url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
ajeshashok.com | 1,603,560,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00414.warc.gz | 216,201,056 | 12,173 | Popular
Finance
Cars
Hobby
Food and drink
Spiritual development
Arts & Entertainment
Spiritual development
Finance
Hobby
# Absolute and relative magnitude
In economic science, statistical disciplinesare in priority positions. This is due to various reasons. First of all, in the framework of general economic specialties, statistical research serves as a basis for the development and improvement of analytical methods. In addition, they are an independent direction, having its own subject.
## Absolute and relative values
These concepts act as key elements in thestatistical science. They are used to determine the quantitative characteristics, the dynamics of their change. Absolute and relative values reflect different characteristics, but without some, there can not be others. The former express the quantitative dimensions of a phenomenon without regard to others. On them it is impossible to estimate the occurring changes and deviations. They express the volume and level of the process or phenomenon. Absolute values are always named numbers. They have a dimension or a unit of measurement. They can be natural, labor, money and so on. For example, the norm-hours, pcs., Thousand rubles. and so on. The average and relative values, on the contrary, express the ratio of several exact dimensions. It can be established for several phenomena or for one, but taken in a different volume and in a different period. These elements act as a quotient of statistical numbers, which characterizes their quantitative relationship. To determine relative values, one size must be divided into another, accepted as the base one. The latter can be planned data, actual information from previous years or another enterprise, and so on. The relative value of the comparison can be expressed as a percentage (with a base taken as 100) or as coefficients (if base is one).
## Classification of statistical numbers
Absolute values are presented in two types:
1. Individual. They characterize the size of the trait in specific units. For example, it can be the salary of an employee, a deposit in a bank, and so on. These dimensions are found directly in the course of statistical observation. They are recorded in the primary accounting records.
2. Total. Values of this type reflect the total indicator of the characteristic for the totality of objects. These dimensions act as the sum of the number of units (population size) or the volume of the variable characteristic.
## Units
Natural absolute values can besimple. This, for example, tons, liters, rubles, pieces, kilometers. They can also be complex, characterize a combination of several quantities. For example, in the statistics, tonne-kilometers are used to establish the freight turnover of the railway transport, kilowatt-hours to estimate the production of electricity and so on. In research, conditionally-natural units are also used. For example, a tractor fleet can be recalculated into standard cars. Value units are used to characterize a heterogeneous commodity in monetary terms. This form, in particular, is used in assessing the income of the population, the gross output. Using value units, statisticians take into account the dynamics of prices over time, and overcome the disadvantage due to "comparable" or "unchanged" prices for the same period. Labor values take into account the overall costs of work, the labor intensity of certain operations that make up the technological cycle. They are expressed in man-days, man-hours and so on.
## Relative values
The main condition for their calculation is thethe comparability of units and the existence of a real connection between the phenomena under investigation. The value with which the comparison is made (the denominator in fractions) appears, as a rule, as the basis or basis of the relationship. Depending on its choice, the result can be expressed in different parts of unity. It can be tenths, hundredths (percentages), thousandth (10th part% - per mille), ten thousandths (one hundredth percent - prodecimille). The units that are to be compared can be either single-sided or heteronymous. In the second case, their names are formed from the units used (q / ha, rubles / person, etc.).
## Types of Relative Values
In statistics, several types of these units are used. So, there is a relative value:
1. Structures.
3. Intensity.
4. Dynamics.
5. Coordination.
6. Comparisons.
7. Degrees of economic development.
The relative value of the job expresses the ratioplanned for the forthcoming period to actually existing for the current period. Similarly, a plan unit is calculated. Relative value of the structure is a characteristic of the share of specific parts of the studied population in its total volume. Their calculation is carried out by dividing the number in separate parts by the total number (or volume). These units are expressed as a percentage or a simple multiple ratio. For example, this is how the share of the urban population is calculated.
## Dynamics
The relative value reflects in this casethe ratio of the level of the object in a particular period to its status in the past tense. In other words, the change in the phenomenon during any period is characterized. The relative magnitude that characterizes dynamics is called the rate of growth. The choice of the base for the calculation is carried out depending on the purpose of the study.
## Intensity
The relative value may reflect the degreethe development of a phenomenon in a particular environment. In this case we speak of intensity. Their calculation is performed by comparing the opposite quantities, which are in communication with each other. They are installed, as a rule, per 1000, 100 and so on units of the population under study. For example, per 100 hectares of land, per thousand people and so on. These relative values are named numbers. For example, this is how the population density is calculated. It is expressed by the average number of citizens per 1 sq. Km. km of the territory. The characteristics of the degree of economic development serve as a subtype of such units. For example, they include such types of relative values as GNP, GDP, VID, and so on. per capita. These characteristics play an important role in analyzing the economic situation in the country.
## Coordination
The relative values cancharacterize the proportionality of the individual elements of the whole to each other. The calculation is carried out by dividing one part into another. Relative values in this case act as a subtype of intensity units. The difference is that they reflect the level of distribution of dissimilar parts of one set. The basis may be one or another feature, depending on the goal. In this connection, several relative coordination values can be calculated for the same whole.
## Comparison
Relative values of comparison are units,which are separate subdivisions of the same statistical characteristics that act as characteristics for different objects, but refer to a single moment or period. For example, the ratio of the cost level of a particular type of product produced by two enterprises, labor productivity for different industries, and so on is calculated.
## Economic evaluation
In this study,absolute and relative units. The former are used to establish the ratio of reserves and costs to sources of financing and to assess the enterprise by the level of monetary stability. Relative indicators reflect the structure of funds with the state of fixed and working capital. Economic evaluation uses horizontal analysis. As the most general absolute value, characterizing the financial stability of the company, there is a shortage or surplus of sources of financing of costs and stocks. The calculation is done by subtraction. The result is a difference in the size of sources (minus non-current assets), with the resources of which inventories are formed, and their quantity. The key elements in this are the following statistical units:
1. Own current assets.
2. The total indicator of planned sources.
3. Long-term borrowed funds and own funds.
## Deterministic Factor Study
This analysis is a specificmethod of studying the impact of factors, the interaction of which with the results has a functional character. This study is carried out by the creation and evaluation of deterministic models. In this analysis, relative indicators are widely used. In most cases, multiplicative models are used in factor analysis. For example, the profit can be expressed by the product of the quantity of goods per unit cost. Part of the analysis in this case is conducted in 2 ways:
1. The absolute difference method assumes a chainsubstitution. The change in the result due to the factor is calculated as the product of the deviation of the studied trait to the base of the other in the selected sequence.
2. The method of relative differences is used when measuring the effect of factors on the increase in the result. It is applied when the initial data contains the previously calculated deviations in percent.
## Dynamic series
They represent a change in the numericalindicators of social phenomena over time. As one of the most important directions in this analysis is the study of the features of the development of events for specific periods. Among them:
1. Rates of growth. This is a relative indicator, which is calculated by dividing the two levels in one row on top of each other. They can be calculated as chain or as basic. In the first case, the comparison of each level of the series with the preceding one is made. In the second case, the base is selected. All levels in the series are compared with one that acts as the basis. Growth rates are expressed in terms of coefficients or percentages.
2. Absolute increase. It is the difference between the two levels of the dynamic series. Depending on the method of choice of the base, it is basic and chain. This indicator has the same dimension as the levels of the series.
3. Rates of growth. This relative indicator reflects the amount of interest that one level of the dynamic series is greater / less than the other, which is taken as the base.
## Conclusion
Undoubtedly, the relative quantities possesshigh scientific value. However, in practice they can not be used apart. They are always in interrelation with absolute indicators, expressing the relations of the latter. If this is not taken into account, it is impossible to accurately characterize the phenomena under investigation. Using relative values, it is necessary to show which particular absolute units are hidden behind them. Otherwise, you can draw wrong conclusions. Only the comprehensive use of relative and absolute values can act as an important means of information and analysis in the study of various phenomena occurring in socio-economic life. In general, the transition to the calculation of deviations allows us to compare the economic potential and the result of the activity of enterprises that differ significantly in terms of the resources used or other characteristics. Relative values, in addition, can smooth out some processes (force majeure, inflation, etc.), which can distort the absolute units in the financial statements.
Similar news | 2,206 | 11,279 | {"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} | 3.5625 | 4 | CC-MAIN-2020-45 | latest | en | 0.912949 |
https://notstatschat.rbind.io/2017/11/25/secret-santa-collisions/ | 1,716,970,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00281.warc.gz | 360,036,444 | 4,715 | # Secret Santa collisions
Attention Conservation Notice: while this probability question actually came up in in real life, that’s just because I’m a nerd.
“Secret Santa” is a Christmas tradition for taming the gift-giving problem in offices, groups of acquaintances, etc. Instead of everyone wondering which subset of people they should give a gift to, each person is randomly assigned one recipient and has to give a gift (with a relatively low upper bound on cost) to that one person. The ‘Secret’ part is that you don’t know who is going to give you the gift.
It’s not completely trivial (though it’s not that hard) to come up with a random process that assigns everyone exactly one recipient and ensures that no-one is left finding a gift for themselves. One procedure is to sample recipients without replacement to get a list that’s guaranteed to be one-to-one, and then just repeat the sampling until you get an allocation where no-one is assigned themselves.
So, a probability question: how likely is it that a random permutation will have ‘collisions’ where someone is their own Secret Santa? Or, equivalently, how many tries would you expect to need to get a working allocation? Does it depend on the number of people $$n$$? What about for 3600 people, as in the scheme hosted by NZ Post on Twitter.
I knew the answer, but I didn’t know a proof, so this is a reasonably honest exploration of how you might find the answer.
First, is there some simple bound? Well, the chance that you, Dear Reader, get assigned yourself is $$1/n$$, so the simple Bonferroni bound is $$n\times 1/n$$. That doesn’t look very helpful, because we already knew 1 was an upper bound; this is a probability. However, we can recast the Bonferroni bound as the expected number of collisions. If the expected number of collisions is 1, then it’s reasonable to expect that the probability of no collisions is appreciable, and that as $$n$$ increases it converges to some useful number that isn’t 1 or 0. At this point, if you had to make a wild guess, a reasonable guess would be that the number of collisions has approximately a Poisson distribution for large $$n$$, so that the probability of no collisions will be approximately $$e^{-1}$$.
Next, try simulation. Doing $$10^5$$ replicates in R gives
The probability is 1 if you try to do this by yourself, 1/2 if you have one friend, and converges astonishingly quickly to a fixed value. The red line is $$1-e^{-1}$$.
Now, can we prove this? The collisions aren’t independent, so we can’t quite just use a Poisson or Binomial argument. We could try the higher-order Bonferroni bounds, eg
$P(\cup\_i A\_i)\geq \sum\_i P(A\_i) - \sum_{i,j} P(A\_i\cap A\_j).$
The second-order bound gives $$n\times 1/n-{n\choose 2}\times 1/n^2=1/2$$, so we’re making progress. The third-order bound gives
$n\times 1/n-{n\choose 2}\times 1/n^2+{n\choose 3}\times 1/n^3=1-1/2+1/6=4/6$
We’re definitely making progress now, and these bounds look suspiciously like the values of the probability for $$n=1,\,2,\,3$$. Continuing the pattern, we will end up with
$\sum_{k=1}^\infty (-1)^{k+1}{n\choose k} n^{-k}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{n^{-k}}$
Comparing that to the infinite series for $$\exp(x)$$, it’s $$1-e^{-1}$$. What we haven’t got this way is whether the number of collisions is Poisson, which was where the guess originally came from. This seems to be a lot harder.
First, simulation confirms that the Poisson is a good approximation. That’s reassuring: it’s typically easier to prove things that are true than things that are not true.
I should next say that I spent quite a long time looking for ‘coupling’ arguments to show that this assignment method gave the same large-sample distribution as some other assignment method that was obviously Poisson. I didn’t get anywhere with this, but it’s a fruitful approach for problems similar to this one. Since that didn’t work, we’re back to combinatorics.
Now, suppose we have 1 collision. That means we have an assignment of the $$n-1$$ other people with no collisions. So, the number of ways to have 1 collision in $$n$$ people is $$n$$ times the number of ways to have no collisions in $$n-1$$ people. Since the number of permutations of $$n$$ people is also $$n$$ times the number of permutations of $$n-1$$ people, the fraction of permutations with 1 collision in $$n$$ people is the same as the fraction of permutations with 0 collisions in $$n-1$$ people, ie, for large $$n$$ it’s $$e^{-1}$$. That agrees with the Poisson formula, so we’re definitely making progress.
If we have $$k$$ collisions in $$n$$ people then we have a set of $$n-k$$ people with no collisions. The number of ways we can have $$n-k$$ people with no collisions is about $$(n-k)!e^{-1}$$ and the number of ways to pick the $$k$$ people who have to buy themselves gifts is $${n\choose k}= n!/(k!(n-k)!)$$. So, the number of ways of having $$k$$ collisions in $$n$$ is about $$n!e^{-1}/k!$$, and the probability of this is $$e^{-1}/k!$$, matching the Poisson distribution. We’ve done it!
Ok. Technically we’re not quite done, since we’d need to show the approximation error in the argument actually gets smaller with increasing $$n$$. But we’re done enough for me.
PS: If you want an easy way not to have to worry about collisions, just arrange the people in random order and assign each person the following person in the list (with the last person being assigned the first). | 1,386 | 5,455 | {"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} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.952198 |
http://oeis.org/A318359 | 1,555,793,219,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530040.33/warc/CC-MAIN-20190420200802-20190420222802-00454.warc.gz | 132,037,292 | 3,963 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A318359 a(1) = 4; for n > 1, a(n) is the least positive number not yet in the sequence such that Sum_{k=1..n} a(k) divides Sum_{k=1..n} a(k)^3. 2
4, 1, 5, 8, 9, 12, 6, 20, 10, 3, 39, 65, 52, 11, 7, 42, 147, 441, 294, 366, 222, 35, 514, 257, 1285, 771, 3084, 672, 99, 925, 608, 291, 2061, 229, 495, 140, 81, 288, 12088, 1511, 750, 476, 209, 2603, 752, 7645, 1079, 5816, 2210, 2830, 1996, 1162, 328, 3690 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Is this sequence infinite? For any v > 0, let b_v be the variant of this sequence starting with v: - b_4 = a (this sequence), - b_1 = A000027 (and indeed, A000217(n) divides A000537(n) for any n > 0), - for v in {1, 2, 3, 5}: b_v(n) = n for all n > 0 except a finite number. This sequence is a variant of A318358. LINKS Rémy Sigrist, Table of n, a(n) for n = 1..500 EXAMPLE For n = 3: - (4^3 + 1^3 + 2^3) == 3 mod (4 + 1 + 2), - (4^3 + 1^3 + 3^3) == 4 mod (4 + 1 + 3), - (4^3 + 1^3 + 5^3) == 0 mod (4 + 1 + 5), - hence a(3) = 5. PROG (PARI) s=0; s3=0; p=0; v=4; for (n=1, 54, print1 (v ", "); s+=v; s3+=v^3; p+=2^v; for (w=1, oo, if (!bittest(p, w) && (s3+w^3)%(s+w)==0, v=w; break))) CROSSREFS Cf. A000027, A000217, A000537, A318358. Sequence in context: A011443 A016687 A139356 * A209297 A243525 A300071 Adjacent sequences: A318356 A318357 A318358 * A318360 A318361 A318362 KEYWORD nonn AUTHOR Rémy Sigrist, Aug 24 2018 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified April 20 16:17 EDT 2019. Contains 322310 sequences. (Running on oeis4.) | 783 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-18 | latest | en | 0.594346 |
https://codeforces.com/blog/rahulkhairwar | 1,575,614,739,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00229.warc.gz | 319,947,839 | 18,537 | ### rahulkhairwar's blog
By rahulkhairwar, history, 20 months ago, ,
Hi, Is anyone else also facing the issue where in Problemset[Last Unsolved], problems that we had a wrong submission for earlier, but solved it later on, are still shown?
MikeMirzayanov, could you look into the issue please?
Thank you.
Upd [23/04/2018] : The bug still exists. Upd 2 [23/04/2018] : FIXED! Thank you Mike! :D
• +28
By rahulkhairwar, history, 2 years ago, ,
I was trying to solve this problem. Found it under the DP category on a2oj. But, after trying for a long time, I couldn't come up with any solution to the problem.
• +5
By rahulkhairwar, history, 3 years ago, ,
There was a contest by Codenation today (13th Nov, 2016) and there were 3 questions to be solved in 1.5 hours.
Q1) Given a string s, 2 players can remove a single character turn by turn where player 1 starts first. But the condition to remove is that each player can remove a character which is at an index >= the index which was removed previously, and the string gets re indexed after each turn. In the 1st turn player 1 can remove any character. If the final string is lexicographically greater than the original string s, player 1 wins, else player 2 wins. We had to tell who will be the winner.
Constraints :
number of tests cases <= 10
length of string <= 100
I couldn't come up with a complete logic during the contest. How can we solve this question?
Q3) We are given n points on a straight line : a0, a1, ...., a(n — 1) (a[i] > 0). Also, there is a starting point, s = 0, and an ending point t (> a[n]). Out of the n + 2 points, we have to choose K + 2 points such that the minimum distance between any 2 consecutive points from the chosen ones is maximum. Also, s and t are always chosen, so essentially we have to choose K out of the n points a[0....n — 1].
For this, I thought of a dp solution :
First I added 0(the starting point) at a0, and shifted the rest by 1 place. Then,
Let dp(i, j, k) = min. dist. b/w any 2 consecutive points by using up to the ith point, with jth point being the previous one, and choosing k points so far.
And the base condition will be that to choose exactly 1 point when i >= 1 and j >= 0, dp[i][j][1] = a[1]. I calculated this dp in iterative manner.
Then, I keep track of answer as :
ans = INFINITY;
for (int i = 0; i < n; i++)
for (int j = 0; j < i; j++)
ans = min(ans, dp[i][j][K]);
Constraints :
1 <= n <= 600 1 <= k <= 100 1 <= t <= 11000 1 <= a[i] < t
But this could clear only 2 out of 3 sample cases. Can someone tell me what's wrong with this logic?
Thank You.
• +7
By rahulkhairwar, history, 3 years ago, ,
I was trying this question and this is the solution I came up with, but it timed out as it doesn't use memoization.
But I'm not getting the idea about how to memoize this solution. And I can't even find any good explanation for the question anywhere. Can someone please help?
Thanks.
• +1
By rahulkhairwar, history, 4 years ago, ,
I was trying to learn Mo's Algorithm from here and this was one of the suggested problems. But, I'm getting TLE verdict with Java, while my code is very similar to the one that the tutorial's author has written, in C++.
Can someone please tell me what am I doing wrong(slow)? Thanks.
UPDATE : Sorry, forgot to post a link for my code
• 0
By rahulkhairwar, history, 4 years ago, ,
I was wondering why does codeforces have registration for contests, unlike some other sites like codechef, hackerearth allowing participants to participate even in ongoing competitions.
• +18
By rahulkhairwar, history, 4 years ago, ,
I was solving this question on dp and I use Java as my preferred language for programming.
But, I got a stack overflow error when I submitted the solution in Java, while the same code when converted into C++, ran perfectly fine.
Here is the Java submission : link
and here is the C++ submission : link
So, I searched on the internet and found out that Java's stack space is very less. :/
So, is there any way I could solve this question recursively in Java? I don't want to switch to C++ as I am very comfortable with Java and really like to code in it.
Java users, do you always implement an iterative version of an algorithm which could be solved recursively? | 1,134 | 4,269 | {"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} | 3.15625 | 3 | CC-MAIN-2019-51 | longest | en | 0.944571 |
https://www.physicsforums.com/threads/quiky-question-before-my-test-tommorow-morning.43957/ | 1,511,417,804,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00281.warc.gz | 784,387,320 | 14,048 | # Quiky question before my test tommorow morning
1. Sep 21, 2004
### EasyStyle4747
quiky question before my test tommorow morning!!!
i can't believe i duno where to find out how to do this:
1) find intersections of the graphs:
x^2+y^2=41
x+2y=6
and
2)which of the following function is equivalent to the following piece function.
x-3, if x>=3
f(x)={
3-x, if x<3
a) g(x)=|x-3| b) g(x)=|x+3| c) g(x)=-|x-3| d)-|x+3| e) none of these
this is off like a study guide checklist for the test.
2. Sep 21, 2004
### recon
Rearrange x + 2y = 6 to get x = 6 - 2y. Then plug this into x2 + y2 = 41. You get:
(6 - 2y)2 + y2 - 41 = 0 | 244 | 632 | {"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} | 3.453125 | 3 | CC-MAIN-2017-47 | longest | en | 0.757145 |
https://static.tinkutara.com/question/0-1-Log-Log-2-x-x-x-5-x-4-x-3-x-2-x-1-dx-Anyone-.htm | 1,721,479,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00652.warc.gz | 455,839,162 | 2,509 | # Question and Answers Forum
Others Questions
Question Number 171665 by Mastermind last updated on 19/Jun/22
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} {Log}\left(\frac{{Log}^{\mathrm{2}} \left({x}\right)}{{x}^{{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}} }\right){dx} \\$$$$\\$$$${Anyone}? \\$$
Answered by mindispower last updated on 19/Jun/22
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{ln}\left(−{ln}\left({x}\right)\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}\right)\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {ln}\left({x}\right){e}^{−{x}} {dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\left(\frac{{x}^{\mathrm{6}} }{\mathrm{6}}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right){dx} \\$$$$=\mathrm{2}\partial_{{a}} \int_{\mathrm{0}} ^{\infty} {x}^{{a}−\mathrm{1}} {e}^{−{x}} {dx}\mid_{{a}=\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{36}}−\frac{\mathrm{1}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1} \\$$$$=\mathrm{2}\partial_{{a}} \Gamma\left({a}\right)\mid_{{a}=\mathrm{1}} −\frac{\mathrm{5}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{25}} \\$$$$=\mathrm{2}\Gamma'\left(\mathrm{1}\right)−\frac{\mathrm{5}}{\mathrm{6}}+\frac{\mathrm{9}}{\mathrm{400}}=−\mathrm{2}\gamma+\frac{−\mathrm{2000}+\mathrm{54}}{\mathrm{2400}} \\$$$$=−\mathrm{2}\gamma−\frac{\mathrm{973}}{\mathrm{1200}} \\$$$$\\$$
Commented by Mastermind last updated on 19/Jun/22
$${is}\:{this}\:{my}\:{question}? \\$$$${if}\:{yes},\:{you}\:{did}'{nt}\:{compose}\:{it}\:{well} \\$$
Commented by mindispower last updated on 19/Jun/22
$${ln}\left(−{ln}\left({x}\right)\right)\:{value}\:{of}\:{integral}\:{is}\:{coorect} \\$$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com | 861 | 2,075 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.188368 |
https://www.scribd.com/document/257657776/Edexcel-AlevelPHY4-Waves-and-Our-Universe | 1,547,972,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583700734.43/warc/CC-MAIN-20190120062400-20190120084400-00300.warc.gz | 918,056,409 | 46,304 | You are on page 1of 4
# råáí=QW
t~îÉë=~åÇ=çìê=råáîÉêëÉ
An understanding of the principles in this unit and in the general requirements is expected in familiar and
unfamiliar situations. In examination questions where the context is beyond the content of the
specification it will be fully described.
