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http://math.stackexchange.com/questions/269678/solving-this-inequality-without-trial-and-error/269692 | 1,464,281,477,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276131.38/warc/CC-MAIN-20160524002116-00053-ip-10-185-217-139.ec2.internal.warc.gz | 192,749,676 | 16,493 | # Solving this inequality without trial and error
"What is the smallest value of $n$ such that an algorithm whose running time is $100n^2$ runs faster than an algorithm whose running time is $2^n$ on the same machine?"
I know the answer is $n = 15$, but is there any way to solve this without trial and error?
-
have you tried with logarithms? – dineshdileep Jan 3 '13 at 12:10
It should be $n=15$ since $2^{14}-100\cdot 14^2=-3216.$ – coffeemath Jan 3 '13 at 12:25
You can start by plotting the function $x\mapsto 2^x - 100x^2$. After this, you just have to check that the value you found is correct. – Siméon Jan 3 '13 at 12:41
Trial and error is necessary, unless you use Lambert's W-function. In maple, the request for a solution to $2^x=100x^2$ gave two complicated expressions involving the Lambert function $W(x)$. Numeric evaluation of these gave the real values $0.103657...$ and $14.324727...$, making the answer $15$. | 273 | 932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2016-22 | latest | en | 0.916168 |
https://sports.stackexchange.com/questions/18208/what-would-i-have-to-track-during-each-point-to-get-detailed-tennis-match-statis | 1,702,204,524,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00345.warc.gz | 576,803,872 | 40,898 | # What would I have to track during each point to get detailed tennis match statistics
I am looking to keep stats for upcoming tennis tournament and I am wondering what would I have to track play by play to get good match statistics.
For example should I track every shot in the rally or maybe note down just where serve landed and where it was returned and at the end of the play note if it was winner, forced error or unforced error.
For example http://tennisabstract.com/ tracks every shot with additional modifiers like slice, drop-shot etc, along with other data. This is too much data for one person to track during a live match...so I am looking for most important metrics that would yield good match stats but would be feasible for a single person to manage during a match.
The most important stats to track are serves, returns, break points and double faults. Depending on how far you want to go you can go into more or less details:
Serves:
• Aces
• 1st serve (succeeded + points won / percentage)
• 2nd serve (dito)
• service points won in total / percentage
Same for Returns:
• 1st serve returns won
• 2nd serve returns won
• return points won in total
Break points:
• break points won
• break points saved
Double Faults are easy to track as well:
• total
• percentage
• per set
Other stats that might be handy:
• total games played / won (percentage)
• total points won
Those stats should be easy to track by one person.. Watch the service, wait for the outcome, write it down, repeat. After the match you can do the math (e.g. calculate the serve and return ratings) and go on with the next one.
• This seems pretty comprehensive, and easy. If we look at this site: atpworldtour.com/en/scores/2018/717/MS004/match-stats , we can see an example of which statistics are important to keep for a match. You covered all of them, except for Serve and Return rating. However, those can be calculated after the fact. Serve Rating is (1st Serve %)+(1st Serve Pts Won %)+(2nd Serve Pts Won %)+(% Service Games Won)+(Aces)-(Double Faults). Return Rating is the same but for returns, and doesn't include aces or double faults.
– Pawr
Apr 23, 2018 at 19:47
• @Pawr yes, the ratings should be calculated after the match or between the sets. When collecting live stats doing the math during the match is too much work for one person and you could miss something important.
– dly
Apr 23, 2018 at 20:43
• There's of course the development of an app or spreadsheet that does the calculation live, with varying degrees of protection from user error.
– Nij
Apr 23, 2018 at 20:59
• For a detailed statistics is it important for instance to track where each server landed (down the middle, wide or T) and where subsequent return(s) landed - are they closer to the line, middle or closer to the net... Apr 24, 2018 at 7:20 | 680 | 2,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | longest | en | 0.959218 |
https://staging-pubs.nctm.org/search?access=all&pageSize=10&q=%22CCSS.Math.Content.HSS-ID.A.4%22&sort=relevance&t=probability_6 | 1,653,382,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00705.warc.gz | 600,204,568 | 30,297 | # Search Results
## You are looking at 1 - 10 of 11items for :
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## Illustrating the Central Limit Theorem
Sampling experiments with different types of beads give students a memorable hands-on experience.
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## American I Do's – The More the Merrier?
Students analyze items from the media to answer mathematical questions related to the article. The mathematics in these clips includes interpretation of graphs, computing percentages, making conjectures, and analyzing data. The first clip concerns college admission, a relevant topic for many students.
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## Activities for Students: Dealing Cards with Confidence
When understood and applied appropriately, mathematics is both beautiful and powerful. As a result, students are sometimes tempted to extend that power beyond appropriate limits. In teaching statistics at both the high school and college level, I have found that one of students' biggest struggles is applying their understanding of probability to make appropriate inferences.
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## Delving Deeper: Differentiated Insurance Contracts
By examining pricing for insurance for a moped, students can explore the theory of systems of inequalities and the topic of distributions in statistics. Fair systems for determining the premium (taking into account cautious and reckless drivers) are considered.
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## Technology Tips: Teaching Sampling Distributions Using Autograph
The author uses Autograph, a powerful software program, to illustrate sampling distributions and to demonstrate the central limit theorem.
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## Analyzing Highway Speeding Data in the Statistics Classroom
Students bring the real world into the classroom by studying speeding data collected on two Pennsylvania highways.
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## Using Data to Motivate the Use of Empirical Sampling Distributions
Research shows that students often struggle with understanding empirical sampling distributions. Using hands-on and technology models and simulations of problems generated by real data help students begin to make connections between repeated sampling, sample size, distribution, variation, and center. A task to assist teachers in implementing research-based strategies is included.
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## A Question Library for Classroom Voting
Voting in the classroom can engage students and promote discussion. All you need is a good set of questions.
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## Statistical Simulation with Web-Based Apps
While looking for an inexpensive Web application to illustrate the Central Limit theorem, I found the Rossman/Chance Applet Collection, a group of free Web-based statistics apps. In addition to illustrating the Central Limit theorem, the apps could be used to cover many classic statistics concepts, including confidence intervals, regression, and a virtual version of the popular Reese's® Pieces problem. The apps allow users to investigate concepts using either preprogrammed or original data.
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## Reader Reflections – August 2013
Readers comment on published articles or offer their own ideas. | 586 | 3,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | latest | en | 0.891596 |
https://www.britannica.com/topic/Albert-Einstein-on-Space-Time-1987141/Euclidean-Geometry | 1,726,168,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00441.warc.gz | 632,697,734 | 21,099 | ## Euclidean Geometry
inAlbert Einstein on space-time inSPACE-TIME
If we consider Euclidean geometry we clearly discern that it refers to the laws regulating the positions of rigid bodies. It turns to account the ingenious thought of tracing back all relations concerning bodies and their relative positions to the very simple concept “distance” (Strecke). Distance denotes a rigid body on which two material points (marks) have been specified. The concept of the equality of distances (and angles) refers to experiments involving coincidences; the same remarks apply to the theorems on congruence. Now, Euclidean geometry, in the form in which it has been handed down to us from Euclid, uses the fundamental concepts “straight line” and “plane” which do not appear to correspond, or at any rate, not so directly, with experiences concerning the position of rigid bodies. On this it must be remarked that the concept of the straight line may be reduced to that of the distance.1 Moreover, geometricians were less concerned with bringing out the relation of their fundamental concepts to experience than with deducing logically the geometrical propositions from a few axioms enunciated at the outset.
Let us outline briefly how perhaps the basis of Euclidean geometry may be gained from the concept of distance.
We start from the equality of distances (axiom of the equality of distances). Suppose that of two unequal distances one is always greater than the other. The same axioms are to hold for the inequality of distances as hold for the inequality of numbers.
Three distances AB1, BC1, CA1 may, if CA1 be suitably chosen, have their marks BB1, CC1, AA1 superposed on one another in such a way that a triangle ABC results. The distance CA1 has an upper limit for which this construction is still just possible. The points A, (BB’) and C then lie in a “straight line” (definition). This leads to the concepts: producing a distance by an amount equal to itself; dividing a distance into equal parts; expressing a distance in terms of a number by means of a measuring-rod (definition of the space-interval between two points).
When the concept of the interval between two points or the length of a distance has been gained in this way we require only the following axiom (Pythagoras’ theorem) in order to arrive at Euclidean geometry analytically.
To every point of space (body of reference) three numbers (co-ordinates) x, y, z may be assigned—and conversely—in such a way that for each pair of points A (x1, y1, z1) and B (x2, y2, z2) the theorem holds:
measure-number AB = sqroot{(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2}.
All further concepts and propositions of Euclidean geometry can then be built up purely logically on this basis, in particular also the propositions about the straight line and the plane.
These remarks are not, of course, intended to replace the strictly axiomatic construction of Euclidean geometry. We merely wish to indicate plausibly how all conceptions of geometry may be traced back to that of distance. We might equally well have epitomised the whole basis of Euclidean geometry in the last theorem above. The relation to the foundations of experience would then be furnished by means of a supplementary theorem.
The co-ordinate may and must be chosen so that two pairs of points separated by equal intervals, as calculated by the help of Pythagoras’ theorem, may be made to coincide with one and the same suitably chosen distance (on a solid).
The concepts and propositions of Euclidean geometry may be derived from Pythagoras’ proposition without the introduction of rigid bodies; but these concepts and propositions would not then have contents that could be tested. They are not “true” propositions but only logically correct propositions of purely formal content.
## Difficulties
A serious difficulty is encountered in the above represented interpretation of geometry in that the rigid body of experience does not correspond exactly with the geometrical body. In stating this I am thinking less of the fact that there are no absolutely definite marks than that temperature, pressure and other circumstances modify the laws relating to position. It is also to be recollected that the structural constituents of matter (such as atom and electron, q.v.) assumed by physics are not in principle commensurate with rigid bodies, but that nevertheless the concepts of geometry are applied to them and to their parts. For this reason consistent thinkers have been disinclined to allow real contents of facts (reale Tatsachenbestände) to correspond to geometry alone. They considered it preferable to allow the content of experience (Erfahrungsbestände) to correspond to geometry and physics conjointly.
This view is certainly less open to attack than the one represented above; as opposed to the atomic theory it is the only one that can be consistently carried through. Nevertheless, in the opinion of the author it would not be advisable to give up the first view, from which geometry derives its origin. This connection is essentially founded on the belief that the ideal rigid body is an abstraction that is well rooted in the laws of nature. | 1,076 | 5,187 | {"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} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.940843 |
https://sciencenotes.org/period-of-a-simple-pendulum/ | 1,713,141,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00152.warc.gz | 477,192,911 | 40,018 | # How to Find the Period of a Simple Pendulum – Example Problem
A simple pendulum is a mass hanging from a massless string of length L swinging from a central pivot point. As the mass is pulled out at a small angle theta and released, the mass will swing back and forth in periodic motion. This example problem will show how to calculate the period of a simple pendulum.
The period of a simple pendulum refers to the time it takes for the mass to complete one complete cycle of its swinging motion. This time can be calculated using the formula
where
T = period
L = length of the pendulum
g = acceleration due to gravity
### Simple Pendulum Period Example Problem
Question: What is the period of a simple pendulum with a length of 1 meter?
Use 9.8 m/s2 for gravity | 173 | 769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | longest | en | 0.847737 |
https://stats.stackexchange.com/questions/83309/p-value-verification | 1,702,302,299,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00400.warc.gz | 606,847,797 | 40,757 | # P-value verification
Question
In some bank, the time it takes from the moment a customer arrives until a clerk is available is distributed normally with $\mu=15,\sigma=2$
a. What's the probability for the next client to wait more than 18 minutes?
b. What's the prob. that the average waiting time of the next 100 customers will be greater than 15.1 minutes (assuming that the waiting times are independent)
c. The bank management hired a new manager for customer service and he claims that since he arrived at this job, the time it takes to be treated by a clerk decreased. To check his claim he measured the waiting time of 30 random customers and got an average of 14.3 minutes. Assuming that the s.d. hasn't changed - Check if his claim is true, with confidence level of a=0.1.
a. P(X>18)=(normalizing) P(Z>1.5)=0.0668
b. $P(\bar X>15.1)=P(Z>0.5)=0.3085$
c. $H_0:\mu=15, H_A:\mu<15$ so $CI=(-\infty,15-Z_0.99 \frac 2{\sqrt{30}}]=(-\infty,14.15]$ but 14.13<14.15 so H_0 is true and not rejected.
Also calculating the p-value: $P_{H_0}(\bar X<14.3)=P(Z<-1.91)=0.0281>0.01$therefore $H_0$ is not rejected.
Since this is the first time I've been doing it on my own I'd love if someone take a look, and confirm that it's correct (or not?) | 378 | 1,248 | {"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": 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} | 3.671875 | 4 | CC-MAIN-2023-50 | latest | en | 0.924188 |
http://www.chegg.com/homework-help/questions-and-answers/box-bananas-weighing400rests-horizontal-surface-coefficient-static-friction-box-surface-is-q2680047 | 1,474,864,842,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660602.38/warc/CC-MAIN-20160924173740-00260-ip-10-143-35-109.ec2.internal.warc.gz | 381,765,496 | 15,383 | A box of bananas weighing40.0rests on a horizontal surface. The coefficient of static friction between the box and the surface is0.40and the coefficient of kinetic friction is0.20.
If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?
What is the magnitude of the friction force if a monkey applies a horizontal force of6.0to the box and the box is initially at rest?
What minimum horizontal force must the monkey apply to start the box in motion?
What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?
If the monkey applies a horizontal force of18.0,what is the magnitude of the friction force ?
If the monkey applies a horizontal force of18.0,what is the box's acceleration? | 177 | 816 | {"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-2016-40 | latest | en | 0.85466 |
https://gis.stackexchange.com/questions/242455/intersection-between-an-angle-and-a-polygon-postgresql-postgis-or-python2-7 | 1,720,950,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00542.warc.gz | 244,033,181 | 39,007 | # Intersection between an angle and a polygon (postgreSQL + postGIS or Python2.7 + shapely + psycopg2)
I have a point geom feature (in wkb) in a PostgreSQL database table (with postGIS extension). There is also an azimuth value (degrees), as float, in an other field of the same table.
How to perform, as simple as it would be to cut a part of a cake, an intersection between a x[m] buffer around the point and a 20° angle around the azimuth (±10° on both side, the azimuth should be the bisecting line of the angle)?
From now I've calculated the buffer as an other geom (as a polygon also in wkb) which is stored in an other field of the table.
I wonder what would be the faster and more CPU effective calculation to intersect this circle with the angle?
The only postGIS function I have found yet is `ST_azimuth`: http://www.postgis.org/docs/ST_Azimuth.html but it may be rather "complicated" to implement as it would need to calculate two "virtual secondary points" that are "far enough" from the buffer and apart respectively from -10 and +10° from the existing azimuth and then make the difference between these two new azimuth angles... And after, I really can't figure out how to intersect an angle, which is some kind of "not clearly defined" geometry, with the circular buffer.
An other point; at the end I would need to implement this in a Python script (I'm basically using `shapely` and `psycopg2` modules yet). But for now it could be done in an SQL query directly on the postgreSQL DB.
--
http://toblerity.org/shapely/manual.html
http://initd.org/psycopg/
In fact, one don't really have to calculate (unless high precision is needed) the resulting intersection between the buffer around the point and an angle of ± 10° (or whatever angle) on both sides from the azimuth direction.
On can simply build the triangle. Let's have a look...
### Code:
``````import math
from shapely import geometry
OFFSET = 1000 # Set a big enough offset from the initial point to suit your needs
azim = 148 # degrees
point0 = geometry.Point(0,0) # center initial point geometry
E1 = OFFSET*math.sin(math.pi*(azim-10)/180.) # east coord. of 1st secondary point
E2 = OFFSET*math.sin(math.pi*(azim+10)/180.) # east coord. of 2nd secondary point
N1 = OFFSET*math.cos(math.pi*(azim-10)/180.) # north coord. of 1st secondary point
N2 = OFFSET*math.cos(math.pi*(azim+10)/180.) # north coord. of 2nd secondary point
point1 = geometry.point.Point((point0.x+E1, point0.y+N1))
point2 = geometry.point.Point((point0.x+E2, point0.y+N2))
coords = ((point0.x, point0.y),(P1.x, P1.y),(P2.x, P2.y),(point0.x, point0.y))
triangle = geometry.Polygon(coords)
# If you really need the intersection with a circular buffer around the
# initial point:
BUFFSIZE = 600 # must be lower than OFFSET, otherwise useless.
buff = point0.buffer(BUFFSIZE)
intersec = triangle.intersection(buff)
``````
That's it.
Result look nice, here some intersections between circular buffers around some points and some given azimuths, with a ± 20° "angular buffer" around these azimuths: | 785 | 3,095 | {"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-2024-30 | latest | en | 0.942517 |
http://www.virtualnerd.com/pre-algebra/rational-numbers/solve-inequalities-fractions.php | 1,495,522,536,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607591.53/warc/CC-MAIN-20170523064440-20170523084440-00390.warc.gz | 694,998,752 | 5,415 | # Solving Inequalities with Fractions
### Popular Tutorials in Solving Inequalities with Fractions
#### How Do You Solve an Inequality Word Problem Where You're Multiplying Positive Fractions?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality Where You're Multiplying Positive Fractions?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality Word Problem Where You're Multiplying Negative Fractions?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality Where You're Multiplying Negative Fractions?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality By Subtracting Fractions?
To solve an inequality containing fractions, focus on isolating the variable on one side of the inequality. In this tutorial, you'll see how to subtract fractions with unlike denominators in order to isolate the variable and find the answer to the inequality!
#### How Do You Solve an Inequality By Adding Fractions?
To solve an inequality containing fractions, focus on isolating the variable on one side of the inequality. In this tutorial, you'll see how to add fractions with unlike denominators in order to isolate the variable and find the answer to the inequality! | 478 | 2,429 | {"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} | 4.65625 | 5 | CC-MAIN-2017-22 | latest | en | 0.925803 |
http://www.cheresources.com/content/articles/calculation-tips/solving-integral-equations-in-ms-excel | 1,369,163,310,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700438490/warc/CC-MAIN-20130516103358-00012-ip-10-60-113-184.ec2.internal.warc.gz | 376,221,925 | 19,360 | • Create Account
## Solving Integral Equations in MS Excel
Nov 08 2010 01:10 PM | Steve Hall in Calculations and Tips Share this topic:
|
TankVolume, as it appears in the online store, calculates the liquid volume of partially filled tanks, horizontal or vertical, with various types of heads. One of these, the torispherical head, requires that integral equations be solved. This article shows how the integrals are evaluated.
Torispherical heads are made up of two parts. The rounded end has a spherical shape. It has a radius that is described with the numerical value “f” which is the ratio of the dish radius to the vessel shell diameter.
So, if the diameter of the vessel, D, is 48 inches and the value of “f” is 1.0 then the radius of the dish is 48 inches. Unless f=0.5, there would be an abrupt angle between the shell and head unless a transition piece is installed. This is the second part of the torispherical head, a transition piece shaped like a donut or torus. It is called the “knuckle” and there is a knuckle radius that again is related to the vessel diameter with the factor, k. Valid values for f are anything greater than 0.5; k must be greater or equal to 0 and less than or equal to 0.5.
Torispherical heads are also called “F&D” or “flanged and dished”. The standard F&D head has f = 1.0 and kD = 3 times the metal thickness of the head. ASME F&D heads require a dish radius no greater than the diameter (f <= 1.0) and knuckle radius no less than 6% of the diameter (k >= 0.06) or three times the metal thickness, whichever is greater.
When calculating the volume of fluid in a torispherical head, the head is divided into three parts. The first is the lower portion occupied by the knuckle (liquid height 0 to h1 as seen in the diagram). The second part is the portion covered by the spherical dish section, which also includes the knuckle at the perimeter. The third portion is that which is entirely within the knuckle but above the spherical section.
Each of the three parts are calculated separately and they each require solving an integral. I’ll use the integral for the first part for an example in this article.
Eq. (1)
where
D = vessel diameter
h = fill height of the vessel
n = R – kD + (k2D2 – x2)^0.5, where R = vessel radius (= D/2)
w = R – h
It looks complicated, but solving in Excel is straightforward if it is done sequentially as described below.
Integrals that are too complex to express implicitly are solved using “numerical methods”. A number of numerical methods have been used for integrals with Simpson’s Rule being one of the best. I chose to use Simpson’s Rule for this problem. (Other methods include the trapezoidal rule, Riemann sums, Romberg integration, Gaussian quadratures and the Monte Carlo method). There are many webpages that give the derivation of Simpson’s Rule and explain it in detail; I won’t repeat that work here.
Simpson’s rule requires that the integral be broken into intervals. Since the integral is evaluating the area under a curve, from x=0 to x= (2kDh-h2)^.5, the method first calculates the maximum value for x, divides that by the number of intervals, and then evaluates the function for each value of x. In other words, if the maximum value of x was 10 and there were 10 intervals, then the function would be evaluated with x = 0, 1, 2, 3, 4, etc. Notice that the term called “n” above includes x in its formula.
An even number of intervals is required. The more the better. I found through trial and error that a good number to use for this particular problem is 1000 intervals.
Implementation could be done in a tabular form on an Excel worksheet. However, it is much more elegant to solve Simpson’s Rule in a Visual Basic for Applications function subroutine. Listing 1 gives the function. This is located in a VBA Module within TankVolume. Each place the calculation is required, the function is called using a cell formula of the form:
= Toris_V1(k, D, h)
where the variables k, D, and h are as defined above. Since recalculation takes time, which can be noticable, I put the function call in a conditional statement so it is only used when the tank is horizontal with torispherical heads. Assume the variable “head_type” refers to the type of head and a value of head_type=4 refers to torispherical, the conditional function call becomes:
In this case, the function is called only if the heads are torispherical. Otherwise, a value of “0” is returned. This is perfectly fine since the result of this cell in the spreadsheet is only used for torispherical heads.
Function Toris_V1(k, D, H) As Double ' ' Integral solution using Simpson's Rule ' Dim interval As Double Dim i As Integer Dim n As Double, n1 As Double, n2 As Double Dim X As Double, xmax As Double Dim steps As Integer Dim func As Double, sumfunc As Double ' On Error GoTo err_TorisV1 steps = 1000 ' '----------------------------------------------------------- ' Procedure: ' evaluate the maximum value for x (minimum value=0) ' for each value of x from min to max, at each interval, ' calculate n ' calculate the function ' sum results according to Simpson's Rule ' calculate result and return ' ' Derived function (arcsin is not an intrinsic function) ' Arcsin(x) = Atn (x / Sqr(-x * x + 1)) '----------------------------------------------------------- ' ' maximum value of x xmax = (2 * k * D * H - H ^ 2) ^ 0.5 R = D / 2 w = R - H interval = xmax / steps sumfunc = 0 X = 0 n1 = R - k * D n2 = k ^ 2 * D ^ 2 For i = 0 To steps n = n1 + Sqr(n2 - X ^ 2) func = Sqr(n ^ 2 - w ^ 2) / n func = n ^ 2 * Atn(func / Sqr(-func * func + 1)) - w * Sqr(n ^ 2 - w ^ 2) sumfunc = sumfunc + func If i > 0 And i < steps Then If (i / 2) = Int(i / 2) Then sumfunc = sumfunc + func Else sumfunc = sumfunc + 3 * func End If X = X + interval If X > xmax Then X = xmax Next i ' Toris_V1 = sumfunc * interval / 3 Exit Function ' err_TorisV1: Toris_V1 = 0 ' End Function | 1,539 | 6,006 | {"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-2013-20 | longest | en | 0.933952 |
https://howkgtolbs.com/convert/2.91-kg-to-lbs | 1,631,992,751,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00629.warc.gz | 343,719,763 | 12,118 | # 2.91 kg to lbs - 2.91 kilograms into pounds
Before we get to the practice - this is 2.91 kg how much lbs calculation - we want to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 2.91 kg to lbs? 2.91 kilograms it is equal 6.4154518242 pounds, so 2.91 kg is equal 6.4154518242 lbs.
## 2.91 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, that is International System of Units (in abbreviated form SI).
From time to time the kilogram could be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but difficult to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.
## 2.91 kilogram to pounds
You know something about kilogram, so now let's go to the pound. The pound is also a unit of mass. We want to emphasize that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. Of course, this unit is in use also in another systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is exactly 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 2.91 kg?
2.91 kilogram is equal to 6.4154518242 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 2.91 kg in lbs
The most theoretical section is already behind us. In this section we will tell you how much is 2.91 kg to lbs. Now you know that 2.91 kg = x lbs. So it is time to know the answer. Have a look:
2.91 kilogram = 6.4154518242 pounds.
That is a correct outcome of how much 2.91 kg to pound. It is possible to also round off the result. After rounding off your outcome is as following: 2.91 kg = 6.402 lbs.
You know 2.91 kg is how many lbs, so have a look how many kg 2.91 lbs: 2.91 pound = 0.45359237 kilograms.
Of course, this time it is possible to also round off the result. After it your outcome will be exactly: 2.91 lb = 0.45 kgs.
We are also going to show you 2.91 kg to how many pounds and 2.91 pound how many kg results in tables. Have a look:
We are going to start with a table for how much is 2.91 kg equal to pound.
### 2.91 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
2.91 6.4154518242 6.4020
Now look at a table for how many kilograms 2.91 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
2.91 0.45359237 0.45
Now you learned how many 2.91 kg to lbs and how many kilograms 2.91 pound, so it is time to move on to the 2.91 kg to lbs formula.
### 2.91 kg to pounds
To convert 2.91 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 6.4154518242 result in pounds
The first version of a formula will give you the most accurate outcome. In some situations even the smallest difference can be considerable. So if you need an accurate outcome - first formula will be the best for you/option to calculate how many pounds are equivalent to 2.91 kilogram.
So move on to the another formula, which also enables calculations to know how much 2.91 kilogram in pounds.
The shorter version of a formula is as following, look:
Number of kilograms * 2.2 = the result in pounds
As you see, this version is simpler. It can be the best option if you need to make a conversion of 2.91 kilogram to pounds in quick way, for example, during shopping. You only have to remember that your result will be not so correct.
Now we want to learn you how to use these two formulas in practice. But before we will make a conversion of 2.91 kg to lbs we want to show you easier way to know 2.91 kg to how many lbs without any effort.
### 2.91 kg to lbs converter
Another way to learn what is 2.91 kilogram equal to in pounds is to use 2.91 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on longer version of a formula which we showed you above. Due to 2.91 kg pound calculator you can effortless convert 2.91 kg to lbs. Just enter number of kilograms which you need to convert and click ‘convert’ button. The result will be shown in a flash.
So try to convert 2.91 kg into lbs using 2.91 kg vs pound converter. We entered 2.91 as an amount of kilograms. Here is the outcome: 2.91 kilogram = 6.4154518242 pounds.
As you see, our 2.91 kg vs lbs converter is so simply to use.
Now we are going to our main topic - how to convert 2.91 kilograms to pounds on your own.
#### 2.91 kg to lbs conversion
We are going to begin 2.91 kilogram equals to how many pounds conversion with the first version of a formula to get the most accurate result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 6.4154518242 the result in pounds
So what have you do to check how many pounds equal to 2.91 kilogram? Just multiply number of kilograms, this time 2.91, by 2.20462262. It gives 6.4154518242. So 2.91 kilogram is exactly 6.4154518242.
It is also possible to round off this result, for example, to two decimal places. It is exactly 2.20. So 2.91 kilogram = 6.4020 pounds.
It is time for an example from everyday life. Let’s calculate 2.91 kg gold in pounds. So 2.91 kg equal to how many lbs? And again - multiply 2.91 by 2.20462262. It is exactly 6.4154518242. So equivalent of 2.91 kilograms to pounds, if it comes to gold, is 6.4154518242.
In this example it is also possible to round off the result. Here is the outcome after rounding off, in this case to one decimal place - 2.91 kilogram 6.402 pounds.
Now we can go to examples converted using a short version of a formula.
#### How many 2.91 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 6.402 the outcome in pounds
So 2.91 kg equal to how much lbs? And again, you need to multiply amount of kilogram, this time 2.91, by 2.2. Look: 2.91 * 2.2 = 6.402. So 2.91 kilogram is 2.2 pounds.
Let’s make another calculation with use of shorer version of a formula. Now convert something from everyday life, for example, 2.91 kg to lbs weight of strawberries.
So convert - 2.91 kilogram of strawberries * 2.2 = 6.402 pounds of strawberries. So 2.91 kg to pound mass is equal 6.402.
If you learned how much is 2.91 kilogram weight in pounds and can convert it with use of two different formulas, let’s move on. Now we want to show you these outcomes in charts.
#### Convert 2.91 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in charts for your convenience. Due to this you can easily make a comparison 2.91 kg equivalent to lbs outcomes.
Begin with a 2.91 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
2.91 6.4154518242 6.4020
And now have a look at 2.91 kg equal pound chart for the second version of a formula:
Kilograms Pounds
2.91 6.402
As you can see, after rounding off, if it comes to how much 2.91 kilogram equals pounds, the results are the same. The bigger amount the more significant difference. Keep it in mind when you need to do bigger number than 2.91 kilograms pounds conversion.
#### How many kilograms 2.91 pound
Now you know how to calculate 2.91 kilograms how much pounds but we are going to show you something more. Do you want to know what it is? What do you say about 2.91 kilogram to pounds and ounces conversion?
We will show you how you can convert it little by little. Let’s start. How much is 2.91 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 2.91, by 2.20462262. So 2.91 * 2.20462262 = 6.4154518242. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To check how much 2.91 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final outcome is equal 2 pounds and 33 ounces.
As you can see, conversion 2.91 kilogram in pounds and ounces quite simply.
The last calculation which we will show you is calculation of 2.91 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate it you need another formula. Before we give you this formula, let’s see:
• 2.91 kilograms meters = 7.23301385 foot pounds,
• 2.91 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 2.91 foot pounds to kilograms meters you need to multiply 2.91 by 0.13825495. It is 0.13825495. So 2.91 foot pounds is exactly 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 2.91 foot pounds is 0.14 kilogram meters.
We hope that this calculation was as easy as 2.91 kilogram into pounds conversions.
This article was a huge compendium about kilogram, pound and 2.91 kg to lbs in conversion. Thanks to this conversion you learned 2.91 kilogram is equivalent to how many pounds.
We showed you not only how to do a conversion 2.91 kilogram to metric pounds but also two another calculations - to check how many 2.91 kg in pounds and ounces and how many 2.91 foot pounds to kilograms meters.
We showed you also other solution to make 2.91 kilogram how many pounds conversions, it is with use of 2.91 kg en pound calculator. It is the best option for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you are able to make 2.91 kilogram equal to how many pounds conversion - on your own or using our 2.91 kgs to pounds calculator.
Don’t wait! Convert 2.91 kilogram mass to pounds in the best way for you.
Do you want to make other than 2.91 kilogram as pounds calculation? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 2.91 kilogram equal many pounds.
### How much is 2.91 kg in pounds
We want to sum up this topic, that is how much is 2.91 kg in pounds , we prepared one more section. Here you can see all you need to remember about how much is 2.91 kg equal to lbs and how to convert 2.91 kg to lbs . It is down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 2.91 kg to pound conversion formula look? . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 2.91 kilogram to pounds? The correct answer is 6.4154518242 lb.
There is also another way to calculate how much 2.91 kilogram is equal to pounds with second, shortened version of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So now, 2.91 kg equal to how much lbs ? The result is 6.4154518242 lb.
How to convert 2.91 kg to lbs quicker and easier? You can also use the 2.91 kg to lbs converter , which will make whole mathematical operation for you and give you a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,530 | 13,317 | {"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-2021-39 | longest | en | 0.943489 |
https://www.cleariitmedical.com/2021/07/sets-1-quiz-5.html | 1,725,831,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00211.warc.gz | 687,059,455 | 121,582 | ## SETS-1 Quiz-5
In mathematics a set is a collection of distinct elements. The elements that make up a set can be any kind of things: people, letters of the alphabet, numbers, points in space, lines, other geometrical shapes, variables, or even other sets. Two sets are equal if and only if they have precisely the same elements. Sets are ubiquitous in modern mathematics. Indeed, set theory, more specifically Zermelo–Fraenkel set theory, has been the standard way to provide rigorous foundations for all branches of mathematics since the first half of the 20th century..
Q1. In a rehabilitation programme, a group of 50 families were assured new houses and compensation by the government. Number of families who got both is equal to the number of families who got neither of the two. The number of families who got new houses is 6 greater than the number of families who got compensation. How many families got houses?
• 22
• 28
• 23
• 25
Solution
(b) Let A and B denote respectively the sets of families who got new houses and compensation It is given that n(A∩B)=n((A∪B) ̅) ⇒n(A∩B)=50-n(A∪B) ⇒n(A)+n(B)=50 ⇒n(B)+6+n(B)=50 [∵n(A)=n(B)+6 (given)] ⇒n(B)=22⇒n(A)=28
Q2.If A and B are two sets, then A-(A-B) is equal to
• B
• A∪B>
• A∩B
• B-A
Solution
(C)
Q3. Let A and B be two non-empty subsets of a set X such that A is not a subset of B. Then,
• A is a subset of complement of B
• B is a subset of A
• A and B are disjoint
• A and the complement of B are non-disjoint
Solution
(D)
Q4. If A={a,b,c},B={b,c,d} and C={a,d,c}, then (A-B)×(B∩C) is equal to
• If A={a,b,c},B={b,c,d} and C={a,d,c}, then (A-B)×(B∩C) is equal to {(a,c),(a,d)}
• {(a,b),(c,d)}
• {(c,a),(d,a)}
• {(a,c),(a,d),(b,d)}
Solution
(a) Given, A={a,b,c}, B={b,c,d} and C={a,d,c} Now, A-B={a,b,c}-{b,c,d}={a} And B∩C={b,c,d}∩{a,d,c}={c,d} ∴(A-B)×(B∩C)={a}×{c,d} ={(a,c),(a,d)}
Q5. If A and B are two sets such that n(A)=7,n(B)=6 and (A∩B)≠Ï•. The least possible value of n(A Δ B),
• 1
• 7
• 6
• 13
Solution
(a) We have, A ∆ B=(A∪B)-(A∪B) ⇒n(A ∆ B)=n(A)+n(B)-2 n(A∩B) So, n(A ∆ B) is greatest when n(A∩B) is least It is given that A∩B≠Ï•. So, least number of elements in A ∩B can be one ∴ Greatest possible value of n(A ∆ B) is 7+6-2×1=11
Q6. If A={1,2,3,4}, then the number of subsets of A that contain the element 2 but not 3, is
• 16
• 4
• 8
• 24
Solution
(b) Required number of subsets is equal to the number of subsets containing 2 and any number of elements from the remaining elements 1 and 4 So, required number of elements =2^2=4
Q7. If A={p∈N:p is a prime and p=(7n^2+3n+3)/n for some n∈N}, then the number of elements in the set A, is
• 1
• 2
• 3
• 4
Solution
(a) We have, p=(7n^2+3n+3)/n⇒p=7n+3+3/n It is given that n∈N and p is prime. Therefore, n=1 ∴n(A)=1
Q8. Let Y={1,2,3,4,5},A{1,2},B={3,4,5} and Ï• denotes null set. If (A×B) denotes cartesian product of the sets A and B; then (Y×A)∩(Y×B) is
• Y
• TA
• B
• Ï•
Solution
(d) (Y×A)={(1,1),(1,2),(2,1),(2,2), (3,1),(3,2),(4,1),(4,2),(5,1),(5,2)} And(Y×B)={(1,3),(1,4),(1,5),(2,3), (2,4),(2,5),(3,3),(3,4),(3,5),(4,3), (4,4),(4,5),(5,3),(5,4),(5,5)} ∴(Y×A)∩(Y×B)=Ï•
Q9. In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is
• 80%
• 40%
• 60%
• 70%
Solution
(c) Clearly, Required percent =20+50-10=60% [∵n(A∪B)=n(A)+n(B)-n(A∩B)]
Q10. If A={x∶x is a multiple of 4} and, B={x∶x is a multiple of 6}, then A∩B consists of multiples of
• 16
• 12
• 8
• 4
Solution
(b) Let x∈A∩B. Then, x∈A and x∈B ⇒x is a multiple of 4 and x is a multiple of 6 ⇒x is a multiple of 4 and 6 both ⇒x is a multiple of 12
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http://www.sciencepublishinggroup.com/journal/paperinfo?journalid=239&doi=10.11648/j.jeee.20140201.15 | 1,601,320,370,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401604940.65/warc/CC-MAIN-20200928171446-20200928201446-00002.warc.gz | 209,760,460 | 9,406 | Archive
Special Issues
Development of MATLAB-Based Software for the Design of the Magnetic Circuit of Three-Phase Power Transformer
Journal of Electrical and Electronic Engineering
Volume 2, Issue 1, February 2014, Pages: 28-35
Received: Feb. 12, 2014; Published: Mar. 30, 2014
Authors
Obinwa Christian Amaefule, Department of Electrical Engineering, Imo State University (IMSU), Owerri, Nigeria
Afolayan Jimoh Jacob, Department of Electrical/Electronic and Computer Engineering, University of Uyo, AkwaIbom, Nigeria
Akaninyene Bernard Obot, Department of Electrical/Electronic and Computer Engineering, University of Uyo, AkwaIbom, Nigeria
Article Tools
Abstract
Essentially, transformers consist of two electrical conductors called the primary winding and the secondary winding which are coupled magnetically together by a magnetic circuit. Transformers work based on the principle that energy can be efficiently transferred by magnetic induction from one winding to another winding by a varying magnetic field produced by alternating current. The magnetic circuit or core of a transformer is designed to provide a path for magnetic field, which is necessary for induction of voltages between the windings. In this paper, MATLAB-based software was developed for automatic computation of the magnetic circuit parameters of a three phase power transformer once the input specifications are supplied. A sample design problem was used to demonstrate the effectiveness of the program. Apart from its flexibility and speed, the program removed the drudgery involved in the design. In addition, the MATLAB-based software presented in this paper will serve as a useful teaching and laboratory tool for undergraduate courses in transformer design
Keywords
Transformer, Magnetic Circuit, Power Transformer, Three Phase, Single Phase, MATLAB
Obinwa Christian Amaefule, Afolayan Jimoh Jacob, Akaninyene Bernard Obot, Development of MATLAB-Based Software for the Design of the Magnetic Circuit of Three-Phase Power Transformer, Journal of Electrical and Electronic Engineering. Vol. 2, No. 1, 2014, pp. 28-35. doi: 10.11648/j.jeee.20140201.15
References
[1]
Amoiralis, E. I., Tsili, M. A., & Kladas, A. G. (2009). Transformer design and optimization: a literature survey. Power Delivery, IEEE Transactions on, 24(4), 1999-2024.
[2]
Amoiralis, E. I., Tsili, M. A., and Georgilakis, P. S. (2008). The state of the art in engineering methods for transformer design and optimization: a survey. Journal of Optoelectronics and Advanced Materials, 10(5), 1149.
[3]
IEEE (2002) IEEE Standard Terminology for Power and Distribution Transformers, IEEE Std C57.12.80-2002.
[4]
Mittle VN, Mittal A (1996) Design of electrical machines, 4th edn. Standard Publishers Distributors, Nai Sarak, Delhi
[5]
Amoiralis. E. I., Georgilakis. P. S., Tsili. M. A., Kladas A.G.and Souflaris A. T. (2011) A complete software package for transformer design optimization and economic evaluation analysis . Materials Science Forum Vol. 670 (2011) pp 535-546) Trans Tech Publications, Switzerland
[6]
Judd F. F., Kressler D. R. (1977), IEEE Trans. Magn., vol. MAG-13, pp. 1058-1069.
[7]
Poloujadoff M., Findlay R. D. (1986), IEEE Trans. Power Sys., vol. PWRS-1.
[8]
Jewell W. T. (1990) IEEE Trans. Power Sys., vol. 5 , pp. 499-505.
[9]
Grady W. M., Chan R., Samotyj M. J., Ferraro R. J., Bierschenk J. L. (1992) IEEE Trans. Power Sys.,vol. 7 (1992), pp. 709-717.
[10]
Rubaai A. (11994), IEEE Trans. Power Sys., vol. 9 , pp. 1174-1181.
[11]
Andersen O. W(1991), IEEE Comput. Appl. Power, vol. 4 ,pp. 11–15.
[12]
Hernandez C., Arjona M. A., Shi-Hai Dong (2008) IEEE Trans. Magn., vol. 44 , pp. 2332-2337.
[13]
Ozuomba, S. et al (2004) “Software Development for the Design of Electric Circuit of 3-Phase Power Transformer” University of Uyo - A draft paper
[14]
Oboma, S. O (2003) Development of Computer Program for the Design of 3-phase Power Transformer, University of Uyo.
PUBLICATION SERVICES | 1,072 | 3,952 | {"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-2020-40 | latest | en | 0.903475 |
https://www.vanessabenedict.com/gold-weight/ | 1,680,288,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00286.warc.gz | 1,155,407,452 | 15,387 | How to measure gold weight?
ByVanessa
Jun 8, 2022
Gold has an atomic weight of 196.96657 u, though it’s typically measured in troy ounces. (For reference, a troy ounce is 31.1 grams, or 0.07 pounds.) Compared to other metals, it’s rather heavy. This is because of gold’s high density of 19.32 grams per cubic centimetre.
Untitled Document
How to measure gold weight
To accurately measure excess gold, follow these steps: First, calibrate the balance by checking the size with an object of known mass.
Now that you have calibrated the scale yourself, place the main weighing item on the current scale and read the result.
To avoid confusion, list each section to be weighed and indicate each section by weight or against it, down to gold or caratage.
Untitled Document
How can I weigh gold at home
Buy a jewelry scale. Such a ladder is available online for less than \$50.
Use large size products if you can’t buy scales from another jeweler. If you have a food scale at home, most people can use it to weigh their own gold.
If you are unable or unwilling to purchase ascension yourself, take your scrap gold to a jeweler for weighing.
You may not find the Seal of Approval in some court cases for several reasons: The Seal has worn off
Stamp must be illegible
The gemstone was too large and a stamp was placed in the process.
Gold jewelry cannot be real gold.
It is important for you to get your expensive gold coins or jewelry completely online. You will most likely be misled when buying studded jewelry, almost as much as the weight can be disguised as the weight of the stones. .You .will not .enter .the .to .
What does 1 pound of gold weigh
Thus, a pound of feathers weighs 453.59 grams, while an ideal gold pound weighs about 373.24 grams.
How much does a gold bar weigh in kg
The standard gold bar held as gold in reserve banks and traded between bullion dealers is the 400 troy ounce (12.4 kg, 438.9 oz) vintage Good Delivery watch bar.
