the power set for that set. Sets also have properties that govern the interaction between sets. The most important of these properties are listed here: Commutative Law: The intersection of set A with set B is equal to the intersec tion of set B with Set A. The same is true for unions. Put another way, when considering intersections and unions of sets, the order the sets are presented in is irrelevant. That is symbolized as seen here: (a) A Ç B = B Ç A (b) A È B = B È A Associative Law: Basically, if you have three sets and the relationships between all three are all unions or all intersections, then the order does not matter. This is symbolized as shown here: (a) ( A Ç B) Ç C = A Ç ( B Ç C ) (b) ( A È B ) È C = A È ( B È C ) Distributive Law: The distributive law is a bit different than the associative, and order does not matter. The union of set A with the intersection of B and C is the same as taking the union of A and B intersected with the union of A and C. This is symbolized as you see here: (a ) A È ( B Ç C ) = ( A È B) Ç ( A È C) (b) A Ç ( B È C ) = ( A Ç B) È ( A Ç C) De Morgan’s Laws: These govern issues with unions and intersections and the compliments thereof. These are more complex than the previously dis cussed properties. Essentially, the compliment of the intersection of set A and set B is the union of the compliment of A and the compliment of B. The symbolism of De Morgan’s Laws are shown here: (a ) ( A Ç B)c = Ac È B c (b) ( A È B)c = Ac Ç B c These are the basic elements of set theory. You should be familiar with them before proceeding.
39
Overview of Statistics
basic ProbabiLity ruLes The following is a brief list of basic probability rules. • The probability of any event will be between zero and one, 0 = 90: print(“It’s really hot out there, be careful!”) if Fahrenheit
Greater than
!=
Not equal to
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Machine Learning for Neuroscience
You will use these later when working with if statements and any conditional statement. Another common task in Python is retrieving a date. Getting the current date is quite simple and is shown here: import datetime currdate = datetime.datetime.now() print(currdate)
object-oriented PrograMMing Python supports object-oriented programming. Therefore, you should be at least basi cally familiar with object-oriented programming concepts. An object is a programming abstraction that groups data with the code that operates on it. All programs contain data of different types. An object simply wraps all that up in a single place, called an object. In object-oriented programming, there are four concepts that are integral to the entire process of object-oriented programming: Abstraction is basically the ability to think about concepts in an abstract way. You can create a class for an employee, without having to think about a specific employee. It is abstract and can apply to any employee. Encapsulation is really the heart of object-oriented programming. This is simply the act of taking the data, and the functions that work on that data, and putting them together in a single class. Think back to our coverage of strings, and the string class. The string class has the data you need to work on (i.e., the particular string in question) as well as the various functions you might use on that data, all wrapped into one class. Inheritance is a process whereby one class inherits, or gets, the public properties and methods of another class. The classic example is to create a class called ani mal. This class has properties such as weight, and methods such as move and eat. All animals would share these same properties and methods. When you wish to create a class for, say, a monkey, you then have class monkey inherit from class animal, and it will have the same methods and properties that animal has. This is one way in which object-oriented programming supports code reuse. Polymorphism literally means “many forms.” When you inherit the proper ties and methods of a class, you need not leave them as you find them. You can alter them in your own class. This will allow you to change the form those methods and properties take. The term class is one you see a lot in object-oriented programming. It is a template for instantiating objects. Think of it as the blueprint for the objects you will need. It defines the properties and methods of that object.
IDE One normally uses an Integrated Development Environment (IDE) to write code. Most IDEs will also provide some additional features that can be helpful. For
Introduction to Python Programming
51
example, IDLE is the IDE that is installed when you install Python. In addition to letting you write scripts, it also has some useful features. We will explore IDLE as well as other IDEs in the following subsections.
idLe You may have noticed in the previous code sample (see figure 3.1) that the lines were color-coded. This makes it very easy to note comments, function names, etc., in the code. In addition, the menu bar at the top of IDLE has several tools for you. Let us start with looking at the Options menu, which you can see in figure 3.3. Showing line numbers is often quite useful. The Configure IDLE will also be use ful. The settings allows you to configure IDLE. This is shown in figure 3.4. These settings allow you to configure the code colors, extensions, fonts, etc., to whatever is most convenient for you. The Run menu option, obviously, allows you to run your script. In this book, the examples will usually be executed from a command line/shell. Also under the Run menu is an option to check the code. This is not meant as a substitute for good programming practices, but can find issues with your code. You can see this menu option in figure 3.5. Obviously, there are other menu options in IDLE. These will be explored as they come up in the natural progression through machine learning scripts in this book.
other ides There are, of course, other IDEs one can use. Each has its own advantages and dis advantages, as well as proponents and detractors. Python Anywhere is a web-based IDE. There are free versions and paid versions. You can find the website at www. pythonanywhere.com/. Figure 3.6 shows the landing page for Python Anywhere. Given the cost for at least some of the options, it is not quite as popular with Python developers. Micro Python is another web-based Python IDE, but this one is free. The website is: https://micropython.org/unicorn/. You can see the landing page for Micro Python in figure 3.7. Sublime is a free editor that one downloads to one’s computer. The website is found at www.sublimetext.com/. The interface is shown in figure 3.8.
FIGURE 3.3
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FIGURE 3.4 IDLE settings.
FIGURE 3.5
Introduction to Python Programming
FIGURE 3.6
Python Anywhere.
FIGURE 3.7
Micro Python.
53
The IDE you use is not a critical issue. You should find one that is comfortable for you and use that one. However, in this book, the IDLE IDE will be used in the code examples you see.
PYTHON TROUBLESHOOTING There are a number of issues that will arise when using Python. Some of the more common issues are addressed in this section. The objective is to make your use of Python go smoothly. However, no text can account for every issue you may encoun ter. Fortunately, the Python community is quite helpful. Should you encounter an issue that is not covered in this section, simply searching the internet for the error message will often yield very helpful results.
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FIGURE 3.8 Sublime.
The first issue is for Windows users. When using Windows, you need to ensure that Python is in your environment variables. If not, then you can only execute scripts from the directory where the Python.exe is installed. That can be cumbersome. You can see how to set Windows environment variables in figures 3.9 and 3.10. You can avoid this problem if, during the installation, you make sure you also check to Add Python to environment variables, as shown in figure 3.11. One of Python’s strengths is the huge number of packages available. There are packages for almost any task one can imagine. And we will use several of these fre quently when writing machine learning code. However, if you don’t have a package your script needs, when you run the script you will get an error message. The obvious thing to do is to attempt to install the package. An example of installing a package is shown in figure 3.12.
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FIGURE 3.9 Setting Windows environment variables step 1.
However, you may find this does not work. You could get an error like this one: EnvironmentError: [WinError 5] Access is denied: It is easy to fix the command prompt as Administrator. If in Linux, run as root/superuser. Then attempt the installation. We will be describing several commonly used packages in chapter 4.
generaL tiPs to reMeMber As you proceed through the chapters, certain helpful hints will be included. However, there are some general pointers that you need to keep in mind as you go forward. These are issues that, if you are aware of them from the outset, can make your Python programming experience more productive and less frustrating.
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FIGURE 3.10 Setting Windows environment variables step 2.
FIGURE 3.11
Windows Python installation.
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FIGURE 3.12 Installing Python packages.
• Indentions mean something in Python. Python does not use brackets or braces, it uses indention. • Running scripts will be easier if the script’s name is all one word (no spaces) and no spaces in the path to the script. That is simply easier for you to type in. • Be aware of single and double quotes. • IDLE will start with some header information that may or may not match your version of Python. That may cause problems compiling. Start a new file and there is no header information, and problem solved.
BASIC PROGRAMMING TASKS After reading the previous sections of this chapter, you should be able to install Python and write basic programs, such as the ubiquitous Hello World. In this final section, we will address common programming tasks. The focus here is not so much on entire scripts, but rather techniques you will frequently use in Python programming.
controL stateMents There are times when you will want to execute a portion of your code only if some value is present. Or you may wish to execute a section of code a certain number of times. These processes require control statements. The most basic of these is the if statement. If some value is met, then do something. Here is an example: if variableA > 10 then print(“variableA is greater than 10”) You can also expand this to handle different circumstances such as: if variableA < 10 then print(“variableA is less than 10”) elif variableA < 100 then print(“variableA is between 10 and 100”) else print(“variableA is greater than 100”)
If statements are quite common and you will find them in almost all programming and/or scripting languages. As you go further in this book, you will see such state
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TABLE 3.4 Python Logical Statements Operation
Symbol
Example
Equals
==
1 + 2 == 3
Not equal
!=
1 + 2!= 7
Less than
1
Less than or equal to
= 1
ments in action quite frequently. This naturally leads to the variety of logical state ments that can be used in these control structures. The following table provides a summary of these. Another type of control statement is the loop. There are actually several different types of loops. The most common is the for loop. This can be used to loop through an array. The following example should help elucidate this. colors = [“red”, “blue”, “green”] for x in colors: print(x)
This code will loop through the array of colors and print each color to the screen. You can combine this with an if statement to make slightly more clever code, as seen in this example: colors = ["red", "blue", "green"] for x in colors: if x == "blue": break print(x)
This bit of code combines if statements, for loops, and even uses the break statement to break out of the loop. Another loop you will encounter frequently is the while loop. This loop essentially states that while some conditions are true, execute some piece of code. The following rather elementary example should elucidate this concept: i = 1 while i < 6: print(i) i += 1
You will see these basic control statements throughout this book in many program ming examples.
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WORKING WITH STRINGS As you can probably surmise, strings are commonly used in any programming lan guage. Strings can be used to represent a person’s name, a file path, a domain name, and many other items that are commonly needed in any programming languages. Python provides you a number of techniques to work with strings. Assume you have a string variable named mystring. Table 3.5 has a few examples of what you can do with that variable in Python. Here is code implementing some of the preceding methods: #first let us create a string
mystring = “machine learning for neuroscience”
#now we can print out various string function results:
print(mystring.capitalize())
print(mystring.isdigit())
print(mystring.islower())
When executed, you will see output like in figure 3.13. The items in the preceding table are just the more common string methods. There are many more, though less commonly used. You can find a comprehensive list at www.w3schools.com/python/python_ref_string.asp.
TABLE 3.5 Python String Functions mystring.capitalize()
Converts the first character to upper case.
mystring.upper()
Converts a string into upper case.
mystring.lower()
Converts string into lower case.
mystring.isdigit()
Returns true if all characters in the string are digits.
mystring.islower()
Checks if the string is all lower case.
mystring.strip()
Returns a version of the string with leading and trailing spaces removed.
mystring.isnumeric()
Returns true if the string contains all numeric values.
mystring.find(‘n’)
Finds the first instance of the letter “N.”
mystring.count(‘s’)
Counts the number of “s” that appear in the string.
FIGURE 3.13 String output.
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WorKing With fiLes The first step in working with a file is to open it. In Python that is relatively simple, as you can see here: fMyFile = open(‘mytext’,’r’)
You may notice the ‘r’ and wonder what that is. This means to open the file in readonly mode. The other options are ‘w’ (write mode), and ‘a’ (append mode). For example, if you wish to open a file and read the contents onto the screen, you can combine a while loop with the open statement and the print statement: with open(‘mytext.txt’) as fMyFile: contents = fMyFile.read() print(contents)
A SIMPLE PROGRAM This is a trivial program, but it brings together many things you have learned. It prints to the screen, has if statements, defines a main function, etc. These are all items you will need to do frequently in your code. # converttemp.py
# A program to convert Celsius temperature to Fahrenheit. def main(): Celsius = eval(input("What is the Celsius temperature? ")) Fahrenheit = 9 / 5 * Celsius + 32 print("The temperature is", Fahrenheit, "degrees Fahrenheit.") if Fahrenheit >= 90: print("It’s hot, are you in Texas?”) if Fahrenheit | 26,163 | 122,761 | {"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-2023-50 | latest | en | 0.629727 |
http://mathoverflow.net/revisions/120495/list | 1,369,100,472,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699632815/warc/CC-MAIN-20130516102032-00060-ip-10-60-113-184.ec2.internal.warc.gz | 180,700,699 | 4,862 | 2 added 125 characters in body
I randomly scatter $N$ points on a bounded rectangular plane $P$ with dimensions $A \times B$. To be more specific, for $N$ iterations, I choose a real number $x \in [0, A]$ and a real number $y \in [0, B]$, and place my point at the coordinates $(x, y)$.
Let $R$ be the radius of some circle I place somewhere on the plane $P$ that encloses no points. What is the probability distribution for the maximum permissible size of $R$?
Note - I specified $P$ to be a rectangular plane but, please, let it be any shape you wish if you can answer my question.
1
# The largest circle that encloses no points on a plane with points placed at $N$ random coordinates
I randomly scatter $N$ points on a bounded rectangular plane $P$ with dimensions $A \times B$. To be more specific, for $N$ iterations, I choose a real number $x \in [0, A]$ and a real number $y \in [0, B]$, and place my point at the coordinates $(x, y)$.
Let $R$ be the radius of some circle I place somewhere on the plane $P$ that encloses no points. What is the probability distribution for the maximum permissible size of $R$? | 293 | 1,124 | {"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} | 3.015625 | 3 | CC-MAIN-2013-20 | latest | en | 0.762829 |
cardinal-number.wikiverse.org | 1,571,398,345,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00173.warc.gz | 39,145,602 | 8,627 | # Cardinal Number
• ## In the news
• Stanford Cardinal preview
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... When USC won last year, it was 4-0 against the Cardinal, including another double overtime victory in the ... Saturday, they will attempt to go for number six.
• Why I felt let down by Blair
Telegraph.co.uk, UK -
... vocation, combined with the fact that Cardinal Murphy-O'Connor was openly critical of the war in Iraq, suggests that Sir Stephen may have left Number 10 as a ...
• Arizona Football Notes vs. Washington State
GoAZcats.com (subscription), AZ -
... A key reason for the low number last week was UA’s third-down conversion rate ... Well the early forecast predicts another tough slate for the Cardinal and Navy. ...
• It is time Syria let Lebanon be sovereign, Cardinal Sfeir says
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... Even then Cardinal Sfeir stressed that whilst many Lebanese want to maintain ... from Lebanon.” It suggests that “Damascus will score a number of points by ...
Alternative meaning: number of pitch classes in a set.
In mathematics, cardinal numbers, or cardinals for short, are numbers used to denote the size of a set. Since mathematics is concerned with infinite objects, a study of cardinality tries to discuss the size of infinite sets. Perhaps anti-intuitively, one of the most basic results is that not all infinite objects are of the same size, and there is a formal characterization of how some infinite objects are strictly smaller than other infinite objects. Concepts of cardinality are embedded in most branches of mathematics and are essential to their study. It is also a area studied for its own sake as part of set theory, particularly in trying to describe the properties of large cardinals.
## History
The cardinal numbers were invented by George Cantor, when he was developing the Naïve set theory in 18741884.
He first established cardinality as an instrument to compare finite sets; i.e the sets {1,2,3} and {2,3,4} are not equal, but have the same cardinality.
Cantor invented the one-to-one mapping, which easily showed that two sets had the same cardinality if there was a one-to-one correspondence between the members of the set. Using this one-to-one correspondence, he transferred the concept to infinite sets; i.e the set of natural numbers N = {1, 2, 3, ...}. He called these cardinal numbers transfinite cardinal numbers, and defined all sets that had a one-to-one correspondence with N to be denumerably infinite sets.
Naming this cardinal number , aleph-null, Cantor proved that many subsets of N have the same cardinality as N, even if this might be against intuition at first. He also proved that the set of all ordered pairs of natural numbers is denumerably infinite, and later that the set of all algebraic numbers (every member of the set is a set of numbers of its own , like an extended ordered pair) is denumerably infinite.
At this point, in 1874, there was a curiosity whether all infinite sets are denumerably infinite, and what the use would be in that case.
But, later this year, Cantor succeeded in proving that there were higher-order cardinal numbers using the ingenious but simple Cantor's diagonal argument. This new cardinal number, called the "power of continuum", was termed c by Cantor.
Cantor also developed a lot of the general theory of cardinal numbers; he proved that there is a cardinal number that is the smallest (, aleph-null) and that every cardinal number has a next-larger cardinal ().
The later Continuum hypothesis suggests that c is the same as , but is has been found impossible to prove or disprove.
## Motivation
In informal use, a cardinal number is what is normally referred to as a counting number. They may be identified with the natural numbers beginning with 0 (i.e. 0, 1, 2, ...). The counting numbers are exactly what can be defined formally as the finite cardinal numbers. Infinite cardinals only occur in higher-level mathematics and logic.
More formally, a non-zero number can be used for two purposes: to describe the size of a set, or to describe the position of an element in a sequence. For finite sets and sequences it is easy to see that these two notions co-incide, since for every number describing a position in a sequence we can construct a set which has exactly the right size, e.g. 3 describes the position of 'c' in the sequence <'a','b','c','d',...>, and we can construct the set {a,b,c} which has 3 elements. However when dealing with infinite sets it is essential to distinguish between the two --- the two notions are in fact different for infinite sets. Considering the position aspect leads to ordinal numbers, while the size aspect is generalized by the cardinal numbers described here.
The intuition behind the formal definition of cardinal is the construction of a notion of the relative size or "bigness" of a set without reference to the kind of members which it has. For finite sets this is easy; one simply counts the number of elements a set has. In order to compare the sizes of larger sets, it is necessary to appeal to more subtle notions.
A set Y is at least as big as, or greater than or equal to a set X if there is a one-to-one mapping from the elements of X to the elements of Y. A one-to-one mapping identifies each element of the set X with a unique element of the set Y. This is most easily understood by an example; suppose we have the sets X = {1,2,3} and Y = {a,b,c,d}, then using this notion of size we would observe that there is a mapping:
1 → a
2 → b
3 → c
which is one-to-one, and hence conclude that Y has cardinality greater than or equal to X. Note the element d has no element mapping to it, but this is permitted as we only require a one-to-one mapping, and not necessarily a one-to-one and onto mapping. The advantage of this notion is that it can be extended to infinite sets.
We can then extend this to an equality-style relation. Two sets X and Y are said to have the same cardinality if there exists a bijection between X and Y, or equivalently both a one-to-one mapping from X to Y and a one-to-one mapping from Y to X. We then write | X | = | Y |. The cardinal number of X itself is often defined as the least ordinal number a with | a | = | X |. This is called the von Neumann cardinal assignment; for this definition to make sense, it must be proved that every set has the same cardinality as some ordinal; this statement is the well-ordering principle. It is however possible to discuss the relative cardinality of sets without explicitly assigning names to objects.
The classic example used is that of the infinite hotel paradox, also called Hilbert's paradox of the Grand Hotel. Suppose you are an innkeeper at a hotel with an infinite number of rooms. The hotel is full, and then a new guest arrives. It's possible to fit the extra guest in by asking the guest who was in room 1 to move to room 2, the guest in room 2 to move to room 3, and so on, leaving room 1 vacant. We can explicity write a segment of this mapping:
1 ↔ 2
2 ↔ 3
3 ↔ 4
...
n ↔ n+1
...
In this way we can see that the set {1,2,3,...} has the same cardinality as the set {2,3,4,...} since a one-to-one mapping from the first to the second has been shown. This motivates the definition of an infinite set being any set which has a proper subset of the same cardinality; in this case {2,3,4,...} is a proper subset of {1,2,3,...}.
When considering these large objects, we might also want to see if the notion of counting order coincides with that of cardinal defined above for these infinite sets. It happens that it doesn't; by considering the above example we can see that if some object "one greater than infinity" exists, then it must have the same cardinality as the infinite set we started out with. It is possible to use a different formal notion for number, called ordinals, based on the ideas of counting and considering each number in turn, and we discover that the notions of cardinality and ordinality are divergent once we move out of the finite numbers.
It is provable that the cardinality of the real numbers is greater than that of the natural numbers just described. This can be visualized using Cantor's diagonal argument; classic questions of cardinality (for instance the continuum hypothesis) are concerned with discovering whether there is some cardinal between some pair of other infinite cardinals. In more recent times mathematicians have been describing the properties of larger and larger cardinals.
Since cardinality is such a common concept in mathematics, a variety of names are in use. Sameness of cardinality is sometimes referred to as equipotence, equipollence, or equinumerosity. It is thus said that two sets with the same cardinality are, respectively, equipotent, equipollent, or equinumerous.
## Formal definition
Formally, the order among cardinal numbers is defined as follows: | X | ≤ | Y | means that there exists an injective function from X to Y. The Cantor-Bernstein-Schroeder theorem states that if | X | ≤ | Y | and | Y | ≤ | X | then | X | = | Y |. The axiom of choice is equivalent to the statement that given two sets X and Y, either | X | ≤ | Y | or | Y | ≤ | X |.
A set X is infinite, or equivalently, its cardinal is infinite, if there exists a proper subset Y of X with | X | = | Y |. A cardinal which is not infinite is called finite; it can then be proved that the finite cardinals are just the natural numbers, i.e., that a set X is finite if and only if | X | = | n | = n for some natural number n. It can also be proved that the cardinal (aleph-0, where aleph is the first letter in the Hebrew alphabet, represented by the Unicode character א) of the set of natural numbers is the smallest infinite cardinal, i.e., that any infinite set admits a subset of cardinality . The next larger cardinal is denoted by and so on. For every ordinal a there is a cardinal number , and this list exhausts all cardinal numbers.
Note that without the axiom of choice there are sets which can not be well-ordered, and the definition of cardinal number given above does not work. It is still possible to define cardinal numbers (a mapping from sets to sets such that sets with the same cardinality have the same image), but it is slightly more complicated. One can also easily study cardinality without referring to cardinal numbers.
If X and Y are disjoint, the cardinal of the union of X and Y is called | X | + | Y |. We also define the product of cardinals by | X | × | Y | = | X × Y | (the product on the right hand side is the cartesian product). Also | X |Y | = | XY | where XY is defined as the set of all functionss from Y to X. It can be shown that for finite cardinals these operations coincide with the usual operations for natural numbers. Furthermore, these operations share many properties with ordinary arithmetic:
• addition and multiplication of cardinal numbers is associative and commutative
• |X||Y| + |Z| = |X||Y| × |X||Z|
• |X||Y| × |Z| = (|X||Y|)|Z|
• (|X| × |Y|)|Z| = |X||Z| × |Y||Z|
The addition and multiplication of infinite cardinal numbers (assuming the axiom of choice) is easy: if X or Y is infinite and both are non-empty, then
X | + | Y | = | X | × | Y | = maxX |, | Y
.
On the other hand, 2X | is the cardinality of the power set of the set X and Cantor's diagonal argument shows that 2X | > | X | for any set X. This proves that there exists no largest cardinal. In fact, the class of cardinals is a proper class.
The continuum hypothesis (CH) states that there are no cardinals strictly between and . The latter cardinal number is also often denoted by c; it is the cardinality of the set of real numbers, or the continuum, whence the name. In this case = . The generalized continuum hypothesis (GCH) states that for every infinite set X, there are no cardinals strictly between | X | and 2X |. The continuum hypothesis is independent from the usual axioms of set theory, the Zermelo-Fraenkel axioms together with the axiom of choice (ZFC). | 2,859 | 12,181 | {"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-2019-43 | latest | en | 0.955066 |
https://math.stackexchange.com/questions/4331843/show-that-the-variety-c-is-rational | 1,702,115,777,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00730.warc.gz | 428,856,160 | 36,038 | # Show that the variety C is rational
I have to show that the variety $$C:=\{(x_0 :x_1:x_2)\in \mathbb{P}_k^2\ |\ x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2 = 0\}\subset \mathbb{P}_k^2$$ is rational.
I already found a solution on this but I don't understand it completely. The solution says that we can use the Cremona-transformation $$\phi: \mathbb{P}_k^2 \to \mathbb{P}_k^2,\ (x_0:x_1:x_2)\mapsto (\frac{1}{x_0}:\frac{1}{x_1}:\frac{1}{x_2})=(x_1x_2:x_0x_2:x_1x_2),$$ which is birational with $$\phi=\phi^{-1}.$$ Therefore, $$\phi^*: k(\mathbb{P_k^2)}\to k(\mathbb{P_k^2})$$ is an isomorphism between the function-fields.
It is $$\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$$, now the solution says it holds that $$C\cap\{x_0x_1x_2 \neq0\}$$ is isomorph to $$Q\cap\{x_0x_1x_2 \neq0\}$$, where $$Q=\{x_0^2+x_1^2+x_2^2=0\}$$ is a smooth conic section, which is known to be isomorph to $$\mathbb{P_k^1}$$.
I understand that $$\phi^*$$ is an isomorphism and that $$\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$$. But why can I follow that the two varietys, induced by those polynomials, are isomorph?
• Thanks for your answer, I now see that $\phi$ is an isomorphism between $C_1$ and $Q_1$. Why does this imply that $C$ is birational to $Q$? Dec 13, 2021 at 22:57
Answer: You have constructed a map $$ϕ:C_1:=C−D(x_0x_1x_2)→Q_1:=Q−D(x_0x_1x_2)$$ with an inverse map $$ϕ^{−1}$$. Hence $$C_1≅Q_1$$ are isomorphic. Hence $$C$$ is birational to $$\mathbb{P}^1_k\cong Q_1$$ , and is therefore rational.
Note: By definition two (irreducible) varieties $$X,Y$$ are birational iff there are open subsets $$U\subseteq X, V\subseteq Y$$ and an isomorphism $$U \cong V$$. | 738 | 1,700 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-50 | latest | en | 0.823861 |
https://dsp.stackexchange.com/questions/43525/reliably-matching-a-degraded-image-with-the-original-out-of-dozens-of-candidates | 1,712,944,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00296.warc.gz | 190,438,552 | 42,508 | # Reliably matching a degraded image with the original out of dozens of candidates
I have several collections contaning low-resolution captures of unevenly-lit presentation slides:
and thirty to sixty original high-resolution slide images:
How would you design a similarity measure between the low-resolution captures and the high-resolution slides that would nearly always give the highest score to the correct image pair (or, analogously, a distance measure that approaches zero only for the correct image pair)?