Suggested activities
4.1
Circular motion and
oscillations
Angular speed, period,
frequency.
ω = ∆θ/∆t υ = ωr
T = 2π/ω f = ω/2π
4.2
Acceleration and resultant force.
For a body moving at a constant speed in
a circular path, acceleration a = υ2/r.
Resultant force F towards the centre of
the circle equals mυ2/r.
Idea of apparent weightlessness when
objects are in free fall.
Investigation – effect
of m, υ and r of orbit
on centripetal force.
KW / KPS KN KC KIT
4.3
Simple harmonic motion.
Variation of displacement, velocity and
acceleration with time, treated
graphically. An understanding that
s.h.m. results when acceleration is
proportional to displacement and in the
opposite direction.
Motion sensor to
generate v/t and x/t
graphs of SHM.
Computer modelling.
KN KIT KLP
4.4
Frequency and period for simple
harmonic motion.
The equation a = −(2πf)2x; T = 1/f.
4.5
Undamped simple harmonic
oscillations.
x = x0 cos 2πft
Angles expressed in both degrees and
Maximum speed = 2πfx0
Velocity as gradient of displacementtime graph.
graph.
Computer modelling
or graphic calculator
to explore the effects
of changing x0 and f.
KN KIT
4.6
Mechanical oscillators.
Experimental study of a simple
pendulum.
T = 2π√(l/g) for small amplitude
oscillations.
Investigation – effect
of m, l and amplitude
on T.
KLP
Experimental study of mass-spring
system.
T = 2π=√(m/k) where k is the spring
stiffness.
Investigation – effect
of m and k on T.
KPS KN KC KIT
28
UA006823 – Specification – AS/A GCE in Physics – Issue 3 – September 2002
Plane polarisation. along springs and in air (sound). Research – Sources/ Detectors and uses of various regions of the e. KC KIT The emphasis should be on the generation of waves and the transmission of energy in the medium. c = fλ Wave machine or computer simulation of wave properties. Photoelastic stress analysis. speed.7 Free and forced vibrations. frequency and phase interpreted graphically. Plane polarisation as a property of transverse waves demonstrated using light and microwaves. Research – problems and applications. Suggested activities Vibration generator – forced oscillations of a mass spring system (damped and undamped). Polaroid to illustrate polarisation of light.11 An experimental and qualitative knowledge of mechanical resonance. spectrum. Energy flux or intensity measured in W m-2 Intensity I = P/4πr2 Investigation – inverse square law using a light meter. microscopy. Progressive waves: longitudinal and transverse waves. 4. 4. Research –models of structures (stress concentration). Demonstrate loudspeaker and waves on long spring/ ripple tank. KW / KPS KN KC KIT UA006823 – Specification – AS/A GCE in Physics – Issue 3 – September 2002 29 .10 4. KC KIT The electromagnetic spectrum. Electromagnetic waves. Damping. It is expected that electromagnetic wave phenomena will be demonstrated using visible light and microwaves.m.9 4. An outline of the whole spectrum: the order of magnitude of the wavelengths of the main regions should be known. Mechanical resonance. Inverse square law. Amplitude. KC KIT Conservation of energy for waves in free space from a point source.8 Waves Mechanical waves on water. wavelength.4.
4. Interpretation of the observed patterns using the principle of superposition.14 Stationary waves and resonance. Demonstrated using water waves.16 Two slit interference patterns. 4. Vibration generator with rubber cord or wire loop to demonstrate standing waves. 4.17 Quantum phenomena The photon model of electromagnetic radiation. Measurements of wavelength with both visible light and microwaves. 30 UA006823 – Specification – AS/A GCE in Physics – Issue 3 – September 2002 . Appreciation of dependence of the width of central maximum on relative sizes of slit and wavelength. all points on a wavefront are in phase. Work function and the photoelectric equation. microwaves and light. Graphical treatment. Maximum energy of photoelectrons = hf − ϕ Discharge of coulombmeter using UV lamp. Measure wavelength of light using a laser. Demonstrate using spreadsheets and graphs.13 Wavefronts. KC KW / KPS KIT By definition. Microwaves or sound waves to show ‘free’ standing waves.12 Superposition of waves The principle of superposition. Research – Einstein. Ripple tank to illustrate two source interference. Wavefronts are most effectively illustrated using a ripple tank. The photoelectric effect. Phase difference and path difference. ∆ (phase) = 2 π∆ ( path) λ 4. λ = xs/D for light.15 Diffraction at a slit. The concept of stopping potential and its measurement. illustrated by the overlapping of two sets of circular waves on water. Simple demonstration (eg discharge of negatively charged zinc plate). Coherence. The Planck constant. Laser and adjustable slit. Demonstration using waves on strings and microwaves. the limitations should be understood but the proof is not required. Half-wavelength separation of nodes or antinodes should be known but need not be derived. Knowledge of experimental details and typical dimensions is expected. E = hf Measure the Planck constant using photocell.Suggested activities 4. KC KIT 4.
Doppler shift of spectral lines. Electron diffraction. with H in s-1. Wave-particle duality.Suggested activities Direct vision spectroscope or diffraction grating to observe spectra of gases and vapours.21 Electromagnetic Doppler effect. λ = h/p Research – model development KC KIT Stationary waves in the hydrogen atom. 4. The age of the Universe: uncertainty in d and H. Research – discovery of pulsars. 4. Wave properties of electrons in atoms. The Big Bang. Analogy with trapped standing waves on a wire loop. Observation of electron diffraction using suitable vacuum tube. Possible energy states of an atom as fixed and discrete. ∆f/f = ∆λ/λ = υ/c. Hubble’s law. Research – the discovery of helium. where υ is the speed of a star or galaxy towards or away from the Earth. Model expansion by inflating a balloon. Open and closed Universes.19 Wave properties of free electrons. 4. KC KIT Expansion of the Universe. The average mass-energy density of the Universe: indefinite expansion or final contraction. The light year. Star spectra emission and absorption lines and their relation to chemical composition. The electronvolt. KC KIT 4. The red shift of galaxies υ = Hd. hf = E1 − E2 The emission and absorption line spectrum of atomic hydrogen related to electron energy levels.20 The expanding Universe Optical line spectra.18 Energy levels. The de Broglie wavelength. Demonstrate the Doppler effect by whirling a small loudspeaker on the end of a chain or wire.22 UA006823 – Specification – AS/A GCE in Physics – Issue 3 – September 2002 31 . KC KIT 4. Demonstration of spectra using a diffraction grating. | 1,670 | 7,054 | {"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-2019-04 | latest | en | 0.847517 |
https://www.jianshu.com/p/e4f4ca3f6b52 | 1,618,429,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00229.warc.gz | 924,293,881 | 45,134 | # Pyautogui - Python 操作鼠标键盘
## 安装
``````pip install python3–xlib
apt–get install scrot
apt–get install python3–tk
apt–get install python3–dev
pip install pyautogui
``````
## 鼠标操作
### 鼠标移动
pyautogui 使用 x-y 坐标系，左上角的坐标是`(0, 0)`
#### 获取屏幕分辨率
``````screen_width, screen_height = pyautogui.size()
``````
#### 获取当前鼠标坐标
``````x, y = pyautogui.position()
``````
#### 使用绝对坐标移动(`moveTo`)
``````for i in range(10):
pyautogui.moveTo(300, 300, duration=0.25)
pyautogui.moveTo(400, 300, duration=0.25)
pyautogui.moveTo(400, 400, duration=0.25)
pyautogui.moveTo(300, 400, duration=0.25)
``````
``````screen_width, screen_height = pyautogui.size()
print('screen_width: %s\nscreen_height: %s' % (screen_width, screen_height))
r = 250 # 圆半径
o_x = screen_width / 2 # 圆心X坐标
o_y = screen_height / 2 # 圆心Y坐标
pi = math.pi
for i in range(10):
for angle in range(0, 360, 20):
# 利用圆的参数方程
x = o_x + r * math.sin(angle * pi / 180)
y = o_y + r * math.cos(angle * pi / 180)
pyautogui.moveTo(x, y, duration=0.1)
``````
#### 使用相对坐标移动(`moveRel`)
``````for i in range(10):
pyautogui.moveRel(100, 0, duration=0.25)
pyautogui.moveRel(0, 100, duration=0.25)
pyautogui.moveRel(-100, 0, duration=0.25)
pyautogui.moveRel(0, -100, duration=0.25)
``````
``````try:
while True:
x, y = pyautogui.position()
print(x, y)
except KeyboardInterrupt:
print('\nExit.')
``````
### 鼠标点击
``````pyautogui.click(x=cur_x, y=cur_y, button='left')
``````
• `x`, `y`是要点击的位置，默认是鼠标当前位置
• button 是要点击的按键，有三个可选值： `left`, `middle`, `right`，默认是`left`
``````pyautogui.click(button='right')
``````
``````pyautogui.click(100, 100)
``````
• `pyautogui.doubleClick()`: 双击
• `pyautogui.rightClick()`: 右击
• `pyautogui.middleClick()`: 中击
### 鼠标拖拽
#### 使用绝对坐标拖拽(`dragTo`)
``````# 画一个回字
pyautogui.moveTo(300, 300, duration=0.25)
pyautogui.dragTo(400, 300, duration=0.25)
pyautogui.dragTo(400, 400, duration=0.25)
pyautogui.dragTo(300, 400, duration=0.25)
pyautogui.dragTo(300, 300, duration=0.25)
pyautogui.moveTo(200, 200, duration=0.25)
pyautogui.dragTo(500, 200, duration=0.25)
pyautogui.dragTo(500, 500, duration=0.25)
pyautogui.dragTo(200, 500, duration=0.25)
pyautogui.dragTo(200, 200, duration=0.25)
``````
#### 使用相对坐标拖拽(`dragRel`)
``````# 画一个回字
pyautogui.moveTo(300, 300, duration=0.25)
pyautogui.dragRel(100, 0, duration=0.25)
pyautogui.dragRel(0, 100, duration=0.25)
pyautogui.dragRel(-100, 0, duration=0.25)
pyautogui.dragRel(0, -100, duration=0.25)
pyautogui.moveTo(200, 200, duration=0.25)
pyautogui.dragRel(300, 0, duration=0.25)
pyautogui.dragRel(0, 300, duration=0.25)
pyautogui.dragRel(-300, 0, duration=0.25)
pyautogui.dragRel(0, -300, duration=0.25)
``````
### 滚轮
``````pyautogui.scroll(200)
``````
### 根据像素颜色定位按钮位置
``````img = pyautogui.screenshot() # 截屏
img_color = img.getpixel((300, 300)) # 获取坐标颜色 (48, 10, 36)
is_matched = pyautogui.pixelMatchesColor(300, 300, (48, 10,36)) # 判断屏幕坐标的像素颜色是不是等于某个值
print(is_matched) # True
``````
### 根据按钮图片定位按钮位置
``````corn_locate = pyautogui.locateOnScreen('corn/settings.png') # 找到按钮所在坐标，分别含义是按钮左上角x坐标，左上角y坐标，x方向大小，y方向大小 (5, 560, 54, 54)
corn_center_x, corn_center_y = pyautogui.center(corn_locate) # 找到按钮中心点
pyautogui.click(corn_center_x, corn_center_y) # 点击按钮
``````
## 键盘操作
### 键盘按键
#### 输入普通字符串
``````pyautogui.typewrite('Hello, world!', 0.25) # 0.25表示每输完一个字符串延时0.25秒
``````
#### 输入特殊字符
`enter`/`return`/`\n` 回车
`esc` ESC键
`shiftleft`/`shiftright` 左右SHIFT键
`altleft`/`altright` 左右ALT键
`ctrlleft`/`ctrlright` 左右CTRL键
`tab`/`\t` TAB键
`backspace`/`delete` BACKSPACE/DELETE键
`pageup`/`pagedown` PAGE UP/PAGE DOWN键
`home`/`end` `HOME`/`END`
`up`/`down`/`left`/`right` 上下左右键
`f1`/`f2`/... F1/F2/...
`volumemute`/`volumedown`/`volumeup` ？？
`pause` PAUSE键
`capslock`/`numlock`/`scrolllock` CAPS LOCK/NUM LOCL/SCROLL LOCK键
`insert` INSERT键
`printscreen` PRINT SCREEN键
`winleft`/`winright` 左右Win键
`command` Mac上的command键
``````pyautogui.click(100, 100)
pyautogui.typewrite('Hello, world!', 0.25)
pyautogui.typewrite(['enter', 'a', 'b', 'left', 'left', 'X', 'Y'], '0.1')
``````
#### 键盘的按下和释放
• `keyDown()`: 按下某个键
• `keyUp()`: 松开某个键
• `press()`: 一次完整的按键，前面两个函数的结合
``````pyautogui.keyDown('altleft')
pyautogui.press('f4')
pyautogui.keyUp('altleft')
``````
``````pyautogui.hotkey('altleft', 'f4')
`````` | 1,827 | 4,224 | {"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-2021-17 | latest | en | 0.339912 |
https://www.daenotes.com/electronics/basic-electronics/coulombs-law-of-electrostatics | 1,575,896,876,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518882.71/warc/CC-MAIN-20191209121316-20191209145316-00196.warc.gz | 678,536,463 | 9,754 | # Coulomb Laws
Charles coulomb formulated two laws on the bases of experiments, which are known as electrostatic laws, or coulomb laws of electrostatic.
## COULOMB FIRST LAW OF ELECTROSTATICS
"It state that like charges of electricity repel each other and unlike charges attract each other".
## COULOMB SECOND LAW OF ELECTROSTATICS
It state that the force of attraction b/w unlike charges or force of repulsion between like charges are directly proportional to the product of the both charges and inversely Proportional to the square of the distance b/w center of the charges.
## EXPLANATION
Suppose Q1 and Q2 are two unlike charges placed at a distance ‘d’ apart from each other as shown in the fig,
F ∝ Q1Q2
F ∝ 1 / d2
Combing these two relation we get
Where "k" is a constant depending upon the permittivity of the medium and its value is given by,
Where "ε" is the permittivity of any medium, now
εr = ε / εo
ε = εoεr
Put this value of ε in above eq, we get | 248 | 976 | {"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} | 3.453125 | 3 | CC-MAIN-2019-51 | latest | en | 0.928906 |
https://socratic.org/questions/how-do-you-solve-the-inequality-15-c-4-15 | 1,579,524,803,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598726.39/warc/CC-MAIN-20200120110422-20200120134422-00092.warc.gz | 662,781,490 | 6,295 | # How do you solve the inequality -15 <= c/-4 - 15?
Jan 1, 2016
The solution set is: $S = \left\{c \in \mathbb{R} | c \le 0\right\}$
#### Explanation:
In this specific case, we first make the inequality easier, simplifying it:
$- 15 \le \frac{c}{-} 4 - 15$ multiply it by 4: $- 60 \le \frac{c}{-} 1 - 60$
$\frac{c}{-} 1 = - c$. So, we pass the $- 60$ to the other side of the inequality and solve it:
$- 60 + 60 \le - c$
$0 \le - c$
In order to make the $c$ to be positive, just multiply it by -1. But, when this is made, the inequality signal is reversed - If it is $<$, becomes $>$, etc. Now, in this case:
$0 \le - c$ multiplied by $- 1$:
$0 \ge c$
$c \le 0$
So, the solution set is: $S = \left\{c \in \mathbb{R} | c \le 0\right\}$ | 284 | 742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-05 | longest | en | 0.79735 |
https://www.simscale.com/knowledge-base/predict-darcy-and-forchheimer-coefficients-for-perforated-plates-using-experimental-data/ | 1,709,372,945,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00599.warc.gz | 992,787,729 | 63,591 | Required field
Required field
Required field
Required field
• Set up your own cloud-native simulation in minutes.
• Documentation
# How to Predict Darcy and Forchheimer Coefficients for Perforated Plates Using Experimental Data?
## Curve Fitting Approach
Pressure loss through the porous media is calculated by the following equation:
$$\Delta P = \mu \cdot d \cdot L \cdot u + \frac{1 }{2} \cdot \rho \cdot f \cdot L \cdot u^{2} \tag{1}$$
where:
• $$\Delta P$$ : Pressure drop $$[Pa]$$
• $$\mu$$ : Dynamic viscosity of fluid $$[\frac{kg}{ms}]$$
• $$L$$ : Thickness of perforated plate $$[m]$$
• $$u$$ : Velocity of fluid $$[m/s]$$
• $$\rho$$ : Density of fluid $$[\frac{kg}{m^3}]$$
• $$d$$ : Darcy coefficient
• $$f$$ : Forchheimer coefficient
Using the experimental data for pressure loss vs. velocity one can use the curve fitting method to extract $$d$$ and $$f$$ coefficients.
The equation of the curve should be in the following format:
$$\Delta P = A \cdot u + B \cdot u^{2}$$
where:
• $$A = \mu \cdot d \cdot L \cdot u$$
• $$B = \frac{1 }{2} \cdot \rho \cdot f \cdot L \cdot u^{2}$$
To create a curve, a minimum of three data points are needed. Using the curve-fitting method, $$A$$ and $$B$$ coefficients (followed by $$d$$ and $$f$$) will be predicted. The more data points you have, the more accurate the predictions can be.
One can use spreadsheet solvers or online curve fitting tools or any other curve fitting software. Just type the pressure versus velocity values from experimental data and let the solver predict $$A$$ and $$B$$ coefficients. In the above figure, an online tool MyCurveFit $$^1$$ is used to predict $$A$$ and $$B$$ by user defined equation (y=A*x+B*x^2). While y values represent pressure drop, x values represent velocity. In this example, $$A$$ and $$B$$ coefficients are predicted as 7.4 and 6.4 respectively.
Once $$A$$ and $$B$$ coefficients are found, extract Darcy $$d$$ and Forchheimer $$f$$ coefficients from the following equations:
$$d = \frac{A}{\mu \cdot L}$$
$$f = \frac{2B}{\rho \cdot L}$$
References
Last updated: June 14th, 2021 | 588 | 2,099 | {"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": 2, "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-2024-10 | latest | en | 0.762225 |
http://slidegur.com/doc/37121/mav---mathculator | 1,519,106,114,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00139.warc.gz | 345,724,436 | 8,775 | ### MAV - mathculator
```Making Your Students
Feel Like A genius
( And maybe you )
What level of maths do I need
to survive this session ?
DO I HAVE TO BE REALLY
GOOD AT
MATHS TO SURVIVE THIS
CLASS ???
If you
can add and subtract one digit numbers
Basic knowledge of fractions
Can use a calculator
(to check if you are really correct
and smarter than you think)
Want to learn
Are willing to approach things positively
•
•
•
•
•
•
• Think of a number
• Double it
• Write down a 3 digit number
with digits decreasing, eg. 861
• Reverse this number and
•
•
subtract it from the first
number
the reverse of the number
• Using 1089 from before
•
•
multiply this by ANY 3 digit
number
How many digits in your
Tell me any 5 of your digits
• Write down any 2 1 digit number
•
•
•
them 8 times.
Eg
1.
6
2.
5
3. 11
When you have done this add the
10 numbers up
I need someone to write their
number on the board Without the
total.
I will Guess the total before finish
they are finished writing all the
number up.
• Write down any number from
•
•
•
•
each of the 3 number sets
A
B
C
8573
5646
2565
2966
4179
4581
4586
8544
2673
4874
9624
5625
7655
2595
7236
Select any digit from each of the
select numbers and enter under
Columns headed A, B or C
Repeat this till all digits are used.
Add up the new 4 3-digit numbers
Answer for this total is 2528
Multiplying 2 digit Numbers
Eg.
11
13
x 12
x 13
13 2
16 9
ADD 11 plus 2 ( = 13)
Multiply 1 x 2 ( = 2 )
Multiplying 2 digit Numbers
Eg.
17
19
x 13
x 19
21
81
20
28
221
361
ADD 17 plus 3 ( = 20)
Multiply 7 x 3 ( = 21 )
Squaring a number ending in 5
Eg. 752 = 7 x 8 and 52
= 56 and 25
= 5625
(N5)2 = n x (n+1) and 52
= n x (n+1) and 25
= n(n+1)25
Squaring a number between 30 and 70
Eg. 522 = 27 and 22
= 2704
[27 : 25+2] [04:2places for 4]
592
592
632
=?
= 34 and 92
= 3481
= 38 and 132
169
= 38
= 3969
632
= ?