Untitled Document | 461 | 1,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.94136 |
https://woosterphysicists.scotblogs.wooster.edu/category/mathematics/ | 1,669,863,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710789.95/warc/CC-MAIN-20221201021257-20221201051257-00562.warc.gz | 658,113,087 | 7,870 | # Category Archives: Mathematics
## Alien Suns Reversing in Exoplanet Skies
Not only can suns stand still in the sky, from some exoplanets their motion can apparently reverse! Wooster physics-math double majors Xinchen (Ariel) Xie ’21 and Hwan (Michelle) Bae ’19 and I just published an article elucidating these apparent solar reversals. Michelle … Continue reading
## Slide Rule Examples
Slide rules were widely used in engineering, science, and mathematics until the early 1970s, including during the Gemini and Apollo space programs. Although rendered largely obsolete by the advent of inexpensive electronic calculators, their descendants continue to have specialized applications, such … Continue reading
## Slide Rule Theory
Slide rules were the analog computers that ruled science and engineering for 400 years. Their brilliant innovation was using logarithms to convert multiplication and division to addition and subtraction, and Slide rules feature logarithmic scales that slide past each other. … Continue reading
## 4D Unknot
In four dimensions, you can’t tie your shoelaces — because 4D knots don’t work. Any 1D curve in 4D space can be continuously deformed to the unit circle, which is an unknot. The looping animation below demonstrates how to undo a … Continue reading
Posted in Mathematics | 1 Comment
## Dandelin Spheres
In 1609, Johannes Kepler first described how planets orbit the sun in ellipses. Kepler understood an ellipse as both the locus of points whose distances from two foci sum to a constant and as the intersection of a cone and a plane. But how … Continue reading
## Spinors
Fermions like electrons, protons, and neutrons inhabit a 720° world: 360° rotations negate their quantum states, but 720° rotations restore them. A simple macroscopic model of such spinors is an arrow translating on a Möbius strip: as the center circle rotates, … Continue reading
## Squares & Cubes
Marvelously, the square of the sum of natural numbers is the sum of their cubes! Equivalently, the sum of their cubes is the square of their sum. This mathematical gem is attributed to Nicomachus of Gerasa who lived almost 2000 years … Continue reading
## Archimedes & Euler
A complex function that is its own derivative normalized to one at zero implicitly defines the famous Archimedean and Euler constants of circular motion and exponential growth. Even in a world of strong gravity, where the ratio of a circle’s circumference to its … Continue reading | 550 | 2,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-49 | longest | en | 0.872571 |
https://stackoverflow.com/questions/4305198/compare-ordering-of-two-lists | 1,571,419,751,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00440.warc.gz | 706,741,887 | 30,686 | # Compare ordering of two lists
Is there an algorithm better than O(n^2) for re-ordering List 2 to match List 1
List 1 : A B C D
List 2 : B D C A
note: List 2 can have more, less or even completely different items compared to List 1.
• Do you know anything special about the contents of the two lists? If the elements of list 1 and 2 are the same, it seems like you can just copy list 1 into list 2. If not, what do you do if you encounter an item in list 2 that isn't in list 1? Where does that get sorted to? – Dave McClelland Nov 29 '10 at 15:36
• can clarify? what is re-ordering? is it finding the permutation from list 2 to list 1? – lijie Nov 29 '10 at 15:37
• the edit just made things less clear! how does one reorder a list A to another list B when the elements may be different?! – lijie Nov 29 '10 at 16:29
• What n is? Length of List 1 or List 2 or list1+list2? – Fabio F. Nov 29 '10 at 18:24
If you can create a total-ordering for the type of items in the lists, then you can create an index for list 1 by sorting the items. You can then use this index to re-order list 2. This algorithm is O(n log n) in time, and requires an extra O(n) in space.
One possibility that would result in O(n Log(n)) would be the following. This requires reading the list values into an array/vector/sortable type of structure:
• Sort list 1 based on value and keep the position information. This provides a quick way to look up an item and know its position. The cost for the sort is O(n log n)
• Sort list 2 using the sorted list from the first step as the compare function. Two compare two values, look them up in the sorted list 1 results and use the relative positions as the comparison between the two values. The cost of this sort is also O(n log n).
After the edit, you mention that the second list may have no correlation to the first list. So the comparison function would have to take that into account. If one or both values being compared are not in the first list, then the compare function needs to make a decision on ordering the values (e.g., do they go at the end or the beginning?).
You are looking for the Diff Algorithm! | 559 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-43 | latest | en | 0.907621 |
https://cboard.cprogramming.com/cplusplus-programming/130764-floats.html?s=208756d40c0d4b683eedb37b77848488 | 1,607,048,083,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00080.warc.gz | 223,944,006 | 15,264 | 1. ## Floats
I was writing my own header for OpenGl rotation, but had alot of hiccups so I decided to take all the code to console and just cout my calculations to see where the problems were occuring. I finally started to notice some of the weird things that floats do, and I don't like it. Basicly I take an objects current angle from a Standard 0 that I decided (using the unit circle it is located at 1, 0, 0)
From this angle I add the change in angle to that then use the following formulas;
Code:
```float DeltaZ = CurrentAngle + NewAngle;
The degrees are converted to radians before the cos and sin function calls. (I found that the math wouldn't work if I gave it degrees). When cos and sin are called at angles that should put their value to 0, (ex. 0 degrees should cause Y to equal 0 on coordinates) the float goes into what appears to be scientific notation mode, and for some reason comparisons with the notation don't work consistently.
Using these functions I can't get X or Y to ever equal 0 but instead end up as
something like 3.42456328e-006. How can I stop the float from going into a scientific notation representation?
Second if I tell it to print a table such as
Degrees X Coordinate Y Coordinate Z Coordinate
0 1 0 0
then change the incriment of degrees change to .001 for a little while degrees goes up as normal but the longer this loop is the more likely I'll end up getting
.001999 instead of .002 and once it gets to that point then it keeps counting like that.
Knowing how important significant digits are in mathematic and scientific calculations I'm wondering if there is a way to control these floats to ONLY hold a number of digits equal to the change you are making. So if I'm increasing by
.001 then I never want to hold more than 3 digits after the decimal point. Any suggestions would be greatly appreciated.
2. Have you considered using a fixed-point representation?
3. >>something like 3.42456328e-006. How can I stop the float from going into a scientific notation representation?
I know I'm probably being curt or rude but, you can't, except for display purposes only. Scientific notation does not change something's value. And the internal representation depends on it.
>>then change the incriment of degrees change to .001 for a little while degrees goes
>>up as normal but the longer this loop is the more likely I'll end up getting
>>.001999 instead of .002 and once it gets to that point then it keeps counting like that.
There are a lot of real numbers that have approximations in floating point due to the space restrictions. If you just keep incrementing you are bound to accrue some rounding errors and end up with very different numbers. To add to that, some numbers in decimal do not have an exact representation in binary anyway like how say 0.3333.... isn't exact in decimal. There will always be numbers like that. Because of this, you don't want to use them to count if you can help it.
I can understand your frustration, but at the precision you're seeking you should not have that much of a problem. I think understanding a float's representation will help you fix the problems you have. If only because that is when people take the time to teach about its problems.
This document breaks down what a float actually is:
IEEE Standard 754 Floating-Point
and there is a section on the site about floating point that should be pretty approachable:
Cprogramming.com - Tutorials - Advanced C, C++, Graphics, Computer Science Tutorials
and this document is a lot more math heavy, but it explains basically everything about them:
What Every Computer Scientist Should Know About Floating-Point Arithmetic
But BMJ is right about fixed point being an answer as well. C++ doesn't have it built in I think.
4. Hmm seems from the information I've read that using fixed-point representation on any machine newer than a Pentium II is going to actually slow down the code. Thats concerning. I was hoping there was a built in C++ function for tracking significant digits that I wasn't aware of. Thnx for the info, not sure if I want to get side tracked on trying to write a library for it.
5. Google for "what every computer scientist should know about floating point".
Cutting a long story short:
Fractions are represented in binary. Just as 10 is represented as:
1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0
I.e. 8 + 0 + 2 + 0
A value of 0.625 is represented as:
1 * 2^-1 + 0 * 2^-2 + 1 * 2^-3
I.e. 0.5 + 0 + 0.125
Cause in a computer everything is most efficient using powers of two.
Now try representing 0.001 in binary, like I did with 0.625 above and you'll see where the problem is.
Or if you'd like a more interesting one to work out, try representing one-third, i.e. 0.333333... in binary!
Edit: Damn, too slow.
6. lol.... and we fall to where the real problem lies... my bianary understanding is non-existent. I've taught myself everything I know about C++ through the web and books (which explains the lack of understanding if have for certain areas of computer science) I do appreciate the information. Sucks lately seems none of the questions I ask have simple answers anymore! I'll be wading through the links you posted and attempting to understand binary to the wee hours of the morning I guess.
>>something like 3.42456328e-006. How can I stop the float from going into a scientific notation representation?
I know I'm probably being curt or rude but, you can't, except for display purposes only. Scientific notation does not change something's value. And the internal representation depends on it.
I know scientific notation doesn't change value, but my comparisons aren't working.
Well is it normal that I tried something like
Code:
```if(Vector.x < .0001 && Vector.x > -.0001)
Vector.x = 0;```
and the result still comes out as 3.42356e-006. Is it not possible to use comparisons at this range or am I doing something wrong?
Thanks again ;o).
7. Originally Posted by Lesshardtofind
Well is it normal that I tried something like
Code:
```if(Vector.x < .0001 && Vector.x > -.0001)
Vector.x = 0;```
and the result still comes out as 3.42356e-006.
Thanks again ;o).
I'd guess you're doing something wrong that you haven't shown.
There is the concern that negative powers of 10 (0.1, 0.01, etc) can be represented exactly in binary, however that won't cause what you're describing.
My guess would be that you either have a typo in your code so it isn't quite what you've posted here, or there is some other logic error that is preventing the code you're showing being executed.
Without seeing a small but representative example of actual code (by which I mean something that is a verbatim copy of code fed to the compiler and that then exhibits your problem) we probably can't help further. One of the joys of forums, when you tell us where you think the problem is, is that there is a good chance you leave out something relevant.
8. Yea woke up this morning and the typo became obvious. Thnx for the comment Grumpy. And thnx to everyone else who gave me info on this. | 1,646 | 7,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-50 | latest | en | 0.954487 |
https://www.mrexcel.com/board/threads/wieghted-average.121666/ | 1,638,816,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00106.warc.gz | 967,698,032 | 14,835 | # Wieghted average
#### rsoper
##### New Member
Sorry by re-posting twice, I was not sure how to add additonal inforamtion to my first cry for help.
I have included the text from first request and added addtional text..
I have a table that calculates on time delivery.
Values are:
Items delivered Items delivered on time OTD
A B =1-(A-B)/B
There are many items in this file.
.
.
I am interested in creating a value for weighted average, where a row with more items delivered would have a higher impact on the total weighted average than items with less delivered.
.
.
Example
Items Items
Ordered On-time" Formula
1364 1223 89.66%
937 800 85.38%
879 777 88.40%
703 669 95.16%
28 9 32.14%
18 5 27.78%
1 0 0.00%
13 0 0.00%
Total Item
3943 3483 52.32% Average
As I stated the Formula is = 1 - (A-B)/A
This value gives "on-time" %. But look at the numbers when averaging the formula column. 52.32% is the correct average, but it is being skewed by the last 4 items with low %'s. I want to make a value that considers the fact that the first 4 items had many more items thus should have a higher weighting making the weighted average closer to 80% area.
"weighted average".
### Excel Facts
Test for Multiple Conditions in IF?
Use AND(test, test, test, test) or OR(test, test, test, ...) as the logical_test argument of IF.
#### Barrie Davidson
##### MrExcel MVP
Why not try the formula on your totals? IE, 3483/3943 which is 88.3%.
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Go back | 665 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | longest | en | 0.907448 |
https://www.esaral.com/q/let-a-x-r-x-1-the-inverse-of-the-function-f-a-a-given-by-77312 | 1,720,911,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00677.warc.gz | 667,274,894 | 11,691 | # Let A={x ∈ R : x ≥ 1}. The inverse of the function, f : A→A given by
Question:
Let $A=\{x \in R: x \geq 1\}$. The inverse of the function, $f: A \rightarrow A$ given by $f(x)=2^{x(x-1)}$, is
(a) $\left(\frac{1}{2}\right)^{x(x-1)}$
(b) $\frac{1}{2}\left\{1+\sqrt{1+4 \log _{2} x}\right\}$
(c) $\frac{1}{2}\left\{1-\sqrt{1+4 \log _{2} x}\right\}$
(d) not defined
Solution:
Let $f^{-1}(x)=y \ldots(1)$
$\Rightarrow f(y)=x$
$\Rightarrow 2^{y(y-1)}=x$
$\Rightarrow 2^{y^{2}-y}=x$
$\Rightarrow y^{2}-y=\log _{2} x$
$\Rightarrow y^{2}-y+\frac{1}{4}=\log _{2} x+\frac{1}{4}$
$\Rightarrow\left(y-\frac{1}{2}\right)^{2}=\frac{4 \log _{2} x+1}{4}$
$\Rightarrow y-\frac{1}{2}=\pm \frac{\sqrt{4 \log _{2} x+1}}{2}$
$\Rightarrow y=\frac{1}{2} \pm \frac{\sqrt{4 \log _{2} x+1}}{2}$
$\Rightarrow y=\frac{1}{2}+\frac{\sqrt{4 \log _{2} x+1}}{2}$ $(\because \mathrm{y} \geq 1)$
So, $f^{-1}(x)=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right)$ | 433 | 970 | {"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} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.250062 |
https://forums.nrel.gov/t/query-about-the-tower-motion-in-monopile-wt/4588 | 1,679,627,397,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00740.warc.gz | 297,813,752 | 6,177 | # Query about the Tower motion in monopile WT
Hi everyone.
Recently, I am using FAST V8 to analyze the frequency response of a fixed offshore wind turbine. I found that there is a response peak of 0.15hz in the motion of the tower base when I set a no wind-wave condition. I want to know what caused this frequency response around the 0.15hz? (The 1P is 0.14hz, and the first-order natural frequency of the tower base is about 0.3hz).
Much thanks!
Dear @Chenzi.Yang,
The response peaks are typically found at excitation frequencies (harmonics of the rotor speed, peak wave frequency) and full-system natural frequencies. The excitation frequencies can be identified by the resulting rotor speed and wave input (`WaveTp`). The full-system natural frequencies can be identified through a linearization and subsequent eigenanalysis.
Best regards,
Dear Jason,
Many thanks for your response. In the case shown in the last figure, there exist no excitation frequencies due to the environment set of still wind and still water, and the rotor speed is 0. Can I regard that the frequency response peak (0.15Hz) in the tower base fore-after moment originated from the system’s natural frequency? However, through the free decay analysis, I found that the natural frequency of 1st Tower Fore-Aft is 0.3hz, which is far away from the peak frequency 0.15Hz. This result confuses me. Is it possible that this peak frequency 0.15Hz is caused by the dynamic response of the monopile foundation? How can I get the natural frequency of a monopile-like supported structure？
Best regards,
Dear @Chenzi.Yang,
Presumably your structural model uses a combination of ElastoDyn down to the transition piece and SubDyn between the transition piece and foundation; is that correct? It is not possible to linearize a model with SubDyn enabled in FAST v8, but this is possible in OpenFAST. Linearization followed by eigenanalysis is the most straightforward way of identifying the full-system natural frequencies. So, upgrading your model to OpenFAST may be best.
Otherwise, you can identify full-system natural frequencies through free-decay analyses. You didn’t mention how you performed a decay test, but presumably you gave an initial tower deflection and tracked the time-series response. I would expect multiple modes of the support structure would get initialized by this method that you could identify by computing the PSD of the time series of deflections and/or loads at nodes along the support structure. What do you see when you do this?
Best regards,
Dear Jason,
The initial tower deflection can be set with ElastDyn (as shown in the picture below) when using the free decay analysis.
However, for the support structure, I don’t know where I can initial its deflection. Could you please give me some pointers?
Best regards.
Dear @Chenzi.Yang,
The initial conditions you highlight will initialize the tower mode modes in ElastoDyn. The initial conditions for the platform just below that in the ElastoDyn input file will initialize the Guyan modes within SubDyn (that are solved through the platform DOFs of ElastoDyn). SubDyn does not currently support nonzero initialization of the Craig-Bampton modes. In reality, the full-system modes of the support structure will include a combination of ElastoDyn tower, ElastoDyn platform DOFs/SubDyn Guyan modes, and SubDyn Craig-Bampton modes. So, it would be very difficult to initialize the first support structure mode purely, and in reality, no matter what initial conditions you set, you will initialize several modes of the support structure.
Best regards,
Dear Jason,
Thanks for your reply. Through my test, I found that tower natural frequency shaft is caused by the flexibility of supporting structures, which settles my confusion.
But with free-decay modal analysis of monopile foundation tower, I found that there only exists the 1st natural frequency peak and the 2nd natural frequency response can’t be observed in FFT figure. What may cause such a phenomenon？
（ElastDyn file is set as follows when employing free-decay analysis）
Best regards,
Dear @Chenzi.Yang,
The tower deflection initial conditions set within ElastoDyn will be applied only to the first tower bending mode degrees of freedom. So, the second mode is not being excited in this case.
Best regards, | 922 | 4,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.919977 |
https://mathematica.stackexchange.com/questions/206482/how-to-replace-a-sub-expression-when-it-implicitly-exists-in-an-expression | 1,708,492,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00763.warc.gz | 403,621,599 | 39,404 | # How to replace a sub-expression when it implicitly exists in an expression
In order to make my question clear, I would like to use a simple expression as an toy example.
Suppose we have an expression, which is,
f = a + 2*b + c
Now I want to replace sub-expression (a+b) with term1, and (b+c) with term2, so that I can obtain the result like
f = term1 + term2
However, I tried functions Replace and ReplaceAll in Mathematica and the results are
[in] f = Replace[f, {(a + b) -> term1, (b + c) -> term2}]
[out] f = a + 2b + c
[in] f = ReplaceAll[f, {(a + b) -> term1, (b + c) -> term2}]
[out] f = a +2b + c
which does not make any changes.
So I was wondering if choose the wrong functions? Or there are some other tricks can be used to achieve this goal.
Simplify has an optional second argument which instructs Simplify to consider this additional information during the simplification process. This can help when the replacement you want is simpler than what you have.
f = a + 2*b + c;
Simplify[f,{a+b==term1,b+c==term2}]
instantly gives you
term1+term2
Try this:
f = a + 2*b + c;
f /. {a -> term1 - b, c -> term2 - b}
(* term1 + term2 *)
Have fun!
• This way is tricky but useful! Oct 11, 2019 at 2:29 | 356 | 1,227 | {"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} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.911376 |
https://webapps.stackexchange.com/questions/86711/combine-two-columns-into-a-list-of-all-possible-combinations-of-entries/154532 | 1,718,393,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00144.warc.gz | 562,179,945 | 47,997 | # Combine two columns into a list of all possible combinations of entries
I have two columns such as:
``````a 1
b 2
c
``````
and I need to combine them like:
``````a 1
a 2
b 1
b 2
c 1
c 2
``````
Is it possible with a formula?
• @TomWoodward No, it's a different one. offa wants a Cartesian product of two sets, not row-by-row concatenation.
– user79865
Commented Nov 15, 2015 at 20:43
• Here you can see another way to achieve this by using Script Add-ons. Hope you will find it useful. Commented Nov 15, 2020 at 0:09
Although this is a special case of In a Google Spreadsheet, show all combinations for a selection of columns I think it's good to have a simpler answer specifically for the case of two columns. The technical term is "Cartesian product of two sets".
I use the same method as Rubén, which requires a character that does not appear in the column entries. Rubén used comma in his example. I prefer something more exotic, e.g. `char(9999)`, which is a pencil: ✏.
Here are the formulas for joining columns A and B in a Cartesian product:
In cell C1:
``````=transpose(split(join("", arrayformula(rept(filter(A1:A, len(A1:A))&char(9999), counta(B1:B)))), char(9999)))
``````
In cell D1:
``````=transpose(split(rept(join(char(9999), filter(B1:B, len(B1:B)))&char(9999), counta(A1:A)), char(9999)))
``````
### Explanation
The formula in C:
1. takes nonempty entries in A
2. puts ✏ next to each
3. repeats each such combo as many times as there are entries in B
4. joins them into a✏a✏b✏b✏c✏c✏
5. splits by pencil character into a row a a b b c c
6. transposes the row so that it becomes a column
The formula in D:
1. takes nonempty entries in B
2. joins them, separated by ✏
3. repeats the entire string as many times as there are entries in A, getting 1✏2✏1✏2✏1✏2✏
4. splits by pencil character into a row 1 2 1 2 1 2
5. transposes the row so that it becomes a column
• thanks for the answer and for the explanations. Works well.
– offa
Commented Nov 16, 2015 at 6:45
• It's a good trick but you're likely to run into a cumbersome limitation if you apply it to a large enough data set: the concatenation result cannot exceed 50.000 characters. Faced with the same problem, I ended up creating a custom function to perform a proper SQL JOIN-like operation on two arbitrary ranges/arrays. Commented Aug 18, 2017 at 14:26
## Update March 2024
Google Sheets nowadays has `tocol()` and `torow()` functions that lets you avoid the 50,000 character limitation that bugs solutions that use `join()`. Use this pattern:
``````=arrayformula(split(tocol(A2:A4 & "→" & torow(B2:B3)), "→"))
``````
In the event you do not know the number of rows in the source data in advance, and need to use open-ended range references, use an `ignore` parameter of `1` with the functions, like this:
``````=arrayformula(split(tocol(tocol(A2:A, 1) & "→" & torow(B2:B, 1)), "→"))
``````
This pattern, however, still hard-codes the number of columns, and starts to become unwieldy already when there are three columns to combine:
``````=arrayformula(split(tocol(tocol(A2:A, 1) & "→" & torow(B2:B, 1)) & "→" & torow(C2:C, 1), "→"))
``````
Further, these formulas use text string manipulation where `split()` may cause side effects such as converting the text string `1 2 3` to the date `2 January 2003` and other such unwanted transformations.
To avoid those side effects and implement a true n-ary Cartesian product, use a more dynamic formula such as the one below. It takes a range of N columns and gives all combinations of their non-blank values, column-wise:
``````=let(
table, A2:C,
blank, iferror(1/0),
first_, lambda(array, tocol(choosecols(array, 1), 1)),
rest_, lambda(n, choosecols(table, sequence(1, columns(table) - n, n + 1))),
wrap_, lambda(array, wrapCount, wraprows(tocol(array, 1), wrapCount)),
cartesian_, lambda(a, b, wrap_(
byrow(a, lambda(row,
reduce(blank, sequence(rows(b)), lambda(acc, i,
{ acc, row, chooserows(b, i) }
) )
) ),
columns(a) + columns(b)
) ),
iterate_, lambda(
self, a, b, if(iserror(b), a,
self(self, cartesian_(a, first_(b)), rest_(columns(a) + 1))
)
),
iterate_(iterate_, first_(table), rest_(1))
)
``````
The formula uses recursion and will work with any number of columns:
source data
a 1 X
b 2 Y
c
× Cartesian
a 1 X
a 1 Y
a 2 X
a 2 Y
b 1 X
b 1 Y
b 2 X
b 2 Y
c 1 X
c 1 Y
c 2 X
c 2 Y
Update November 2022
I tried to apply the solutions above, both had their own drawbacks. The first solution has a 50.000 character limitation, and the second solution had performance issues when using the Query function at scale.
So I've updated the last solution to be open-ended through the OFFSET and COUNTA formulas which should be performing better.
=arrayformula( split( flatten( offset(A2,0,0,COUNTA(A2:A)) & "µ" & transpose(offset(B2,0,0,COUNTA(B2:B)) )), "µ" ) )
Please note: this solution will create problems if you have empty cells in your list. For example this will not work with a data set that looks like:
``````a 1
b 2
3
c 4
``````
Hopefully this will be useful for others as well.
• Please cite the answers, as what is "above" or "before" may change over time or the answers may disappear altogether Commented Nov 14, 2022 at 10:53
This is known as the outer product. You may do so fairly straightforwardly, without resorting to weird `SPLIT` special characters, nor limits, nor performance issues, as follows:
## Quick, inelegant solution:
Assuming your data is in A3:A5 and B3:B7... (see elegant solution for a better way)
``````={
FLATTEN(
MAP(A3:A5, LAMBDA(x,
MAP(TRANSPOSE(B3:B7), LAMBDA(y,
x
))
))
),
FLATTEN(
MAP(A3:A5, LAMBDA(x,
MAP(TRANSPOSE(B3:B7), LAMBDA(y,
y
))
))
)
}
``````
Explanation: This uses the fact that `FLATTEN`'s order (row-major order) is the same, so you can flatten the x's and the y's separately, and be confident that when you zipper them back together with `{`xi`,`yi`}` they will be matched correctly.
Of course you should not use a range like A3:A, since you'd get 10000 blank entries, which would then get multiplied by 10000 B3:B entries, to give you 100,000,000 entries and make your sheet slow...
## Elegant solution:
Assuming your data is in A3:A and B3:B...
Define a Named Function `flatouter2d(as,bs,f)`:
``````=LAMBDA(bsT,
FLATTEN(
MYMAP1(as, LAMBDA(a,
MYMAP1(bsT, LAMBDA(b,
f(a,b)
))
))
)
)(TRANSPOSE(bs))
``````
tip: The builtin function `MAP` may SOMETIMES work fine, but will fail if your matrix is small (i.e. if you only had `a` or only had `1` in OP's example, they'd get an error). To workaround this bug with Google Sheets, you can define a Named Function `MYMAP1(xs,f)`=`IF( (ROWS(xs)<>1)+(COLUMNS(xs)<>1), MAP(xs, f), f(xs) )`.
Then define four variables, write a one-liner expression, and you're done; this is what you paste into your cell:
``````=LAMBDA(as,bs,takeA,takeB,
{ flatouter(as,bs,takeA) , flatouter(as,bs,takeB) }
)(
NONBLANKS(A3:A),
NONBLANKS(B3:B),
LAMBDA(a,b, a),
LAMBDA(a,b, b)
)
``````
(you can tell how this would be written like `as = NONBLANKS(A3:A); bs=NONBLANKS(...); takeA=LAMBDA(...)` in a more imperative programming language)
where by `NONBLANKS(xs)` we mean `FILTER(xs, xs<>"")` (you can define another Named Function, or type it out substituting your range for each occurrence of `xs`).
In conclusion, the following one-liner `{ flatOuter2d(as,bs,takeA) , flatOuter2d(as,bs,takeB) }` will do the trick in a performant way. It is a bit verbose unless you add a few Named Functions.
If you don't want to add some Named Functions, you can still squeeze it into a single cell if you really wanted to:
``````=LAMBDA(nonblanks,MYMAP1,
LAMBDA(as,bs,takeA,takeB,flatOuter2d,
{ flatOuter2d(as,bs,takeA) , flatOuter2d(as,bs,takeB) }
)(
NONBLANKS(A3:A),
NONBLANKS(B3:B),
LAMBDA(a,b,
a
),
LAMBDA(a,b,
b
),
LAMBDA(as,bs, f,
LAMBDA(bsT,
FLATTEN(
MYMAP1(as, LAMBDA(a,
MYMAP1(bsT, LAMBDA(b,
f(a,b)
))
))
)
)(TRANSPOSE(bs))
)
)
)(
LAMBDA(xs,
FILTER(xs, xs<>"")
),
LAMBDA(xs,f,
IF( (ROWS(xs)<>1)+(COLUMNS(xs)<>1),
MAP(xs, f),
f(xs)
)
)
)
``````
Then if you really only cared about this problem, you could shorten it a bit, but I wouldn't recommend it. I'd define MYMAP1 on principle as a Named Function, so I omit it below:
``````=LAMBDA(nonblanks,
LAMBDA(as,bs,flatOuterCol,
MAP({1,2}, LAMBDA(col,
flatOuterCol(as,bs, col)) )
)(
NONBLANKS(A3:A),
NONBLANKS(B3:B),
LAMBDA(as,bs, col,
LAMBDA(bsT,
FLATTEN(
MYMAP1(as, LAMBDA(a,
MYMAP1(bsT, LAMBDA(b,
IF(col=1,a,b)
))
))
)
)(TRANSPOSE(bs))
)
)
)(
LAMBDA(xs, FILTER(xs, xs<>"")),
)
``````
This is a "one-liner".
``````=MAP( {0,1}, LAMBDA(x,
FLATTEN(MAP( {"a";"b";"c"}, LAMBDA(a,
MAP( {1,2}, LAMBDA(n,
IF(x=0, a, n)
))
)))
))
``````
The idea is that you 'mappend' aka. 'flatmap'* aka. 'concatmap' i.e. `FLATTEN(MAP(` to keep all our things in one columns, and at the end you `MAP(` to splay things onto multiple rows. Keeping a loop structure is nice since it lets one refer to appropriate elements or indices as your use case may generally require.
• (or whatever it's called in your parlance, namely, the operation of 1) taking the pairs of the cartesian outer product, then 2) considering them as a 1d set ........ rather than just simply 1) considering the pairs as a 2d set. Namely, flatmap would be like `[x*y for x in X's for y in Y's]` in Python, as opposed to `[[x*y for x in X's...] for y in Y's]`)
This interestingly seems to work even if the arrays are of size 1, though I haven't proved it always works.
In general, the formula should be something like this: If you want general "block matrix"-style stuff where you can refer to everything in a closure, then you can do this pattern:
``````=LET(
Xs, {"x","y","z"},
Is, {0,1},
Ns, {1,2},
As, {"a";"b";"c"},
WRAPROWS(
FLATTEN(MAP(Xs,LAMBDA(x,
FLATTEN(MAP(Is,LAMBDA(i,
FLATTEN(MAP(Ns, LAMBDA(n,
MAP(As, LAMBDA(a,
a&n&x&i
))
)))
)))
)))
,COLUMNS(Xs)*COLUMNS(Is))
)
``````
and use whatever `IF`/`IFS` logic in the inner 'loop'. This will give you access to every variable 'index' in the block matrix, e.g.:
``````a1x0 a1y0 a1z0 a1x1 a1y1 a1z1
a2x0 a2y0 a2z0 a2x1 a2y1 a2z1
b1x0 b1y0 b1z0 b1x1 b1y1 b1z1
b2x0 b2y0 b2z0 b2x1 b2y1 b2z1
c1x0 c1y0 c1z0 c1x1 c1y1 c1z1
c2x0 c2y0 c2z0 c2x1 c2y1 c2z1
``````
Of course, you can resort to indexing, which you might need in more complicated cases. The below example should be extendable to most use cases. Here, we could (if we wanted to) access `ni` in the inner loop (second argument of the lambda passed into `forWithIndex`) to refer to the index (which in this case is the same as the element, but might not be). This design is equivalent to Python's `for x,i in enumerate(...])`. The below final version does away with asymmetric `;` rows vs columns and is rewritten in a way that is a bit easier to understand.
## Final Version
The example code below generates the block matrix [A x N] x [X x I], or said another way `[ [... for a in A for n in N] for x in X for i in I ]`. One may use `IF` statements in the inner part of the loop to reference values (use `for` and first argument of lambda), or alternatively, pairs of values and indices (use `forWithIndex` and first and second argument of lambda).
You can extend this to do something like Python's `range`s by using `SEQUENCE(length)` or `SEQUENCE(1,length)`. You must add the appropriate `*COLUMNS(...)` on the last line if adding another dimension which multiplies the cardinality of the columns (another `for` in the outer set of `for`s). You can of course also delete the "for loops" (delete the corresponding `))` and, if a column, the corresponding term of `*COLUMNS(...)`)
``````=LET(
Xs, {"x","y","z"},
Is, {0,1},
Ns, {1,2},
As, {"a","b","c"},
for, LAMBDA(list,f, FLATTEN(MAP(list, f))),
forWithIndex, LAMBDA(list,f, FLATTEN(MAP(list,SEQUENCE(1,COLUMNS(list)), f))),
WRAPROWS(
for(Xs,LAMBDA(x,
for(Is,LAMBDA(i,
forWithIndex(Ns, LAMBDA(n,ni,
for(As, LAMBDA(a,
"("&a&n&", "&x&i&")"
))
))
))
))
,COLUMNS(Xs)*COLUMNS(Is))
)
``````
output:
``````(a1, x0) (a1, y0) (a1, z0) (a1, x1) (a1, y1) (a1, z1)
(a2, x0) (a2, y0) (a2, z0) (a2, x1) (a2, y1) (a2, z1)
(b1, x0) (b1, y0) (b1, z0) (b1, x1) (b1, y1) (b1, z1)
(b2, x0) (b2, y0) (b2, z0) (b2, x1) (b2, y1) (b2, z1)
(c1, x0) (c1, y0) (c1, z0) (c1, x1) (c1, y1) (c1, z1)
(c2, x0) (c2, y0) (c2, z0) (c2, x1) (c2, y1) (c2, z1)
`````` | 4,083 | 12,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.918298 |
https://www.codeguru.com/cpp/g-m/opengl/article.php/c2683/Panning-zooming-and-rotation-in-OpenGL.htm | 1,532,256,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00028.warc.gz | 844,263,992 | 29,973 | # Panning zooming and rotation in OpenGL
### WEBINAR:On-Demand
Application Security Testing: An Integral Part of DevOps
## Pan,Zoom and Rotate the scene
In this program ,I show some simple OpenGL code.There is nothing special ,but I thing it can be useful for someone.
Features:
1. Zoom out,zoom in and pan the scene.
2. Draw the standard primiry like cylinder in ambiguous direction.
3. Draw the 3-D Mesh in a simple way.
In my program ,I use the class to wrap the detail of the realization in OpenGL.So I can use it freely at any other program without consider what will be do first ,and what be second. I have constructed a framework in Visual C++ 6.0,in which you can freely pan and rotate the scene ,without consider what is displaying in the scene. How did I do that ?
Apply the Camera
I construct CCamera and CScene class ,which take care of the rotate and pan of the camera and scene.In some program ,people use the glPushMatrix and glPopMatrix to accomplish the simple transfer.I do this in another way. By using the TransferMatrix class ,which I define to contain the MODELVIEW matrix of a special object(like the CCamera),thing will be simple.
We know that ,the OpenGL using the transform of MODELVIEW matrix to accomplish complex drawing ,and the glTranslate, glRotate is all base on the current MODELVIEW matrix. So if you want to transfer according to an ambiguous direction, it will be difficult. You have to mirror it into the current. Two suggestion to solve the problem.
The first is to use the matrix stack by using glPushMatix, glPopMatrix, but it is easy to be wrong when there are a lot of transform action. The second is to apply a matrix to the object, one matrix per object. Using the glLoadMatrix, glMultMatrix to apply the Object's matrix into the OpenGL.
So the course is like this:
```
DrawScene()
{
Init Camera ; //apply the Matrix of camera.
Init Scene; //apply the Matrix of scene.
for( i=0; i < ObjectNumber; i++)
{
Init Objcet[ i]; //apply the Matrix of object[i].
drawing Object[ i]; //drawing the object[i];
for( j=0; j < SubObjectNumber; i++) //if there is any subobject...
{
Init SubObject[ j];
drawing Subject[ j];
.......
}
......
}
}
// below is what I do in the code
m_Camera.Apply(); // set the camera transfer matrix
gluLookAt(5,5,2,0,0,0,0,0,1); // the initial view angle
m_Scene.Apply(); // set scene transfer matrix
```
Drawing ambiguous direction primiry
When I need a arrow pointer ,I think of the cylinder and cone. But how can I drawing it in ambiguous direction ? Thank's to a programmer I don't remember his name ,in his code , he use the cross product of the original vector and target. Vector as the rotate axis, solve the problem. For detail ,we can see the arrow.cpp.
How can we draw 3-D Mesh
There is no difficulty in doing this. But it is hard to set the normal of the point. So I use a CClixmax class to wrap the detail. And I write a function to get normal of a point by using the points around it. There is no complete project for all modules I have writed. For the space limited ,I only put the Polarize Light in the Archive. But all modules can be easily put into the project, because I have used the class to wrap it. For convenience I use the code downloaded from the URL below as a framwork, so you can see some claim of the copyright reserved. But there is no reason you have to do so, you can just use any framework you like.
http://devcentral.iftech.com/learning/tutorials/mfc-win32/opengl/
• #### good work
Posted by Legacy on 08/06/2003 12:00am
Originally posted by: mdrosdi
very good
• #### Great Help
Posted by Legacy on 10/17/1999 12:00am
Originally posted by: Holger
Thanx man, great work you did, it helped me as an OpenGL-beginner very much.
• #### Disagree
Posted by Legacy on 07/18/1999 12:00am
Originally posted by: Robbie
I disagree with the the above mail. Using vectors is not incorrect, but it does incure a performace hit. Each vector operation is evauated by GL to a matrix op (64 mults) so doing a single matrix op is better than doing several vector ops
• #### suggestion
Posted by Legacy on 06/20/1999 12:00am
Originally posted by: Xing Zhaohui
Vector can do everywork of matrix. You can replace matrix
with vectors in processing translate-scale-rotate.
• #### Good work
Posted by Legacy on 06/18/1999 12:00am
Originally posted by: Xing Zhaohui
^O^
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Let f : AB and g : BC denote two functions. Consider the following two statements:
S1 : If both f and g are injections then the composition function : AC is an injection.
S2 : If the function : A → C is surjection and g is an injection then the function f is a surjection.
S3 : If h(a) = g(f(a)) and h(a) is onto then g must be onto, where ∀a, aA.
Which of the above statements are valid?
all valid?
how??
I couldnt find any contradicting example.
What is your approach to above problem?
How did you started?
Just took 3 sets with cardinalities 3/4 and try to have simple mappings. Just you need to check if you can contadict the statement. There is no specific rule . I hope you get it.
There is some function h: A->C which is onto.
Given that h(a) = g( f(a) ), try to find a function 'g' which isnt onto. If you can find, then you have contradicted the statement.
+1 vote
Let f : A → B and g : B →
S1: if f and g are injection fuction then composition function gof :A → C is an injection. ----> this statement true its well known property.
S2:If the function gof : A → C is surjection and g is an injection then the function f is a surjection.
here keypoint is g is injection fuction.
means that g(f(x) = g(y) -----> f(x) = y so that fuction f is surjection
S3:If h(a) = g(f(a)) and h(a) is onto then g must be onto, where ∀aa ∈ A
all of the statement true. its all property .
answered by Boss (7.1k points) 5 40 149 | 426 | 1,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-43 | latest | en | 0.853806 |
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# Data types
R has several data types, including:
• Vectors
• Matrices
• Dataframes
• Lists
## Vectors
A vector is a simple data structure, where data is stored in one column. The simplest way to define a numeric vector is with the c() statement:
X <- c(1,2,3,5,6) # numeric vector
X
## [1] 1 2 3 5 6
### Colon notation
R has a colon notation to create series of numbers:
X <- c(1:6) # numeric vector
X
## [1] 1 2 3 4 5 6
### Function seq()
An explicit function is seq(from=X, to=Y, by=Z):
X <- seq(from = 1, to = 3, by = 0.25) # numeric vector
X
## [1] 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
seq() can be used with the parameter length:
X <- seq(from = 1, to = 3, length = 9) # numeric vector
X
## [1] 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
### Function rep()
You can create vectors containing repetitions with the function rep():
X <- rep(1:2, times = 3) # numeric vector
X
## [1] 1 2 1 2 1 2
The function rep() may have also the argument each:
X <- rep(1:3, each = 3) # numeric vector
X
## [1] 1 1 1 2 2 2 3 3 3
Both arguments times and each can be used together:
X <- rep(1:3, each = 2 , times = 2) # numeric vector
X
## [1] 1 1 2 2 3 3 1 1 2 2 3 3
### Type of vectors
A vector may contain text:
X <- c("A","B","C") # character vector
X
## [1] "A" "B" "C"
A vector may contain logical values:
X <- c(TRUE,TRUE,TRUE,FALSE,TRUE,FALSE) #logical vector
X
## [1] TRUE TRUE TRUE FALSE TRUE FALSE
A vector contain data of the same type. In the following examples, R interpret all the data as characters:
X <- c(1, "A", TRUE) #R will interpret all these values as text
X
## [1] "1" "A" "TRUE"
### Accessing elements of a vector
You may have access to a n-element of a vector by its index X[n]:
X <- c(1:6)
X[1]
## [1] 1
You may select multiple elements of a vector by specifying multiple indices, like X[x, y, z]:
X <- c(1:6)
X[c(2, 4)]
## [1] 2 4
### Removing elements of a vector
Elements can be also excluded by a negative index, like X[-n]:
X <- c(1:6)
X[-c(1:3)]
## [1] 4 5 6
The indices can be used also to substitute the value of an elements:
X[2] <- 20
X
## [1] 1 20 3 4 5 6
### Searching elements of a vector
Logical indices can be used to search specific elements of a vector:
X <- c(1, 3, 7, 4, 9, 2) # define the vector X
X[X > 4] # select only those elements with values higher than...
## [1] 7 9
## Matrices
Matrices are a collection of vectors, all of the same type. The elements are arranged in a two-dimensional rectangular layout.
### Function cbind()
A simple way to define matrices is with the cbind() function, which bind a series of vectors column-wise:
x <- c(1,2,3,4,5) # numeric vector
y <- c(1,2,3,4,5) # numeric vector
z <- c(1,2,3,4,5) # numeric vector
M <- cbind(x,y,z)
M
## x y z
## [1,] 1 1 1
## [2,] 2 2 2
## [3,] 3 3 3
## [4,] 4 4 4
## [5,] 5 5 5
Similar to cbind() function, there is also the rbind() function, which binds vectors by one row at a time:
M <- rbind(x, y, z)
M
## [,1] [,2] [,3] [,4] [,5]
## x 1 2 3 4 5
## y 1 2 3 4 5
## z 1 2 3 4 5
### Function matrix()
Matrices can be also defined by the matrix() function:
A = matrix(
c(1:6), # the data elements to be filled in
nrow=2, # the number of rows,
ncol=3, # the number of columns, and
byrow = TRUE) # filling the matrix rowwise (one row at a time).
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 4 5 6
### Select elements of a matrix
An element of a n x m matrix can be selected with its index M[n, m], similarly to vectors. Consider the matrix:
M <- matrix(round(runif(12, 5, 10), 0), # generate random integers
nrow = 3, # and fill into a matrix of 3 rows
ncol = 4) # and 4 columns.