# Experimental setup
So far I tried a machine-learning solution to get a better feeling for how difficult the problem is. I assigned each image pair a feature vector consisting of:
1. various grayscale histogram similarity / distance measures (correlation, intersection, $\chi^2$ distance, and Bhattacharyya distance) applied to both the entire images and the image quadrants,
2. Pearson's correlation coefficient between Haralick texture features,
3. Hamming distance between various image hashes (aHash, pHash, dHash), and
4. Levenshtein distance between the image OCRs normalized by the maximum OCR length.
For the training part of a dataset, I take each feature vector $\vec x$, let $\vec a\vec x + q = 1$ for matching images and $\vec a\vec x + q = -1$ for non-matching images. Then I use linear regression to obtain $\vec a$ and $q$. Finally, for the test part of a dataset, I use $\vec a\vec x+q$ as a scoring function to retrieve a sorted list of high-resolution slides for each low-resolution capture and look at the rank of the correct result.
# Experimental results
With a small hand-annotated dataset, this is getting me the average rank of 6, which is better than random draws, but not nearly good enough for an application that would overlay the high-resolution slides over live capture; such an application will require an average rank that approaches 1. There is an opportunity for the application to be smart and only look at a small window of slides around the current slide, but it still needs to correctly guess the initial slide.
As you might expect, the feature vectors are also quite slow to compute (several seconds per a feature vector with a naive implementation on a higher-end quad-core laptop), which makes the solution unsiutable for a real-time application.
• Have you tries feature point matching methods, like sift? Sep 3, 2017 at 16:39
• No, not yet. Do you think these would be well-suited for the task? Sep 3, 2017 at 21:46
• It is too much complex but definitely would work, but wait a minute, the images are just different in colour and size yes? I mean difference is just scale, i.e resolution (not rotation,...)? Sep 3, 2017 at 21:53
• There is no rotation, although there will be some perspective distortion. Sep 4, 2017 at 0:17
• Ok then if images are basically the same except for their size I have idea to solve the problem. Sep 4, 2017 at 9:06
Your question has many solutions which can be good or bad, so I am giving the solution I came up with, hope it is a good one. Since your images are only different in resolution and color I think the best way to compare them is to make their size the same and then to get rid of color and shading differences only compare their edges not the images themselves. This way, you can use comparison criteria that you've tried now independent of color and shading variations. I simply used sum of differences between two images in the code below and for the images you provided seems it works.
Through this code I try to give you an idea idea on how it is working:
I1=rgb2gray(imread('Low.png'));
[h,v]=size(I1);
I2=imresize(I2,[h,v]);
I3=imresize(I3,[h,v]);
d1=edge(I1,'log');
d2=edge(I2,'log');
d3=edge(I3,'log');
score1=sum(sum(abs(d1-d2)))
score2=sum(sum(abs(d1-d3)))
Some low pass filtering before calculating the scores might help.
• While using the edge information to abstract away the differences in color and lighting sounds like a step in the right direction, doing a simple image difference expects the images to be aligned, which they are not. It is curious that the above procedure works. Sep 5, 2017 at 12:39
• @Witiko , Actually that is why I asked if the difference is only in color and size. Sep 5, 2017 at 13:06
• And, as I replied, there is some perspective distortion and the alignment is not perfect. However, I imagine that performing edge detection, segmenting the image (using, say, the watershed algorithm), and then comparing the results of the segmentation might lead to interesting results! Sep 5, 2017 at 13:11 | 1,052 | 4,557 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.91449 |
https://www.jadferrocements.net/building-tips/how-to-draw-a-house-step-by-step-3d.html | 1,653,733,038,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00158.warc.gz | 977,442,520 | 11,668 | # How To Draw A House Step By Step 3d?
## How can I make a 3D drawing at home?
Here’s how to do it:
1. Create your floor plan. Either draw floor plans yourself with our easy-to-use home design software – just draw your walls and add doors, windows and stairs.
2. Furnish and Decorate. Add color and materials to floors and walls.
## How do you draw a simple step by step House?
How to Draw a House Step by Step
1. Step 1 – Draw the First Section of the House.
2. Step 2 – Draw the First Section of the Roof.
3. Step 3 – Draw the Second Section of the House.
4. Step 4 – Draw the Outlines of the Windows.
5. Step 5 – Draw the Door.
6. Step 6 – Draw the Chimney.
7. Step 7 – Draw the Smaller Details.
## How do you draw a simple 3D building?
DIRECTIONS
1. Draw a ground line.
3. Draw 45 degree angle lines at corners shown.
4. Connect ends with straight lines.
5. Start adding doors and windows.
6. Continue adding doors and windows.
7. Finish the doors and windows.
8. Add a tree, cloud and sun.
## How do you draw a house shape?
Kids, learn how to draw the House-with-shapes by following the steps below.
1. Step:1. Draw a triangle.
2. Step:2. Below the triangle, draw a square.
3. Step:3. Inside the triangle, draw 2 small squares.
4. Step:4. Between the windows, draw a lengthy rectangle.
5. Step:5. Put a small dot indide the rectangle.
6. Step:6.
7. Step:7.
8. Step:8.
You might be interested: Question: How To Make Light House?
## How do you make a 3D picture?
3D drawings use optical illusions to make it appear that an image has depth. Method 3 of 3: Using Perspective
1. Study your subject. If you want to draw something in 3D that you have in real life, it can be help to look at its details.
2. Start a drawing with a horizon line.
3. Incorporate a vanishing point in to a drawing. | 492 | 1,807 | {"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} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.867584 |
https://intro.nyuadim.com/2021/02/07/week-3-object-oriented-game-ballz/ | 1,713,438,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00857.warc.gz | 291,332,214 | 22,125 | # Week 3: Object Oriented Game Ballz
Overview:
This weeks task was to apply modularity concepts for object oriented programming through the of classes in either a game or work of art. I decided to try my hand at making knockoff of a (what I though was simple) IOS game Ballz. In this game the player shoots off a series of balls that bounce off tiles until they are destroyed. The tiles must not reach the bottom of the screen otherwise the player loses.
Programming:
To accomplish the object oriented process, I used two custom objects, the ball and the block, and then a central main runner with several assisting methods.
My first task was to simulate an IPhone screen. I did this by maximizing the height or width, and constraining the screen to a 13 by 6 aspect ratio. By only printing in this rectangle. I had to define new variables as the native height and width values would not be applicable to my project. I also needed a game board area within the simulated IPhone screen which had to be just as wide as the phone, but smaller in height as a perfect fit for 7 x 9 tiles. With this area defined, I made a method that would reprint my new phone shaped background.
I then had to determine stages of the game. There are two main stages. First the player aims the ball, and can see where the shot will go based on a preview of balls on the screen, and then the game progresses to where the balls are actually moving and bouncing off the tiles.
The first stage was much more simple. I used polar coordinates determined based on a starting location of the ball at the bottom of the screen, and the players clicked mouse to create a line of balls coming out of the source. I could then save the angle this created and use it for the next stage of the game.
The next stage was far more complicated. I had to manage several balls, blocks, angles, and values all at the same time. For each ball I had to tell if it was hitting a block, and if so where it would go next. I had to determine the interaction each ball had on a collision, as well as other game mechanics such as the ability to get more balls.
This took some time but using equations to calculate which edge of a box a ball had hit, I was able to simulate the ball bouncing off of a box.
Results:
Here is a link to a google drive with the executables as well as the code
Here is the code:
Main Class
```//Inspired by IOS game Ballz
//Created by Cole Beasley for Intro to Interactive Media
//7-8/2/21
import java.util.Iterator; // Import the class of Iterator
//Designed for IPhone X screen res of 13:6 ratio, will be full screnned contained by height
//To do
/*
Initalize board, set squares to be 7 sqaures across, 7 down, plus empty row on top and bottom (9 vertical rows total)
Initialize ball with size proportional 1/6th of square size
Have a difficulty value
Every round:
Include one "new ball" square
Random number of squares 1-7 show up with random values, higher values as rounds go on, color of box based on value of box
On shot line up
Draw line of darker balls to show where shot will go from mouse pull back
*/
//Global Variables
int phoneWidth;
int phoneHeight;
float gameBoardOriginX;
float gameBoardOriginY;
float gameBoardHeight;
float gameBoardWidth;
int timer = 0;
int savedTime = 0;
float gridSize; //Square dimensions without padding
float blockPadding = 0.1; //What percent of a block should be left to padding around it, 0 will be no padding and blocks touch, 1 will be 100% padding and no block visible
float blockSize; //Square dimensions minus padding
float ballSize;
//2D array containing current block elements
Block blocks[][] = new Block[9][7];
ArrayList<Ball> balls = new ArrayList<Ball>();
//Game control variables
boolean newRound = true; //Should the squares advance
boolean gameOver = false; //Did they lose
boolean activeRound = false; //Is a round running is is someone aiming
boolean firstBallDone = false; //Has a ball landed to set the starting pos for next round
float ballStartPosX; //Where the launch will come from
float initTheta; //The initial theta that will be used as a reference
float ballSpeed = 10; //how fast the ball is going (going faster increases hit errors)
int ballTimer = 100; //How frequently the balls are released, 1 = 1ms)
boolean firstBall = true; //Controller as to if the first ball has landed yet to set ballStartPosX
//Mouse control variables
boolean mouseDown = false;
int mouseOriginX;
int mouseOriginY;
//Game score
int gameScore = 1;
int highScore = 1;
int reserveBalls = 1;
//Ball Colors
color ballColor = color(200);
color ballPreviewColor = color(150);
void setup() {
fullScreen();
//Set "phone" screen to 13x6 aspect ratio (IPhone X)
if (height / 13 > width / 6) {
phoneWidth = width;
phoneHeight = int((width / 6) * 13);
} else {
phoneHeight = height;
phoneWidth = int((height / 13) * 6);
}
//Set basic variables defining game area width height and pos
gameBoardWidth = phoneWidth;
gameBoardHeight = (phoneWidth / 7) * 9.1666;
gameBoardOriginX = (width/2) - (gameBoardWidth/2);
gameBoardOriginY = (height/2) - (gameBoardHeight/2);
//Set block sizes
gridSize = gameBoardWidth / 7;
blockSize = gridSize - (gridSize * blockPadding);
ballSize = gridSize/6;
ballStartPosX = width/2;
}
void draw() {
//Draw the game board background
gameBackground();
//Check if it is a new round
if (newRound) {
newRound = false;
}
drawSquares();
//Detect if game is over
if (gameOver()) {
noLoop();
}
//Print out game score, highscore
textAlign(CENTER, BOTTOM);
textSize(30);
fill(255);
text("Score " + gameScore, width/2, gameBoardOriginY);
textAlign(LEFT, BOTTOM);
textSize(20);
text("Highscore: " + highScore, gameBoardOriginX, gameBoardOriginY);
//If round is not yet running
if (!activeRound) {
//Draw ball at start pos
fill(ballColor);
noStroke();
ellipse(ballStartPosX, gameBoardOriginY + gameBoardHeight - ballSize/2, ballSize, ballSize);
//If mouse is down and in the board and below original click, draw preview balls
if (mouseDown && mouseY > mouseOriginY && mouseInBounds()) {
float initX = ballStartPosX;
float initY = gameBoardOriginY + gameBoardHeight - ballSize/2;
float w = mouseX - mouseOriginX;
float h = mouseY - mouseOriginY;
float theta = atan(h/w);
if (theta > 0) {
theta += PI;
}
initTheta = theta;
float r = 50; //Distance between preview balls
float x = r * cos(theta);
float y = r * sin(theta);
fill(ballPreviewColor);
while ((initX + x - (ballSize/2) > gameBoardOriginX && initX + x + (ballSize/2) < gameBoardOriginX + gameBoardWidth) && (initY + y - (ballSize/2) > gameBoardOriginY && initY + y + (ballSize/2) < gameBoardOriginY + gameBoardHeight)) {
pushMatrix();
translate(initX, initY);
ellipse(x, y, ballSize, ballSize);
x = r * cos(theta);
y = r * sin(theta);
initX += x;
initY += y;
popMatrix();
}
}
}
//Round is running
else {
//Spawn in more balls if there are some still and the timer has passed half a second
timer = millis() - savedTime;
if (reserveBalls > 0 && timer >= ballTimer) {
savedTime = millis(); //Reset timer for future ball spawns
balls.add(new Ball(ballColor, ballStartPosX, gameBoardOriginY + gameBoardHeight - ballSize/2, initTheta, ballSize));
reserveBalls--;
}
//Draw balls and advance their pos
Iterator<Ball> ballIterator = balls.iterator();
while (ballIterator.hasNext()) {
Ball ball = ballIterator.next(); // must be called before you can call i.remove()
//Move the ball
Block hitBlock = ball.moveBall(ballSpeed, blocks, gameBoardOriginX, gameBoardOriginX + gameBoardWidth, gameBoardOriginY, gameBoardOriginY + gameBoardHeight, gridSize);
//See if a block was hit
if (hitBlock != null) {
//See if it is a ball increase block
if (hitBlock.freeBall) {
gameScore++;
blocks[hitBlock.gridY(ball.y, gridSize, gameBoardOriginY)][hitBlock.gridX(ball.x, gridSize, gameBoardOriginX)] = null;
} else {
hitBlock.decrease();
//See if block was destroyed
if (hitBlock.strength <= 0) {
blocks[hitBlock.gridY(ball.y, gridSize, gameBoardOriginY)][hitBlock.gridX(ball.x, gridSize, gameBoardOriginX)] = null;
}
}
}
//Check to see if any balls should be removed
if (ball.y > gameBoardOriginY + gameBoardHeight) {
//If it is the first ball to be removed, set the next rounds start position to its x pos
if (firstBall) {
firstBall = false;
ballStartPosX = ball.x;
}
ballIterator.remove();
} else {
ball.drawBall();
}
}
//Check to see if the round ended
if (balls.size() == 0 && reserveBalls == 0) {
newRound = true;
activeRound = false;
reserveBalls = gameScore;
if (gameScore > highScore) {
highScore = gameScore;
}
}
}
}
//Acts like background function, but draws for aspect ratio of phone
void gameBackground() {
//translate to center of screen
pushMatrix();
translate(width/2, height/2);
//draw main game background dark gray
rectMode(CENTER);
noStroke();
fill(30);
rect(0, 0, phoneWidth, phoneHeight);
//draw board area
fill(20);
//Game board width is defined as the max width, gameboard ratio is 9.166/7 height to width
rect(0, 0, gameBoardWidth, gameBoardHeight);
popMatrix();
}
//When called advances the round, moves all the blocks down a row and creates a new row in the block matrix
//First move each square down a row
for (int i = 8; i > 0; i--) {
for (int j = 0; j < 7; j++) {
blocks[i][j] = blocks[i-1][j];
}
}
//Clear first row
for (int i = 0; i < 7; i++) {
blocks[1][i] = null;
}
//Generate new row
for (int i = 0; i < 7; i++) {
//random chance for new block to be added
if (random(1) > 0.5) {
//Get random strength valuse roughly twice current score
int strength = int(random(0.8, 2) * gameScore) + 1;
blocks[1][i] = new Block(strength);
}
}
//Set the one block which gives you a ball
Block ballBlock = new Block();
ballBlock.freeBall = true;
blocks[1][int(random(0, 7))] = ballBlock;
firstBall = true;
}
//Draw the squares
void drawSquares() {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 7; j++) {
if (blocks[i][j] != null) {
blocks[i][j].drawBlock(gameBoardOriginX + (gridSize/2) + (j*gridSize), gameBoardOriginY + (gridSize/2) + (i*gridSize), blockSize);
}
}
}
}
//See if game is over
boolean gameOver() {
for (int i = 0; i < 7; i++) {
if (blocks[8][i] != null) {
//Print game over
textAlign(CENTER, CENTER);
textSize(50);
fill(255);
text("Game Over =(", width/2, height/2);
return true;
}
}
return false;
}
//Mouse processing
void mousePressed() {
//Make sure not in round and not yet counted as clicked
if (!activeRound && !mouseDown) {
mouseDown = true;
mouseOriginX = mouseX;
mouseOriginY = mouseY;
}
}
void mouseReleased() {
//Make sure not in round
if (!activeRound) {
mouseDown = false;
//Start round if mouse pos Y is below origin y
if (mouseY > mouseOriginY) {
activeRound = true;
savedTime = millis() - ballTimer;
}
mouseOriginX = 0;
mouseOriginY = 0;
}
}
//Check to see if mouse is on board
boolean mouseInBounds() {
if ((mouseX > gameBoardOriginX && mouseX < gameBoardOriginX + gameBoardWidth) && (mouseY > 0 && mouseY < height)) {
return true;
}
return false;
}
```
Ball Class
```class Ball {
boolean directionY = true; //true = down
boolean directionX = true; //true = right
float incrementX = 0;
float incrementY = 0;
float theta;
float ballSize;
float x = 0;
float y = 0;
color colorValue;
Ball(color colorVal, float x, float y, float theta, float ballSize) {
colorValue = colorVal;
this.x = x;
this.y = y;
this.theta = theta;
this.ballSize = ballSize;
}
void drawBall() {
ellipseMode(CENTER);
fill(colorValue);
noStroke();
ellipse(x, y, ballSize, ballSize);
}
void moveBall(float speed) {
incrementX = speed * cos(theta);
incrementY = speed * sin(theta);
x += incrementX;
y += incrementY;
}
Block moveBall(float speed, Block blocks[][], float minX, float maxX, float minY, float maxY, float gridSize) {
incrementX = speed * cos(theta);
incrementY = speed * sin(theta);
//If ball has bounced into a vertical wall
if (checkTopBound(incrementY, minY)) {
y = minY + ballSize/2;
incrementY = 0;
}
//Check horizontal walls
if (checkHorizontalLeftBound(incrementX, minX)) {
x = minX + ballSize/2;
incrementX = 0;
if (theta >= 0)
theta = PI - theta;
else
theta = -PI - theta;
}
if (checkHorizontalRightBound(incrementX, maxX)) {
x = maxX - ballSize/2;
incrementX = 0;
if (theta >= 0)
theta = PI - theta;
else
theta = -PI - theta;
}
//Check for a box collision
Block hitBlock = checkBoxCollision(blocks, minX, minY, gridSize);
x += incrementX;
y += incrementY;
if(hitBlock != null){
return hitBlock;
}
return null;
}
//Check vertical bounds of game area
boolean checkTopBound(float incrementY, float minY) {
//Check if gone off top
if ((y + incrementY - ballSize/2) < minY) {
return true;
}
return false;
}
//Check walls of game area
boolean checkHorizontalRightBound(float incrementX, float maxX) {
//Check for too far right
if ((x + incrementX + ballSize/2) > maxX) {
return true;
}
return false;
}
boolean checkHorizontalLeftBound(float incrementX, float minX) {
//Check if gone off left
if ((x + incrementX - ballSize/2) < minX) {
return true;
}
return false;
}
//Check to see if a box has been hit, return box that was hit if hit
Block checkBoxCollision(Block blocks[][], float minX, float minY, float gridSize) {
//First get which grid the ball is in and which grid it will be in
int initGridX = int((x - minX)/gridSize);
int initGridY = int((y - minY)/gridSize);
int newGridX = int(((x + incrementX) - minX)/gridSize);
int newGridY = int(((y + incrementY) - minY)/gridSize);
//Check to see if ball has gone off the bottom
if (initGridY == 9) {
initGridY = 8;
}
if (newGridY == 9) {
newGridY = 8;
}
//Check to see if moving to a different grid
if (initGridX != newGridX || initGridY != newGridY) {
//Check to see if new grid has a block
if (blocks[newGridY][newGridX] != null) {
//Check to see which edge has been hit first
//Edge values
float leftEdge = minX + (newGridX * gridSize);
float rightEdge = minX + ((newGridX+1) * gridSize);
float topEdge = minY + (newGridY * gridSize);
float bottomEdge = minY + ((newGridY+1) * gridSize);
//Which edges have been hit
boolean hitLeft = false;
boolean hitRight = false;
boolean hitTop = false;
boolean hitBottom = false;
//Check if vertical slope first
if ((x+incrementX)-x != 0) {
//Slope of line between oringal point and collision point
float m = ((y+incrementY)-y)/((x+incrementX)-x);
//Check which two edges have been hit
//Top edge
if (((topEdge-y)/m)+x > leftEdge && ((topEdge-y)/m)+x < rightEdge) {
hitTop = true;
}
//Bottom Edge
if (((bottomEdge-y)/m)+x > leftEdge && ((bottomEdge-y)/m)+x < rightEdge) {
hitBottom = true;
}
//Left edge
if (m*(leftEdge-x)+y < bottomEdge && m*(leftEdge-x)+y > topEdge) {
hitLeft = true;
}
//Right edge
if (m*(rightEdge-x)+y < bottomEdge && m*(rightEdge-x)+y > topEdge) {
hitRight = true;
}
//Calculate which of the two is closest
if (hitRight && hitLeft) {
//println("1");
if (dist(x, y, leftEdge, y) < dist(x, y, rightEdge, y)) {
hitRight = false;
} else {
hitLeft = false;
}
} else if (hitRight && hitTop) {
//println("2");
if (dist(x, y, rightEdge, y) < dist(x, y, x, topEdge)) {
hitTop = false;
} else {
hitRight = false;
}
} else if (hitRight && hitBottom) {
//println("3");
if (dist(x, y, rightEdge, y) < dist(x, y, x, bottomEdge)) {
hitBottom = false;
} else {
hitRight = false;
}
} else if (hitLeft && hitTop) {
//println("4");
if (dist(x, y, leftEdge, y) < dist(x, y, x, topEdge)) {
hitTop = false;
} else {
hitLeft = false;
}
} else if (hitLeft && hitBottom) {
//println("5");
if (dist(x, y, leftEdge, y) < dist(x, y, x, bottomEdge)) {
hitBottom = false;
} else {
hitLeft = false;
}
} else if (hitTop && hitBottom) {
//println("6");
if (dist(x, y, x, bottomEdge) < dist(x, y, x, topEdge)) {
hitTop = false;
} else {
hitBottom = false;
}
}
}
//Vertical slope
else {
//println("7");
if (dist(x, y, x, bottomEdge) < dist(x, y, x, topEdge)) {
hitBottom = true;
} else {
hitTop = true;
}
}
//Determine if the block being hit is actually the free ball block
if(!blocks[newGridY][newGridX].freeBall){
//Transform theta if hitting vertal side
if (hitLeft || hitRight) {
if (theta >= 0)
theta = PI - theta;
else
theta = -PI - theta;
}
//Transform theta if hitting horizontal side
else {
}
}
}
return blocks[newGridY][newGridX];
}
return null;
}
}
```
Block Class
```class Block {
color colorValue;
int strength;
color colorArray[] = {color(#efb92b), color(#c0c038), color(#82b54a), color(#c6654c), color(#dd3a4d), color(#df1375), color(#1f76ba), color(#17998c)};
boolean freeBall = false; //Is this a block that when you hit it you get a ball?
Block() {
colorValue = color(int(random(0, 255)), int(random(0, 255)), int(random(0, 255)));
strength = 10;
}
//Block constructor that takes in a strength value and associates a color based on this
Block(int strenth) {
this.strength = strenth;
colorValue = getColor(strength);
}
//Draw the block, parameters are: x position, y position: height of side
void drawBlock(float x, float y, float sideLength) {
if (!freeBall) {
pushMatrix();
translate(x, y);
rectMode(CENTER);
noStroke();
fill(colorValue);
rect(0, 0, sideLength, sideLength);
fill(0);
textAlign(CENTER);
textSize(20);
text(strength, 0, 10);
popMatrix();
} else {
pushMatrix();
translate(x, y);
ellipseMode(CENTER);
noStroke();
fill(255);
ellipse(0, 0, sideLength/4, sideLength/4);
fill(0);
ellipse(0, 0, sideLength/6, sideLength/6);
popMatrix();
}
}
//Decrease value and color, called when a ball hits it
void decrease() {
strength--;
colorValue = getColor(strength);
}
//Get its grid x value, typically 0-6
int gridX(float x, float grid, float minX) {
return int((x - minX)/grid);
}
//Get its grid x value, typically 0-8
int gridY(float y, float grid, float minY) {
return int((y - minY)/grid);
}
//Computes the color of block based on its strength
color getColor(int val) {
//Yellow
if (val <= 5){
return colorArray[0];
}
//Olive
else if(val > 5 && val <= 10){
return colorArray[1];
}
//Green
else if(val > 10 && val <= 15){
return colorArray[2];
}
//Orange
else if(val > 15 && val <= 20){
return colorArray[3];
}
//Red
else if(val > 20 && val <= 30){
return colorArray[4];
}
//Pink
else if(val > 30 && val <= 50){
return colorArray[5];
}
//Blue
else if(val > 50 && val <= 100){
return colorArray[6];
}
//Teal
else{
return colorArray[7];
}
}
}
```
Some screenshots from the game
Problems:
There are several aspects of the game I am not quite happy with. For starters, the way Processing deals with movement is by stepping a shape a certain number of pixels. This is problematic as if a step is too large, it could skip a series of pixels where other shapes lie and as a result a collision is not detected. This is observed in my game when a ball hits a corner of a box, it often behaves as if the box is not there. This Ould be fixed with more rigorous calculations, but would also make the game more resource dependent as every grid would have to be checked for every ball which could quickly get out of hand.
I would also like to implement the ability to restart the round, but I ran out of time. In theory this would be easy as variables would just have to be reset and the game board cleared. | 5,309 | 18,983 | {"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-2024-18 | latest | en | 0.978846 |
https://newproxylists.com/why-is-stability-considered-a-desirable-feature-of-a-sorting-algorithm/ | 1,560,735,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998339.2/warc/CC-MAIN-20190617002911-20190617024911-00520.warc.gz | 528,339,181 | 9,948 | # Why is stability considered a desirable feature of a sorting algorithm?
The current argument of stability in a sort algorithm usually involves an example in which a list is sorted according to two criteria. For example:
``````1,4,5,7,2,6,8,9,15,65,24,27
sort by order of magnitude, then by value
2,4,6,8,24,1,5,7,9,15,27,65
``````
The argument is that by choosing a stable sort algorithm, you can sort this list twice – by value, then by plan – and you will get the sorted list as you wish.