[34 : 25+9] [81:2places for 81]
[38 : 25+13] [169:2places for 169]
Squaring a number between 30 and 70
Eg. 482 = 23 and 22
= 2304
[23 : 25-2] [04:2places for 4]
472
=?
392
=?
472 = 22 and 32 [22 : 25-3] [04:2places for 9]
= 2209
392 = 14 and 112 [14: 25-11] [121:2places for 121]
121
= 14
= 1521
Squaring a number between 80 and 120
Eg. 1032 = 106 and 32
= 10609
[06 : two x 3] [09:2places for 9]
1072 = 11449 [14 : two x 7]
[49:2places for 7x7]
1092 = 11881
1122
144
= 124
= 12544
Multiplying 2 numbers between 80 and 120
97 x 98 = 98 ( -2 : less than 100)
97 ( -3 : less than 100)
= 9506
[95 : 97- 3 : 98 - 2] [06:2places for 6]
91 x 97 = 91
97
= 8827
( -9 : less than 100)
( -3 : less than 100)
[88 : 91- 3 : 97 - 9] [27:2places for 6]
Multiplying 2 numbers between 80 and 120
107 x 112 = 107
112
= 11984
( 7 : morethan 100)
( 12 : more than 100)
[19 : 7+12 : 12 + 7] [84: 7 x 12]
103 x 115 = 103
( 3 : more than 100)
115 ( 15 : more than 100)
= 11845
[18 : 3 + 15 : 15 + 3] [45 : 3 x 15]
Squaring a number between 80 and 120
Eg. 1032 = 106 and 32
= 10609
[06 : two x 3] [09:2places for 9]
1072 = 11449 [14 : two x 7]
[49:2places for 7x7]
1092 = 11881
1122
144
= 124
= 12544
55 times 55 = ???
552 = 5 x 6 and 52
= 30 and 25
= 3025
55 times 75 = ???
55 x 75 = 5 x 7 + 6 and 52
= 41 and 25
= 4125
Multiplying 2 mixed numbers with the
same units and fractions adding to 1
Multiplying a number ending
in digits that add to 10
Eg 75 x 75
= 7 x 8 and 5 x 5
= 56 and 25
= 5625 (as before)
84 x 86 = ?
84 x 86
= 8 x 9 and 4 x 6
= 72 and 24
= 7224
Squaring 2 digit numbers
Eg 732 = 76 x 70 + 32
= 5320 + 9
= 5329
or 732
= 702 +2(70 x 3) + 32
= 4900 + 2(210) + 9
= 5329
(ab)2 = (a+b)(a-b) + b2
or
(ab)2 = a2 + 2(a+b) + b2
Thank you and Have a great holiday
``` | 1,552 | 3,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-09 | latest | en | 0.717856 |
https://www.atlaspm.com/slides2/ | 1,721,309,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00442.warc.gz | 567,270,772 | 13,790 | ``` Slide 201 [Counting only text (non-numeric, non-error, non-empty) entries in a range] ```
``` Slide 202 [Formatting numbers for Indian Rupee.] Slide 203 [Converting seconds to minutes and seconds, or to hours, minutes, and seconds] Slide 204 [Counting numbers in a list that are positive, negative, or zero] Slide 205 [Counting odd and even numbers in a range] Slide 206 [Summing and averaging the last 3 (for example) numbers along a row] Slide 207 [Returning the year's week number of a date] Slide 208 [Verifying a text entry's true upper case] Slide 209 [Conditionally formatting a month's number as even or odd based on dates] Slide 210 [Getting information about a month based on one of its dates] Slide 211 [Analyzing data based on specified alphanumeric characters] Slide 212 [Summing or averaging numbers in alphanumeric strings beginning with specified characters] Slide 213 [Using the spacebar operator to sum the intersection of two named ranges.] Slide 214 [Formatting a zero to look like -0- (with a dash on either side)] Slide 215 [Expanding the formula bar -- two methods] Slide 216 [Averaging numbers between upper and lower limits] Slide 217 [Summing and averaging numbers based on dates in a specified month] Slide 218 [Returning the last row of a filtered table] Slide 219 [Using the name box to enter and get help with functions] Slide 220 [Parsing a decimalized number into its integer and decimal portions] Slide 221 [Toggling to show or hide the Ribbon -- two methods] Slide 222 [Listing a few rules and tips for VBA variable names] Slide 223 [Explaining VBA's OR logical operator] Slide 224 [From a list of values, randomly return any one cell's value] Slide 225 [Explaining the VBA IIF (yes, that's I-I-F) conditional function] Slide 226 [Generating a random whole number between any two numbers] Slide 227 [Conditionally format the largest number in each row] Slide 228 [Creating a list without blanks from a list with blanks] Slide 229 [Filling the space in a cell from its left edge to the start of its numeric contents] Slide 230 [Filling cell contents with a trailing character] Slide 231 [Preventing the addition of new worksheets] Slide 232 [Printing pages with the workbook's name in all footers] Slide 233 [Automatically go to cell A1 for any activated worksheet] Slide 234 [Controlling the times of day a workbook can be saved] Slide 235 [Pre-sorting your UserForm's ListBox and ComboBox items] Slide 236 [Showing the Add-Ins dialog box with a keyboard shortcut] Slide 237 [Parsing a cell's value into 1 character per cell] Slide 238 [Showing your computer's System information with a keyboard shortcut] Slide 239 [Exporting data from each worksheet into its own text file] Slide 240 [Recording the dates and times in a text file that your workbook gets saved] Slide 241 [Formatting cell values to their column width without changing font size or column width] Slide 242 [Showing your File Explorer directory with a keyboard shortcut] Slide 243 [Finding your Excel Templates path] Slide 244 [Converting military time to conventional AM/PM time] Slide 245 [Looking up the column and row header labels of minimum and maximum numbers in a range] Slide 246 [Showing an example of the PMT function] Slide 247 [Returning the first and last numbers in a row, ignoring text and blank cells] Slide 248 [Looking up the oldest date for a specified name having many repeats] Slide 249 [Looking up the cell address of a unique item in a list] Slide 250 [Calculating ages in years, months, and days] Slide 251 [Returning everything to the right of the nth character] Slide 252 [Finding the row of the first empty cell in a list] Slide 253 [Counting multiple criteria items in a list with various other items] Slide 254 [Summing every other cell along a row] Slide 255 [Doing a horizontal lookup to find a cell's address] Slide 256 [Extracting the first word in a string, or the entire text if there are no spaces] Slide 257 [Summing two cells, but if both cells are blank, the sum cell is blank not showing zero] Slide 258 [Creating a named range with the keyboard shortcut Ctrl+Shift+F3] Slide 259 [Searching for a word or string inside another string] Slide 260 [Using the last number in a list in a formula] Slide 261 [Fusing 2 staggered lists into 1 list with Paste Special's Skip blanks] Slide 262 [Bypassing data validation with a custom formatted reminder text for improper data entry.] Slide 263 [Further customizing Social Security Numbers beyond the built-in Special category] Slide 264 [Counting cells in a range that contain an error value] Slide 265 [Calculating the percentage of numbers in a list that are positive, negative, and zero] Slide 266 [Rounding a number to its nearest Nth whole number] Slide 267 [Isolating and/or omitting parentheses and their contents] Slide 268 [Converting a decimal number to an AM/PM time] Slide 269 [Summing numbers in one column depending on text values in 3 other columns on the same row] Slide 270 [Concatenating text with a formatted percentage calculation] Slide 271 [Counting weekdays in a list of dates] Slide 272 [Returning the first positive or negative number in a list] Slide 273 [Counting numbers in a list that are greater than or equal to a specified number] Slide 274 [Summing numbers in one column for a matching criteria date of varying formats] Slide 275 [Counting specified text in two columns on the same row] Slide 276 [Counting dates in a list that fall into a specified year] Slide 277 [Counting occupied or blank cells while disregarding one cell] Slide 278 [Adding or removing borders around selected cells using keyboard shortcuts] Slide 279 [Locating a specified character's first position in a string] Slide 280 [Calculating the absolute sum of differences] Slide 281 [Showing 2 options to round numbers to their nearest hundred thousand] Slide 282 [Showing the Spelling dialog box with a keyboard shortcut] Slide 283 [Showing the Thesaurus in Excel using a keyboard shortcut] Slide 284 [Getting the most-repeated value in a list] Slide 285 [Getting the column number and letter of the last cell in a row that contains a number] Slide 286 [Generating a random time between two times] Slide 287 [Rounding times to their previous, next, and nearest quarter hour] Slide 288 [Filling a range of cells with the value from a single cell] Slide 289 [Finding the row and column with the largest and smallest number] Slide 290 [Ranking to ignore zeroes and blank cells] Slide 291 [Counting for a month, and for a year and month, in a list of dates] Slide 292 [Listing a running count of repeated items] Slide 293 [Counting blank and non-blank cells in a range] Slide 294 [Showing different function results for parsed positive and negative decimal numbers] Slide 295 [Getting your operating system's name and version from Excel] Slide 296 [Counting unique entries in a list, ignoring text] Slide 297 [Getting the first, second, and third most frequent value in a list, considering blank cells] Slide 298 [Returning the last positive or negative number in a list] Slide 299 [Summing a range of numbers formatted as Text] Slide 300 [Showing the HYPERLINK function's link address with descriptive text ] ``` | 1,662 | 7,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.81556 |
http://compcert.inria.fr/doc/html/Fcore_Zaux.html | 1,516,305,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00568.warc.gz | 85,906,119 | 10,900 | # Module Fcore_Zaux
Require Import ZArith.
Require Import Zquot.
Section Zmissing.
Theorem Zopp_le_cancel :
forall x y : Z,
(-y <= -x)%Z -> Zle x y.
Proof.
intros x y Hxy.
apply Zplus_le_reg_r with (-x - y)%Z.
now ring_simplify.
Qed.
Theorem Zgt_not_eq :
forall x y : Z,
(y < x)%Z -> (x <> y)%Z.
Proof.
intros x y H Hn.
apply Zlt_irrefl with x.
now rewrite Hn at 1.
Qed.
End Zmissing.
Section Proof_Irrelevance.
Scheme eq_dep_elim := Induction for eq Sort Type.
Definition eqbool_dep P (h1 : P true) b :=
match b return P b -> Prop with
| true => fun (h2 : P true) => h1 = h2
| false => fun (h2 : P false) => False
end.
Lemma eqbool_irrelevance : forall (b : bool) (h1 h2 : b = true), h1 = h2.
Proof.
assert (forall (h : true = true), refl_equal true = h).
apply (eq_dep_elim bool true (eqbool_dep _ _) (refl_equal _)).
intros b.
case b.
intros h1 h2.
now rewrite <- (H h1).
intros h.
discriminate h.
Qed.
End Proof_Irrelevance.
Section Even_Odd.
Zeven, used for rounding to nearest, ties to even
Definition Zeven (n : Z) :=
match n with
| Zpos (xO _) => true
| Zneg (xO _) => true
| Z0 => true
| _ => false
end.
Theorem Zeven_mult :
forall x y, Zeven (x * y) = orb (Zeven x) (Zeven y).
Proof.
now intros [|[xp|xp|]|[xp|xp|]] [|[yp|yp|]|[yp|yp|]].
Qed.
Theorem Zeven_opp :
forall x, Zeven (- x) = Zeven x.
Proof.
now intros [|[n|n|]|[n|n|]].
Qed.
Theorem Zeven_ex :
forall x, exists p, x = (2 * p + if Zeven x then 0 else 1)%Z.
Proof.
intros [|[n|n|]|[n|n|]].
now exists Z0.
now exists (Zpos n).
now exists (Zpos n).
now exists Z0.
exists (Zneg n - 1)%Z.
change (2 * Zneg n - 1 = 2 * (Zneg n - 1) + 1)%Z.
ring.
now exists (Zneg n).
now exists (-1)%Z.
Qed.
Theorem Zeven_2xp1 :
forall n, Zeven (2 * n + 1) = false.
Proof.
intros n.
destruct (Zeven_ex (2 * n + 1)) as (p, Hp).
revert Hp.
case (Zeven (2 * n + 1)) ; try easy.
intros H.
apply False_ind.
omega.
Qed.
Theorem Zeven_plus :
forall x y, Zeven (x + y) = Bool.eqb (Zeven x) (Zeven y).
Proof.
intros x y.
destruct (Zeven_ex x) as (px, Hx).
rewrite Hx at 1.
destruct (Zeven_ex y) as (py, Hy).
rewrite Hy at 1.
replace (2 * px + (if Zeven x then 0 else 1) + (2 * py + (if Zeven y then 0 else 1)))%Z
with (2 * (px + py) + ((if Zeven x then 0 else 1) + (if Zeven y then 0 else 1)))%Z by ring.
case (Zeven x) ; case (Zeven y).
rewrite Zplus_0_r.
now rewrite Zeven_mult.
apply Zeven_2xp1.
apply Zeven_2xp1.
replace (2 * (px + py) + (1 + 1))%Z with (2 * (px + py + 1))%Z by ring.
now rewrite Zeven_mult.
Qed.
End Even_Odd.
Section Zpower.
Theorem Zpower_plus :
forall n k1 k2, (0 <= k1)%Z -> (0 <= k2)%Z ->
Zpower n (k1 + k2) = (Zpower n k1 * Zpower n k2)%Z.
Proof.
intros n k1 k2 H1 H2.
now apply Zpower_exp ; apply Zle_ge.
Qed.
Theorem Zpower_Zpower_nat :
forall b e, (0 <= e)%Z ->
Zpower b e = Zpower_nat b (Zabs_nat e).
Proof.
intros b [|e|e] He.
apply refl_equal.
apply Zpower_pos_nat.
elim He.
apply refl_equal.
Qed.
Theorem Zpower_nat_S :
forall b e,
Zpower_nat b (S e) = (b * Zpower_nat b e)%Z.
Proof.
intros b e.
rewrite (Zpower_nat_is_exp 1 e).
apply (f_equal (fun x => x * _)%Z).
apply Zmult_1_r.
Qed.
Theorem Zpower_pos_gt_0 :
forall b p, (0 < b)%Z ->
(0 < Zpower_pos b p)%Z.
Proof.
intros b p Hb.
rewrite Zpower_pos_nat.
induction (nat_of_P p).
easy.
rewrite Zpower_nat_S.
now apply Zmult_lt_0_compat.
Qed.
Theorem Zeven_Zpower :
forall b e, (0 < e)%Z ->
Zeven (Zpower b e) = Zeven b.
Proof.
intros b e He.
case_eq (Zeven b) ; intros Hb.
b even *)replace e with (e - 1 + 1)%Z by ring.
rewrite Zpower_exp.
rewrite Zeven_mult.
replace (Zeven (b ^ 1)) with true.
apply Bool.orb_true_r.
unfold Zpower, Zpower_pos. simpl.
now rewrite Zmult_1_r.
omega.
discriminate.
b odd *)rewrite Zpower_Zpower_nat.
induction (Zabs_nat e).
easy.
unfold Zpower_nat. simpl.
rewrite Zeven_mult.
now rewrite Hb.
now apply Zlt_le_weak.
Qed.
Theorem Zeven_Zpower_odd :
forall b e, (0 <= e)%Z -> Zeven b = false ->
Zeven (Zpower b e) = false.
Proof.
intros b e He Hb.
destruct (Z_le_lt_eq_dec _ _ He) as [He'|He'].
rewrite <- Hb.
now apply Zeven_Zpower.
now rewrite <- He'.
Qed.
The radix must be greater than 1
forall r1 r2, radix_val r1 = radix_val r2 -> r1 = r2.
Proof.
intros (r1, H1) (r2, H2) H.
simpl in H.
revert H1.
rewrite H.
intros H1.
apply f_equal.
apply eqbool_irrelevance.
Qed.
Theorem radix_gt_0 : (0 < r)%Z.
Proof.
apply Zlt_le_trans with 2%Z.
easy.
apply Zle_bool_imp_le.
apply r.
Qed.
Theorem radix_gt_1 : (1 < r)%Z.
Proof.
destruct r as (v, Hr). simpl.
apply Zlt_le_trans with 2%Z.
easy.
now apply Zle_bool_imp_le.
Qed.
Theorem Zpower_gt_1 :
forall p,
(0 < p)%Z ->
(1 < Zpower r p)%Z.
Proof.
intros [|p|p] Hp ; try easy.
simpl.
rewrite Zpower_pos_nat.
generalize (lt_O_nat_of_P p).
induction (nat_of_P p).
easy.
intros _.
rewrite Zpower_nat_S.
assert (0 < Zpower_nat r n)%Z.
clear.
induction n.
easy.
rewrite Zpower_nat_S.
apply Zmult_lt_0_compat with (2 := IHn).
apply Zle_lt_trans with (1 * Zpower_nat r n)%Z.
rewrite Zmult_1_l.
now apply (Zlt_le_succ 0).
apply Zmult_lt_compat_r with (1 := H).
Qed.
Theorem Zpower_gt_0 :
forall p,
(0 <= p)%Z ->
(0 < Zpower r p)%Z.
Proof.
intros p Hp.
rewrite Zpower_Zpower_nat with (1 := Hp).
induction (Zabs_nat p).
easy.
rewrite Zpower_nat_S.
apply Zmult_lt_0_compat with (2 := IHn).
Qed.
Theorem Zpower_ge_0 :
forall e,
(0 <= Zpower r e)%Z.
Proof.
intros [|e|e] ; try easy.
apply Zlt_le_weak.
now apply Zpower_gt_0.
Qed.
Theorem Zpower_le :
forall e1 e2, (e1 <= e2)%Z ->
(Zpower r e1 <= Zpower r e2)%Z.
Proof.
intros e1 e2 He.
destruct (Zle_or_lt 0 e1)%Z as [H1|H1].
replace e2 with (e2 - e1 + e1)%Z by ring.
rewrite Zpower_plus with (2 := H1).
rewrite <- (Zmult_1_l (r ^ e1)) at 1.
apply Zmult_le_compat_r.
apply (Zlt_le_succ 0).
apply Zpower_gt_0.
now apply Zle_minus_le_0.
apply Zpower_ge_0.
now apply Zle_minus_le_0.
clear He.
destruct e1 as [|e1|e1] ; try easy.
apply Zpower_ge_0.
Qed.
Theorem Zpower_lt :
forall e1 e2, (0 <= e2)%Z -> (e1 < e2)%Z ->
(Zpower r e1 < Zpower r e2)%Z.
Proof.
intros e1 e2 He2 He.
destruct (Zle_or_lt 0 e1)%Z as [H1|H1].
replace e2 with (e2 - e1 + e1)%Z by ring.
rewrite Zpower_plus with (2 := H1).
rewrite Zmult_comm.
rewrite <- (Zmult_1_r (r ^ e1)) at 1.
apply Zmult_lt_compat2.
split.
now apply Zpower_gt_0.
apply Zle_refl.
split.
easy.
apply Zpower_gt_1.
clear -He ; omega.
apply Zle_minus_le_0.
now apply Zlt_le_weak.
revert H1.
clear -He2.
destruct e1 ; try easy.
intros _.
now apply Zpower_gt_0.
Qed.
Theorem Zpower_lt_Zpower :
forall e1 e2,
(Zpower r (e1 - 1) < Zpower r e2)%Z ->
(e1 <= e2)%Z.
Proof.
intros e1 e2 He.
apply Znot_gt_le.
intros H.
apply Zlt_not_le with (1 := He).
apply Zpower_le.
clear -H ; omega.
Qed.
End Zpower.
Section Div_Mod.
Theorem Zmod_mod_mult :
forall n a b, (0 < a)%Z -> (0 <= b)%Z ->
Zmod (Zmod n (a * b)) b = Zmod n b.
Proof.
intros n a [|b|b] Ha Hb.
now rewrite 2!Zmod_0_r.
rewrite (Zmod_eq n (a * Zpos b)).
rewrite Zmult_assoc.
unfold Zminus.
rewrite Zopp_mult_distr_l.
apply Z_mod_plus.
easy.
apply Zmult_gt_0_compat.
now apply Zlt_gt.
easy.
now elim Hb.
Qed.
Theorem ZOmod_eq :
forall a b,
Z.rem a b = (a - Z.quot a b * b)%Z.
Proof.
intros a b.
rewrite (Z.quot_rem' a b) at 2.
ring.
Qed.
Theorem ZOmod_mod_mult :
forall n a b,
Z.rem (Z.rem n (a * b)) b = Z.rem n b.
Proof.
intros n a b.
assert (Z.rem n (a * b) = n + - (Z.quot n (a * b) * a) * b)%Z.
rewrite <- Zopp_mult_distr_l.
rewrite <- Zmult_assoc.
apply ZOmod_eq.
rewrite H.
apply Z_rem_plus.
rewrite <- H.
apply Zrem_sgn2.
Qed.
Theorem Zdiv_mod_mult :
forall n a b, (0 <= a)%Z -> (0 <= b)%Z ->
(Zdiv (Zmod n (a * b)) a) = Zmod (Zdiv n a) b.
Proof.
intros n a b Ha Hb.
destruct (Zle_lt_or_eq _ _ Ha) as [Ha'|Ha'].
destruct (Zle_lt_or_eq _ _ Hb) as [Hb'|Hb'].
rewrite (Zmod_eq n (a * b)).
rewrite (Zmult_comm a b) at 2.
rewrite Zmult_assoc.
unfold Zminus.
rewrite Zopp_mult_distr_l.
rewrite Z_div_plus by now apply Zlt_gt.
rewrite <- Zdiv_Zdiv by easy.
apply sym_eq.
apply Zmod_eq.
now apply Zlt_gt.
now apply Zmult_gt_0_compat ; apply Zlt_gt.
rewrite <- Hb'.
rewrite Zmult_0_r, 2!Zmod_0_r.
apply Zdiv_0_l.
rewrite <- Ha'.
now rewrite 2!Zdiv_0_r, Zmod_0_l.