M
## [,1] [,2] [,3] [,4]
## [1,] 9 9 6 10
## [2,] 6 7 9 6
## [3,] 7 9 6 8
To get the value of the first column:
M[, 1]
## [1] 9 6 7
To get the value of the second row:
M[2, ]
## [1] 6 7 9 6
To get the values of the first and third column:
M[, c(1, 3)]
## [,1] [,2]
## [1,] 9 6
## [2,] 6 9
## [3,] 7 6
To get the value of a specific element:
M[2, 3]
## [1] 9
To get the elements that satisfy a condition:
M[M > 7] # this returns a vector
## [1] 9 9 9 9 10 8
To get rid of a row or column, you can use a negative index:
M[, -2]
## [,1] [,2] [,3]
## [1,] 9 6 10
## [2,] 6 9 6
## [3,] 7 6 8
### Function dim()
Given a matrix:
M <- matrix(rep(c(1:4), times = 3), 3, 4)
M
## [,1] [,2] [,3] [,4]
## [1,] 1 4 3 2
## [2,] 2 1 4 3
## [3,] 3 2 1 4
The dimension of a matrix is given by the function dim():
dim(M) # return the number of rows and columns
## [1] 3 4
The function dim() can also be used to change dimensions:
dim(M) <- c(4, 3)
M
## [,1] [,2] [,3]
## [1,] 1 1 1
## [2,] 2 2 2
## [3,] 3 3 3
## [4,] 4 4 4
The same synthax is used to transform a matrix to a vector:
dim(M) <- c(12, 1)
M
## [,1]
## [1,] 1
## [2,] 2
## [3,] 3
## [4,] 4
## [5,] 1
## [6,] 2
## [7,] 3
## [8,] 4
## [9,] 1
## [10,] 2
## [11,] 3
## [12,] 4
### Function dimnames():
Sometime, it is more simple to refer to names instead of numerical indices. For this you can define the names of rows and columns by the fnuction dimnames():
M <- matrix(rep(c(1:3), times = 4), 3, 4)
dimnames(M) = list(
c("row1", "row2", "row3"), # row names
c("col1", "col2", "col3", "col4")) # column names
M
## col1 col2 col3 col4
## row1 1 1 1 1
## row2 2 2 2 2
## row3 3 3 3 3
## Lists
A list is the most flexible container of objects in R. Its elements can be unrelated, of any type and size.
mylist <- list(name=c("A", "B", "C"),
numb=c(1,2,3,4),
matr=cbind(c(2,1),c(1,2)),
vect=c(5,3,4,5,6,2))
mylist
## $name ## [1] "A" "B" "C" ## ##$numb
## [1] 1 2 3 4
##
## $matr ## [,1] [,2] ## [1,] 2 1 ## [2,] 1 2 ## ##$vect
## [1] 5 3 4 5 6 2
An object contained in the list can be accessed with the dollar notation:
mylist$name ## [1] "A" "B" "C" ### Dataframes Dataframes are between a list and a matrix. It is like a list since the columns can contain different types of objects (i.e. texts, numbers, factors). It is like a matrix since the output is a table. To create a dataframe from scratch: origin <- c("ITA", "AUT", "FRA") protein <- c(2, 3, 2) sugar <- c(8, 12, 10) mydata <- data.frame(origin,protein,sugar) mydata ## origin protein sugar ## 1 ITA 2 8 ## 2 AUT 3 12 ## 3 FRA 2 10 To edit manually the data, use the command "edit(mydata)" ### Factors With the dataframe "mydata", the variable "origin" should be used as categorical factor: mydata[,1] <- factor(mydata[,1]) This statement stores this vector as (1, 2, 1, 1) and associates it with 1=Type1 and 2=Type2 internally. ### Built in functions to check data structures str(mydata) # structure of an object ## 'data.frame': 3 obs. of 3 variables: ##$ origin : Factor w/ 3 levels "AUT","FRA","ITA": 3 1 2
## $protein: num 2 3 2 ##$ sugar : num 8 12 10
class(mydata) # class or type of an object
## [1] "data.frame"
names(mydata) # names
## [1] "origin" "protein" "sugar" | 2,854 | 7,481 | {"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} | 3.09375 | 3 | CC-MAIN-2024-30 | latest | en | 0.746849 |
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# AP+Statistics+CHapter+10+Section+3-1 - If so you need...
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AP Statistics Chapter 10 Section 3
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Decision or description of evidence? Making a decision is different in spirit from testing significance. Choosing a level of significance in advance makes sense if you must make a decision, but not if you wish only to describe the strength of your evidence.
How much evidence is needed to reject the null hypothesis? How plausible is the null hypothesis? If the null hypothesis represents an assumption that the people you must convince have believed for years, strong evidence (small level of significance) will be needed to persuade them. What are the consequences of rejecting the null hypothesis? Will the result of rejecting mean making an expensive change of some kind?
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Unformatted text preview: If so, you need strong evidence (small level of significance). There is no sharp border between “significant” and “insignificant”, only increasingly strong evidence as the P-value decreases. Hawthorne effect • Almost any change in the work place together with knowledge that a study is under way will produce a short-term productivity increase. Statistical Significance Works when: • You decide what effect you are seeking • Design a study to search for it • And use a test of significance to weigh the evidence. P-values and their importance P-values are more informative than the reject-or-not result of a fixed level of significance test....
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## This note was uploaded on 04/10/2011 for the course MATH 1070 taught by Professor Akbas during the Fall '08 term at Georgia State.
### Page1 / 8
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# Excel statistical functions: RSQ
Support for Office 2003 has ended
Microsoft ended support for Office 2003 on April 8, 2014. This change has affected your software updates and security options. Learn what this means for you and how to stay protected.
##### SUMMARY
This article describes the RSQ function in Microsoft Office Excel 2003 and in later versions of Excel. This article discusses how the function is used and compares the results of RSQ in these later versions of Excel with the results of RSQ in earlier versions of Excel.
The RSQ(array1, array2) function returns the Square of the Pearson Product-Moment Correlation Coefficient between two arrays of data.
### Syntax
``RSQ``
The arguments, array1 and array2, must be either numbers or names, array constants, or references that contain numbers.
The most common usage of RSQ includes two ranges of cells that contain the data, such as RSQ(A1:A100, B1:B100).
### Example of usage
To illustrate the RSQ function, follow these steps:
1. Create a blank Excel worksheet, and then copy the following table.
1 = 3 + 10^\$D\$2 Power of 10 to add to data 2 =4 + 10^\$D\$2 0 3 =2 + 10^\$D\$2 4 =5 + 10^\$D\$2 5 =4+10^\$D\$2 6 =7+10^\$D\$2 pre-Excel 2003 =RSQ(A1:A6,B1:B6) when D2 = 7.5 =PEARSON(A1:A6,B1:B6)^2 RSQ = PEARSON^2 0.492857142857143 =CORREL(A1:A6,B1:B6)^2 CORREL^2 0.509470304975923 when D2 = 8 RSQ = PEARSON^2 #DIV/0! CORREL^2 0.509470304975923
2. Select cell A1 in your blank Excel worksheet, and then paste the entries so that the table fills cells A1:D13 in your worksheet.
3. After you paste the table into your new Excel worksheet, click the Paste Options button, and then click Match Destination Formatting. With the pasted range still selected, use one of the following procedures, as appropriate for the version of Excel that you are running:
• In Microsoft Office Excel 2007, click the Home tab, click Format in the Cells group, and then click AutoFit Column Width.
• In Excel 2003, point to Column on the Format menu, and then click AutoFit Selection.
Note You may want to format cells B1:B6 as Number with 0 decimal places.
Cells A1:A6 and B1:B6 contain the two data arrays that are used in this example to call RSQ, PEARSON, and CORREL in cells A8:A10. RSQ is calculated by essentially calculating PEARSON and squaring the result. Because PEARSON and CORREL both compute the Pearson Product-Moment Correlation Coefficient, their results should agree. RSQ could have been (but was not) implemented as essentially calculating CORREL and squaring the result.
In versions of Excel that are earlier than Excel 2003, PEARSON may exhibit round-off errors. This behavior leads to round-off errors in RSQ. The behavior of PEARSON, and therefore of RSQ, has been improved for Excel 2003 and for later versions of Excel. CORREL has always been implemented by using the improved procedure that is found in Excel 2003 and in later versions of Excel. Therefore, an alternative to RSQ for an earlier version of Excel is to use CORREL instead and then to square the result.
In versions of Excel that are earlier than Excel 2003, you can use the worksheet in this article to run an experiment and to discover when round-off errors occur. If you add a constant to each of the observations in B1:B6, the values of RSQ, PEARSON^2, and CORREL^2 in cells A7:A9 should not be affected. If you increase the value in D2, a larger constant is added to B1:B6. If D2 <= 7, there are no round-off errors that appear in A7:A9. Now change the value of 7.25, 7.5, 7.75, and then 8. CORREL^2 in A9 is unaffected, but RSQ and PEARSON^2 ( (these expressions always agree with each other) show round-off errors in A7:A8. D6:D13 show values of RSQ = PEARSON^2 and CORREL^2 when D2 = 7.5 and 8, respectively.
Note that CORREL is still well-behaved, but round-off errors in PEARSON have become so severe that division by 0 occurs in RSQ and PEARSON^2 when D2 = 8.
Earlier versions of Excel exhibit incorrect answers in these cases because the effects of round-off errors are more profound with the computational formula that is used by these versions of Excel. Still, the cases that are used in this experiment may be viewed as extreme.
If you have Excel 2003 or a later version of Excel, you see no changes in values of RSQ and PEARSON^2 if you try the experiment. However, cells D6:D13 show round-off errors that you would have obtained with earlier versions of Excel.
### Results in earlier versions of Excel
If you name the two data arrays X's and Y's, earlier versions of Excel used a single pass through the data to compute the sum of squares of X's, the sum of squares of Y's, the sum of X's, the sum of Y's, the sum of XY's, and the count of the number of observations in each array. These quantities were then combined in the computational formula that is given in the Help file in earlier versions of Excel. The Help file for RSQ shows the formula for the Pearson Product-Moment Correlation Coefficient. This result is squared to obtain RSQ.
### Results in Excel 2003 and in later versions of Excel
The procedure that is used in Excel 2003 and in later versions of Excel uses a two-pass process through the data. First, the sums of X's and Y's and the count of the number of observations in each array are computed, and from these the means (averages) of X and Y observations can be computed. Then, on the second pass, the squared difference between each X and the X mean is found, and these squared differences are summed. The squared difference between each Y and the Y mean is found, and these squared differences are summed. Additionally, the products (X – X mean) * (Y – Y mean) are found for each pair of data points and summed. These three sums are combined in the formula for PEARSON. Notice that none of the three sums is affected if you add a constant to each value in the Y array (or in the X array). This behavior occurs because that same value is added to the Y mean (or to the X mean). In the numeric examples, even with a high power of 10 in cell D12, these three sums are not affected, and the results of the second pass are independent of the entry in cell D2. Therefore, the results in Excel 2003 and in later versions of Excel are more stable numerically.
### Conclusions
Replacing a one-pass approach by a two-pass approach guarantees better numeric performance of PEARSON, and therefore RSQ, in Excel 2003 and in later versions of Excel. The results that you obtain in Excel 2003 and in later versions of Excel will never be less accurate than results that you obtained in earlier versions of Excel.
In most practical examples, you are not likely to see a difference between the results in later versions of Excel and the results in earlier versions of Excel. This behavior occurs because typical data is unlikely to exhibit the kind of unusual behavior that this experiment illustrates. Numeric instability is most likely to appear in earlier versions of Excel when data contains a high number of significant digits combined with relatively little variation between data values.
The procedure of finding the sum of squared deviations about a sample mean by finding the sample mean, by computing each squared deviation, and by summing the squared deviations is more accurate than the alternative procedure. This alternative procedure was frequently named the "calculator formula" because it was suitable for use of a calculator on a small number of data points. The alternative procedure used the following procedure:
• Found the sum of squares of all observations, the sample size, and the sum of all observations
• Computed the sum of squares of all observations minus ([sum of all observations]^2)/sample size).
There are many other functions that have been improved for Excel 2003 and for later versions of Excel. These functions are improved because later versions of Excel replace the one-pass procedure with the two-pass procedure that finds the sample mean on the first pass and then computes the sum of squared deviations about the sample mean on the second pass.
The following list is a list of such functions:
• VAR
• VARP
• STDEV
• STDEVP
• DVAR
• DVARP
• DSTDEV
• DSTDEVP
• FORECAST
• SLOPE
• INTERCEPT
• PEARSON
• RSQ
• STEYX
Similar improvements were made in each of the three Analysis of Variance tools in the Analysis ToolPak.
Properties
Article ID: 828131 - Last Review: 01/17/2007 19:58:23 - Revision: 2.2
Microsoft Office Excel 2007, Microsoft Office Excel 2003
• kbformula kbhowto kbexpertisebeginner kbinfo KB828131 | 2,083 | 8,601 | {"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} | 3.765625 | 4 | CC-MAIN-2016-44 | latest | en | 0.824963 |
https://mathematica.stackexchange.com/questions/156866/sorting-elements-in-a-table-with-respect-to-another-table | 1,723,519,945,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00654.warc.gz | 298,416,745 | 42,113 | # Sorting elements in a table with respect to another table
I have two tables, each having three rows (R1, R2 & R3). The elements of R2 & R3 are same in two tables but in different orders. The elements in row R1 are different in two tables. I want to sort the elements of second table such that rows R2 & R3 become similar. For example, If
T1 = {{t1, 1, 2}, {t2, 2, 3}, {t3, 3, 2}, {t4, 2, 1}}
T2 = {{u1, 3, 2}, {u2, 1, 2}, {u3, 2, 1}, {u4, 2, 3}}
How can I sort elements of T2 such that second and third elements of each row are same as T1. That is
T3 = {{u2, 1, 2}, {u4, 2, 3}, {u1, 3, 2}, {u3, 2, 1}}
thanks
• Are there duplicate values in the 2nd and 3rd columns of the table? Commented Sep 30, 2017 at 8:52
• For example {t1, 1, 2} and {t2, 2, 1} are possible at same time, so are {t1, 1, 1} and {t2, 2, 2}. But {t1, 1, 2 } and {t2, 1, 2} are not possible. For a set of last two elements, first row can't have two different elements. thanks Commented Sep 30, 2017 at 9:07
• Then the association method will work. Commented Sep 30, 2017 at 9:17
There are multiple ways. One is to use Ordering.
Let p1 and p2 be the permutations that order T1 and T2 by their R2 & R3.
p1 = Ordering[T1[[All, 2 ;; 3]]]
(* {1, 4, 2, 3} *)
p2 = Ordering[T2[[All, 2 ;; 3]]]
(* {2, 3, 4, 1} *)
Let's invert p1:
invp1 = Ordering[p1]
(* {1, 3, 4, 2} *)
Then we sort T2, then transform that to the order seen in T1:
T2[[p2]][[invp1]]
(* {{u2, 1, 2}, {u4, 2, 3}, {u1, 3, 2}, {u3, 2, 1}} *)
Another way is to use Association
a1 = AssociationThread[
T1[[All, 2 ;; 3]],
T1[[All, 1]]
]
(* <|{1, 2} -> t1, {2, 3} -> t2, {3, 2} -> t3, {2, 1} -> t4|> *)
T2[[All, 2 ;; 3]],
T2[[All, 1]]
]
(* <|{3, 2} -> u1, {1, 2} -> u2, {2, 1} -> u3, {2, 3} -> u4|> *)
This assumes that there were no duplicates {R2, R3} pairs.
KeyTake[a2, Keys[a1]]
(* <|{1, 2} -> u2, {2, 3} -> u4, {3, 2} -> u1, {2, 1} -> u3|> *)
If you have Mathematica 10.0+ and no trouble with duplicate keys, I would use the association-based method, and I would keep the data structures as associations instead of tables.
• Thanks. Association is working fine for me..One more thing, how can I convert <|{1, 2} -> u2, {2, 3} -> u4, {3, 2} -> u1, {2, 1} -> u3|> back to table form like {{u2, 1, 2}, {u4, 2, 3}, {u1, 3, 2}, {u3, 2, 1}}. Commented Sep 30, 2017 at 10:21
• I tried - MapThread[Append, {Values[KeyTake[a2, Keys[a1]]], Keys[KeyTake[a2, Keys[a1]]]}] but coming back with error Append::normal: Nonatomic expression expected at position 1 in Append[u2,{1,2}]. Commented Sep 30, 2017 at 10:44
• @user49535 KeyValueMap[Prepend, ...]. But if you need the table form, use the full lists as association values instead of just the first element. Then you can just use Values in the end. Commented Sep 30, 2017 at 11:31
• I am struggling with this small issue, can you please elaborate. Not able to get back {{u2, 1, 2}, ....} by any method. Could this be because of u2 not being a numerical value ? Commented Oct 1, 2017 at 10:44
In addition to @Szabolcs's method, a more straightforward approach might be SortBy:
SortBy[T2, Position[T1[[All, 2 ;; 3]], #[[2 ;; 3]]] &]
(* {{u2, 1, 2}, {u4, 2, 3}, {u1, 3, 2}, {u3, 2, 1}} *)
which gives the answer you want. | 1,278 | 3,217 | {"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} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.871747 |
https://web2.0calc.com/questions/help_81667 | 1,603,196,654,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00089.warc.gz | 581,083,634 | 6,505 | +0
# help
0
86
2
Two points A and B are chosen at random on a circle of radius 1. What is the probability that AB is greater than 1?
Jun 8, 2020
#1
+31087
+1
Two points A and B are chosen at random on a circle of radius 1. What is the probability that AB is greater than 1?
Jun 8, 2020
#2
0
Two points A and B are chosen at random on a circle of radius 1. What is the probability that AB is greater than 1?
I think that no explanation is needed.
Jun 8, 2020 | 149 | 470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.939736 |
https://e-adventure.net/how-many-ounces-in-a-liter-discover-the-conversion-here/ | 1,685,840,751,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00550.warc.gz | 258,177,376 | 16,116 | # How Many Ounces in a Liter? Discover the Conversion Here!
If you’re in the United States, you’re likely used to thinking of volumes in terms of ounces, gallons, and quarts. But if you’re in Europe or elsewhere in the world, liters are the standard measure of volume. So what happens when you’re trying to convert between the two? In this article, we’ll explore how many ounces are in a liter and provide you with a handy conversion chart to use anytime you need it.
## The Basic Conversion: 1 Liter = 33.814 Ounces
Before we dive into the nitty-gritty details of converting liters to ounces, it’s helpful to understand the basic conversion. One liter is equivalent to 33.814 ounces. This means that if you have a liter of water, for example, it would weigh 33.814 ounces.
### Why Do We Need to Convert Between Liters and Ounces?
While liters are the standard measure of volume in much of the world, ounces are still commonly used in the United States. This means that if you’re trying to convert a recipe from a European cookbook to use in the United States, for example, you’ll need to know how many ounces are in a liter in order to accurately measure your ingredients.
## Converting Liters to Ounces: The Math
Now that we’ve covered the basics, let’s dive into the math behind converting liters to ounces. To convert liters to ounces, you simply need to multiply the number of liters by 33.814.
For example, imagine you have a liter of milk that you want to convert to ounces. You would simply multiply 1 (the number of liters) by 33.814 to get 33.814 ounces.
### Converting Ounces to Liters
Of course, sometimes you’ll need to go the other way and convert ounces to liters. To do this, you’ll need to divide the number of ounces by 33.814.
For example, imagine you have 50 ounces of soda that you want to convert to liters. You would divide 50 by 33.814 to get 1.48 liters.
## Common Conversions: A Handy Chart
While it’s helpful to understand the math behind converting liters to ounces, it’s also helpful to have a handy chart to reference when you’re cooking or measuring other liquids. Here’s a chart that provides you with some of the most common conversions:
Liters Ounces
0.25 L 8.45 oz
0.5 L 16.91 oz
1 L 33.81 oz
2 L 67.63 oz
3 L 101.44 oz
4 L 135.26 oz
While we’ve been talking about ounces as a measure of volume, it’s worth noting that there are actually two different types of ounces: weight ounces and fluid ounces.
Weight ounces are used to measure the weight of a particular substance, while fluid ounces are used to measure the volume of a particular liquid. When we talk about converting liters to ounces, we’re talking about fluid ounces.
### How Many Fluid Ounces are in a Liter?
To be precise, there are 33.814 fluid ounces in a liter. This is the same as regular ounces, as we’ve been discussing, but it’s important to make the distinction between weight ounces and fluid ounces.
## Common Questions About Liters and Ounces
• What is the easiest way to convert liters to ounces?
• How many ounces are in a liter of water?
• Why do Americans use ounces instead of liters?
• How many fluid ounces are in a liter?
• Do I need to worry about converting between liters and ounces if I’m just cooking for myself?
• The easiest way to convert liters to ounces is to use the basic conversion of 1 liter = 33.814 ounces.
• A liter of water weighs approximately 33.814 ounces.
• Americans use ounces instead of liters because the imperial system, which includes ounces, was adopted before the metric system was popularized.
• There are 33.814 fluid ounces in a liter.
• If you’re just cooking for yourself, you may not need to worry about converting between liters and ounces. However, it can be helpful to have a basic understanding of the conversion in case you want to try a recipe from a cookbook that uses liters.
## Conclusion
Now that you know how to convert between liters and ounces, you’ll be able to confidently measure liquids for any recipe, no matter where it originated. Whether you’re cooking up a storm or simply curious about how different units of measure work, understanding the conversion between liters and ounces is an important part of being able to communicate about volume. | 968 | 4,240 | {"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} | 4.15625 | 4 | CC-MAIN-2023-23 | latest | en | 0.918383 |
https://blog.prepscholar.com/browns-chemistry-the-central-science-15-8-exercise-1 | 1,723,273,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00295.warc.gz | 108,608,867 | 16,943 | This posts contains a Teaching Explanation.
You can buy Chemistry: The Central Science here.
Why You Should Trust Me: I’m Dr. Fred Zhang, and I have a bachelor’s degree in math from Harvard. I’ve racked up hundreds and hundreds of hours of experience working with students from 5th grade through graduate school, and I’m passionate about teaching. I’ve read the whole chapter of the text beforehand and spent a good amount of time thinking about what the best explanation is and what sort of solutions I would have wanted to see in the problem sets I assigned myself when I taught.
Exercise: 15.8 Practice Exercise 1
Question: … When 9.2g of frozen \$N_2O_4\$ is added to a .50L reaction vessel … [What is the value of \$K_c\$]
Part 1: Approaching the Problem
The question is asking for an equilibrium constant (\$K_c\$). We want to know \$K_c\$.
Generally, we can know the equilibrium constant ONLY IF we can figure out the equilibrium concentrations of the species (nitrous oxide and dinitrogen tetraoxide):
\$\$K_c = [NO_2]^2/[N_2O_4]\$\$
Thus, the entire game to figuring out the equilibrium constant here is to figure out the equilibrium concentrations.
We are already given that in equilibrium, the concentration of \$[N_2O_4]\$=.057 molar. So we have half the puzzle:
\$\$K_c = [NO_2]^2/.057\$\$
The other half of the puzzle if figuring out the equilibrium concentration \$[NO_2]\$. Sadly, the question doesn’t just give us this. But we have a piece of information nearly as good, which is the starting (initial) amount of \$[N_2O_4]\$. Because we know the reaction equation, the key now is to go from initial amount of \$[N_2O_4]\$ to the final (equilibrium) concentration \$[NO_2]\$.
Part 2: Converting Grams to Molar
We are given that the reaction started out with 9.2g of \$N_2O_4\$ in a 0.50L reaction vessel. For equilibrium calculations, we generally want to know concentrations of types molecules, instead of actual mass or volume.
We apply stoichiometry here and convert grams per liter to molarity using molar mass. We use the periodic table to look up the molar mass of \$N_2O_4\$ is 92.01 grams per mole.
We get that:
\$\$(9.2g N_2O_4)/(0.50L) *(1 mol)/(92.01 g N_2O_4) = (0.100mol)/L = 0.200 molar\$\$
Thus the initial concentration of \$N_2O_4\$ is 0.200 molar, and written as [\$N_2O_4\$]=.200
Part 3: Running the Reaction
Now that we know the starting concentration, we want to get to final concentrations. The algebraic equation that links the two is the equation of reaction:
\$\$N_2O_4 (g) ↔ 2 NO_2 (g)\$\$
This means that for every molecule of \$N_2O_4\$ we get two molecules of \$NO_2\$. As the reaction goes forward, when \$N_2O_4\$ decreases by \$x\$ molar, \$NO_2\$ increases by \$2x\$ molar.
The concentration table is then:
\$N_2O_4 (g)\$ \$2 NO_2 (g)\$ Initial Concentration (M) 0.200 0 Change in Concentration (M) -x +2x Equilibrium Concentration (M) 0.200-x 2x
Part 4: Calculating the Equilibrium
We are given that the equilibrium concentration of [\$N_2O_4\$]=.057 molar. The concentration table above gives the equilibrium concentration of [\$N_2O_4\$]=0.200-x, so we just equate the two and solve for x.
0.200-x = 0.057
x = .143
Now that we know x,
2x = .268
Or that in equilibrium, \$[NO_2]=.268\$
To calculate the equilibrium constant Kc, we plug in the information above:
\$\$K_c = [NO_2]^2/[N_2O_4]=.268^2/.057 = 1.43\$\$
Therefore, the right answer is d) 1.4
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## Ask a Question Below
Have any questions about this article or other topics? Ask below and we'll reply! | 1,217 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-33 | latest | en | 0.871996 |
https://stage.geogebra.org/m/WYgnT8eu | 1,713,255,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00522.warc.gz | 517,879,966 | 27,299 | GeoGebra Classroom
# Reflection from a real concave mirror
Author:
Ray Tuck
This simulation looks at the reflection of a single incident ray from a concave spherical mirror. The ray is reflected back from the mirror so that the angle of incidence and the angle of reflection are the same.
• Drag point P to change the position of the incident ray.
• The dotted line CP is the normal to the mirror (it is the radius).
• The angle between the incident ray and the normal is the same as the angle between the normal and the reflected ray according to the law of reflection.
• Drag point C to change the curvature of the mirror.
• The point F is half way between the centre of curvature (C) and the centre of the mirror. F is the focal point. | 162 | 740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-18 | latest | en | 0.910057 |
https://worldofzero.com/videos/exploring-dot-products-in-unity/ | 1,701,340,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00846.warc.gz | 714,980,797 | 29,462 | ## Exploring Dot Products in Unity
· · ☕ 2 min read
Dot Products are a really easy and fun way to compare Vectors in space. When comparing two normalized with a dot product the result will be the cosine of the angle between the two Vectors. This means that for normalized Vectors the results will be:
AlignmentResult
Parallel1
Anti-Parallel-1
Orthogonal0
Taking the ArcCosine of these values will return the angle between the two Vectors (Note that Unity uses radians for most calculations so you may need to convert to degrees yourself).
Lets explore some applications of Dot Products and show off how to take advantage of them in your games or other projects. For example, how can you detect if a point is inside of a cone? Perhaps as a way to check if an enemy AI can see the player or another reason. A second example may be detecting the difference in orientation between joysticks and a vehicles orientation to detect if a turn or thrust is the appropriate action to perform when moving.
The formula for dot products is the sum of the product of each axis. For example the dot product of two vectors A and B would be:
• In 2D space: (A.x * B.x) + (A.y * B.y)
• In 3D space: (A.x * B.x) + (A.y * B.y) + (A.z * B.z)
• In 4D space: (A.x * B.x) + (A.y * B.y) + (A.z * B.z) + (A.w * B.w)
• This formula can be expanded as the dimensions continues to increase if you need it to.
Note: Because of this formula the length of the Vectors being compared in the dot product does matter!
A similar application of dot products but in shader form in our revealing flashlight shader: https://youtu.be/b4utgRuIekk
An example of using dot products to help control how a top down spaceship uses it’s thrusters and rotation: https://youtu.be/TUIHY-5MvWQ
Source Code for this series is available on GitHub: https://github.com/WorldOfZero/top-down-spaceship
Join the World of Zero Discord server: https://discord.gg/hU5Kq2u
WRITTEN BY
Sam Wronski
Maker of things and professional software engineer. Lets make something awesome together! | 506 | 2,037 | {"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} | 3.171875 | 3 | CC-MAIN-2023-50 | longest | en | 0.861365 |
https://www.enotes.com/homework-help/find-numbers-b-from-relation-3x-1-x-2-4-x-2-b-x-2-216313 | 1,516,442,991,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00240.warc.gz | 917,232,429 | 9,385 | # Find the numbers a and b from in relation (3x+1)/(x^2-4)=a/(x-2)+b/(x+2)
giorgiana1976 | Student
To determine a and b, we'll have to calculate the least common denominator of the 3 ratios.
LCD = (x-2)(x+2) = x^2 - 4
LCD = x^2 - 4
Now, we'll multiply the first ratio from the right side by x+2 and the second ratio by x-2. The ratio from the left side has the denominator x^2 - 4, so it won't be multiplied.
We'll re-write the equation, all 3 quotients having the denominator x^2 - 4.
3x + 1 = a(x+2) + b(x-2)
We'll remove the brackets:
3x + 1 = ax + 2a + bx - 2b
We'll combine the terms from the right side with respect to x:
3x + 1 = x(a + b) + 2a - 2b
The expressions from both sides are equals if the correspondent coefficients are equal:
The coefficient of x from the left side has to be equal to the coefficient of x, from the right side:
a + b = 3 (1)
2a - 2b = 1 (2)
We'll multiply (1) by 2 and we'll get:
2a + 2b = 6 (3)
2a - 2b + 2a + 2b = 1 + 6
We'll combine and eliminate like terms:
4a = 7
a = 7/4
We'll substitute a in (1):
7/4 + b = 3
b = 3 - 7/4
b = (12-7)/4
b = 5/4
neela | Student
To find a and b in (3x+1)/(x^2-4)=a/(x-2)+b/(x+2).
We notice that the denominators on both sides have LCM (x-2)(x+2) = x^2-4.
So multiply both sides by (x-2)(x+2) and get:
3x+1 = a(x+2) +b(x-2).
3x+1 = ax+2a+bx-2b
3x = (a+b)x +2(a-b).
Now equate the like terms on both sides:
(a+b)x = 3x . Or a+b = 3.....(1).
2(a-b) = 1. Or a-b = 1/2......(2).
Eq(1) +Eq(2) gives: 2a = 3+1/2. So a = 7/2*2 = 7/4 .
Eq(1)- eq(2) gives: 2b = 3-1/2 = 5/2. Or b = 5/4.
Therefore a = 7/4 and b = 5/2. | 674 | 1,623 | {"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} | 4.59375 | 5 | CC-MAIN-2018-05 | latest | en | 0.827201 |
https://doubtnut.com/question-answer/show-that-the-line-joining-the-points-45-and-12-is-parallel-to-the-line-joining-the-points-9-2-and-1-31343456 | 1,585,766,383,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00316.warc.gz | 438,491,948 | 42,980 | or
# Show that the line joining the points (4,5) and (1,2) is parallel to the line joining the points (9,-2) and (12,1).
Question from Class 11 Chapter Straight Lines
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Solution :
Slope of line joining the points `(4,5)` and `(1,2)` is <br> `m_(1)=(2-5)/(1-4)=1` <br> Slope of line joining the points `(9,-2)` and `(12,1)` is <br> `m_(2)=(1-(-2))/(12-9)=1` <br> `:' m_(1)=m_(2)` <br> `:.` Given two lines are parallel to each other.
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24.2 K+ Views | 42.9 K+ Likes | 340 | 864 | {"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} | 3.375 | 3 | CC-MAIN-2020-16 | latest | en | 0.730614 |
http://learn.parallax.com/tutorials/robot/activitybot/regular-polygons-and-circles/circles-regular-polygons | 1,558,269,566,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254882.18/warc/CC-MAIN-20190519121502-20190519143502-00025.warc.gz | 126,658,301 | 7,168 | # Circles as Regular Polygons
As we said before, the circle is special because we can approach it as a regular polygon with 36 (or more) vertices. We will use exactly 36 vertices for this example.
So, if we want the robot to do a circle (clockwise) we need to give the following commands (note: the exterior angle values may be difficult to see in the image but each is 10 degrees):
Keep in mind that the S mustn’t be a big number because otherwise your circle will be very large! Choose small numbers for the S values. For example, in the next example S = 8 ticks.
The following SimpleIDE code programs the ActivityBot to make a circle with radius of approximately 15 cm:
• Start a new SimpleIDE Project
• Enter in the code below and make sure load it to your robot with the "Load EEPROM & Run" button
The question that might arises from this example is: what length should the S be in order to make a circle with a specific radius (R)? In other words, can I calculate the S as a function of R? Math theory shows us that “Cosines Law” is the answer to this. Looking at the triangle OA1A2 (below) we can see that:
So now we are almost ready for the final algorithm that will be able to create any circle with a given radius R, or create a polygon within a circle of radius R. | 299 | 1,283 | {"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-2019-22 | latest | en | 0.897847 |
https://www.studyadda.com/sample-papers/mathematics-sample-paper-3_q24/414/320091 | 1,597,491,145,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00573.warc.gz | 845,662,273 | 17,972 | • question_answer Find the three rational numbers between $\frac{1}{2}$and $-\text{ }2.$
A rational number between $\frac{1}{2}$and $-\text{ }2$ $=\left[ \frac{1}{2}+(-2) \right]\div 2$ $=\left[ \frac{1-4}{2} \right]\div 2$ $=\left[ -\frac{3}{2} \right]\times \frac{1}{2}=-\frac{3}{4}$ A rational number between $\frac{1}{2}$ and $\left( \frac{-3}{4} \right)$ $=\left[ \frac{1}{2}+\left( -\frac{3}{4} \right) \right]\div 2$ $=\left[ \frac{2-3}{4} \right]\times \frac{1}{2}$ $=-\frac{1}{4}\times \frac{1}{2}=-\frac{1}{8}$
A rational number between $\left( -\frac{3}{4} \right)$ and $\left( -\text{ }2 \right)$ $=\left[ \left( -\frac{3}{4} \right)+(-2) \right]\div 2$ $=\left[ \frac{(-3)+(-8)}{4} \right]\times \frac{1}{2}$ $=\frac{-11}{4}\times \frac{1}{2}=\frac{-11}{8}$
Thus, the three rational numbers are $\left( -\frac{3}{4} \right),\left( -\frac{1}{8} \right)$ and $\left( -\frac{11}{8} \right).$ | 387 | 905 | {"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} | 4.40625 | 4 | CC-MAIN-2020-34 | latest | en | 0.386994 |
https://procivilengineer.com/how-to-determine-water-content-of-soil/ | 1,652,751,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00572.warc.gz | 552,700,994 | 21,583 | # HOW TO DETERMINE WATER CONTENT OF SOIL?
## What is the water content of soil?
The water content of the soil mass is defined as the ratio of the weight of the water present in a given soil mass to the weight of dry soil.
In other words Water content = (weight of water in mass of soil) / (weight of dry soil)
The amount of water is usually expressed in percentage (%).
## How much sample is needed for the test?
The minimum amount of sample required for testing depends mainly on the maximum particle size present in that soil mass. The table below can be used as a reference to determine the minimum weight of a moist soil sample required.
Maximum particle size in soil (mm) Minimum weight of soil sample (g) 0.425 20 2.0 50 4.75 100 10.0 500 19.0 2500
## necessary equipment
• Moisture cans (should be made of heat-resistant material)
• oven with temperature. Control (105 .)0C to 1100C)
• Desiccator (used for cooling)
• Balance (readability of 0.01 g for samples with a mass of 200 g or less and 0.1 g for samples with a mass greater than 200 g.
## Standard Testing Procedures to Follow
1. Determine Weight (g) of Empty Moisture Can Plus Cap (W.)1), and also record the CAN number.
2. Keep the required amount of moist soil in the can with its cap on to avoid moisture loss.
3. Determine the combined weight of the can and moist soil (W.)2,
4. Remove the cap from the top of the can and place it on the bottom of the can.
5. Put the can with the moist soil sample in the oven for 24 hours.
6. After the clay mass is completely dry, remove it from the oven and let it cool to room temperature inside the desiccator. Determine the combined weight of the can and its cap along with the dry soil sample (W.)3,
## Calculations Involved
Calculate Moisture Weight = W2W3
Calculate mass of dry soil = W3W1
Calculate the quantity (%) of water as given below
W (%) = [(W2-W3)/(W3-W1)]*100
## How to report exam result?
At least three samples of the same sample should be tested. The average value of 3 observations should be reported. The water content of the soil is described as two important figures.
Er. Mukesh Kumar
Er. Mukesh Kumar is Editor in Chief and Co-Funder at ProCivilEngineer.com Civil Engineering Website. Mukesh Kumar is a Bachelor in Civil Engineering From MIT. He has work experience in Highway Construction, Bridge Construction, Railway Steel Girder work, Under box culvert construction, Retaining wall construction. He was a lecturer in a Engineering college for more than 6 years. | 610 | 2,518 | {"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} | 3.875 | 4 | CC-MAIN-2022-21 | latest | en | 0.910469 |
http://mathhelpforum.com/differential-geometry/177260-finite-union-compact-sets-compact-print.html | 1,529,703,846,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00094.warc.gz | 208,113,733 | 3,447 | # Finite union of compact sets is compact
• Apr 8th 2011, 09:47 AM
Sheld
Finite union of compact sets is compact
I think i have the proof, however I am worried about some assertions.
Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i
\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$
\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n
\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$
\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$
I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.
And the union of the finite subcovers remains finite?
Thanks for helping.
• Apr 8th 2011, 09:57 AM
Plato
Quote:
Originally Posted by Sheld
Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$
\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$
\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$
I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists. And the union of the finite subcovers remains finite?
I think that I would point out that any open covering of \$\displaystyle K\$ must also by definition be an open covering for each \$\displaystyle K_i\$. So in each case there is a finite subcover, as you said. Also point out that the finite union of finite collections of sets is also a finite collect of sets.
• Apr 8th 2011, 12:20 PM
Sheld
• Apr 8th 2011, 07:43 PM
hatsoff
Quote:
Originally Posted by Sheld
I think i have the proof, however I am worried about some assertions.
Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i
\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$
\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n
\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$
\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$
Sure, that's fine. Maybe add Plato's comments if you want to be even more explicit, but either way it works.
Quote:
I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.
Two things...
First of all, an open cover of any set ALWAYS exists because the whole space is open by definition. For instance, suppose \$\displaystyle X\$ is a space and \$\displaystyle A\subset X\$. Then \$\displaystyle \{X\}\$ is an open cover of \$\displaystyle A\$.
Second, even if an open cover didn't exist, we don't need one to show that a set is compact. In fact, that would GUARANTEE that it was compact!
Recall that a set A is compact if and only if it satisfies the following:
For all Y, if Y is an open cover of A, then there is a finite subcover.
In other words, a set A is non-compact if and only if it satisfies the following:
Not for all Y, if Y is an open cover of A, then there is a finite subcover.
This is true if and only if
There exists Y such that Y is an open cover of A, and it has no finite subcover.
If there is no open cover of A, then it cannot satisfy this property and hence is compact.
Quote:
And the union of the finite subcovers remains finite?
Yup. Any finite union of finite sets is again finite. | 1,166 | 4,093 | {"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} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.866684 |
https://nebusresearch.wordpress.com/tag/reply-all/ | 1,638,265,325,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00419.warc.gz | 513,595,396 | 33,673 | Reading the Comics, January 14, 2017: Redeye and Reruns Edition
So for all I worried about the Gocomics.com redesign it’s not bad. The biggest change is it’s removed a side panel and given the space over to the comics. And while it does show comics you haven’t been reading, it only shows one per day. One week in it apparently sticks with the same comic unless you choose to dismiss that. So I’ve had it showing me The Comic Strip That Has A Finale Every Day as a strip I’m not “reading”. I’m delighted how thisbreaks the logic about what it means to “not read” an “ongoing comic strip”. (That strip was a Super-Fun-Pak Comix offering, as part of Ruben Bolling’s Tom the Dancing Bug. It was turned into a regular Gocomics.com feature by someone who got the joke.)
Comic Strip Master Command responded to the change by sending out a lot of comic strips. I’m going to have to divide this week’s entry into two pieces. There’s not deep things to say about most of these comics, but I’ll make do, surely.
Julie Larson’s Dinette Set rerun for the 8th is about one of the great uses of combinatorics. That use is working out how the number of possible things compares to the number of things there are. What’s always staggering is that the number of possible things grows so very very fast. Here one of Larson’s characters claims a science-type show made an assertion about the number of possible ideas a brain could hold. I don’t know if that’s inspired by some actual bit of pop science. I can imagine someone trying to estimate the number of possible states a brain might have.
And that has to be larger than the number of atoms in the universe. Consider: there’s something less than a googol of atoms in the universe. But a person can certainly have the idea of the number 1, or the idea of the number 2, or the idea of the number 3, or so on. I admit a certain sameness seems to exist between the ideas of the numbers 2,038,412,562,593,604 and 2,038,412,582,593,604. But there is a difference. We can out-number the atoms in the universe even before we consider ideas like rabbits or liberal democracy or jellybeans or board games. The universe never had a chance.
Or did it? Is it possible for a number to be too big for the human brain to ponder? If there are more digits in the number than there are atoms in the universe we can’t form any discrete representation of it, after all. … Except that we kind of can. For example, “the largest prime number less than one googolplex” is perfectly understandable. We can’t write it out in digits, I think. But you now have thought of that number, and while you may not know what its millionth decimal digit is, you also have no reason to care what that digit is. This is stepping into the troubled waters of algorithmic complexity.
Bob Weber Jr’s Slylock Fox and Comics for Kids for the 9th is built on soap bubbles. The link between the wand and the soap bubble vanishes quickly once the bubble breaks loose of the wand. But soap films that keep adhered to the wand or mesh can be quite strangely shaped. Soap films are a practical example of a kind of partial differential equations problem. Partial differential equations often appear when we want to talk about shapes and surfaces and materials that tug or deform the material near them. The shape of a soap bubble will be the one that minimizes the torsion stresses of the bubble’s surface. It’s a challenge to solve analytically. It’s still a good challenge to solve numerically. But you can do that most wonderful of things and solve a differential equation experimentally, if you must. It’s old-fashioned. The computer tools to do this have gotten so common it’s hard to justify going to the engineering lab and getting soapy water all over a mathematician’s fingers. But the option is there.
Gordon Bess’s Redeye rerun from the 28th of August, 1970, is one of a string of confused-student jokes. (The strip had a Generic Comedic Western Indian setting, putting it in the vein of Hagar the Horrible and other comic-anachronism comics.) But I wonder if there are kids baffled by numbers getting made several different ways. Experience with recipes and assembly instructions and the like might train someone to thinking there’s one correct way to make something. That could build a bad intuition about what additions can work.
Corey Pandolph’s Barkeater Lake rerun for the 9th just name-drops algebra. And that as a word that starts with the “alj” sound. So far as I’m aware there’s not a clear etymological link between Algeria and algebra, despite both being modified Arabic words. Algebra comes from “al-jabr”, about reuniting broken things. Algeria comes from Algiers, which Wikipedia says derives from `al-jaza’ir”, “the Islands [of the Mazghanna tribe]”.
Guy Gilchrist’s Nancy for the 9th is another mathematics-cameo strip. But it was also the first strip I ran across this week that mentioned mathematics and wasn’t a rerun. I’ll take it.
Donna A Lewis’s Reply All for the 9th has Lizzie accuse her boyfriend of cheating by using mathematics in Scrabble. He seems to just be counting tiles, though. I think Lizzie suspects something like Blackjack card-counting is going on. Since there are only so many of each letter available knowing just how many tiles remain could maybe offer some guidance how to play? But I don’t see how. In Blackjack a player gets to decide whether to take more cards or not. Counting cards can suggest whether it’s more likely or less likely that another card will make the player or dealer bust. Scrabble doesn’t offer that choice. One has to refill up to seven tiles until the tile bag hasn’t got enough left. Perhaps I’m overlooking something; I haven’t played much Scrabble since I was a kid.