I could not be more at odds with this ideology, though. First of all, sorting is performed at the back (value, regularity, regularity being the main criterion), which is not intuitive. Second, by doing this, you call sort () two times.
Let's see a documentation now. We have qsort (3) of JavaScript C and Array.prototype.sort. As far as I know, these two functions implement unstable sort algorithms …
If two members compare equally, their order in the sorted array is not defined.
and
If compareFunction (a, b) returns 0, leave a and b unchanged relative to each other, but sorted by all items. Note: The ECMAscript standard does not guarantee this behavior and all browsers (for example, Mozilla versions dating back to at least 2003) do not respect it.
… and both accept a function as an argument. I believe this function calls a comparator – a function that takes two values A and B and returns -1, 0 or 1 depending on whether A is considered "less than", "equal to" or "greater than" B ", based on the arbitrary criteria chosen by the implementer.
That said, what I found is that whatever I attach to the respective sort functions, whether I implement them myself or use it with the standard library, is that the stability has absolutely no effect on the sort result when the sort function is used correctly.
Let's use the C `qsort` for example. `qsort` implements fast sorting and is known to be unstable.
If two members compare equally, their order in the sorted array is not defined.
To clarify, this does not mean that the implementation is unstable in itself. It means that stability is not guaranteedSo relying on these semantics is a very bad idea. Which is close enough.
``````#understand
#understand
#understand
#define INT (p)
(* ((int *) (p)))
#define ISEVEN (p)
(INT (p)% 2 == 0)
empty
randomize (int * list, size_t len)
{
for (size_t i = 0; i <len; ++ i)
listing[i] = rand ()% (len * 10);
}
empty
printlist (int * list, size_t len)
{
for (size_t i = 0; i <len; ++ i)
printf ("% i,", list[i])
putchar (# 39; n);
}
int
by_even (void const * a, void const * b)
{
return (ISEVEN (a) &&! ISEVEN (b))? (-1): (ISEVEN (b) &&! ISEVEN (a));
}
int
by_value (void const *, void const * b)
{
return (INT (a) < INT(b)) ? (-1) : (INT(a) > INT (b));
}
int
by_even_and_value (void const * a, void const * b)
{
return by_even (a, b)! = 0? by_even (a, b): by_value (a, b);
}
int
main (empty)
{
static size_t const listsz = 20;
int list[listsz];
srand (time (NULL));
randomize (list, listsz);
print list (list, listsz);
qsort (list, listsz, sizeof list[0], by_even_and_value);
print list (list, listsz);
returns 0;
}
``````
Here's the result:
``````\$ cc qsort.c
\$ ./a.out
100, 111, 12, 122, 96, 50, 52, 96, 173, 125, 135, 173, 78, 144, 108, 60, 75, 116, 24, 180,
12, 24, 50, 52, 60, 78, 96, 96, 100, 108, 116, 122, 144, 180, 75, 111, 125, 135, 173, 173,
``````
So put all the sort criteria in the comparator and sort once gave me the sorted list that I wanted. It only took one sort, and the criteria could be given in order (even first, second value).
Since this makes the stability seem irrelevant to the result, why deal with the stability of a sorting algorithm? | 1,065 | 3,691 | {"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-2019-26 | latest | en | 0.92824 |
http://www.ask.com/web?q=How+Many+Face+Cards+in+a+Deck+of+Cards%3F&o=2603&l=dir&qsrc=3139&gc=1 | 1,476,997,119,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00089-ip-10-171-6-4.ec2.internal.warc.gz | 320,425,553 | 16,420 | Web Results
## Standard 52-card deck - Wikipedia
en.wikipedia.org/wiki/Standard_52-card_deck
The deck of 52 French playing cards is the most common deck of playing cards used today. It includes thirteen ranks of each of the four French suits: clubs (♧), diamonds (♢), hearts (♥...
## How many face cards are in a deck of cards? - Quora
www.quora.com/How-many-face-cards-are-in-a-deck-of-cards
1 deck = 4 suits. 1 suit = 3 face cards ( King, Queen & Jack ). Joker is normally not considered a face card. So 4 x 3 = 12. Hence 12 face cards. Ace can be used ...
## How many face cards are in a deck of 52 cards? | Reference.com
www.reference.com/hobbies-games/many-face-cards-deck-52-cards-33b8dafa982f1a04
A deck of 52 cards contains 12 face cards. Face cards are those with a king, queen or jack on them. Since there are four suits and each suit contains one of each ...
## SOLUTION: In an ordinary deck of 52 cards, how many cards are ...
www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.958153.html
SOLUTION: In an ordinary deck of 52 cards, how many cards are either RED or FACE CARDS? Algebra -> Probability-and-statistics -> SOLUTION: In an ...
## Trivia - The Playing Card Factory
www.theplayingcardfactory.com/trivia
Test your knowledge of a standard deck of cards for card games. How many face cards are in a standard deck? 12. The King, Queen, and Jack of each suit are ...
## probability - Drawing a card from a deck - Mathematics Stack ...
math.stackexchange.com/questions/33745/drawing-a-card-from-a-deck
Apr 19, 2011 ... If each suit has three face cards, how many ways could the drawn card be either a club of any kind or anything else besides a face card?
## Playing Card Frequencies - Milefoot
www.milefoot.com/math/discrete/counting/cardfreq.htm
The jack, queen, and king are often referred to as face cards. In many card games, a player has a number of cards, and this is referred to as his hand. In a few ...
## Deck of Cards Hearts – Red Diamonds – Red Clubs – Black Spades ...
www.asu.edu/courses/mat142ej/probability/notes/card_pictures_description.pdf
Deck of Cards: 52 total, 4 suits of 13 each. Clubs (♧) and Spades (♤) are black, Hearts. (♢) and Diamonds (♥) are red. Jack (J), Queen (Q) and King (K) are face ...
## Design History: The Art of Playing Cards | Design Shack
designshack.net/articles/layouts/design-history-the-art-of-playing-cards/
Nov 7, 2011 ... Playing cards have been around in some form or another dating all the way back to ... The simple French suits were much easier and cheaper to ... into the “court” cards, which modern players commonly refer to as face cards.
www.math.ttu.edu/~drager/Classes/02Spring/m1430/anse3.pdf
Apr 21, 2001 ... C. How many of the residents in the survey read the Gazette but not the. Journal? Answer: ... B. What is the probability that a face card is drawn? Answer: ... A five card hand is drawn from a standard deck of 52 cards. 100 pts. 4 ...
### How many cards are in a deck? Also, how many of each card come ...
www.quora.com
Apr 20, 2015 ... There are 52 cards in a deck, if you include the wild cards(joker), 54. In 52 cards .... How many possible outcomes would there be if three cards were drawn from the same deck of cards and laid face-up simultaneously?
### 1.What is the probability of choosing a face card from a deck of 52 ...
www.algebra.com
What is the probability of choosing a face card from a deck of 52 cards (face cards are jacks, queens, and kings)? What is the probability of the 2nd card being a ...
### Playing Cards Probability | Basic Concept on Drawing a Card ...
www.math-only-math.com
Playing cards probability problems based on a well-shuffled deck of 52 cards. In a pack or deck of ... So, there are 12 face cards in the deck of 52 playing cards. | 991 | 3,844 | {"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} | 3.5 | 4 | CC-MAIN-2016-44 | longest | en | 0.917849 |
https://blog.desmos.com/articles/new-activity-release-functions-and-their/ | 1,721,771,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00699.warc.gz | 118,472,704 | 2,948 | # New Activity Release: Functions and Their Derivatives!
One of the most important understandings in calculus is that functions have values which can be positive and negative but that those values are also changing, and that change can be in a positive or negative direction. Slope isn’t just for straight lines!
For example, when you’re getting out of student loan debt, the total value in your bank accounts might be negative, but the rate of change of your money is positive.
Or for another example, the value of the gross domestic product of the United States is always positive and the rate of change of the GDP is almost always positive so it makes more sense here to look at the rate of change of the rate of change. What is the rate of change of the increase? How does it compare to the increase of previous decades or other countries?
Because of the importance of these questions, calculus teachers frequently ask students questions about rate of change. Given a function, what is its derivative? Give a second derivative, what might the first derivative look like?
We were extremely impressed with a functions and derivatives activity developed by Sandi Yoder, especially the conversation it generated in her classroom. (Filmed here!) Inspired by Sandi’s work, we created Functions and Their Derivatives.
We give students a function and its first and second derivative, without revealing which is which. We ask them to label the derivatives accurately and then we give them feedback on their thinking.
But then we bring in a Challenge Creator and invite students to create their own function and label its derivatives. If they do that successfully, they can enter it into the gallery to challenge their classmates.
You get one function from us and then dozens more from your classmates. A calculus class that is social and creative! That’s why we’re here. | 372 | 1,873 | {"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} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.944997 |
https://cityhikerapp.com/2016/09/30/map-search-optimization/ | 1,674,861,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00655.warc.gz | 182,983,503 | 28,515 | Map Search Optimization
CityHiker uses a TON of data. To give some perspective, the city of San Francisco contains a little over 50k street intersections. If we assume they are on an even square grid (they’re not), that means roughly 225 x 225. According to Wolfram, a grid graph Gmn has 2*mn-m-n edges. For a square grid, m = n; so there are 2(n^2-n) edges. That means there are a minimum of 100k line segments required to model the city of San Francisco. That does not account for dead end streets or grid edges comprised of multiple segments in a chain.
Map Search
Searching for line segments in a rectangular space seems easy enough. We do a bounding box query for both start and end points to get everything inside the bounding region. If we provide our query in terms of precise floating point values, that means our database must check every line segment in the entire collection each time a request is received. That doesn’t scale at all. As the database grows, requests will be slower and slower. We could do some kind of shard-based strategy, where we introduce a set of discrete collections of line segments and restrict requests to regions. That requires us to maintain some overlap between regions, so we can avoid the edge cases (literally).
None of the bounding box approaches really enable caching at the web service layer. In order to support caching in a more meaningful way, we need to have a query that has a 1-1 deterministic relationship with the results. Let’s dive into the nuances of this.
Bounding Box Approach
```find all segments
where (lat0, lng0) <= startLat/startLng <= (lat1, lng1)
or (lat0, lng0) <= endLat/endLng <= (lat1, lng1)```
As mentioned above, this requires a comparison of every item in the segments table. This is very slow for large data sets. Databases use indices to speed up the lookup of these results. However, the primary input here is four floating point numbers, representing the bottom-left and top-right of the lat/lng bounding box. If these numbers are conveyed as strings in the query, it looks something like this:
`GET /segments?lat0=38.79912&lng0=-122.41456&lat1=38.79965&lng1=-122.41447`
It’s fairly clear that subtle changes in the values result in different query strings, each with their own copy of results in the cache. This is bad, since minor differences in the query have potentially no effect on the results, filling the cache with duplicate result sets. Additionally, the likelihood that multiple clients send the same exact query is very low.
Discrete Grid Approach
```find all segments
where (startLat * 1000) = gridLat
or (startLng * 1000) = gridLng
or (endLat * 1000) = gridLat
or (endLng * 1000) = gridLng```
This approach dramatically increases the likelihood of multiple identical requests. This is because the input values are three orders of magnitude larger, collapsing many requests for possible high precision values into the same low precision query. Here’s an example.
Consider this input:
`(38.79912, -122.41456 -> 38.79965, -122.41447)`
With the discrete grid approach, the above resolves like this:
`(38799, -122414 -> 38799, -122414)`
Note that both corners are now the same value, so we can consider the input as a single grid cell, indexed on the lat/lng pair.
Now, consider input nearby the input:
`(38.79936, -122.41498 -> 38.79912, -122.41466)`
Both input values resolve to the same discrete grid values, yet they are nearly a hundred meters apart. That’s the key. This discrete grid approach consolidates all the results in a discrete rectangular bounding box into one result set. By essentially rounding up to the nearest hundred, all segments in the rounded bounding box can be cached for repeat requests. We are no longer querying a range, so the data source can simply look for results based on the lat/lng pair. The query might look something like this:
`GET /segments/38799/-122414`
Side Effects
There are some beneficial side effects to using the discrete grid approach. Adjacent regions can be queried by incrementing or decrementing the input values, allowing easy precaching of neighboring data sets. Also, we can stratify the discrete grid into layers, representing collections of results at different grid sizes.
Since we used a discretization factor of 1000, we are effectively defining a hundred-meter grid. This is effective for wide zoom levels and has large results data sets. We can use a factor of 10000 to define a finer grid, which is more effective at supporting narrow zoom levels.
Important note: The results in one hundred-meter grid cell contain all of the results for all contained ten-meter grid cells. These are identical data points, so it’s unnecessary to make a ten-meter request after receiving results from a hundred-meter request that contains the ten-meter cell.
Data Optimization
Given the restrictions of a JSON web service, all data values are stringified. This is inconvenient for large result sets, as it means a large number of bytes per query. This has impacts on infrastructure budget, since bandwidth is usually not free. Therefore, we have strong motivation to optimize the way we represent the line segment data.
Computing Discrete Keys
A line segment is represented by start and end coordinates (latitude, longitude, altitude), a globally unique id, and a value for percentGrade. Each segment also has a computed value representing the hundred-meter grid index. Let’s dive into computation of this index.
Both latitude and longitude are given in values with 5 digits of precision, like this:
`(38.79912, -122.41456)`
This roughly equates to a one-meter grid. Let’s focus on latitude.
`38.79912 (32-bit float)`
Multiplying that by 100,000 gives a max absolute value of 18,000,000. That fits into a 26-bit signed integer, but we will use a 30-bit unsigned integer to represent it. For clarity, that’s a 25-bit unsigned value in bits 0-24, plus a sign bit at 29.
`3879912 (25-bit unsigned), 0 (sign bit)`
Now, let’s look at longitude.
`-122.41456 (32-bit float)`
`12241456 (25-bit unsigned), 1 (sign bit)`
We use 30-bit values because they can be represented by five Base64 characters. This way, we concatenate the Base64 value for latitude and longitude together to compute the hundred-meter grid cell index as a ten-character Base64 string.
Using this approach, we can construct a set of hundred-meter indices for a given arbitrary bounding box. Each index can be looked up in a local hash map, mapping Base64 string index to segment array.
Data Transport
The above approach is actually very useful for compressing JSON data for transport. Consider the following serialized segment:
```{
"id": 123,
"start_lat": 38.79912,
"start_lng": -122.41425,
"end_lat": 38.79923,
"end_lng": -122.41428,
}
// 115 bytes minimum```
If we apply the above compression strategy, we can render this string with substantially fewer bytes, as follows:
```{
"id": 123,
"start": "aArzpJuImF",
"end": "aAryqJuIna",
}
// 60 bytes minimum```
If we’re really squeezing the life out of it, we can render the percentGrade value as an 8-bit signed int in hexadecimal, but that really only saves a couple bytes per segment.
Overkill?
If we’re really keen on trimming everything down to the bare metal, we can represent a segment purely as a Base64 string, as long as we pack the data into 6-bit boundaries. That might look something like this:
0-29: start latitude (30-bit)
30-59: start longitude (30-bit)
60-89: end latitude (30-bit)
90-119: end longitude (30-bit)
120-179: guid (60-bit)
This results in 31-character Base64 values, which can either be concatenated or provided as array elements, like this:
```["aArzpJuImFaAryqJuInaaaaaaaaaGcN", ... ]
// 33 bytes fixed```
Observant readers will note there is another beneficial side effect of this strategy. The ten-character prefix of this string is the one-meter key. Built-in index!
At this point, we barely save anything by abandoning JSON in favor of a pure string response. Also, the concatenated string approach is more susceptible to in-transit corruption, whereas a JSON-formatted string would fail to parse in that case.
All told, for an average cell with 1000 segments, we save 70%. That’s the difference between 115KB and 35KB. Bear in mind, this is the response body size. To fetch all segments for the city of San Francisco, we need at least 100 requests (max 1000 per request). The total response size is 3.5MB, compared to the uncompressed 11.5MB. This has a measurable effect on mobile device bandwidth. | 2,013 | 8,540 | {"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 | longest | en | 0.895229 |
https://www.javatpoint.com/output-of-one-way-anova-in-spss | 1,603,770,787,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893011.54/warc/CC-MAIN-20201027023251-20201027053251-00079.warc.gz | 755,589,920 | 10,769 | # Output of One-Way ANOVA
In this section, we will learn the output of One-Way ANOVA. The output of One-Way ANOVA is given below:
In the above output, we can see only three job categories are appearing, i.e., Clerical, Custodial, and Manager. The missing value 0 is not available in the output. 30 people belong to job category 1, i.e., Clerical, 5 belonging to job category 2, i.e., Custodial, and 15 belonging to job category 3, i.e., Managers. We are having a slightly lesser number of people in job category 2.
Now we will see it has any recursion for our analysis. Average salaries are 29,125 dollars, 53,600 dollars, and 94,616 dollars. If we look at these Means scores, we will see the impression that there are differences between people's salaries from three different job categories. But if we move from Clerks to Custodial employees, the amount of difference is not huge. But if we move from the Custodial employees to the Manager, there is a huge significant difference, more than double the salary difference. So, we are expecting a significant difference between the groups. So that's how we can guess about the significant differences just by looking at the Mean scores.
The Standard deviation in the case of Managers is quite high compared to the rest of two groups. So, managers are drawing on an average more salary, but there is a huge variation in the managerial category compared to other groups.
The Standard error is shown in the following image. Standard error refers to the standard deviation of the sampling distribution of mean. So, it's an indication of the amount of error in measurement. So, the smaller it is better for us.
There is 95% confidence information. So, we can see none of the confidence information is having 0. So, we don't have a positive or negative value on either side, and these are the minimum and maximum amount of salary drawn by different groups. So, in the case of Manager, the difference is very high, 31 thousand to 123 thousand that's the descriptive scenario of output. | 455 | 2,033 | {"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-2020-45 | latest | en | 0.937365 |
http://nmr-software.blogspot.com/2010/12/ | 1,495,933,106,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00248.warc.gz | 321,577,910 | 8,534 | ## Wednesday, December 01, 2010
### Routine Prediction
Not everybody remembers how easy it is and how effective it can be to add (or correct) a few points at the beginning of the FID. Two years ago I explained how Linear Prediction works and how we can extrapolate the FID in both directions. This time I will show a simple practical application.
I have observed that, in recent years, C-13 spectra acquired on Varian instruments require a much-larger-then-it-used-to-be phase correction. When I say correction, I mean first-order phase correction, because the zero-order correction is merely a different way of representing the same thing (a different perspective).
A large first-order phase correction can be substituted with linear prediction. I will show the advantage with a F-19 example, yet the concept is general.
The spectrum, after FT and before phase correction, looks well acquired. Now we apply the needed correction, which amounts to no less than 1073 degrees.
Have you noticed what happened to the baseline? It's all predictable. When you increase the phase correction, the baseline starts rolling. The higher the phase correction, the shorter the waves. With modern method for correcting the baseline we eliminate all the waves, yet there are two potential problems: 1) The common methods for automatic phase correction will have an hard time. 2) If you prefer manual phase correction, you need an expert eye to assess the symmetry of the peaks over such a rolling baseline. Anyway, just to show you that linear prediciton is not a necessity, here is the spectrum after applying standard baseline correction:
Now let's start from the FID again, this time applying linear prediction. one way to use it is to add the 3 missing points at the beginning. The result, after a mild phase correction (<10°) and before the baseline correction is:
The lesson is: by adding the missing points we correct the phase.
Alternatively we can both add the 3 missing points and recalculate the next 4 points. In this way the baseline improves a lot:
The lesson is: by recalculating the first few points of the FID we can correct the baseline of the frequency domain spectrum. | 453 | 2,179 | {"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} | 3.125 | 3 | CC-MAIN-2017-22 | longest | en | 0.892705 |
https://stackoverflow.com/questions/30265375/how-to-draw-orthographic-projection-from-equirectangular-projection | 1,534,736,236,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215523.56/warc/CC-MAIN-20180820023141-20180820043141-00638.warc.gz | 762,265,621 | 28,514 | # How to draw orthographic projection from equirectangular projection
I have this image :
I don’t know exactly what kind on projection it is, I guess equirectangular or mercator by the shape. It's the texture for an attitude indicator, b.
I want to draw a orthographic projection, b or maybe a General Perspective projection (which one looks better) of it according to a direction vector defined by two angles (heading and pitch). This direction define a point on the sphere, this point should be the center of the projection.
I want it to look from the pilot point of view, so only half of the sphere should be drawn.
I use python, and I have not yet chosen a graphic library, I will probably be using pygame though.
I’ve found something related : http://www.pygame.org/project-Off-Center+Map+Projections-2881-.html but it uses OpenGL and I have no experience with it, but I can try if needed.
How should I do that ? I probably can draw it manually by calculating every pixel from the calculation formulas but I think there are some kind of library tools to do that efficiently (hardware accelerated probably ?).
For an all-Python solution (using numpy/scipy array ops, which will be faster than any explicit per-pixel looping), this:
``````#!/usr/bin/env python
import math
import numpy as np
import scipy
import scipy.misc
import scipy.ndimage.interpolation
import subprocess
size=256
frames=50
for frame in xrange(0,frames):
# Image pixel co-ordinates
px=np.arange(-1.0,1.0,2.0/size)+1.0/size
py=np.arange(-1.0,1.0,2.0/size)+1.0/size
hx,hy=scipy.meshgrid(px,py)
# Compute z of sphere hit position, if pixel's ray hits
r2=hx*hx+hy*hy
hit=(r2<=1.0)
hz=np.where(
hit,
-np.sqrt(1.0-np.where(hit,r2,0.0)),
np.NaN
)
# Some spin and tilt to make things interesting
spin=2.0*np.pi*(frame+0.5)/frames
cs=math.cos(spin)
ss=math.sin(spin)
ms=np.array([[cs,0.0,ss],[0.0,1.0,0.0],[-ss,0.0,cs]])
tilt=0.125*np.pi*math.sin(2.0*spin)
ct=math.cos(tilt)
st=math.sin(tilt)
mt=np.array([[1.0,0.0,0.0],[0.0,ct,st],[0.0,-st,ct]])
# Rotate the hit points
xyz=np.dstack([hx,hy,hz])
xyz=np.tensordot(xyz,mt,axes=([2],[1]))
xyz=np.tensordot(xyz,ms,axes=([2],[1]))
x=xyz[:,:,0]
y=xyz[:,:,1]
z=xyz[:,:,2]
# Compute map position of hit
latitude =np.where(hit,(0.5+np.arcsin(y)/np.pi)*src.shape[0],0.0)
longitude=np.where(hit,(1.0+np.arctan2(z,x)/np.pi)*0.5*src.shape[1],0.0)
latlong=np.array([latitude,longitude])
# Resample, and zap non-hit pixels
dst=np.zeros((size,size,3))
for channel in [0,1,2]:
dst[:,:,channel]=np.where(
hit,
scipy.ndimage.interpolation.map_coordinates(
src[:,:,channel],
latlong,
order=1
),
0.0
)
# Save to f0000.png, f0001.png, ...
scipy.misc.imsave('f{:04}.png'.format(frame),dst)
# Use imagemagick to make an animated gif
subprocess.call('convert -delay 10 f????.png anim.gif',shell=True)
``````
will get you
.
OpenGL is really the place to be doing this sort of pixel wrangling though, especially if it's for anything interactive.
• Amazing really nice work ! – luxcem May 16 '15 at 13:56
I glanced at the code in the "Off-Center Map Projections" stuff you linked...
As a starting point for you, I'd say it was pretty good, especially if you want to achieve this with any sort of efficiency in PyGame as offloading any sort of per-pixel operations to OpenGL will be much faster than they'll ever be in Python.
Obviously to get any further you'll need to understand the OpenGL; the projection is implemented in `main.py`'s GLSL code (the stuff in the string passed to `mod_program.ShaderFragment`) - the atan and asin there shouldn't be a surprise if you've read up on equirectangular projections.
However, to get to what you want, you'll have to figure out how to render a sphere instead of the viewport-filling quad (rendered in main.py at `glBegin(GL_QUADS);`). Or alternatively, stick with the screen-filling quad and do a ray-sphere intersection in the shader code too (which is effectively what the python code in my other answer does).
• Thank you for help, I’ll definitively look at OpenGL. – luxcem May 15 '15 at 23:04 | 1,157 | 4,060 | {"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-2018-34 | longest | en | 0.803683 |
https://kidsworksheetfun.com/dilations-worksheet-answer-key/ | 1,708,956,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00066.warc.gz | 344,805,284 | 25,386 | Dilations Worksheet Answer Key. Dilations worksheet answer key kuta software. 30 dilations worksheet answer key education template from smithfieldjustice.com geometry cp 6 7 dilations worksheet name state whether a.
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30 Dilations Worksheet Answer Key Education Template From Smithfieldjustice.com Geometry Cp 6 7 Dilations Worksheet Name State Whether A.
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Cc 2 triangle pqr has vertices p 2 2 q 3 2 and r o 2. Worksheets are dilations translations work answer. Free worksheets on adding subtracting multiplying and dividing fractions high school math printouts real life quadratic equations englishwork sheets substitution calculator calculating greatest. | 486 | 2,353 | {"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-2024-10 | latest | en | 0.82765 |
https://www.packtpub.com/books/content/adjusting-key-parameters-r | 1,440,732,895,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644060173.6/warc/CC-MAIN-20150827025420-00014-ip-10-171-96-226.ec2.internal.warc.gz | 915,580,381 | 20,379 | # Adjusting Key Parameters in R
#### R Graphs Cookbook
\$29.99
Detailed hands-on recipes for creating the most useful types of graphs in R – starting from the simplest versions to more advanced applications
## R Graph Cookbook
Detailed hands-on recipes for creating the most useful types of graphs in R – starting from the simplest versions to more advanced applications Learn to draw any type of graph or visual data representation in R Filled with practical tips and techniques for creating any type of graph you need; not just theoretical explanations All examples are accompanied with the corresponding graph images, so you know what the results look like Each recipe is independent and contains the complete explanation and code to perform the task as efficiently as possible
(For more resources on R, see here.)