Qed.
Theorem ZOdiv_mod_mult :
forall n a b,
(Z.quot (Z.rem n (a * b)) a) = Z.rem (Z.quot n a) b.
Proof.
intros n a b.
destruct (Z_eq_dec a 0) as [Za|Za].
rewrite Za.
now rewrite 2!Zquot_0_r, Zrem_0_l.
assert (Z.rem n (a * b) = n + - (Z.quot (Z.quot n a) b * b) * a)%Z.
rewrite (ZOmod_eq n (a * b)) at 1.
rewrite Zquot_Zquot.
ring.
rewrite H.
rewrite Z_quot_plus with (2 := Za).
apply sym_eq.
apply ZOmod_eq.
rewrite <- H.
apply Zrem_sgn2.
Qed.
Theorem ZOdiv_small_abs :
forall a b,
(Zabs a < b)%Z -> Z.quot a b = Z0.
Proof.
intros a b Ha.
destruct (Zle_or_lt 0 a) as [H|H].
apply Zquot_small.
split.
exact H.
now rewrite Zabs_eq in Ha.
apply Zopp_inj.
rewrite <- Zquot_opp_l, Zopp_0.
apply Zquot_small.
generalize (Zabs_non_eq a).
omega.
Qed.
Theorem ZOmod_small_abs :
forall a b,
(Zabs a < b)%Z -> Z.rem a b = a.
Proof.
intros a b Ha.
destruct (Zle_or_lt 0 a) as [H|H].
apply Zrem_small.
split.
exact H.
now rewrite Zabs_eq in Ha.
apply Zopp_inj.
rewrite <- Zrem_opp_l.
apply Zrem_small.
generalize (Zabs_non_eq a).
omega.
Qed.
Theorem ZOdiv_plus :
forall a b c, (0 <= a * b)%Z ->
(Z.quot (a + b) c = Z.quot a c + Z.quot b c + Z.quot (Z.rem a c + Z.rem b c) c)%Z.
Proof.
intros a b c Hab.
destruct (Z_eq_dec c 0) as [Zc|Zc].
now rewrite Zc, 4!Zquot_0_r.
apply Zmult_reg_r with (1 := Zc).
rewrite 2!Zmult_plus_distr_l.
assert (forall d, Z.quot d c * c = d - Z.rem d c)%Z.
intros d.
rewrite ZOmod_eq.
ring.
rewrite 4!H.
rewrite <- Zplus_rem with (1 := Hab).
ring.
Qed.
End Div_Mod.
Section Same_sign.
Theorem Zsame_sign_trans :
forall v u w, v <> Z0 ->
(0 <= u * v)%Z -> (0 <= v * w)%Z -> (0 <= u * w)%Z.
Proof.
intros [|v|v] [|u|u] [|w|w] Zv Huv Hvw ; try easy ; now elim Zv.
Qed.
Theorem Zsame_sign_trans_weak :
forall v u w, (v = Z0 -> w = Z0) ->
(0 <= u * v)%Z -> (0 <= v * w)%Z -> (0 <= u * w)%Z.
Proof.
intros [|v|v] [|u|u] [|w|w] Zv Huv Hvw ; try easy ; now discriminate Zv.
Qed.
Theorem Zsame_sign_imp :
forall u v,
(0 < u -> 0 <= v)%Z ->
(0 < -u -> 0 <= -v)%Z ->
(0 <= u * v)%Z.
Proof.
intros [|u|u] v Hp Hn.
easy.
apply Zmult_le_0_compat.
easy.
now apply Hp.
replace (Zneg u * v)%Z with (Zpos u * (-v))%Z.
apply Zmult_le_0_compat.
easy.
now apply Hn.
rewrite <- Zopp_mult_distr_r.
apply Zopp_mult_distr_l.
Qed.
Theorem Zsame_sign_odiv :
forall u v, (0 <= v)%Z ->
(0 <= u * Z.quot u v)%Z.
Proof.
intros u v Hv.
apply Zsame_sign_imp ; intros Hu.
apply Z_quot_pos with (2 := Hv).
now apply Zlt_le_weak.
rewrite <- Zquot_opp_l.
apply Z_quot_pos with (2 := Hv).
now apply Zlt_le_weak.
Qed.
End Same_sign.
Boolean comparisons
Section Zeq_bool.
Inductive Zeq_bool_prop (x y : Z) : bool -> Prop :=
| Zeq_bool_true_ : x = y -> Zeq_bool_prop x y true
| Zeq_bool_false_ : x <> y -> Zeq_bool_prop x y false.
Theorem Zeq_bool_spec :
forall x y, Zeq_bool_prop x y (Zeq_bool x y).
Proof.
intros x y.
generalize (Zeq_is_eq_bool x y).
case (Zeq_bool x y) ; intros (H1, H2) ; constructor.
now apply H2.
intros H.
specialize (H1 H).
discriminate H1.
Qed.
Theorem Zeq_bool_true :
forall x y, x = y -> Zeq_bool x y = true.
Proof.
intros x y.
apply -> Zeq_is_eq_bool.
Qed.
Theorem Zeq_bool_false :
forall x y, x <> y -> Zeq_bool x y = false.
Proof.
intros x y.
generalize (proj2 (Zeq_is_eq_bool x y)).
case Zeq_bool.
intros He Hn.
elim Hn.
now apply He.
now intros _ _.
Qed.
End Zeq_bool.
Section Zle_bool.
Inductive Zle_bool_prop (x y : Z) : bool -> Prop :=
| Zle_bool_true_ : (x <= y)%Z -> Zle_bool_prop x y true
| Zle_bool_false_ : (y < x)%Z -> Zle_bool_prop x y false.
Theorem Zle_bool_spec :
forall x y, Zle_bool_prop x y (Zle_bool x y).
Proof.
intros x y.
generalize (Zle_is_le_bool x y).
case Zle_bool ; intros (H1, H2) ; constructor.
now apply H2.
destruct (Zle_or_lt x y) as [H|H].
now specialize (H1 H).
exact H.
Qed.
Theorem Zle_bool_true :
forall x y : Z,
(x <= y)%Z -> Zle_bool x y = true.
Proof.
intros x y.
apply (proj1 (Zle_is_le_bool x y)).
Qed.
Theorem Zle_bool_false :
forall x y : Z,
(y < x)%Z -> Zle_bool x y = false.
Proof.
intros x y Hxy.
generalize (Zle_cases x y).
case Zle_bool ; intros H.
elim (Zlt_irrefl x).
now apply Zle_lt_trans with y.
apply refl_equal.
Qed.
End Zle_bool.
Section Zlt_bool.
Inductive Zlt_bool_prop (x y : Z) : bool -> Prop :=
| Zlt_bool_true_ : (x < y)%Z -> Zlt_bool_prop x y true
| Zlt_bool_false_ : (y <= x)%Z -> Zlt_bool_prop x y false.
Theorem Zlt_bool_spec :
forall x y, Zlt_bool_prop x y (Zlt_bool x y).
Proof.
intros x y.
generalize (Zlt_is_lt_bool x y).
case Zlt_bool ; intros (H1, H2) ; constructor.
now apply H2.
destruct (Zle_or_lt y x) as [H|H].
exact H.
now specialize (H1 H).
Qed.
Theorem Zlt_bool_true :
forall x y : Z,
(x < y)%Z -> Zlt_bool x y = true.
Proof.
intros x y.
apply (proj1 (Zlt_is_lt_bool x y)).
Qed.
Theorem Zlt_bool_false :
forall x y : Z,
(y <= x)%Z -> Zlt_bool x y = false.
Proof.
intros x y Hxy.
generalize (Zlt_cases x y).
case Zlt_bool ; intros H.
elim (Zlt_irrefl x).
now apply Zlt_le_trans with y.
apply refl_equal.
Qed.
Theorem negb_Zle_bool :
forall x y : Z,
negb (Zle_bool x y) = Zlt_bool y x.
Proof.
intros x y.
case Zle_bool_spec ; intros H.
now rewrite Zlt_bool_false.
now rewrite Zlt_bool_true.
Qed.
Theorem negb_Zlt_bool :
forall x y : Z,
negb (Zlt_bool x y) = Zle_bool y x.
Proof.
intros x y.
case Zlt_bool_spec ; intros H.
now rewrite Zle_bool_false.
now rewrite Zle_bool_true.
Qed.
End Zlt_bool.
Section Zcompare.
Inductive Zcompare_prop (x y : Z) : comparison -> Prop :=
| Zcompare_Lt_ : (x < y)%Z -> Zcompare_prop x y Lt
| Zcompare_Eq_ : x = y -> Zcompare_prop x y Eq
| Zcompare_Gt_ : (y < x)%Z -> Zcompare_prop x y Gt.
Theorem Zcompare_spec :
forall x y, Zcompare_prop x y (Zcompare x y).
Proof.
intros x y.
destruct (Z_dec x y) as [[H|H]|H].
generalize (Zlt_compare _ _ H).
case (Zcompare x y) ; try easy.
now constructor.
generalize (Zgt_compare _ _ H).
case (Zcompare x y) ; try easy.
constructor.
now apply Zgt_lt.
generalize (proj2 (Zcompare_Eq_iff_eq _ _) H).
case (Zcompare x y) ; try easy.
now constructor.
Qed.
Theorem Zcompare_Lt :
forall x y,
(x < y)%Z -> Zcompare x y = Lt.
Proof.
easy.
Qed.
Theorem Zcompare_Eq :
forall x y,
(x = y)%Z -> Zcompare x y = Eq.
Proof.
intros x y.
apply <- Zcompare_Eq_iff_eq.
Qed.
Theorem Zcompare_Gt :
forall x y,
(y < x)%Z -> Zcompare x y = Gt.
Proof.
intros x y.
apply Zlt_gt.
Qed.
End Zcompare.
Section cond_Zopp.
Definition cond_Zopp (b : bool) m := if b then Zopp m else m.
Theorem abs_cond_Zopp :
forall b m,
Zabs (cond_Zopp b m) = Zabs m.
Proof.
intros [|] m.
apply Zabs_Zopp.
apply refl_equal.
Qed.
Theorem cond_Zopp_Zlt_bool :
forall m,
cond_Zopp (Zlt_bool m 0) m = Zabs m.
Proof.
intros m.
apply sym_eq.
case Zlt_bool_spec ; intros Hm.
apply Zabs_non_eq.
now apply Zlt_le_weak.
now apply Zabs_eq.
Qed.
End cond_Zopp.
Section fast_pow_pos.
Fixpoint Zfast_pow_pos (v : Z) (e : positive) : Z :=
match e with
| xH => v
| xO e' => Z.square (Zfast_pow_pos v e')
| xI e' => Zmult v (Z.square (Zfast_pow_pos v e'))
end.
Theorem Zfast_pow_pos_correct :
forall v e, Zfast_pow_pos v e = Zpower_pos v e.
Proof.
intros v e.
rewrite <- (Zmult_1_r (Zfast_pow_pos v e)).
unfold Z.pow_pos.
generalize 1%Z.
revert v.
induction e ; intros v f ; simpl.
- rewrite <- 2!IHe.
rewrite Z.square_spec.
ring.
- rewrite <- 2!IHe.
rewrite Z.square_spec.
apply eq_sym, Zmult_assoc.
- apply eq_refl.
Qed.
End fast_pow_pos.
Section faster_div.
Lemma Zdiv_eucl_unique :
forall a b,
Zdiv_eucl a b = (Zdiv a b, Zmod a b).
Proof.
intros a b.
unfold Zdiv, Zmod.
now case Zdiv_eucl.
Qed.
Fixpoint Zpos_div_eucl_aux1 (a b : positive) {struct b} :=
match b with
| xO b' =>
match a with
| xO a' => let (q, r) := Zpos_div_eucl_aux1 a' b' in (q, 2 * r)%Z
| xI a' => let (q, r) := Zpos_div_eucl_aux1 a' b' in (q, 2 * r + 1)%Z
| xH => (Z0, Zpos a)
end
| xH => (Zpos a, Z0)
| xI _ => Z.pos_div_eucl a (Zpos b)
end.
Lemma Zpos_div_eucl_aux1_correct :
forall a b,
Zpos_div_eucl_aux1 a b = Z.pos_div_eucl a (Zpos b).
Proof.
intros a b.
revert a.
induction b ; intros a.
- easy.
- change (Z.pos_div_eucl a (Zpos b~0)) with (Zdiv_eucl (Zpos a) (Zpos b~0)).
rewrite Zdiv_eucl_unique.
change (Zpos b~0) with (2 * Zpos b)%Z.
rewrite Z.rem_mul_r by easy.
rewrite <- Zdiv_Zdiv by easy.
destruct a as [a|a|].
+ change (Zpos_div_eucl_aux1 a~1 b~0) with (let (q, r) := Zpos_div_eucl_aux1 a b in (q, 2 * r + 1)%Z).
rewrite IHb. clear IHb.
change (Z.pos_div_eucl a (Zpos b)) with (Zdiv_eucl (Zpos a) (Zpos b)).
rewrite Zdiv_eucl_unique.
change (Zpos a~1) with (1 + 2 * Zpos a)%Z.
rewrite (Zmult_comm 2 (Zpos a)).
rewrite Z_div_plus_full by easy.
apply f_equal.
rewrite Z_mod_plus_full.
apply Zplus_comm.
+ change (Zpos_div_eucl_aux1 a~0 b~0) with (let (q, r) := Zpos_div_eucl_aux1 a b in (q, 2 * r)%Z).
rewrite IHb. clear IHb.
change (Z.pos_div_eucl a (Zpos b)) with (Zdiv_eucl (Zpos a) (Zpos b)).
rewrite Zdiv_eucl_unique.
change (Zpos a~0) with (2 * Zpos a)%Z.
rewrite (Zmult_comm 2 (Zpos a)).
rewrite Z_div_mult_full by easy.
apply f_equal.
now rewrite Z_mod_mult.
+ easy.
- change (Z.pos_div_eucl a 1) with (Zdiv_eucl (Zpos a) 1).
rewrite Zdiv_eucl_unique.
now rewrite Zdiv_1_r, Zmod_1_r.
Qed.
Definition Zpos_div_eucl_aux (a b : positive) :=
match Pos.compare a b with
| Lt => (Z0, Zpos a)
| Eq => (1%Z, Z0)
| Gt => Zpos_div_eucl_aux1 a b
end.
Lemma Zpos_div_eucl_aux_correct :
forall a b,
Zpos_div_eucl_aux a b = Z.pos_div_eucl a (Zpos b).
Proof.
intros a b.
unfold Zpos_div_eucl_aux.
change (Z.pos_div_eucl a (Zpos b)) with (Zdiv_eucl (Zpos a) (Zpos b)).
rewrite Zdiv_eucl_unique.
case Pos.compare_spec ; intros H.
now rewrite H, Z_div_same, Z_mod_same.
now rewrite Zdiv_small, Zmod_small by (split ; easy).
rewrite Zpos_div_eucl_aux1_correct.
change (Z.pos_div_eucl a (Zpos b)) with (Zdiv_eucl (Zpos a) (Zpos b)).
apply Zdiv_eucl_unique.
Qed.
Definition Zfast_div_eucl (a b : Z) :=
match a with
| Z0 => (0, 0)%Z
| Zpos a' =>
match b with
| Z0 => (0, 0)%Z
| Zpos b' => Zpos_div_eucl_aux a' b'
| Zneg b' =>
let (q, r) := Zpos_div_eucl_aux a' b' in
match r with
| Z0 => (-q, 0)%Z
| Zpos _ => (-(q + 1), (b + r))%Z
| Zneg _ => (-(q + 1), (b + r))%Z
end
end
| Zneg a' =>
match b with
| Z0 => (0, 0)%Z
| Zpos b' =>
let (q, r) := Zpos_div_eucl_aux a' b' in
match r with
| Z0 => (-q, 0)%Z
| Zpos _ => (-(q + 1), (b - r))%Z
| Zneg _ => (-(q + 1), (b - r))%Z
end
| Zneg b' => let (q, r) := Zpos_div_eucl_aux a' b' in (q, (-r)%Z)
end
end.
Theorem Zfast_div_eucl_correct :
forall a b : Z,
Zfast_div_eucl a b = Zdiv_eucl a b.
Proof.
unfold Zfast_div_eucl.
intros [|a|a] [|b|b] ; try rewrite Zpos_div_eucl_aux_correct ; easy.
Qed.
End faster_div.
Section Iter.
Context {A : Type}.
Variable (f : A -> A).
Fixpoint iter_nat (n : nat) (x : A) {struct n} : A :=
match n with
| S n' => iter_nat n' (f x)
| O => x
end.
Lemma iter_nat_plus :
forall (p q : nat) (x : A),
iter_nat (p + q) x = iter_nat p (iter_nat q x).
Proof.
induction q.
now rewrite plus_0_r.
intros x.
rewrite <- plus_n_Sm.
apply IHq.
Qed.
Lemma iter_nat_S :
forall (p : nat) (x : A),
iter_nat (S p) x = f (iter_nat p x).
Proof.
induction p.
easy.
simpl.
intros x.
apply IHp.
Qed.
Fixpoint iter_pos (n : positive) (x : A) {struct n} : A :=
match n with
| xI n' => iter_pos n' (iter_pos n' (f x))
| xO n' => iter_pos n' (iter_pos n' x)
| xH => f x
end.
Lemma iter_pos_nat :
forall (p : positive) (x : A),
iter_pos p x = iter_nat (Pos.to_nat p) x.
Proof.
induction p ; intros x.
rewrite Pos2Nat.inj_xI.
simpl.
rewrite plus_0_r.
rewrite iter_nat_plus.
rewrite (IHp (f x)).
apply IHp.
rewrite Pos2Nat.inj_xO.
simpl.
rewrite plus_0_r.
rewrite iter_nat_plus.
rewrite (IHp x).
apply IHp.
easy.
Qed.
End Iter. | 7,215 | 19,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-05 | latest | en | 0.520837 |
https://www.sololearn.com/Discuss/1964843/why-is-the-difference-in-result/ | 1,582,212,480,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00091.warc.gz | 925,714,974 | 6,679 | +11
# Why is the difference in result
Among the tests there is such a question. def f(q, my list =[]): mylist = mylist + q return mylist a = [1] print(f(a)) a = [2] print(f(a)) >>>[1] [2] If we change the code, then the result changes: def f(q, my list =[]): mylist += q return mylist a = [1] print(f(a)) a = [2] print(f(a)) >>>[1] [1, 2] and now what doesn't fit in my head: def f(q, my list =[]): print(mylist) mylist = mylist + q return mylist a = [1] f(a) a = [2] f(a) >>>[] [] And here is another result, and I don't understand this. After all, print goes on the first line and the result should be the same: def f(q, my list =[]): print(mylist) mylist += q return mylist a = [1] f(a) a = [2] f(a) >>>[] [1] Is this something that just needs to be taken as an axiom, or is there some kind of explanation?
9/12/2019 3:57:07 AM
Mikhail Gorchanyuk | 280 | 853 | {"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-2020-10 | latest | en | 0.882327 |
https://ccssmathanswers.com/into-math-grade-5-module-10-lesson-3-answer-key/ | 1,702,005,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00050.warc.gz | 187,121,579 | 59,203 | # Into Math Grade 5 Module 10 Lesson 3 Answer Key Use Representations of Division of Unit Fractions by Whole Numbers
We included HMH Into Math Grade 5 Answer Key PDF Module 10 Lesson 3 Use Representations of Division of Unit Fractions by Whole Numbers to make students experts in learning maths.
## HMH Into Math Grade 5 Module 10 Lesson 3 Answer Key Use Representations of Division of Unit Fractions by Whole Numbers
I Can create a story context and use a visual fraction model to interpret the division of a unit fraction by a whole number.
Darren thinks about how he can use division of unit fractions by whole numbers in his cooking class. He considers the expression $$\frac{1}{2}$$ ÷ 8. What word problem can be written for $$\frac{1}{2}$$ ÷ 8? Draw a visual model to show the quotient, and then explain what it represents.
Darren prepares soup from $$\frac{1}{2}$$ of flour and
divides among 8 people, each people will get how much portion
of flour soup?
Explanation:
Given Darren thinks about how he can use division of
unit fractions by whole numbers in his cooking class.
He considers the expression $$\frac{1}{2}$$ ÷ 8.
word problem can be written for $$\frac{1}{2}$$ ÷ 8 is
Darren prepares soup from $$\frac{1}{2}$$ of flour and
divides among 8 people, each people will get how much portion
of flour soup?
Drawn a visual model above to show the quotient as
$$\frac{1}{2}$$ ÷ 8 =$$\frac{1}{2}$$ X $$\frac{1}{8}$$ =
$$\frac{1}{2 X 8}$$ = $$\frac{1}{16}$$.
Turn and Talk When dividing a fraction by a whole number, what do you notice about the values of the dividend and the quotient? Explain how they compare.
The divided and quotient value will be less,
Explanation:
Dividing fractions with whole numbers results in a value
less than the given fraction. It is because we are dividing
it into smaller equal parts and the quotient obtained will
be the value of one part out of those.
Build Understanding
Question 1.
Darren considers situations in which a fractional quantity is divided into equal groups. Darren thinks about the expression $$\frac{1}{4}$$ ÷ 2.
A. Describe a situation in which $$\frac{1}{4}$$ of an item can be divided into 2 equal parts. How do the quantities represent the dividend, the divisor, and the quotient?
Situation:
Darren prepares tea using $$\frac{1}{4}$$ jug of 1000 ml milk and
distributes among two persons each person will get
how much quantity of milk?
Dividend – 2 persons,
Divisor – $$\frac{1}{4}$$ jug of milk,
Quotient – $$\frac{1}{8}$$ of 1,000 ml milk,
Explanation:
Given Darren considers situations in which a fractional quantity is divided into equal groups. Darren thinks about the expression $$\frac{1}{4}$$ ÷ 2.