Perhaps we can take the strip as portraying the folk belief that mathematicians get to know secret, barely-explainable advantages on ordinary folks. That itself reflects a folk belief that experts of any kind are endowed with vaguely cheating knowledge. I’ll admit being able to go up to a blackboard and write with confidence a bunch of integrals feels a bit like magic. This doesn’t help with Scrabble.
Gordon Bess’s Redeye continued the confused-student thread on the 29th of August, 1970. This one’s a much older joke about resisting word problems.
Ryan North’s Dinosaur Comics rerun for the 10th talks about multiverses. If we allow there to be infinitely many possible universes that would suggest infinitely many different Shakespeares writing enormously many variations of everything. It’s an interesting variant on the monkeys-at-typewriters problem. I noticed how T-Rex put Shakespeare at typewriters too. That’ll have many of the same practical problems as monkeys-at-typewriters do, though. There’ll be a lot of variations that are just a few words or a trivial scene different from what we have, for example. Or there’ll be variants that are completely uninteresting, or so different we can barely recognize them as relevant. And that’s if it’s actually possible for there to be an alternate universe with Shakespeare writing his plays differently. That seems like it should be possible, but we lack evidence that it is. | 1,595 | 7,179 | {"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} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.931149 |
https://www.physicsforums.com/threads/factoring-help.674230/ | 1,521,806,875,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00464.warc.gz | 868,188,977 | 13,628 | # Factoring Help
1. Feb 24, 2013
### JNeutron2186
Hey everyone first time poster here, I need help with some factoring of cubes. I know this might tie closely to Diophantine equations but here goes.
Under what condition is the expression x^3+y^3+z^3 factorable? Where x,y,z are positive whole numbers.
2. Feb 25, 2013
### mathwonk
well it seems to factor if x=1, y=2, z=3, and if x=y=z ≥ 2, and more generally if gcd(x,y,z) > 1.
3. Feb 27, 2013
### JNeutron2186
My bad for not stating before. Let's assume the gcd(x,y,z) =1 and gcd(x,y)>1 | 178 | 548 | {"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-2018-13 | latest | en | 0.885707 |
https://eanswers.in/math/question720303 | 1,606,208,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00514.warc.gz | 265,910,839 | 14,123 | Sita bought a cycle for rupees 2400 and sold it for rupees 2100. find gain or loss%
, 17.10.2019 23:00, dhvani6272
# Sita bought a cycle for rupees 2400 and sold it for rupees 2100. find gain or loss%
### Other questions on the subject: Math
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Math, 18.08.2019 02:00, dhvani6272
Aman sold two articles for rs.12 each, on the first he gains 20% and on the second he losses 20%. find the amount of profit or loss on the whole transaction | 198 | 547 | {"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-2020-50 | latest | en | 0.824229 |
http://www.lyricsrocket.com/Classes/CalcI/OneSidedLimits.aspx | 1,618,134,025,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00345.warc.gz | 149,650,288 | 20,564 | Paul's Online Notes
Home / Calculus I / Limits / One-Sided Limits
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Section 2-3 : One-Sided Limits
In the final two examples in the previous section we saw two limits that did not exist. However, the reason for each of the limits not existing was different for each of the examples.
We saw that
$\mathop {\lim }\limits_{t \to 0} \,\,\cos \left( {\frac{\pi }{t}} \right)$
did not exist because the function did not settle down to a single value as $$t$$ approached $$t = 0$$. The closer to $$t = 0$$ we moved the more wildly the function oscillated and in order for a limit to exist the function must settle down to a single value.
However, we saw that
$\mathop {\lim }\limits_{t \to 0} H\left( t \right)\hspace{0.25in}{\mbox{where,}}\hspace{0.25in}H\left( t \right) = \left\{ \begin{array}{ll}0 & {\mbox{if }}t < 0\\ 1 & {\mbox{if }}t \ge 0\end{array} \right.$
did not exist not because the function didn’t settle down to a single number as we moved in towards $$t = 0$$, but instead because it settled into two different numbers depending on which side of $$t = 0$$ we were on.
In this case the function was a very well-behaved function, unlike the first function. The only problem was that, as we approached $$t = 0$$, the function was moving in towards different numbers on each side. We would like a way to differentiate between these two examples.
We do this with one-sided limits. As the name implies, with one-sided limits we will only be looking at one side of the point in question. Here are the definitions for the two one sided limits.
Right-handed limit
We say
$\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L$
provided we can make $$f\left( x \right)$$ as close to $$L$$ as we want for all $$x$$ sufficiently close to $$a$$ with $$x > a$$ without actually letting $$x$$ be $$a$$.
Left-handed limit
We say
$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L$
provided we can make $$f\left( x \right)$$ as close to $$L$$ as we want for all $$x$$ sufficiently close to $$a$$ with $$x < a$$ without actually letting $$x$$ be $$a$$.
Note that the change in notation is very minor and in fact might be missed if you aren’t paying attention. The only difference is the bit that is under the “lim” part of the limit. For the right-handed limit we now have $$x \to {a^ + }$$(note the “+”) which means that we know will only look at $$x > a$$. Likewise, for the left-handed limit we have $$x \to {a^ - }$$(note the “-”) which means that we will only be looking at $$x < a$$.
Also, note that as with the “normal” limit (i.e. the limits from the previous section) we still need the function to settle down to a single number in order for the limit to exist. The only difference this time is that the function only needs to settle down to a single number on either the right side of $$x = a$$ or the left side of $$x = a$$ depending on the one?sided limit we’re dealing with.
So, when we are looking at limits it’s now important to pay very close attention to see whether we are doing a normal limit or one of the one-sided limits. Let’s now take a look at the some of the problems from the last section and look at one-sided limits instead of the normal limit.
Example 1 Estimate the value of the following limits. $\mathop {\lim }\limits_{t \to {0^ + }} H\left( t \right)\hspace{0.25in}\,{\rm{and}}\hspace{0.25in}\mathop {\lim }\limits_{t \to {0^ - }} H\left( t \right)\hspace{0.5in}{\rm{where}}\hspace{0.25in}H\left( t \right) = \left\{ \begin{array}{ll}0 & {\mbox{if }}t < 0\\ 1 & {\mbox{if }}t \ge 0\end{array} \right.$
Show Solution
To remind us what this function looks like here’s the graph.
So, we can see that if we stay to the right of $$t = 0$$ (i.e. $$t > 0$$) then the function is moving in towards a value of 1 as we get closer and closer to $$t = 0$$, but staying to the right. We can therefore say that the right-handed limit is,
$\mathop {\lim }\limits_{t \to {0^ + }} H\left( t \right) = 1$
Likewise, if we stay to the left of $$t = 0$$ (i.e $$t < 0$$) the function is moving in towards a value of 0 as we get closer and closer to $$t = 0$$, but staying to the left. Therefore, the left-handed limit is,
$\mathop {\lim }\limits_{t \to {0^ - }} H\left( t \right) = 0$
In this example we do get one-sided limits even though the normal limit itself doesn’t exist.
Example 2 Estimate the value of the following limits. $\mathop {\lim }\limits_{t \to {0^ + }} \,\,\cos \left( {\frac{\pi }{t}} \right)\hspace{0.5in}\mathop {\lim }\limits_{t \to {0^ - }} \,\,\cos \left( {\frac{\pi }{t}} \right)$
Show Solution
From the graph of this function shown below,
we can see that both of the one-sided limits suffer the same problem that the normal limit did in the previous section. The function does not settle down to a single number on either side of $$t = 0$$. Therefore, neither the left-handed nor the right-handed limit will exist in this case.
So, one-sided limits don’t have to exist just as normal limits aren’t guaranteed to exist.
Let’s take a look at another example from the previous section.
Example 3 Estimate the value of the following limits. $\mathop {\lim }\limits_{x \to {2^ + }} g\left( x \right)\hspace{0.25in}{\rm{and}}\hspace{0.25in}\mathop {\lim }\limits_{x \to {2^ - }} g\left( x \right)\hspace{0.25in}\hspace{0.25in}{\rm{where}}\hspace{0.25in}g\left( x \right) = \left\{ \begin{array}{ll}\displaystyle \frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}} & {\mbox{if }}x \ne 2\\ 6 & {\mbox{if }}x = 2\end{array} \right.$
Show Solution
So, as we’ve done with the previous two examples, let’s remind ourselves of the graph of this function.
In this case regardless of which side of $$x = 2$$ we are on the function is always approaching a value of 4 and so we get,
$\mathop {\lim }\limits_{x \to {2^ + }} g\left( x \right) = 4\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to {2^ - }} g\left( x \right) = 4$
Note that one-sided limits do not care about what’s happening at the point any more than normal limits do. They are still only concerned with what is going on around the point. The only real difference between one-sided limits and normal limits is the range of $$x$$’s that we look at when determining the value of the limit.
Now let’s take a look at the first and last example in this section to get a very nice fact about the relationship between one-sided limits and normal limits. In the last example the one-sided limits as well as the normal limit existed and all three had a value of 4. In the first example the two one-sided limits both existed, but did not have the same value and the normal limit did not exist.
The relationship between one-sided limits and normal limits can be summarized by the following fact.
Fact
Given a function f(x) if,
$\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L$
then the normal limit will exist and
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$
Likewise, if
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$
then,
$\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L$
This fact can be turned around to also say that if the two one-sided limits have different values, i.e.,
$\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)$
then the normal limit will not exist.
This should make some sense. If the normal limit did exist then by the fact the two one-sided limits would have to exist and have the same value by the above fact. So, if the two one-sided limits have different values (or don’t even exist) then the normal limit simply can’t exist.
Let’s take a look at one more example to make sure that we’ve got all the ideas about limits down that we’ve looked at in the last couple of sections.
Example 4 Given the following graph,
compute each of the following.
1. $$f( - 4)$$
2. $$\mathop {\lim }\limits_{x \to - {4^ - }} f\left( x \right)$$
3. $$\mathop {\lim }\limits_{x \to - {4^ + }} f\left( x \right)$$
4. $$\mathop {\lim }\limits_{x \to - 4} f\left( x \right)$$
1. $$f\left( 1 \right)$$
2. $$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)$$
3. $$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$$
4. $$\mathop {\lim }\limits_{x \to 1} f\left( x \right)$$
1. $$f\left( 6 \right)$$
2. $$\mathop {\lim }\limits_{x \to {6^ - }} f\left( x \right)$$
3. $$\mathop {\lim }\limits_{x \to {6^ + }} f\left( x \right)$$
4. $$\mathop {\lim }\limits_{x \to 6} f\left( x \right)$$
Show Solution
a$$f( - 4)$$ doesn’t exist. There is no closed dot for this value of $$x$$ and so the function doesn’t exist at this point.
b$$\mathop {\lim }\limits_{x \to - {4^ - }} f\left( x \right) = 2$$ The function is approaching a value of 2 as $$x$$ moves in towards -4 from the left.
c$$\mathop {\lim }\limits_{x \to - {4^ + }} f\left( x \right) = 2$$ The function is approaching a value of 2 as $$x$$ moves in towards -4 from the right.
d$$\mathop {\lim }\limits_{x \to - 4} f\left( x \right) = 2$$ We can do this one of two ways. Either we can use the fact here and notice that the two one-sided limits are the same and so the normal limit must exist and have the same value as the one-sided limits or just get the answer from the graph.
Also recall that a limit can exist at a point even if the function doesn’t exist at that point.
e$$f\left( 1 \right) = 4$$. The function will take on the $$y$$ value where the closed dot is.
f$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 4$$ The function is approaching a value of 4 as $$x$$ moves in towards 1 from the left.
g$$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = - 2$$ The function is approaching a value of -2 as $$x$$ moves in towards 1 from the right. Remember that the limit does NOT care about what the function is actually doing at the point, it only cares about what the function is doing around the point. In this case, always staying to the right of $$x = 1$$, the function is approaching a value of -2 and so the limit is -2. The limit is not 4, as that is value of the function at the point and again the limit doesn’t care about that!
h$$\mathop {\lim }\limits_{x \to 1} f\left( x \right)$$ doesn’t exist. The two one-sided limits both exist, however they are different and so the normal limit doesn’t exist.
i$$f\left( 6 \right) = 2$$. The function will take on the $$y$$ value where the closed dot is.
j$$\mathop {\lim }\limits_{x \to {6^ - }} f\left( x \right) = 5$$ The function is approaching a value of 5 as $$x$$ moves in towards 6 from the left.
k$$\mathop {\lim }\limits_{x \to {6^ + }} f\left( x \right) = 5$$ The function is approaching a value of 5 as $$x$$ moves in towards 6 from the right.
l$$\mathop {\lim }\limits_{x \to 6} f\left( x \right) = 5$$ Again, we can use either the graph or the fact to get this. Also, once more remember that the limit doesn’t care what is happening at the point and so it’s possible for the limit to have a different value than the function at a point. When dealing with limits we’ve always got to remember that limits simply do not care about what the function is doing at the point in question. Limits are only concerned with what the function is doing around the point.
Hopefully over the last couple of sections you’ve gotten an idea on how limits work and what they can tell us about functions. Some of these ideas will be important in later sections so it’s important that you have a good grasp on them. | 3,556 | 12,021 | {"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} | 3.65625 | 4 | CC-MAIN-2021-17 | latest | en | 0.938962 |
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# NIFS PROC 88
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```The transport equation of particles is
\u2202
\u2202t
n(r, t) +\u2207 · (n(r, t)V (r, t)) = 0 (7.1)
provided processes of the ionization of neutrals and the recombination of ions are negligible (see
ch.5.1). The particle \ufb02ux \u393 = nV is given by
n(r, t)V (r, t) = \u2212D(r, t)\u2207n(r, t)
in many cases, where D is di\ufb00usion coe\ufb03cient. (Additional terms may be necessary in more general
cases.)
Di\ufb00usion coe\ufb03cient D and particle con\ufb01nement time \u3c4p are related by the di\ufb00usion equation of
the plasma density n as follows:
\u2207 · (D\u2207n(r, t)) = \u2202
\u2202t
n(r, t).
Substitution of n(r, t) = n(r) exp(\u2212t/\u3c4p) in di\ufb00usion equation yields
\u2207 · (D\u2207n(r)) = \u2212 1
\u3c4p
n(r).
When D is constant and the plasma column is a cylinder of radius a, the di\ufb00usion equation is
reduced to
1
r
\u2202
\u2202r
(
r
\u2202n
\u2202r
)
+
1
D\u3c4p
n = 0.
The solution satisfying the boundary condition n(a) = 0 is
n = n0J0
(
2.4r
a
)
exp
(
\u2212 t
\u3c4p
)
and the particle con\ufb01nement time is
\u3c4p =
a2
2.42D
=
a2
5.8D
, (7.2)
where J0 is the zeroth-order Bessel function. The relationship (7.2) between the particle con\ufb01ne-
ment time \u3c4p and D holds generally, with only a slight modi\ufb01cation of the numerical factor. This
formula is frequently used to obtain the di\ufb00usion coe\ufb03cient from the observed values of the plasma
70
7 Di\ufb00usion of Plasma, Con\ufb01nement Time 71
The equation of energy balance is given by (A.19), which will be derived in appendix A, as
follows:
\u2202
\u2202t
(
3
2
n\u3baT
)
+\u2207 ·
(
3
2
\u3baTnv
)
+\u2207 · q = Q\u2212 p\u2207 · v \u2212
\u2211
ij
\u3a0ij
\u2202vi
\u2202xj
. (7.3)
The \ufb01rst term in the right-hand side is the heat generation due to particle collisions per unit
volume per unit time, the second term is the work done by pressure and the third term is viscous
heating. The \ufb01rst term in the left-hand side is the time derivative of the thermal energy per
unit volume, the second term is convective energy loss and the third term is conductive energy
loss. Denoting the thermal conductivity by \u3baT, the thermal \ufb02ux due to heat conduction may be
expressed by
q = \u2212\u3baT\u2207(\u3baT ).
If the convective loss is neglected and the heat sources in the right-hand side of (7.3) is zero, we
\ufb01nd that
\u2202
\u2202t
(
3
2
n\u3baT
)
\u2212\u2207 · \u3baT\u2207(\u3baT ) = 0.
In the case of n = const., this equation reduces to
\u2202
\u2202t
(
3
2
\u3baT
)
= \u2207 ·
(
\u3baT
n
\u2207(\u3baT )
)
.
When the thermal di\ufb00usion coe\ufb03cient \u3c7T is de\ufb01ned by
\u3c7T =
\u3baT
n
,
the same equation on \u3baT is obtained as (7.1). In the case of \u3c7T = const., the solution is
\u3baT = \u3baT0J0
(
2.4
a
r
)
exp
(
\u2212 t
\u3c4E
)
, \u3c4E =
a2
5.8(2/3)\u3c7T
. (7.4)
The term \u3c4E is called energy con\ufb01nement time.
7.1 Collisional Di\ufb00usion (Classical Di\ufb00usion)
7.1a Magnetohydrodynamic Treatment
A magnetohydrodynamic treatment is applicable to di\ufb00usion phenomena when the electron-to-
ion collision frequency is large and the mean free path is shorter than the connection length of the
inside regions of good curvature and the outside region of bad curvature of the torus; i.e.,
vTe
\u3bdei
<\u223c
2\u3c0R
\u3b9
,
\u3bdei >\u223c \u3bdp \u2261
1
R
\u3b9
2\u3c0
vTe =
1
R
\u3b9
2\u3c0
(
\u3baTe
me
)1/2
where vTe is electron thermal velocity and \u3bdei is electron to ion collision frequency. From Ohm\u2019s
law (5.28)
E + v ×B \u2212 1
en
\u2207pi = \u3b7j,
the motion of plasma across the lines of magnetic force is expressed by
nv\u22a5 =
1
B
((
nE \u2212 \u3baTi
e
\u2207n
)
× b
)
\u2212 me\u3bdei
e2
\u2207p
B2
71
72 7 Di\ufb00usion of Plasma, Con\ufb01nement Time
Fig.7.1 Electric \ufb01eld in a plasma con\ufb01ned in a toroidal \ufb01eld. The symbols \u2297 and \ufffd here show the
direction of the P\ufb01rsch-Schlu¨ter current.
=
1
B
((
nE \u2212 \u3baTi
e
\u2207n
)
× b
)
\u2212 (\u3c1\u3a9e)2\u3bdei
(
1 +
Ti
Te
)
\u2207n (7.5)
where \u3c1\u3a9e = vTe/\u3a9e, vTe = (\u3baTe/me)1/2 and \u3b7 = me\u3bdei/e2ne (see sec.2.8).
If the \ufb01rst term in the right-hand side can be neglected, the particle di\ufb00usion coe\ufb03cient D is
given by
D = (\u3c1\u3a9e)2\u3bdei
(
1 +
Ti
Te
)
. (7.6)
The classical di\ufb00usion coe\ufb03cient Dei is de\ufb01ned by
Dei \u2261 (\u3c1\u3a9e)2\u3bdei = n\u3baTe
\u3c3\u22a5B2
=
\u3b2e\u3b7\u2016
\u3bc0
, (7.7)
where \u3c3\u22a5 = nee2/(me\u3bdei), \u3b7\u2016 = 1/2\u3c3\u22a5.
However the \ufb01rst term of the right-hand side of (7.5) is not always negligible. In toroidal con-
\ufb01guration, the charge separation due to the toroidal drift is not completely cancelled along the
magnetic \ufb01eld lines due to the \ufb01nite resistivity and an electric \ufb01eld E arises (see \ufb01g.7.1). Therefore
the E×b term in (7.5) contributes to the di\ufb00usion. Let us consider this term. From the equilibrium
equation, the diamagnetic current
j\u22a5 =
b
B
×\u2207p, j\u22a5 =
\u2223\u2223\u2223\u2223 1B \u2202p\u2202r
\u2223\u2223\u2223\u2223
\ufb02ows in the plasma. From \u2207 · j = 0, we \ufb01nd \u2207 · j\u2016 = \u2212\u2207 · j\u22a5. By means of the equation
B = B0(1\u2212 (r/R) cos \u3b8), the P\ufb01rsch-Schlu¨ter current j\u2016 is given by (refer to (6.43))
j\u2016 = 2
2\u3c0
\u3b9
1
B0
\u2202p
\u2202r
cos \u3b8. (7.8)
If the electric conductivity along the magnetic lines of force is \u3c3\u2016, the parallel electric \ufb01eld is
E\u2016 = j\u2016/\u3c3\u2016. As is clear from \ufb01g.7.1, the relation
E\u3b8
E\u2016
\u2248 B0
B\u3b8
holds. From B\u3b8/B0 \u2248 (r/R)(\u3b9/2\u3c0), the \u3b8 component of the electric \ufb01eld is given by
E\u3b8 =
B0
B\u3b8
E\u2016 =
R
r
2\u3c0
\u3b9
1
\u3c3\u2016
j\u2016 =
2
\u3c3\u2016
R
r
(
2\u3c0
\u3b9
)2 1
B0
\u2202p
\u2202r
cos \u3b8. (7.9)
Accordingly (7.5) is reduced to
nVr = \u2212nE\u3b8
B
\u2212 (\u3c1\u3a9e)2\u3bdei
(
1 +
Ti
Te
)
\u2202n
\u2202r
72
7.1 Collisional Di\ufb00usion (Classical Di\ufb00usion) 73
Fig.7.2 Magnetic surface (dotted line) and drift surfaces (solid lines).
= \u2212
(
R
r
· 2
(
2\u3c0
\u3b9
)2 n\u3baTe
\u3c3\u2016B20
cos \u3b8
(
1 +
r
R
cos \u3b8
)
+
n\u3baTe
\u3c3\u22a5B20
(
1 +
r
R
cos \u3b8
)2)
×
(
1 +
Ti
Te
)
\u2202n
\u2202r
.
Noting that the area of a surface element is dependent of \u3b8, and taking the average of nVr over \u3b8,
we \ufb01nd that
\u3008nVr\u3009 = 12\u3c0
\u222b 2\u3c0
0
nVr
(
1 +
r
R
cos \u3b8
)
d\u3b8
= \u2212 n\u3baTe
\u3c3\u22a5B20
(
1 +
Ti
Te
)(
1 +
2\u3c3\u22a5
\u3c3\u2016
(
2\u3c0
\u3b9
)2) \u2202n
\u2202r
. (7.10)
Using the relation \u3c3\u22a5 = \u3c3\u2016/2, we obtain the di\ufb00usion coe\ufb03cient of a toroidal plasma:
DP.S. =
nTe
\u3c3\u22a5B20
(
1 +
Ti
Te
)(
1 +
(
2\u3c0
\u3b9
)2)
. (7.11)
This di\ufb00usion coe\ufb03cient is (1 + (2\u3c0/\u3b9)2) times as large as the di\ufb00usion coe\ufb03cient of (7.2). This
value is called P\ufb01rsch-Schlu¨ter factor (ref.[1]). When the rotational tranform angle \u3b9/2\u3c0 is about
0.3, P\ufb01rsch-Schlu¨ter factor is about 10.
7.1b A Particle Model
The classical di\ufb00usion coe\ufb03cient of electrons
Dei = (\u3c1\u3a9e)2\u3bdei
is that for electrons which move in a random walk with a step length equal to the Larmor radius.
Let us consider a toroidal plasma. For rotational transform angle \u3b9, the displacement \u394 of the
electron drift surface from the magnetic surface is (see \ufb01g.7.2)
\u394 \u2248 ±\u3c1\u3a9e 2\u3c0
\u3b9
. (7.12)
The ± signs depend on that the direction of electron motion is parallel or antiparallel to the
magnetic \ufb01eld (see sec.3.5). As an electron can be transferred from one drift surface to the other
by collision, the step length across the magnetic \ufb01eld is
\u394 =
(
2\u3c0
\u3b9
)
\u3c1\u3a9e. (7.13)
73
74 7 Di\ufb00usion of Plasma, Con\ufb01nement Time
Consequently, the di\ufb00usion coe\ufb03cient is given by
DP.S. = \u3942\u3bdei =
(
2\u3c0
\u3b9
)2
(\u3c1\u3a9e)2\u3bdei, (7.14)
thus the P\ufb01rsch-Schlu¨ter factor has been reduced (|2\u3c0/\u3b9| \ufffd 1 is assumed).
7.2 Neoclassical Di\ufb00usion of Electrons in Tokamak
The magnitude B of the magnetic \ufb01eld of a tokamak is given by
B =
RB0
R(1 + \ufffdt cos \u3b8)
= B0(1\u2212 \ufffdt cos \u3b8), (7.15)
where
\ufffdt =
r
R
. (7.16)
Consequently, when the perpendicular component v\u22a5 of a electron velocity is much larger than
the parallel component v\u2016, i.e., when(
v\u22a5
v
)2
>
R
R+ r
,
that is,
v\u22a5
v\u2016
>
1
\ufffd
1/2
t
, (7.17)
the electron is trapped outside of the torus, where the magnetic \ufb01eld is weak. Such an electron
drifts in a banana orbit (see \ufb01g.3.9). In order to complete a circuit of the banana orbit, the e\ufb00ective
collision time \u3c4e\ufb00 = 1/\u3bde\ufb00 of the trapped electron must``` | 3,771 | 8,937 | {"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-2020-10 | latest | en | 0.801565 |
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## 문제
Marcy is attending Pride Fest and has signed up to participate in the Rainbow Road Race. On each street, there are volunteers who have colored chalk powder. As contestants walk along a street, volunteers throw chalk onto the contestants. The chalk thrown on each street is one of the seven colors of the rainbow (Red, Orange, Yellow, Green, Blue, Indigo, Violet). Once a person starts walking along a street, they must walk to the end of that street.
The race starts at the Festival Tent. The goal of the race is to get covered by chalk of every color and come back to the tent. Help Marcy determine the shortest distance she has to travel to get every color and make it back to the tent.
Figure J.1: The left picture illustrates Sample Input 1 and the right picture illustrates Sample Input 2. The triangle is the Festival Tent.
## 입력
The first line of input contains two integers $n$ ($7 \leq n \leq 7^3$), which is the number of fun locations at the festival, and $m$ ($7 \leq m \leq 7^4$), which is the number of streets connecting the fun locations. The fun locations are numbered $1, \dots , n$ and the Festival Tent is location $1$.
The next $m$ lines describe the streets. Each of these lines contains three integers $\ell_1$, $\ell_2$ ($1 \leq \ell_1 < \ell_2 \leq n$) and $d$ ($1 \leq d \leq 7^5$), followed by a single character $c$ ($c$ is one of R, O, Y, G, B, I, V). This specifies that this street connects locations $\ell_1$ and $\ell_2$, is $d$ meters long and the chalk thrown is color $c$.
It is always possible to travel between any pair of fun locations. There is at most one street between any two pairs of locations and each color appears at least once.
## 출력
Display the shortest distance Marcy has to travel to get every color and make it back to the Festival Tent.
## 예제 입력 1
7 7
1 2 1 R
2 3 1 O
3 4 1 Y
4 5 1 G
5 6 1 B
6 7 1 I
1 7 1 V
## 예제 출력 1
7
## 예제 입력 2
8 7
1 2 1 R
1 3 1 O
1 4 1 Y
1 5 1 G
1 6 1 B
1 7 1 I
1 8 1 V
## 예제 출력 2
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# While Governor Verdant has been in office, the state s
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04 Jul 2010, 09:39
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While Governor Verdant has been in office, the state’s budget has increased by an average of 6 percent each year. While the previous governor was in office, the state’s budget increased by an average of 11.5 percent each year. Obviously, the austere budgets during Governor Verdant’s term have caused the slowdown in the growth in state spending.
Which of the following, if true, would most seriously weaken the conclusion drawn above?
(A) The rate of inflation in the state averaged 10 percent each year during the previous governor’s term in office and 3 percent each year during Verdant’s term.
(B) Both federal and state income tax rates have been lowered considerably during Verdant’s term in office.
(C) In each year of Verdant’s term in office, the state’s budget has shown some increase in spending over the previous year.
(D) During Verdant’s term in office, the state has either discontinued or begun to charge private citizens for numerous services that the state offered free to citizens during the previous governor’s term.
(E) During the previous governor’s term in office, the state introduced several so-called “austerity” budgets intended to reduce the growth in state spending.
[Reveal] Spoiler: OA
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04 Jul 2010, 10:24
EnterMatrix wrote:
While Governor Verdant has been in office, the state’s budget has increased by an average of 6 percent each year. While the previous governor was in office, the state’s budget increased by an average of 11.5 percent each year. Obviously, the austere budgets during Governor Verdant’s term have caused the slowdown in the growth in state spending.
Which of the following, if true, would most seriously weaken the conclusion drawn above?
(A) The rate of inflation in the state averaged 10 percent each year during the previous governor’s term in office and 3 percent each year during Verdant’s term.
(B) Both federal and state income tax rates have been lowered considerably during Verdant’s term in office.
(C) In each year of Verdant’s term in office, the state’s budget has shown some increase in spending over the previous year.
(D) During Verdant’s term in office, the state has either discontinued or begun to charge private citizens for numerous services that the state offered free to citizens during the previous governor’s term.
(E) During the previous governor’s term in office, the state introduced several so-called “austerity” budgets intended to reduce the growth in state spending.
Definitely A. Choice A simply refutes what is stated in the original prompt, showing that Verdant has actually been increasing the rate of growth in state spending.
Governor Verdant = 6% - 3% = 3% growth per year
Previous governor = 11.5% - 10% = 1.5% growth per year
This new data most definitely weakens the conclusion.
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05 Jul 2010, 11:40
Yup...makes sense! Thanks
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05 Jul 2010, 13:39
Definitely A
B - decrease in taxation does not necessarily affect budgets
C - the degree of increase must be taken into account
D - changes in government revenue do not necessarily affect budgets
E - supports the argument
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06 Jul 2010, 11:14
OA is A
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06 Jul 2010, 19:48
marked C ...
Re: Critical Reasoning [#permalink] 06 Jul 2010, 19:48
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Display posts from previous: Sort by | 1,380 | 5,279 | {"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} | 3.46875 | 3 | CC-MAIN-2017-22 | latest | en | 0.947485 |
https://documen.tv/question/10-1-7-37-this-is-very-hard-to-comprehend-because-the-answers-i-have-a-choice-to-choose-from-are-19774972-67/ | 1,653,302,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00440.warc.gz | 268,690,687 | 15,755 | ## -10x +1 +7x =37 This is very hard to comprehend because the answers I have a choice to choose from are -15 -12 12
Question
-10x +1 +7x =37
This is very hard to comprehend because the answers I have a choice to choose from are
-15
-12
12
15
in progress 0
7 months 2021-08-09T08:04:33+00:00 2 Answers 0 views 0 | 109 | 314 | {"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-2022-21 | longest | en | 0.931021 |
http://mathhelpforum.com/discrete-math/84497-one-more-proof.html | 1,480,810,313,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00487-ip-10-31-129-80.ec2.internal.warc.gz | 173,708,642 | 10,427 | 1. ## One more proof
Prove that for all natural numbers k,m, Nk X Nm is finite.
I know the lemma "for every k belong to N, every subset of Nk is finite", could we just say that Nk X Nm is a subset of Nk, so it is finite....?
The hint is "consider the function f: NkXNm--->Nkm given by f(a,b)=(a-1)m+b. Use the Division Algorithm to show f is a one-to-one correspondence. This is a formal version of the product rule that says the number of elements in NkXNm is km
This makes it more complicated than I thought.
2. I suggest this mapping: $\Phi :N_k \times N_m \mapsto N_{\left( {2^k } \right)\left( {3^m } \right)} \,,\,\Phi (g,h) = \left( {2^g } \right)\left( {3^h } \right)$.
Of course $N_{\left( {2^k } \right)\left( {3^m } \right)} \,$ is finite.
It is easy to show that $\Phi$ is one to one.
Here are theorems that you need to do these sorts of cardinality problems.
T1 There is an injection from A to B if and only if there is a surjection from B to A.
T2 If there is an injection from A to B then the cardinality of A at most the cardinality of B.
i.e. $|A|\le |B|$.
T3 If there is an surjection from B to A then the cardinality of B at least the cardinality of A.
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#1
05-29-2015, 02:04 PM
abdurrehmaanseba Junior Member Join Date: May 2015 Posts: 3
Help answering this puzzle logic.
There are three trains in front of you: two steam and diesel. However, you are blindfolded, so you cannot see which one is which.
You can ask one yes-no question to any one of the three trains. If you ask the question to a steam, he will answer your question truthfully. If you ask the question to the Diesel, he will answer randomly (he could answer yes or no).
What single question can you ask to identify, with certainty, the location of one of the steams?
Second,
A group of dogs are stranded on Island, with assorted coloured furs. They are all perfect logicians - if a conclusion can be logically deduced, they will do it instantly. Each of them can see out of his own muzzle, but cannot see himself. This means each can see the colour fur of everyone else, but does not know the colour of his own fur. Every night at midnight, a ferry stops at the island, to take the dogs to safety. But, there is a catch. Dogs can only leave the island if they know what colour fur they have. Other than sight, they cannot communicate (being dogs of course). There are no mirrors, reflecting surfaces on the island, and everyone knows the rules in this paragraph.
On this island, there are 20 white coloured dogs, and 20 black coloured dogs, although the dogs do not know this. So any given white dog can see 19 white dogs and 20 black dogs, but this doesn't tell himself what colour he has. For all he knows, he could have a black paint and there are 21 black dogs in total, or he could have a red, or orange paint.
Suddenly, on one day, a booming voice calls from the sky, and shouts:
"There is a white dog on the island"
Who leaves, and on which following night?
Thanks.
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11-12-2010, 10:07 PM #49
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Re: Looking for a special equation...
Quote:
Originally Posted by pianoman2011 The point of knowing the answer to this question isn't to be able to figure out how to round numbers... it's more of a "how can I mathematically do that" kinda' thing.
Good for you for wanting to know the "why" and the "how" of something most people take for granted. It is fun to pose a problem, then look for a solution.
BTW, have you seen this site? How a Calculator Does Division
That site also links to a page that discusses many different rounding algorithms: http://www.diycalculator.com/sp-round.shtml
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11-13-2010, 02:05 PM #50
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Re: Looking for a special equation...
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Originally Posted by BShanahan14rulz lim x->oo (2a^x) = 0 if -0.5<=a<=0.5, oo otherwise lim x->oo (1/1+(2a^x))=1 if -0.5<=a<=0.5, else oo because lim x->oo (1/1) = 1, regardless of x, and limit of 2a^x is 0. ...? I failed Calc 2 twice, and had to drop out... Don't think they even offer calc at the comm. college Limits are all very interesting and everything, but simplifying integrals is where I fell on my face. That, and I couldn't understand my professor, my teaching assistant, or my tutor.
it's not 1/1, its 1/(all the other)
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11-13-2010, 02:12 PM #51
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Re: Looking for a special equation...
Quote:
Originally Posted by BShanahan14rulz lim x->oo (2a^x) = 0 if -0.5<=a<=0.5, oo otherwise lim x->oo (1/1+(2a^x))=1 if -0.5<=a<=0.5, else oo because lim x->oo (1/1) = 1, regardless of x, and limit of 2a^x is 0. ...? I failed Calc 2 twice, and had to drop out... Don't think they even offer calc at the comm. college Limits are all very interesting and everything, but simplifying integrals is where I fell on my face. That, and I couldn't understand my professor, my teaching assistant, or my tutor.
it's not 1/1, its 1/(all the other)
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If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0)....
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If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0).... [#permalink]
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If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0), what is the area of the triangle?
(1) |x| = y = 20
(2) x = |y| = 20
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If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0).... [#permalink]
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03 Jul 2018, 05:10
CAMANISHPARMAR wrote:
If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0), what is the area of the triangle?
(1) |x| = y = 20
(2) x = |y| = 20
D..
look at the sketch figure...
AC = 30-10=20..
area = $$\frac{1}{2}*20*altitude=10*altitude$$
altitude depends on B(x,y) and just y component of B
(1) |x| = y = 20
so y= 20
area = 10*20=200
suff
(2) x = |y| =20
irrespective of y in any quadrant be it -20 or 20, the altitude will be 20 as area or altitude cannot be NEGATIVE
area = 10*20=200
suff
D
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Re: If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0).... [#permalink]
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03 Jul 2018, 07:20
CAMANISHPARMAR wrote:
If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0), what is the area of the triangle?
(1) |x| = y = 20
(2) x = |y| = 20
This is a copy of our own question: https://gmatclub.com/forum/d01-183495.html
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Re: If vertices of a triangle are A (10, 0), B (x, y) and C (30, 0).... &nbs [#permalink] 03 Jul 2018, 07:20
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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®. | 1,030 | 3,214 | {"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} | 4.1875 | 4 | CC-MAIN-2018-39 | latest | en | 0.815902 |
https://serverlogic3.com/how-do-i-convert-numeric-to-data-type-in-r/ | 1,695,717,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00202.warc.gz | 561,670,256 | 15,617 | # How Do I Convert Numeric to Data Type in R?
//
Heather Bennett
In R, converting a numeric value to a different data type can be done using various functions and methods. This article will explore some common techniques to convert numeric values to other data types in R.
## Using the as.factor() Function
If you want to convert a numeric value to a factor, you can use the as.factor() function. Factors are used to represent categorical variables in R.
Here’s an example:
``````
# Create a numeric vector
numeric_vector <- c(1, 2, 3, 4)
# Convert the numeric vector to a factor
factor_vector <- as.factor(numeric_vector)
```
```
The factor_vector variable now contains the converted factor values.character() Function
To convert a numeric value to character type, you can use the as.character() function. Character data type represents strings in R.
``````
# Create a numeric variable
numeric_value <- 123
# Convert the numeric value to character type
character_value <- as.character(numeric_value)
```
```
The character_value variable now holds the converted character representation of the original numeric value.
## Using the paste() Function
If you want to combine or concatenate a numeric value with other characters or strings, you can use the paste() function. This function converts all input values into character type and concatenates them into a single string.
``````
# Create a numeric value
numeric_value <- 123
# Concatenate the numeric value with a string
combined_value <- paste("The numeric value is:", numeric_value)
```
```
The combined_value variable now contains the concatenated string.integer() Function
If you need to convert a numeric value to an integer, you can use the as.integer() function. The integer data type represents whole numbers without decimal places.
``````
# Create a numeric variable
numeric_value <- 3.14
# Convert the numeric value to an integer
integer_value <- as.integer(numeric_value)
```
```
The integer_value variable now stores the converted integer representation of the original numeric value.
## Using the format() Function
If you want to format a numeric value into a specific style, such as adding commas for thousands separators or specifying decimal places, you can use the format() function.
``````
# Create a numeric variable
numeric_value <- 1234567.89
# Format the numeric value with commas and two decimal places
formatted_value <- format(numeric_value, big.mark = ",", decimal.mark = ".", digits = 2)
```
```
The formatted_value variable now holds the formatted version of the original numeric value.
### In Conclusion
This article explored various techniques to convert a numeric value to different data types in R. You can use functions like as.factor(), as.character(), paste(), and as.integer() to achieve the desired conversions. Additionally, the format() function can be used to format numeric values in a specific style.
By understanding these conversion methods, you can effectively manipulate and transform numeric data in R for your analytical and programming needs. | 628 | 3,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.572053 |
https://kg-to-lbs.appspot.com/9/zht/2920-kg-to-lbs.html | 1,642,761,238,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00219.warc.gz | 395,608,280 | 6,128 | # 2920kg至lbs2920千克至磅
kg
=
lbs
## 如何轉換2920千克至磅?
2920 kg * 2.2046226218 lbs = 6437.4980558 lbs 1 kg
2920000.0 g
6437.4980558 lbs
## 另一種拼寫
2920 千克至磅, 2920 千克成磅, 2920 kg至lbs, 2920 kg成lbs, 2920 kg至磅, 2920 kg成磅, 2920 千克至lbs, 2920 千克成lbs, 2920 千克至lb, 2920 千克成lb | 164 | 262 | {"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-2022-05 | longest | en | 0.219219 |
https://www.mathworks.com/matlabcentral/cody/problems/554-is-the-point-in-a-circle/solutions/489842 | 1,582,594,734,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00551.warc.gz | 816,562,170 | 15,429 | Cody
# Problem 554. Is the Point in a Circle?
Solution 489842
Submitted on 23 Aug 2014 by bainhome
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
%% circle = [0, 0, 1]; Points = [0, 0.5] y_correct = 1; assert(isequal(your_fcn_name(Points,circle),y_correct))
Points = 0 0.5000 ans = 1
2 Pass
%% circle = [0, 0, 1]; Points = [0, 2] y_correct = 0; assert(isequal(your_fcn_name(Points,circle),y_correct))
Points = 0 2 ans = 0
3 Pass
%% circle = [1, 1, 1]; Points = [0, 1; 0, 0] y_correct = [1; 0]; assert(isequal(your_fcn_name(Points,circle),y_correct))
Points = 0 1 0 0 ans = 1 0 | 261 | 709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-10 | latest | en | 0.663438 |
https://help.scilab.org/docs/6.1.0/ru_RU/find.html | 1,611,456,009,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00304.warc.gz | 377,583,111 | 7,544 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Français - Português - 日本語 -
# find
find indices of boolean vector or matrix true elements
### Syntax
```[ii]=find(x [,nmax])
[i1,i2,..]=find(x [,nmax])```
### Arguments
x
may be a boolean vector, a boolean matrix, a boolean hypermatrix, a "standard" matrix or hypermatrix
nmax
an integer giving the maximum number of indices to return. The default value is -1 which stands for "all". This option can be used for efficiency, to avoid searching all indices.
ii, i1, i2, ..
integer vectors of indices or empty matrices
### Description
If `x` is a boolean matrix,
`ii=find(x)` returns the vector of indices `i` for which `x(i)` is "true". If no true element found find returns an empty matrix.
`[i1,i2,..]=find(x)` returns vectors of indices `i1` (for rows) and `i2` (for columns),.. such that `x(i1(n),i2(n),..)` is "true". If no true element found find returns empty matrices in `i1`, `i2`, ...
if `x` is a standard matrix or hypermatrix `find(x)` is interpreted as `find(x<>0)`
`find([])` returns `[]`
### Examples
```beers=["Desperados", "Leffe", "Kronenbourg", "Heineken"];
find(beers=="Leffe") // OK
find(beers=="1664") // KO
find(beers=="Foster") // KO
beers=[beers, "Foster"]
find(beers=="Foster") // OK
A=rand(1,20);
w=find(A<0.4)
A(w)
w=find(A>100)
B=rand(1,20);
w=find(B<0.4,2) //at most 2 returned values
H=rand(4,3,5); //an hypermatrix
[i,j,k]=find(H>0.9)
H(i(1),j(1),k(1))```
• boolean — Объекты Scilab'а, логические переменные и операторы & | ~
• extraction — выделение элемента матрицы или списка
• insertion — частичное присвоение или модификация значения переменной
• vectorfind — ищет расположение вектора (отмеченный символом групповой подстановки) в матрице или гиперматрице | 574 | 1,840 | {"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} | 2.765625 | 3 | CC-MAIN-2021-04 | longest | en | 0.230103 |
https://bizfluent.com/are-biweekly-payroll-income-taxes-calculated-differently.html | 1,713,820,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00353.warc.gz | 105,990,768 | 28,492 | Federal and state income tax-withholding calculation requirements apply to all relevant employees in the United States. The withholding amount depends on multiple factors, including on the employee’s pay frequency. A biweekly payroll occurs every other week, on the same day. If you have a biweekly payroll, income taxes are withheld according to that frequency.