# Setting colors of points, lines, and bars
In this recipe we will learn the simplest way to change the colors of points, lines, and bars in scatter plots, line plots, histograms, and bar plots.
All you need to try out this recipe is to run R and type the recipe at the command prompt. You can also choose to save the recipe as a script so that you can use it again later on.
## How to do it...
The simplest way to change the color of any graph element is by using the col argument. For example, the plot() function takes the col argument:
`plot(rnorm(1000), col="red")`
If we choose plot type as line, then the color is applied to the plotted line. Let’s use the dailysales.csv example dataset. First, we need to load it:
`Sales <- read.csv("dailysales.csv",header=TRUE)plot(sales\$units~as.Date(sales\$date,"%d/%m/%y"),type="l", #Specify type of plot as l for linecol="blue")`
Similarly, the points() and lines() functions apply the col argument's value to the plotted points and lines respectively.
barplot() and hist() also take the col argument and apply them to the bars. So the following code would produce a bar plot with blue bars:
`barplot(sales\$ProductA~sales\$City,col="blue")`
The col argument for boxplot() is applied to the color of the boxes plotted.
## How it works...
The col argument automatically applies the specified color to the elements being plotted, based on the plot type. So, if we do not specify a plot type or choose points, then the color is applied to points. Similarly, if we choose plot type as line then the color is applied to the plotted line and if we use the col argument in the barplot() or histogram() commands, then the color is applied to the bars.
col accepts names of colors such as red, blue, and black. The colors() (or colours()) function lists all the built-in colors (more than 650) available in R. We can also specify colors as hexadecimal codes such as #FF0000 (for red), #0000FF (for blue), and #000000 (for black). If you have ever made any web pages¸ you would know that these hex codes are used in HTML to represent colors.
col can also take numeric values. When it is set to a numeric value, the color corresponding to that index in the current color palette is used. For example, in the default color palette the first color is black and the second color is red. So col=1 and col=2 refers to black and red respectively. Index 0 corresponds to the background color.
## There's more...
In many settings, col can also take a vector of multiple colors, instead of a single color. This is useful if you wish to use more than one color in a graph. The heat.colors() function takes a number as an argument and returns a vector of those many colors. So heat.colors(5) produces a vector of five colors.
Type the following at the R prompt:
`heat.colors(5)`
You should get the following output:
`[1] "#FF0000FF" "#FF5500FF" "#FFAA00FF" "#FFFF00FF" "#FFFF80FF"`
Those are five colors in the hexadecimal format.
Another way of specifying a vector of colors is to construct one:
`barplot(as.matrix(sales[,2:4]), beside=T,legend=sales\$City,col=c("red","blue","green","orange","pink"),border="white")`
In the example, we set the value of col to c("red","blue","green","orange","pink"), which is a vector of five colors.
We have to take care to make a vector matching the length of the number of elements, in this case bars we are plotting. If the two numbers don’t match, R will 'recycle’ values by repeating colors from the beginning of the vector. For example, if we had fewer colors in the vector than the number of elements, say if we had four colors in the previous plot, then R would apply the four colors to the first four bars and then apply the first color to the fifth bar. This is called recycling in R:
`barplot(as.matrix(sales[,2:4]), beside=T,legend=sales\$City,col=c("red","blue","green","orange"),border="white")`
In the example, both the bars for the first and last data rows (Seattle and Mumbai) would be of the same color (red), making it difficult to distinguish one from the other.
One good way to ensure that you always have the correct number of colors is to find out the length of the number of elements first and pass that as an argument to one of the color palette functions. For example, if we did not know the number of cities in the example we have just seen; we could do the following to make sure the number of colors matches the number of bars plotted:
`barplot(as.matrix(sales[,2:4]), beside=T,legend=sales\$City,col=heat.colors(length(sales\$City)),border="white")`
We used the length() function to find out the length or the number of elements in the vector sales\$City and passed that as the argument to heat.colors(). So, regardless of the number of cities we will always have the right number of colors.
In the next four recipes, we will see how to change the colors of other elements. The fourth recipe is especially useful where we look at color combinations and palettes.
(For more resources on R, see here.)
# Setting plot background colors
The default background color of all plots in R is white, which is usually the best choice as it is least distracting for data analysis. However, sometimes we may wish to use another color. We will see how to set background colors in this recipe.
All you need to try out this recipe is to run R and type the recipe at the command prompt. You can also choose to save the recipe as a script so that you can use it again later on.
## How to do it...
To set the plot background color to gray we use the bg argument in the par() command:
`par(bg="gray")plot(rnorm(100))`
## How it works...
The par() command's bg argument sets the background color for the entire plotting area including the margins for any subsequent plots on the same device. Until the plotting device is closed or a new device is initiated, the background color stays the same.
## There's more...
It is more likely that we want to set the background color only for the plot region (within the axes) but there is no straightforward way to do this in R. We must draw a rectangle of the desired color in the background and then make our graph on top of it:
`plot(rnorm(1000),type="n")x<-par("usr")rect(x[1],x[3],x[2],x[4],col="lightgray ")points(rnorm(1000))`
First we draw the plot with type set to "n" so that the plotted elements are invisible. This does not show the graph points or lines but sets the axes up, which we need for the next step. par("usr") gets us the co-ordinates of the plot region in a vector of form c(xleft, xright, ybottom, ytop). We then use the rect() function to draw a rectangle with a fill color that we wish to use for the plot background. Note that rect() takes a set of arguments representing the xleft, ybottom, xright, ytop co-ordinates. So we must pass the values we obtained from par("usr") in the correct order. Then, finally we redraw the graph with the correct type (points or lines).
# Setting colors for text elements: axis annotations, labels, plot titles, and legends
Axis annotations are the numerical or text values placed beside tick marks on an axis. Axis labels are the names or titles of axes, which tell the reader what the values on a particular axis represent. In this recipe, we will learn how to set the colors for these elements and legends.
All you need to try out this recipe is to run R and type the recipe at the command prompt. You can also choose to save the recipe as a script so that you can use it again later on.
## How to do it...
Let's say we want to make the axis value annotations black, the labels of the axes gray, and the plot title dark blue, you should do the following:
`plot(rnorm(100),main="Plot Title",col.axis="blue",col.lab="red",col.main="darkblue")`
## How it works...
Colors for axis annotations, labels, and plot titles can be set either using the par() command before making the graph or in the graph command such as plot() itself. The arguments for setting the colors for axis annotations, labels, and plot titles are col.axis, col.lab, and col.main respectively.
They are similar to the col argument and take names of colors or hex codes as values, but do not take a vector of more than one color.
## There's more...
If we use the par() command, the difference is that par() will apply these settings to every subsequent graph, until it is reset either by specifying the settings again or starting a new graphics device:
`par(col.axis="black",col.lab="#444444",col.main="darkblue")plot(rnorm(100),main="plot")`
The col.axis argument can also be passed to the axis() function, which is useful for making a custom axis if you do not want to use the default axis. The col.lab argument does not work with axis() and must be specified in par() or the main graph function such as plot() or barplot().
The col.main argument can also be passed to the title() function, which is useful for adding a custom plot title if you do not want to use the default title:
`title("Sales Figures for 2010", col.main="blue")`
Axis labels can also be specified with title():
`title(xlab="Month",ylab="Sales",col.lab="red")`
This is handy because you can specify two different colors for the X and Y axes:
`title(xlab="X axis",col.lab="red")title(ylab="Y axis",col.lab="blue")`
When setting the axis titles with the title() command, we must set xlab and ylab to empty strings "" in the original plot command to avoid overlapping titles.
# Choosing color combinations and palettes
We often need more than one color to represent various elements in graphs. Palettes are combinations of colors which are a convenient way to use multiple colors without choosing individual colors separately. R provides inbuilt color palettes as well as the ability to make our own custom palettes. Using palettes is a good way to avoid repeatedly choosing or setting colors in multiple locations, which can be a source of error and confusion. It helps in separating the presentation settings of a graph from the construction.
All you need to try out this recipe is to run R and type the recipe at the command prompt. You can also choose to save the recipe as a script so that you can use it again later on. One new library needs to be installed, which is also explained.
## How to do it...
We can change the current palette by passing a character vector of colors to the palette() function. For example:
`palette(c("red","blue","green","orange"))`
To use the colors in the current palette, we can refer to them by the index number. For example, palette()[1] would be red.
## How it works...
R has a default palette of colors which can be accessed by calling the palette() function. If we run the palette() command just after starting R, we get the default palette:
`palette()[1] "black" "red" "green3" "blue" "cyan" "magenta""yellow"[8] "gray"`
To revert back to the default palette type:
`palette("default")`
When a vector of color names is passed to the palette() function, it sets the current palette to those colors. We must enter valid color names otherwise we will get an invalid color name error.
## There's more...
Besides the default palette provided by the palette() function, R has many more built-in palettes and additional palette libraries. One of the most commonly used palettes is the heat.colors() palette, which provides a range of colors from red through yellow to white, based on the number of colors specified by the argument n. For example, heat.colors(10) produces a palette of 10 warm colors from red to white.
Other palettes are rainbow(), terrain.colors(), cm.colors(), and topo.colors which take the number of colors as an argument.
RColorBrewer is a very good color palette package that creates nice looking color palettes especially for thematic maps. It is an R implementation of the RColorBrewer palettes, which provides three types of palettes: sequential, diverging, and qualitative. More information is available at Colorbrewer 2.0
To use RColorBrewer, we need to install and load it:
`install.packages("RColorBrewer")library(RColorBrewer)`
To see all the RColorBrewer palettes run the following command at the R prompt:
`display.brewer.all()`
The names of the palettes are displayed in the left-hand margin and the colors in each palette are displayed in each row running to the right.
To use one of the palettes, let's say YlOrRd (which as the names suggests is a combination of yellows and reds), we can use the brewer.pal() function:
`brewer.pal(7,"YlOrRd")[1] "#FFFFB2" "#FED976" "#FEB24C" "#FD8D3C" "#FC4E2A" "#E31A1C""#B10026"`
The brewer.pal function takes two arguments: the number of colors we wish to choose and the name of the palette. The minimum number of colors is three but the maximum varies from palette to palette.
We can view the colors of an individual palette by using the display.brewer.pal() command:
`display.brewer.pal(7,"YlOrRd")`
To use a specific color of the palette we can refer to it by its index number. So the first color in the palette is brewer.pal(7,"YlOrRd")[1], the second is brewer.pal(7,"YlOrRd")[2], and so on.
We can set the current palette to the previous one by using the palette() function:
`palette(brewer.pal(7,"YlOrRd"))`
Now we can refer to the individual colors as palette()[1], palette()[2], and so on. We can also store the palette as a vector:
`pal1<- brewer.pal(7,"YlOrRd")`
# Setting fonts for annotations and titles
For most data analysis we can just use the default fonts for titles. However, sometimes we may want to choose different fonts for presentation and publication purposes. Selecting fonts can be tricky as it depends on the operating system and the graphics device. We will see some simple ways to choose fonts in this recipe.
All you need to try out this recipe is to run R and type the recipe at the command prompt. You can also choose to save the recipe as a script so that you can use it again later on.
## How to do it...
The font family and face can be set with the par() command:
`par(family="serif",font=2)`
## How it works...
A font is specified in two parts: a font family (such as Helvetica or Arial) and a font face within that family (such as bold or italic).
The available font families vary by operating system and graphics devices. So R provides some proxy values which are mapped on to the relevant available fonts irrespective of the system. Standard values for family are "serif", "sans", and "mono".
The font argument takes numerical values: 1 corresponds to plain text (the default), 2 to bold face, 3 to italic, and 4 to bold italic.
For example, par(family="serif",font=2) sets the font to a bold Times New Roman on Windows. You can check the other font mappings by running the windowsFonts() command at the R prompt.
The fonts for axis annotations, labels, and plot main title can be set separately using the font.axis, font.lab, and font.main arguments respectively.
## There's more...
The choice of fonts is very limited if we just use the proxy family names. However, we can use a wide range of fonts if we are exporting our graphs in the PostScript or PDF formats. The postscriptFonts() and pdfFonts() functions show all the available fonts for those devices. To see the PDF fonts, run the following command:
`names(pdfFonts())[1] "serif" "sans" "mono"[4] "AvantGarde" "Bookman" "Courier"[7] "Helvetica" "Helvetica-Narrow" "NewCenturySchoolbook"[10] "Palatino" "Times" "URWGothic"[13] "URWBookman" "NimbusMon" "NimbusSan"[16] "URWHelvetica" "NimbusSanCond" "CenturySch"[19] "URWPalladio" "NimbusRom" "URWTimes"[22] "Japan1" "Japan1HeiMin" "Japan1GothicBBB"[25] "Japan1Ryumin" "Korea1" "Korea1deb"[28] "CNS1" "GB1"`
To use one of these font families in a PDF, we can pass the family argument to the pdf() function:
`pdf(family="AvantGarde") pdf(paste(family="AvantGarde")`
# Summary
In this article we saw how to set colors of points, lines, and bars. We also took a look at setting plot background colors and setting colors for text elements, choosing color combinations and palettes, and setting fonts for annotations and titles.
In the next article, we will see how to choose styles for various graph elements in R.
Further resources on this subject:
### Books to Consider
R Object-oriented Programming
\$ 26.99
Instant R Starter
\$ 14.99
Learning Data Mining with R
\$ 29.99
R for Data Science
\$ 29.99 | 4,108 | 17,379 | {"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} | 2.6875 | 3 | CC-MAIN-2015-35 | longest | en | 0.871621 |
https://web2.0calc.com/questions/what_18 | 1,553,547,892,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00297.warc.gz | 652,887,814 | 5,757 | +0
# What???
0
436
2
+445
Cybil and Ronda are sisters. The 10 letters from their names are placed on identical cards so that each of 10 cards contains one letter. Without replacement, two cards are selected at random from the 10 cards. What is the probability that one letter is from each sister's name? Express your answer as a common fraction
Nov 11, 2017
#1
+98196
+1
We can count sets here.....
No. of possible sets = C(10,2) = 45
No. of sets containing only letters from Cybil's name = C(5,2) = 10
No. of sets containing only letters from Ronda's name = C(5,2) = 10
So the number of sets containing one letter from each name = 45 - 10 - 10 = 25
So....the probability of choosing one of these sets = 25 / 45 = 5 / 9
Nov 11, 2017
#2
+445
+1
Thanks so much!
Mr.Owl Nov 11, 2017 | 264 | 808 | {"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} | 4.03125 | 4 | CC-MAIN-2019-13 | latest | en | 0.897457 |
https://math.answers.com/Q/What_is_8800_cm_in_meters | 1,708,682,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00397.warc.gz | 387,330,669 | 45,302 | 0
# What is 8800 cm in meters?
Updated: 9/22/2023
Wiki User
11y ago
100 cm = 1 metre so 8800 cm = 8800/100 = 88 metres. Simple!
Wiki User
11y ago | 60 | 152 | {"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-2024-10 | latest | en | 0.645686 |
https://convertoctopus.com/309-cubic-meters-to-fluid-ounces | 1,590,813,759,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00079.warc.gz | 305,998,486 | 8,149 | ## Conversion formula
The conversion factor from cubic meters to fluid ounces is 33814.022558919, which means that 1 cubic meter is equal to 33814.022558919 fluid ounces:
1 m3 = 33814.022558919 fl oz
To convert 309 cubic meters into fluid ounces we have to multiply 309 by the conversion factor in order to get the volume amount from cubic meters to fluid ounces. We can also form a simple proportion to calculate the result:
1 m3 → 33814.022558919 fl oz
309 m3 → V(fl oz)
Solve the above proportion to obtain the volume V in fluid ounces:
V(fl oz) = 309 m3 × 33814.022558919 fl oz
V(fl oz) = 10448532.970706 fl oz
The final result is:
309 m3 → 10448532.970706 fl oz
We conclude that 309 cubic meters is equivalent to 10448532.970706 fluid ounces:
309 cubic meters = 10448532.970706 fluid ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 9.5707215817152E-8 × 309 cubic meters.
Another way is saying that 309 cubic meters is equal to 1 ÷ 9.5707215817152E-8 fluid ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred nine cubic meters is approximately ten million four hundred forty-eight thousand five hundred thirty-two point nine seven one fluid ounces:
309 m3 ≅ 10448532.971 fl oz
An alternative is also that one fluid ounce is approximately zero times three hundred nine cubic meters.
## Conversion table
### cubic meters to fluid ounces chart
For quick reference purposes, below is the conversion table you can use to convert from cubic meters to fluid ounces
cubic meters (m3) fluid ounces (fl oz)
310 cubic meters 10482346.993 fluid ounces
311 cubic meters 10516161.016 fluid ounces
312 cubic meters 10549975.038 fluid ounces
313 cubic meters 10583789.061 fluid ounces
314 cubic meters 10617603.084 fluid ounces
315 cubic meters 10651417.106 fluid ounces
316 cubic meters 10685231.129 fluid ounces
317 cubic meters 10719045.151 fluid ounces
318 cubic meters 10752859.174 fluid ounces
319 cubic meters 10786673.196 fluid ounces | 548 | 2,150 | {"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-2020-24 | latest | en | 0.782983 |
http://stackoverflow.com/questions/6175558/assign-weights-based-on-frequency-of-occurrence-of-values | 1,464,147,093,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274119.11/warc/CC-MAIN-20160524002114-00031-ip-10-185-217-139.ec2.internal.warc.gz | 273,598,620 | 18,767 | # Assign weights based on frequency of occurrence of values
I would like to ask you for help with my data frame. It is a vector of many phases and for every one we have names of variables. Lets say
``````vec<-data.frame(phase1= c("var1","var2","var3","var4","var5","var6"),
phase2= c("var1","var3","var4","var2","var6","var5"),
phase3= c("var4","var3","var2","var1","var6","var5"))
vec
phase1 phase2 phase3
1 var1 var1 var4
2 var2 var3 var3
3 var3 var4 var2
4 var4 var2 var1
5 var5 var6 var6
6 var6 var5 var5
``````
Now, lets say we are interested for the first 3 rows and therefore the weight of variable in one of them is 1/3, zero otherwise. My function would ideally output sth like that:
`````` phase1 phase2 phase3
var1 0.33 0.33 0
var2 0.33 0 0.33
var3 0.33 0.33 0.33
var4 0 0.33 0.33
var5 0 0 0
var6 0 0 0
``````
The function should also be applicable for the first 4, 5 or all 6 rows (ie the weights will change then). Regards, Alex
-
I believe you are looking for this:
``````n<-3
l<-dim(vec)[1]
wghts<-c(rep(1/n, n), rep(0, l-n))
result<-do.call(cbind, lapply(vec, function(curcol){
wghts[match(curcol, vec\$phase1)]
}))
``````
If need be, you could add:
``````rownames(result)<-vec\$phase1
``````
-
You can use `%in%` to find matches and `ifelse` to set weigths:
``````set_weigth <- function(x, v, w) ifelse(x%in%v,w,0)
as.data.frame(lapply(vec, set_weigth, v=vec\$phase1[1:3], w=0.33))
``````
-
You are essentially setting the weight of `var_i` in `phase_i` as the fraction of rows `var_i` occurs in `phase_i`. The simplest way is to use the `table()` function: given a vector of discrete values, it produces a frequency-count of the different values. If you want to get your desired weights based on the first 3 rows of the data-frame `vec`, you simply do:
``````> sapply(vec[1:3,],table)/3
phase1 phase2 phase3
var1 0.3333333 0.3333333 0.0000000
var2 0.3333333 0.0000000 0.3333333
var3 0.3333333 0.3333333 0.3333333
var4 0.0000000 0.3333333 0.3333333
var5 0.0000000 0.0000000 0.0000000
var6 0.0000000 0.0000000 0.0000000
``````
Similarly if you want to use the first 4 rows you do:
``````> sapply(vec[1:4,],table)/4
phase1 phase2 phase3
var1 0.25 0.25 0.25
var2 0.25 0.25 0.25
var3 0.25 0.25 0.25
var4 0.25 0.25 0.25
var5 0.00 0.00 0.00
var6 0.00 0.00 0.00
``````
- | 948 | 2,445 | {"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-2016-22 | latest | en | 0.618256 |
https://ca.answers.yahoo.com/question/index?qid=20191117161225AAXt6ua | 1,600,411,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00455.warc.gz | 427,809,209 | 29,186 | Lily asked in Science & MathematicsMathematics · 10 months ago
# Can you find a 5 numbers with a mean of 6 , a median of 7 and a mode of 8 ?
Relevance
• geezer
Lv 7
10 months ago
5 numbers ... mean of 6 ... the total of the numbers added together must be 30
5 numbers ... median of 7 ... the number in the middle must be 7
5 numbers ... mode of 8 ... so 8 must be used the most number of times
So ..
If there are 5 numbers and we know it's xx7xx then the 8 is there twice .. xx788
7 + 8 + 8 = 23 ... so the first 2 numbers add up to 7.
So .. the number could be 16788 .. 25788 .. 34788
• 10 months ago
8, 6, 8, 1, 7
2, 8, 7, 5, 8
3, 4, 8, 7, 8
• 10 months ago
3,4,7,8,8
fits the description
Lv 7
10 months ago
There are five numbers and the median is 7, so 7 is the middle (third) number.
There have to be more 8s than any other number, so the fourth and fifth numbers must be 8.
The sum of all the numbers must be 5*6=30.
Sum of first number and second number = 30 - 7 - 8 - 8 = 7.
Since the mode is 8, the first and second numbers must be different and neither can be 7.
There is more than one solution.
{1,6,7,8,8}, {2,5,7,8,8}, {3,4,7,8,8}
• 10 months ago
The median of 5 numbers will be the third number when sorted. We know that's a 7 so from that alone we know this:
a b 7 x y
Two numbers smaller and two numbers larger than 7.
We are told the mode is 8, which means there are at least 2 8's in the set and it has the most numbers in the set. There is room for 2 numbers larger than 7's, so these are both 8s. Since there are two of them, no other number can have more than 1 for 8 to be the only mode:
a b 7 8 8
we now need a mean of 6. There are an infinite number of solutions here, but presuming you want only integers, we can figure this out.
The mean is the sum of the data point values divided by the number of data points. Do this and set it equal to 6 and we can come up with an expression that compares a and b:
(a + b + 7 + 8 + 8) / 5 = 6
(a + b + 23) / 5 = 6
a + b + 23 = 30
a + b = 7
So we know that the sum of a and b is 7, and both must be less than 7 and not the same. Keeping these as integers, we don't have them completely narrowed down to one possible set, but a small set of sets:
The solution can be either:
1 6 7 8 8
2 5 7 8 8
3 4 7 8 8
All three of these sets satisfy the conditions of your set. We would need more information in order to narrow it down further.
• 10 months ago
a , b , c , d , e
These are sorted from least to greatest. We know that c = 7
a , b , 7 , d , e
We know that the most common number is 8, so d and e are 8
a , b , 7 , 8 , 8
We know that a and b can't be equal to each other, or else 8 wouldn't be the mode
a < b
We know that the mean is 5
5 = (a + b + 7 + 8 + 8) / 5
25 = a + b + 23
2 = a + b
As long as a + b = 2, we're good. Are there any other restrictions? Can a be 0? If so, we have:
0 , 2 , 7 , 8 , 8
That fits everything you need.
Do you see how straightforward this stuff is? There's nothing crazy or ridiculous here. All you have to do is work through it methodically, piece-by-piece, until you have an answer.
EDIT:
Mean of 6. Whatever. Same thing.
6 = (a + b + 7 + 8 + 8) / 5
30 = a + b + 23
7 = a + b
a = 1 , b = 6
a = 2 , b = 5
a = 3 , b = 4
There you go.
1 , 6 , 7 , 8 , 8
2 , 5 , 7 , 8 , 8
3 , 4 , 7 , 8 , 8 | 1,159 | 3,366 | {"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} | 4.375 | 4 | CC-MAIN-2020-40 | latest | en | 0.887911 |
https://answer-helper.com/engineering/question514604704 | 1,642,368,535,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00664.warc.gz | 179,250,671 | 16,373 | , 17.10.2021 19:30 glocurlsprinces
# Can someone give me an example of a machine that uses a simple pulley ramp system? I would really appreciate multiple examples :)
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https://crypto.stackexchange.com/questions/77716/compute-statistical-distance-between-two-distributions-over-tuples | 1,695,937,009,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00323.warc.gz | 206,634,020 | 39,624 | # Compute statistical distance between two distributions over tuples
Let $$X$$ denote one distribution. Let $$f,g, \text{ and } h$$ denote three functions. If we have the results: $$g(X)$$ is within a negligible statistical distance of $$h(X)$$. Is it possible to prove
$$(f(X),g(X)) \text{ is within negligible statistical distance of } (f(X),h(X))$$
I was struggling with this problem for a long time. Any hints are welcomed.
• Now that I think about it, this question is probably off-topic. Feb 20, 2020 at 12:10
No, you cannot prove that, since it is not generally true. Consider the following counter example.