Situation:
Darren prepares tea using $$\frac{1}{4}$$ jug of 1000 ml milk and
distributes among two persons each person will get
how much quantity of milk? Here Dividend is 2 persons,
Divisor is $$\frac{1}{4}$$ ju of milk and
Quotient is $$\frac{1}{8}$$ of 1,000 ml milk.
B. What word problem can you write that can be modeled by $$\frac{1}{4}$$ ÷ 2, and includes a question about the size of each part?
Draw a visual model to show the quotient.
Word problem:
Darren gives $$\frac{1}{4}$$ of his pencils among
his two freinds, each freind will get how much part,
$$\frac{1}{4}$$ is dividend,
2 is divisor and $$\frac{1}{8}$$ is
quotient,
Explanation:
Darren gives $$\frac{1}{4}$$ of his pencils among
his two freinds, each freind will get how much part,
$$\frac{1}{4}$$ of pencils is dividend, 2 friends are
divisor and each will get $$\frac{1}{8}$$ is quotient.
Drawn a visual model to show the quotient above.
C. Use your visual model to write a division equation. Then explain what the quotient represents.
Division equation : $$\frac{1}{4}$$ ÷ 2,
quotient $$\frac{1}{2}$$ represents what
quanity we will get,
Explanation:
Used the visual model and wrote a
division equations as $$\frac{1}{4}$$ ÷ 2,
we have rule to divide a fraction by a whole number,
multiply the given fraction by the reciprocal of whole numbers.
So $$\frac{1}{4}$$ ÷ 2 is $$\frac{1}{4}$$ X $$\frac{1}{2}$$ =
$$\frac{1}{4 X 2}$$ = $$\frac{1}{8}$$.
The quotient represents $$\frac{1}{4}$$ is been divided
by 2 we get each part as $$\frac{1}{8}$$.
Turn and Talk Discuss different situations that can be represented by $$\frac{1}{4}$$ ÷ 2. What types of quantities can be modeled by the dividend?
Situation 1:
Darren has $$\frac{1}{4}$$ bag of coffee beans he gives
them to 2 people how much quantity will each person receives.
Situation 2:
Darren has $$\frac{1}{4}$$ shelf of books he gives
them to 2 people how much number of books
Quantities can be pizzas, candies etc.
Explanation:
Situation 1:
Darren has $$\frac{1}{4}$$ bag of coffee beans he gives
them to 2 people how much quantity will each person receives.
Situation 2:
Darren has $$\frac{1}{4}$$ shelf of books he gives
them to 2 people how much number of books
Quantities can be pizzas, candies etc.
Question 2.
Write and solve a word problem that can be modeled by the expression $$\frac{1}{2}$$ ÷ 6. Draw a visual model to show the quotient, and then explain what it represents.
A. What word problem can you write that can be modeled by $$\frac{1}{2}$$ ÷ 6?
Draw a visual model to show the quotient.
Word problem:
Karl has $$\frac{1}{2}$$ pizza he divides it among
his 6 freinds how much quantity will each friend will get.
quotient is $$\frac{1}{12}$$,
Explanation:
Karl has $$\frac{1}{2}$$ pizza he divides it among
his 6 freinds how much quantity will each friend will get.
Drawn a visual model to show the quotient as
$$\frac{1}{12}$$ as shown above.
B. Describe the size of each part and what it represents.
The size of each part represents each person,
Explanation:
The size of each part represents the quantity how much
each person gets the pizza.
C. Solve your word problem. What does the quotient represent?
The quotient represents the quantity of size each receive,
Explantion:
Solving $$\frac{1}{2}$$ ÷ 6 =
$$\frac{1}{2}$$ X $$\frac{1}{6}$$ =
$$\frac{1}{12}$$, the quotient $$\frac{1}{12}$$
represents the quantity of pizza size each person will receive.
Turn and Talk How did you decide the type of visual model to use for your problem? Explain.
Upon the given situation or problem,
we use tape diagrams, area models, number lines,
Explanation:
Based on the situation or problem we use visuals or
visual model, A math visual model consists of materials
(mostly with pictures or images) to help interpret the numbers.
which ever is suitable for the particular situation can be tape diagrams,
area models, number lines etc.
Step It Out
Question 3.
Look at the visual model.
A. Write a division word problem that can be represented by the visual model.
Word problem:
Sara has $$\frac{1}{4}$$ part of candies which she distributes
among 3 freinds each freind will get how much part of candies,
Explanation:
The word problem that can be represented by the visual model is
Sara has $$\frac{1}{4}$$ part of candies which she distributes
among 3 freinds each freind will get how much part of candies.
B. Write a division equation to model your problem. Then solve the problem.
Division equation:
$$\frac{1}{4}$$ ÷ 3,
quotient = $$\frac{1}{12}$$,
Explanation:
The division equation to model the given problem is
$$\frac{1}{4}$$ ÷ 3 solving
$$\frac{1}{4}$$ X $$\frac{1}{3}$$ =
$$\frac{1}{4 X 3}$$ = $$\frac{1}{12}$$.
Check Understanding
Question 1.
Amy writes a word problem that can be modeled by the expression $$\frac{1}{4}$$ ÷ 5. Complete her word problem.
There are _________ truck(s) that need to share the total amount of recycled plastic bottles equally. If there is a collection of _________ ton of plastic bottles, then how much does each truck carry?
There are 5 truck(s) that need to share the total
amount of recycled plastic bottles equally.
If there is a collection of $$\frac{1}{4}$$ ton of plastic bottles,
then how much does each truck carry?
Explanation:
Given Amy writes a word problem that can be modeled
by the expression $$\frac{1}{4}$$ ÷ 5.
Completed her word problem as there are 5 truck(s)
that need to share the total amount of recycled
plastic bottles equally.
If there is a collection of $$\frac{1}{4}$$ ton of plastic bottles,
then how much does each truck carry?.
Question 2.
A doctor has $$\frac{1}{2}$$ hour to see 3 patients. If the doctor divides the time equally among the patients, how much time will be spent with each patient? Draw a visual model to represent the problem.
Time spent with each patient is $$\frac{1}{6}$$,
Explanation:
Given a doctor has $$\frac{1}{2}$$ hour
to see 3 patients. If the doctor divides the time
equally among the patients, time will be spent with
each patient is $$\frac{1}{2}$$ ÷ 3,
$$\frac{1}{2}$$ X $$\frac{1}{3}$$ =
$$\frac{1}{6}$$, Drawn a visual model to
represent the problem as shown above.
Question 3.
Write a word problem that involves a plot of land and a number of different crops that can be modeled by the expression $$\frac{1}{2}$$ ÷ 5.
Will the quotient be less than $$\frac{1}{2}$$ or greater than $$\frac{1}{2}$$. Justify your thinking.
Word problem:
A plot of $$\frac{1}{2}$$ land has been used to grow
5 different crops how much part of land each crop will get,
quotient will be less than $$\frac{1}{2}$$,
Explanation:
Wrote a word problem that involves a plot of land and
a number of different crops that can be modeled by the
expression $$\frac{1}{2}$$ ÷ 5 is
Word problem:
A plot of $$\frac{1}{2}$$ land has been used to grow
5 different crops how much part of land each crop will get,
The quotient will be less than $$\frac{1}{2}$$ as it is
again divided by 5 so it will be divided again into parts
theerefore the quotient will be less than $$\frac{1}{2}$$.
Question 4.
Use Tools Write a word problem for the expression $$\frac{1}{5}$$ ÷ 4. Then draw a visual model and solve your problem.
Word problem:
A wall of $$\frac{1}{5}$$ of portion is completed by
4 persons each person has built how much portion of the wall,
Each person has buitl $$\frac{1}{20}$$,
Explanation:
Word problem for the expression $$\frac{1}{5}$$ ÷ 4 is
Word problem:
A wall of $$\frac{1}{5}$$ of portion is completed by
4 persons each person has built how much portion of the wall,
drawn a visual model as shown above and solving
$$\frac{1}{5}$$ ÷ 4 = $$\frac{1}{5}$$ X $$\frac{1}{4}$$ =
$$\frac{1}{5 X 4}$$ = $$\frac{1}{20}$$,
therefore each person has buitl $$\frac{1}{20}$$.
Question 5.
Attend to Precision Margie writes a word problem for the expression $$\frac{1}{3}$$ ÷ 6: “A bottle of mouthwash is $$\frac{1}{3}$$ empty. If the remaining amount is divided equally over 6 weeks, what fraction of the bottle is used each week?”
What mistake did she make in her problem?
mouth wah and remaining amount is division,
Explanation:
Given Margie writes a word problem for the
expression $$\frac{1}{3}$$ ÷ 6:
“A bottle of mouthwash is $$\frac{1}{3}$$ empty.
If the remaining amount is divided equally over 6 weeks,
what fraction of the bottle is used each week?”
The mistake did she make in her problem instead of empty
$$\frac{1}{3}$$ and remaining amount is divided among
6 weeks.
Correct her word problem. Then find the answer using a visual model.
Correct word problem:
“A bottle of mouthwash is $$\frac{1}{3}$$ filled.
If the amount is divided equally over 6 weeks,
what fraction of the bottle is used each week?”,
$$\frac{1}{18}$$ of the bottle is used each week,
Explanation:
Correct word problem for the expression $$\frac{1}{3}$$ ÷ 6 is
“A bottle of mouthwash is $$\frac{1}{3}$$ filled.
If the amount is divided equally over 6 weeks,
what fraction of the bottle is used each week?”,
Drawn a visual model as shown above solving
$$\frac{1}{3}$$ ÷ 6 = $$\frac{1}{3}$$ X $$\frac{1}{6}$$ =
$$\frac{1}{3 X 6}$$ = $$\frac{1}{18}$$,
therefore $$\frac{1}{18}$$ of the bottle is used each week.
Question 6.
Model with Mathematics Write a division word problem that can be represented by the visual model. Then write a division equation and solve your problem.
Word problem:
“Mary has shared her
$$\frac{1}{2}$$ of jam bottle among
4 of her friends, how much portion of the jam
each of friend will get”,
Division equation: $$\frac{1}{2}$$ ÷ 4,
Quotient is $$\frac{1}{8}$$,
Explanation:
The word problem that can be represented by
the given visual model is “Mary has shared her
$$\frac{1}{2}$$ of jam bottle among
4 of her friends, how much portion of the jam
each of friend will get”.
Division equation: $$\frac{1}{2}$$ ÷ 4,
Solving $$\frac{1}{2}$$ ÷ 4 =
$$\frac{1}{2}$$ X $$\frac{1}{4}$$ =
$$\frac{1}{2 X 4}$$ = $$\frac{1}{8}$$.
Quotient is $$\frac{1}{8}$$ so each friend
will get $$\frac{1}{8}$$ portion of jam.
Question 7.
Open-Ended Consider the expression $$\frac{1}{5}$$ ÷ 3.
Write two different word problems that can be represented by this expression.
Word problem 1:
“Rosy has $$\frac{1}{5}$$ portion of apple juice
with her she distributes among 3 friends how much juice
will each friend will get?”
Word problem 2:
“Three friends finishes $$\frac{1}{5}$$ portion of
planting flower plants in a garden, How much does each
one has planted?”,
Explanation:
Considering the expression $$\frac{1}{5}$$ ÷ 3.
The two different word problems that can be represented
by this expression is Word problem 1:
“Rosy has $$\frac{1}{5}$$ portion of apple juice
with her she distributes among 3 friends how much juice
will each friend will get?”
Word problem 2:
“Three people finishes $$\frac{1}{5}$$ portion of
planting flower plants in a garden, How much does each
one has planted?”.
Draw a visual model to represent the problems and then solve.
Quotient is $$\frac{1}{15}$$,
Explanation:
Drawn a visual model to represent the problems and then solved
$$\frac{1}{5}$$ ÷ 3 = $$\frac{1}{5}$$ X $$\frac{1}{3}$$ =
$$\frac{1}{5 X 3}$$ = $$\frac{1}{15}$$.
What does the quotient represent in each problem?
Problem 1 quotient represents each friend got
$$\frac{1}{15}$$ portion of apple juice,
Problem 2 quotient represents each person got
$$\frac{1}{15}$$ portion of plants planted,
Explanation:
The quotient is same in both the problems but
it represents differently as in problem 1 quotient
epresents each friend got $$\frac{1}{15}$$ portion
of apple juice, In problem 2 quotient represents
each person got $$\frac{1}{15}$$ portion of
plants planted,
I’m in a Learning Mindset!
How can I work with others to write word problems? | 4,045 | 14,346 | {"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} | 4.9375 | 5 | CC-MAIN-2023-50 | latest | en | 0.884988 |
https://aptitude.gateoverflow.in/tag/logical-reasoning?start=40 | 1,618,176,879,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00466.warc.gz | 225,983,496 | 21,432 | # Recent questions tagged logical-reasoning
41
Fill in the blanks by choosing the best option. $\text{ELFA GLHA ILJA ______ MLNA}$ $\text{OLPA}$ $\text{KLMA}$ $\text{LLMA}$ $\text{KLLA}$
42
Fill in the blanks by choosing the best option. $\text{JAK KBL LCM MDN}$_________ $\text{OEP}$ $\text{NEO}$ $\text{MEN}$ $\text{PFQ}$
43
Fill in the blanks by choosing the best option. $\text{QPO NML KJI}$ _____ $\text{EDC}$ $\text{HGF}$ $\text{CAB}$ $\text{JKL}$ $\text{GHI}$
44
Complete the series by choosing the appropriate missing term from the options given. $8, 43, 11, 41, ?, 39, 17.$ $8$ $14$ $43$ $44$
45
Fill in the blanks by choosing the best option. $\text{CMM EOO GQQ}$ ________ $\text{KUU}$ $\text{GRR}$ $\text{GSS}$ $\text{ISS}$ $\text{ITT}$
46
Fill in the blanks by choosing the best option. $P_5QR \: \: P_4QS \: \: P_3QT \: \:$ _______ $PQV$ $PQW$ $PQV_2$ $P_2QU$ $PQ_3U$
47
Fill in the blanks by choosing the best option. $B_2 CD \: \:$ _______ $\: \: BCD_4 \: \: B_5 CD \: \: BC_6 D$ $B_2C_2D$ $BC_3D$ $B_2C_3D$ $BCD_7$
48
Given: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday What day comes three days after the day which comes two days after the day which comes immediately after the day which comes two days after Monday? Sunday Tuesday Thursday Saturday
49
Choose the most appropriate option. $70, 71, 76, \underset{\text{_____}}{?}, 81, 86, 70, 91$ $70$ $71$ $80$ $96$
50
Choose the most appropriate option. $4, 7, 25, 10, \underset{\text{_____}}{?}, 20, 16, 19$ $13$ $15$ $20$ $28$
51
Choose the most appropriate option. $72, 76, 73, 77, 74, \underset{\text{____}}{?}, 75$ $70$ $71$ $75$ $78$
1 vote
52
Nutritionists have recommended that people eat more fiber. Advertisements for a new fiber-supplement pill state only that it contains $44$ percent fiber. The advertising claim is misleading in its selection of information on which to focus if which one of the following is true? ... pills every day. The recommended daily intake of fiber is $20$ to $30$ grams, and the pill contains one-third gram.
1 vote
53
Saurabh: Airlines have made it possible for anyone to travel around the world in much less time than was formerly possible. Monica: That is not true. Many flights are too expensive for all but the rich. Monica's response shows that she interprets Saurabh's ... routes serviced by airlines most forms of world travel are not affordable for most people anyone can afford to travel long distances by air
1 vote
54
In a certain code language $\text{‘1 3 4’}$ means ‘you are well’, $\text{‘7 5 8’}$ means ‘they go home’, $\text{‘8 3 9’}$ means ‘we are home’, Which of the statement can be dispensed with while answering the above question? (a) only (b) only (a) or (b) only (b) and (c) only
1 vote
55
Find the missing term: $60,40,55,45,50,50,?$ $45$ $50$ $55$ $60$
1 vote
56
Monisha is working with a real estate agent to find a location for the kids' toy store she plans to open in her town. She is looking for a place that is either in the centre or not too far from the centre of town. It should also be attractive for the right ... town's center a storefront in a small strip mall located on the outskirts of town that is also occupied by a pharmacy and a dry cleaner
1 vote
57
Here are some words translated from an artificial language. qmelaqali means fruitcake qalitiimmeo means cakewalk useguamao means buttercup Which word could mean “cupcake”? qalitiiqali amaotiimmeo pakitreft amaoqali
1 vote
58
There is a queue in a ticketing office. Amanda is $10^{th}$ from the front while Murthy is $25^{th}$ from behind and Marta is just in the middle of the two. If there be $50$ persons in the queue. What position does Marta occupy from the front? $16$ $18$ $15$ $17$
1 vote
59
If the consonants in the word ‘DROVE’ are first arranged alphabetically and the vowels are put in between two pairs of consonants in the alphabetical order, which of the following will be the fourth from the right end after the rearrangement? $D$ $E$ $R$ $O$
1 vote
60
Find the missing alphabet: $T,r,O,m,J,?$ $h$ $i$ $I$ $g$
1 vote
61
Count the number of squares in the given figure. $32$ $30$ $29$ $28$
1 vote
62
There are five janitors. Pali, Qureshi, Rohan, Sant and Timber. They all have a different height, Qureshi is shorter than only Timber and Sant is shorter than Pali and Qureshi. Who among them is the shortest? Rohan Sant Pali Data inadequate
63
$A\$B$means$A$is the father of$B;A\#B$means$A$is the sister of$B;A\ast B$means$A$is the daughter of$B$and [email protected]$ means $A$ is the brother of $B.$ Which of the following indicates that $M$ is the wife of $Q?$ $Q \$ R \# [email protected] MQ \$[email protected] \#M$ $Q \$ R\ast T \#MQ \$R @T \ast M$
1 vote
64
Fact $1:$ All chickens are birds. Fact $2:$ Some chickens are hens. Fact $3:$ Female birds lay eggs. If the first three statements are facts, which of the following statements must also be a fact? All birds lay eggs. Hens are birds. Some chickens are not hens. II only II and III only I,II and III None of the statements is a known fact
1 vote
65
Pointing to a photograph, a man said, “ I have no brother or sister but that man’s father is my father’s son.” Whose photograph was it? His Own His Son His Father His Grandfather
1 vote
66
Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.” How is the woman related to this man? Mother Aunt Sister Daughter
1 vote
67
Count the number of convex pentagons in the adjoining figure. $16$ $12$ $8$ $4$
1 vote
68
Find the number of quadrilaterals in the given figure. $6$ $7$ $9$ $11$
1 vote
69
There is a family of six people whose nick names are Pat, Qat, Rat, Sat, Tat and Uat. Their professions are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an ... and Tat. The Doctor, Sat is married to the Manager. How many male members are there in the family? Two Three Four Data Inadequate
70
There is a family of six people whose nick names are Pat, Qat, Rat, Sat, Tat and Uat. Their professions are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the ... the two married couples in the family? Pat-Qat and Sat-Rat Rat-Uat and Sat-Tat Pat-Tat and Sat-Rat Pat-Sat and Rat-Qat
1 vote
71
There is a family of six people whose nick names are Pat, Qat, Rat, Sat, Tat and Uat. Their professions are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an ... and Tat. The Doctor, Sat is married to the Manager. What is the profession of Pat? Lawyer Lawyer or Teacher Manager None of these
1 vote
72
There is a family of six people whose nick names are Pat, Qat, Rat, Sat, Tat and Uat. Their professions are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an ... the mother of Uat and Tat. The Doctor, Sat is married to the Manager. How Pat is related to Tat? Father Grandfather Mother Grandmother
1 vote
73
Below question presents a situation and asks you to make a judgement regarding that particular circumstance. Choose an answer based on given information. Zineb Fez has invited her three friends over to watch the basketball game on her wide-screen television. They are all hungry, but ... of the numbers is most likely the telephone number of the pizzeria? $995-9266$ $995-9336$ $995-9268$ $995-8266$
1 vote
74
Each question presents a situation and asks you to make a judgement regarding that particular circumstance. Choose an answer based on given information. A Hollywood film director wants a woman for the lead role of Leez who perfectly fits the description that appears in the original screenplay. However he is not ... thirties. She's of very slight build and stands at ${5}'$ $1,2$ $2,3$ $1,4$ $2,4$
1 vote
75
In the following question, various terms of a letter series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives. $\text{C,F,J,O,?,B}$ $\text{S}$ $\text{T}$ $\text{U}$ $\text{V}$
1 vote
76
In the following question, various terms of a letter series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives. $\text{ABP, CDQ, EFR, }(\dots?\dots)$ $\text{GHS}$ $\text{GHT}$ $\text{HGS}$ $\text{GHR}$
1 vote
77
In the following question, various terms of a letter series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives. $\text{T,r,O,m,J,?}$ $\text{h}$ $\text{i}$ $\text{I}$ $\text{g}$
1 vote
In the following question, various terms of a letter series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives. $\text{C,G,L,R,?}$ $\text{Y}$ $\text{S}$ $\text{U}$ $\text{Z}$
Statement: A large number of people in ward $X$ of the city are diagnosed to be suffering from a fatal malaria type. Courses of Action: The city municipal authority should take immediate steps to carry out extensive fumigation in ward $X$. The people in the area should be advised to take steps to avoid mosquito bites. Only I follows. Only II follows. Either I or II follows. Both I and II follows. | 2,684 | 9,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.745948 |
https://math.answers.com/math-and-arithmetic/What_is_19_out_of_50_as_a_percantage | 1,726,188,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00750.warc.gz | 350,788,159 | 46,342 | 0
# What is 19 out of 50 as a percantage?
Updated: 9/20/2023
Wiki User
12y ago
38 per cent
Wiki User
12y ago
Earn +20 pts
Q: What is 19 out of 50 as a percantage?
Submit
Still have questions?