## Calculation Criteria
Federal income tax withholding depends on the number of allowances and filing status an employee claims on her W-4 and the Internal Revenue Service Circular E tax-withholding table that matches the W-4 and the employee’s wages and pay period. To calculate a biweekly payroll, use the respective biweekly tax table instead of the weekly, semi-monthly or monthly tables. Many states use a withholding system that is similar to federal income tax for state income tax withholding purposes; if your state does this, apply the relevant state biweekly tax table. A few states, such as Arizona and Pennsylvania, require withholding at a flat percentage. If so, multiply the employee’s taxable wages for the pay period by the required percentage to arrive at the withholding. Taxable wages are an employee’s earnings after deducting pretax benefits, such as Section 125 health insurance and 401(k) contributions, from gross pay.
## Allowance Value
Each allowance an employee claims on the W-4 gives a sum that reduces her taxable wages. The IRS gives a specific amount based on the employee’s pay period. For example, in 2012, one allowance for a biweekly payroll equaled \$146.15. This is the equivalent of \$73.08 for a weekly payroll, \$158.33 for a semi-monthly payroll, and \$316.67 for a monthly payroll. Assume an employee’s taxable wages equal \$750 biweekly and that she claims two allowances on the W-4. Multiply \$146.15 times two allowances, which equals \$292.30. Subtract \$750 from \$292.30 to arrive at \$457.70, which represents her wages subject to taxation. You may use the IRS percentage method to figure the withholding. Apply the Circular E percentage method tax table that matches the employee’s biweekly pay period, filing status and wages after allowances.
## Considerations
You typically do not have to figure allowance values or use the percentage method when calculating federal income tax. In most cases, you may use the Circular E’s wage-bracket tax table that matches the employee’s wages, allowances, pay period and filing status. This method gives the exact withholding amount. If an employee earns more than the wage bracket method’s income limit, apply the percentage method. If the state gives allowances for state income tax withholding purposes, apply the required calculation to figure the withholding.
## Pay Frequency Difference
A biweekly-paid employee might appear to pay more income taxes than if she were paid weekly. That’s only because a biweekly payroll happens less frequently than a weekly payroll. In the end, it balances out. For example, an employee claims married filing status and three allowances on the W-4 and earns \$900 biweekly. According to Page 44 of the 2012 Circular E, her withholding would be \$16. As an employee paid weekly who earns half of \$900, which is \$450, her withholding would be \$8, according to Page 40 of the 2012 Circular E. | 679 | 3,295 | {"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-2024-18 | latest | en | 0.913289 |
https://dev.thep.lu.se/yat/changeset/494/trunk/doc/Statistics.tex | 1,675,375,408,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00860.warc.gz | 214,052,668 | 13,049 | # Changeset 494 for trunk/doc/Statistics.tex
Ignore:
Timestamp:
Jan 10, 2006, 2:44:14 PM (17 years ago)
Message:
r492 \documentstyle[12pt]{article} \documentclass[12pt]{article} \usepackage{html} \flushbottom \topmargin 0pt \newcommand{\bea} {\begin{eqnarray}} \newcommand{\eea} {\end{eqnarray}} \newcommand{\beq} {} \newcommand{\eeq} {} \newcommand{\bibl}[5] {#1, {\it #2} {\bf #3} (#4) #5} \newcommand{\ol}{\overline} \renewcommand{\baselinestretch} {1.0} \renewcommand{\textfraction} {0.1} \newcommand{\ovr}[2]{\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} % Use these to include comments and remarks into the text, these will % obviously appear as footnotes in the final output. \newcommand{\CR}[1]{\footnote{CR: #1}} \newcommand{\JH}[1]{\footnote{JH: #1}} \newcommand{\PE}[1]{\footnote{PE: #1}} \newcommand{\PJ}[1]{\footnote{PJ: #1}} \begin{document} {\bf Weighted Statistics} \normalsize \begin{htmlonly} This document is also available in \htmladdnormallink{PDF}{Statistics.pdf}. \end{htmlonly} \tableofcontents \section{Introduction} There are several different reasons why a statistical analysis needs to adjust for weighting. In literature reasons are mainly diveded in to groups. The first group is when some of the measurements are known to be more precise than others. The more precise a measuremtns is the larger weight it is given. The simplest case is when the weight are given before the measurements and they can be treated as deterministic. It becomes more complicated when the weight can be determined not until afterwards, and even more complicated if the weight depends on the value of the observable. The second group of situations is when calculating averages over one distribution and sampling from another distribution. Compensating for this discrepency weights are introduced to the analysis. A simple example may be that we are interviewing people but for economical reasons we choose to interview more people from the city than from the countryside. When summarizing the statistics the answers from the city are given a smaller weight. In this example we are choosing the proportions of people from countryside and people from city being intervied. Hence, we can determine the weights before and consider them to be deterministic. In other situations the proportions are not deterministic, but rather a result from the sampling and the weights must be treated as stochastic and only in rare situations the weights can be treated as independent of the observable. Since there are various origin for a weight occuring in a statistical analysis, there are various way to treat the weights and in general the analysis should be tailored to treat the weights correctly. We have not chosen one situation for our implementations, so see specific function documentation for what assumtions are made. Though, common for implementationare the following: There are several different reasons why a statistical analysis needs to adjust for weighting. In literature reasons are mainly diveded in to groups. The first group is when some of the measurements are known to be more precise than others. The more precise a measuremtns is the larger weight it is given. The simplest case is when the weight are given before the measurements and they can be treated as deterministic. It becomes more complicated when the weight can be determined not until afterwards, and even more complicated if the weight depends on the value of the observable. The second group of situations is when calculating averages over one distribution and sampling from another distribution. Compensating for this discrepency weights are introduced to the analysis. A simple example may be that we are interviewing people but for economical reasons we choose to interview more people from the city than from the countryside. When summarizing the statistics the answers from the city are given a smaller weight. In this example we are choosing the proportions of people from countryside and people from city being intervied. Hence, we can determine the weights before and consider them to be deterministic. In other situations the proportions are not deterministic, but rather a result from the sampling and the weights must be treated as stochastic and only in rare situations the weights can be treated as independent of the observable. Since there are various origin for a weight occuring in a statistical analysis, there are various way to treat the weights and in general the analysis should be tailored to treat the weights correctly. We have not chosen one situation for our implementations, so see specific function documentation for what assumtions are made. Though, common for implementationare the following: \begin{itemize} \item Setting all weights to unity yields the same result as the non-weighted version. \item Setting all weights to unity yields the same result as the non-weighted version. \item Rescaling the weights does not change any function. \item Setting a weight to zero is equivalent to removing the data point. \end{itemize} An important case is when weights are binary (either 1 or 0). Then we get same result using the weighted version as using the data with weight not equal to zero and the non-weighted version. Hence, using binary weights and the weighted version missing values can be treated in a proper way. An important case is when weights are binary (either 1 or 0). Then we get same result using the weighted version as using the data with weight not equal to zero and the non-weighted version. Hence, using binary weights and the weighted version missing values can be treated in a proper way. \section{AveragerWeighted} \subsection{Mean} For any situation the weight is always designed so the weighted mean is calculated as $m=\frac{\sum w_ix_i}{\sum w_i}$, which obviously fulfills the conditions above. In the case of varying measurement error, it could be motivated that the weight shall be $w_i = 1/\sigma_i^2$. We assume measurement error to be Gaussian and the likelihood to get our measurements is $L(m)=\prod (2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}}$. We maximize the likelihood by taking the derivity with respect to $m$ on the logarithm of the likelihood $\frac{d\ln L(m)}{dm}=\sum \frac{x_i-m}{\sigma_i^2}$. Hence, the Maximum Likelihood method yields the estimator $m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2}$. For any situation the weight is always designed so the weighted mean is calculated as $m=\frac{\sum w_ix_i}{\sum w_i}$, which obviously fulfills the conditions above. In the case of varying measurement error, it could be motivated that the weight shall be $w_i = 1/\sigma_i^2$. We assume measurement error to be Gaussian and the likelihood to get our measurements is $L(m)=\prod (2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}}$. We maximize the likelihood by taking the derivity with respect to $m$ on the logarithm of the likelihood $\frac{d\ln L(m)}{dm}=\sum \frac{x_i-m}{\sigma_i^2}$. Hence, the Maximum Likelihood method yields the estimator $m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2}$. \subsection{Variance} In case of varying variance, there is no point estimating a variance since it is different for each data point. Instead we look at the case when we want to estimate the variance over $f$ but are sampling from $f'$. For the mean of an observable $O$ we have $\widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum w_i}$. Hence, an estimator of the variance of $X$ is In case of varying variance, there is no point estimating a variance since it is different for each data point. Instead we look at the case when we want to estimate the variance over $f$ but are sampling from $f'$. For the mean of an observable $O$ we have $\widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum w_i}$. Hence, an estimator of the variance of $X$ is \begin{eqnarray} \sigma^2=-^2= \\\frac{\sum w_i(x_i-m)^2}{\sum w_i} \end{eqnarray} This estimator fulfills that it is invariant under a rescaling and having a weight equal to zero is equivalent to removing the data point. Having all weight equal to unity we get $\sigma=\frac{\sum (x_i-m)^2}{N}$, which is the same as returned from Averager. Hence, this estimator is slightly biased, but still very efficient. This estimator fulfills that it is invariant under a rescaling and having a weight equal to zero is equivalent to removing the data point. Having all weight equal to unity we get $\sigma=\frac{\sum (x_i-m)^2}{N}$, which is the same as returned from Averager. Hence, this estimator is slightly biased, but still very efficient. \subsection{Standard Error} The standard error squared is equal to the expexted squared error of the estimation of $m$. The squared error consists of two parts, the variance of the estimator and the squared bias. $^2=+-\mu>^2=>^2+(-\mu)^2$. In the case when weights are included in analysis due to varying measurement errors and the weights can be treated as deterministic ,we have \begin{eqnarray} The standard error squared is equal to the expexted squared error of the estimation of $m$. The squared error consists of two parts, the variance of the estimator and the squared bias. $^2=+-\mu>^2=>^2+(-\mu)^2$. In the case when weights are included in analysis due to varying measurement errors and the weights can be treated as deterministic ,we have Var(m)=\frac{\sum w_i^2\sigma_i^2}{\left(\sum w_i\right)^2}= \\\frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2} \frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2}= \frac{\sigma_0^2}{\sum w_i}, \end{eqnarray} \end{equation} where we need to estimate $\sigma_0^2$. Again we have the likelihood $L(\sigma_0^2)=\prod\frac{1}{\sqrt{2\pi\sigma_0^2/w_i}}\exp{-\frac{w_i(x-m)^2}{2\sigma_0^2}}$ and taking the derivity with respect to $\sigma_o^2$, $\frac{d\ln L}{d\sigma_i^2}=\sum -\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2}$ which yields an estimator $\sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2$. This estimator is not ignoring weights equal to zero, because deviation is most often smaller than the expected infinity. Therefore, we modify the expression as follows $\sigma_i^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^2}\sum w_i(x-m)^2$ and we get the following estimator of the variance of the mean $\sigma_i^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^3}\sum w_i(x-m)^2$. This estimator fulfills the conditions above: adding a weight zero does not change it: rescaling the weights does not change it, and setting all weights to unity yields the same expression as in the non-weighted case. In a case when it is not a good approximation to treat the weights as deterministic, there are two ways to get a better estimation. The first one is to linearize the expression $\left<\frac{\sum w_ix_i}{\sum w_i}\right>$. The second method when the situation is more complicated is to estimate the standard error using a bootstrapping method. $L(\sigma_0^2)=\prod\frac{1}{\sqrt{2\pi\sigma_0^2/w_i}}\exp{-\frac{w_i(x-m)^2}{2\sigma_0^2}}$ and taking the derivity with respect to $\sigma_o^2$, $\frac{d\ln L}{d\sigma_i^2}=\sum -\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2}$ which yields an estimator $\sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2$. This estimator is not ignoring weights equal to zero, because deviation is most often smaller than the expected infinity. Therefore, we modify the expression as follows $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^2}\sum w_i(x-m)^2$ and we get the following estimator of the variance of the mean $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^3}\sum w_i(x-m)^2$. This estimator fulfills the conditions above: adding a weight zero does not change it: rescaling the weights does not change it, and setting all weights to unity yields the same expression as in the non-weighted case. In a case when it is not a good approximation to treat the weights as deterministic, there are two ways to get a better estimation. The first one is to linearize the expression $\left<\frac{\sum w_ix_i}{\sum w_i}\right>$. The second method when the situation is more complicated is to estimate the standard error using a bootstrapping method. \section{AveragerPairWeighted} Here data points come in pairs (x,y). We are sampling from $f'_{XY}$ but want to measure from $f_{XY}$. To compensate for this decrepency, averages of $g(x,y)$ are taken as $\sum \frac{f}{f'}g(x,y)$. Even though, $X$ and $Y$ are not independent $(f_{XY}\neq f_Xf_Y)$ we assume that we can factorize the ratio and get $\frac{\sum w_xw_yg(x,y)}{\sum w_xw_y}$ Here data points come in pairs (x,y). We are sampling from $f'_{XY}$ but want to measure from $f_{XY}$. To compensate for this decrepency, averages of $g(x,y)$ are taken as $\sum \frac{f}{f'}g(x,y)$. Even though, $X$ and $Y$ are not independent $(f_{XY}\neq f_Xf_Y)$ we assume that we can factorize the ratio and get $\frac{\sum w_xw_yg(x,y)}{\sum w_xw_y}$ \subsection{Covariance} Following the variance calculations for AveragerWeighted we have $Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}$ where $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$ Following the variance calculations for AveragerWeighted we have $Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}$ where $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$ \subsection{correlation} As the mean is estimated as $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$, the variance is estimated as $\sigma_x^2=\frac{\sum w_xw_y(x-m_x)^2}{\sum w_xw_y}$. As in the non-weighted case we define the correlation to be the ratio between the covariance and geomtrical avergae of the variances $\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sqrt{\sum w_xw_y(x-m_x)^2\sum w_xw_y(y-m_y)^2}}$. As the mean is estimated as $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$, the variance is estimated as $\sigma_x^2=\frac{\sum w_xw_y(x-m_x)^2}{\sum w_xw_y}$. As in the non-weighted case we define the correlation to be the ratio between the covariance and geomtrical avergae of the variances $\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sqrt{\sum w_xw_y(x-m_x)^2\sum w_xw_y(y-m_y)^2}}$. This expression fulfills the following \begin{itemize} \item Having N weights the expression reduces to the non-weighted expression. \item Adding a pair of data, in which one weight is zero is equivalent to ignoring the data pair. \item Correlation is equal to unity if and only if $x$ is equal to $y$. Otherwise the correlation is between -1 and 1. \item Adding a pair of data, in which one weight is zero is equivalent to ignoring the data pair. \item Correlation is equal to unity if and only if $x$ is equal to $y$. Otherwise the correlation is between -1 and 1. \end{itemize} \section{Score} \subsection{ROC} An interpretation of the ROC curve area is the probability that if we take one sample from class $+$ and one sample from class $-$, what is the probability that the sample from class $+$ has greater value. The ROC curve area calculates the ratio of pairs fulfilling this \beq An interpretation of the ROC curve area is the probability that if we take one sample from class $+$ and one sample from class $-$, what is the probability that the sample from class $+$ has greater value. The ROC curve area calculates the ratio of pairs fulfilling this \frac{\sum_{\{i,j\}:x^-_i)^N$, where $$is the linear kenrel (usual scalar product). For weights we define the linear kernel to be =\frac{\sum w_xw_yxy}{\sum w_xw_y} and the polynomial kernel can be calculated as before (1+)^N. Is this kernel a proper kernel (always being semi positive definite). Yes, because$$ is obviously a proper kernel as it is a scalar product. Adding a positive constant to a kernel yields another kernel so$1+$is still a proper kernel. Then also$(1+)^N$is a proper kernel because taking a proper kernel to the$Nth$power yields a new proper kernel (see any good book on SVM). The polynomial kernel of degree$N$is defined as$(1+)^N$, where $$is the linear kenrel (usual scalar product). For weights we define the linear kernel to be =\frac{\sum w_xw_yxy}{\sum w_xw_y} and the polynomial kernel can be calculated as before (1+)^N. Is this kernel a proper kernel (always being semi positive definite). Yes, because$$ is obviously a proper kernel as it is a scalar product. Adding a positive constant to a kernel yields another kernel so$1+$is still a proper kernel. Then also$(1+)^N$is a proper kernel because taking a proper kernel to the$Nth$power yields a new proper kernel (see any good book on SVM). \subsection{Gaussian Kernel} We define the weighted Gaussian kernel as$\exp\left(-\frac{\sum w_xw_y(x-y)^2}{\sum w_xw_y}\right)$, which fulfills the conditions listed in the introduction. Is this kernel a proper kernel? Yes, following the proof of the non-weighted kernel we see that$K=\exp\left(-\frac{\sum w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum w_xw_y}\right)\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$, which is a product of two proper kernels.$\exp\left(-\frac{\sum w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum w_xw_y}\right)$is a proper kernel, because it is a scalar product and$\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$is a proper kernel, because it a polynomial of the linear kernel with positive coefficients. As product of two kernel also is a kernel, the Gaussian kernel is a proper kernel. We define the weighted Gaussian kernel as$\exp\left(-\frac{\sum w_xw_y(x-y)^2}{\sum w_xw_y}\right)$, which fulfills the conditions listed in the introduction. Is this kernel a proper kernel? Yes, following the proof of the non-weighted kernel we see that$K=\exp\left(-\frac{\sum w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum w_xw_y}\right)\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$, which is a product of two proper kernels.$\exp\left(-\frac{\sum w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum w_xw_y}\right)$is a proper kernel, because it is a scalar product and$\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$is a proper kernel, because it a polynomial of the linear kernel with positive coefficients. As product of two kernel also is a kernel, the Gaussian kernel is a proper kernel. \section{Distance} We have the model \beq \begin{equation} y_i=\alpha+\beta (x-m_x)+\epsilon_i, \eeq \end{equation} where$\epsilon_i$is the noise. The variance of the noise is model parameters, we minimimize the sum of quadratic errors. \beq \begin{equation} Q_0 = \sum \epsilon_i^2 \eeq \end{equation} Taking the derivity with respect to$\alpha$and$\beta$yields two conditions \beq \frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - \beta (x_i-m_x)=0 \eeq \frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - \beta (x_i-m_x)=0 and \beq \frac{\partial Q_0}{\partial \beta} = -2 \sum w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 \eeq \begin{equation} \frac{\partial Q_0}{\partial \beta} = -2 \sum w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 \end{equation} or equivalently \beq \begin{equation} \alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y \eeq \end{equation} and \beq \beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} \eeq \begin{equation} \beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} \end{equation} Note, by having all weights equal we get back the unweighted$\alpha$and$\beta$. \beq \begin{equation} \textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}= \frac{\sigma^2}{\sum w_i} \eeq \end{equation} and \beq \begin{equation} \textrm{Var}(\beta )= \frac{w_i^2(x_i-m_x)^2\frac{\sigma^2}{w_i}} {(\sum w_i(x_i-m_x)^2)^2}= \frac{\sigma^2}{\sum w_i(x_i-m_x)^2} \eeq \end{equation} Finally, we estimate the level of noise,$\sigma^2$. Inspired by the unweighted estimation \beq \begin{equation} s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2} \eeq \end{equation} we suggest the following estimator \beq s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum w_i-2\frac{\sum w_i^2}{\sum w_i}} \eeq s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum w_i-2\frac{\sum w_i^2}{\sum w_i}} \section{Outlook} \subsection{Hierarchical clustering} \label{hc} A hierarchical clustering consists of two things: finding the two closest data points, merge these two data points two a new data point and calculate the new distances from this point to all other points.\\ and calculate the new distances from this point to all other points. In the first item, we need a distance matrix, and if we use Euclidean distanses the natural modification of the expression would be \beq d(x,y)=\frac{\sum w_i^xw_j^y(x_i-y_i)^2}{\sum w_i^xw_j^y} \eeq \\ d(x,y)=\frac{\sum w_i^xw_j^y(x_i-y_i)^2}{\sum w_i^xw_j^y} For the second item, inspired by average linkage, we suggest \beq d(xy,z)=\frac{\sum w_i^xw_j^z(x_i-z_i)^2+\sum w_i^yw_j^z(y_i-z_i)^2}{\sum w_i^xw_j^z+\sum w_i^yw_j^z} \eeq d(xy,z)=\frac{\sum w_i^xw_j^z(x_i-z_i)^2+\sum w_i^yw_j^z(y_i-z_i)^2}{\sum w_i^xw_j^z+\sum w_i^yw_j^z} to be the distance between the new merged point$xy$and$z\$, and we | 5,952 | 20,948 | {"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} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.781882 |
https://developer.apple.com/documentation/simd/simd_float3?language=objc | 1,586,200,638,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00100.warc.gz | 419,549,283 | 12,463 | Type Alias
# simd_float3
A vector of three 32-bit single-precision values.
## Topics
### Vector Creation Functions
`simd_make_float3`
Returns a new vector by truncating the specified vector.
`simd_make_float3`
Returns a new vector by truncating the specified vector.
`simd_make_float3`
Returns a new vector with the first element set to a scalar value, and other elements set to zero.
`simd_make_float3`
Returns a new vector with the first and second elements from a two-element vector, and other elements set to zero.
`simd_make_float3`
Returns a new vector from a three-element vector.
`simd_make_float3`
Returns a new vector from the first, second, and third elements of a four-element vector.
`simd_make_float3`
Returns a new vector from 3 scalar values.
`simd_make_float3`
Returns a new vector from a scalar value and a two-element vector.
`simd_make_float3`
Returns a new vector from a scalar value and a two-element vector.
`simd_make_float3_undef`
Returns a new vector with the first element set to a scalar value, and other elements undefined.
`simd_make_float3_undef`
Returns a new vector with the first and second elements from a two-element vector, and other elements undefined.
### Common Functions
`simd_abs`
Returns the absolute value of each element in a vector.
`simd_clamp`
Returns each element in a vector clamped to a specified range.
`simd_sign`
Returns the sign of each element in a vector.
`simd_fract`
Returns the fractional part of each element in a vector.
`simd_step`
Returns zero for each element in a vector less than a specified edge; one otherwise.
`simd_equal`
Returns true if all elements of a vector are equal; false otherwise.
### Reduce Functions
`simd_reduce_add`
Returns the sum of all elements in a vector.
`simd_reduce_max`
Returns the maximum value in a vector.
`simd_reduce_min`
Returns the miniumum value in a vector.
### Interpolation Functions
`simd_mix`
Returns an elementwise linearly interpolated value between two vectors.
`simd_smoothstep`
Returns an elementwise smoothly interpolated value between two vectors.
### Extrema Functions
`simd_max`
Returns the maximum value of each element in a vector.
`simd_min`
Returns the minimum value of each element in a vector.
### Reciprocal and Reciprocal Square Root Functions
`simd_recip`
Returns the reciprocal of each element in a vector.
`simd_rsqrt`
Returns the reciprocal square root of each element in a vector.
`simd_precise_recip`
Returns the precise reciprocal of each element in a vector.
`simd_precise_rsqrt`
Returns the precise reciprocal square root of each element in a vector.
`simd_fast_recip`
Returns the fast reciprocal of each element in a vector.
`simd_fast_rsqrt`
Returns the fast reciprocal square root of each element in a vector.
### Exponential and Logarithmic Functions
`exp`
Returns e raised to the power of each element in a vector.
`exp2`
Returns 2 raised to the power of each element in a vector.
`exp10`
Returns 10 raised to the power of each element in a vector.
`expm1`
Returns eˣ-1 for each element in a vector.
`log`
Returns the natural logarithm of each element in a vector.
`log2`
Returns the base 2 logarithm of each element in a vector.
`log10`
Returns the base 10 logarithm of each element in a vector.
`log1p`
Returns log(1+x) for each element in a vector.
### Geometry Functions
`simd_cross`
Returns the cross product of two vectors.
`simd_dot`
Returns the dot product of two vectors.
`simd_insphere`
Returns true if a point is within a sphere; false otherwise.
`simd_normalize`
Returns a vector pointing in the same direction of the supplied vector with a length of 1.
`simd_orient`
Tests the orientation of the three supplied vectors.
`simd_orient`
Tests the orientation of the four supplied vectors.
`simd_project`
Returns the first vector projected onto the second vector.
`simd_precise_normalize`
Returns the precise normalized vector.
`simd_precise_project`
Returns the precise projected vector.
`simd_fast_normalize`
Returns the fast normalized vector.
`simd_fast_project`
Returns the fast projected vector.
`simd_distance_squared`
Returns the square of the distance between two vectors.
`simd_length`
Returns the length of a vector.
`simd_length_squared`
Returns the square of the length of a vector.
`simd_norm_inf`
Returns the inf-norm (the maximum absolute value) of a vector.
`simd_norm_one`
Returns the one-norm (the sum of absolute values) of a vector.
`simd_refract`
Returns the refraction direction of an incident vector, a unit normal vector, and an index of refraction eta.
`simd_reflect`
Returns the reflection direction of an incident vector and a unit normal vector.
### Hyperbolic Functions
`acosh`
Returns the inverse hyperbolic cosine of each element in a vector.
`asinh`
Returns the inverse hyperbolic sine of each element in a vector.
`atanh`
Returns the inverse hyperbolic tangent of each element in a vector.
`cosh`
Returns the hyperbolic cosine of each element in a vector.
`sinh`
Returns the hyperbolic sine of each element in a vector.
`tanh`
Returns the hyperbolic tangent of each element in a vector.
### Logic Functions
`simd_select`
Returns a vector containing elements from either the first or second parameter, depending on the high-order bit of the corresponding element in the third parameter.
`simd_bitselect`
Returns a vector containing elements from either the first or second parameter, depending on the corresponding element in the third parameter.
### Math Functions
`cbrt`
Returns the cube root of each element in a vector.
`copysign`
Returns each element of a vector, with the sign of the corresponding element in a second vector.
`erf`
Returns the error function for each element in a vector.
`erfc`
Returns the complementary error function for each element in a vector.
`fabs`
Returns the absolute value of each element in a vector.
`fdim`
Returns the positive difference between corresponding elements in two vectors.
`fma`
Returns the multiply-add result for corresponding elements in three vectors.
`fmod`
Returns the modulus after dividing each element in a vector by the corresponding element in a second vector.
`hypot`
Returns the hypotenuse of a right-angled triangle with the sides that are adjacent to the right angle defined by two vectors.
`nextafter`
Returns the next representable value of each element in a vector in the direction of the corresponding element in a second vector.
`pow`
Returns each element in a vector raised to the power of the corresponding element in a second vector.
`remainder`
Returns the remainder after dividing each element in an array by the corresponding element in a second array of double-precision values.
`rint`
Returns each element in a vector rounded to the nearest integer in the specified direction.
`round`
Returns each element in a vector rounded to the nearest integer.
`simd_muladd`
Returns the multiply-add result for corresponding elements in three vectors.
`sqrt`
Returns the square root of each element in a vector.
`tgamma`
Returns the gamma function for each element in a vector.
### Trigonometric Functions
`acos`
Returns the arccosine of each element in a vector.
`asin`
Returns the arcsine of each element in a vector.
`atan`
Returns the arctangent of each element in a vector.
`atan2`
Returns the arctangent of each pair of corresponding elements in two vectors.
`cospi`
Returns the cosine of each element in a vector multiplied by pi.
`sin`
Returns the sine of each element in a vector.
`sinpi`
Returns the sine of each element in a vector multiplied by pi.
`tan`
Returns the tangent of each element in a vector.
`tanpi`
Returns the tangent of each element in a vector multiplied by pi.
`simd_float2`
`simd_float4` | 1,707 | 7,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-16 | latest | en | 0.675951 |
http://math.stackexchange.com/questions/431999/proving-the-normed-linear-space-v-a-b-is-a-metric-space-symmetry | 1,469,485,069,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00014-ip-10-185-27-174.ec2.internal.warc.gz | 165,881,331 | 17,714 | # Proving the normed linear space, $V, ||a-b||$ is a metric space (Symmetry)
The following theorem is given in Metric Spaces by O'Searcoid
Theorem: Suppose $V$ is a normed linear space. Then the function $d$ defined on $V \times V$ by $(a,b) \to ||a-b||$ is a metric on $V$
Three conditions of a metric are fairly straight-forward.
By the definitions of a norm, I know that $||x|| \ge 0$ and only holds with equality if $x=0$. Thus $||a-b||$ is non-negative and zero if and only if $a=b$.
The triangle inequality of a normed linear space requires: $||x+y|| \le ||x|| + ||y||$. Let $x = a - b$ and $y = b - c$. Then $||a - c|| \le || a - b || + || b - c||$ satisfying the triangle inequality for a metric space.
What I am having trouble figuring out is symmetry. The definition of a linear space does not impose any condition of a symmetry. I know from the definition of a linear space that given two members of $V$, $u$ and $v$ they must be commutative, however, I do not see how that could extend here.
Thus what I would like to request help with is demonstrating $||a - b|| = ||b - a||$.
-
Well, which clause in the definition of a norm have you not used? – Chris Eagle Jun 29 '13 at 0:25
@PeterTamaroff I knew it had to be something really simple. Thanks! – GovEcon Jun 29 '13 at 0:26
$$\|a-b\|=\|(-1)(b-a)\|=|-1|\cdot\|b-a\|=\|b-a\|$$
Use $\lVert c\cdot x\rVert =|c|\lVert x\rVert$. | 432 | 1,395 | {"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} | 3.9375 | 4 | CC-MAIN-2016-30 | latest | en | 0.857869 |
https://www.gurobi.com/documentation/current/quickstart_mac/cpp_example_mip1_cpp_cpp.html | 1,695,980,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00487.warc.gz | 851,933,593 | 121,226 | Filter Content By
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## Example mip1_c++.cpp
This is the complete source code for our example (also available in <installdir>/examples/c++/mip1_c++.cpp)...
/* Copyright 2023, Gurobi Optimization, LLC */
/* This example formulates and solves the following simple MIP model:
maximize x + y + 2 z
subject to x + 2 y + 3 z <= 4
x + y >= 1
x, y, z binary
*/
#include "gurobi_c++.h"
using namespace std;
int
main(int argc,
char *argv[])
{
try {
// Create an environment
GRBEnv env = GRBEnv(true);
env.set("LogFile", "mip1.log");
env.start();
// Create an empty model
GRBModel model = GRBModel(env);
// Create variables
GRBVar x = model.addVar(0.0, 1.0, 0.0, GRB_BINARY, "x");
GRBVar y = model.addVar(0.0, 1.0, 0.0, GRB_BINARY, "y");
GRBVar z = model.addVar(0.0, 1.0, 0.0, GRB_BINARY, "z");
// Set objective: maximize x + y + 2 z
model.setObjective(x + y + 2 * z, GRB_MAXIMIZE);
// Add constraint: x + 2 y + 3 z <= 4
model.addConstr(x + 2 * y + 3 * z <= 4, "c0");
// Add constraint: x + y >= 1
model.addConstr(x + y >= 1, "c1");
// Optimize model
model.optimize();
cout << x.get(GRB_StringAttr_VarName) << " "
<< x.get(GRB_DoubleAttr_X) << endl;
cout << y.get(GRB_StringAttr_VarName) << " "
<< y.get(GRB_DoubleAttr_X) << endl;
cout << z.get(GRB_StringAttr_VarName) << " "
<< z.get(GRB_DoubleAttr_X) << endl;
cout << "Obj: " << model.get(GRB_DoubleAttr_ObjVal) << endl;
} catch(GRBException e) {
cout << "Error code = " << e.getErrorCode() << endl;
cout << e.getMessage() << endl;
} catch(...) {
cout << "Exception during optimization" << endl;
}
return 0;
}
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Request free trial hours, so you can see how quickly and easily a model can be solved on the cloud. | 655 | 2,176 | {"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-2023-40 | latest | en | 0.569666 |
https://ebs.sakarya.edu.tr/DersDetay/DersinDetayliBilgileri/25408/33058?Disaridan= | 1,576,312,268,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00322.warc.gz | 343,252,987 | 23,246 | Ders Bilgileri
#### Ders Tanımı
Ders Kodu Yarıyıl T+U Saat Kredi AKTS
HISTORY OF MATHEMATICS IME 103 1 2 + 0 2 4
Dersin Dili Türkçe Dersin Seviyesi Lisans Dersin Türü ZORUNLU Dersin Koordinatörü Dr.Öğr.Üyesi AYŞE ZEYNEP AZAK Dersi Verenler Dr.Öğr.Üyesi AYŞE ZEYNEP AZAK Dersin Yardımcıları Research Assistan .Mithat TAKUNYACI Dersin Kategorisi Alanına Uygun Öğretim Dersin Amacı The aim of this course: introduce historical process of mathematics and important people on this field from the century of old Egypt and Mesopotamia to Islamic Civilisation. Dersin İçeriği Development process and operations about arithmetic from B.C. 50 000 to wadays. Studies about geometry, solids, analytic geometry, modern geometry, geometry tools, algebra, equations, Binom theorem, logarithm, trigonometry, measures, metric system, sets, integral, computers, numbers, structures, solving problem, vectors and graphs. Bibliographies about mathematics scientist.
Dersin Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri 1 - Express the process of history of mathematics. 1 - A - C - 2 - Explain the calculus technic and number systems in old Egyptions. 1 - C - A - 3 - Explain the sumerian Calculus technic and sixty based system 1 - A - C - 4 - Intrepret the Zero and Pi number. 1 - C - A - 5 - Explain Babylon mathematics, babylon algebra and geometry. 1 - A - C - 6 - Explain Mathematics in Islamic Civilisation. 1 - C - A -
Öğretim Yöntemleri: 1:Lecture Ölçme Yöntemleri: A:Testing C:Homework
#### Ders Akışı
Hafta Konular ÖnHazırlık
1 Calculus technic, number systems, digits and calculus art in old Egyptions
2 Old Egypt Geometry
3 Sumerian Calculus technic, sixty based system
4 Babylon mathematics, babylon algebra and geometry
5 Old Greece mathematics, Tales
6 Pythagorean, Zeno, Demokritus
7 Archytas, Platon, Eudoxus, Aristo
8 Euclidean and elements of euclidean
9 Mid – Term Exam
10 Archimedes, Eratosthenes and Apolonyus, cause of Old Greece Civilisation collapse
11 Heron, Batlamyus, Diyafont, Pappus, Hypatia
12 Mathematics in Islamic Civilisation, Harezmi
13 Abdülhamid İbni Türk, Sabit Bin Kurra
14 Ömer Hayyam, Nasıreddin-i Tusi
Ders Notu
Ders Kaynakları
#### Dersin Program Çıktılarına Katkısı
No Program Öğrenme Çıktıları KatkıDüzeyi
1 2 3 4 5
1 Have general information about basic concept, theory and applicaitons on mathematics X
2 Have ability of mathematical thinking and apply to real life X
3 Classify a problem systematically also create comprehensible, understandable and objective solutions X
4 Establish relationship among events looked different X
5 Get clear and exact ideas about relationships among time, place and numbers X
6 Use principles of scientific method on problem solving X
7 Have character on explorer, impartial, disinterested, give sound decisions, open mind and believe that spreading information is needed X
8 Think creative and critical X
9 Improve methods as fast, understandable and practical to faced problems X
10 Have information about national and international modern problems
11 Get life long learning behaviour X
12 Know and apply approach, aim, goal, principle and techniques of teaching program on special field with basic value and principles of Turkish National Education System
13 Evaluate development and learning of students, provide to evaluate self-assessment and other students. Use the assessment results to better instruction; share the results with student, parent, administrators and teachers.
14 Endeavor for constant development by making self-evaluation. Open the new knowledge and ideas, have a role to develop himself/herself and his/her institution. Have enough knowledge and conscious to protection subjects on society values and environment.
15 Interrogator ( get habit to find source of knowledge by asking “Why?” question instead of direct acceptance to given knowledges ) X
#### Değerlendirme Sistemi
YARIYIL İÇİ ÇALIŞMALARI SIRA KATKI YÜZDESİ
AraSinav 1 40
KisaSinav 1 20
Odev 1 20
KisaSinav 2 20
Toplam 100
Yıliçinin Başarıya Oranı 50
Finalin Başarıya Oranı 50
Toplam 100
#### AKTS - İş Yükü
Etkinlik Sayısı Süresi(Saat) Toplam İş yükü(Saat)
Course Duration (Including the exam week: 16x Total course hours) 16 2 32
Hours for off-the-classroom study (Pre-study, practice) 16 1 16
Mid-terms 1 15 15
Quiz 2 8 16
Assignment 1 5 5
Final examination 1 20 20
Toplam İş Yükü 104
Toplam İş Yükü /25(s) 4.16
Dersin AKTS Kredisi 4.16
; ; | 1,221 | 4,415 | {"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-2019-51 | latest | en | 0.600774 |
http://community.wolfram.com/groups/-/m/t/247262?p_p_auth=W55OM9Iq | 1,521,546,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00209.warc.gz | 66,445,521 | 16,621 | # Simple question about Simplify[] with assumptions
GROUPS:
Hello,If I try to simplify Sqrt[x^2] - Abs under the assumptions x>0 and x<=0 separately, it works correctly, but if I combine them into one condition it does not:Simplify[Sqrt[x^2] - Abs[x],x<=0]0Simplify[Sqrt[x^2] - Abs[x],x>0]0Simplify[Sqrt[x^2] - Abs[x],(x>0) || (x<=0)]Sqrt[x^2] - Abs[x]Why did it leave the expression unevaluated for (x>0) || (x<=0) case? It should evaluate to 0 like in the other two cases. What am I missing? Thank you.Curiously, if I use Element[x,Reals] as a condition then it works:Simplify[Sqrt[x^2]-Abs[x],Element[x,Reals]]0So, apparently, Element[x,Reals] and (x>0) || (x<=0) is not the same thing, oddly enough... | 234 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-13 | longest | en | 0.855415 |
https://optimization.mccormick.northwestern.edu/index.php?title=Adaptive_robust_optimization&oldid=4618 | 1,670,641,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00289.warc.gz | 470,273,582 | 14,462 | Author: Woo Soo Choe (ChE 345 Spring 2015)
Steward: Professor You, Dajun Yue
## Methodology
In order to investigate how Adaptive Robust Optimization problem, numerous techniques may be used. However, given the scope of this page, only three of the techniques will be introduced. The three algorithms are Bender's Decomposition, Trilevel Optimization, and column-and-Constraint Generation Algorithm and for the Benders Decomposition and Trilevel . When using Benders Decomposition approach, the algorithm essentially breaks down the original problem into the outer and inner problems. Once the problem is divided into two parts, the outer problem is solved using the Benders Decomposition and the inner problem is solved using the Outer Approximation. The detailed steps are as follows.
Benders Decomposition
The Outer Problem: Benders Decomposition
Step 1: Initialize, by denoting the lower bound as $LB = - \infty$ and the upper bound as $UB=\infty$ and set the iteration count as $C=0$. Then choose the termination tolerance $\epsilon$.
Step 2: Solve the master problem
$\begin{array}{llr} max_{x,\zeta} &c^T x + \zeta \\ s.t. &Fx \le f &\\ &{\zeta} \ge -h^T \alpha_l + (Ax-g)^T \beta_l + d_l^T \lambda_l , \forall \le C \end{array}$
In this case, $(x_c, \zeta_c)$ denote the optimum solution.
Step 3: Update the lower bound $LB=c^T x_c + \zeta_c$
Step 4: Increase $C$, the iteration count by 1
Step 5: Solve $I(x_c)$, the inner problem and denote the optimal solution as $(d_c, \alpha_c, \beta_c, \lambda_c)$. Update $UB=min(UB, c^T x_c + I(x_c))$, where $UB$ stands for the upper bound.
Detailed procedure of Step 5 is as follows.
if $UB-LB \ge \epsilon$ then
Go to step 2
else
Calculate $y_c$, the dispatch variable given $x_c$ and $d_c$
end
The Inner Problem : Outer Approximation
Step 1: Initialize by using the commitment decision from the outer problem $x_c$ from the outer problem. Then, find an initial uncertainty realization $d_1 \in \mathbb{D}$, set the lower bound $LOA = - \infty$ and the upper bound $UOA = \infty$, set iteration count j=1 and then termination tolerance which is denoted as $\theta$
Step 2: Solve the sub-problem below.
$\begin{array}{llr} S(x_c,d_j) &= max -h^T \alpha + (Ax_c - g ) ^T \Beta + d_j^T \lambda \\ s.t. & -H^T \alpha - B^T \beta + J^T \lambda = b \\ &alpha \ge 0, \beta \ge 0 \\ \end{array}$
In the inner problem, the optimal solution is denoted as $(\alpha_j, \beta_j, \lambda_j)$. Furthermore, define $d_j^T \lambda_j$ as $L_j(d_j,\lambda_j) + (d-d_j)^T \lambda_j + (\lambda - \lambda_j)^T d_j$. Then, update $LOA = max (S(x_c,d_j)), LOA)$
Step 3: Solve the master problem
$\begin{array}{llr} U(d_j, \lambda_j) &= max -h^T \alpha + (Ax_c-g)^T \beta + \zeta \\ s.t &\zeta \le L_i (d_i, \lambda_i), \forall \le j \\ &-H^T \alpha -B^T \beta +J^T \lambda =b \\ &Dd \le k &\\ &\alpha \ge 0 , \beta \ge 0 \\ \end{array}$
Increase the iteration of j by 1. While the optimal solution is denoted as $(d_j, \alpha_j, \beta_j, \lambda_j, \zeta)$, update the upper bound as $UOA = min(UOA, U(d_j, \lambda_j))$
if $UB - LB \ge 0$ then
Go to Step 2
else
Return optimal solution as the output
end
As seen from the algorithms, Benders Decomposition divides an Adaptive Robust Optimization problem into outer and inner problems and solves them using two algorithms. While this may cause some confusion for ones who have no previous exposures to Benders Decomposition, the approach to solving the outer problem is also called the Benders Decomposition and the approach to solving the inner problem is called the Outer Approximation method. Fundamentally, this algorithm works by first solving the outer problem until $UB -LB \ge \epsilon$ condition is met and then use the $x. \zeta$ from the outer problem to plug into the inner problem and solve for the optimum solution until $UB -LB \ge \ 0$ condition is met. This method has an advantage over traditional Robust Optimization in a sense that it does not sacrifice as much optimality in the solution at the cost of obtaining a conservative answer. Unfortunately, Benders Decomposition method has three problems. First problem is the fact that the master problem relies on the dual variables of the inner and outer problems, which means that the sub-problems cannot have integer variables. Second problem is that the solution does not guarantee a global optimal solution, and it means that the algorithm may not return the absolute worst case scenario before returning the solution. Third problem is that it takes a long time to compute the answer and this might pose a problem when solving a large scale problem. In order to resolve this issue, another algorithm called Trilevel Optimization was proposed by Bokan Chen of Iowa University. Before iterative Trilevel Optimization algorithm applied, the problem needs to be reformulated in an appropriate form as shown below.