Let $$X$$ be a distribution over $$\{0,1\}$$ with $$\Pr_{b\gets X}[b=0] = \Pr_{b\gets X}[b=1]=\frac{1}{2}.$$ Let $$f,g,h : \{0,1\} \to \{0,1\}$$ be defined as $$f : b \mapsto b,\quad g : b \mapsto b, \quad \text{and} \quad h : b \mapsto b\oplus 1.$$
The statistical distance between the distributions $$g(X)$$ and $$h(X)$$ is zero and thereby negligible.
\begin{align}&\frac{1}{2} \sum_{b\in \{0,1\}} \left|\Pr_{b'\gets X}[g(b')=b] - \Pr_{b'\gets X}[h(b')=b]\right|\\ =&\frac{1}{2} \sum_{b\in \{0,1\}} \left|\Pr_{b'\gets X}[b'=b] - \Pr_{b'\gets X}[b'\oplus 1=b]\right|\\ =&\frac{1}{2} \sum_{b\in \{0,1\}} \left|\frac{1}{2} - \frac{1}{2}\right| = 0 \end{align}
However, the support of the two distributions $$(f(X),g(X)) \in \{(0,0),(1,1)\}\quad \text{and} \quad (f(X),h(X)) \in \{(0,1),(1,0)\}$$ are completely disjoint and their statistical distance is thus $$1$$ which is non-negligible. | 537 | 1,513 | {"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": 15, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | longest | en | 0.806389 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-7-section-7-3-double-angle-half-angle-and-product-sum-formulas-7-3-exercises-page-561/6 | 1,560,868,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998755.95/warc/CC-MAIN-20190618143417-20190618165417-00414.warc.gz | 786,607,644 | 13,207 | ## Precalculus: Mathematics for Calculus, 7th Edition
$sin(2x) = -\sqrt {15}/8$ $cos(2x) = 7/8$ $tan(2x) -\sqrt{15}/7$
1. Find cos(x) and sin(x): $$sin(x) = \frac{1}{csc(x)} = \frac 14$$ $$sin^2(x) + cos^2(x) = 1$$ $$cos(x) = \pm \sqrt{1 - sin^2(x)}$$ $$cos(x) = \pm \sqrt{1 - (1/4)^2} = \pm \sqrt{15/16}$$ $$tan(x) = \frac{sin(x)}{cos(x)}$$ Since $tan(x) \lt 0$, $\frac{sin(x)}{cos(x)}$ must be negative. sin(x) = 1/4, which is positive. Thus, cos(x) must be negative. $$cos(x) = -\sqrt{15/16} = \frac{-\sqrt {15}}{4}$$ 2. Calculate cos(2x) and sin(2x) $$sin(2x) = 2sin(x)cos(x) = 2(1/4)(\frac{-\sqrt {15}} 4)$$ $$sin(2x) = \frac{-2\sqrt {15}}{16} = -\frac {\sqrt {15}} 8$$ $$cos(2x)= cos^2(x) - sin^2(x) = 15/16 - 1/16 = 7/8$$ 3. Calculate tan(2x) $$tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{\frac{-\sqrt {15}}{8}}{7/8} = -\frac{\sqrt {15}}{7}$$ | 416 | 848 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-26 | latest | en | 0.517721 |
https://www.mrexcel.com/board/threads/a-quick-way-to-place-numbers-into-a-banding-group.1156907/ | 1,620,763,339,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00327.warc.gz | 975,119,648 | 16,883 | # A quick way to place numbers into a banding group?
#### TheWennerWoman
##### Board Regular
Is there a quick way, either via Excel or via VBA? I have two columns of data, column A contains the numbers 1 - 20000 in numerical order.
What I need in column B is to place them in a band of 50.......so numbers 1 to 50 in column A will have "1 to 50" in column B, numbers 51-100 in column A will have "51 to 100" in column B.
I started off with
Code:
``=IF(A2<51,"1 to 50",IF(AND(A2>50,A2<101),"51 to 100",IF(AND(A2>100,A2<151),"101 to 150",IF(AND(A2>150,A2<201),"151 to 200",IF(AND(A2>200,A2<251),"201 to 250",IF(AND(A2>250,A2<301),"251 to 300",IF(AND(A2>300,A2<351),"301 to 350",IF(AND(A2>350,A2<401),"351 to 400",IF(AND(A2>400,A2<451),"401 to 450",IF(AND(A2>450,A2<501),"451 to 500",IF(AND(A2>500,A2<551),"501 to 550")))))))))))``
but that's so time consuming to write plus I believe Excel has a limit on nested IF statements.
Any help appreciated and seasons greetings to you all.
### Excel Facts
Can a formula spear through sheets?
Use =SUM(January:December!E7) to sum E7 on all of the sheets from January through December
#### Phuoc
##### Active Member
Try this:
=FLOOR(A1-1,50)+1&" to "&CEILING(A1,50)
#### TheWennerWoman
##### Board Regular
Try this:
=FLOOR(A1-1,50)+1&" to "&CEILING(A1,50)
That is ridiculously brilliant! Thank you!
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Go back | 756 | 2,409 | {"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-2021-21 | latest | en | 0.835461 |
https://www.sanfoundry.com/python-program-implement-insertion-sort/ | 1,701,177,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00415.warc.gz | 1,103,697,015 | 19,736 | # Python Program to Implement Insertion Sort
What is Insertion Sort?
Insertion Sort in Python is basically the insertion of an element from a random set of numbers, to its correct position where it should actually be, by shifting the other elements if required.
Problem Description
Write a Python program to sort a list by using insertion sort.
Insertion Sort Algorithm
```function insertion_sort(arr):
for i in range(1, length(arr)):
key = arr[i]
j = i - 1
while j >= 0 and arr[j] > key:
arr[j + 1] = arr[j]
j = j - 1
arr[j + 1] = key```
The insertion_sort function takes an array as input and performs the insertion sort algorithm. It starts iterating from the second element (i = 1) and compares it with the elements in the sorted portion (j = i – 1) from right to left. If an element is greater than the key, it is shifted one position to the right. Finally, the key is inserted into its correct position within the sorted portion.
Insertion Sort Working Example
Here’s a step-by-step example of how the Insertion Sort algorithm works on the list [2, 4, 1, 5, 8, 0]:
• Step 1: Initial state: [2, 4, 1, 5, 8, 0]
• Step 2: Start with the second element (4) and compare it with the first element (2). Since 4 is greater, no swap is needed.
State: [2, 4, 1, 5, 8, 0]
• Step 3: Move to the next unsorted element (1). Compare it with the sorted portion (2, 4) and insert it in the correct position.
State: [1, 2, 4, 5, 8, 0]
• Step 4: Repeat the process for the next unsorted element (5).
State: [1, 2, 4, 5, 8, 0]
• Step 5: Repeat the process for the next unsorted element (8).
State: [1, 2, 4, 5, 8, 0]
• Step 6: Repeat the process for the next unsorted element (0). Compare it with the sorted portion (1, 2, 4, 5, 8) and insert it in the correct position.
State: [0, 1, 2, 4, 5, 8]
• Step 7: The entire list is now sorted: [0, 1, 2, 4, 5, 8].
By iteratively considering one element at a time and inserting it into the correct position within the sorted portion of the list, the Insertion Sort algorithm gradually builds a sorted list.
Program/Source Code
Here is the source code of a Python program to implement insertion sort. The program output is shown below.
```def insertion_sort(alist):
for i in range(1, len(alist)):
temp = alist[i]
j = i - 1
while (j >= 0 and temp < alist[j]):
alist[j + 1] = alist[j]
j = j - 1
alist[j + 1] = temp
alist = input('Enter the list of numbers: ').split()
alist = [int(x) for x in alist]
insertion_sort(alist)
print('Sorted list: ', end='')
print(alist)```
Program Explanation
1. Define the function insertion_sort that takes a list (alist) as input.
2. Start a loop that iterates from the second element (i = 1) to the end of the list (len(alist)).
3. Assign the value of the current element to a temporary variable (temp).
4. Set the initial index for comparison as the previous element (j = i – 1).
5. Enter a nested loop that continues while the index is greater than or equal to 0 and the temporary value is less than the element at the current index (temp < alist[j]).
6. Within the nested loop, shift the element at the current index one position to the right (alist[j + 1] = alist[j]).
7. Decrement the index by 1 (j = j – 1).
8. After exiting the nested loop, insert the temporary value into its correct position in the sorted portion of the list (alist[j + 1] = temp).
9. Outside the function, prompt the user to enter a list of numbers, which are stored as a string and split into individual elements.
10. Convert the elements from strings to integers using a list comprehension.
11. Call the insertion_sort function, passing the converted list as an argument.
12. Print the sorted list.
Time Complexity: O(n2)
The time complexity of the code is O(n2), where n is the number of elements in the input list.
Space Complexity: O(1)
The space complexity is O(1) since it uses a constant amount of additional space.
Runtime Test Cases
Testcase 1: Here, the elements are entered in random order.
```
Enter the list of numbers: 2 4 1 5 8 0
Sorted list: [0, 1, 2, 4, 5, 8]```
Testcase 2: In this scenario, the elements are entered in reverse sorted order.
```Enter the list of numbers: 5 4 3 2 0 -1
Sorted list: [-1, 0, 2, 3, 4, 5]```
Testcase 3: In this case, the user enters the list of numbers as “3 4 1 4 5 0 7”, and the program will output:
```Enter the list of numbers: 3 4 1 4 5 0 7
Sorted list: [0, 1, 3, 4, 4, 5, 7]```
To practice programs on every topic in Python, please visit “Programming Examples in Python”.
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! | 1,341 | 4,721 | {"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-2023-50 | latest | en | 0.822006 |
http://exxamm.com/QuestionSolution11/Aptitude/The+area+of+a+trapezium+is+336+cm+2+If+its+parallel+sides+are+in+the+ratio+5+7+and+the+perpendicular+distance+b/1780145917 | 1,556,002,083,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578593360.66/warc/CC-MAIN-20190423054942-20190423080942-00448.warc.gz | 66,469,447 | 9,602 | The area of a trapezium is 336 cm^2. If its parallel sides are in the ratio 5 : 7 and the perpendicular distance b
### Question Asked by a Student from EXXAMM.com Team
Q 1780145917. The area of a trapezium is 336 cm^2. If its parallel
sides are in the ratio 5 : 7 and the perpendicular
distance between them is sqrt 4 cm, then the smaller
of the parallel sides is
CDS 2015 Paper-2
A
20 cm
B
22 cm
C
24 cm
D
26 cm
#### HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
#### Access free resources including
• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis | 211 | 747 | {"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} | 2.65625 | 3 | CC-MAIN-2019-18 | latest | en | 0.849037 |
https://www.mathworks.com/matlabcentral/cody/problems/351-back-to-basics-8-matrix-diagonals/solutions/241162 | 1,506,301,958,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690268.19/warc/CC-MAIN-20170925002843-20170925022843-00682.warc.gz | 817,680,763 | 11,556 | Cody
# Problem 351. Back to basics 8 - Matrix Diagonals
Solution 241162
Submitted on 7 May 2013 by J-G van der Toorn
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1:4]; y_correct = [1 0 0 0; 0 2 0 0; 0 0 3 0; 0 0 0 4]; assert(isequal(diag_array(x),y_correct))
``` ```
2 Pass
%% x = [100 23 3.14 12 200]; y_correct = [100.0000 0 0 0 0; 0 23.0000 0 0 0; 0 0 3.1400 0 0; 0 0 0 12.0000 0; 0 0 0 0 200.0000] assert(isequal(diag_array(x),y_correct))
``` y_correct = 100.0000 0 0 0 0 0 23.0000 0 0 0 0 0 3.1400 0 0 0 0 0 12.0000 0 0 0 0 0 200.0000 ``` | 316 | 682 | {"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-2017-39 | latest | en | 0.505701 |
https://brainiak.in/239/b-c-odd-positive-integers-number-integral-solutions-b-c-13 | 1,652,964,382,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00318.warc.gz | 200,179,282 | 8,962 | # If a, b, c are odd positive integers, then number of integral solutions of a + b + c = 13, is
more_vert
If a, b, c are odd positive integers, then number of integral solutions of a + b + c = 13, is
(a) 14 , (b) 21 , (c) 28 , (d) 56
more_vert
verified
The correct answer is option b) 21
explaination ::
Let a = 2k + 1, b = 2l + 1, c = 2m + 1
where klm are whole number
a + b + c = 13
2k + 1 + 2l + 1 + 2z + 1 = 13
or, x + y + z = 5
The number of integrals solutions
$$=^{5+3-1}C_{3-1}=^7C_2$$
$$={7.6\over 1.2}=21$$ | 257 | 562 | {"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} | 3.859375 | 4 | CC-MAIN-2022-21 | longest | en | 0.283171 |
http://support.sas.com/documentation/cdl/en/statug/65328/HTML/default/statug_lifereg_examples05.htm | 1,718,619,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00719.warc.gz | 28,619,610 | 6,199 | ### Example 51.5 Probability Plotting—Right Censoring
The following statements create a SAS data set containing observed and right-censored lifetimes of 70 diesel engine fans (Nelson, 1982):
```data Fan;
datalines;
450 0 460 1 1150 0 1150 0 1560 1
1600 0 1660 1 1850 1 1850 1 1850 1
1850 1 1850 1 2030 1 2030 1 2030 1
2070 0 2070 0 2080 0 2200 1 3000 1
3000 1 3000 1 3000 1 3100 0 3200 1
3450 0 3750 1 3750 1 4150 1 4150 1
4150 1 4150 1 4300 1 4300 1 4300 1
4300 1 4600 0 4850 1 4850 1 4850 1
4850 1 5000 1 5000 1 5000 1 6100 1
6100 0 6100 1 6100 1 6300 1 6450 1
6450 1 6700 1 7450 1 7800 1 7800 1
8100 1 8100 1 8200 1 8500 1 8500 1
8500 1 8750 1 8750 0 8750 1 9400 1
9900 1 10100 1 10100 1 10100 1 11500 1
;
```
Some of the fans had not failed at the time the data were collected, and the unfailed units have right-censored lifetimes. The variable LIFETIME represents either a failure time or a censoring time, in thousands of hours. The variable CENSOR is equal to 0 if the value of LIFETIME is a failure time, and it is equal to 1 if the value is a censoring time. The following statements use the LIFEREG procedure to produce the probability plot with an inset for the engine lifetimes:
```ods graphics on;
proc lifereg data=Fan;
model Lifetime*Censor( 1 ) = / d = Weibull;
probplot
ppout
npintervals=simul;
inset;
run;
ods graphics off;
```
The resulting graphical output is shown in Output 51.5.1. The estimated CDF, a line representing the maximum likelihood fit, and pointwise parametric confidence bands are plotted in the body of Output 51.5.1. The values of right-censored observations are plotted along the bottom of the graph. The Cumulative Probability Estimates table is also created in Output 51.5.2.
Output 51.5.1: Probability Plot for the Fan Data
Output 51.5.2: CDF Estimates
Cumulative Probability Estimates
Probability
Simultaneous 95%
Confidence Limits
Kaplan-Meier
Estimate
Kaplan-Meier
Standard Error
Lower Upper
0.45 0.0071 0.0007 0.2114 0.0143 0.0142
1.15 0.0215 0.0033 0.2114 0.0288 0.0201
1.15 0.0360 0.0073 0.2168 0.0433 0.0244
1.6 0.0506 0.0125 0.2304 0.0580 0.0282
2.07 0.0666 0.0190 0.2539 0.0751 0.0324
2.07 0.0837 0.0264 0.2760 0.0923 0.0361
2.08 0.1008 0.0344 0.2972 0.1094 0.0392
3.1 0.1189 0.0436 0.3223 0.1283 0.0427
3.45 0.1380 0.0535 0.3471 0.1477 0.0460
4.6 0.1602 0.0653 0.3844 0.1728 0.0510
6.1 0.1887 0.0791 0.4349 0.2046 0.0581
8.75 0.2488 0.0884 0.6391 0.2930 0.0980 | 1,118 | 2,529 | {"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} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.673686 |
https://www.clutchprep.com/chemistry/practice-problems/73817/the-equilibrium-constant-for-the-chemical-equation-n2-g-3h2-g-2nh3-g-is-kp-13-0- | 1,611,632,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704795033.65/warc/CC-MAIN-20210126011645-20210126041645-00574.warc.gz | 735,633,016 | 31,157 | # Problem: The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) is Kp = 13.0 at 191°C. Calculate the value of the Kc for the reaction at 191°C. Kc =
88% (10 ratings)
###### Problem Details
The equilibrium constant for the chemical equation
N2(g) + 3H2(g) ⇌ 2NH3(g)
is Kp = 13.0 at 191°C. Calculate the value of the Kc for the reaction at 191°C.
Kc = | 130 | 377 | {"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-2021-04 | latest | en | 0.75421 |
http://www.informatyczna-pomoc.eu/download-pdf-solutions-manual-for-the-dynamics-of-heat-book-by-springer-science-business-media.pdf | 1,477,273,475,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719463.40/warc/CC-MAIN-20161020183839-00506-ip-10-171-6-4.ec2.internal.warc.gz | 512,797,748 | 3,173 | # Solutions Manual for The Dynamics of Heat
4.11 - 1251 ratings - Source
This manual contains detailed solutions of slightly more than half of the end of chapter problems in The Dynamics of Heat. The numbers of the problems includ ed here are listed on the following page. A friend who knows me well noticed that I have included only those problems which I could actually solve myself. Also, to make things more interesting, I have built random errors into the solutions. If you find any of them, please let me know. Also, if you have different ways of solving a problem, I would be happy to hear from you. Any feedback, also on the book in general, would be greatly appreciated. There is an Errata sheet for the first printing of The Dynamics of Heat. By the time you read this, it should be available on the Internet for you to download. A reference to the URL of the sheet can be found in the announcement of my book on Springer's WWWpages (www.springer-ny.com). Winterthur, 1996 Hans Fuchs vi Numbers of Problems Solved Prologue 1, 2, 4, 5, 6, 8, 12, 13, 17, 19, 23, 25, 27, 30, 32, 33, 34, 38, 39, 40, 42, 44, 47, 49, 50, 53, 55, 60, 61, 62 Chapter 1 2, 4, 5, 8, 9, 11, 13, 15, 16, 17, 18, 20, 21, 24, 26, 27, 29, 31, 33, 34, 37, 39, 41, 42, 44, 45, 47, 49, 51, 53, 55, 57, 58, 60, 62 Chapter 2 1, 3, 5, 6, 7, 9, 10, 12, 14, 15, 16, 17, 19, 20, 22, 23, 24, 26, 27, 29, 30, 32, 33, 36, 37, 38, 41, 42, 46, 47, 49 Interlude 2, 3, 4, 5, 6, 8, 10, 11, 12, 13, 18, 19, 20, 21, 23, 24, 28 Chapter 3 2, 4, 6, 8, 10, 12, 15, 16, 17, 18, 22, 24, 25, 28, 30, 31, 35, 36 Chapter 4 1, 2, 4, 6, 8, 9, 11, 12, 13, 15, 18, 20, 21, 22, 25, 27, 28, 29, 30, 31, 33, 34, 35, 39, 40, 43, 44, 46 Epilogue 1, 2, 11 PROLOGUE Solutions of Selected Problems 2 PROLOGUE: Problem 1 Calculate the hydraulic capacitance of a glass tube used in a mercury pressure gauge. The inner diameter of the tube is 8.0 mm.Is) # - = 0 || W/k b) *k, wa towof alalai M., softh Asked h wiAA (, a. hion. availull 9 k. We trull (alalai Mua#39;awsaquot; wival as a raw|| | la fins (a klow), l. H/ki, |awk isa#39; hof equal lo Mt rak ac| 04 { | #!, * of Mu #14, 6 aamp;AMA(6. a#39;s A. * of ... . awa, powt), wi sks haveanbsp;...
Title : Solutions Manual for The Dynamics of Heat Author : Hans U. Fuchs Publisher : Springer Science & Business Media - 2012-12-06 | 938 | 2,325 | {"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} | 3.078125 | 3 | CC-MAIN-2016-44 | latest | en | 0.901642 |
http://psych.fullerton.edu/mbirnbaum/decisions/prior_dom.htm | 1,550,511,957,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487595.4/warc/CC-MAIN-20190218155520-20190218181520-00140.warc.gz | 218,714,159 | 2,824 | ### Decision-Making Experiment: Choices between Gambles: Part 1
```Email address:
Country of Birth (e.g., USA):
Age: You must be over 18 years to participate.
Are you Male or Female?
Female
OR
Male
Education (in years).
If you are a college graduate, put 16.
If you have a Ph.D., put 20.
Education: Years.
```
In each choice, click the button beside the gamble that you would rather play.
Now, look at the first choice below. Would you rather play:
``` A: 50 tickets to win \$100
50 tickets to win \$0 (nothing)
OR
B: 50 tickets to win \$45
50 tickets to win \$35
Think of probability as the number of tickets in a bag containing 100 tickets,
divided by 100. Gamble A has 50 tickets that say \$100 and 50 that say \$0, so
the probability to win \$100 is .50 and the probability to get \$0 is .50. If
someone reaches in bag A, half the time they might win \$0 and half the time
\$100. But in this study, you only get to play a gamble once, so the prize will
be either \$0 or \$100. Gamble B's bag has 100 tickets also, but 50 of them say \$45,
and 50 of them say \$35. Bag B thus guarantees you win at least \$35, but
the most you can win is \$45. Some will prefer A and others will prefer B.
To mark your choice, click the button next to A or B. Notice that the dot
next to No. 1 will empty and fill in the button next to your choice.
For each choice below, click the button beside the gamble you would rather play.
After people have finished their choices, three people will be selected randomly
to play one gamble for real money. One trial will be selected randomly from the
20 trials, and if you were the lucky person, you will get to play the gamble
you chose on the trial selected. You might win as much as \$100. Any one of
the 20 choices might be the one you get to play, so choose carefully.
Winners will be notified by email. Offer void where prohibited by law.
```
```1. Which do you choose?
A: 50 tickets to win \$100
50 tickets to win \$0
OR
B: 50 tickets to win \$35
50 tickets to win \$25
2. Which do you choose?
C: 50 tickets to win \$100
50 tickets to win \$0
OR
D: 50 tickets to win \$45
50 tickets to win \$35
3. Which do you choose?
E: 20 tickets to win \$100
30 tickets to win \$96
50 tickets to win \$50
OR
F: 20 tickets to win \$100
30 tickets to win \$62
50 tickets to win \$50
4. Which do you choose?
G: 10 tickets to win \$108
50 tickets to win \$12
40 tickets to win \$2
OR
H: 10 tickets to win \$108
50 tickets to win \$96
40 tickets to win \$2
5. Which do you choose?
I: 10 tickets to win \$100
90 tickets to win \$0
OR
J: 10 tickets to win \$98
90 tickets to win \$20
6. Which do you choose?
K: 10 tickets to win \$100
90 tickets to win \$55
OR
L: 90 tickets to win \$60
10 tickets to win \$0
7. Which do you choose?
M: 90 tickets to win \$60
10 tickets to win \$22
OR
N: 90 tickets to win \$100
10 tickets to win \$20
8. Which do you choose?
O: 01 tickets to win \$100
99 tickets to win \$0
OR
P: 01 tickets to win \$22
99 tickets to win \$20
9. Which do you choose?
Q: 99 tickets to win \$100
01 tickets to win \$0
OR
R: 10 tickets to win \$22
90 tickets to win \$20
10. Which do you choose?
S: 10 tickets to win \$100
90 tickets to win \$0
OR
T: 90 tickets to win \$90
10 tickets to win \$0
11. Which do you choose?
U: 99 tickets to win \$60
01 tickets to win \$55
OR
V: 01 tickets to win \$100
99 tickets to win \$0
12. Which do you choose?
W: 99 tickets to win \$100
01 tickets to win \$0
OR
X: 99 tickets to win \$22
01 tickets to win \$20
13. Which do you choose?
Y: 10 tickets to win \$100
90 tickets to win \$0
OR
Z: 10 tickets to win \$98
90 tickets to win \$20
14. Which do you choose?
a: 10 tickets to win \$100
90 tickets to win \$55
OR
b: 90 tickets to win \$60
10 tickets to win \$0
15. Which do you choose?
c: 99 tickets to win \$60
01 tickets to win \$55
OR
d: 01 tickets to win \$100
99 tickets to win \$0
16. Which do you choose?
e: 99 tickets to win \$100
01 tickets to win \$0
OR
f: 99 tickets to win \$22
01 tickets to win \$20
17. Which do you choose?
g: 99 tickets to win \$100
01 tickets to win \$0
OR
h: 10 tickets to win \$22
90 tickets to win \$20
18. Which do you choose?
i: 10 tickets to win \$100
90 tickets to win \$0
OR
j: 90 tickets to win \$90
10 tickets to win \$0
19. Which do you choose?
k: 90 tickets to win \$60
10 tickets to win \$22
OR
l: 90 tickets to win \$100
10 tickets to win \$20
20. Which do you choose?
m: 01 tickets to win \$100
99 tickets to win \$0
OR
n: 01 tickets to win \$22
99 tickets to win \$20
-----------
21. Have you ever read a scientific paper
(i.e., a journal article or book) on the theory of
decision making or on the psychology of decision making?
No. Never.
OR
Yes, I have.
Please check to make sure that you have answered all of the Questions.
When you are finished, push this button to send your data:
Thank you!
```
```
``` | 1,460 | 4,866 | {"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-2019-09 | latest | en | 0.9491 |
https://www.lidolearning.com/questions/m-bb-selina7-ch3-ex3c-q12/q12a-motorcycle-runs-km-consum/ | 1,601,168,044,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00302.warc.gz | 923,410,977 | 11,397 | Selina solutions
# Question 12
Q12)A Motorcycle runs 31\frac{1}{4} km consuming 1 litre of petrol. How much distance will it run consuming 1\frac{3}{5} litre of petrol ?
Solution :
It is given that the Distance covered by the motorcycle in 1 litre of petrol = 31\frac{1}{4}km =\frac{125}{4} km
∴ Distance covered by the motorcycle in 1\frac{3}{5} litre of petrol
=\frac{125}{4}\times1\frac{3}{5}
=\frac{125}{4}\times\frac{8}{5}
=\frac{10000}{20}
=50km
Hence,50 km will be covered in 1\frac{3}{5} litre of petrol.
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Chemistry | 271 | 842 | {"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} | 3.421875 | 3 | CC-MAIN-2020-40 | latest | en | 0.734514 |
https://puzzling.stackexchange.com/questions/52079/no-more-kings-puzzle | 1,653,099,726,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00459.warc.gz | 550,861,736 | 67,637 | # No More Kings puzzle
Inspired by the game No More Kings
Rules:
• You start as the black piece.
• Your goal is to capture every piece on the board.
• The king must be the last piece captured.
• Captured pieces are used as the next move.
Example: If a rook captures a bishop, play continues from the position of the bishop with piece movement of the bishop.