Related questions
58/50 = 116%
56%
### What is 19 percent out of 50?
19% of 50= 19% * 50= 0.19 * 50= 9.5
2.5%
56%
14.55
.125
### What is 19 over 50?
19/50 is a fraction in simplest terms as there is no common factor other than 1 between 19 and 50. A a decimal it is 19 ÷ 50 = 0.38
### What percent of 50 equals 19?
It is: 38% of 50 = 19
4 percent
66.667%.
83.73 percent | 220 | 591 | {"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} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.909618 |
https://brainly.in/question/99913 | 1,484,996,681,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281069.89/warc/CC-MAIN-20170116095121-00305-ip-10-171-10-70.ec2.internal.warc.gz | 796,229,031 | 11,011 | Prove that... cos(a)cos(60-a)cos(60+a)=(cos(3a))/4
2
by ajngamer420
2015-04-24T17:49:04+05:30
This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We use the identities
can you derive the 2nd identity and thanks if you did
should be easy with angle sum identity : cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) = (2cos^2(x)-1)cos(x) - 2sin(x)cos(x)sin(x) = ...
2015-04-25T05:58:53+05:30
This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We know that standard identity:
2 Cos A Cos B = Cos (A+B) + Cos (A-B)
here, A = 60-a and B = 60+a
Cos (a) Cos (60 - a ) Cos (60+a)
= Cos (a) * 1/2 * [ Cos (60-a+60+a) + Cos (60-a- 60-a) ]
= Cos (a) * 1/2 * [ Cos 120 + Cos (-2a) ]
= 1/2 * Cos (a) * [ -1/2 + Cos (2a) ]
= -1/4 * Cos (a) + 1/2 Cos (a) Cos (2a) --- apply the above identity again
= -1/4 * Cos (a) + 1/2 * 1/2 [ Cos 3a + cos a]
= 1/4 * Cos (3a)
click on thank you button above pls | 484 | 1,415 | {"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} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.796217 |
https://www.physicsforums.com/threads/heat-loss-through-walls.392975/ | 1,518,897,546,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807660.32/warc/CC-MAIN-20180217185905-20180217205905-00570.warc.gz | 916,126,923 | 15,697 | # Heat Loss Through Walls
1. Apr 6, 2010
### sdoyle1
1. The problem statement, all variables and given/known data
I'm not sure if this problem should go here.. it's a bit of math and a bit of physics. I know that the concepts should be simple but I don't really know where to start.. Here's the question:
a) Determine the breakeven point for heating oil in dollars per gallon – the point where it would be worth enduring the cost of going to 2 x6 walls from 2 x4 walls in a house of 2000 square feet. We are looking at a payback of 3 years as interest rates have risen to 12 percent and with the high unemployment rate there is little cash around. State any assumptions made in addition to those given.
2. Relevant equations
There were a number of assumptions given to us:
Assume all heat loss is thru walls (no windows, doors etc.) No effect via basement or roof.
Assume 10 studs per 100 sq feet.
Assume all heat movement is due to conduction, no radiation or convection.
Assume oil initially at \$4.00 per liter, fuel oil number 2 averages 140,000 Btu per Gallon(US) and a furnace efficiency of 92%
Median season length of 7 months.
Median outside temperature of 27 degrees C
Indoor temperature kept at 70 degrees
3. The attempt at a solution
I am assuming that you need to use R=deltaT/Q but I'm really not sure. I don't want someone to do the question for me... just kind of guide me in the right direction :) | 354 | 1,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-09 | longest | en | 0.939698 |
http://search.cpan.org/~brianski/Number-Range-Regex-0.32/lib/Number/Range/Regex.pm | 1,498,592,608,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00346.warc.gz | 361,588,913 | 7,419 | Brian Szymanski > Number-Range-Regex-0.32 > Number::Range::Regex
Number-Range-Regex-0.32.tgz
Dependencies
Annotate this POD
# CPAN RT
Open 0
View/Report Bugs
Module Version: 0.32
# NAME
Number::Range::Regex - create regular expressions that check for integers in a given range
# SYNOPSIS
``` use Number::Range::Regex;
my \$lt_20 = range( 0, 19 );
print "foo(\$foo) contains an integer < 20" if \$foo =~ /\$lt_20/;
print "foo(\$foo) is an integer < 20" if \$foo =~ /^\$lt_20\$/;
if( \$line =~ /^\S+\s+(\$lt_20)\s/ ) {
print "the second field (\$1) is an integer < 20";
}
my \$nice_numbers = rangespec( "42,175..192" );
my \$my_values = \$lt_20->union( \$nice_numbers );
if( \$line =~ /^\S+\s+(\$my_values)\s/ ) {
print "the second field has one of my values (\$1)";
}
my \$lt_10 = rangespec( "0..9" );
my \$primes_lt_30 = rangespec( "2,3,5,7,11,13,17,19,23,29" );
my \$primes_lt_10 = \$lt_10->intersection( \$primes_lt_30 );
my \$nonprimes_lt_10 = \$lt_10->minus( \$primes_lt_30 );
print "nonprimes under 10 contains: ".join",", \$nonprimes_lt_10->to_string;
if( \$something =~ /^\$nonprimes_lt_10\$/ ) {
print "something(\$something) is a nonprime less than 10";
}
if( \$nonprimes_lt_10->contains( \$something ) ) {
print "something(\$something) is a nonprime less than 10";
}
my \$octet = range(0, 255);
my \$ip4_match = qr/^\$octet\.\$octet\.\$octet\.\$octet\$/;
my \$range_96_to_127 = range(96, 127);
my \$my_slash26_match = qr/^192\.168\.42\.\$range_96_to_127\$/;
my \$my_slash19_match = qr/^192\.168\.\$range_96_to_127\.\$octet\$/;
my \$in_a_or_in_b_but_not_both = \$a->xor(\$b);
my \$it = rangespec("-20..42,47..52")->iterator();
\$it->first;
do { print \$it->fetch } while (\$it->next);
\$it->last;
do { print \$it->fetch } while (\$it->prev);```
# DESCRIPTION
Number::Range::Regex lets you manage sets of integers and generate regular expressions matching them. For example, here is one way to match number ranges in a regular expression:
` \$date =~ m/^0*(?:[1-9]|[12][0-9]|3[01])\/0*(?:[0-9]|1[012])\$/;`
here is another:
``` my \$day_range = range(1, 31);
my \$month_range = range(1, 12);
\$date =~ m/^\$day_range\/\$month_range\$/;```
which is more legible? (bonus points if you spotted the bug)
# METHODS
## RANGE CREATION
range
` \$range = range( \$min, \$max );`
Create a range between the first argument and the last. For example, \$min==8 and \$max==12, corresponds to the list containing 8, 9, 10, 11, and 12. This method is exported by default.
rangespec
` \$range = rangespec( '8..12,14,19..22' );`
Create a "compound" range given the range specification passed. For example, the range above would consist of 8, 9, 10, 11, 12, 14, 19, 20, 21, and 22. This method is exported by default.
## RANGE INTERROGATION
contains
` \$range->contains( \$number );`
Returns a true value if \$range contains \$number - otherwise, it returns a false value.
overlaps
` \$range->overlaps( \$another_range );`
Returns a true value if \$range overlaps with \$another_range - otherwise, it returns a false value. e.g.
``` rangespec('7..9')->overlaps( rangespec('4..6') ) => false
rangespec('7..9')->overlaps( rangespec('10..12') ) => false
rangespec('7..9')->overlaps( rangespec('6..10') ) => true
rangespec('7..9')->overlaps( rangespec('6..7') ) => true```
## RANGE DISPLAY
to_string
` \$range->to_string();`
Return a compact representation of the range suitable for consumption by a human, perl(1), or rangespec(). For example:
``` \$range = range( 6, 22 );
print \$range->to_string;```
will output: "6..22", which can be parsed by perl(1) or rangespec().
regex
` \$range->regex();`
Return a regular expression matching members of this range. For example:
``` \$range = range( 6, 22 );
print \$range->regex;```
will output something equivalent to:
` qr/0*(?:[6-9]|1\d|2[0-2])/`
which, on my machine with perl v5.14.2 and a development version of Number::Range::Regex between v0.12 and v0.13, is:
` (?^:(?# begin Number::Range::Regex::SimpleRange[6..22] )[+]?0*(?:(?^:[6-9])|(?^:1\d)|(?^:2[0-2]))(?# end Number::Range::Regex::SimpleRange[6..22] ))`
Please note that range objects are overloaded so that in regex context, \$range will be equivalent to \$range->regex(). This works in all versions of perl >= v5.6.0. When it is further possible to distinguish regex context from string context (as in overload v1.10 or higher, available in perl >= v5.12.0), range objects will display in string but not regex context as the terser, more legible \$range->to_string() instead.
## UNBOUNDED (aka "infinite") RANGES
It is also possible to specify unbounded ranges, ie the set of all integers less than 17. This may be specified in any of the following ways:
``` range( undef, 17 );
range( '-inf', 17);
rangespec('-inf..17');```
Similarly the set of all integers greater than 17:
``` range( 17, undef );
range( 17, '+inf' );
rangespec('17..+inf');```
Note carefully that, in order to prevent errors, it is not possible to specify the set of all integers via range(undef, undef). If you try to do so, Number::Range::Regex will complain that you "must specify either a min or a max or use the allow_wildcard argument":
` range( undef, undef ); #boom`
If you really want range() with no defined arguments to mean the set of all possible integers, you can use one of the below:
``` range( undef, undef, {allow_wildcard => 1} ) ;
range( '-inf', '+inf' );```
To test if a range is infinite, you can call is_infinite():
``` 1 == ! rangespec('3..7')->is_infinite();
1 == rangespec('16..+inf')->is_infinite();
1 == rangespec('')->not->is_infinite();```
## SET OPERATIONS
given \$range2 = rangespec( '0,2,4,6,8' ) and \$range3 = rangespec( '0,3,6,9' )
union
` \$range = \$range2->union( \$range3 );`
Return the union of one range with another. In the example above, \$range would consist of: 0, 2, 3, 4, 6, 8, and 9. Note that union() can take more than one argument, e.g. \$range2->union( \$range3, \$range5, \$range7, \$range11, ... );
intersect
` \$range = \$range2->intersect( \$range3 );`
Return the intersection of one range with another. In the example above, \$range would consist of: 0 and 6. This method is also available via the alias intersection.
xor
` \$range = \$range->xor( \$another_range );`
Return the symmetric difference of \$range2 and \$range3 (elements in one or the other range, but not in both). In the example above, \$range would consist of 2, 3, 4, 8, and 9.
subtract
` \$range = \$range2->subtract( \$range3 );`
Return the relative complement of \$range2 in \$range3. In the example above, \$range would consist of: 2, 4, and 8. Note carefully that this method is not symmetric - \$range3->subtract( \$range2 ) would be a different range consisting of 3 and 9. This method is also available via the aliases subtraction and minus.
invert
` \$range = \$range2->invert();`
Return the absolute complement of \$range2. In the example above, \$range would include:
``` any integer less than or equal to -1
1, 3, 5, and 7, and
any integer greater than or equal to 9.```
that is, \$range->to_string would be '-inf..-1,1,3,5,7,9..+inf'. This method is also available via the alias not.
## ITERATORS
iterators let you examine large or infinite sets with minimal memory:
``` \$it = \$range->iterator();
\$it->first();
do {
do_something_with_value( \$it->fetch );
} while (\$it->next);```
## OTHER METHODS
regex_range
` \$regex = regex_range( \$min, \$max );`
This is a shortcut for range( \$min, \$max )->regex(). Useful for one-off use when overload.pm does not support regex context. It should only be used with perl 5.10.X or lower where regex context overloading is not possible. This method may be deprecated in a future release of Number::Range::Regex. This method is not exported by default.
# NOTES
It's usually better to check for number-ness only in the regular expression and verify the range of the number separately, eg: \$line =~ /^\S+\s+(\d+)/ && \$1 > 15 && \$1 < 32; but it's not always practical to refactor in that way.
If you like one-liners, something like the following may suit you... m{^\${\( range(1, 31) )}\/\${\( range(1, 12) )}\$} but, for readability's sake, please don't do that!
# BUGS AND LIMITATIONS
Please report any bugs or feature requests through the web interface at http://rt.cpan.org.
# AUTHOR
Brian Szymanski <ski-cpan@allafrica.com> -- be sure to put Number::Range::Regex in the subject line if you want me to read your message. | 2,582 | 8,499 | {"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-2017-26 | longest | en | 0.376873 |
https://zbmath.org/?q=an%3A0963.30001 | 1,628,114,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00160.warc.gz | 1,117,359,651 | 10,097 | # zbMATH — the first resource for mathematics
An application of certain integral operators. (English) Zbl 0963.30001
Let $$A$$ denote the class of functions of the form $f(z)= z+ \sum^\infty_{n=2} a_nz^n$ which are analytic in the open disk $$U= \{z\in \mathbb{C};|z|< 1\}$$. Consider the following integral operators $$P^\alpha$$ and $$Q^\alpha_\beta$$ defined by the formulas $P^\alpha f(z)= {2^\alpha\over z\Gamma(\alpha)} \int^z_0 \Biggl(\log{1\over p}\Biggr)^{\alpha- 1} f(t) dt$ and $Q^\alpha_\beta f(z)= {\alpha+\beta\choose \beta}{\alpha\over z^\beta} \int^z_0 \Biggl(1-{t\over z}\Biggr)^{\alpha- 1} t^{\beta- 1}f(t) dt,$ where $$\alpha> 0$$, $$\beta>-1$$ and $$\Gamma$$ is the familiar Euler’s function. In the paper the authors prove, under suitable assumptions, that $$|P^\alpha f(z)|< 1$$ and $$|Q^\alpha_\beta f(z)|< 1$$ for all $$z\in U$$, and all $$f\in A$$.
In the opinion of the reviewer the title of the work is not adequate to the contents of this paper.
##### MSC:
30A10 Inequalities in the complex plane 30C99 Geometric function theory
##### Keywords:
integral operators
Full Text:
##### References:
[1] Bernardi, S., Convex and starlike univalent functions, Trans. amer. math. soc., 135, 429-446, (1968) · Zbl 0172.09703 [2] Bernardi, S., The radius of univalence of certain analytic functions, Proc. amer. math. soc., 24, 312-318, (1974) · Zbl 0191.08803 [3] Jung, I.B.; Kim, Y.C.; Srivastava, H.M., The Hardy space of analytic functions associated with certain one-parameter families of integral operators, J. math. anal. appl., 176, 138-147, (1993) · Zbl 0774.30008 [4] Miller, S.S.; Mocanu, P.T., Second order differential inequalities in the complex plane, J. math. anal. appl., 65, 289-305, (1978) · Zbl 0367.34005 [5] Owa, S., Properties of certain integral operators, Georgian math. J., 2, 535-545, (1995) · Zbl 0837.30013 [6] Owa, S.; Srivastava, H.M., Some applications of the generalized libera operator, Proc. Japan acad., 62, 125-128, (1986) · Zbl 0583.30016 [7] Srivastava, H.M.; Owa, S., New characterization of certain starlike and convex generalized hypergeometric functions, J. natl. acad. math. India, 3, 198-202, (1985) · Zbl 0603.30017 [8] Srivastava, H.M.; Owa, S., A certain one-parameter additive family of operators defined on analytic functions, J. math. anal. appl., 118, 80-87, (1986) · Zbl 0579.30012
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 859 | 2,727 | {"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} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.618103 |
http://www.aimsuccess.in/2016/11/ibps-rrb-v-2016-pre-officer-scale-i_6.html | 1,521,463,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646914.32/warc/CC-MAIN-20180319120712-20180319140712-00249.warc.gz | 336,726,828 | 72,116 | # IBPS RRB V 2016 Pre Officer Scale- I Exam Analysis - 6th November 2016 - 1st Slot
The Preliminary phase of the IBPS RRB Scale – I Exam had started from 04th November 2016. The Third & Final set of the exam is scheduled today. The first slot of the exam was scheduled for 9:00 am – 09:45 am. We are now going to provide you the analysis of this slot of the exam.
The level of the exam was Moderate.
S.N NAME OF TESTS MEDIUM OF EXAM NO OF QUESTIONS MAXIMUM MARKS TOTAL TIME ALLOWED 1 Reasoning English/Hindi 40 40 Composite time of 45 Minutes 2 Numerical Ability English/ Hindi 40 40 Total 80 80
Here is the analysis of the 4th slot of the exam as shared by our Regular readers-
Here is the analysis of the first slot of the exam as shared by our Regular readers –
### Quantitative Aptitude (40 Qs)
Level –Easy to Moderate
(i) Number Series – 5 Questions (Moderate)
• 4, 16, 26, 34, 40, ?
• 825, 582, 501, 474, 465, ?
• 8, 12, 24, 60, 180, ?
• 3, 4, 10, 33, 136, ?
• 16, 17, 13, 22, 6, ?
(ii) Approximation/ Simplification – 5 Questions (Easy )
(iii) Data Interpretation (3 Sets) – 15 Questions (Easy to Moderate)
• Tabular graph – based on total no. of college students, ratio of Assistant professors
• Pie Chart – calculative
• Bar graph – based on the companies of M & N selling scafs
(iv) Quadratic Equation – 5 Questions (Easy)
(v) Miscellaneous Questions – 10 Questions (Moderate)
Miscellaneous questions include – Age, Probability, Partnership, Profit Loss, Average, Boats & Stream, Compound/Simple Interest, Ratio and Proportion, Percentage, Time & Work, etc.
### Reasoning Ability (40 Qs)
Level – Moderate
• (i) Syllogism – 5 Questions (Easy)
• (ii) Inequality – 5 Questions (Easy)
• (iii) Puzzle- Order Ranking – (Floor) 10 Questions (one was floor based), (other was based on the days of the week with 7 Companies)
• (iv) Seating Arrangement – Square – 5 Questions (Moderate) (people sitting on corners facing outside & people sitting in between facing inward)
• (v) Linear Seating Arrangement – 5 Questions (All Facing North)
• (vi) Direction – 2 Questions
• (vii) Blood Relation – 3 questions
• (viii) Miscellaneous Questions – 5 Questions (Easy to Moderate)
• (Alphabetical Series, Coding – Decoding etc.)
AimSuccess team is working on providing detailed analysis of today’s paper, which will include good attempt, section-wise difficulty level and expected cutoff. Thus, we would like you all the participate in small poll on your attempts in IBPS RRB V Officer Scale I Exam.
Note: Only recommendation for aspirants who are yet to attempt IBPS RRB V 2016 Pre Officer Scale- I Exam in the upcoming days/shifts would be – “AVOID NEGATIVE MARKING”.
We hope this IBPS RRB V 2016 Pre Officer Scale- I Exam Analysis helps in getting an estimate about your performance in the paper. If you are yet to attempt the Tier 1 Examination, improve your chances by giving the following article a thorough read.
Best BANKING INSTITUTE LIST
Stay connected for more information regarding IBPS Exam Recruitment 2016. You may share your queries in the comment section below.
### IBTS INSTITUTE
India's Leading Institution for Competitive Exams in Chandigarh visit www.ibtsindia.com | 871 | 3,193 | {"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 | 0.899749 |
https://gmatclub.com/forum/man-cat-4-14-base-7-remainder-78684.html?fl=similar | 1,498,457,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00253.warc.gz | 773,299,597 | 48,551 | It is currently 25 Jun 2017, 23:09
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Man Cat 4 #14-Base 7 remainder
Author Message
Manager
Joined: 04 Dec 2008
Posts: 102
Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
19 May 2009, 23:57
1
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
100% (01:02) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
SVP
Joined: 17 Jun 2008
Posts: 1547
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
20 May 2009, 00:56
It should be 0 since the unit digit of expression is 0.
Senior Manager
Joined: 08 Jan 2009
Posts: 327
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
20 May 2009, 01:09
We know x > 1
we know that 7^1/5 = 2, 7^2/5 = 4, 7^3/5 = 4,7^4/5 = 4. This goes on in a cycle for every 4.
so x = 1 7^15/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1
so x = 2 7^27/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1
so x = 3 7^39/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1
I think Ans. B
Manager
Joined: 10 May 2009
Posts: 65
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
20 May 2009, 08:57
Let x=1.
7^15 + 3
7 follows cyclicity of 4 and hence 7^15 will have 3 as Unit digit. So sum will be 3+3 =6
Hence remainder is 1.
Hence B
Director
Joined: 23 May 2008
Posts: 806
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
20 May 2009, 10:41
agree with B
7,9,3,1 is the cycle for powers of 7
so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3
3+3=6 if we divide by 5 the remainder will be 1
Manager
Joined: 04 Dec 2008
Posts: 102
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
21 May 2009, 20:40
bigtreezl wrote:
agree with B
7,9,3,1 is the cycle for powers of 7
so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3
3+3=6 if we divide by 5 the remainder will be 1
Now, I understand. I just have to deal with the last digit. Thanks!
Manager
Joined: 02 Oct 2009
Posts: 193
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
18 Oct 2009, 17:12
B
bigtreezl has it good at generalization.
Senior Manager
Joined: 01 Mar 2009
Posts: 367
Location: PDX
Re: Man Cat 4 #14-Base 7 remainder [#permalink]
### Show Tags
21 Oct 2009, 11:56
bigtreezl wrote:
agree with B
7,9,3,1 is the cycle for powers of 7
so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3
3+3=6 if we divide by 5 the remainder will be 1
Well 7^15 = 7^5 (taking last digit into account) will result in 7
7^27 = 7^7 will result in 3
7^9 wil result in 7 .. so we have end results of 7 and 3 both .. isn't it ?