$\begin{array}{llr} \min\limits_x &c^T x + b^T y \\ s.t &Fx \ le f \\ &max_d & b^T y \\ &s.t. &Dd \le k \\ \min\limits_y &b^T y \\ s.t. &Ax + By \le g \\ &Hy \le h \\ &Jy = d \\ \end{array}$
Equation $Fx \ le f$ represents the first constraints on the first stage commitment variables and equation $Dd \le k$ represents the uncertainty set of demand. Equation $Ax + By \le g$ represents the constraints that couples the commitment and dispatch variables. $Hy \le h$ constrains the dispatch variable and $Jy =d$ is the constraint that couples the demand variables and the dispatch variables. Then, the reformulated model can be further refined into the following model.
$\begin{array}{llr} \min_{x,\phi} &c^T x + \phi \\ s.t. &Fx \le f \\ & \phi \ge b^T y, \forall y \in \mathbb{Y_D} \end{array}$
When $\mathbb{Y_D} = \big\{ y| Ax + By \le g Hy \le h \big\} \cap \big\{y | Jy = d, d \in \mathbb{D} \big\}$. Assuming $\Omega \subset \mathbb{D}$, we have $\mathbb{Y_\Omega} = \big\{ y| Ax + By \le g Hy \le h \big\} \cap \big\{y | Jy = d, d \in \Omega \big\}$. This implies $\mathbb{Y}_\Omega \subset \mathbb{Y_D}$. This relaxes the problem into the following form.
$\begin{array}{llr} \min_{x,\phi} &c^T x + \phi \\ s.t. &Fx \le f \\ &\phi \ge b^T y, \forall y \in \mathbb{Y_\Omega} \end{array}$
This allows the trilevel problem to be split into the master problem and a sub-problem. Following is the relaxation of the master problem M of the trilevel problem as follows.
$\begin{array}{llr} \min_{x,\phi} &c^T x + \phi &\\ s.t. &Fx \le f &\\ &\phi \ge b^T y, \forall y \in \mathbb{Y_D} \end{array}$
The master problem M is a relaxation of the trilevel problem as follows:
$\begin{array}{llr} \min_{x,\phi} &c^T x + \phi \\ s.t. &Fx \le f \\ &\phi \ge b^T y^i, \forall i = 1,...,| \Omega | \\ &H y^i \le h, \forall i = 1, ..., \ \Omega | \\ &Ax + By^i \le g, \forall i = 1,..., | \Omega | \\ &J y^i = d^i, \forall i = 1,..., | \Omega | \end{array}$
Following is the bilevel sub-problem which yields the dispatch cost under the worst-case scenario
$\begin{array}{llr} \max\limits_d &b^T y \\ \text{s.t.} &Dd \le k \\ &min_y b^T y \\ &s.t. Hy \le h \\ &By \le g - Ax \\ &Jy = d &\\ \end{array}$
An Iterative Algorithm For The Trilevel Optimization Problem Optimization Problem
Step 1: Initialize by denoting lower bound as $LB= -\infty$ and the upper bound as $UB= \infty$. Then create and empty set $\Omega$.
Step 2: Solve the master problem M. Where the solution of the problem is $(x^M, \zeta^M, y^M)$. Then update the lower bound of the algorithm $LB=max(LB, c^Tx^M+\zeta^M)$.
Step 3: Solve the sub-problem $S(x^M)$. The solution to the problem is $(y^S, d^S)$. Update the upper bound which is $UB=min(UB, c^Tx^M+b^Ty^S)$ and the set
$\Omega = d^S \cap \Omega$.
if Failed to parse(unknown function '\g'): UB-LB \g 0
then
Go to Step 2
else
Find $text{argmax}_i b^T y^i$. Calculate the total cost $\zeta = c^T x^M + b^T y^i$, return the optimal solution as $x^M, d^i, y^i, \zeta$
end
## Example
In order to illustrate how Adaptive Robust Optimization works, a numerical example is given in this section. This example involves 3 factories and 5 customers and a detailed information if provided through the table below.
Before solving the problem, the basic set up is as follows.
$\begin{array}{llr} & v_j = \min\limits_{i \in O(y)} \big\{c_{ij}\big\} \\ & w_{ij} \begin{cases} 0 &i \in O(y) \\ max_{i \in C(y)} \big\{(v_j - c_{ij}),0 \big\} &i \in C(y) \\ \end{cases} \end{array}$
In this case, $v_{ij}$ are the dual variables associated with the demand constraints and $w_{ij}$ represent the dual variables associated with the setup constraints. Furthermore the dual variable can be represented as $u$ and it means the combination of $(v,w)$. From the proposition, the following Benders cut is derived.
$\beta_y(y)=u(b-By)+f^ty$
For this specific problem, the Benders cut can be rewritten as follows.
$\beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i$
Returning to the problem, we denote factory 1, factory 2, and factory 3 as $y_1, y_2, y_3$ and to start the problem, we only assume factory 1 is open and in this case, $v_j = min_{i \in O(y)} \big\{ c_{ij} \big\} , j=1,...,m$ would become $v_j=(2,3,4,5,7)$. Based on the proposition, $w_{ij}$ may be found as follows. $\begin{array}{llr} w_{1j}=0 \\ w_{2j}=(0,0,3,3,1) \\ w_{3j}=(0,0,2,4,4) \end{array}$
Then, solving Benders Cut, we get the following result.
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)=2+3+4+5+7+(2-0)y_1+(3-(3+3+1))y_2+(3-(2+4+4+4))y_3 \\ \beta_y(y)=21+2y_1-4y_2-7y_3 \end{array}$
From this, the upper bound on the solution, 23, obtained. Then, the Benders cut is used to solve the master problem and by inserting the Benders cut into the master problem, we get the problem in the following form.
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &y \in \mathbb{B}^3 \end{array}$
In the above problem, the optimal solution is y=(0,1,1), meaning it is best to keep factory 1 closed and open factories 2 and 3. This yield a solution of 10, which becomes new y and one more iteration of the algorithm may be done with this. When the $v_j$ and $w_{ij}$ are found again with the solution, following values are obtained
$\begin{array}{llr} v_j = (4,3,1,1,3) \\ w_{1j} = (2,0,0,0,0)\\ w_{2j}=w{3j}=0 \end{array}$
Then the Benders cut was calculated again as follows.
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)= (4+3+1+1+3)+(2-2)y_1+(3-0)y_2+(3-0)y_3 \\ \beta_y(y)=12+3y_2+3y_3 \end{array}$
From this, we get a new upper bound which is 18 and the master problem looks like:
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &z \ge 12+ 3y_2+3y_3 \\ &y \in \mathbb{B}^3 \end{array}$
As the solution to the problem, we get $y=(0,0,1)$. Then, we repeat the iteration process, which yields:
$\begin{array}{llr} v_j = (5,4,2,1,3) \\ w_{1j} = (3,1,0,0,0) \\ w_{2j} = (1,1,1,0,0) \\ w_{3j} = 0 \end{array}$
Then the Benders cut becomes:
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)=(5+4+2+1+3)+(2-4)y_1+(3-3)y_2+(3-0)y_3 \\ \beta_y(y)=15-2y_1+3y_3 \end{array}$
Now, there is no better upper bound and the master problem becomes:
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &z \ge 12+ 3y_2+3y_3 \\ &z \ge 15-2y_1+3y_3 &y \in \mathbb{B}^3 \end{array}$
This yields the optimal solution of $y=(1,0,1)$ and the new lower bound is 16. when the iteration process is repeated until the upper and lower bound are the same, we obtain the optimal solution value of 16 and come to the decision of opening factories 1 and 3 only.
As seen from the iterative procedure, Trilevel Optimization also breaks an optimization problem into smaller parts and use iterative algorithm to close in the difference between the upper and the lower bound. However, the Trilevel Optimization addresses the issues with the Benders Decomposition approach.
## Model Formulation
Adaptive Robust Optimization implements different techniques to improve on the original static robust optimization by incorporating multiple stages of decision into the algorithm. Currently, in order to minimize the complexity of algorithm, most of the studies on adaptive robust optimization have focused on two-stage problems. Generally, Adaptive Robust Optimization may be formulated in various different forms but for simplicity, Adaptive Robust Optimization in convex case was provided.
$\begin{array}{llr} \max\limits_{x\in \mathit{S}} &f(x) + \max\limits_{b\in \mathit{B}} Q(x,b) &\\ \end{array}$
In the equation $x$ is the first stage variable and $y$ is the second stage variable, where S and Y are all the possible decisions, respectively.$b$ represents a vector of data and when $\mathit{B}$ represents uncertainty set.
In order for the provided convex case formulation to work, the case must satisfy five conditions:
1. $\mathit{S}$ is a nonempty convex set
2. $f(x)$ is convex in $x$
3. $\mathit{Y}$ is a nonempty convex set
4. $h(y)$ is convex in $y$
5. For all i=1,...,n, $H_i (x,y,b)$ is convex in $(x,y), \forall b \in \mathit{B}$
Clearly, not every Adaptive Robust Optimization problem may be solved using exactly one model. However, key features that need to be present in a model of Adaptive Robust Optimization are the variables which respectively represent the multiple stages, uncertainty sets whether in ellipsoidal form, polyhedral form, or other novel way, and general layout of the problem which solves for the minimum loss at the worst case scenario. Furthermore, another key feature is that second stage variables are not known. Another form of Adaptive Robust Optimization formulation is provided below.
$\begin{array}{llr} \ min_x &c^T x + \max\limits_{d\in \mathbb{D}} \min\limits_{y\in {\Omega}} b^T y &\\ \text{s.t.} &Fx \le f &\\ &{\Omega} (x,d)= \big\{y: Hy \le h, Ax+By \le g, Jy=d \big\} &\\ &\mathbb{D} = \big\{ d: Dd \le k \big\} \end{array}$
Similarly as in the first formulation provided, $x$ and $y$ represent the first stage variable and the second stage variable respectively. In this case the, $\mathbb{D}$ is the polyhedron uncertainty set of demand $d$and $\Omega$ represents the uncertainty set for the second stage variable $y$. In this case, H, A, B, g, J, D, and k are numerical parameters which could represent different parameters under different circumstances.
## Introduction
Traditionally, robust optimization has solved problems based on static decisions which are predetermined by the decision makers. Once the decisions were made, the problem was solved and whenever a new uncertainty was realized, the uncertainty was incorporated to the original problem and the entire problem was solved again to account for the uncertainty.[1] Generally, robust optimization problem is formulated as follows.
In the equation $x\epsilon\mathbb{R}^n$ is a vector of decision variables and $f_o,f_i$are functions and are the uncertainty parameters which take random value in the uncertainty sets Failed to parse(unknown function '\subseteqmathbb'): \mathcal{U}_i\subseteqmathbb{R}^k . When robust optimization is utilized to solve a problem, three implicit assumptions are made.
1. All entries need in the decision vector$x$ get specific numerical values prior to the realization of the actual data.
2. When the real data is within the range of the uncertainty set $\mathcal{U}$, the decision maker is responsible for the result obtained through the robust optimization algorithm
3. The constraints are hard and the violation of the constraints may not be tolerated when the real data is within the uncertainty set $mathcal{U}$
The three assumptions grant robust optimization technique immunity from uncertainties. There are other types of optimization techniques such as Stochastic Optimization which may be used to handle problems with uncertainties. However, because Stochastic Optimization has its own drawback because it requires the probability distribution of the events. By having the decision makers make guesses about the probability distribution, Stochastic Optimization method often yield results that are less conservative than the ones by Robust Optimization method.
Robust Optimization certainly may have advantages over other optimization methods, but unfortunately, most robust optimization problems for real life applications require multiple stages to account for uncertainties and traditional static robust has shown limitations. In order to improve the pre-existing technique, Adaptive Robust Optimization was studied and advances in the field was made to address the problems which could not be easily handled with previous methods.[2] | 4,946 | 16,664 | {"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": 109, "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} | 3.75 | 4 | CC-MAIN-2022-49 | latest | en | 0.781427 |
https://es.mathworks.com/matlabcentral/cody/problems/2593-polite-numbers-n-th-polite-number/solutions/670868 | 1,603,425,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00583.warc.gz | 318,331,596 | 17,309 | Cody
Problem 2593. Polite numbers. N-th polite number.
Solution 670868
Submitted on 17 May 2015 by Ted
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; y_correct = 3; assert(isequal(Nth_polite(x),y_correct))
pol = 3
2 Pass
%% x = 2; y_correct = 5; assert(isequal(Nth_polite(x),y_correct))
pol = 5
3 Pass
%% x = 4; y_correct = 7; assert(isequal(Nth_polite(x),y_correct))
pol = 7
4 Pass
%% x = 5; y_correct = 9; assert(isequal(Nth_polite(x),y_correct))
pol = 9
5 Pass
%% x = 7; y_correct = 11; assert(isequal(Nth_polite(x),y_correct))
pol = 11
6 Pass
%% x = 11; y_correct = 15; assert(isequal(Nth_polite(x),y_correct))
pol = 15
7 Pass
%% x = 12; y_correct = 17; assert(isequal(Nth_polite(x),y_correct))
pol = 17
8 Pass
%% x = 14; y_correct = 19; assert(isequal(Nth_polite(x),y_correct))
pol = 19
9 Pass
%% x = 19; y_correct = 24; assert(isequal(Nth_polite(x),y_correct))
pol = 24
10 Pass
%% x = 21; y_correct = 26; assert(isequal(Nth_polite(x),y_correct))
pol = 26
11 Pass
%% x = 27; y_correct = 33; assert(isequal(Nth_polite(x),y_correct))
pol = 33
12 Pass
%% x = 64; y_correct = 71; assert(isequal(Nth_polite(x),y_correct))
pol = 71
13 Pass
%% x = 1e6; y_correct = x+20; assert(isequal(Nth_polite(x),y_correct))
pol = 1000020
14 Pass
%% x = 1e7; y_correct = x+24; assert(isequal(Nth_polite(x),y_correct))
pol = 10000024
15 Pass
%% x = 999999999; y_correct = x+30; assert(isequal(Nth_polite(x),y_correct))
pol = 1.0000e+09
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Start Hunting! | 618 | 1,730 | {"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} | 3.625 | 4 | CC-MAIN-2020-45 | latest | en | 0.4749 |
https://stats.stackexchange.com/questions/619430/adding-a-new-insignificant-coefficient-to-the-regression-that-is-collinear | 1,721,224,133,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00656.warc.gz | 472,603,274 | 42,476 | # Adding a new insignificant coefficient to the regression, that is collinear
## Problem
This is more of a theoretical question than a practical one...
Let's say I have three Random Variables: $$Y$$, $$X_1$$, and $$X_2$$. With the following properties:
• Cor($$Y$$,$$X_1$$) > 0
• Cor($$Y$$,$$X_2$$) = 0
• Cor($$X_1$$, $$X_2$$) > 0
You run two regressions:
$$Y = \alpha * X_1 + \epsilon$$
and:
$$Y = \beta_1 * X_1 + \beta_2 * X_2 + \epsilon$$
## Approach 1: Dumb One
My gut instinct says the following, if the formula for the coefficient of a simple OLS, $$\alpha$$ is usually:
$$\tag{1} \alpha = \frac{Cov(X_1,Y)}{Var(X_1)}$$
We now have a multiple regression beta when we include $$X_2$$:
$$\tag{2} \hat{\beta}_{X_1+X_2} = \frac{Cov(X_1+X_2,Y)}{Var(X_1+X_2)}$$
Then the numerator doesn't change: $$Cov(X_1+X_2,Y) = Cov(X_1,Y) + Cov(X_2,Y) = Cov(X_1, Y)$$
The denominator changes because the variance now becomes: $$Var(X_1) + Var(X_2) + 2 Cov(X_1,X_2)$$
Which, because $$Cov(X_1, X_2) > 0$$, it means that we have this inequality:
$$Var(X_1) + Var(X_2) + 2 Cov(X_1,X_2) > Var(X_1)$$
Intuitively, $$\beta_1$$ is less stable so it will be larger because of the instability.
## Approach 2: Google It.
I thought I'd check my answer and found a formula for $$\beta_i$$ along the lines of Equation (1) that takes into account the correlation between coefficients. It's from this website (written once in their notation, and another in mine):
$$\tag{3} \beta_1 = \frac{r_{1y} - r_{2y}r_{12}}{1 - r_{12}^2}$$
$$\beta_1 = \frac{Cor(X_1,Y) - Cor(X_2,Y) * Cor(X_1,X_2)}{1-Cor(X_1,X_2)^2}$$
Their notation is cleaner. So I'll stick with using it: $$Cov(X_1,Y) = s_{1y}$$, $$Cor(X_1,Y)=r_{1y}$$, $$SD(X) = s_x$$, and $$Var(X) = s_x^2$$
Now I rewrite Equation (1) in terms of correlations, in their notation:
$$\tag{1b} \alpha = \frac{s_{1y}}{s_x^2} = r_{1y} \frac{s_y}{s_x}$$
So comparing Equation (3) with Equation (1b), we have:
$$\alpha > \beta_1$$
$$r_{1y} \frac{s_y}{s_x} > \frac{r_{1y} - r_{2y}r_{12}}{1 - r_{12}^2}$$
if and only if:
$$\frac{1}{1-r_{12}^2} > \frac{s_y}{s_x}$$
$$\tag{4} 1 - r_{12}^2 < \frac{s_x}{s_y}$$
## Questions
1 - Are Approach 2 and Equation (4) correct? The answer is it depends? I kinda thought it was cut and dry and the answer would either be larger or smaller
2 - I'm not sure that Equation (2) makes sense... what does the coefficient of a sum of two random variables even mean? Am I writing that correctly?
3 - most importantly: how do I derive something like Equation (3)? I don't want to use Google every time I'm stuck on problems like these (this is for personal practice)
## Edit: Approach 3 - Frisch-Waugh-Lovell Theorem
Can't I also use the FWL Theoerem to do:
$$resid_{Y \sim X_2} = Y - \alpha_2 * X_2$$
$$= Y - \frac{Cov(X_2,Y)}{Var(X_2)} * X_2$$ $$= Y - 0$$ $$resid_{Y \sim X_2} = Y$$
And then run the regression of:
$$resid_{Y \sim X_2} \sim X_1 = T \sim X_1$$
...which is just means that $$\beta_1 = \alpha$$?
I'm so lost now, I feel like I've solved the answer in three different ways and they all sound reasonable enough...
• In your second approach you involve the variances $s$ when computing $\alpha$ but not when computing $\beta$ you should end up with $$\alpha = r_{1y} < \frac{r_{1y}}{1 - r_{12}^2} =\beta$$ when you use $s_y = s_{x_1} = 1$ and $r_{2y} = 0$. Commented Jun 22, 2023 at 7:24
• Also useful can be Intuition behind $(X^TX)^{-1}$ in closed form of w in Linear Regression. This you may also relate to FWL theorem which is about the way that the coefficients transform due to correlations between the different $X_i$. (Your application of it in your third approach is not very clear so I can not comment much more on it) Commented Jun 22, 2023 at 7:34
• In your first approach it is not true that the numerator doesn't change. You can have Cov(X_1+X_2,Y) ≠ Cov(X_2, Y) Commented Jun 22, 2023 at 7:39
• My first criticism on your first approach was not correct. Indeed we have $$Cov(X_1+X_2,Y) = Cov(X_1,Y) + Cov(X_2,Y)$$ Commented Jun 22, 2023 at 15:31
• @qdread consider $$X = \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ -1 & 0 \\ -1 & -1 \\ \end{bmatrix} \qquad Y = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0\end{bmatrix}$$ Commented Jun 22, 2023 at 19:18
Consider the matrix equation for finding the coefficients
$$\hat\beta = (X^TX)^{-1} X^T Y$$
If you assume the variables to have variance equal to one and centered at zero then you can express this in terms of the correlations
$$\begin{bmatrix}\hat\beta_1 \\ \hat\beta_2 \end{bmatrix} = \begin{bmatrix} 1 & \rho_{X_1,X_2} \\ \rho_{X_1,X_2} & 1\end{bmatrix}^{-1} \cdot \begin{bmatrix} \rho_{X_1,Y}\\ \rho_{X_2,Y}\end{bmatrix} = \frac{1}{1-\rho_{X_1,X_2}^2}\begin{bmatrix} 1 & -\rho_{X_1,X_2} \\ -\rho_{X_1,X_2} & 1\end{bmatrix} \cdot \begin{bmatrix} \rho_{X_1,Y}\\ \rho_{X_2,Y}\end{bmatrix}$$
And for the coefficient $$\beta_1$$ you get
$$|\hat\beta_1| = \frac{|\rho_{X_1,Y}-\overbrace{\rho_{X_2,Y}}^{\llap{\text{is zero by }}\rlap{\text{assumption}}}\rho_{X_1,X_2}|}{1-\rho_{X_1,X_2}^2} =\frac{|\rho_{X_1,Y}|}{1-\rho_{X_1,X_2}^2} > |\rho_{X_1,Y}| = |\hat\alpha|$$
Example
$$X = \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ -1 & 0 \\ -1 & -1 \\ \end{bmatrix} \qquad Y = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0\end{bmatrix}$$
Will give $$\alpha = 0.5$$ and $$\beta_1 = 1$$
Intuitively: the $$X_1$$ variable correlates with $$Y$$ but also 'brings in some noise'. The $$X_2$$ variable, that doesn't directly correlate with $$Y$$ but does correlate with $$X_1$$ can cancel some of this noise, allowing the $$X_1$$ variable to fit better (and that increases the coefficient).
Say that we have two uncorrelated variables $$Z$$, $$Y$$.
• Then $$X_1 = Y$$ would fit the model with a coefficient equal to one.
• But $$X_1 = Y + Z$$ will be related with a coefficient less than one.
By adding a variable in the regression $$X_2 = Z$$ we can cancel some of the noise that 'prevents' $$X_1$$ form being fitted with a coefficient equal to one.
• How do you get the matrices: $\begin{bmatrix} 1 & \rho_{X_1,X_2} \\ \rho_{X_1,X_2} & 1\end{bmatrix}^{-1} \cdot \begin{bmatrix} \rho_{X_1,Y}\\ \rho_{X_2,Y}\end{bmatrix}$ in your solution? You lost me at that jump from the beta vector to the two matrices
– Joe
Commented Jun 22, 2023 at 15:52
• @JoeVictor you get those matrices for standardized $X$ and $Y$. Such that $X^TX$ is the correlation matrix. In my example the variables are not standardized. Commented Jun 22, 2023 at 15:55
• if the variables are not standardized, wouldn't that change the answer?
– Joe
Commented Jun 22, 2023 at 15:58
• @JoeVictor not when they remain centered, in that case the correlation and covariance are just a difference in the scale of the variables. But for a shift the result is different. Commented Jun 22, 2023 at 16:07
• @JoeVictor an example is when we add to both the vectors $X_2$ and $Y$ some fixed value, say 10. Then the covariance/correlation is not influenced, but the product $X_2 \cdot Y$ changes a lot. Commented Jun 22, 2023 at 16:31 | 2,489 | 7,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 59, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.754306 |
https://www.ic.unicamp.br/index.php/node/1196 | 1,547,926,444,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583680452.20/warc/CC-MAIN-20190119180834-20190119202834-00514.warc.gz | 830,054,579 | 11,269 | # Genome Matrices and the Median Problem
Data
Add to Calendin 2017-11-10 00:00:00 2017-11-10 00:00:00 Genome Matrices and the Median Problem The genome median problem is an important problem in phylogenetic reconstruction under rearrangement models. It can be stated as follows: Given three genomes, find a fourth that minimizes the sum of the pairwise rearrangement distances between it and the three input genomes. In this talk, we model genomes as matrices and study the matrix median problem using the rank distance. We present the first polynomial-time algorithms to compute medians with respect to a sophisticated distance. In addition to the approximation algorithm, we suggest heuristics to get a genome from an arbitrary square matrix. This is useful to translate the results of our median algorithms back to genomes, and it shows good results in our tests. To assess the relevance of our approach in the biological context, we ran simulated evolution tests and compared our to those of an exact DCJ median solver. The results show that our method is capable of solving practical problems of a thousand genes within seconds. Auditório do IC Auditório do IC Auditório do IC America/Sao_Paulo public
Horário
14:00
Local
Auditório do IC
Palestrante
Prof. João Meidanis
Descrição
The genome median problem is an important problem in phylogenetic reconstruction under rearrangement models. It can be stated as follows: Given three genomes, find a fourth that minimizes the sum of the pairwise rearrangement distances between it and the three input genomes. In this talk, we model genomes as matrices and study the matrix median problem using the rank distance. We present the first polynomial-time algorithms
to compute medians with respect to a sophisticated distance. In addition to the approximation algorithm, we suggest heuristics to get a genome from an arbitrary square matrix. This is useful to translate the results of our median algorithms back to genomes, and it shows good results in our tests. To assess the relevance of our approach in the biological context, we ran simulated evolution tests and compared our to those of an exact DCJ median solver. The results show that our method is capable of solving practical problems of a thousand genes within seconds.
Sobre o Palestrante
João Meidanis possui graduação em Bacharelado em Matemática pela Universidade de São Paulo (1980), mestrado em Ciência da Computação - University of Wisconsin - Madison (1989), mestrado em Matemática pela Universidade de São Paulo (1985) e doutorado em Ciência da Computação - University of Wisconsin - Madison (1992). Atualmente é diretor da Scylla Informática e professor titular da Universidade Estadual de Campinas. Suas áreas de atuação incluem análise de algoritmos, teoria dos grafos, biologia computacional e otimização. | 619 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-04 | latest | en | 0.855753 |
http://davin.50webs.com/J.T.W/tutorial-04-DoForWhile.html | 1,500,554,414,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423183.57/warc/CC-MAIN-20170720121902-20170720141902-00229.warc.gz | 95,878,951 | 4,420 | davin.50webs.com/J.T.W a GNU world order – your home of everything that is free
Main Menu Research Projects Photo Album Curriculum Vitae The Greatest Artists Email Address Computer Games Web Design Java Training Wheels The Fly (A Story) Political Activism Bob Dylan Quotes+ My Life Story Smoking Cessation Other Links Tutorial 1 Tutorial 2 Tutorial 3 Tutorial 4 Tutorial 5 Tutorial 6 Tutorial 7 Tutorial 8 Tutorial 9 Tutorial 10 Tutorial 11 Tutorial 12 Tutorial 13 Tutorial 14 Tutorial 15 Tutorial 16 Tutorial 17 Tutorial 18 Using Emacs Download Links
# J.T.W. tutorial 4: Four kinds of loops
## § 4 Tutorial 4
Study the following code:
class LoopTest
begin
function int powerOf2A(int n)
begin
var int counter = n;
var int result = 1;
while (counter != 0)
begin
result = 2 * result;
counter = counter - 1;
end
return result;
end
function int powerOf2B(int n)
begin
var int counter = n;
var int result = 1;
do
begin
result = 2 * result;
counter = counter - 1;
end while (counter != 0);
return result;
end
function int powerOf2C(int n)
begin
var int result = 1;
for (var int counter = n; counter != 0; counter = counter - 1)
begin
result = 2 * result;
end
return result;
end
function int powerOf2D(int n)
begin
var int result = 1;
superfor (var int counter = n downto 1)
begin
result = 2 * result;
end
return result;
end
/**
* Prints a row of stars of a given length.
*/
function void printLineC(int length)
begin
for (var int i = 0; i<length; i=i+1)
begin
System.out.print("#");
end
System.out.println();
end
beginMain
// For question 4.1 add some code here...
endMain
end
Question 4.2: To the main function add some code to call the functions powerOf2A, powerOf2B, powerOf2C and powerOf2D to verify that they all return the same result. To inspect the result you will need to apply the System.out.println() statement to the values returned by those functions.
Question 4.3: There is a bug in the powerOf2B method because it does not behave correctly in the case when n is zero. Put an if statement at the top of this method to make it handle the case of zero properly.
Question 4.4: By copying the pattern of powerOf2A, powerOf2B, powerOf2C and powerOf2D, write methods printLineA, printLineB and printLineD that work identically to the method printLineC, except that they use while loops, do loops and superfor loops, respectively. Add some code to the main function to test them out.
Question 4.5: Based on the previous three questions, is there a best looping construct? Or does it depend on what the looping construct is going to be used for?
Back to J.T.W | 690 | 2,578 | {"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-2017-30 | longest | en | 0.551279 |
https://scicomp.stackexchange.com/questions/8639/recursive-multiply-analysis | 1,627,261,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00664.warc.gz | 513,507,608 | 32,167 | # Recursive-Multiply Analysis [closed]
I am not sure if I can post these kinds of questions on this site, but I was wondering if you could help me with a question that's been bothering me.
We have started learning about analysis of recursive algorithms and I got the gist of it. However there are some questions, like the one I'm going to post, that confuse me a little. Can you please help me?
Question:
Consider the problem of multiplying two big integers, i.e. integers represented by a large number of bits that cannot be handled directly by the ALU of a single CPU. This type of multiplication has applications in data security where big integers are used in encryption schemes. The elementary-school algorithm for multiplying two n-bit integers has a complexity of . To improve this complexity, let x and y be the two n-bit integers, and use the following algorithm
Recursive-Multiply(x,y)
Write x = x1 * 2^(n/2)+x0 //x1 and x0 are high order and low order n/2 bits
y = y1 * 2^(n/2)+y0//y1 and y0 are high order and low order n/2 bits
Compute x1+x0 and y1+y0
p = Recursive-Multiply (x1+x0,y1+y0)
x1y1 = Recursive-Multiply (x1,y1)
x0y0 = Recursive-Multiply (x0,y0)
Return x1y1*2^n + (p-x1y1-x0y0)*2^(n/2)+x0y0
(a)Explain how the above algorithm works and provides the correct answer.
(b)Write a recurrence relation for the number of basic operations for the above algorithm.
(c)Solve the recurrence relation and show that its complexity is O(n^lg3)
• This is the Karatsuba algorithm for integer multiplication. Thematically your Question might be better suited for the Theoretical Computer Science site, but I see related Questions posed at crypto.SE and math.SE. See specifically this Question. – hardmath Sep 30 '13 at 5:46
• That said, you should explain what you've tried so far to solve this problem, and narrow it down from simply repeating what looks like a course exercise to the specific issue which you need help with. Just putting homework out and hoping for someone to do it for you will provoke feelings of resentment. – hardmath Sep 30 '13 at 6:09
• @hardmath: I'd recommend Computer Science Stack Exchange instead of Theoretical Computer Science, because Karatsuba is a topic that could be covered at the undergraduate level. (For instance, MIT covers it in their first undergraduate course on algorithms.) For recurrence relations, Cormen, Leiserson, Rivest and Stein (also known as CLRS) is one classic reference. – Geoff Oxberry Sep 30 '13 at 17:35 | 618 | 2,486 | {"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-2021-31 | latest | en | 0.927282 |
https://www.convertunits.com/from/nanometer/to/light-week | 1,620,414,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00185.warc.gz | 740,128,187 | 16,528 | ## ››Convert nanometre to light week
nanometer light-week
How many nanometer in 1 light-week? The answer is 1.813144785984E+23.
We assume you are converting between nanometre and light week.
You can view more details on each measurement unit:
nanometer or light-week
The SI base unit for length is the metre.
1 metre is equal to 1000000000 nanometer, or 5.5152793518213E-15 light-week.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between nanometres and light weeks.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from light-week to nanometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Nanometre
The SI prefix "nano" represents a factor of 10-9, or in exponential notation, 1E-9.
So 1 nanometre = 10-9 metre.
## ››Definition: Light-week
A light-week (also written light week) is a unit of length. It is defined as the distance light travels in an absolute vacuum in one week (seven days of 86,400 seconds each) or 181,314,478,598,400 metres (~181 Tm).
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 418 | 1,627 | {"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} | 2.625 | 3 | CC-MAIN-2021-21 | longest | en | 0.861921 |
https://listserv.gmu.edu/cgi-bin/wa?A3=ind1103&L=ECJ-INTEREST-L&E=0&P=16272&B=--&T=text%2Fplain;%20charset=us-ascii&header=1 | 1,721,025,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00690.warc.gz | 347,557,502 | 7,715 | ```This is a very good idea, but it may make the whole algorithm hefty.
Best Regards,
----- Original Message ----
Sent: Wed, March 9, 2011 6:30:27 PM
Subject: Re: Help on a general GA issue
A while back, i was using a GA to evolve logic puzzles, and it created many
invalid puzzles. But I didn't exclude invalid puzzles from the search, because
I was afraid that they would include good building blocks for valid puzzles.
So, my algorithm pulled error counts into the fitness function as a way to guide
the invalid puzzles towards valid ones.
I wonder if there is a good way to determine if a search is better or worse off
by doing this.
-Dave
On Mar 9, 2011, at 11:56 AM, Robert Baruch wrote:
> I, too, am interested in the answer to this question.
>
> In my own work, I've modified the crossover or mutation algorithm to explicitly
>generate valid individuals. For example, if a given element in a GA individual
>must be within a certain range, I could clip the value to the upper or lower
>limit if the value goes out of range.
>
> GP has retries when it generates individuals that are too big -- it just tries
>again. If it can't generate a valid individual after a certain number of
>retries, it gives up and copies the parent.
>
> On the other hand, sometimes you don't know that an individual is valid until
>you evaluate it. For example, perhaps an individual based on code will throw an
>exception. Then you just have to score that individual as very poor.
>
> --Rob
>
> On Mar 9, 2011, at 11:33 AM, Paul Fisher wrote:
>
>> Hello everyone
>>
>> This is a plea for help on a general point regarding the genetic algorithm
>>method (rather than a technical ECJ issue). I hope my question makes sense to
>>someone who can point me in the right direction.
>>
>> Simply put, if you have a large solution space (c.9000 element matrix) with a
>>set of constraints that make a large proportion of the possible solutions
>>invalid, how do you treat invalid solutions generated by the reproduction
>>process to ensure the population is composed of only (or mostly) valid
>>solutions? For example, what happens when you take two good solutions from the
>>initial population, cross them over and mutate them according to some standard
>>method, and the result is two solutions which happen to violate the constraints
>>of the solution space and therefore render the new solutions invalid? How can
>>you take two good solutions and mate them in such a way to produce only valid
>>solutions according to the problem constraints?
>>
>> The only way I have known how to treat invalid solutions so far is to tolerate
>>them in the population but score them out of the selection process. The problem
>>with this is that 9 times out of 10 the population will be swamped by these duds
>>and never get going.
>> | 662 | 2,810 | {"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-2024-30 | latest | en | 0.926402 |
http://mathoverflow.net/revisions/39733/list | 1,369,454,945,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705502703/warc/CC-MAIN-20130516115822-00000-ip-10-60-113-184.ec2.internal.warc.gz | 174,421,224 | 4,907 | 2 corrected the condition k >deg(P+Q) to k > max(deg(P),deg(Q))
Let t be a positive real number. Let P(x) and Q(x) be two random polynomials with integer coefficients. If P(t) = Q(t), then what is the probability that P(x) is not identical to Q(x)?
Will it make a difference if I restrict t to be an integer?
Suppose I had a set T ={t0,t1,…tk}, can we answer a similar question --- If P(ti) = Q(ti) for all ti in the set T, what is the probability that P is identical to Q? If k > deg(P +Q), max(deg(P), deg(Q)), the probability is 1. But can we say something about how many points we need to check before we can be fairly certain that the polynomials are identical?
Thanks
1
# How to tell if two random polynomials are identical
Let t be a positive real number. Let P(x) and Q(x) be two random polynomials with integer coefficients. If P(t) = Q(t), then what is the probability that P(x) is not identical to Q(x)?
Will it make a difference if I restrict t to be an integer?
Suppose I had a set T ={t0,t1,…tk}, can we answer a similar question --- If P(ti) = Q(ti) for all ti in the set T, what is the probability that P is identical to Q? If k > deg(P +Q), the probability is 1. But can we say something about how many points we need to check before we can be fairly certain that the polynomials are identical?
Thanks | 350 | 1,328 | {"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} | 3.640625 | 4 | CC-MAIN-2013-20 | latest | en | 0.909348 |
http://www.edupil.com/question/salim-rahim-karim-started-business-partnership-amount-rs-2000-rs-2500-rs-3000/ | 1,495,681,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607963.70/warc/CC-MAIN-20170525025250-20170525045250-00010.warc.gz | 501,179,744 | 18,032 | # Calculate the Share of Salim in profit :
Salim, Rahim and Karim started a business in partnership with amount of Rs. 2,000, Rs.2,500 and Rs.3,000 respectively. They earned profit of Rs.6,000 in the end of year. Calculate the Share of Salim in profit.
1. Rs.2,000
2. Rs.1,600
3. Rs.2,800
4. Rs.2,400
Anurag Mishra Professor Asked on 14th December 2015 in
Explanation:-
Salim. Rahim and Karim started a business in partnership with amount of Rs. 2, 000, Rs. 2, 500 and Rs. 3, 000 respectively.
Then, the ratio of the share in the business = 2, 000: 2, 500: 3, 000
= 4: 5: 6
Total earned profit = Rs. 6, 000
Then, the share of Salim in profit = 6, 000 x 4/15
= 400 x 4
= Rs. 1, 600
Hence, the answer is (2) Rs. 1, 600.
Anurag Mishra Professor Answered on 15th December 2015. | 286 | 781 | {"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} | 3.96875 | 4 | CC-MAIN-2017-22 | longest | en | 0.944052 |
http://www.slideshare.net/HeenaModi/subtraction-problem | 1,475,255,461,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662321.82/warc/CC-MAIN-20160924173742-00251-ip-10-143-35-109.ec2.internal.warc.gz | 735,235,754 | 32,905 | Upcoming SlideShare
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# Subtraction problem
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### Subtraction problem
1. 1. Can you remember the four steps?
2. 2. 1. Understand the problem 2. Plan how to solve the problem 3. Carry out the plan 4. Look Back
3. 3. On the 3rd of December 532 people went to see the new Harry Potter movie. Half way through the movie 67 people left because they didn’t like it. There were 213 kids in the audience. How many people saw the whole film?
4. 4. 1. Understand the problem On the 3rd of December 532 people went to see the new Harry Potter movie. Half way through the movie 67 people left because they didn’t like it. There were 213 kids in the audience. How many people saw the whole film? What information don’t we need? What is the question? What is the important information?
5. 5. 2. Plan how to solve the problem What strategy will we use? Solving an equation? Subtraction - 3. Carry out the plan
6. 6. 532 - 67 2-7 = can’t do 12-7 = 5 5 2 12 20-60 = can’t do 120-60 = 60 4 64 400-0 = 400 1
7. 7. 4. Look Back Is our Answer reasonable? YES Write your answer in a sentence. 465 people saw the whole Harry Potter Movie. | 428 | 1,426 | {"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-2016-40 | latest | en | 0.909506 |
https://sciencenotes.org/error-in-science/ | 1,713,525,955,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00322.warc.gz | 458,772,006 | 43,039 | # Sources of Error in Science Experiments
Science labs usually ask you to compare your results against theoretical or known values. This helps you evaluate your results and compare them against other people’s values. The difference between your results and the expected or theoretical results is called error. The amount of error that is acceptable depends on the experiment, but a margin of error of 10% is generally considered acceptable. If there is a large margin of error, you’ll be asked to go over your procedure and identify any mistakes you may have made or places where error might have been introduced. So, you need to know the different types and sources of error and how to calculate them.
### How to Calculate Absolute Error
One method of measuring error is by calculating absolute error, which is also called absolute uncertainty. This measure of accuracy is reported using the units of measurement. Absolute error is simply the difference between the measured value and either the true value or the average value of the data.
absolute error = measured value – true value
For example, if you measure gravity to be 9.6 m/s2 and the true value is 9.8 m/s2, then the absolute error of the measurement is 0.2 m/s2. You could report the error with a sign, so the absolute error in this example could be -0.2 m/s2.
If you measure the length of a sample three times and get 1.1 cm, 1.5 cm, and 1.3 cm, then the absolute error is +/- 0.2 cm or you would say the length of the sample is 1.3 cm (the average) +/- 0.2 cm.
Some people consider absolute error to be a measure of how accurate your measuring instrument is. If you are using a ruler that reports length to the nearest millimeter, you might say the absolute error of any measurement taken with that ruler is to the nearest 1 mm or (if you feel confident you can see between one mark and the next) to the nearest 0.5 mm.
### How to Calculate Relative Error
Relative error is based on the absolute error value. It compares how large the error is to the magnitude of the measurement. So, an error of 0.1 kg might be insignificant when weighing a person, but pretty terrible when weighing a apple. Relative error is a fraction, decimal value, or percent.
Relative Error = Absolute Error / Total Value
For example, if your speedometer says you are going 55 mph, when you’re really going 58 mph, the absolute error is 3 mph / 58 mph or 0.05, which you could multiple by 100% to give 5%. Relative error may be reported with a sign. In this case, the speedometer is off by -5% because the recorded value is lower than the true value.
Because the absolute error definition is ambiguous, most lab reports ask for percent error or percent difference.
### How to Calculate Percent Error
The most common error calculation is percent error, which is used when comparing your results against a known, theoretical, or accepted value. As you probably guess from the name, percent error is expressed as a percentage. It is the absolute (no negative sign) difference between your value and the accepted value, divided by the accepted value, multiplied by 100% to give the percent:
% error = [accepted – experimental ] / accepted x 100%
### How to Calculate Percent Difference
Another common error calculation is called percent difference. It is used when you are comparing one experimental result to another. In this case, no result is necessarily better than another, so the percent difference is the absolute value (no negative sign) of the difference between the values, divided by the average of the two numbers, multiplied by 100% to give a percentage:
% difference = [experimental value – other value] / average x 100%
### Sources and Types of Error
Every experimental measurement, no matter how carefully you take it, contains some amount of uncertainty or error. You are measuring against a standard, using an instrument that can never perfectly duplicate the standard, plus you’re human, so you might introduce errors based on your technique. The three main categories of errors are systematic errors, random errors, and personal errors. Here’s what these types of errors are and common examples.
### Systematic Errors
Systematic error affects all the measurements you take. All of these errors will be in the same direction (greater than or less than the true value) and you can’t compensate for them by taking additional data.
Examples of Systematic Errors
• If you forget to calibrate a balance or you’re off a bit in the calibration, all mass measurements will be high/low by the same amount. Some instruments require periodic calibration throughout the course of an experiment, so it’s good to make a note in your lab notebook to see whether the calibrations appears to have affected the data.