Solution:
I didn't have a method. It happened to be the first path I tried.
• Since it was the first path I tried, I wonder if there are more solutions. This would be a cool puzzle type to refine and use the moves to draw a picture of something. May 24, 2017 at 13:58
• This is the cleanest demonstration of the solution so far. In regards to there being more solutions, I figured out the same path by realising only 6 could capture 7. Using retrograde analysis, there was only one piece that could capture 6, etc... So it does seem to be a unique solution. May 24, 2017 at 16:59
• 9 and 10 can only capture each other, so one of them must be the final move. Since 8 can only capture 9, the final three moves therefore must be as above. From there, it's straightforward to trace backward with only one possible move available at each step. May 30, 2017 at 15:04
There's nothing wrong with Forklift's solution, but here's how to be a bit more methodical. First of all
the possible captures form a directed graph (with a bit of subtlety when pieces' possible paths pass through other pieces that may or may not have been captured by the relevant point):
Now there are some captures we can identify with confidence.
The Ne7 can be captured only by the Nc6, after which the only thing it can do is to capture the Rc8. And the Pa3 can only capture the king. If the Rc8 is captured by the Ne7 then it isn't captured by the Bg4, whose only other available capture is the Bf3; it then has nothing to do but capture the Nc6. What captures the Bg4? The only thing that can is the Ne5, so we know how everything begins: so we have the sequence Ne5-Bg4-Bf3-Nc6-Ne7-Rc8. The Rc8 can now capture the Bh8 and nothing else can, and then the Bh8 has nothing to do but capture the Ra1. So Ne5-Bg4-Bf3-Nc6-Ne7-Rc8-Bh8-Ra1.
And now
it's obvious that the only way to finish is Ra1-Bc1-Pa3-Kb4.
• I knew I should have used red arrows. Way better. May 24, 2017 at 15:02
• It's the artistry and sheer geometrical precision in my presentation that really lifts it to the highest level, I think. May 24, 2017 at 15:08
• "bit more methodical" I guess unintelligible mass of red lines and arrows is a bit more methodical. @Forklift Freehand red circles would be even better. May 24, 2017 at 22:47
• lol. Sorry. I meant that comment to be completely in jest, but I don't think I quite got that tone across. May 25, 2017 at 0:32
• @Forklift: The key is freehand red arrows. :-) Jul 9, 2017 at 22:43
Answer using one non-capturing move (pawns aren't pieces in Chess):
... xB (KN6)
... xB
... xR
... xB
... xR
... xB
... -KN4
... xN
... xN
... xK
If you can't make free moes I'm absolutely certain the given answer is the only one.
• Pawns aren't pieces in chess? What would you call them? May 24, 2017 at 17:10
• @DavidStarkey: Pawns are pawns. Pieces are everything else but the kings. May 24, 2017 at 17:27
The exists only one unique solution to this problem. Here is how to solve it-
Draw an arrow moving away from every piece, towards all pieces that the piece can kill. After you are done drawing a the complete network of arrows, connect them in head to tail basis, starting from the knight and ending at the king. There will be clues within on how to proceed, like the pawn can kill only the king, so it must kill the king, etc. | 1,019 | 3,691 | {"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} | 3.828125 | 4 | CC-MAIN-2022-21 | longest | en | 0.961061 |
https://qanda.ai/en/solutions/BXxVF26KKB-nf(x)begincasesdfracx2-1x-1kendcases-when-x1 | 1,623,507,329,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487584018.1/warc/CC-MAIN-20210612132637-20210612162637-00149.warc.gz | 423,094,112 | 13,086 | Symbol
Problem
$3.$ $nf\left(x\right)= \begin{cases} \dfrac {x^{2-1}} {x-1} \\ k \end{cases}$ when $x≠1$ If the function when $x=1$ is given to be continuous at $x=1$ then the value of $k$ is
Physics
Solution
Qanda teacher - Kommineni | 89 | 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": 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} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.391722 |
https://questions.llc/questions/1141002 | 1,716,234,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00438.warc.gz | 426,342,302 | 4,960 | # A car driving cross-country traveled 180 miles in 4 hours. At this rate of speed, how many miles further will the car travel in the next 3 hours?
## To find the number of miles further the car will travel in the next 3 hours, we need to calculate the car's rate of speed.
Given that the car traveled 180 miles in 4 hours, we can find the rate of speed by dividing the distance traveled by the time taken:
Rate of speed = Distance / Time
Rate of speed = 180 miles / 4 hours
Rate of speed = 45 miles per hour
Now we can find how far the car will travel in the next 3 hours using the rate of speed:
Distance traveled = Rate of speed * Time
Distance traveled = 45 miles per hour * 3 hours
Distance traveled = 135 miles
Therefore, your answer of 135 miles is correct. The car will travel an additional 135 miles in the next 3 hours. | 205 | 835 | {"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-2024-22 | latest | en | 0.926913 |
https://scicomp.stackexchange.com/questions/24931/0-hz-quite-sharp-peak-in-fft-and-division-by-0 | 1,718,841,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00176.warc.gz | 439,112,219 | 39,946 | # 0 Hz (quite sharp) peak in FFT and division by 0
In a previous question, link, I asked about how I could most effectively do a Fourier Transform of a radial function given at certain values and which we knew the asymptotical behaviour of. The Fourier transform reading $$\frac{4\pi}{q}\int^{\infty}_0 dr\, r \sin(qr) f(r).$$ I tried several ways and ended up choosing FFT, which approximates it by calculating the DFT over the interval $[0,\Delta r]$ with $N$ points. (I used more particularly the NAG subroutine C06FAF)
I still get some issue around $q=0$. Indeed, as can be seen on the figure , I have some weird peak at very low frequencies. The black curve that is flat in q=0 is the analytical result while other curves are FFT calculations with increasing $\Delta r$. ($\color{blue}{\Delta r} >\color{red}{\Delta r} > {\bf \color{black}{\Delta r}} > \color{magenta}{\Delta r}$ and the dashed one being the highest $\Delta r$). As can be seen, this peak is narrower when $\Delta r$ gets larger.
The question is from where this peak does come ? And how could I get rid of it ? I already tried to take the mean of the function and substract it to the function but it does not change anything.
One subsidiary question is also the following : the subroutine calculates the integral. I then have to divide by $q$ to get this $4\pi/q$ factor. Though, at $q=0$, this can't be done for obvious reasons. So, what can be done instead ?
EDIT : the problem of the wide peak was simply due to the fact that I was doing a bad conversion between $q$ and the frequency from the routine. As far as the problem of the division by $q=0$ is concerned, I'm happy with Endulum's answer.
First, do you really need to divide by $q$? In the problems where these transforms come up, you can often get away with working with $rf(r)$ and $q\hat{f}(q)$ in their own right. Decide if you can put off dividing by $q$ until later on in the problem.
At such a time that you actually do need $q=0$, you can use the fact that $$\lim_{q\to 0} {\sin{(q r)} \over {q r}} = 1$$ for any finite $r$. Thus you have $$\hat{f}(0)\approx \lim_{q\to 0} 4\pi\int_0^R dr~{\sin{(q r)} \over {q r}} r^2 f(r) = 4\pi\int_0^R dr~r^2 f(r)$$ assuming $r^2 f(r)$ is well-behaved on $[0, R]$. | 639 | 2,248 | {"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": 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} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.951955 |
www.erudicat.com | 1,721,254,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00794.warc.gz | 654,513,728 | 35,942 | # PMP formulas you have to know
When taking the PMP exam, you can expect to answer different types of questions: formula-based questions, situation-based questions, definition-based questions, ITTO-based questions, interpretational questions and PMP ethics questions.
Different types of questions need a different approach. For me personally, it's easier to calculate, and formula-based questions were my favorite during the preparation for the exam. I made a list of all formulas used in project management and learned them by heart. I will share this list and also, I will show you my cheat sheet of the formulas that I actually used during the exam.
PV = Planned % Complete * BACWhat the project should be worthPV - planned value
BAC - budget at completion
EV = Actual % Complete * BACWhat the project is worthEV - earned value
BAC - budget at completion
SD = (P - O) / 6SD measures the Variation from AverageSD - standard deviation
P - pessimistic estimate
O - optimistic estimate
EAD = (O + 4R + P) / 6EAD calculated by beta distributionEAD - estimated activity duration
R - realistic estimate
P - pessimistic estimate
O - optimistic estimate
EAD = (O + R + P) / 3EAD calculated by triangular distributionEAD - estimated activity duration
R - realistic estimate
P - pessimistic estimate
O - optimistic estimate
CV = EV - ACCV represents the amount of budget deficit or surplus at a given point in timeCV - cost variance
EV - earned value
AC - actual cost
SV = EV - PVSV aims to measure schedule performance through the difference between the earned value and the planned valueSV - schedule variance
EV - earned value
PV - planned value
CPI = EV / ACCPI measures the cost efficiency of budgeted resourcesCPI - cost performance index
EV - earned value
AC - actual cost
SPI = EV / PVSPI measures schedule efficiencySPI - schedule performance index
EV - earned value
PV - planned value
ETC = EAC - ACETC predicts how much more the remainder of the project will costETC - estimate to complete
EAC - estimate at completion
AC - actual cost
EAC = BAC / CPIPredicts final project costs based on current performanceEAC - estimate at completion
BAC - budget at completion
CPI - cost performance index
EAC = AC + BAC - EVPredicts final project costs based on current performanceEAC - estimate at completion
BAC - budget at completion
AC - actual cost
EV - earned value
EAC = AC + [(BAC - EV) / (CPI * SPI)]Predicts final project costs based on current performanceEAC - estimate at completion
BAC - budget at completion
AC - actual cost
EV - earned value
CPI - cost performance index
SPI - schedule performance index
VAC = BAC - EACVAC is a projection of the amount of budget deficit or surplusVAC - variance at completion
BAC - budget at completion
EAC - estimate at completion
TCPI = (BAC - EV) / (BAC - AC)Predicts likelihood of reaching BACTCPI - to-complete performance index
BAC - budget at completion
EV - earned value
AC - actual cost
TCPI = (BAC - EV) / (EAC - AC)Predicts likelihood of reaching EACTCPI - to-complete performance index
BAC - budget at completion
EV - earned value
AC - actual cost
EAC - estimate at completion
N * (N - 1) / 2Number of communication channelsN - number of team members
Slack = LS - ESDate constraints, within which we can plan our activity without affecting the overall length of the projectLS - late start
ES - early start
Slack = LF - EFDate constraints, within which we can plan our activity without affecting the overall length of the projectLF - late finish
EF - early finish
EF = ES + Duration - 1The earliest you can finish the activity without delaying the project end dateEF - early finish
ES - early start
LS = LF - Duration + 1The latest you can start the activity without delaying the project end dateLS - late start
LF - late finish
EMV = Probability * ImpactEMV represents the expected money to be made from a specific decisionEMV - expected monetary value
ROI = (Net Profit / COI) * 100Measures profit or loss gained through an investment based on the amount investedROI - return on investment
COI - cost of investment
CBR = NPV / initial investment costCBR tells about the profit to be received from an investmentCBR - cost-benefit ratio
NPV - net present value
NPV = Value / (1 + r) ^ tNet present value of investmentValue - value of benefits
r - rate of discount
t - defined time frame
This is the most complete list of formulas I could make for myself. I knew all of them and I definitely recommend you learning them all without exception. However, some of them are used more often than others.
On the PMP exam, it's highly effective to write a cheat sheet as soon as the exam starts. It's a concise set of notes used for quick reference. In my cheat sheet I included the most commonly used formulas. From my personal experience, all the formulas that I needed on the exam were on this list. But this is only my experience and you never know what sorts of questions you will have and what formulas you will have to use. That's why, even though most probably you will use the formulas mentioned in my cheat sheet, you should not limit yourself to only them, but learn all the formulas used in project management.
## PMP Formulas Cheat Sheet
• CV = EV - AC
• SV = EV - PV
• EAC = BAC / CPI
• VAC = BAC - EAC
• CPI = EV / AC
• SPI = EV / PV
• ETC = EAC - AC
• TCPI = (BAC - EV) / (BAC - AC)
• PV = Planned % Complete * BAC
• EV = Actual % Complete * BAC
Author: Yury Shkoda
Date: 4/8/2020 | 1,332 | 5,445 | {"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} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.876629 |
https://studysoup.com/tsg/calculus/141/calculus-early-transcendental-functions/chapter/16949/8-6 | 1,606,647,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00536.warc.gz | 506,042,770 | 10,636 | ×
×
# Solutions for Chapter 8.6: Integration by Tables and Other Integration Techniques
## Full solutions for Calculus: Early Transcendental Functions | 6th Edition
ISBN: 9781285774770
Solutions for Chapter 8.6: Integration by Tables and Other Integration Techniques
Solutions for Chapter 8.6
4 5 0 379 Reviews
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##### ISBN: 9781285774770
Chapter 8.6: Integration by Tables and Other Integration Techniques includes 74 full step-by-step solutions. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions, edition: 6. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9781285774770. This expansive textbook survival guide covers the following chapters and their solutions. Since 74 problems in chapter 8.6: Integration by Tables and Other Integration Techniques have been answered, more than 48023 students have viewed full step-by-step solutions from this chapter.
Key Calculus Terms and definitions covered in this textbook
• Additive inverse of a real number
The opposite of b , or -b
• Associative properties
a + (b + c) = (a + b) + c, a(bc) = (ab)c.
• Categorical variable
In statistics, a nonnumerical variable such as gender or hair color. Numerical variables like zip codes, in which the numbers have no quantitative significance, are also considered to be categorical.
• Center
The central point in a circle, ellipse, hyperbola, or sphere
• Coefficient of determination
The number r2 or R2 that measures how well a regression curve fits the data
• Definite integral
The definite integral of the function ƒ over [a,b] is Lbaƒ(x) dx = limn: q ani=1 ƒ(xi) ¢x provided the limit of the Riemann sums exists
• Degree of a polynomial (function)
The largest exponent on the variable in any of the terms of the polynomial (function)
• Expanded form
The right side of u(v + w) = uv + uw.
• Focal length of a parabola
The directed distance from the vertex to the focus.
• Graph of a function ƒ
The set of all points in the coordinate plane corresponding to the pairs (x, ƒ(x)) for x in the domain of ƒ.
• Horizontal Line Test
A test for determining whether the inverse of a relation is a function.
• Horizontal translation
A shift of a graph to the left or right.
• Identity
An equation that is always true throughout its domain.
• Matrix, m x n
A rectangular array of m rows and n columns of real numbers
• Measure of center
A measure of the typical, middle, or average value for a data set
• n-set
A set of n objects.
• Ordinary annuity
An annuity in which deposits are made at the same time interest is posted.
• Reference triangle
For an angle ? in standard position, a reference triangle is a triangle formed by the terminal side of angle ?, the x-axis, and a perpendicular dropped from a point on the terminal side to the x-axis. The angle in a reference triangle at the origin is the reference angle
• Rigid transformation
A transformation that leaves the basic shape of a graph unchanged.
• Solution set of an inequality
The set of all solutions of an inequality
× | 731 | 3,100 | {"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} | 3.84375 | 4 | CC-MAIN-2020-50 | latest | en | 0.849123 |
https://proofwiki.org/wiki/Category:Product_Rule_for_Derivatives | 1,679,302,576,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00565.warc.gz | 540,330,038 | 11,159 | # Category:Product Rule for Derivatives
Jump to navigation Jump to search
This category contains pages concerning Product Rule for Derivatives:
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x \map k x$.
Then:
$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$
## Subcategories
This category has only the following subcategory.
## Pages in category "Product Rule for Derivatives"
The following 8 pages are in this category, out of 8 total. | 243 | 807 | {"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} | 2.859375 | 3 | CC-MAIN-2023-14 | longest | en | 0.801983 |
https://www.icr.org/article/214/245 | 1,545,128,767,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00312.warc.gz | 932,382,141 | 18,646 | Starlight and the Age of the Universe
The Problem
Light, traveling at 186,000 miles per second, will travel about 6 trillion miles in one year. This distance is called one light-year. There are galaxies that are alleged to be billions of light-years distant from us in space. This means that the light, which left the galaxies 5 billion years ago, should just now be reaching us. This would seem to indicate that the Universe and the creation must be at least 5 billion years old or else we wouldn't be seeing this light. In other words, if the stars were only 6 - 10,000 years old, the light from these distant galaxies would not have even reached us yet.
Four Possible Solutions
(1) Distances in space cannot be accurately measured. Obviously we cannot stretch a string into outer space or measure these distances with a yardstick, and so distances are calculated, rather than measured. This is accomplished by a technique known as triangulation, or parallax. Surveyors use this method using the laws of trigonometry which state that if the baseline and two angles of a triangle are known, then the height of that triangle can be calculated.
Short distances of a few hundred thousand miles can be measured by triangulating the simultaneous observations of observatories on opposite sides of the earth, but as the ratio of the unknown to the known distance increases, the baseline angles become greater and greater, so that beyond an altitude-to-baseline ratio of 28.5 to 1, the angle becomes greater than 89º and must be further divided into minutes and seconds of arc. The limitations of this method are evident even within our own solar system, as the apex angle to our sun would be only 10 seconds of arc (1/360 of a degree). The distances to even the nearest stars are so great that a greater triangulational baseline is needed, and so the earth's orbit around our sun is used, allowing a baseline of about 186 million miles. Sightings are taken 6 months apart, the angles are compared, and the distance is computed with trigonometry.
Because the distances to the stars are so great, the sides of the triangle are virtually perpendicular and so only the nearest stars (up to about 200 light-years) can be measured by this technique. For example, our sun is 8 light minutes from us, so the baseline of the triangle would be 16 light minutes. But our nearest star, Alpha Centauri, is 4 1/2 light-years or 2,365,000 light minutes from earth, for a ratio of approximately 148,000 to 1. At that ratio, an 8 1/2" line drawn across this page would have the apex of its triangle 20 miles away!
Greater distances are determined by the presumed sizes and intensity of stars, red shift, and many questionable factors which may have nothing whatever to do with distance. 1 In fact, some astronomers feel that it is possible that the entire universe could fit into an area within a 200 light-year radius from the earth! Therefore, there is no guarantee that the actual distances in space are as great as we have been told, and light from the farthest point in the universe could have reached us in only a few hundred years.
(2) Light may take a "shortcut" as it travels through space. This is difficult to illustrate, but suffice it to say that there are two concepts of the "shape" of outer space. One is that it is straight-line (Euclidean), and the other is that it is curved (Riemannian). Based on observations of 27 binary star systems, it appears that light in deep space travels in curved paths on Riemannian surfaces. 2
The formula for converting straight-line to curved space is:
where r is the Euclidean or straight-line distance, and R is the radius of curvature of Riemannian space. Using this formula, and a radius of curvature of 5 light-years for Riemannian space, the time for light to reach us from points in our own solar system is practically the same for either Euclidean or Riemannian distances, and there is not much of a change even out to the nearest star (4 1/2 light-years). But if we insert an infinite Euclidean distance for the farthest conceivable star, it would take only 15.71 years for light to reach us from that distance! The following table gives an idea of the distance-to-time conversions:
Euclidean Distance
in light years
Riemannian Distance
in Actual Time
1
.997 years
4
3.81
30
12.5
100
14.7
1,000
15.6
10,000
15.7
infinite
15.71
Notice that, by the nature of the formula, the upper limit in time in Riemannian space has a definite limit, and even if the radius of curvature is modified by new discoveries, it will never get very large.
Conclusion: It appears that light may take a "shortcut" as it travels through deep space, and even if we grant that the uniformitarian distances are valid, the time for light to get here from the uttermost part of the universe would be only about 15 years.
(3) It is possible that the speed of light was considerably faster in the past.
There are different approaches to this theory. One suggestion is that the speed of light has been slowing consistently over the last 300 years, which extrapolates to a speed 5 x 1011 (500 billion) times faster 6000 years ago. 3 If this is true, light from a 5 billion light-year star (assuming the distances actually are that great) would have reached us in 3 days! Another suggestion is that the speed of light at the time of the Creation was infinite, and that a "shock wave" went out from the earth at the time of the Curse, slowing the speed of light down to its present value. 4 Light currently being emitted from these distant stars would take a long time to reach us, but at the time of the creation it would have arrived here almost instantaneously. Yet another suggestion is that the permittivity and permeability of free space changed at the time of the Flood, slowing the speed of light by a factor of 513 from its previous value. 5
Conclusion: Creationists differ on their interpretations of how, when, and by what factor the speed of light changed in the past, but there does seem to be some agreement that this is a real possibility. This third factor can either be used independently or in conjunction with the first two points. In other words, it may not only be that the distances are smaller than commonly claimed, but it may also be true at the same time that light travelled faster in the past and took a "shortcut" as well.
(4) There are Biblical indications that the earth and the universe were created with the appearance of age. There are several examples of this:
The plants (Gen. 1: 11-12)—They were created mature and bearing fruit at the moment of their creation. What would have taken years to accomplish by uniformitarian processes took place in seconds.
The animals (Gen. 1:20-25)—Fish, birds, and the three categories of land animals were created fully mature, having the appearance of age, and were immediately capable of reproduction on the first day of their existence (v. 22). The Bible therefore allows us to answer the otherwise-unanswerable question: "Which came first, the chicken or the egg?" Evolutionists would take the endless trail back to the first life forms, but the Creationist can say that the chicken was created first, and then laid its eggs.
Man (Gen. 2:7)—Adam was created as an adult, with an inherent storehouse of knowledge and vocabulary, and was capable of articulate speech and reproduction on the first day of his existence (Gen. 1:28-29; 2:8, 16-20, 24). Whether Adam gave the appearance of being 20 or 50 years of age is irrelevant—a person walking into Eden five minutes after Adam's creation would have been able to converse intelligently with him and would probably conclude, on uniformitarian assumptions, that Adam had been around for many years.
Eve (Gen. 2:21-23)—Likewise, Eve was created fully mature and ready for marriage to Adam immediately (Gen. 1:27-28; 2:22-25).
The stars (Gen. 1:14-19)—The sun, moon, and stars were created on the fourth day of the creation week. Individually and collectively they were to have different functions: dividing the day from the night, serving as navigational aids, as chronological indicators, for illuminating the earth, as well as for declaring the glory of God (Psalm 19:1). What is not often noticed is that "it was so" on the very day of their creation (Gen. 1:15). Granted, the Biblical word "star" (Heb: kokab; Gr: aster) is a broader term than our English usage of "star" as an energy source, and includes just about anything in space, but the point is that the stars—and the nearest is 4 1/2 light-years distant—were seen on the first day of their existence. This means that even if the distances are correct, the stars would merely have given the appearance of having been here longer. Therefore, the stars and the light beams connecting them visually to the Earth were both created at the same time.
This concept raises several questions. First, does this not mean that God—like some magician—is intentionally deceiving us by making things appear to be older than they actually are? The question really goes back to the matter of intent: did God intend to fool us, or did He intend primarily to make things fully functional but we are fooled only because we view them with certain uniformitarian assumptions? Therefore, while it is true that the earth and the universe were created with the appearance of age, I think we do better to speak of the creation of a fully functional universe that, as a secondary feature, merely gives the appearance of age.
There are two red herrings that are usually drawn across the trail at this point of the discussion: "Were the trees created with rings?" and, "Did Adam have a navel?" First, the question about the tree rings depends partially on the purpose for which God originally intended the tree rings. We have observed this phenomenon and have noticed a correlation between the number of rings and the age of the tree, but chronology may be a secondary function of the rings, as indicated by certain trees that grow several rings in one year, or none at all, depending on weather conditions. Another possibility here is that God could have accelerated natural processes, so that He did in seconds what would normally have taken years to accomplish. Whether the trees had rings or not, the fact should not be overlooked that they were indeed mature and bearing fruit before the day was over (Gen. 1:11-12).
Now, did Adam have a navel? Probably not, simply because the navel is a remnant of the normal birth process, and not of maturity or functionality. Therefore, for God to have created him with one would indeed be deception, because it would convey misleading information.
The Biblical concept of a fully-functional creation of materials that give the appearance of age can be extended as well to the New Testament. The creation of wine in John 2:1-10 is one example. The fact that the water was not created ex nihilo, but was already in existence is beside the point. Wine is chemically greater than water, and so it does involve a certain amount of ex nihilo creation, and furthermore it gave the appearance of having been aged (John 2:8-10). So it is with the creation of the bread and the fish in John 6:1-13. Normally, bread and fish are the end result of a long growth process before they are finally prepared for consumption. The bread and the fish were created with the appearance of maturity, not because Jesus was doing parlor magic and trying to deceive His audience, but because He wanted the materials to be functional and immediately usable.
Conclusion
There are three "secular" or non-Biblical possibilities to the problem of harmonizing a young universe with the allegedly-great distances of the outer galaxies: (1) the distances may not be that great after all; (2) light may take a "shortcut" as it travels through deep space; (3) the speed of light may have been considerably faster in the past. These three are not mutually exclusive, and may in fact be used in conjunction with each other. The fourth solution, which may be used independently or in conjunction with the above three, is that God created the light beams as well as the stars so that they could be—as indeed they were—seen on the fourth day of the creation week.
References
l. Many astronomical assumptions are like a house of cards. For example, Hubble's constant, a mainstay in uniformitarian distance calculations, has recently been determined to be considerably different than previously thought, thus resulting in the overnight "collapse" of the size of the universe to 50% of its former value.
2. Harold S. Slusher, "Travelling of Light in Space," in Age of the Cosmos (San Diego: ICR, 1980), pp. 25-37. For an additional view on the hyperbolic or non-Euclidean geometry of space, see Wayne M. Zage, "The Geometry of Binocular Visual Space," Mathematics Magazine 53 (Nov. 1980), pp. 289-293.
3. Barry Setterfield, "The Velocity of Light and the Age of the Universe," Ex Nihilo 1 (1982): 52-93.
4. David M. Harris, "A Solution to Seeing Stars," Creation Research Society Quarterly, 15 (Sep. 1978): 112-115. He also offers an explanation for red-shift that does not require an expanding universe.