_________________
In the land of the night, the chariot of the sun is drawn by the grateful dead
Re: Man Cat 4 #14-Base 7 remainder [#permalink] 21 Oct 2009, 11:56
Display posts from previous: Sort by | 1,288 | 3,591 | {"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} | 4.4375 | 4 | CC-MAIN-2017-26 | longest | en | 0.862842 |
https://www.onlinemathlearning.com/put-together-total-unknown.html | 1,539,739,181,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510932.57/warc/CC-MAIN-20181017002502-20181017024002-00273.warc.gz | 1,043,523,020 | 10,891 | # Put Together with Total Unknown Word Problems
Related Topics:
Lesson Plans and Worksheets for Grade 1
Lesson Plans and Worksheets for all Grades
Eureka Math/EngageNY grade 1 module 2 lesson 11 Worksheets
New York State Common Core Math Grade 1 Module 2, Lesson 11
Eureka Math grade 1 module 2 lesson 11 Worksheets (pdf)
Worksheets, solutions, and videos to help Grade 1 students learn how to share and critique peer solution strategies for put together with total unknown word problems.
Common Core Standards: 1.OA.2, 1.OA.3, 1.OA.6
Lesson 11 Exit Ticket
John thinks the problem below should be solved using 5-group drawings and Sue thinks
it should be solved using a number bond. Solve both ways and circle the strategy you
think is the most efficient.
1. Kim scores 5 goals in her soccer game and 8 runs in her softball game. How many
points does she score altogether?
Lesson 11 Homework
Look at the student work. Correct the work. If the answer is incorrect, show a correct solution in the space below the student work.
1. Todd has 9 red cars and 7 blue cars. How many cars does he have altogether?
2. Jill has 8 beta fish and 5 goldfish. How many fish does she have in total?
3. Dad baked 7 chocolate and 6 vanilla cupcakes. How many cupcakes did he bake in all?
4. Mom caught 9 fireflies and Sue caught 8 fireflies. How many fireflies did they catch altogether?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 414 | 1,779 | {"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-2018-43 | longest | en | 0.905415 |
https://www.physicsforums.com/threads/work-and-rotational-kinetic-energy.232647/ | 1,709,275,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00187.warc.gz | 929,698,453 | 16,294 | # Work and rotational Kinetic Energy
• frig0018
In summary, the problem involves a 32.0 kg wheel with a radius of 1.20 m, rotating at 280 rev/min. To bring it to a stop in 15.0 s, the angular acceleration must be determined. The formula for work as a change in kinetic energy can be used to solve for the amount of work needed.
## Homework Statement
A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?
/
## The Attempt at a Solution
I=mr^2=46.08 kg*m^2
W=(1/2)I$$\omega$$f^2 - (1/2)I$$\omega$$i^2
W=$$\tau$$($$\Delta$$$$\theta$$)
$$\tau$$=I$$\alpha$$
Rev in 15 sec = 70 rev => 140$$\Pi$$ radians in 15 sec.
I'm not sure which equation(s) to use. I've tried plugging numbers into all of them and getting stuck or wrong answers. Thanks!
Remember, work is a change in kinetic energy. Do you know the formula for kinetic energy of a rotating object?
i thought it was
W = delta K = 1/2(I)(omega final)^2 - 1/2(I)(omega initial)^2...?
or do i use the formula
W= (integral from theta initial to theta final) torque d-theta?
… one step at a time …
frig0018 said:
A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?
Hi frig0018!
First step: what angular acceleration is needed to stop it in 15 s?
## 1. What is work in the context of rotational kinetic energy?
Work is the force applied to an object multiplied by the distance over which the force is applied. In rotational kinetic energy, work is the force applied to an object that is rotating about an axis, multiplied by the distance from the axis to the point where the force is applied.
## 2. How is rotational kinetic energy different from linear kinetic energy?
Rotational kinetic energy is the energy an object possesses due to its rotation about an axis, while linear kinetic energy is the energy an object possesses due to its motion in a straight line. The main difference between the two is the axis of rotation, as rotational kinetic energy depends on the distance from the axis to the point where the force is applied, while linear kinetic energy depends on the speed and mass of the object.
## 3. What is the formula for calculating rotational kinetic energy?
The formula for rotational kinetic energy is ½ Iω², where I is the moment of inertia and ω is the angular velocity. The moment of inertia is a measure of an object's resistance to changes in its rotation, and the angular velocity is the rate at which an object is rotating.
## 4. How does rotational kinetic energy affect an object's stability?
Rotational kinetic energy can affect an object's stability by influencing its center of mass. If an object has a high rotational kinetic energy, it will have a larger moment of inertia and a higher center of mass, making it less stable. On the other hand, a lower rotational kinetic energy will result in a smaller moment of inertia and a lower center of mass, making the object more stable.
## 5. What are some real-life applications of rotational kinetic energy?
Rotational kinetic energy is involved in many everyday activities, such as throwing a ball, riding a bike, and driving a car. In engineering, it is important in designing and analyzing rotating machinery, such as turbines and engines. It is also crucial in understanding the movement and stability of objects in sports, such as gymnastics and figure skating. | 849 | 3,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-10 | latest | en | 0.915203 |
http://gogeometry.com/problem/p310_semicircle_circle_tangent_angle.htm | 1,553,265,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202671.79/warc/CC-MAIN-20190322135230-20190322161230-00402.warc.gz | 91,170,582 | 3,187 | Problem 310: Circle Inscribed in a Semicircle, 45 Degrees Angle.
Level: High School, SAT Prep, College geometry.
In the figure below, circle C is inscribed in a semicircle with diameter AB. If D and E are the tangency points, prove that the measure of angle AED is 45 degrees.
Sketch and Typography of Problem 310
In the figure below, circle C is inscribed in a semicircle with diameter AB. If D and E are the tangency points, prove that the measure of angle AED is 45 degrees. | 122 | 480 | {"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-2019-13 | latest | en | 0.934354 |
http://compaland.com/trial-and/what-is-the-trial-and-error-method-in-factoring.html | 1,511,395,605,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806708.81/warc/CC-MAIN-20171122233044-20171123013044-00489.warc.gz | 63,656,798 | 4,803 | ## How To Repair What Is The Trial And Error Method In Factoring Tutorial
Home > Trial And > What Is The Trial And Error Method In Factoring
# What Is The Trial And Error Method In Factoring
## Contents
There are strengths and weaknesses to both approaches. x2 - 5x + 6 = (x ... )(x ... ) Step 2: The last term is 6. The more you practice factoring, the less error you'll run into, because you'll learn to see which trials will work without having to write down all the steps. yaymath 275,294 views 8:41 Loading more suggestions... his comment is here
I'd love to start with the proof, however it's a little much for intro Algebra. Rebecca Savell 247,177 views 7:38 Loading more suggestions... slomathteacher 1,463 views 14:55 Factoring Quadratic Trinomials: Part 1 [fbt] - Duration: 13:42. Not necessarily a bad thing when you're searching for the right answer.
## Factor By Trial And Error Calculator
Loading... Add to Want to watch this again later? Wow...it's like we're psychic.
Steps to Factoring Trinomials Back to Top Below you can see the steps for factoring trinomials:Steps for factoring trinomials:Step 1: Factor out common factor, if possible.Step 2: Factorized the trinomial using Both methods build on previous techniques and topics, and therefore can be used to help students increase their conceptual understanding. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Trial And Error Method Formula shana donohue | June 18, 2010 at 3:01 pm Ok, I made the animation on factoring trinomials….
Transcript The interactive transcript could not be loaded. Trial And Error Method In Mathematics Factor trinomial by unfoiling (trial and error) 4x2 + 15x + 9 Factor trinomial by unfoiling (trial and error) 4x2 − 15x + 9 Factor trinomial by unfoiling (trial and Close Yeah, keep it Undo Close This video is unavailable. http://www.onlinemathlearning.com/factor-trinomials-unfoil.html Mathispower4u 17,930 views 9:45 Complex Trinomial Factoring - CrissCross - Duration: 13:33.
Reply 2. Trial And Error Method Of Problem Solving Neither m nor n make an appearance alongside the first term in the final polynomial, which is probably just as well, since that x appears to be busy squaring itself.Let's see MathemAddicts 7,133 views 21:28 Factoring Trinomials (A quadratic Trinomial) by Trial and Error - Duration: 7:36. Jermaine Gordon 479 views 6:11 15 videos Play all Polynomials : FactoringpatrickJMT factor a trinomial using the criss-cross method - Duration: 5:35.
## Trial And Error Method In Mathematics
This feature is not available right now. more info here Up next Factoring Trinomials Using Trial and Error - Duration: 15:27. Factor By Trial And Error Calculator Assume m and n are integers. Factoring By Trial And Error Worksheet Question2: Factorize x2 - 18x + 81 Solution: Given x2 - 18x + 81x2 - 18x + 81 = x2 - 9x - 9x + 81= x(x - 9) - (x
Another super fun example! this content MrAcostaMath 772 views 9:20 Factoring Trinomials: Trial and Error and Grouping - Duration: 9:45. The constant term of the original polynomial is 3, so we need mn = 3.What integers multiply together to give 3? Each Wednesday I post an article related to general teaching on my blog. Trial And Error Method Calculator
In fact, it will benefit us to use some factorization organization. Like this:Like Loading... Trinomial equations can be solved by solving their factors. weblink Please try again later.
MrCaryMath 77,635 views 5:38 factoring trinomials with "a" greater than 1 - Duration: 7:38. What Property Is Used In Solving Quadratic Equations Working... Loading... | 950 | 3,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-47 | latest | en | 0.860601 |
https://www.prepadviser.com/integrated-reasoning/gmat-tutorials-integrated-reasoning-part-1-video | 1,718,792,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00655.warc.gz | 832,357,903 | 8,980 | This video from the new GMAT tutorial series produced by PrepAdviser and examPAL gives you an overview of Graphics Interpretation.
Integrated Reasoning (IR) is a section in, and of itself, on the GMAT. It is unique in that it combines both mathematical skills and verbal skills. In the Integrated Reasoning section, we will face 12 questions in 30 minutes – which means we have an average of 2.5 minutes per question. This may sound like a lot, but the truth is IR questions are a lot of work, and hence, this is the GMAT section in which we are most pressed for time. The Integrated Reasoning section gets its own score – between 1 and 8, in single digit increments.
There are two other special things about this section: one is that we get to use a calculator – and we will need it! The other is that this is the only section on the test which is not computer-adaptive: the questions do not change based on how well or poorly we are doing.
Questions on the Integrated Reasoning section fall into four separate categories: Graphics Interpretation, Two-Part Analysis, Table Analysis, and Multi-Source Reasoning.
In this video, we are going to discuss Graphics Interpretation.
In Graphics Interpretation questions, we are presented with a graph of some sort.
When first approaching these questions, we need to study the graph and try to see what TALE it tells us. This means we will look at the following aspects:
• Text – We will read it, including the title – if there is one – and take notes.
• Axes – What does each axis represent? In which units?
• Legend – What does each symbol, colour or texture represent in the graph?
• Example – We will pick a random point and ask ourselves: what does it represent?
Next to the graph, we will find a passage of text, which we have to read. Not only do we need to read the text in order to answer the question, but the text is often crucial to understand the information presented in the table as well. For example, in this case, only by reading the text will we understand that the increments of 10 on the y-axis represent thousands of kilometres. Similarly, the text is not enough without the graph – there is lots of information which can be found in the graph only.
Finally, and most importantly, at the end of the text will be two sentences, in which a word or a section is left blank. Our job is to complete each sentence by choosing the right option from a drop-down menu.
It is important to note that both of these sentences make up a single question, and there is no partial credit – in order to get the points for the question we have to get them both right.
In this case, the first statement asks us to choose the amount of kilometres per hour a satellite was flying at a specific time, while the second statement asks us to determine which date this satellite last sent a signal.
When looking at the statements, it is important to ask ourselves several questions:
• What are we looking for? These include things like:
• Words that point us to formulas: such as “probability”, “profit”, “mean” and so on.
• Subjects that demand that we ask another question. For example, if the question mentions “ratio”, we should immediately ask – is a quantity given in the question or only ratios?
• Asking – Is that really the subject? Some questions seem at first to be about a particular topic, but actually are about something else.
• If the question says “products are produced” – does that mean it is a work problem? It could be, but it could be something else.
• If “distance” is mentioned – is the question about figuring out a rate? That is not necessarily the case.
• Next, we need to ask ourselves – how can we find what we are looking for? Visually? Using a formula? Where in the graph should we look for the data? And do we need a calculator?
• Finally, we need to ask – how should we approach the question? We should remember the different tools at our disposal:
• If the question calls for a tough calculation, we can take the precise approach and use the calculator. IR has some crazy calculations just because there is a calculator – so use it or lose it!
• If there are no numbers at all, we can take the alternative approach and just pick numbers to use in the question.
• Finally, we have to use logic and learn all we can from the graph visually. Many IR questions do not require any calculation at all.
So this is the first part of Integrated Reasoning, out of four. Do not miss the other three! | 967 | 4,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.955483 |
http://mathhelpforum.com/business-math/index48.html | 1,513,458,623,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589177.70/warc/CC-MAIN-20171216201436-20171216223436-00146.warc.gz | 166,534,828 | 14,542 | 1. ### Application of Black-Scholes
• Replies: 0
• Views: 628
Nov 23rd 2009, 09:57 AM
2. ### Math of Finance annuities
• Replies: 3
• Views: 912
Nov 22nd 2009, 08:37 PM
3. ### [Interest]
• Replies: 1
• Views: 523
Nov 22nd 2009, 01:37 AM
4. ### Interest Rate and principal calculations
• Replies: 2
• Views: 666
Nov 20th 2009, 08:35 PM
5. ### Net Present Value
• Replies: 9
• Views: 1,088
Nov 19th 2009, 05:01 PM
6. ### Nonconstant Growth (FINC 3105)
• Replies: 2
• Views: 2,695
Nov 19th 2009, 11:19 AM
7. ### Total Cost Funtion
• Replies: 1
• Views: 789
Nov 19th 2009, 07:53 AM
8. ### Help with simple microeconomic problems
• Replies: 0
• Views: 1,076
Nov 18th 2009, 04:02 PM
9. ### Choice of Two Options for Paying off a Loan
• Replies: 2
• Views: 763
Nov 18th 2009, 12:45 PM
10. ### Utilization
• Replies: 1
• Views: 614
Nov 18th 2009, 11:24 AM
• Replies: 1
• Views: 765
Nov 18th 2009, 11:05 AM
12. ### problems with markdowns
• Replies: 1
• Views: 573
Nov 18th 2009, 10:46 AM
13. ### math Review help
• Replies: 1
• Views: 643
Nov 17th 2009, 09:09 PM
14. ### Brownian motion. Asset pricing. Portfolio dynamics in continuous time
• Replies: 0
• Views: 878
Nov 17th 2009, 05:48 AM
15. ### Bonus and taxes: graphing
• Replies: 5
• Views: 870
Nov 16th 2009, 02:29 PM
16. ### Simple GDP Proof with One-Variable Calculus - Macroeconomics
• Replies: 0
• Views: 660
Nov 15th 2009, 08:05 AM
17. ### Need an MicroEcon wiz - SOS!
• Replies: 1
• Views: 623
Nov 12th 2009, 01:34 PM
18. ### Calculating the GDP and Net Domestic Income
• Replies: 0
• Views: 1,057
Nov 12th 2009, 01:13 PM
19. ### Simple Interest Calculation Help
• Replies: 9
• Views: 748
Nov 12th 2009, 12:41 PM
20. ### Discounts and Simple Interest Rates.
• Replies: 5
• Views: 711
Nov 12th 2009, 09:44 AM
• Replies: 0
• Views: 788
Nov 12th 2009, 12:36 AM
22. ### Finding the sellling price of a bond
• Replies: 0
• Views: 700
Nov 11th 2009, 10:09 PM
23. ### Solving for interesting rate in PV formula
• Replies: 2
• Views: 689
Nov 10th 2009, 08:04 PM
24. ### Optimization problem
• Replies: 1
• Views: 516
Nov 10th 2009, 07:55 PM
25. ### Simple math problem need help to solve
• Replies: 3
• Views: 1,118
Nov 10th 2009, 11:17 AM
26. ### Could someone tell me how to set this up?
• Replies: 3
• Views: 759
Nov 9th 2009, 01:03 PM
27. ### Annuity Due
• Replies: 4
• Views: 709
Nov 8th 2009, 03:20 PM
• Replies: 4
• Views: 876
Nov 7th 2009, 02:17 PM
29. ### PortfolioTheory
• Replies: 0
• Views: 473
Nov 7th 2009, 02:36 AM
,
,
,
# statistical displays page 171
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted.
Note: when sorting by date, 'descending order' will show the newest results first.
## » Pre University Math Help
Math Discussion | 1,107 | 2,895 | {"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-2017-51 | longest | en | 0.631883 |
https://www.howiswhat.org/what/what-is-the-cheapest-form-of-electricity-generation/ | 1,679,990,707,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00055.warc.gz | 912,359,022 | 13,535 | # What is the cheapest form of electricity generation?
## What is the cheapest form of electricity generation?
And there is some very good news for the planet: Solar and wind power, at the scale that a major utility would deploy them, are now the cheapest form of power. They’re a bit less expensive than natural gas-fired power plants and considerably cheaper than coal and nuclear.16 Oct 2021
## How much fuel does it take to power the world?
For every kilogram of LNG that undergoes combustion, 5.36 × 107 Joules of energy can be gained, meaning it would take a mere 10.4 billion tonnes of gas to power the world.20 Sept 2017
## How much does it cost to generate electricity?
Hydroelectric power is the cheapest source of renewable energy, at an average of \$0.05 per kilowatt hour (kWh), but the average cost of developing new power plants based on onshore wind, solar photovoltaic (PV), biomass or geothermal energy is now usually below \$0.10/kWh.
## Which energy source is the cheapest?
The report follows the International Energy Agency’s (IEA) conclusion in its World Energy Outlook 2020 that solar power is now the cheapest electricity in history. The technology is cheaper than coal and gas in most major countries, the outlook found.5 Jul 2021
## How many Litres of diesel are in a kWh?
One litre of diesel fuel (auto) has an energy content of approximately 38 MJ which approximates to 10 kWh (using a ballpark figure) but the efficiency of conversion into kinetic energy is only about 30% – that is better than petrol which is typically 25% depending on the design.
READ What was William Shakespeare's first published work?
10 kWh
## How much diesel does it take to produce 1 kWh?
A general rule of thumb is that a diesel generator will use 0.4 L of diesel per kWh produced. The diesel engine used is essentially an internal combustion engine.3 Sept 2015
## How many Litres are in a kWh?
Convert LPG kWh to litres: 1 kWh = 0.145 L of LPG.Dec 8, 2021
## How long does it take to produce 1 kWh?
The basic unit of electricity is the Kilowatt hour (kWh). In simple terms, 1 kWh is the amount of energy used by a 1kW (1000 watt) electric heater for 1 hour.
## How much does it cost to generate 1 megawatt of electricity?
The cost of producing one megawatt-hour of electricity — a standard way to measure electricity production — is now around \$50 for solar power, according to Lazard’s math. The cost of producing one megawatt-hour of electricity from coal, by comparison, is \$102 — more than double the cost of solar.8 May 2018
## What is 1kw equal to?
Much like one kilowatt is equal to 1,000-watts of power, one kilowatt-hour is equivalent to 1,000-watts, or joules, of energy use over one hour.
## How much fuel does it take to generate electricity?
One metric ton of coal can generate 1,927 kilowatt hours of electricity, in comparison to 1,000 cubic feet of natural gas which can generate 99 kilowatt hours. The largest power plant in the United States is the hydroelectric power plant, Grand Coulee. It has a summer capacity of 7.08 gigawatts.
## What is the cheapest electricity to produce?
What is the cheapest renewable energy source? Hydroelectric power is currently the cheapest renewable energy source, costing \$0.05 per kilowatt-hour on average2. Hydroelectric power is the cheapest because the infrastructure has been in place for a long time, and it produces electricity consistently.
1,000 watts
## How much does it cost to produce 1 kWh?
Individual projects in wind power regularly supply electricity at \$0.04 per kWh without financial support, while for power plants running on fossil fuels, the cost interval is \$0.04–0.14 per kWh (IRENA, 2018). | 877 | 3,708 | {"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-2023-14 | latest | en | 0.940837 |
http://mathbyvemuri.blogspot.com/2011/10/arithmetic-4-snap-2008.html | 1,508,561,568,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00509.warc.gz | 208,324,561 | 17,027 | ## Saturday, 22 October 2011
### Arithmetic -5 (SNAP-2008)
A cyclist drove one kilometer with the wind in his back, in three minutes and drove the same way back against the wind in four minutes. If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one kilometer without wind?
(A) 7/3 (B) 24/7 (C) 17/7 (D) 43/12
Solution follows here:
Solution:
Let the speed of cyclist be ‘c’ and speed of wind be ‘w’.
When the wind is at his back, the resultant speed becomes c+w
Speed = distance/time => c+w = 1/3 ---(1)
When the wind is against him, the resultant speed becomes c-w
Speed = distance/time => c-w = 1/4 ---(2)
Adding (1) and (2), 2c = 1/3 + 1/4 = 7/12 => c = 7/24
Without wind, the speed is his sole speed ie.., ‘c’.
Time to drive 1 kilometer = distance/speed
= 1/c = 1/(7/24) = 24/7 minutes | 282 | 889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-43 | longest | en | 0.931353 |
http://goszechuanhouse.com/rice/readers-ask-how-ro-cook-rice.html | 1,716,683,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00179.warc.gz | 10,790,159 | 22,362 | ## What is the ratio of water to rice?
The basic water to white rice ratio is 2 cups water to 1 cup rice. You can easily, double and even triple the recipe; just make sure you are using a pot large enough to hold the rice as it cooks and expands.
## How long does it take to boil rice?
Instructions
1. Add rice and water to a medium saucepan and bring to a boil over high heat.
2. Simmer until water is completely absorbed and rice is tender – about 15-25 minutes (will depend on size and freshness of rice ).