• Another example is measuring volume by reading a meniscus (parallax). You likely read a meniscus exactly the same way each time, but it’s never perfectly correct. Another person taking the reading may take the same reading, but view the meniscus from a different angle, thus getting a different result. Parallax can occur in other types of optical measurements, such as those taken with a microscope or telescope.
• Instrument drift is a common source of error when using electronic instruments. As the instruments warm up, the measurements may change. Other common systematic errors include hysteresis or lag time, either relating to instrument response to a change in conditions or relating to fluctuations in an instrument that hasn’t reached equilibrium. Note some of these systematic errors are progressive, so data becomes better (or worse) over time, so it’s hard to compare data points taken at the beginning of an experiment with those taken at the end. This is why it’s a good idea to record data sequentially, so you can spot gradual trends if they occur. This is also why it’s good to take data starting with different specimens each time (if applicable), rather than always following the same sequence.
• Not accounting for a variable that turns out to be important is usually a systematic error, although it could be a random error or a confounding variable. If you find an influencing factor, it’s worth noting in a report and may lead to further experimentation after isolating and controlling this variable.
### Random Errors
Random errors are due to fluctuations in the experimental or measurement conditions. Usually these errors are small. Taking more data tends to reduce the effect of random errors.
Examples of Random Errors
• If your experiment requires stable conditions, but a large group of people stomp through the room during one data set, random error will be introduced. Drafts, temperature changes, light/dark differences, and electrical or magnetic noise are all examples of environmental factors that can introduce random errors.
• Physical errors may also occur, since a sample is never completely homogeneous. For this reason, it’s best to test using different locations of a sample or take multiple measurements to reduce the amount of error.
• Instrument resolution is also considered a type of random error because the measurement is equally likely higher or lower than the true value. An example of a resolution error is taking volume measurements with a beaker as opposed to a graduated cylinder. The beaker will have a greater amount of error than the cylinder.
• Incomplete definition can be a systematic or random error, depending on the circumstances. What incomplete definition means is that it can be hard for two people to define the point at which the measurement is complete. For example, if you’re measuring length with an elastic string, you’ll need to decide with your peers when the string is tight enough without stretching it. During a titration, if you’re looking for a color change, it can be hard to tell when it actually occurs.
### Personal Errors
When writing a lab report, you shouldn’t cite “human error” as a source of error. Rather, you should attempt to identify a specific mistake or problem. One common personal error is going into an experiment with a bias about whether a hypothesis will be supported or rejects. Another common personal error is lack of experience with a piece of equipment, where your measurements may become more accurate and reliable after you know what you’re doing. Another type of personal error is a simple mistake, where you might have used an incorrect quantity of a chemical, timed an experiment inconsistently, or skipped a step in a protocol. | 1,731 | 8,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-18 | latest | en | 0.930751 |
https://www.wyzant.com/resources/answers/topics/pi?page=1 | 1,606,826,012,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00712.warc.gz | 886,172,531 | 14,067 | 35 Answered Questions for the topic Pi
08/14/19
#### What is the area A of the cross section of the column?
A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18π meters.Give your answer in terms of pi
06/03/19
#### What is the fastest way to get the value of π?
I'm looking for the fastest way to obtain the value of π, as a personal challenge. More specifically, I'm using ways that don't involve using #define constants like M_PI, or hard-coding the... more
04/18/19
#### Math help please!
shown below, .Prove .
04/18/19
#### Help,Please! Extra math help
shown below has diameter which intersects at point and the measurements shown. Calculate the following measures:
04/14/19
#### Help, Please! Math ...
 is an equilateral triangle with side length 23 in. What is the area of the shaded segment? Round your answer to the nearest hundreth.
04/14/19
What is ?
03/13/19
#### Why are turns not used as the default angle measure?
Why is $2\\pi$ radians not replaced by $1$ turn in formulas? The majority of them would be simpler. If such a replacement was proposed earlier, why was it declined?
Pi Length
10/11/18
#### help me please i cant figure this out
A large pizza at Jake's Pizza has a radius of 7 inches. Find the length of the crust around the outside of the pizza. Use 3.14 for pi and round your answer to the nearest tenth. Please label your... more
08/28/18
#### How do I calculate pi starting with number close to pi like 2.57 or 3.11 or any random number as long as it is close to pi?
I know that there was a way to calculate, but i forgot it. Can you please help me.
03/19/18
#### Pi Day Problem
If the top surface of a particular slice of pie is a sector of a circle with a vertex angle measuring 20 degrees and the length of the arc is 3.14π units, what is the perimeter(left in terms of π)... more
Pi Algebra
03/14/18
#### How do you find the area of 2 inches
I need to figure out how to find the area of 2 in for a circle.
11/27/17
#### sin theta = 1/4 find theta+pi/6 in quadrant 2
i think the question said find exact solution?
Pi
08/17/17
#### Is it true that it is impossible to divide a circle in half?
Given that pi cannot be divided in half, does it follow that circle can not be divided in half? If so, what implications does this have?
05/25/17
#### Solve the following in the interval of [0,2π)
1. 4 cos x * 2 √ 3 = 0 2. 2sin^2 x - sin x = 1 3. cos^2 x = cos x
02/20/17
#### Solve the equation: 2 sin(3θ)- √3=0, 0 ≤ θ< 2π.
Solve the equation 2 sin(3θ)- √3=0, 0 ≤ θ< 2π
02/12/17
#### find the exact value?
sin 450°, cos(-780°), and tan (37π/6)
Pi
01/27/17
#### Rotation of a cirle with the diameter of 34"
The diameter of the rear tire of this bike is 34". In low gear, you need to rotate the pedals 3x to make the rear tire rotate 360 degrees. How far will you travel in low gear each time you rotate... more
09/05/16
#### The surface area of a sphere is 46(pi) cm². what is the radius of the sphere?
This is an algebra 2 question, and I don't know how to approach it.
Pi
04/25/16
#### The circumference of a circle is 282.54282.54 m. What is the approximate area of the? circle? Use 3.14 for p.
i need to know the answer for this problem
Pi
03/30/16
#### A circle has a diameter of 40 inches. What is the circumference? Write your answer using pi
I know the answer is 125.6, but I don't have an idea of how to write it using pi.
Pi
03/30/16
#### What is the circumference with the radius 3? Write your answer using pi.
I know that the circumference is 125.6, but how do I write it using pi?
1
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 1,095 | 3,891 | {"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} | 3.265625 | 3 | CC-MAIN-2020-50 | latest | en | 0.901774 |
http://forums.wolfram.com/mathgroup/archive/2006/Aug/msg00121.html | 1,526,817,895,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00451.warc.gz | 115,181,523 | 9,321 | Re: Re: Re: Finding the Number of Pythagorean Triples below a bound
• To: mathgroup at smc.vnet.net
• Subject: [mg68415] Re: [mg68382] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 4 Aug 2006 03:59:29 -0400 (EDT)
• References: <eaeqa3\$53v\$1@smc.vnet.net><200607300848.EAA25171@smc.vnet.net> <eakfgm\$rl6\$1@smc.vnet.net> <200608020923.FAA28520@smc.vnet.net> <79C36C70-E091-4A82-8EC5-0EDC743D081D@mimuw.edu.pl> <200608031007.GAA15743@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```On 3 Aug 2006, at 12:07, Andrzej Kozlowski wrote:
> On 2 Aug 2006, at 20:01, Andrzej Kozlowski wrote:
>
>>
>> On 2 Aug 2006, at 11:23, titus_piezas at yahoo.com wrote:
>>
>>> Hello all,
>>>
>>> My thanks to Peter and Andrzej, as well as those who privately
>>> emailed
>>> me.
>>>
>>> To recall, the problem was counting the number of solutions to a
>>> bivariate polynomial equal to a square,
>>>
>>> Poly(a,b) = c^2
>>>
>>> One form that interested me was the Pythagorean-like equation:
>>>
>>> a^2 + b^2 = c^2 + k
>>>
>>> for {a,b} a positive integer, 0<a<=b, and k any small integer. I was
>>> wondering about the density of solutions to this since I knew in the
>>> special case of k=0, let S(N) be the number of primitive solutions
>>> with
>>> c < N, then S(N)/N = 1/(2pi) as N -> inf.
>>>
>>> For k a squarefree integer, it is convenient that any solution is
>>> also
>>> primitive. I used a simple code that allowed me to find S(10^m) with
>>> m=1,2,3 for small values of k (for m=4 took my code more than 30
>>> mins
>>> so I aborted it). The data is given below:
>>>
>>> Note: Values are total S(N) for *both* k & -k:
>>>
>>> k = 2
>>> S(N) = 4, 30, 283
>>>
>>> k = 3
>>> S(N) = 3, 41, 410
>>>
>>> k = 5
>>> S(N) = 3, 43, 426
>>>
>>> k = 6
>>> S(N) = 3, 36, 351
>>>
>>> Question: Does S(N)/N for these also converge? For example, for the
>>> particular case of k = -6, we have
>>>
>>> S(N) = 2, 20, 202
>>>
>>> which looks suspiciously like the ratio might be converging.
>>>
>>> Anybody know of a code for this that can find m=4,5,6 in a
>>> reasonable
>>> amount of time?
>>>
>>>
>>> Yours,
>>>
>>> Titus
>>>
>>
>>
>> Here is a piece code which utilises the ideas I have described in
>> my previous posts:
>>
>> ls = Prime /@ Range[3, 10];
>>
>> test[n_] :=
>> Not[MemberQ[JacobiSymbol[n, ls], -1]] && Element[Sqrt[n],
>> Integers]
>>
>> f[P_, k_] := Sum[If[(w =
>> a^2 + b^2 - k) < P^2 && test[w], 1, 0], {a, 1, P}, {b, a,
>> Floor[Sqrt[P^2 - a^2]]}]
>>
>> g[m_, k_] := f[10^m, k] + f[10^m, -k]
>>
>> We can easily confirm the results of your computations,.e.g. for k=2.
>>
>>
>> Table[g[i,2],{i,3}]
>>
>>
>> {4,30,283}
>>
>>
>> Since you have not revealed the "simple code" you have used it is
>> hard to tell if the above one is any better. It is however,
>> certainly capable of solving the problem for m=4:
>>
>>
>> g[4,2]//Timing
>>
>>
>> {4779.39 Second,2763}
>>
>> The time it took on my 1 Gigahertz PowerBook was over 70 minutes,
>> which is longer than you thought "reasonable", so I am still not
>> sure if this is any improvement on what you already have. The time
>> complexity of this algorithm seems somewhat larger than exponential
>> so I would expect that it will take about 6 hours to deal with n=5
>> on my computer, and perhaps 2 weeks to deal with n=6.
>>
>> Andrzej Kozlowski
>>
>>
>>
>
>
> I mistakenly copied and pasted a wrong (earlier) definition of f.
> Here is the correct one:
>
> f[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
> 1 && (w = a^2 +
> b^2 - k) < P^2 && test[w], 1, 0], {a, 1,
> P}, {b, a, Floor[Sqrt[P^2 - a^2]]}]]
>
> The definition of g is as before:
>
> g[m_, k_] := f[10^m, k] + f[10^m, -k]
>
> Andrzej Kozlowski
>
Below is a faster version of the above code. (It owes a significant
improvement to Daniel Lichtblau, which I borrowed from him without
his knowledge ;-))
test[n_] :=
JacobiSymbol[n,
Prime[Random[Integer, {2, 20}]]] -1 && Element[Sqrt[n],
Integers]
f[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
1 && (w = a^2 +
b^2 - k) < P^2 && test[w], 1, 0], {a, 1,
Floor[Sqrt[(P^2+k)/2]]}, {b, a, Floor[Sqrt[P^2 - a^2+k]]}]]
g[m_, k_] := f[10^m, k] + f[10^m, -k]
The improvement is in the upper bound on a in the sum. Since a is the
smaller of the two squares whose sum is equal to P^2+k it can't be
larger than Floor[Sqrt[(P^2+k)/2]].
Note that you can improve the performance by loosing some accuracy if
you use a cruder test for a perfect square:
test1[n_] := With[{w = Sqrt[N[n]]}, w == Round[w]
]
f1[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
1 && (w = a^2 +
b^2 - k) < P^2 && test1[w], 1, 0], {a, 1,
Floor[Sqrt[(P^2+k)/2]]}, {b, a, Floor[Sqrt[P^2 - a^2+k]]}]]
g1[m_, k_] := f1[10^m, k] + f1[10^m, -k]
Let's compare the two cases.
In[7]:=
g[3,2]//Timing
Out[7]=
{89.554 Second,283}
In[8]:=
g1[3,2]//Timing
Out[8]=
{37.376 Second,283}
So we see that we get the same answer and the second approach is
considerably faster. However:
In[9]:=
g[1,6]//Timing
Out[9]=
{0.008863 Second,3}
In[10]:=
g1[1,6]//Timing
Out[10]=
{0.005429 Second,5}
The correct answer is 3 (returned by the first method). The faster
method found two false solutions. This should not matter if you are
interested only in approximate answers (as you seem to be) but it is
worth keeping in mind.
Andrzej Kozlowski
```
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By Harold M. Edwards
This ebook is an creation to algebraic quantity conception through the recognized challenge of "Fermat's final Theorem. The exposition follows the ancient improvement of the matter, starting with the paintings of Fermat and finishing with Kummer's conception of "ideal" factorization, via which the theory is proved for all top exponents lower than 37. The extra effortless subject matters, comparable to Euler's evidence of the impossibilty of x+y=z, are handled in an easy means, and new suggestions and strategies are brought merely after having been inspired by means of particular difficulties. The e-book additionally covers intimately the applying of Kummer's perfect thought to quadratic integers and relates this thought to Gauss' conception of binary quadratic types, a fascinating and demanding connection that's not explored in the other e-book.
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The most function of those lectures is first to in brief survey the elemental con nection among the illustration idea of the symmetric crew Sn and the speculation of symmetric services and moment to teach how combinatorial equipment that come up evidently within the concept of symmetric features bring about effective algorithms to specific a variety of prod ucts of representations of Sn by way of sums of irreducible representations.
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Sample text
In Section 3, we prove the theorem of Segre. Essentially the same argument, with minor modifications to be made afterwards, proves the following stronger theorem of Manin (1966). 1 For an arbitrary normal projective variety the Picard number is defined as the rank of the N´eron–Severi group, the group of Cartier divisors up to numerical equivalence. For varieties with h 1 (X, O X ) = 0 the two definitions coincide. 2. Two smooth cubic surfaces defined over a perfect field, each of Picard number one, are birationally equivalent if and only if they are projectively equivalent.
For arbitrary X , the spaces (X, ⊗m X ) are usually hard to compute because the ⊗m X have quite high rank. Therefore it is important to have similar criteria which involve line bundles only. The natural candidate is the canonical bundle ω X = ∧n X of highest degree K¨ahler differential forms, which is always defined over the fixed ground field. For smooth X , the canonical bundle is represented by a divisor K X defined over the given ground field, and it is convenient to denote it by O X (K X ).
Prove that X is rational. 7 Numerical criteria for nonrationality Rationality and unirationality force strong numerical constraints on a variety. Let X = X/k be the sheaf of regular differential forms (K¨ahler differentials) on a variety X over k. 52. If a smooth projective variety X is rational, then it has no nontrivial global K¨ahler one-forms. In fact, the space of global sections ⊗m (X, ⊗m X ) of the sheaf X is zero for all m ≥ 1. The same holds for unirational X , provided the ground field has characteristic zero. | 684 | 3,195 | {"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} | 2.578125 | 3 | CC-MAIN-2018-51 | longest | en | 0.908109 |
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A square shape photo with a side length of 20 cm is framed by a 4 cm wide bar. Find the external frame size of this photo (the frame is from all sides)
Result
b = 28 cm
#### Solution:
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2 chickens give 2 eggs in 2 days. How many eggs can give 8 chickens for 8 days? | 718 | 2,798 | {"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-2019-26 | latest | en | 0.932095 |
https://edvoy.com/exams/ielts-score-guide/ | 1,726,468,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00770.warc.gz | 194,996,565 | 56,910 | # Your quick guide to IELTS scores
Updated on: Jul 15, 2024
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Universities in most top study destinations like the UK, Canada, Ireland and the US have English as their official medium of instruction. So, they want to make sure that international students like you will be able to understand and communicate in English.
These universities therefore use Standard English Language tests to assess their applicants' language skills. And IELTS is one of the most commonly used English Language Assessment tests.
Sign up now to achieve a Score of 6 or higher with our free IELTS Masterclass!
Now, you need to take the IELTS, but how can you prepare well and ensure a good score?
Entrance test coaching experts say that you are more likely to prepare well and get a good score if you completely understand the test's structure and scoring system, so we have come up with this detailed IELTS marking guide.
## 1. What are the IELTS overall band score and the individual sectional band scores?
When you get your IELTS result, you will get 5 different band scores, one for each section (listening, reading, writing, and speaking), and one overall band score. The overall band score is the average of all the sectional band scores.
All band scores will lie between 0 and 9, with each score increasing by 0.5. For example, you can get a 6.5 or an 8 or any such number based on how you performed.
So, how can you calculate your IELTS overall band score?
Simply add all your sectional band scores and divide the sum by 4.
If the answer is a decimal number, just round it off to the closest whole number or .5 number.
Register now and start your free IELTS Masterclass!
It might sound a bit confusing, but just look at the following examples, and you will definitely get a clear idea.
Example 1:
6 + 8 + 6 + 8 = 28. Dividing 28 by 4 will result in an average of 7, so 7 is the overall band score.
Example 2:
7.5 + 7.5 + 8 + 7.5 = 30. Dividing 30 by 4 will result in an average of 7.5, so 7.5 is the overall band score.
Example 3:
6 + 6 + 6.5 + 6 = 24.5. Dividing 24.5 by 4 will result in an average of 6.12. Since 6 is the closest whole number, the overall band score will be 6.
Example 4:
7 + 6 + 7 + 7 = 27. Dividing 27 by 4 will result in an average of 6.75. Since 6.75 is exactly in between 6.5 and 7, you can round off your overall score as 7.
## 2. How is the listening and reading section scored?
Both the listening and reading sections have 40 questions each. You will be awarded one mark for every correct answer. There is no negative marking and your band will be given based on your marks on the test.
The breakdown of IELTS scores and bands in listening and reading sections are as follows:
### Listening section:
Also read: Describe a daily routine that you enjoy
Note:
While taking the IELTS, you need to either book for an Academic or General Training IELTS. Most international universities will require you to submit the academic IELTS test scores.
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## 3.How is the writing and speaking section scored?
You will have two tasks in the writing section. You will have to answer the first one by writing at least 150 words and the second one by writing at least 250 words. | 820 | 3,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.93134 |
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# AC Waveform and AC Circuit Theory - PowerPoint PPT Presentation
AC Waveform and AC Circuit Theory. Md Shahabul Alam Dept: EEE. AC Waveform and AC Circuit Theory. DC Circuit and Waveform AC Waveform Characteristics Types of Periodic Waveform Relationship Between Frequency and Periodic Time Amplitude of an AC Waveform The Average Value of an AC Waveform
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### AC Waveform and AC Circuit Theory
MdShahabulAlam
Dept: EEE
AC Waveform and AC Circuit Theory
• DC Circuit and Waveform
• AC Waveform Characteristics
• Types of Periodic Waveform
• Relationship Between Frequency and Periodic Time
• Amplitude of an AC Waveform
• The Average Value of an AC Waveform
• The RMS Value of an AC Waveform
• Form Factor and Crest Factor
DC Circuit and Waveform
• a form of current or voltage that flows around an electrical circuit in one direction only, making it a “Uni-directional” supply
• do not change their value with regards to time, a constant uni-directional DC supply never changes or becomes negative unless its connections are physically reversed
• both DC currents and voltages are produced by power supplies, batteries, dynamos and solar cells
The AC Waveform
• An alternating function or AC Waveform varies in both magnitude and with respect to time making it a “Bi-directional” waveform
• The term AC or Alternating Current, generally refers to a time-varying waveform with the most common of all being called a Sinusoid better known as a Sinusoidal Waveform
• An AC waveform is constantly changing its polarity every half cycle alternating between a positive maximum value and a negative maximum value respectively with regards to time with a common example of this being the domestic mains voltage supply we use in our homes.
• it is much easier and cheaper to generate them using alternators and waveform generators
AC Waveform Characteristics
• The Period, (T) is the length of time in seconds that the waveform takes to repeat itself from start to finish. This can also be called the Periodic Time of the waveform for sine waves, or the Pulse Width for square waves.
• The Frequency, (ƒ) is the number of times the waveform repeats itself within a one second time period. Frequency is the reciprocal of the time period, ( ƒ = 1/T ) with the unit of frequency being the Hertz, (Hz).
• The Amplitude (A) is the magnitude or intensity of the signal waveform measured in volts or amps.
Types of Periodic Waveform
• Waveforms are basically a visual representation of the variation of a voltage or current plotted to a base of time
Relationship Between Frequency and Periodic Time
What will be the periodic time of a 50Hz waveform and what is the frequency of an AC waveform that has a periodic time of 10mS.
Amplitude of an AC Waveform
• another important parameter of the AC waveform is Amplitude, better known as its Maximum or Peak value represented by the terms, Vmax for voltage or Imax for current
• The peak value is the greatest value of either voltage or current that the waveform reaches during each half cycle measured from the zero baseline
• For pure sinusoidal waveforms this peak value will always be the same for both half cycles ( +Vm = -Vm ) but for non-sinusoidal or complex waveforms the maximum peak value can be very different for each half cycle
The Average Value of an AC Waveform
• The average or mean value of a continuous DC voltage will always be equal to its maximum peak value as a DC voltage is constant
• In a pure sine wave if the average value is calculated over the full cycle, the average value would be equal to zero as the positive and negative halves will cancel each other out
• average or mean value of an AC waveform is calculated or measured over a half cycle only and this is shown below.
Average Value of a Non-sinusoidal Waveform
The zero axis base line is divided up into any number of equal parts and in our simple example above this value was nine, ( V1 to V9 ). The more ordinate lines that are drawn the more accurate will be the final average or mean value. The average value will be the addition of all the instantaneous values added together and then divided by the total number.
Where: n equals the actual number of mid-ordinates used.
For a pure sinusoidal waveform this average or mean value will always be equal to 0.637 x Vmax and this relationship also holds true for average values of current.
The RMS Value of an AC Waveform
• The average value of an AC waveform is NOT the same value as that for a DC waveforms average value. This is because the AC waveform is constantly changing with time and the heating effect given by the formula ( P = I 2.R ), will also be changing producing a positive power consumption. The equivalent average value for an alternating current system that provides the same power to the load as a DC equivalent circuit is called the “effective value”.
• The effective or RMS value of an alternating current is measured in terms of the direct current value that produces the same heating effect in the same value resistance. The RMS value for any AC waveform can be found from the following modified average value formula.
The RMS Value of an AC Waveform
For a pure sinusoidal waveform this effective or R.M.S. value will always be equal to 1/√2 x Vmax which is equal to 0.707 x Vmax and this relationship holds true for RMS values of current. The RMS value for a sinusoidal waveform is always greater than the average value except for a rectangular waveform. In this case the heating effect remains constant so the average and the RMS values will be the same
One final comment about R.M.S. values. Most multimeters, either digital or analogue unless otherwise stated only measure the R.M.S. values of voltage and current and not the average. Therefore when using a multimeter on a direct current system the reading will be equal to I = V/R and for an alternating current system the reading will be equal to Irms = Vrms/R.
Form Factor and Crest Factor
• both Form Factor and Crest Factor can be used to give information about the actual shape of the AC waveform
For a pure sinusoidal waveform the Form Factor will always be equal to 1.11
For a pure sinusoidal waveform the Crest Factor will always be equal to 1.414.
AC Waveform Example
A sinusoidal alternating current of 6 amps is flowing through a resistance of 40Ω. Calculate the average voltage and the peak voltage of the supply
The R.M.S. Voltage value is calculated as:
The Average Voltage value is calculated as:
The Peak Voltage value is calculated as: | 1,590 | 7,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-43 | latest | en | 0.805209 |
https://www.physicsforums.com/threads/surface-area-and-friction.462651/ | 1,582,455,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00175.warc.gz | 862,606,720 | 16,862 | # Surface Area and Friction
## Homework Statement
How does the surface area of an object affect the force of static friction? I'm trying to figure out whether two different objects with equal mass and different surface areas requires the same amount of applied force or different amounts.
## Homework Equations
I know of this one.
Ff = coefficient of friction(Fn)
## The Attempt at a Solution
I'm not sure how to answer this.
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
rock.freak667
Homework Helper
Does Fn depend on area?
Well each object in contact with a flat surface will have equal masses just different surface areas. I'm using a same flat surface for each object with equal masses but different surface areas.
rock.freak667
Homework Helper
Well each object in contact with a flat surface will have equal masses just different surface areas. I'm using a same flat surface for each object with equal masses but different surface areas.
Right, well in most cases, the normal force would be the weight right? So would the frictional force formula have area in it?
Ya it would have area.
rock.freak667
Homework Helper
Ya it would have area.
No, it would not, if Fn=mg and Ff=μFn then area does not appear.
Once your materials are the same, then μ is the same and if the masses are the same, then the frictional force produced by them are the same, irregardless of area.
Ah ok. Thanks so much.
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
No, it would not, if Fn=mg and Ff=μFn then area does not appear.
Once your materials are the same, then μ is the same and if the masses are the same, then the frictional force produced by them are the same, irregardless of area.
As long as you can treat the coefficient of friction, μ, as simply being a constant which depends only upon the two materials which are present, then surface area will not make any difference. That's the simple model used in physics courses.
However, coefficient of friction, μ, actually does depend on the temperature of the materials. We seldom include that in our model. The greater the surface area, the less the increase in temperature (The thermal energy is dissipated over a wider area.) so that μ will tend to change less than in the case less surface area. Of course, this is more a factor with sliding friction, which tends to produce heat.
It's also the case that for materials which deform rather easily, like rubber used in tires, the pressure at the point of contact affects μ in a rather complicated way.
This is just "scratching the surface". - pun intended
Generally, when solving a physics problem with friction involved, frictional force does not depend on surface area. In any case, Ff = μ Fn works very well. | 610 | 2,780 | {"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} | 3.640625 | 4 | CC-MAIN-2020-10 | longest | en | 0.934725 |
http://mathhelpforum.com/math-software/125862-matlab-help-cant-write-script-print.html | 1,513,527,283,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948596115.72/warc/CC-MAIN-20171217152217-20171217174217-00632.warc.gz | 183,957,471 | 3,188 | # Matlab help can't write script
Printable View
• Jan 27th 2010, 08:23 PM
Mathhelpz
Matlab help can't write script
Q-2e One type of candy costs 60 cents a pound while a second type costs 80 cents a pound. How many pounds of each type must be combined in order to produce 20 pounds of a mixture worth 75 cents a pound?
I got the answer though,
type 1 =x
type 2 =y
x+y=20
y=20-x
60x+80y=75(x+y)
60x+80y=75x+75y
simplify
5y=15x
sub in y=20-x
5(20-y)=15x
100-5x=15x
100=20x
x=5
plug 5 into y=20-x
y=15
Could someone help me form a matlab script?
• Jan 28th 2010, 12:07 AM
CaptainBlack
Quote:
Originally Posted by Mathhelpz
Q-2e One type of candy costs 60 cents a pound while a second type costs 80 cents a pound. How many pounds of each type must be combined in order to produce 20 pounds of a mixture worth 75 cents a pound?
I got the answer though,
type 1 =x
type 2 =y
x+y=20
y=20-x
60x+80y=75(x+y)
60x+80y=75x+75y
simplify
5y=15x
sub in y=20-x
5(20-y)=15x
100-5x=15x
100=20x
x=5
plug 5 into y=20-x
y=15
Could someone help me form a matlab script?
You have the pair of equations:
$x+y=20$
$60x+80y=75\times 20$
or in matrix form:
$\left[ \begin{array}{cc}1&1\\60&80\end{array}\right] \left[ \begin{array}{c}x\\y \end{array} \right] =$ $\left[ \begin{array}{c}20\\1500 \end{array} \right]$
So is we set:
$
\bold{x}= \left[ \begin{array}{c}x\\y \end{array} \right]
$
We can write:
Code:
A=[1,1;60,80]; Z=[20;1500] X=A\Z
or of you prefer:
Code:
A=[1,1;60,80]; Z=[20;1500] X=inv(A)*Z
CB
• Jan 28th 2010, 06:00 AM
Mathhelpz
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# Aristotle Prep NEW RC Passages
Author Message
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Joined: 12 Sep 2010
Posts: 224
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Aristotle Prep NEW RC Passages [#permalink]
### Show Tags
25 Feb 2012, 12:50
I just found out that Aristotle Prep has discontinued RC99 because the creators have admitted that some of the passages are not a good representation of the GMAT passages. So now, Aristotle Prep has two new sets of RC passages. Has anyone purchase the new sets? How are the new set compare with RC99?
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Re: Aristotle Prep NEW RC Passages [#permalink]
### Show Tags
22 Mar 2012, 23:13
Yep, I bought one of the RC sets. It contains 15 passages, with 3 or 4 questions each. I think it's pretty good, with the length of the passage reflective of the real GMAT and not too many questions for each passage.
But I still want to get my hands on the RC99. Heard too many good things about it!
D
Re: Aristotle Prep NEW RC Passages [#permalink] 22 Mar 2012, 23:13
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# Aristotle Prep NEW RC Passages
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 658 | 2,453 | {"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-2017-13 | longest | en | 0.869802 |
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In this lesson, you will watch the PBS news story “Why it will take more than just recycling to cut back on plastic” and explore how new ideas or...
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Through the invention process, you will explore ways to use household items or routines to monitor health conditions for at-risk patients. You will...
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These resources provide students with guided practice (problem-solving process) and independent practice (formative/informative assessment). | 623 | 2,727 | {"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} | 2.984375 | 3 | CC-MAIN-2020-50 | latest | en | 0.802287 |
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Note: You can use any kind of loop for this section's independent project provided that you correctly solve the problem. We recommend using `Array.prototype.forEach()` or `for` loops for the independent project.
Over the course of this section, we've covered many different looping techniques. It may feel overwhelming to decide which loop is best for a specific use case. This lesson will hopefully narrow things down and make life a little easier.
Also, it's important to note that at this point in your learning, you shouldn't worry too much about which loop is "best" for the job at hand. You should feel free to practice them all - and also to experiment with code and solutions that interest you. For the upcoming independent project, you will be required to loop - but how you implement a loop is up to you.
All of these examples assume that you are starting with the following array:
``````const array = [0,1,2,3,4,5];
``````
Yep, we are returning to the doubled array example - but this time all of our looping examples are in one place!
`for`
Example
``````let doubledArray = [];
for (let index = 0; index < array.length; index +=1) {
doubledArray.push(array[index] * 2);
}
``````
When to Use
This is a great "starter" loop. You should practice it frequently - at least until you get the hang of it. At that point, you should favor other kinds of loops instead!
`Array.prototype.forEach()`
Example
``````let doubledArray = [];
array.forEach(function(element) {
doubledArray.push(element * 2);
});
``````
When to Use
Use this when both conditions are met:
1. You want to loop through every element in an array without breaking out of the loop;
2. You don't want a transformed array.
`for...of`
Example
``````let doubledArray = [];
for (const element of array) {
doubledArray.push(element * 2);
}
``````
When to Use
Use this whenever you want to loop through an array or a string but you don't want to transform the elements into a new string or array. It comes down to preference, but generally you can favor this one over `Array.prototype.forEach()`.
`Array.prototype.map()`
Example
``````const doubledArray = array.map(function(element) {
return element * 2;
});
``````
When to Use
Use this whenever you want to iterate through an array and create a new array with all of its elements transformed.
`while`
Example
``````let index = 0;
let doubledArray = [];
while
(index < array.length) {
doubledArray.push(array[index] * 2);
index ++;
}
``````
When to Use
You won't use this one often - practice it until you have the hang of it and then use it sparingly. It can be useful when you want to loop only until a specific condition is met - or when you are writing code that interacts with users.
What About a Use Case That Doesn't Fit Perfectly?
Let's look at a use case that doesn't fit perfectly - and then solve it with two different kinds of loops. Let's say we want to use a loop to create a transformed string. Should we solve the problem with `Array.prototype.map()` if we need to transform the string into an array first? Or should we solve it with `for...of` even though we should favor `Array.prototype.map()` for transformations?
Well, neither approach is wrong. Ultimately, we can try both solutions and see which feels more elegant and concise. Let's return to our `for...of` vowelized loop example:
``````const consonantString = "bdfmxtgl";
let vowelizedString = "";
for (const letter of consonantString) {
vowelizedString = vowelizedString.concat(letter + "a");
}
vowelizedString;
``````
Here's how we'd solve the same problem with `Array.prototype.map()`:
``````const consonantString = "bdfmxtgl";
const consonantArray = consonantString.split("");
const vowelizedArray = consonantArray.map(function(element) {
return element + "a";
});
vowelizedArray.join("");
``````
Both of these solutions come up with the same answer in the same number of lines. Ultimately, the approach we take here comes down to preference. There are valid reasons for both approaches.
You can use just about any kind of loop to solve many problems - and while there are best practices and ways to improve our code, there's no one right approach.
Lesson 14 of 15
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# Renaissance/STAR Inspired Math Practice Tests 2nd Grade PDFs
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Need help getting your students to show growth and progress on the STAR test? I did too! That's why I made this packet. This is a zip file that contains 4 tests (Operations and Algebraic Thinking, Number and Operations in Base Ten, Geometry, and Measurement and Data) created as a Study Guide or Practice Test for Star testing by Renaissance Learning. It is 78 pages and contains 241 questions. I used the Core Progress to help me create these tests and I discovered that a lot of the questions on the STAR test are hard for students if they haven't been taught how to do them. My students are making great progress after using these tests. This is written for 2nd grade, but if you have high 1st graders or low 3rd graders you need this test too. Get it before the price increases!
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https://www.physicsforums.com/threads/uniform-circular-motion-problem-please-help.536373/ | 1,519,583,616,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00014.warc.gz | 929,949,921 | 15,735 | 1. Oct 3, 2011
naham1866
1. At a distance of 25km from the eye of a hurricane, the wind is moving at 180km/h in a circle. What is the magnitude of the centripetal acceleration, in meters per second squared, of the particles that make up the wind?
2. Relevant equations
Ac=V^2/r = 4∏^2r/T^2 = 4∏^2rf^2
3. The attempt at a solution
Ac=V^2/r
The answer is 0.10m/s but I just cant seem to get it right.
2. Oct 3, 2011
Staff: Mentor
You have the right equation in your attempt. Fill in the numbers for velocity V and radius r, and be sure to stay consistent in your units. Show us that work...
3. Oct 3, 2011
naham1866
Okay so I got Ac=180km/h^2/25km
Problem is that it doesn't give me 0.10m/s
4. Oct 3, 2011
Staff: Mentor
You need to fix the units. You need to convert everything into meters, kilograms and seconds (the mks standard SI system of units). If you mix units like seconds and hours up, you won't get the right answers.
So convert everything you are given into mks units, and plug those quantities into the equation. Carry units along in the equation, and if you have the same units in numerator and denominator, you can cancle them out. Like m/m = 1, and s^2/s = s, and so on.
Now show us what you get...
5. Oct 3, 2011
naham1866
Haha YES!!! I Did it!! I went out and on my way back home I remembered that I had to change from kilometers to meters, so here's what I did.
180km/h / 60 = 3m/min /60 = 0.05m/s
25km / 1000 = 0.025m
Ac = V^2/r
Ac = 0.05m/s^2/0.025m
Ac= .10m/s
YES! :D
6. Oct 4, 2011
naham1866
Thank you vey much for the help
7. Oct 5, 2011
Staff: Mentor
You are welcome. Learning to carry units along in your calculations is a huge trick, IMO. I still remember the first time I learned that wey back in undergrad many years ago. | 556 | 1,775 | {"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} | 3.953125 | 4 | CC-MAIN-2018-09 | longest | en | 0.928871 |
https://www.lmfdb.org/EllipticCurve/Q/47040/hb/ | 1,606,815,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00658.warc.gz | 733,115,938 | 43,954 | # Properties
Label 47040.hb Number of curves $6$ Conductor $47040$ CM no Rank $1$ Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("47040.hb1")
sage: E.isogeny_class()
## Elliptic curves in class 47040.hb
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
47040.hb1 47040hc6 [0, 1, 0, -52684865, -147207093825] [2] 2359296
47040.hb2 47040hc4 [0, 1, 0, -3292865, -2300844225] [2, 2] 1179648
47040.hb3 47040hc5 [0, 1, 0, -3073345, -2620597057] [4] 2359296
47040.hb4 47040hc3 [0, 1, 0, -1160385, 454345023] [2] 1179648
47040.hb5 47040hc2 [0, 1, 0, -219585, -30919617] [2, 2] 589824
47040.hb6 47040hc1 [0, 1, 0, 31295, -2971585] [2] 294912 $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curves in class 47040.hb have rank $$1$$.
## Modular form 47040.2.a.hb
sage: E.q_eigenform(10)
$$q + q^{3} + q^{5} + q^{9} + 4q^{11} - 2q^{13} + q^{15} - 2q^{17} + 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 4 & 8 & 4 & 8 \\ 2 & 1 & 2 & 4 & 2 & 4 \\ 4 & 2 & 1 & 8 & 4 & 8 \\ 8 & 4 & 8 & 1 & 2 & 4 \\ 4 & 2 & 4 & 2 & 1 & 2 \\ 8 & 4 & 8 & 4 & 2 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 670 | 1,509 | {"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} | 3.171875 | 3 | CC-MAIN-2020-50 | longest | en | 0.455237 |
http://www.adras.com/Mask-Account-Number.t135938-7.html | 1,638,354,989,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00503.warc.gz | 82,566,924 | 2,561 | From: Need help Need on 6 Apr 2010 17:19 I have a spreadsheet with tons of accounts numbers, and I need them to show up as 1234567 on my spreadsheet, but need them to show up XXX4567 how do I do this? From: Mike H on 6 Apr 2010 17:27 Hi, One way. Say these numbers are in column A starting in A2. Put this in B2 ="XXXX"&RIGHT(A2,4) Drag down as required and then hide column A You could even have the numbers on another sheet and use the formula ="XXXX"&RIGHT(Sheet2!A2,4) You could then hide the entire worksheet with the numbers on. -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "Need help" wrote: > I have a spreadsheet with tons of accounts numbers, and I need them to show > up as 1234567 on my spreadsheet, but need them to show up XXX4567 > > how do I do this? | Pages: 1 Prev: Combine DescriptionsNext: COUNTIF, Sorting, Two Sheets | 249 | 959 | {"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} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.92472 |
https://en.wikipedia.org/wiki/Torsion_balance | 1,537,897,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161902.89/warc/CC-MAIN-20180925163044-20180925183444-00433.warc.gz | 493,561,905 | 25,385 | Torsion spring
(Redirected from Torsion balance)
Video of a model torsion pendulum oscillating
A torsion spring is a spring that works by torsion or twisting; that is, a flexible elastic object that stores mechanical energy when it is twisted. When it is twisted, it exerts a force (actually torque) in the opposite direction, proportional to the amount (angle) it is twisted. There are various types. For example, clocks use a spiral wound torsion spring sometimes called a "clock spring" or colloquially called a mainspring. Those types of torsion springs are also used for attic stairs, clutches, and other devices that need near constant torque for large angles or even multiple revolutions.
A torsion bar is a straight bar of metal or rubber that is subjected to twisting (shear stress) about its axis by torque applied at its ends. A more delicate form used in sensitive instruments, called a torsion fiber consists of a fiber of silk, glass, or quartz under tension, that is twisted about its axis. The other type, a helical torsion spring, is a metal rod or wire in the shape of a helix (coil) that is subjected to twisting about the axis of the coil by sideways forces (bending moments) applied to its ends, twisting the coil tighter. This terminology can be confusing because in a helical torsion spring the forces acting on the wire are actually bending stresses, not torsional (shear) stresses.[1] [2]
Torsion coefficient
As long as they are not twisted beyond their elastic limit, torsion springs obey an angular form of Hooke's law:
${\displaystyle \tau =-\kappa \theta \,}$
where ${\displaystyle \tau \,}$ is the torque exerted by the spring in newton-meters, and ${\displaystyle \theta \,}$ is the angle of twist from its equilibrium position in radians. ${\displaystyle \kappa \,}$ is a constant with units of newton-meters / radian, variously called the spring's torsion coefficient, torsion elastic modulus, rate, or just spring constant, equal to the change in torque required to twist the spring through an angle of 1 radian. It is analogous to the spring constant of a linear spring. The negative sign indicates that the direction of the torque is opposite to the direction of twist.
The energy U, in joules, stored in a torsion spring is:
${\displaystyle U={\frac {1}{2}}\kappa \theta ^{2}}$
Uses
Some familiar examples of uses are the strong, helical torsion springs that operate clothespins and traditional spring-loaded-bar type mousetraps. Other uses are in the large, coiled torsion springs used to counterbalance the weight of garage doors, and a similar system is used to assist in opening the trunk (boot) cover on some sedans. Small, coiled torsion springs are often used to operate pop-up doors found on small consumer goods like digital cameras and compact disc players. Other more specific uses:
• A torsion bar suspension is a thick, steel torsion-bar spring attached to the body of a vehicle at one end and to a lever arm which attaches to the axle of the wheel at the other. It absorbs road shocks as the wheel goes over bumps and rough road surfaces, cushioning the ride for the passengers. Torsion-bar suspensions are used in many modern cars and trucks, as well as military vehicles.
• The sway bar used in many vehicle suspension systems also uses the torsion spring principle.
• The torsion pendulum used in torsion pendulum clocks is a wheel-shaped weight suspended from its center by a wire torsion spring. The weight rotates about the axis of the spring, twisting it, instead of swinging like an ordinary pendulum. The force of the spring reverses the direction of rotation, so the wheel oscillates back and forth, driven at the top by the clock's gears.
• Torsion springs consisting of twisted ropes or sinew, were used to store potential energy to power several types of ancient weapons; including the Greek ballista and the Roman scorpio and catapults like the onager.
• The balance spring or hairspring in mechanical watches is a fine, spiral-shaped torsion spring that pushes the balance wheel back toward its center position as it rotates back and forth. The balance wheel and spring function similarly to the torsion pendulum above in keeping time for the watch.
• The D'Arsonval movement used in mechanical pointer-type meters to measure electric current is a type of torsion balance (see below). A coil of wire attached to the pointer twists in a magnetic field against the resistance of a torsion spring. Hooke's law ensures that the angle of the pointer is proportional to the current.
• A DMD or digital micromirror device chip is at the heart of many video projectors. It uses hundreds of thousands of tiny mirrors on tiny torsion springs fabricated on a silicon surface to reflect light onto the screen, forming the image.
Torsion balance
Drawing of Coulomb's torsion balance. From Plate 13 of his 1785 memoir.
Torsion balance used by Paul R. Heyl in his measurements of the gravitational constant G at the U.S. National Bureau of Standards (now NIST) between 1930 and 1942.