5. Gletin R. Morton, "Electromagnetics and the Appearance of Age," Creation Research Society Quarterly 18 (Mar. 1982): 227-232.
* Richard Niessen is Associate Professor of Bible & Apologetics at Christian Heritage College, El Cajon, CA 92021.
Cite this article: Richard Niessen. 1983. Starlight and the Age of the Universe. Acts & Facts. 12 (7).
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# Region R is defined as the region in the first quadrant satisfying the
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Region R is defined as the region in the first quadrant satisfying the [#permalink]
Given 3x + 4y < 12
In Slope Intercept Form: y < -(3/4)x + 3
The Region represented by this inequality is all the Area UNDERNEATH the downward sloping line BUT ONLY in quadrant 1.
This region described above underneath the Line 3x + 4y = 12 ends up forming a Right Triangle with:
Y Axis from (0 , 0) up to (0 , 3) as One Leg
and
X Axis from (0 , 0) up to (4 , 0) as Second Leg
The Region in which the X-Coordinate of Point P (r) is > 2 is defined by the Region to the RIGHT of the Vertical Line X = 2 but INSIDE the Right Triangle.
Since this Vertical Line divides the Whole Right Triangle into 2 Similar Triangles:
Probability = (Area of Smaller Right Triangle) / (Area of ENTIRE Right Triangle)
Corresponding Side of Smaller Triangle / Corresponding Side of ENTIRE Triangle = 2/4 = 1/2
Therefore, because the triangles are Similar Triangles:
(Area of Smaller Triangle) / ( Area of ENTIRE Triangle) = (1)^2 / (2)^2 = 1 /4
Probability = 1/4
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Re: Region R is defined as the region in the first quadrant satisfying the [#permalink]
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Re: Region R is defined as the region in the first quadrant satisfying the [#permalink]
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https://desklib.com/document/question-1-the-researcher-is-interested/ | 1,716,160,281,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00473.warc.gz | 173,082,369 | 36,033 | # Research On Normal Distribution Of Energy & Vitamin C Intake
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QUESTION 11.The researcher is interested to know if energy intake (ENERGY) and vitamin C intake (VITC) have a Normal distribution. Use the following Table as a guide.MeasuresCriteria/Cut-off points for supporting normalityHistogramSymmetrical, bell-shaped curveBoxplotMedian in the centre of the box with whiskers at equal length at both ends of the box and no outliersNormal Q-Q plotMost observations appear on the straight lineSkewness and kurtosis coefficientSTATA users: Skewness and (kurtosis-3) are between -1 and 1; (SPSS users: Skewness and kurtosis are between -1 and 1) 2.3.Which of the following would be appropriate?a.ENERGYandVITCboth have a Normal distribution, and hencea natural logarithm transformation is not necessary for bothENERGYandVITCb.ENERGYandVITCboth do not have a Normal distribution, and hencea natural logarithm transformation is necessary for bothENERGYandVITCc.ENERGYhas a Normal distribution andVITChas a right (positively) skewed distribution, and hencea natural logarithm transformation is necessaryonlyforVITCd.ENERGYhas a Normal distribution andVITChas a left (negatively) skewed distribution, and hencea natural logarithm transformation is necessaryonlyforENERGYe.None of the above1 points QUESTION 21.Based on previous question, you now understandwhether the variablesENERGYandVITCare normally distributed. What should be the most appropriate measures of centrality and variability to report for variableENERGYandVITC? (Hint: different measures of centrality and variability need to be reported for data that have a Normal or a skewed distribution).a.Mean and standard deviation forENERGY. The reason is that variableENERGYhas a normal (symmetric) distributionb.Median and interquartile range forVITC. The reason is that variableVITCdoes not have a normal distribution but a skewed distributionc.Median and interquartile range forENERGY. The reason is that variableENERGYhas a normal (symmetric) distributiond.Mean and standard deviation forVITC. The reason is that variableVITCdoes not have a normal distribution and it has a skewed distributione.Answer (a) and (b) are correctf.Answer (c) and (d) are correct1 points QUESTION 31.Obtain summary statistics forvariableENERGY.Which of the following isNOTCORRECT?a.The sample mean energy intake of these children is 4764.879 kJ
b.There were 50% of the children, whose energy intake is higher than 4804.95 kJ in this samplec.There was no any child in this sample whose energy intake lower than 2816.8 kJd.The sample standard deviation ofenergy intakeis 850.676kJ, then we can concludethat99% of the children in this samplewhose energy intakeranged from 3063.528(i.e., mean-2*SD) to 6466.232 (i.e., mean+2*SD)kJ1 points QUESTION 41.Obtain 95% confidence interval forvariableENERGY.Which of the following statement is correct about the estimation of the mean energy intake in the population of the2 to 3 year old children in WA?a.According to the sample information, the average daily energy intake in the population of the2 to 3 year old children in WA was estimated to be between 4555.757 and 4974.001 kJb.Based on the sample information, we are 95% confident that the sample mean daily energyintake of 2 to 3 year old children was between 4555.757 and 4974.001 kJc.Based on the sample information, the mean daily energy intake in the population of the 2 to 3 year old children in WA was estimated with 95% certainty to be between 4555.757 and 4974.001 kJd.Based on the sample information, there are 95% of the 2 to 3 year old children in WA population having a daily energy intake between 4555.757 and 4974.001 kJe.None of the above is correct1 points QUESTION 51.Which of the following statement related to confidence interval is correct?a.If the sample size of this study increased from 66 to 660, we will expect the 95% CI to becomewider as there is a larger variation now with a larger sample sizeb.If the sample size of this study increased from 66 to 660, we will expect the 95% CI to becomenarrower and be more precise than when the sample size was 66c.The higher the confidence levels (e.g. from 90% to 95% to 99%), the more confident we are about capturing the actual population parameter and therefore the correspondinglengths of the CIs tend to be shorterd.The higher the confidence levels (e.g. from 90% to 95% to 99%), the more confident we are about capturing the actual population parameter and therefore the corresponding CIs tend tobe widere.Answers(b) and (d) are both correct1 points QUESTION 61.The dietician now wants to investigate whether there is an association in energy intake between the children who live in the country and those that live in the city. You need to first recode the variableENERGYinto a categorical variableENERGYCataccording to the following table(Hint: Give the recoded variable a new name and rememberto assign value labels to the new recoded variable).Values of theoriginal variableENERGY to be recodedinto following levelsValues of thenew recodedvariableENERGYCatLess than or equal to 4500 kJ (<= 4500)1
Greater than 4500 kJ & less than or equal to 5000 kJ (>4500 & <=5000)2Greater than 5000 kJ (>5000)3Which of the following would be appropriate to describe the frequency distribution of this categorical variableENERGYCat?a.Frequency and percentage, and the percentage of kids having energy intakegreater than 4500 kJ and less than or equal to 5000 kJ (>4500 & <=5000) is 30.30% (n=20)b.Mean (=2.06) and standard deviation (=0.839)c.Median (=2) and interquartile range (=2)d.Skewness (=-0.114) and kurtosis (=1.451)e.Minimum (=1) and maximum (=3)1 points QUESTION 71.Based on the new variable you recoded inQuestion 6, obtain a cross-tabulationofENERGYCatandLOCATION.Which of the following statements is appropriate to describe the levels of energy intake between the children who live in city and country?a.More than half of the boys (53.85% of them) lived in city while slightly more of the girls (55.56% of them) lived in city toob.Of the children who lived in country, more of them tend to have daily energy intake equal or less than 4500 kJ (40.00%), compared to those who lived in city (25.00%)c.Half (50%) of the country children had daily energy intake between 4500 to 5000 kJ, and the percentage is the same for city childrend.Children who lived in city were more likely to have total daily energy intake >5000 kJ than those kids who lived in country (City 47.22% vs Country 26.67%)e.Both (b) and (d) are correct1 points QUESTION 81.Assuming all relevant assumptions are met, how would you test whether there is any association between the levels of energy intake of childrenENERGYCatand the location they livedLOCATION?a.Pearson Correlation Coefficient can be used asENERGYandLOCATIONboth are continuousb.Chi-square test can be used asENERGYCat andLOCATIONboth are categoricalc.An independent (two samples) samples t-test can be used asENERGYis continuous,andLOCATIONis categorical having two levelsd.One-way ANOVA is suitable for this research question asLOCATIONis continuousand ENERGYCatis categorical with 3 levelse.Paired samples t-test is fine to answer this research question, asLOCATION and ENERGYCatare repeated variables1 points QUESTION 9
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Jan 16, 2017
See full solution process below:
#### Explanation:
First, we need to factor the quadratic equation:
$\left(x - 7\right) \left(x + 1\right)$
Now we need to solve each term for $0$:
Solution 1)
$x - 7 = 0$
$x - 7 + 7 = 0 + 7$
$x - 0 = 7$
$x = 7$
Solution 2)
$x + 1 = 0$
$x + 1 - 1 = 0 - 1$
$x + 0 = - 1$
$x = - 1$
The solution is:
$x = 7$ and $x = - 1$ | 176 | 440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-34 | latest | en | 0.638224 |
https://sciencing.com/temperature-physics-definition-formula-examples-13722755.html | 1,656,493,033,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00170.warc.gz | 554,421,588 | 88,878 | # Temperature (Physics): Definition, Formula & Examples
Print
## Heat
You may already have an intuitive sense that temperature is a measure of the "coldness" or "hotness" of an object. Many people are obsessed with checking the forecast so they know what the temperature will be for the day. But what does temperature really mean in physics?
## Definition of Temperature
Temperature is a measure of average kinetic energy per molecule in a substance. It is different from heat, although the two quantities are intimately related. Heat is the energy transferred between two objects at different temperatures.
Any physical substance to which you might attribute the property of temperature is made of atoms and molecules. Those atoms and molecules do not stay still, even in a solid. They are constantly moving and jiggling around, but the motion happens on such a small scale, that you can’t see it.
As you likely recall from your study of mechanics, objects in motion have a form of energy called kinetic energy that is associated with both their mass and how fast they are moving. So when temperature is described as average kinetic energy per molecule, it is the energy associated with this molecular motion that is being described.
## Temperature Scales
There are many different scales by which you might measure temperature, but the most common ones are Fahrenheit, Celsius and Kelvin.
The Fahrenheit scale is what those who live in the United States and a few other countries are most familiar with. On this scale water freezes at 32 degrees Fahrenheit, and the temperature of boiling water is 212 F.
The Celsius scale (sometimes also referred to as centigrade) is used in most other countries around the world. On this scale the freezing point of water is at 0 C and the boiling point of water is at 100 C.
The Kelvin scale, named for Lord Kelvin, is the scientific standard. Zero on this scale is at absolute zero, which is where all molecular motion stops. It is considered an absolute temperature scale.
## Converting Between Temperature Scales
To convert from Celsius to Fahrenheit, use the following relationship:
T_F = \frac{9}{5}T_C + 32
Where TF is the temperature in Fahrenheit, and TC is the temperature in Celsius. For example, 20 degrees Celsius is equivalent to:
T_F = \frac{9}{5}20 + 32 = 68\text{ degrees Fahrenheit.}
To convert in the other direction, from Fahrenheit to Celsius, use the following:
T_C = \frac{5}{9}(T_F - 32)
To convert from Celsius to Kelvin, the formula is even simpler because the increment size is the same, and they just have different starting values:
T_K=T_C+273.15
#### Tips
• In many expressions in thermodynamics, the important quantity is ΔT (the change in temperature) as opposed to the absolute temperature itself. Because the Celsius degree is the same size as an increment on the Kelvin scale, ΔTK = ΔTC, meaning these units can be used interchangeable in those cases. However, anytime an absolute temperature is required, it must be in Kelvin.
## Heat Transfer
When two objects at different temperatures are in contact with each other, heat transfer will occur, with heat flowing from the object at the higher temperature to the object at the lower temperature until thermal equilibrium is reached.
This transfer occurs due to collisions between the higher-energy molecules in the hot object with the lower-energy molecules in the cooler object, transferring energy to them in the process until enough random collisions between molecules in the materials have occurred that the energy becomes equally distributed between the objects or substances. As a result, a new final temperature is achieved, which lies between the original temperatures of the hot and the cool objects.
Another way to think of this is that the total energy contained in both substances eventually becomes equally distributed between the substances.
The final temperature of two objects at different initial temperatures once they reach thermal equilibrium can be found by using the relationship between heat energy Q, specific heat capacity c, mass m and the temperature change given by the following equation:
Q = mc\Delta T
Example: Suppose 0.1 kg of copper pennies (cc = 390 J/kgK) at 50 degrees Celsius are dropped into 0.1 kg of water (cw = 4,186 J/kgK) at 20 degrees Celsius. What will the final temperature be once thermal equilibrium is achieved?
Solution: Consider that the heat added to the water from the pennies will equal the heat removed from the pennies. So if the water absorbs heat Qw where:
Q_w = m_wc_w\Delta T_w
Then for the copper pennies:
Q_c=-Q_w = m_cc_c\Delta T_c
This allows you to write the relationship:
m_cc_c\Delta T_c=-m_wc_w\Delta T_w
Then you can make use of the fact that both the copper pennies and the water should have the same final temperature, Tf, such that:
\Delta T_c=T_f-T_{ic}\\\Delta T_w=T_f-T_{iw}
Plugging these ΔT expressions into the previous equation, you can then solve for Tf. A little algebra gives the following result:
T_f = \frac{m_cc_c T_{ic}+m_wc_w T_{iw}}{m_cc_c+m_wc_w}
Plugging in the values then gives:
Note: If you're surprised that the value is so close to the water's initial temperature, consider the significant differences between the specific heat of water and the specific heat of copper. It takes a lot more energy to cause a temperature change in water than it does to cause a temperature change in copper.
## How Thermometers Work
Old-fashioned glass-bulb mercury thermometers measure temperature by making use of the thermal expansion properties of mercury. Mercury expands when warm and contracts when cool (and to a much larger degree than the glass thermometer which contains it does.) So as the mercury expands, it rises inside the glass tube, allowing for measurement.
Spring thermometers – those that usually have a circular face with a metal pointer – also work off of the principle of thermal expansion. They contain a piece of coiled metal that expands and cools based on temperature, causing the pointer to move.
Digital thermometers make use of heat-sensitive liquid crystals to trigger digital temperature displays.
## Relationship Between Temperature and Internal Energy
While temperature is a measure of the average kinetic energy per molecule, internal energy is the total of all of the kinetic and potential energies of the molecules. For an ideal gas, where potential energy of the particles due to interactions is negligible, the total internal energy E is given by the formula:
E = \frac{3}{2}nRT
Where n is the number of moles and R is the universal gas constant = 8.3145 J/molK.
Not surprisingly, as temperature increases, thermal energy increases. This relationship also makes it clear why the Kelvin scale is important. The internal energy should be any value 0 or greater. It would never make sense for it to be negative. Not using the Kelvin scale would complicate the internal energy equation and require the addition of a constant to correct it. The internal energy becomes 0 at absolute 0 K.
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We Have More Great Sciencing Articles! | 1,534 | 7,182 | {"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} | 3.828125 | 4 | CC-MAIN-2022-27 | longest | en | 0.957186 |
http://woodenbooks.com/browse/qed/QED.00025.php | 1,545,203,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00276.warc.gz | 304,944,731 | 3,882 | # Q.E.D.
42 43 the numbers oF nature the geometry of growth A spiral of squares growing around a unit square as on the top left consists of squares the lengths of whose sides are the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, ..., named after Leonardo Fibonacci 1170 1250. Every number in the sequence is the sum of the two preceding it, so that 2 1 1, 3 1 2, 5 2 3, and so on. Fibonacci numbers are connected in many wonderful ways. For example, the tiling of the rectangle on the top left demonstrates that 12 12 22 32 52 82 132 13.13 8 13.21. In general, the sum of the squares of the first n Fibonacci numbers equals the product of the nth and n1st such numbers. Similarly, the tiling of the square on the right shows that 1.1 2.1 3.2 5.3 8.5 13.8 21.13 212. This equality also generalises easily. The Fibonacci numbers often show up in the same phenomena as the golden ratio f see previous page and it can be proved that the nth Fibonacci number is the closest natural number to f n5. This implies that the rectangles we come across when building our spiral of squares become indistinguishable from golden rectangles. Fibonacci numbers are hidden in many growth processes. For example, the numbers of clockwise and counterclockwise spirals apparent in sunflower heads opposite, middle are usually consecutive Fibonacci numbers. Pascals triangle lower, opposite also grows, here row by row, with neighboring entries in one row adding up to the number below them. Since the sums of the first two diagonals of this triangle are both 1, and the sums of any two consecutive diagonals add up to the sum of the next, our golden sequence will appear yet again. | 417 | 1,659 | {"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} | 3.53125 | 4 | CC-MAIN-2018-51 | latest | en | 0.905701 |
https://mathpoem.wordpress.com/2012/01/17/157/ | 1,498,745,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329344.98/warc/CC-MAIN-20170629135715-20170629155715-00069.warc.gz | 780,589,770 | 38,215 | Given two integers n and m, their Greatest Common Divisor or GCD is defined as the largest number that divides into both n and m. So, given 6 and 35, for example, we have that the GCD(6,35) = (2·3,5·7) = 1, and likewise, the GCD of 121 and 44, or GCD(121,44) = GCD(11·11,4·11)= 11 because 11 is the largest number that divides into both 121 and 44.
But there is a very unexpected application of the GCD going on every second inside our own heads, as Manfred Schroder* goes on to say:
“An interesting and most surprising application of the GCD occurs in human perception of pitch: the brain, upon being presented with a set of harmonically related frequencies, will perceive the GCD of these frequencies as the pitch. Thus, the subjective pitch of the two-tone chord (320 Hz and 560 Hz) is (320,560) = 80Hz, and not the difference frequency (240 Hz).
Upon a frequency shift of +5Hz applied to both frequencies, the GCD drops to 5 Hz; and for an irrational frequency shift, the GCD even drops to 0 Hz. But that is not what the ear perceives as the pitch. Rather it tries to find a close match in the range of pitches above 50 Hz. For the frequencies 325 Hz and 565 Hz such a match is given by 81 Hz, which is the GCD of 324 Hz and 567 Hz – close to the two given frequencies.
Note that the concept that pitch is given by the difference frequency or “beat” frequency has been beaten: if both frequencies are shifted by the same amount, their difference remains unchanged. Yet psycho-acoustic experiments clearly show that the perceived pitch is increased, from 80 Hz to about 81 Hz in our example, just as the amplified GCD model predicts.
What this tells us is that the human brain switches on something like a GCD spectral matching computer program when listening to tone complexes. Fascinating? Indeed. Unbelievable? Well, the brain has been caught doing much trickier things than that.”
[ Technical Notes ]
The GCD appears in the solution to many, seemingly unrelated, problems. For example, take n jugs with capacities of L1, L2, . . . , Ln liters. What amounts k of water (or wine) can be dispensed by these n/1 jugs?
Answer: k must be a multiple of the GCD. Now, to better understand the GCD and its counterpart, the LCM, we will need more math.
Fundamental theorem of arithmetic. Each natural number n > 1 is either a prime or can be uniquely factored into primes
$n=p_{1}^{{{e}_{1}}}p_{2}^{{{e}_{2}}}p_{3}^{{{e}_{3}}}...p_{k}^{{{e}_{k}}}...p_{r}^{{{e}_{r}}}=\prod\limits_{i}{p_{i}^{{{e}_{i}}}}$
where the order of the prime factors makes no difference.
Proof. There are only two cases to consider. Either n is prime or n is composite. If n is prime, then the theorem is obviously true, by definition. Therefore, the only case we need to consider is when n is composite.
If n is composite, then there exists an integer p where p divides into n so that and 1 < p < n must be true because n is, by definition, greater than 1, and so if p divides into n, then p must be smaller than n, hence 1 < p < n as just stated. [this proof is still being written, and will be completed by the Jan 25 weekend]
It is extremely relevant to note that because there is no corresponding theorem for the additive decomposition of natural numbers into primes, this lack explains why additive number theory, when dealing with things like partitions, is so difficult.
Now, two integers n and m have a least common multiple (LCM) [n,m], and we need this every time we want to add two fractions, because we then need to find the LCM to combine two fractions with denominators n and m into a single fraction, and that is where the expression least common denominator comes from.
So, the LCM of 6 and 9 is LCM[6,9] = 18 so that:
$\frac{1}{6}+\frac{2}{9}=\frac{3}{18}+\frac{4}{18}=\frac{7}{18}$
So that we have, in general:
Definition. The Lowest Common Multiple or LCM of two integers n and m is defined as:
${}[n,m]=\prod\limits_{i}{p_{i}^{\max ({{e}_{i}},{{f}_{i}})}}$
and likewise, given two integers, we have:
Definition. the Greatest Common Divisor or GDC is defined as:
${}(n,m)=\prod\limits_{i}{p_{i}^{\min ({{e}_{i}},{{f}_{i}})}}$
and there are interesting properties between the GDC and LCM, such as that, for any two numbers n and m, the product of the GCD and the LCM equals the
product of n and m.
$(n,m)[n,m]=nm$
because whenever the GCD picks the exponent e of i for p of i, the LCM picks the exponent f of i, and vice versa and so:
$(n,m)[n,m]=\prod\limits_{i}{p_{i}^{{{e}_{i}}+{{f}_{i}}}=nm}$
and this works for three or more integers as well
$(n,m,k)[n,m,k]=nmk$
$(n,m,k,j)[n,m,k,j]=nmkj$
$(n,m,k,j,l)[n,m,k,j,l]=nmkjl$
Also, the direct result of the Min and Max functions in the definition of the GCD and LCD yields a “distributive law”
$(k[n,m])=[(k,n),(k,m)]$
$[k(n,m)]=([k,n],[k,m])$
And a beautiful “self-duel” relationship
$([k,n],[k,m],[n,m])=[(k,n),(k,m),(n,m)]$
Tuesday 17 January 2012 | 1,364 | 4,926 | {"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": 12, "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-2017-26 | longest | en | 0.953294 |
https://www.numbersaplenty.com/51213624 | 1,621,186,072,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00180.warc.gz | 964,274,689 | 3,549 | Search a number
51213624 = 233721137107
BaseRepresentation
bin1100001101011…
…1010100111000
310120100220212220
43003113110320
5101102313444
65025404040
71161210600
oct303272470
9116326786
1051213624
11269aa610
1215199620
13a7c1927
146b31c00
154769619
hex30d7538
51213624 has 192 divisors, whose sum is σ = 168428160. Its totient is φ = 12821760.
The previous prime is 51213601. The next prime is 51213641. The reversal of 51213624 is 42631215.
51213624 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (24).
It is a junction number, because it is equal to n+sod(n) for n = 51213594 and 51213603.
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 478579 + ... + 478685.
It is an arithmetic number, because the mean of its divisors is an integer number (877230).
Almost surely, 251213624 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 51213624, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (84214080).
51213624 is an abundant number, since it is smaller than the sum of its proper divisors (117214536).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
51213624 is a wasteful number, since it uses less digits than its factorization.
51213624 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 178 (or 167 counting only the distinct ones).
The product of its digits is 1440, while the sum is 24.
The square root of 51213624 is about 7156.3694706185. The cubic root of 51213624 is about 371.3600397089.
Adding to 51213624 its reverse (42631215), we get a palindrome (93844839).
The spelling of 51213624 in words is "fifty-one million, two hundred thirteen thousand, six hundred twenty-four". | 617 | 2,187 | {"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} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.871961 |
https://muzic-ivan.info/what-is-the-percent-composition-of-oxygen-in-h2o/ | 1,656,112,810,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00764.warc.gz | 456,520,710 | 5,330 | Explanation:Mass Percent of OxygenThe second method is the exact same as for finding the mass percent of hydrogen. From previous calculations, you know the total molar mass of oxygen in water is 16.00. Divide 16.00 by the full molar mass of water, 18.016, to gain 0.8881. Multiply by 0.8881 by 100 to acquire the percentage: 88.81 percent."},"id":19872662,"content":"
## The mass percent of oxygen in water is 88.88 %Explanation:To calculate the mass percent of oxygen in water, we usage the equation:\textMass percent of oxygen=\frac\textMass of oxygen\textMass of water\times 100Mass of oxygen = 16 gMass of water = <(2 u00d71) + 16> = 18 gPutting values in over equation, we get:\textMass percent of oxygen=\frac16g18g\times 100=88.88\%Learn even more about mass percentage:https://muzic-ivan.info.com/question/14613800https://muzic-ivan.info.com/question/14030523#learnwithmuzic-ivan.info">" data-test="answer-box-list">You are watching: What is the percent composition of oxygen in h2oAnswer:Explanation:Mass Percent of OxygenThe second method is the very same as for finding the mass percent of hydrogen. From previous calculations, you know the total molar mass of oxygen in water is 16.00. Divide 16.00 by the complete molar mass of water, 18.016, to gain 0.8881. Multiply by 0.8881 by 100 to obtain the percentage: 88.81 percent.AdvertisementAdvertisementCarlynBronkCarlynBronkThe mass percent of oxygen in water is 88.88 %
Explanation:
To calculate the mass percent of oxygen in water, we use the equation:
Mass of oxygen = 16 g
Mass of water = <(2 ×1) + 16> = 18 g
Putting values in over equation, we get:
Find Out even more around mass percentage:
https://muzic-ivan.info.com/question/14613800
https://muzic-ivan.info.com/question/14030523
#learnwithmuzic-ivan.info
### New concerns in Chemistry
identify branch of chemistry: facets radium decays by emitting a-pwrite-ups and is converted right into one more radon
The maximum and also minimum wavelengths are obtained by electron jumping from the initially cell of hydrogen to which cell ?analyze to English:-হাইড্রোজেনের…প্রথম কক্ষ থেকে কোন কোন কক্ষে ইলেকট্রন লাফ দিলে সর্বোচ্চ ও সর্বনিম্ন তরঙ্গদৈর্ঘ্য পাওয়া যায়?
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See more: What Is The Difference Between A Widow And A Widower, What Is The Difference Between Widow And Widower
LONG ANSWER TYPE QUESTIONS1. Why is the subsoil not suitable for thriving crops ?2. Why does some land also appear shimmering on a hot day?3. Mention the pr…operties of the soil that decide the form of crops to be cultivated on it.4. How would you display that soil has actually moisture content ? Exsimple via suitable instance.5. What is soil erosion ? List the reasons of soil erosion. | 886 | 2,953 | {"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} | 4.40625 | 4 | CC-MAIN-2022-27 | latest | en | 0.849871 |
https://oeis.org/A091668 | 1,631,986,659,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00552.warc.gz | 479,875,293 | 4,267 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A091668 Decimal expansion of exp(2*Pi/v) * (v/(1+(5^(3/4)/((1+v)/2)^(5/2)-1)^(1/5))-(1+v)/2), where v = sqrt(5). 1
9, 9, 9, 9, 9, 9, 2, 0, 8, 7, 3, 2, 9, 0, 0, 7, 9, 3, 1, 2, 7, 4, 7, 3, 0, 4, 0, 9, 3, 3, 7, 1, 5, 7, 8, 6, 5, 1, 5, 1, 5, 9, 4, 1, 5, 0, 0, 5, 4, 0, 9, 4, 7, 8, 9, 4, 4, 7, 8, 4, 1, 2, 5, 3, 6, 9, 9, 2, 1, 5, 6, 7, 5, 7, 8, 5, 0, 4, 2, 0, 6, 3, 9, 3, 3, 5, 7, 4, 4, 3, 0, 4, 8, 1, 1, 0, 7, 9, 9, 3, 8, 4, 8, 8, 7 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Has a nice (non-simple) continued fraction due to Ramanujan. Continued fraction is 1/(1+q/(1+q^2/(1+q^3/(1+...)))) where q=exp(-2*Pi*sqrt(5)). - Michael Somos, Sep 12 2005 REFERENCES K. S. Rao, Srinivasa Ramanujan, a Mathematical Genius, pp. 42, Eastwest Books Chennai Madras 2000. G. H. Hardy, Ramanujan: Twelve Lectures on subjects as suggested by his Life and Work, pp. 8 section (1.12), AMS Chelsea Providence RI 1999. LINKS I. E. S. Cartuja, Srinivasa Ramanujan(Text in Spanish) H. Gierhardts, Three Famous Formulas Of Ramanujan S. Sarvotham, Ramanujan Eric Weisstein's World of Mathematics, Ramanujan Continued Fractions FORMULA Equals 1/A091900. EXAMPLE 0.999999208... PROG (PARI) {a(n)=local(s); s=sqrt(5); x=exp(2*Pi/s)*(s/(1+(5^(3/4)/((1+s)/2)^(5/2)-1)^(1/5))-(1+s)/2); floor(x*10^(n+1))%10} /* Michael Somos, Sep 12 2005 */ (PARI) {a(n)= x=exp(-2*Pi*sqrt(5)); x=contfracpnqn(matrix(2, oo, i, j, if(j==1, i==1, if(i==1, x, 1)^(j-2)))); x=t[1, 1]/t[2, 1]; floor(x*10^(n+1))%10} /* Michael Somos, Sep 12 2005 */ CROSSREFS Sequence in context: A346435 A346449 A290665 * A196498 A277535 A292905 Adjacent sequences: A091665 A091666 A091667 * A091669 A091670 A091671 KEYWORD nonn,cons AUTHOR Eric W. Weisstein, Jan 27 2004 STATUS approved
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Last modified September 18 12:35 EDT 2021. Contains 347527 sequences. (Running on oeis4.) | 1,010 | 2,284 | {"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-2021-39 | latest | en | 0.585483 |
https://gauravtiwari.org/2011/08/10/two-interesting-math-problems/ | 1,501,065,488,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426133.16/warc/CC-MAIN-20170726102230-20170726122230-00047.warc.gz | 633,787,232 | 24,259 | # Two Interesting Math Problems
## Problem1: Smallest Autobiographical Number:
A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on.
For example: 42101000 is autobiographical.
Find, with explanation, the smallest autobiographical number.
Solution of Problem 1
## Problem 2: Fit Rectangle:
A rectangle has dimensions $39.375$ cm $\times 136.5$ cm.
• Find the least number of squares that will fill the rectangle.
• Find the least number of squares that will fill the rectangle, if every square must be the same size and Find the largest square that can be tiled to completely fill the rectangle.
Solution of Problem 2
## Solutions of Problem 1:
The restrictions which define an autobiographical number make it straightforward to find the lowest one. It cannot be 0, since by
definition the first digit must indicate the number of zeros in the number. Presumably then, the smallest possible autobiographical number will contain only one 0.If this is the case, then the first digit must be 1. 10 is not a candidate because the second digit must indicate the number of 1s in the number–in this case, 1. So If the
number contains only one zero, it must contain more than one 1.
(If it contained one 1 and one 0, then the first two digits
would be 11, which would be contradictory since it actually contains two 1s).
Again, presumably the lowest possible such number will contain the lowest
possible number of 1s, so we try a number with one 0 and two 1s. It will be of the form: 12-0–..
Now, there is one 2 in this number, so the first three digits must be 121. To meet all the conditions discussed above, we can simply take a 0 onto the end of this to obtain 1210, which is
the smallest auto-biographical number.
## Solution of Problem 2:
We solve the second and third parts of the question
first:
We convert each number to a fraction and get a common denominator, then find the gcd (greatest common divisor) of the numerators.
That is, with side lengths $39.375$ cm and $136.5$ cm , we convert those numbers to fractions (with a common
denominator):
$39.375 = \dfrac{315}{8}$ .
$136.5 = \dfrac{273}{2} = \dfrac{1092}{8}$ .
Now we need to find the largest common factor of 315 and 1092.
Which is 21. So $\dfrac{21}{8}=2.625$ is the largest number that divides evenly into the two numbers $39.375$ and $136.5$ .
There will be $\dfrac{1092}{21} \times \dfrac{315}{21} = 52 \times 15 = 780$ squares, each one a $2.625$ cm $\times 2.625$ cm square needed to fill the rectangle (52 in each row,with 15 rows).
Now we shall solve the first part.
Number of squares lengthwise is 52 and breadthwise is 15. Now we will combine these squares in order to find least number of squares to fill the rectangle. First three squares would be of
dimension 15 by 15. In this way length of 45 units is utilized. Now the rectangle which is left with us excluding three squares is 7 by 15. Again in the same way we can make two squares of dimension 7 by 7. In this way breadth of 14 units is utilized.
Now we are left with the rectangle of dimension 7 by 1.
These can further be subdivided into seven squares each of
dimension 1 by 1. In this way the least number of squares to fill the
rectangle is 3 + 2+ 7 = 12. The required answer is 12.
Note that the three numbers 3, 2, and 7 are involved in the Euclidean Algorithm for finding the g.c.d.!
Source: Internet
Gaurav Tiwari
A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant.
### 1 Response
1. Very interesting problems.. | 979 | 3,855 | {"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} | 4.78125 | 5 | CC-MAIN-2017-30 | longest | en | 0.858701 |
https://electronics.stackexchange.com/questions/429500/debouncer-with-led-inside | 1,701,771,361,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00742.warc.gz | 267,940,585 | 42,590 | Debouncer with LED inside
How It's possible to add a LED to a line that it's debounced (switch) with an RC circuit. If I place 2 LED, one on ground and one on VCC to know button position. Will they remove my charge/discharge rate of Capacitor since they have a low R load?
/* EDIT */
Of course with schematic is easier.
simulate this circuit – Schematic created using CircuitLab
Switch closed, D2 ON and D1 OFF Switch open, D2 OFF and D1 ON
With debouncing on Input
This schematics is wrong, how can I correctly set it up?
/* EDIT 2 */
It's a reference for a Micro input.
• You can remove LED D2, it is shorted and will never light up. With debouncing on Input You added a capacitor which charges and discharges via resistors. That is filtering and not really debouncing. It does work though. Does "INPUT" connect to another circuit? You need to be more clear what you're trying to achieve with this circuit. Why the are LEDs there? Mar 28, 2019 at 13:58
• If the user has tactile feedback and proper debouncing on the switch, why do you need optical feedback to screw up the debouncing then add more parts to fix that and two LED's just in case the LED fails. It's the wrong approach. Mar 28, 2019 at 17:31
• It's not a button, but a switch, so I want to see the result of switch. Mar 28, 2019 at 17:36
simulate this circuit – Schematic created using CircuitLab
Figure 1. D1 is normally on. When SW1 is pressed D1 turns off and D2 turns on.
Note that the input will fall to 0.7 V minimum. Check that this is OK. I've changed the R values so it will affect your debounce timing. Re-calculate those too.
I've split R1 in two: R1 + R4. This raises the INPUT voltage by $$\ \frac {V_{cc}-V_{D1}}{2} \$$ to address Dave Tweed's comment.
simulate this circuit
Figure 2. A better design?
How it works:
• With SW1 open 'B' is pulled high via R2 and D2. Meanwhile D1 is lit via R1. VA will be whatever Vf of the LED is so D3 will be reverse biased.
• When S1 is pressed 'B' will be pulled to 0 V. D3 will now be in parallel with D1 but it's 0.7 V forward voltage will steal the current from D1 so it will turn off.
• Note that Vih is now limited to the Vf of D1. This may or may not be enough, depending on what "INPUT" is connected to. Mar 28, 2019 at 17:05
• Thanks, @Dave. This was a lunchtime schematic sketch and that was in the back of my mind. I ran out of time to give the final checkover. I've added R4 which will help. Mar 28, 2019 at 17:10
• Or you could just add another diode, switching D1 and INPUT separately. R2 becomes the pullup for INPUT. Mar 28, 2019 at 17:12
• There's a way to reduce INPUT voltage to 0v? I've updated my post, it's a reference for a Micro. Mar 28, 2019 at 17:26
• What is the threshold voltage for your microcontroller to guarantee that it reads a '0'? Mar 28, 2019 at 17:54 | 788 | 2,823 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-50 | longest | en | 0.930829 |
http://slideplayer.com/slide/3387185/ | 1,500,847,288,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00153.warc.gz | 305,769,403 | 21,192 | # Physics 2112 Unit 8: Capacitors
## Presentation on theme: "Physics 2112 Unit 8: Capacitors"— Presentation transcript:
Physics 2112 Unit 8: Capacitors
Today’s Concept: Capacitors in a circuits Dielectrics Energy in capacitors
Where we are…….
Simple Capacitor Circuit
Direction of arrows is opposite of the direction of electron motion Sorry. It’s historical. There is nothing I can do. Q C V Q = VC V C This “Q” really means that the battery has moved charge Q from one plate to the other, so that one plate holds +Q and the other -Q.
Parallel Capacitor Circuit
Qtotal C1 C2 V Q1 = C1V Q2 = C2V Key point: V is the same for both capacitors Key Point: Qtotal = Q1 + Q2 = VC1 + VC2 = V(C1 + C2) Ctotal = C1 + C2
Series Capacitor Circuit
Q Q = VCtotal C1 C2 V1 V2 V V Q Key point: Q is the same for both capacitors Key point: Q = VCtotal = V1C1 = V2C2 Also: V = V1 + V2 Q/Ctotal = Q/C1 + Q/C2 1 Ctotal C1 C2 + =
Capacitor Summary Series Parallel Wiring Voltage Current Capacitance
Every electron that goes through one must go through the other Each resistor on a different wire. Wiring Same for each capacitor. Vtotal = V1 = V2 Voltage Different for each capacitor. Vtotal = V1 + V2 Same for each capacitor Itotal = I1 = I2 Different for each capacitor Itotal = I1 + I2 Current Decreases 1/Ceq = 1/C1 + 1/C2 Increases Ceq = C1 + C2 Capacitance
Example 8.1 (Capacitors in Series)
Given the circuit to the left: What is Q1? What is V1? C1 =3uF V =12V C2 =9uF Conceptual Idea: Plan: Find equivalent capacitance Use knowledge that in series Q1 = Q2 = Qtot Find V using Q = CV
Example 8.2 (Capacitors in Parallel)
Given the circuit to the left: What is Q1? What is Q2? C2 =9uF C1 =3uF V =12V Conceptual Idea: Plan: Find equivalent capacitance Use knowledge that in parallel V1 = V2 Find Q using Q = CV
CheckPoint: Three Capacitor Configurations
The three configurations shown below are constructed using identical capacitors. Which of these configurations has lowest total capacitance? B C C A 1/Ctotal = 1/C + 1/C = 2/C Ctotal = C Ctotal = 2C Ctotal = C/2
CheckPoint: Two Capacitor Configurations
The two configurations shown below are constructed using identical capacitors. Which of these configurations has the lowest overall capacitance? B C A C Cright = C/2 Cleft = C/2 Ctotal = C Ctotal = Cleft + Cright Ctotal = C A B Both configurations have the same capacitance
CheckPoint: Capacitor Network
A circuit consists of three unequal capacitors C1, C2, and C3 which are connected to a battery of voltage V0. The capacitance of C2 is twice that of C1. The capacitance of C3 is three times that of C1. The capacitors obtain charges Q1, Q2, and Q3. C1 C2 C3 V0 V1 Q1 V2 Q2 V3 Q3 Compare Q1, Q2, and Q3. Q1 > Q3 > Q2 Q1 > Q2 > Q3 Q1 > Q2 = Q3 Q1 = Q2 = Q3 Q1 < Q2 = Q3
Example 8.3 (Capacitor Network)
C3 =12uF Given the circuit to the left: What is Q1? What is Q3? C1 =3uF C2 =11uF V =12C C5 =9uF C4 =6uF Conceptual Idea: Find V at each capacitor and then Q . Plan: Break circuit down into elements that are in parrallel or in series. Find equivalent capacitance Work backwards to find DV across each one Fine Q using Q = CV
Example 8.1 (Capacitor Network)
Energy in a Capacitor +Q C V U = 1/2QV = 1/2CV2 -Q = 1/2Q2/C
In Prelecture 7 we calculated the work done to move charge Q from one plate to another: C +Q V U = 1/2QV = 1/2CV2 Since Q = VC = 1/2Q2/C -Q This is potential energy waiting to be used…
Messing with Capacitors
If connected to a battery V stays constant If isolated then total Q stays constant V V1 = V Q1 = C1V1 = k CV = k Q C1 = k C Q1 = Q V1 = Q1/C1 = Q/k C = V /k C1 = k C
Dielectrics C1 = k C0 V Q1 = VC1 C0 V Q0 = VC0 By adding a dielectric you are just making a new capacitor with larger capacitance (factor of k)
CheckPoint: Capacitors and Dielectrics 1
Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected, it is filled with a dielectric. Compare the voltages of the two capacitors. V1 > V2 V1 = V2 V1 < V2
CheckPoint: Capacitors and Dielectrics 2
Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected, it is filled with a dielectric. Compare the potential energy stored by the two capacitors. U1 > U2 U1 = U2 U1 < U2
CheckPoint: Capacitors and Dielectrics 3
The two capacitors are now connected to each other by wires as shown. How will the charge redistribute itself, if at all? The charges will flow so that the charge on C1 will become equal to the charge on C2. The charges will flow so that the energy stored in C1 will become equal to the energy stored in C2 The charges will flow so that the potential difference across C1 will become the same as the potential difference across C2. No charges will flow. The charge on the capacitors will remain what it was before they were connected.
Example 8.4 (Partial Dielectric)
V C0 x0 An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. V k x0/4 A dielectric (k ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? Conceptual Analysis: The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Strategic Analysis: Think of new capacitor as two capacitors in parallel Calculate new capacitance C Apply definition of capacitance to determine Q
Calculation A1 =3/4 Ao A2 =1/4 Ao k k
The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. | 1,626 | 5,700 | {"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-2017-30 | latest | en | 0.847243 |
http://desero.tk/trigonometry-rules-sheet.html | 1,582,470,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00291.warc.gz | 39,296,340 | 3,979 | # Trigonometry rules sheet
Sheet rules
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The triangle of most interest is the right- angled triangle. Trigonometry is a mathematical method used to define relations between elements of a triangle. Our maths trigonometry worksheets with answers will help your child or student to grasp and understand basic and more advanced ways of solving trigonometric equations. | 924 | 4,418 | {"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-2020-10 | latest | en | 0.90669 |
https://studysoup.com/tsg/1840/physics-principles-with-applications-6-edition-chapter-8-problem-16pe | 1,631,958,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00333.warc.gz | 586,408,498 | 12,760 | ×
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 8 - Problem 16pe
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 8 - Problem 16pe
×
Calculate the final speed of a 110-kg rugby player who is
ISBN: 9780130606204 3
Solution for problem 16PE Chapter 8
Physics: Principles with Applications | 6th Edition
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Problem 16PE
Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.76×104 N for 5.50×10–2 s .
Step-by-Step Solution:
Step-by-step solution In this problem calculate the final speed of a rugby player who experience a backward force when collides with a goal post using the data find his final speed respectively: Step 1 of 5 Running speed of rugby player is, However, one collide head on padded goal post and experiences the backward force of .
Step 2 of 5
Step 3 of 5
ISBN: 9780130606204
Since the solution to 16PE from 8 chapter was answered, more than 671 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 16PE from chapter: 8 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. This full solution covers the following key subjects: backward, calculate, collides, experiences, final. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.76×104 N for 5.50×10–2 s .” is broken down into a number of easy to follow steps, and 35 words. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204.
Related chapters
Unlock Textbook Solution | 544 | 2,183 | {"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-2021-39 | latest | en | 0.88755 |
http://complementarytraining.net/stats-playbook-what-is-anscombes-quartet-and-why-is-it-important/ | 1,503,516,613,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123359.11/warc/CC-MAIN-20170823190745-20170823210745-00549.warc.gz | 89,621,287 | 339,265 | # Stats Playbook: What is Anscombe’s Quartet and why is it important?
Stats Playbook: What is Anscombe’s Quartet and why is it important?
# Stats Playbook: What is Anscombe’s Quartet and why is it important?
The following paragraph is take from Wikipedia
“Anscombe’s quartet comprises four datasets that have nearly identical simple statistical properties, yet appear very different when graphed. Each dataset consists of eleven (x,y) points. They were constructed in 1973 by the statistician Francis Anscombe to demonstrate both the importance of graphing data before analyzing it and the effect of outliers on statistical properties”
Here is the famous Anscombe’s Quartet
``````library(xtable)
# Create the interactive table
ansombe.table <- xtable(anscombe)
print(ansombe.table, "html")
``````
x1 x2 x3 x4 y1 y2 y3 y4
1 10.00 10.00 10.00 8.00 8.04 9.14 7.46 6.58
2 8.00 8.00 8.00 8.00 6.95 8.14 6.77 5.76
3 13.00 13.00 13.00 8.00 7.58 8.74 12.74 7.71
4 9.00 9.00 9.00 8.00 8.81 8.77 7.11 8.84
5 11.00 11.00 11.00 8.00 8.33 9.26 7.81 8.47
6 14.00 14.00 14.00 8.00 9.96 8.10 8.84 7.04
7 6.00 6.00 6.00 8.00 7.24 6.13 6.08 5.25
8 4.00 4.00 4.00 19.00 4.26 3.10 5.39 12.50
9 12.00 12.00 12.00 8.00 10.84 9.13 8.15 5.56
10 7.00 7.00 7.00 8.00 4.82 7.26 6.42 7.91
11 5.00 5.00 5.00 8.00 5.68 4.74 5.73 6.89
Here it how it looks like graphically
``````library(ggplot2)
anscombe.1 <- data.frame(x = anscombe[["x1"]], y = anscombe[["y1"]], Set = "Anscombe Set 1")
anscombe.2 <- data.frame(x = anscombe[["x2"]], y = anscombe[["y2"]], Set = "Anscombe Set 2")
anscombe.3 <- data.frame(x = anscombe[["x3"]], y = anscombe[["y3"]], Set = "Anscombe Set 3")
anscombe.4 <- data.frame(x = anscombe[["x4"]], y = anscombe[["y4"]], Set = "Anscombe Set 4")
anscombe.data <- rbind(anscombe.1, anscombe.2, anscombe.3, anscombe.4)
gg <- ggplot(anscombe.data, aes(x = x, y = y))
gg <- gg + geom_point(color = "black")
gg <- gg + facet_wrap(~Set, ncol = 2)
gg
``````
Let’s do the simple descriptive statistics on each data set
Here is mean of x and y
``````aggregate(cbind(x, y) ~ Set, anscombe.data, mean)
``````
``````## Set x y
## 1 Anscombe Set 1 9 7.501
## 2 Anscombe Set 2 9 7.501
## 3 Anscombe Set 3 9 7.500
## 4 Anscombe Set 4 9 7.501
``````
And SD
``````aggregate(cbind(x, y) ~ Set, anscombe.data, sd)
``````
``````## Set x y
## 1 Anscombe Set 1 3.317 2.032
## 2 Anscombe Set 2 3.317 2.032
## 3 Anscombe Set 3 3.317 2.030
## 4 Anscombe Set 4 3.317 2.031
``````
And correlation between x and y
``````library(plyr)
correlation <- function(data) {
x <- data.frame(r = cor(data\$x, data\$y))
return(x)
}
ddply(.data = anscombe.data, .variables = "Set", .fun = correlation)
``````
``````## Set r
## 1 Anscombe Set 1 0.8164
## 2 Anscombe Set 2 0.8162
## 3 Anscombe Set 3 0.8163
## 4 Anscombe Set 4 0.8165
``````
As can be seen they are pretty much the same for every data set.
Let’s perform linear regression model for each
``````model1 <- lm(y ~ x, subset(anscombe.data, Set == "Anscombe Set 1"))
model2 <- lm(y ~ x, subset(anscombe.data, Set == "Anscombe Set 2"))
model3 <- lm(y ~ x, subset(anscombe.data, Set == "Anscombe Set 3"))
model4 <- lm(y ~ x, subset(anscombe.data, Set == "Anscombe Set 4"))
``````
Here are the summaries
``````summary(model1)
``````
``````##
## Call:
## lm(formula = y ~ x, data = subset(anscombe.data, Set == "Anscombe Set 1"))
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9213 -0.4558 -0.0414 0.7094 1.8388
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.000 1.125 2.67 0.0257 *
## x 0.500 0.118 4.24 0.0022 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.24 on 9 degrees of freedom
## Multiple R-squared: 0.667, Adjusted R-squared: 0.629
## F-statistic: 18 on 1 and 9 DF, p-value: 0.00217
``````
``````summary(model2)
``````
``````##
## Call:
## lm(formula = y ~ x, data = subset(anscombe.data, Set == "Anscombe Set 2"))
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.901 -0.761 0.129 0.949 1.269
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.001 1.125 2.67 0.0258 *
## x 0.500 0.118 4.24 0.0022 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.24 on 9 degrees of freedom
## Multiple R-squared: 0.666, Adjusted R-squared: 0.629
## F-statistic: 18 on 1 and 9 DF, p-value: 0.00218
``````
``````summary(model3)
``````
``````##
## Call:
## lm(formula = y ~ x, data = subset(anscombe.data, Set == "Anscombe Set 3"))
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.159 -0.615 -0.230 0.154 3.241
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.002 1.124 2.67 0.0256 *
## x 0.500 0.118 4.24 0.0022 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.24 on 9 degrees of freedom
## Multiple R-squared: 0.666, Adjusted R-squared: 0.629
## F-statistic: 18 on 1 and 9 DF, p-value: 0.00218
``````
``````summary(model4)
``````
``````##
## Call:
## lm(formula = y ~ x, data = subset(anscombe.data, Set == "Anscombe Set 4"))
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.751 -0.831 0.000 0.809 1.839
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.002 1.124 2.67 0.0256 *
## x 0.500 0.118 4.24 0.0022 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.24 on 9 degrees of freedom
## Multiple R-squared: 0.667, Adjusted R-squared: 0.63
## F-statistic: 18 on 1 and 9 DF, p-value: 0.00216
``````
And graphically
``````gg <- gg + geom_smooth(formula = y ~ x, method = "lm", se = FALSE, data = anscombe.data)
gg
``````
It can be seen both graphically and from regression summary that each data set resulted in same statistical model!
Intercepts, coeficients and their p values are the same. SEE (standard error of the estimate, or SD of residuals), F-value and it’s p values are the same.
What is the conclusion: ALWAYS plot your data! And always do model diagnostics by plotting the residuals.
``````par(mfrow = c(2, 2))
plot(model1, main = "Model 1")
``````
``````plot(model2, main = "Model 2")
``````
``````plot(model3, main = "Model 3")
``````
``````plot(model4, main = "Model 4")
``````
In the next part I will be covering outliers and influential cases and their difference
I am a physical preparation coach from Belgrade, Serbia, grow up in Pula, Croatia (which I consider my home town). I was involved in physical preparation of professional, amateur and recreational athletes of various ages in sports such as basketball, soccer, volleyball, martial arts and tennis. Read More »
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Your Information will never be shared with any third party | 2,678 | 7,149 | {"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} | 3.84375 | 4 | CC-MAIN-2017-34 | longest | en | 0.64842 |
https://www.playtaptales.com/numbers/billiseptemoctogintaquadringentillion/ | 1,544,703,365,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00066.warc.gz | 990,118,273 | 4,344 | A Billiseptemoctogintaquadringentillion (1 Billiseptemoctogintaquadringentillion) is 10 to the power of 7464 (10^7464). This is an extremely astronomical number!
## How many zeros in a Billiseptemoctogintaquadringentillion?
There are 7,464 zeros in a Billiseptemoctogintaquadringentillion.
A Billiseptemoctogintaquadringentillionaire is someone whos assets, net worth or wealth is 1 or more Billiseptemoctogintaquadringentillion. It is unlikely anyone will ever be a true Billiseptemoctogintaquadringentillionaire. If you want to be a Billiseptemoctogintaquadringentillionaire, play Tap Tales!
## Is Billiseptemoctogintaquadringentillion the largest number?
Billiseptemoctogintaquadringentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
## Big Numbers
This is just one of many really big numbers! | 5,224 | 10,878 | {"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-2018-51 | longest | en | 0.625851 |
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