The torsion balance, also called torsion pendulum, is a scientific apparatus for measuring very weak forces, usually credited to Charles-Augustin de Coulomb, who invented it in 1777, but independently invented by John Michell sometime before 1783.[3] Its most well-known uses were by Coulomb to measure the electrostatic force between charges to establish Coulomb's Law, and by Henry Cavendish in 1798 in the Cavendish experiment[4] to measure the gravitational force between two masses to calculate the density of the Earth, leading later to a value for the gravitational constant.
The torsion balance consists of a bar suspended from its middle by a thin fiber. The fiber acts as a very weak torsion spring. If an unknown force is applied at right angles to the ends of the bar, the bar will rotate, twisting the fiber, until it reaches an equilibrium where the twisting force or torque of the fiber balances the applied force. Then the magnitude of the force is proportional to the angle of the bar. The sensitivity of the instrument comes from the weak spring constant of the fiber, so a very weak force causes a large rotation of the bar.
In Coulomb's experiment, the torsion balance was an insulating rod with a metal-coated ball attached to one end, suspended by a silk thread. The ball was charged with a known charge of static electricity, and a second charged ball of the same polarity was brought near it. The two charged balls repelled one another, twisting the fiber through a certain angle, which could be read from a scale on the instrument. By knowing how much force it took to twist the fiber through a given angle, Coulomb was able to calculate the force between the balls. Determining the force for different charges and different separations between the balls, he showed that it followed an inverse-square proportionality law, now known as Coulomb's law.
To measure the unknown force, the spring constant of the torsion fiber must first be known. This is difficult to measure directly because of the smallness of the force. Cavendish accomplished this by a method widely used since: measuring the resonant vibration period of the balance. If the free balance is twisted and released, it will oscillate slowly clockwise and counterclockwise as a harmonic oscillator, at a frequency that depends on the moment of inertia of the beam and the elasticity of the fiber. Since the inertia of the beam can be found from its mass, the spring constant can be calculated.
Coulomb first developed the theory of torsion fibers and the torsion balance in his 1785 memoir, Recherches theoriques et experimentales sur la force de torsion et sur l'elasticite des fils de metal &c. This led to its use in other scientific instruments, such as galvanometers, and the Nichols radiometer which measured the radiation pressure of light. In the early 1900s gravitational torsion balances were used in petroleum prospecting. Today torsion balances are still used in physics experiments. In 1987, gravity researcher A.H. Cook wrote:
The most important advance in experiments on gravitation and other delicate measurements was the introduction of the torsion balance by Michell and its use by Cavendish. It has been the basis of all the most significant experiments on gravitation ever since.[5]
Torsional harmonic oscillators
Definition of terms
Term Unit Definition
${\displaystyle \theta \,}$ rad Angle of deflection from rest position
${\displaystyle I\,}$ kg m2 Moment of inertia
${\displaystyle C\,}$ joule s rad−1 Angular damping constant
${\displaystyle \kappa \,}$ N m rad−1 Torsion spring constant
${\displaystyle \tau \,}$ ${\displaystyle \mathrm {N\,m} \,}$ Drive torque
${\displaystyle f_{n}\,}$ Hz Undamped (or natural) resonant frequency
${\displaystyle T_{n}\,}$ s Undamped (or natural) period of oscillation
${\displaystyle \omega _{n}\,}$ ${\displaystyle \mathrm {rad\,s^{-1}} \,}$ Undamped resonant frequency in radians
${\displaystyle f\,}$ Hz Damped resonant frequency
${\displaystyle \omega \,}$ ${\displaystyle \mathrm {rad\,s^{-1}} \,}$ Damped resonant frequency in radians
${\displaystyle \alpha \,}$ ${\displaystyle \mathrm {s^{-1}} \,}$ Reciprocal of damping time constant
${\displaystyle \phi \,}$ rad Phase angle of oscillation
${\displaystyle L\,}$ m Distance from axis to where force is applied
Torsion balances, torsion pendulums and balance wheels are examples of torsional harmonic oscillators that can oscillate with a rotational motion about the axis of the torsion spring, clockwise and counterclockwise, in harmonic motion. Their behavior is analogous to translational spring-mass oscillators (see Harmonic oscillator#Equivalent systems). The general equation of motion is:
${\displaystyle I{\frac {d^{2}\theta }{dt^{2}}}+C{\frac {d\theta }{dt}}+\kappa \theta =\tau (t)}$
If the damping is small, ${\displaystyle C\ll {\sqrt {\frac {\kappa }{I}}}\,}$, as is the case with torsion pendulums and balance wheels, the frequency of vibration is very near the natural resonant frequency of the system:
${\displaystyle f_{n}={\frac {\omega _{n}}{2\pi }}={\frac {1}{2\pi }}{\sqrt {\frac {\kappa }{I}}}\,}$
Therefore, the period is represented by:
${\displaystyle T_{n}={\frac {1}{f_{n}}}={\frac {2\pi }{\omega _{n}}}=2\pi {\sqrt {\frac {I}{\kappa }}}\,}$
The general solution in the case of no drive force (${\displaystyle \tau =0\,}$), called the transient solution, is:
${\displaystyle \theta =Ae^{-\alpha t}\cos {(\omega t+\phi )}\,}$
where:
${\displaystyle \alpha =C/2I\,}$
${\displaystyle \omega ={\sqrt {\omega _{n}^{2}-\alpha ^{2}}}={\sqrt {\kappa /I-(C/2I)^{2}}}\,}$
Applications
Animation of a torsion spring oscillating
The balance wheel of a mechanical watch is a harmonic oscillator whose resonant frequency ${\displaystyle f_{n}\,}$ sets the rate of the watch. The resonant frequency is regulated, first coarsely by adjusting ${\displaystyle I\,}$ with weight screws set radially into the rim of the wheel, and then more finely by adjusting ${\displaystyle \kappa \,}$ with a regulating lever that changes the length of the balance spring.
In a torsion balance the drive torque is constant and equal to the unknown force to be measured ${\displaystyle F\,}$, times the moment arm of the balance beam ${\displaystyle L\,}$, so ${\displaystyle \tau (t)=FL\,}$. When the oscillatory motion of the balance dies out, the deflection will be proportional to the force:
${\displaystyle \theta =FL/\kappa \,}$
To determine ${\displaystyle F\,}$ it is necessary to find the torsion spring constant ${\displaystyle \kappa \,}$. If the damping is low, this can be obtained by measuring the natural resonant frequency of the balance, since the moment of inertia of the balance can usually be calculated from its geometry, so:
${\displaystyle \kappa =(2\pi f_{n})^{2}I\,}$
In measuring instruments, such as the D'Arsonval ammeter movement, it is often desired that the oscillatory motion die out quickly so the steady state result can be read off. This is accomplished by adding damping to the system, often by attaching a vane that rotates in a fluid such as air or water (this is why magnetic compasses are filled with fluid). The value of damping that causes the oscillatory motion to settle quickest is called the critical damping ${\displaystyle C_{c}\,}$:
${\displaystyle C_{c}=2{\sqrt {\kappa I}}\,}$ | 2,922 | 12,406 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 42, "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} | 3.390625 | 3 | CC-MAIN-2018-39 | latest | en | 0.924092 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-3-differentiation-3-5-higher-derivatives-exercises-page-135/23 | 1,585,422,185,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370492125.18/warc/CC-MAIN-20200328164156-20200328194156-00099.warc.gz | 964,682,065 | 12,660 | ## Calculus (3rd Edition)
$-\frac{558}{81}$.
Since $f(x)=\frac{12}{x}-x^3$, then we get $$f'(x)=-\frac{12}{x^2}-3x^2, \quad f''(x)=\frac{24}{x^3}-6x, \quad f'''(x)=-\frac{72}{x^4}-6.$$ Hence, $f'''(-3)=-\frac{72}{81}-6=-\frac{558}{81}$. | 121 | 237 | {"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} | 3.875 | 4 | CC-MAIN-2020-16 | latest | en | 0.29364 |
http://users.humboldt.edu/rpaselk/C109.S13/C109_Notes/C109_lec09.htm | 1,508,258,536,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822145.14/warc/CC-MAIN-20171017163022-20171017183022-00659.warc.gz | 347,910,536 | 3,610 | # Reaction Stoichiometry: Chemical Equations, cont.
Notice that a chemical equation gives both qualitative information (what things react to give what products) and quantitative information (how much stuff is produced if a particular amount of stuff reacts). This gives rise to various practical applications.
• Example: How many grams of carbon dioxide are produced during the complete combustion of 475.5 g of natural gas (methane, CH4)
First need to find moles of CH4:
MW = 12.01 + 4 (1.008) = 16.04
Moles of CH4 = 475.5 g / 16.04 g/mol = 29.64 mole.
Next need balanced equation:
CH4 + 2 O2 CO2 + 2 H2O
Note from the equation we have a 1:1 ratio of moles methane to mole carbon dioxide
So we have 29.64 moles CO2
& MW of CO2 = 12.01 + 2 (16.00) = 44.01
and (29.64 mol)(44.01 g/mol) = 1.304 kg
Let's look at a slightly more complicated reaction.
• Example: How many moles of oxygen will be needed for the complete combustion of 275.5 g of butane (C4H10).
First we need the MW of butane: MW = 4 (12.01) + 10 (1.008) = 58.12 g/mol
and the number of moles of butane = (275.5 g) / (58.12 g/mol) = 4.740 mole
Then from the balanced equation:
2 C4H10 + 13 O2 8 CO2 + 10 H2O
we can find the number of moles of oxygen required
O2 = (4.740 moles butane)(13 moles O2 / 2 moles butane) = 30.81 moles
# Limiting Problems
Asking question of what is the maximum amount of something which can be produced from a given mixture of stuff. This is a fairly straight-forward sort of problem in the day-to-day world, but seems to cause a great deal of difficulty for lots of folks in chemistry. Let's start by looking at a non-chemical problem:
Consider you have to make a bunch of sandwichs for a party. The equation for the sandwichs (in slices) is:
2 Bread + 1 Cheese + 1 Meat 1 Sandwich
You have a 32 oz loaf of bread, 22 oz of sliced cheese and 32 oz of sliced meat. If bread slices weigh 1/2 oz, cheese slices 3/4 oz and meat slices 1 oz, how many sandwichs can you make?
Look at how many sandwichs can be made from each ingredient:
• If cheese limits: (22 oz cheese)(1 slice cheese/3/4 oz cheese)(1 sandwich/slice cheese) = 29.3 sandwiches
• If meat limits: (32 oz meat)(1 slice meat/oz meat)(1 sandwich/slice meat) = 32 sandwiches
Cheese limits and we can make 29 sandwichs.
• Example: Consider the reaction:
Fe3O4 + 4 C 3 Fe + 4 CO
What is the maximum mass of Fe which could be made from 115.0 g Fe3O4 and 24.00 g C?
The trick here is to find the maximum amount of iron which could be made from each reactant.
The lesser amount will then be the max possible:
C: (3 mol Fe/ 4 mol C)(24.00 g C/ 12.01 g C/mol C) = 1.499 mole
Fe3O4: (3 mol Fe/ mol Fe3O4)(115.0 g Fe3O4/231.6 g Fe3O4/mol Fe3O4) = 1.490 mole
therefore Fe3O4 limits, can only make 1.490 moles Fe.
Grams = (1.490 moles Fe)(55.85 g Fe/mol Fe) = 83.217 g Fe = 83.22 g Fe
• Example: Consider the reaction of zinc metal with acid:
Zn + 2H+ Zn2+ + H2(g)
What is the maximum amount (moles) of hydrogen gas which may be produced by reacting 0.50 g of Zinc with 0.800 mole hydrogen ion? Show work!
For Zn limiting: (1 mole H2/1 mole Zn)(0.50 g Zn) / (65.39 g Zn/mol) = 7.646 x 10-3 mol H2
For H+ limiting: (1 mole H2/2 mole H+)(0.800 mole H+) = 0.400 mol H2
Much less with Zn therefore, 7.6 x 10-3 mol H2
## Percent Yield
Another frequent question arising in chemical processes is the percent yield. This deals with the question of how effective was a given process in producing a product. Its an important consideration because chemical reactions rarely go completley to products. The maximum possible yield for a reaction is known as the Theoretical Yield.
• Example: Looking at the Zn limiting case above, the amount of hydrogen generated would be the theoretical yield = 7.6 x 10-3 mol. Now if the actual yield turned out for a particular experiment to be 6.75 x 10-3 mol, the percent yield would be calculated to be:
(6.75 x 10-3 mol)(100 %) / (7.6 x 10-3 mol) = 88.8% = 89%
### NEXT
Syllabus / Schedule C109 Home
© R A Paselk | 1,225 | 4,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-43 | longest | en | 0.903372 |
https://davida.davivienda.com/viewer/number-line-1-to-100-printable.html | 1,709,621,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00648.warc.gz | 201,278,029 | 9,206 | # Number Line 1 To 100 Printable
As this is a blank number line, it also provides a good opportunity to. Web a number line with no numbers is a great visual aid to use to support your children with the recognition of numerals all the way to 100. You can also download the pdf version with the link given below the respective charts. There is a starting point and an ending point on the number line. Perfect for skip counting/multiples patterns. Choose from three options below, including a. Here are the number lines from 1.
A number line from 1 to 100 can be useful in the classroom for practicing counting to 100 or addition and subtraction with larger numbers. Standard number line up to 100 (portrait version). You will find several various designs in this set, so pick one that works the best for your needs. Web a number line with no numbers is a great visual aid to use to support your children with the recognition of numerals all the way to 100. Perfect for skip counting/multiples patterns.
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You can also download the pdf version with the link given below the respective charts. Choose from three options below, including a. Preview images of the first and second (if there is one) pages are shown. Cut the number line into seperate pieces for children to arrange them in the right order. Perfect for skip counting/multiples patterns.
As this is a blank number line, it also provides a good opportunity to. Number lines are a great way to understand numbers and see them visually on a line. Each number line worksheet has been saved at a high resolution. They are all easy to download and print pdf files.
## As This Is A Blank Number Line, It Also Provides A Good Opportunity To.
Web or print the a3 version for a number line to 100 poster to display in your math centre.featuring colourful numbers and cute illustrations, this number line is sure to capture children's interest. Web number line printable pdfs in various ranges, including 1 to 20, 1 to 100, blank number line templates, double number lines and more. Get these free printable 100 number line worksheets to help with your math curriculum in kindergarten and first grade. These printable charts will help your child learn to read and write numbers to 100.
## Web This Colorful Number Line To 100 Worksheet Is A Handy Tool You Can Use To Help Engage Your Youngsters And Get Them Excited About Counting!Number Lines Are A Great Visual Aid You Can Use To Help Children Recognize Numerals All The Way Up To 10, 100, Or Even Higher.
Choose from three options below, including a. There are two templates to choose from. Cut the number line into seperate pieces for children to arrange them in the right order. My number line 0 to 100 (portrait version).
## Place Small Objects Next To The Number Line And Count Along With Quantities.
The size of the pdf file is 13537 bytes. You can also download the pdf version with the link given below the respective charts. Here you will find a range of free printable first grade place value charts. Therefore, the number line is also known as a closed line. | 844 | 3,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.908377 |
https://lobste.rs/s/y2gfgg/case_for_learned_sorting_algorithm | 1,638,318,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00569.warc.gz | 450,233,119 | 5,098 | 1. 2
blog.acolyer.org
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I think there are three interesting ideas in this kind of system:
The first is the observation that underlies bloom filters, skip lists, and a lot of other data structures: If you have a very cheap way of getting a half-arsed almost answer, then you can often do an O(n) thing to get from your approximation to the right answer and still outperform a formally correct algorithm because your guess reduces n to a much smaller value. That’s not a particularly new observation but it is often overlooked.
The second is that most data is not a uniform random distribution. If you know something about the distribution of the data then you can often do better than an algorithm that is optimal for a uniform distribution. Again, this is not particularly novel, and is often used in real-world sorting algorithms in systems software, to special-case common values. Outperforming a radix sort is nice, but it’s quite common in systems software to do a bucket sort, which is O(n), to get the most common values out of the way and then do something slower on the values that end up in the slow-path unusual-case bucket.
The third, which was actually the conclusion of some of my PhD work so one that I’m quite biased towards, is that you can often apply some fairly simple and cheap machine learning to figure out enough of the characteristics of the data that you can apply the first to observations to get a rough guess that’s mostly good-enough and fall back to something slower for when the guess is wrong. If your data has characteristics that you can look at and figure out how to design an optimised data structure then you may well be able to have a machine infer these characteristics.
The thing I really like about some of the recent work in this area is that they’re starting to separate these three aspects.
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Do you have any suggested materials or papers for number three? I’ve seen simple techniques applied in my problem domain very successfully a handful of times, but trying to develop intuition around what set of techniques to consider and when is difficult since so much of the literature is devoted to much grander forms of ML.
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Not sure if they fit your need, but here are 2 articles
1. 1
I don’t, unfortunately, other than the ones cited in the article. | 489 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | longest | en | 0.956933 |
http://people.wku.edu/charles.smith/music/stats7.htm | 1,670,110,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00671.warc.gz | 40,635,005 | 2,686 | Musical Influences Statistics In the attached pdf some additional statistics are offered in an effort to further summarize the influences data given in the "Composers" section on a composer-by-composer basis. In that section the relative removed age of influence is conveyed in the last two, "Influences", fields by preceding the names in each list with '+' and '-' signs. In each 'Musical Influences' field an individual's name is tagged as follows: (1) no tag: indicates that the influencing composer is older than the subject composer, and was alive during the latter's life; (2) one minus ( - ): indicates that the influencing composer died 0 to 25 years before the birth of the subject composer; (3) two minuses ( -- ): indicates that the influencing composer died 25 to 100 years before the birth of the subject composer; (4) three minuses ( --- ): indicates that the influencing composer died more than 100 years before the birth of the subject composer; (5) one plus ( + ): indicates that the influencing composer was younger than the subject composer. In each 'Has Influenced' field the reverse pattern is applied to identify relative temporal remoteness of influences generated *by* the subject composer. The first six columns in the pdf statistically summarize the influences *on* the subject composers. A mean temporal remoteness index ('Mean1') characterizing the influencing composers is calculated by assigning the values 0, -1, -2, -3, and +1 to the conditions described above, and then taking sums and means. For example, it turns out that the multiple composer influences on Carl Orff code temporally as: -3 0 -3 -3 -3 -3 0 0, which summed comes to a value of -15., which divided by the number of influences involved (8, as shown in the third column) produces a mean value of -1.88 (fifth column). The first column relays an unusual statistic: the number of composers influencing the subject composer who were *younger* than he/she was; the second column, the total number of minuses (for Orff, as above, 15). 'Mean2' (column four) for each subject composer is the mean of the index values associated with the individuals identified as influences on the subject (in the case of Orff, this is the mean of 6.9, 4.0, 2.6, 1.9, 2.1, 2.2, 3.7 and 3.8, or 3.40). Column six is the associated standard deviation (actually, the square root of the variance). The second group of six columns summarize each subject composers' influence on *other* composers. The first two of these parallel the first two in the first group, now totalling the number of composers influenced who are *older* than the subject ('Sum-'), and all the plusses connected with the composers influenced ('Sum+'; in the case of J. S. Bach this is a remarkable 274, a testimony to the extent of his lasting influence). The third column ('n/n+') gives the number of composers influenced by the subject, the first value being the raw total and the second, when present, a total weighted by instances of major influence. Column four ('Sum') for each subject composer is the mean of the index values associated with the individuals identified as influenced *by* the subject (in the case of Orff, this is 1.90, as he influenced only one other person, whose index value is 1.9). Column six is again the associated standard deviation; for Orff this is 0. because of the n of 1. The final column gives the overall rank of each composer in the list, for comparative purposes. Hyphens are placed in the table when no values can be calculated because no influences have been identified. retrieve PDF file here | 815 | 3,577 | {"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-2022-49 | latest | en | 0.9305 |
https://mail.queryxchange.com/q/21_184053/how-to-find-lim-limits-x-to0-frac-e-x-1-x-x-2-without-using-l-39-hopital-39-s-rule-nor-any-series-expansion/ | 1,537,332,303,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155924.5/warc/CC-MAIN-20180919043926-20180919063926-00379.warc.gz | 557,257,374 | 8,931 | # How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?
by enzotib Last Updated March 11, 2017 11:20 AM
Is it possible to determine the limit
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
without using l'Hopital's rule nor any series expansion?
For example, suppose you are a student that has not studied derivative yet (and so not even Taylor formula and Taylor series).
Tags :
Define $f(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$. One possibility is to take $f(x)$ as the definition of $e^x$. Since the OP has suggested a different definition, I will show they agree.
If $x=\frac{p}{q}$ is rational, then \begin{eqnarray*} f(x)&=&\lim_{n\to\infty}\left(1+\frac{p}{qn}\right)^n\\ &=&\lim_{n\to\infty}\left(1+\frac{p}{q(pn)}\right)^{pn}\\ &=&\lim_{n\to\infty}\left(\left(1+\frac{1}{qn}\right)^n\right)^p\\ &=&\lim_{n\to\infty}\left(\left(1+\frac{1}{(qn)}\right)^{(qn)}\right)^{p/q}\\ &=&\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^{p/q}\\ &=&e^{p/q} \end{eqnarray*} Now, $f(x)$ is clearly non-decreasing, so $$\sup_{p/q\leq x}e^{p/q}\leq f(x)\leq \inf_{p/q\geq x}e^{p/q}$$ It follows that $f(x)=e^x$.
Now, we have \begin{eqnarray*} \lim_{x\to0}\frac{e^x-1-x}{x^2}&=&\lim_{x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^n-1-x}{x^2}\\ &=&\lim_{x\to0}\lim_{n\to\infty}\frac{n-1}{2n}+\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-2}\\ &=&\frac{1}{2}+\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\\ \end{eqnarray*}
We want to show that the limit in the last line is 0. We have $\frac{{n\choose k}}{n^k}\leq\frac{1}{k!}\leq 2^{-(k-3)}$, so we have \begin{eqnarray*} \left|\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\right|&\leq&\lim_{x\to0}|x|\lim_{n\to\infty}\sum_{k=3}^n \left(\frac{|x|}{2}\right)^{k-3}\\ &=&\lim_{x\to0}|x| \frac{1}{1-\frac{|x|}{2}}\\ &=&0 \end{eqnarray*}
Julian Rosen
August 18, 2012 17:43 PM
Let us call our limit $\ell$.
I was considering the following identity
$$4\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}=\left(\frac{e^x-1}{x}\right)^2\quad\forall x\ne0$$
If $\mathbf{\ell}$ exists and is not infinite, taking the limit of the above identity we have
$$4\ell-2\ell=1\implies\ell=\frac{1}{2}$$
but I am not able to prove the bold part above (if at all possible, in a simple way).
enzotib
August 19, 2012 10:36 AM
Consider fundamental limit: $e = \lim\limits_{n\to \infty}(1+\frac{1}{n})^n$ and $e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$
Proof
$e^x = [\lim\limits_{k\to\infty}(1+1/k)^k]^x = \lim\limits_{k\to \infty}((1+1/k)^{kx})\Rightarrow kx = n \Rightarrow e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$.
Understand the first expression:
$P = \large\frac{e^x-1}{x}$
Note that $e^x - 1 - x = x.[\large\frac{(e^x-1)}{x} - 1]\,\,\therefore\,\,$ $\boxed{\lim\limits_{x\to 0}\frac{e^x-1-x}{x^2}=\lim\limits_{x\to 0}\frac{P-1}{x}}$
Lets go to understand the expression $\,\,P-1$.
$P - 1= \frac{e^x - 1}{x} - 1 = \lim\limits_{n\to\infty}\left(\large\frac{[(1+\frac{x}{n})^n - 1]}{x} - 1\right)=$
Using that tool:
$\boxed{b^n - 1 = (b-1).(b^{n-1}+b^{n-2}+...+1)}$
$=\lim\limits_{n\to\infty}\left((1+\frac{x}{n}-1).\large\frac{[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + {1+x/n}]}{x}-1 \right) =\\ \\ = \lim\limits_{n\to\infty}\left(\frac{1}{n}.[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)]-1\right) = \\ \\ =\lim\limits_{n\to\infty}\frac{1}{n}.\left((1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)-n\right)$
Writing the last "$n$" as $\underbrace{1+1+1...+1}_{n\,\, times}$ and inputing these $1`s$ into it:
$P-1 = \lim\limits_{n\to\infty} (1/n).[((1+x/n)^{n-1} - 1)+ ((1+x/n)^{n-2} - 1) + ... + ((1+x/n) - 1)]$
Using again that tool in each expression:
$=\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n}) [((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... +1)+((1+x/n)^{n-3}+...+1)+...+1]$
Finally,
$L = \lim\limits_{x\to 0}\frac{P-1}{x} =\lim\limits_{x\to 0}\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}]=$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to0}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}] =\\$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to 0}\left(\frac{1}{n^2}\right).((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1) =$
$=\lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1 + n-2 + n-3 + ... + 1) = \lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1)(\frac{n}{2}) = \lim\limits_{n\to\infty}\frac{n-1}{2n} = \boxed{\large\frac{1}{2}}$.
miguel747
January 13, 2014 03:22 AM
$$\displaylines{ \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x - x - 1}}{{x^2 }} = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2t - 1}}{{t^2 }} \cr = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1 - 1 - 2t - 1}}{{t^2 }} \cr = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1}}{{t^2 }} - 2\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t - t - 1}}{{t^2 }} \cr \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}$$
$$m = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1}}{{t^2 }} - 2m \Leftrightarrow m = \frac{1}{2}$$
sabachir
January 05, 2015 11:58 AM
I thought it might be useful to present a way forward that relies on an integral representation of the numerator along with the mean-value theorem for integrals. To that end, we now proceed.
Note that we can write the numerator as
\begin{align} e^x-x-1&=\int_0^x \int_0^t e^s \,ds\,dt\\\\ &=\int_0^x \int_s^x e^s\,dt\,ds\\\\ &=\int_0^x (x-s)e^s\,ds \end{align}
Next, we apply the Mean-Value-Theorem for integrals to reveal
\begin{align} e^x-x-1&=e^{s^*}\int_0^x(x-s)\,ds\\\\ &=\frac12 x^2e^{s^*} \end{align}
for some value of $s^*\in (0,x)$.
Finally, exploiting the continuity of the exponential function yields the coveted limit
\begin{align} \lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{\frac12 x^2e^{s^*}}{x^2}\\\\ &=\frac12 \end{align}
as expected!
Dr. MV
November 14, 2016 22:14 PM
Accidentally I came across this post and I thought of how to prove the statement $$\lim_{x\rightarrow 0} \frac{e^x-1-x}{x^2} = \frac12 .$$ assuming only that the function $e^x$ satisfies the two properties: $$e^{x+y}=e^x e^y \ \mbox{and} \ \lim_{x\rightarrow 0} \frac{e^x-1}{x} = 1$$
It turns out that it is possible using $\sum_{k=0}^{n-1} = \frac{n(n-1)}{2}$ and elementary algebraic properties of limits, but being very careful with the uniform bounds for these limits. The proof, although elementary, is not simple so is probably not of much practical use. Also, all the difficulties are hidden in the existence of the function $e^x$ verifying the functional equation. Anyway, I post it for the curios reader.
To start, note that the second property for $e^x$ is equivalent to the following: Write $R(x) = e^x-1-x$. Then for $\delta>0$ and $|x|\leq \delta$ we have the uniform bound: $|R(x)|\leq \Delta(\delta)$ with a function $\Delta$ that verifies: $$\lim_{\delta\rightarrow 0} \frac{\Delta(\delta)}{\delta} \rightarrow 0.$$
By the above definitions we also have $|e^x|\leq M(\delta) \equiv 1+\delta+\Delta(\delta) <+\infty$.
Fix $x\neq 0$, $L=\Delta(|x|)/|x|$, $M=M(|x|)$ and let $n\geq 1$. Using the functional equation for $e^x$ we may rewrite $e^x-1=e^{nx/n}-1$ as a telescopic sum:
$$e^x-1= \sum_{k=0}^{n-1} e^{\frac{k}{n} x} \left( e^{\frac{x}{n}} -1\right)= \sum_{k=0}^{n-1} \left( 1+ \frac{k}{n}x + R(\frac{k}{n}x) \right) \left( \frac{x}{n} + R(\frac{x}{n}) \right)$$ Developping the RHS and using $\sum_{k=0}^{n-1} k = \frac{n^2-n}{2}$ we get the expression $x + \frac{n-1}{2n} x^2$ plus an error term which is bounded by $$\sum_{k=0}^{n-1} \left[ \Delta( \frac{k}{n}|x|) \times (1+L) \frac{|x|}{n}+ e^{\frac{k}{n} x} \Delta(\frac{|x|}{n}) \right] \leq n \Delta(|x|) \times (1+L) \frac{|x|}{n} + M n \times \Delta(\frac{|x|}{n})$$ Therefore, $$\left| \frac{e^x-(1+x+x^2/2)}{x^2} \right| \leq \frac{x^2}{2n} + (1+L) \frac{\Delta(|x|)}{|x|} + M \frac{1}{|x|} \frac{\Delta(|x|/n)}{|x|/n}$$
Now let $n\rightarrow \infty$ (keeping $x\neq 0$ fixed). By the properties of the function $\Delta$, the first and the last terms on the RHS goes to zero and as the LHS is independent of $n$ we deduce: $$\left|\frac{e^x-(1+x+x^2/2)}{x^2} \right| \leq (1+L(|x|)) \frac{\Delta(|x|)}{|x|} .$$ The RHS goes to zero as $|x|$ goes to zero, and this implies the stated limit.
Remark: Incidently one may use the same telescopic procedure, i.e. without binomial expansion, to show that for $x$ fixed, $e^x - (1+\frac{x}{n})^n \rightarrow 0$ as $n\rightarrow \infty$.
H. H. Rugh
March 11, 2017 11:00 AM | 3,839 | 8,812 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-39 | latest | en | 0.582081 |
https://justaaa.com/physics/396875-a-pair-of-figure-skaters-glide-holding-hands-on-a | 1,713,001,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00074.warc.gz | 302,885,800 | 10,092 | Question
# A pair of figure skaters glide holding hands on a frictionless ice rink, at a speed...
A pair of figure skaters glide holding hands on a frictionless ice rink, at a speed of 3.0 m/s and heading due East in a straight line. They then push away from each other. Skater A, of mass 48 kg, glides off at a speed of 3.8 m/s and at angle of 29 degrees counterclockwise with respect to the original direction. Skater B has a mass of 72 kg.
a. After they push off from each other, what is the velocity of skater B?
b. What is the velocity of their combined center of mass after they separate?
Here we have given that,
Their initial speed of Vi = 3.0 m/s
Skater A mass 48 kg
final speed of A Vf= 3.8 m/s
angle = 29° degrees counterclockwise
Skater B mass = 72kg.
a. Now using conservation of momentum here we will get,
Conserving the momentum along east direction we will get
Pi = pf
Vi(mA+mB) = mAVAfCos29° + mBVbf
Here only
On plugging the value we will get,
Vbf = 2.78429 m/s
b. Now the velocity of centre of mass after they seperated is given as,
Vcm = mAvAf +mBvBf/mA+mA
on plugging the values we will get,
Vcm = 3.190 m/s | 337 | 1,149 | {"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} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.832199 |
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# Direct conversion RF MIxer
Status
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#### Foufou
##### Member level 5
Hi, I am DEsigning an RF mixer (Gilbert Mixer) using ADS of Agilent but the problem is that in ADS the IF output can't be 0Hz (DC output) and an error message is appearing.
What is the solution to Do Direct conversion by ADS?
Thanks
make a very samll difference between 2 , abot 1 Khz or even less , and in this case they are very close
khouly
Foufou
at DC there is no information.
Consider the bandwidth of your received signal, let's say BW at center freq CF
After the RF downmixer you will have CF= 0 Hz and the signal bandwidth will be [-BW/2,BW/2]. So all your signal analysis (ac, noise, linearity...) have to be performed in [0,BW/2] bandwidth (we are neglecting negative frequencies because IQ signals are helping us!).
I hope it will help.
Mazz
yes Mazz
I know, but my problem is in ADS and it Dc output, I think I will try the solution of khouly.
And I have an other question. And always using ADS, when we define the correct DC biais values of the RF and the LO port to have transistors working in saturation for exemple Vdclo=1.2V and VdcRF=1.5V, when I finish the hole design, how could I replace the DCs tensions used ? using Vdd tension of my circuits
thinks
Iknow that to bias the DC comonent we sould add the transistors like in figure attached,
but the question is how could I manipulate W/L to have the Desired tensions
Help and exemple
thanks[/img][/GVideo]
W/L ratio is depended in technology,desired Vgs, current etc.
There isn't enough information here..
You need to design a bias for your circuit.
Bias can be a voltage or a current and usually stays some chapter BEFORE the design of amplifier and some chapter BEFORE RF design. OK, I'm joking, but it is true.
You should always consider the equations that keep MOS in saturation and properly fix some variable; for example, if you fix current and W/L you will obtain a certain Vgs (this is a very simplified explanation).
I hope it can help.
Mazz
Status
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Data Structures And Algorithm Notes
Questions and Topics
```NOT - Bits that are 0 become 1, and those that are 1 become 0. Used to unset/turn off bit flags.
AND - Multiplication of bits; i.e., 1 × 0 = 0 and 1 × 1 = 1. Used to check state of a bit flag.
OR - 1 if either bit is 1, otherwise 0. Used to set/turn on bit flags regardless of whether already on or not.
XOR - 1 if both bits are different, 0 if they are the same. Used to toggle flag bits of a mask.
Arithmetic Shifts: left << (multiplies by 2, 0 shifted on as least significant bit), right >> (divides by 2, left-most sign bit is retained)
Logical Shift: right >>> (divides by 2, 0 shifted on as most significant bit)```
Q. What's the difference between a LinkedList and an ArrayList (or between a linked list and an array/vector)?
```An ArrayList is a List implementation backed by a Java array. With a LinkedList, the List implementation is backed by a
Q. How can you detect a cycle in a linked list?
```Turtle and Rabbit: The idea is to have two references to the list and move them at different speeds. Move one forward by 1
node and the other by 2 nodes. If the linked list has a loop they will meet. If there is no loop either of the references
will become null when it reaches the end of the list. O(n)```
Q. What is a weighted round robin load balancing algorithm?
```Round robin distribution is used by DNS servers, peer-to-peer networks, and many other multiple-node clusters/networks.
In a weighted round-robin algorithm, each destination (in this case, server) is assigned a value that signifies, relative
to the other servers in the list, how that server performs. This "weight" determines how many more (or fewer) requests are
sent that server's way; compared to the other servers on the list. ```
```root, children (enqueue/add), children of children (dequeue/remove first and enqueue), etc... Use a fifo queue
Typical applications:
- Shortest path between two vertices.```
```public class BreadthFirstPaths {
private boolean[] marked;
private int[] edgeTo;
...
private void bfs(Graph G, int s) {
Queue<Integer> q = new Queue<Integer>();
q.enqueue(s);
marked[s] = true;
while (!q.isEmpty()) {
int v = q.dequeue();
for (int w : G.adj(v)) {
if (!marked[w]) {
q.enqueue(w);
marked[w] = true;
edgeTo[w] = v;
}
}
}
}
}
```
```root, children (push) to leaf nodes, left, then up (pop) and right... Use a stack
+
3 4
preorder/prefix (Polish notation, e.g. + 3 4), postorder/postfix (reverse Polish notation, e.g. 3 4 +),
inorder/infix (e.g. 3 + 4) - expression tree
Recursive algorithm.
Typical applications:
- Find all vertices connected to a given source vertex.
- Find a path between two vertices.
After DFS, should be able to:
- Find vertices connected to root/source in constant time
- Find a path back to root/source in time proportional to its length```
```public class DepthFirstPaths {
private boolean[] marked;
private int[] edgeTo;
private int s;
public DepthFirstSearch(Graph G, int s) {
...
dfs(G, s);
}
private void dfs(Graph G, int v) {
marked[v] = true;
if (!marked[w]) {
dfs(G, w);
edgeTo[w] = v;
}
}
}
```
Q. Name some self balancing binary trees
`Red/black tree, splay tree, AVL tree`
Q. What is Big-O Notation?
```Defines best, average and typically worst case performance of an algorithm.
Given N we might have, N, N log N, N^2 performance, for example.```
```Split in 2, recurse left, recurse right, merge sorted halves.
Can be "stable" - i.e. retains existing sorts if order of equal items is not changed, and no long distance exchanges are made.
Java sorts Objects with a merge sort algorithm. N log N.```
```public class MergeSort {
public static void sort(Object[] a, Comparator comparator) {
Object[] aux = new Object[a.length];
sort(a, aux, comparator, 0, a.length - 1);
}
private static void sort(Object[] a, Object[] aux, Comparator c, int lo, int hi) {
// Optimize sort for small subarrays (CUTOFF = 7 perhaps):
if (hi <= lo + CUTOFF - 1) {
Insertion.sort(a, c, lo, hi);
return;
}
int mid = lo + (hi - lo) / 2;
sort(a, aux, c, lo, mid);
sort(a, aux, c, mid + 1, hi);
if (!less(c, a[mid + 1], a[mid])) return;
merge(a, aux, c, lo, mid, hi);
}
private static void merge(Object[] a, Object[] aux, Comparator c, int lo, int mid, int hi) {
assert isSorted(a, c, lo, mid); // precondition: a[lo..mid] sorted
assert isSorted(a, c, mid + 1, hi); // precondition: a[mid+1..hi] sorted
// Copy to working array:
for (int k = lo; k <= hi; k++)
aux[k] = a[k];
// Merge two arrays:
int i = lo, j = mid + 1;
for (int k = lo; k <= hi; k++) {
if (i > mid) a[k] = aux[j++];
else if (j > hi) a[k] = aux[i++];
else if (less(c, aux[j], aux[i])) a[k] = aux[j++];
else a[k] = aux[i++];
}
assert isSorted(a, c, lo, hi); // postcondition: a[lo..hi] sorted
}
...
}
```
```Randomly shuffle array to prep process and avoid worst case comparisons (~1/2 * N^2) which becomes very unlikely to happen (if shuffle actually
sorted the array!)
Then pick a partitioning element as a divide/pivot point, less vs. greater, exchange elements that are out of place, recurse.
Is not "stable".
Faster than Merge sort because less movement of items even though more compares in average case. N log N. Too many duplicate keys can also
affect efficiency.
Java sorts primitive types with a quick sort algorithm.```
```public class QuickSort {
private static int partition(Comparable[] a, int lo, int hi) {
int i = lo, j = hi + 1;
while (true) {
while (less(a[++i], a[lo])) // find item on left to swap
if (i == hi) break;
while (less(a[lo], a[--j])) // find item on right to swap
if (j == lo) break;
if (i >= j) break; // check if pointers cross
exch(a, i, j); // swap items
}
exch(a, lo, j); // swap with partitioning item
return j; // return index of item now known to be in place
}
public static void sort(Comparable[] a) {
StdRandom.shuffle(a); // shuffle ensures performance is good
sort(a, 0, a.length - 1);
}
private static void sort(Comparable[] a, int lo, int hi) {
// Optimize sort for small subarrays (CUTOFF between 10 and 20):
if (hi <= lo + CUTOFF - 1) {
Insertion.sort(a, lo, hi);
return;
}
int j = partition(a, lo, hi);
sort(a, lo, j - 1);
sort(a, j + 1, hi);
}
}
```
Q. Sorting Summary
Q. Implement a hashtable (dictionary)
`An array of linked lists, indexed by a hash, where hash collisions are placed in buckets in the linked list.`
`left subtree nodes < parent node <= right subtree nodes. Efficient for sorting, searching in-order.`
```public class BST<Key extends Comparable<Key>, Value> {
private Node root; // root of BST
private class Node {
private Key key; // sorted by key
private Value val; // associated data
private Node left, right; // left and right subtrees
public Node(Key key, Value val, int N) {
this.key = key;
this.val = val;
this.N = N;
}
}
// return value associated with the given key, or null if no such key exists
public Value get(Key key) {
return get(root, key);
}
private Value get(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) return get(x.left, key);
else if (cmp > 0) return get(x.right, key);
else return x.val;
}
/***********************************************************************
* Insert key-value pair into BST
* If key already exists, update with new value
***********************************************************************/
public void put(Key key, Value val) {
root = put(root, key, val);
}
private Node put(Node x, Key key, Value val) {
if (x == null) return new Node(key, val, 1);
int cmp = key.compareTo(x.key);
if (cmp < 0) x.left = put(x.left, key, val);
else if (cmp > 0) x.right = put(x.right, key, val);
else x.val = val;
return x;
}
}
```
Q. Heap (priority queue)
```Complete binary tree (i.e. only bottom level might not be complete, left to right),
heap ordered (i.e. parent values > children values)```
Q. Graphs
`trees with interconnecting nodes`
Q. Graph algorithms, such as Dijkstra and A*
```Best-first search is a search algorithm which explores a graph by expanding the most promising node chosen
according to a specified rule. A* uses a best-first search and finds the least-cost path from a given initial
node to one goal node (out of one or more possible goals). It uses a distance-plus-cost heuristic function
(usually denoted f(x)) to determine the order in which the search visits nodes in the tree.
Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1959,[1] is a graph search
algorithm that solves the single-source shortest path problem for a graph with nonnegative edge path costs,
producing a shortest path tree. This algorithm is often used in routing.```
Q. NP-complete problems, such as traveling salesman and the knapsack problem
```Traveling Salesman problem: Given a list of cities and their pairwise distances, the task is to find a shortest
possible tour that visits each city exactly once.
Knapsack problem: Given a set of items, each with a weight and a value, determine the number of each item to
include in a collection so that the total weight is less than a given limit and the total value is as large as
possible.``` | 2,456 | 9,397 | {"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} | 3.34375 | 3 | CC-MAIN-2019-30 | longest | en | 0.920329 |
https://newproxylists.com/calculus-how-to-solve-0-frac-a-2-2-x-2-y-2-x-2-x-2-y-2-a-2-y-only-in-terms-of/ | 1,560,859,463,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998716.67/warc/CC-MAIN-20190618103358-20190618125358-00419.warc.gz | 532,105,574 | 8,782 | # calculus – How to solve \$ 0 = frac {(a ^ {2} -2 (x ^ {2} + y ^ {2})) x} {(2 (x ^ {2} + y ^ {2}) + a ^ {2}) y} \$ only in terms of?
I find the derivative of $${y}$$ w.r.t. $${x}$$ of $$(x ^ {2} + y ^ {2}) ^ {2} = a ^ {2} (x ^ {2} + y ^ {2})$$ be
$${y} (x) = frac {(a ^ {2} -2 (x ^ {2} + y ^ {2})) x} {(2 (x ^ {2} + y ^ 2)) + a ^ {2}) y}$$
where y is a function of x and a is a positive constant. The question then asks for points where the original function is parallel to the x-axis. After setting the derivative above to 0, I end with a result that has x in terms of y and a. Solutions give $$x = pm frac {1} {4} a sqrt {6}$$. How can you solve this problem only in terms of? Thank you in advance for any help! (apologize for any poor latex code) | 278 | 753 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-26 | latest | en | 0.802877 |
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