## How much dry rice does it take to make 2 cups of cooked rice?
Rice Conversions & Equivalents
1 cup uncooked white rice yields 3 cups of cooked white rice
¾ cup (96 grams) uncooked rice serves 2 persons
1 cup (180 grams) uncooked rice serves 3 persons
2 cups (360 grams) uncooked rice serves 6 persons
Assuming 1/ 2 cup of cooked rice per person.
You might be interested: Question: How To Make Steam Rice Cake?
8
## What is the healthiest way to eat rice?
1. Steaming or boiling rice is the best way of cooking them, due to the elimination of any high-fat vegetable oils. 2. Team your cooked rice with blanched or stir-fried high-fibre vegetables to make your meal more satiating and healthy.
## Do you boil water before adding rice?
Once you bring the water to a boil you need to turn it down to a simmer. Adding cold rice at this point can thriw off this process. Rice is typically cooked based on ratios of volume. Boiling the water first won’t effect this unless you lose track of it and the boiling causes your water content to reduce.
## Is 1 cup of rice enough for 2?
Measure one cup of long grain white rice into a cup and level it off. One cup of dry rice will make enough cooked rice for two to three adult servings. (Or two adults and two small children.) The cool thing about this recipe is it is proportional.
## How do you know if rice is done?
Once the rice has been cooking for around 12 minutes, all of the water should be soaked up by the grains, and the rice will be al dente. To determine if the rice is cooked, taste it and see what you think about the texture. If it is chewy or hard in the center then add 1/8 to 1/4 cup of water.
## Why do we boil rice?
This process hardens the starch in the grains so they remain firmer, less sticky, and separate when cooked. It also forces the vitamins and minerals from the outer layer of the grains into the endosperm, which is the part we eat.
You might be interested: Question: How Much Is A Serving Of Brown Rice Cooked?
## How much rice does 2 cups make?
RICE CONVERSIONS & EQUIVALENTS 1 cup uncooked white rice, or wild rice, yields 3 cups of cooked rice. 1 cup brown whole grain rice, yields 4 cups of cooked rice.. 1 cup long grain rice, yields 3 cups cooked rice. 1 cup dry pre-cooked instant rice, yields 2 cups cooked rice.
## How much does 1 cup rice make cooked?
One cup of uncooked rice will yield approximately three cups cooked.
## How long should I cook 3 cups of rice?
Add 3 cups of fresh water, butter (or oil) and salt to the rice in the saucepan. Over medium-high heat, bring rice and water to a boil. Reduce heat to medium-low, cover with lid, and let simmer for 15 minutes.
## How can I lose weight eating rice?
Here are some ways to safely eat rice on a weight loss diet:
1. Exercise Portion Control: Take only one helping of rice during a single meal to restrict the amount of calories you eat with rice.
2. Pair It With Lots Of Vegetables: Rice may make you feel hungry sooner than other carbs.
3. Opt For Low-Calorie Cooking Methods:
## Does rice make you fat?
Rice is one of the most widely consumed grains in the world. White rice is a refined, high-carb food that’s had most of its fiber removed. Brown Versus White Rice.
White Brown
Fat 0 grams 1 gram
Manganese 19% RDI 55% RDI
Magnesium 3% RDI 11% RDI
Phosphorus 4% RDI 8% RDI
6
## Is Rice better than bread?
If your goal is to lose fat and lean out – bread is probably the better choice for you pound for pound vs white rice. This is of course if you equate for the same calories. It’ll make you fuller, for longer than white rice due to its protein and fiber content. It also has more protein to increase your metabolic rate. | 1,041 | 4,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.935669 |
https://linearalgebras.com/elementary-analysis-08-08.html | 1,721,700,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00464.warc.gz | 307,531,387 | 13,579 | If you find any mistakes, please make a comment! Thank you.
## Prove the limits equal to the corresponding numbers
We shall use the following useful formula.
$\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, which is a variation of $(a-b)(a+b)=a^2-b^2$.
Solution:
### Part a
Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
$$s_n=\dfrac{(n^2+1)-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}.$$ Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$
\begin{align*}
|s_n-0|
=&\ \frac{1}{\sqrt{n^2+1}+n}\\
<&\ \frac{1}{n+n}=\frac{2}{n}\\
<&\ \frac{1}{n} <\frac{1}{N}=\epsilon
\end{align*} as desired. Hence $\lim s_n=0$.
### Part b
Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
$$\sqrt{n^2+n}-n=\dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}$$ and
\begin{align*}
s_n-\dfrac{1}{2}
=&\ \dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}-\frac{1}{2}\\
=&\ \frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}\\
=&\ \frac{n-\sqrt{n^2+n}}{2(\sqrt{n^2+n}+n)}\\
=&\ -\frac{n}{2(\sqrt{n^2+n}+n)^2}.
\end{align*} Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$
\begin{align*}
\left|s_n-\dfrac{1}{2}\right|
=&\ \frac{n}{2(\sqrt{n^2+n}+n)^2}\\
<&\ \frac{n}{2(n+n)^2}=\frac{1}{8n}\\
<&\ \frac{1}{n} <\frac{1}{N}=\epsilon
\end{align*} as desired. Hence $\lim s_n=\dfrac{1}{2}$.
### Part c
Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
$$\sqrt{4n^2+n}-2n=\dfrac{(4n^2+n)-(2n)^2}{\sqrt{4n^2+n}+2n}=\frac{n}{\sqrt{4n^2+n}+2n}$$ and
\begin{align*}
s_n-\dfrac{1}{4}
=&\ \frac{n}{\sqrt{4n^2+n}+2n}-\frac{1}{4}\\
=&\ \frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}\\
=&\ -\frac{n}{4(\sqrt{4n^2+n}+2n)^2}.
\end{align*} Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$
\begin{align*}
\left|s_n-\dfrac{1}{4}\right|
=&\ \frac{n}{4(\sqrt{4n^2+n}+2n)^2}\\
<&\ \frac{n}{4(2n+2n)^2}=\frac{1}{64n}\\
<&\ \frac{1}{n} <\frac{1}{N}=\epsilon
\end{align*} as desired. Hence $\lim s_n=\dfrac{1}{4}$. | 951 | 1,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-30 | latest | en | 0.448459 |
https://pedestrianobservations.com/2019/04/21/stop-spacing-and-route-spacing/?replytocom=59587 | 1,618,726,377,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00183.warc.gz | 554,812,349 | 38,519 | # Stop Spacing and Route Spacing
Six months ago I blogged a model for optimal stop spacing on an urban transit route. These models exist in the published literature, but they assume that the speed benefit of stop consolidation reduces operating costs, which requires introducing new variables for the value of time. My model assumes the higher speed of stop consolidation is plugged into higher frequency, which means only five variables are needed, and only two of them vary substantially between different cities and their networks. The formula is a square root.
In this post, I’m going to extend this formula to optimizing route spacing on a grid.
I’m using mode-neutral language like “vehicle,” but this is really just about buses, because to a good approximation, urban rail networks are never grids. I’m sorry, Mexico City, I know your Metro network does its best to pretend you have an isotropic city, but your three core radial lines are just far busier than the tangential ones.
Optimal stop spacing: a recap
My previous post uses words rather than symbolic language, since there are only five relevant parameters. Here I’m going to use symbols for the variables to make the calculation even somewhat tractable. All units I’m using are base SI units, so speed is expressed in meters per second rather than kilometers per hour, but the dimensional analysis works out so that it’s not necessary to pick units in advance.
• s: stop spacing
• v: walk speed
• p: stop penalty
• d: average distance traveled
• w: walk/wait penalty, expressed as a ratio of perceived walk or wait time to in-vehicle time
• λ: average distance between successive vehicles, or in other words headway in units of distance, not time
The variables v and p are fairly consistent from place to place. The variable w is as well, but may well differ by circumstance, e.g. people with luggage may have a higher walk penalty and a lower wait penalty, and people who are more familiar with the system usually have lower w. The parameter λ is a function of how much service runs on the line, as we will see when we expand to cover route spacing.
A key assumption in this model is that d does not change based on the network. This is a simplification: if s is too low then it will drag down d with it, as people who are discouraged by the slow in-vehicle speed avoid long trips or choose other modes of travel, whereas if s is too high then it will drag d up, as people who have to walk too long to the stop may just walk all the way to their destination if it’s nearby. In Carlos Daganzo’s textbook this situation is resolved by replacing an empirically determined d with the size of the city, assuming travel is isotropic, but the effect is essentially the same as just setting d to be half the length of a square city.
The formula for perceived travel time is
$\frac{sw}{2v} + \frac{dp}{s} + \frac{\lambda wp}{2s}$
if travel along the line is isotropic, or
$\frac{sw}{4v} + \frac{dp}{s} + \frac{\lambda wp}{2s}$
if one end of the travel (e.g. the residential end) is isotropic and the other is at a fixed node (e.g. a subway transfer). In either case, in-vehicle time excluding stops is omitted, as it is constant.
The minimum travel time occurs at
$s = \sqrt{2\cdot \frac{v}{w}\cdot p\cdot(d + \frac{\lambda w}{2})}$
if travel is isotropic and
$s = \sqrt{4\cdot \frac{v}{w}\cdot p\cdot(d + \frac{\lambda w}{2})}$
if there is a distinguished node at one end of the trip.
Observe that there is negative interaction between stop consolidation and other aspects of bus modernization. First, higher frequency, as expressed in concentrating service on strong routes, reduces the value of λ and therefore slightly reduces the optimal stop spacing. Second, the model assumes the same penalty w for walking and waiting, but sometimes these two activities have distinct penalties, and then the walk penalty is responsible for the occurrence of w in the denominator in the formula whereas the wait penalty supplies the appearance of w in the numerator. Improving bus stop facilities reduces the wait penalty, pushing the optimal s farther down, even though at the same time it’s cheaper to improve bus stops if there are fewer of them.
The empirically determined values of the five variables in the formula are as follows:
• v is 1.45 m/s in Forde-Daniel, 1.3-1.4 m/s in Bohannon, and 1.38 in TRB Part 4, PDF-p. 16; I take v = 4/3
• p is 25 seconds based on examining the differences in schedules between local and limited buses in New York and Vancouver
• d is 3,360 meters per unlinked trip per the NTD
• w is around 2 for waiting in Fan-Guthrie-Levinson, 2 in general for buses in Teulings-Ossokina-de Groot, PDF-p. 25, 1.75 in the New York MTA’s internal model, 2.25 in the MBTA’s (as mentioned in one of Reinhard Clever’s papers), and a range of 2-3 in Lago-Mayworm-McEnroe; I take w = 2
• λ is single-lane network length (that is, twice the route-length, modulo one-way loops) divided by fleet size in actual use, which is 1,830 meters in Brooklyn today and 1,160 based on what Eric Goldwyn and I recommend
This leads to optimal stop spacing equal to
$s = \sqrt{2\cdot \frac{4/3}{2}\cdot 25\cdot(3360 + \frac{1160\cdot 2}{2})} = 388 \mbox{ meters}$
if travel is isotropic and
$s = \sqrt{4\cdot \frac{4/3}{2}\cdot 25\cdot(3360 + \frac{1160\cdot 2}{2})} = 549 \mbox{ meters}$
if there is a distinguished node. The numbers are slightly lower than in my older post since I’m using a slightly lower walk speed, 1.33 m/s rather than 1.5.
Optimal route spacing: stops at intersection points
Studying route spacing has to incorporate stop spacing for a simple reason: there should be a stop at every intersection between routes, and therefore the route spacing should be an integer multiple of the stop spacing. There are three modifications required to the above formula, of which the first is easy, the second requires defining more parameters but is mathematically still easy, and the third is very hard:
1. Passengers need to walk not just along the route to their stop but also from their origin to the route, which increases walk time
2. The value of λ may change, since fewer routes imply more vehicles per route and thus denser vehicle spacing, and in particular wait time depends not just on how many stops are on the way but also on the speed net of stops
3. Increasing the route and stop spacing in tandem reduces the number of stops involved in waiting for the bus (this is λ again) twice, that is quadratically
The first modification means that instead of traveling an average distance of s/4 to the stop at each end, assuming isotropy, people have to travel a distance of s/4 along the route and also s/4 to the route itself. In the travel time formula, we replace sw/2v with just sw/v with isotropic travel.
To deal with the second modification, we define the following variables, in addition to the ones from the section above on stop spacing:
• f: fleet size in independent vehicles in actual revenue operation (buses or trains, not train cars)
• a: area of the network to be covered by the grid, e.g. a city, metro area, or borough
• u: speed assuming there are no stops along the route
If the area is a, then we can approximate it as a square of side $\sqrt{a}$, which has $\sqrt{a}/s$ north-south and $\sqrt{a}/s$ east-west routes, each of length $\sqrt{a}$, and thus the total two-way network length is 2a/s. Since the value of λ is the one-way length divided by fleet size, we write
$\lambda = \frac{4a}{sf}$
Moreover, people wait an additional λw/2u; in the previous section this wait existed as well but was ignored in the formula as it did not depend on s, but here it does, and thus we need to add this wait factor.
We deal with the third modification by replacing λ with 4a/sf in the formula for wait time. If people travel isotropically and do not transfer, the travel time formula is now
$\frac{sw}{v} + \frac{dp}{s} + \frac{d}{u} + \frac{2aw}{sfu} + \frac{2awp}{fs^{2}}$
The summand d/u is constant but is included for completeness here, in analogy with the no-longer-constant summand 2aw/sfu.
But it’s the last summand that gives the most problems: it turns the optimization problem from extracting a square root to solving a cubic. This is technically possible, but the formula is opaque and does not really help showcase how the parameters affect the final outcome. We need to solve for s:
$\frac{w}{v}s^{3} - (dp + \frac{2aw}{fu})s - \frac{4apw}{f} = 0$
We can plug in the above values of w, v, d, and p, as well as the following values of the new variables, and use any cubic solver:
• f = 612 buses in Brooklyn, excluding vehicles in turnaround, non-revenue service, etc. (it’s actually slightly lower today, around 600, but our network is a bit more efficient with depot moves)
• a = 180,000,000 m^2 for Brooklyn
• u = 5.3 m/s net of stops, assuming our other proposals, such as bus lanes, are implemented
The cubic formula turns into
$1.5s^{3} - 305976s - 58823529 = 0$
for which the positive solution is s = 528 meters.
We can complicate this formula in two ways.
First, we can let go of the assumption of isotropy. If there is a distinguished node at one end, then walk time is halved, as in the formula for stop spacing on a given route. The overall travel time is equal to
$\frac{sw}{2v} + \frac{dp}{s} + \frac{d}{u} + \frac{2aw}{sfu} + \frac{2awp}{fs^{2}}$
and this is optimized when
$\frac{w}{2v}s^{3} - (dp + \frac{2aw}{fu})s - \frac{4apw}{f} = 0.$
Plugging the usual values of the parameters, we get
$0.75s^{3} - 305976s - 58823529 = 0,$
for which the positive solution is s = 719 meters. The ratio between the results with isotropy and a distinguished node is 1.36, close to the square root of 2 that we get in the formula for stop spacing on a predetermined route; the reason is that in the cubic formula the linear term is much larger than the constant term near the root, so the effect of changing the cubic term is much closer to the square root than to the cube root.
The second complication is introducing transfers. Transfers do not change the walk time – the walking time between platforms or curbside waiting areas is small and constant – but introduce additional wait time, which means we need to double both terms that include waits. But if we have transfers we need to restore the assumption of isotropic travel, since for the most part the distinguished nodes for Brooklyn buses involve subway transfers.
In that case, the travel time formula is
$\frac{sw}{v} + \frac{dp}{s} + \frac{d}{u} + \frac{4aw}{sfu} + \frac{4awp}{fs^{2}}$
which is minimized at the positive root of the cubic
$\frac{w}{v}s^{3} - (dp + \frac{4aw}{fu})s - \frac{8apw}{f} = 0.$
We need to figure out the value of d, which is difficult in this case – the New York bus network discourages bus-to-bus transfers through low frequency and poor bus stop amenities. That the formulas I’m using do not allow for how the shape of the network influences d is a real drawback here. But if we let d be the usual 3,360 meters that it is for unlinked trips, and plug the usual values of the other parameters, we get,
$1.5s^{3} - 527951s - 117647059 = 0$
to which the solution is s = 683 meters.
Optimal route spacing: the general case
The above section makes a critical assumption about route spacing and stop spacing: they must be equal, making every stop a transfer. However, this assumption is not strictly necessary. Indeed, if we assume isotropy, and let the route spacing be 860 meters, then it’s better for passengers to double the density of stops to one every 430 meters just from looking at the formula for stop spacing.
In this section, we look at the optimal formulas assuming route spacing is twice or thrice the stop spacing. Then in the next section we will compare everything together.
We keep all the variable names from before, and set s to be the stop spacing, not the route spacing. Instead, we will find formulas for route spacing equal to 2s and 3s and compare their optima with that for the special case in which stop and route spacing are equal.
We need to modify the formula in the previous section in two ways. First, walk time is, in the isotropic case, half the stop spacing plus half the route spacing. And second, the dependence of λ on the shape of the network comes from route spacing rather than stop spacing. If route spacing is 2s, the formula for travel time is
$\frac{3sw}{2v} + \frac{dp}{s} + \frac{d}{u} + \frac{aw}{sfu} + \frac{awp}{fs^{2}}$
and its minimum is at the positive solution to
$\frac{3w}{2v}s^{3} - (dp + \frac{aw}{fu})s - \frac{2apw}{f} = 0.$
We retain the New York- and Brooklyn-oriented variables from the above sections and obtain
$2.25s^{3} - 194989s - 29411765 = 0.$
The solution is s = 352 meters, i.e. routes are to be spaced 704 meters apart, with one intermediate station on each route between each pair of successive crossing routes.
If we have three interstation segments between two successive routes, then we need to solve the cubic
$\frac{2w}{v}s^{3} - (dp + \frac{2aw}{3fu})s - \frac{4apw}{3f} = 0$
or
$3s^{3} - 157992s - 19607843 = 0$
to which the solution is s = 276 meters.
In the above section we also looked at two potential complications: introducing transfers, and introducing non-isotropy. Non-isotropy, expressed as an isotropic origin and a distinguished destination, halves the cubic term; transfers double the wait times and thus double the constant term and the larger of the two summands adding up to the linear term.
If the route spacing is exactly twice the stop spacing, then the non-isotropic formula is
$\frac{3w}{4v}s^{3} - (dp + \frac{aw}{fu})s - \frac{2apw}{f} = 0$
or, using the same parameters as always,
$1.125s^{3} - 194989s - 29411765 = 0.$
The solution is s = 420 meters, with routes spaced 840 meters apart.
The isotropic cubic with transfers is
$\frac{3w}{2v}s^{3} - (dp + \frac{2aw}{fu})s - \frac{4apw}{f} = 0$
and with the usual parameters, again sticking with d = 3,360 even though in practice it is likely to be higher, this is
$2.25s^{3} - 305976s - 58823529 = 0$
and then the root is s = 442 meters, with routes spaced 884 meters apart.
We conclude this section with the same formulas assuming the route spacing is not 2s but 3s. The non-isotropic, one-seat ride formula is
$\frac{w}{v}s^{3} - (dp + \frac{2aw}{3fu})s - \frac{4apw}{3f} = 0$
or with the usual parameters
$1.5s^{3} - 157992s - 19607843 = 0,$
of which the positive root is s = 374 meters, with routes spaced 1,123 meters apart,
The transfer-based isotropic formula is,
$\frac{2w}{v}s^{3} - (dp + \frac{4aw}{3fu})s - \frac{8apw}{3f} = 0$
or
$3s^{3} - 231984s - 39215686 = 0.$
The positive root is s = 340 meters, with routes spaced 1,021 meters apart.
What’s the best route spacing?
We have optimums based on assumptions about the interaction between stop and route spacing, but so far we have not compared these assumptions with each other. In this section, we do. For each scenario – isotropic, transfer-free travel; a distinguished node along transfer-free travel; and isotropic travel with a transfer – we look at the optimal values of route spacing equal to one, two, or three times the stop spacing.
In the table below, the walk and wait times are without penalty; but the penalty is applied to them when summed with in-vehicle time.
Scenario Component Route spacing = s Route spacing = 2s Route spacing = 3s Isotropy; 1-seat ride Optimal s 528 352 276 Walk time 396 396 414 Wait time 262.954 216.997 198.394 In-vehicle time 793.053 872.599 938.31 Total time 2110.962 2098.593 2163.097 Distinguished node; 1-seat ride Optimal s 719 420 374 Walk time 269.625 236.25 280.5 Wait time 182.811 173.812 133.965 In-vehicle time 750.791 833.962 858.561 Total time 1655.663 1654.086 1687.49 Isotropy; 2-seat ride Optimal s 683 442 340 Walk time 512.25 497.25 510 Wait time 388.05 326.378 302.432 In-vehicle time 756.949 824.008 881.021 Total time 2557.549 2471.263 2505.885
The table implies that in all scenarios it’s optimal to have two interstations between parallel routes, though if there’s a distinguished node the difference with having just one interstation between parallel routes is very small. The three-interstation option is never optimal, but is also never far from the optimum, only half a minute to a minute worse.
But please interpret the table with caution, especially the two-seat ride section. The total time for a 3.36-kilometer trip without applying the walk or wait penalty is about 28 minutes regardless of whether the route to stop spacing ratio is 1, 2, or 3. This is still faster than walking, but not by much, and riders may well be so discouraged as to walk the entire way. If the trip is much shorter than 3.36 kilometers or the rider’s particular disutility of walking is much lower than 2 then transit will not be competitive with walking. In turn, a network set up with the stop spacing implied by the above formulas will only get transfer trips if they’re much longer, which should raise the optimal interstation somewhat. If d = 6,000 then in the transfer scenario the optimum if stop and route spacing are equal is 711 meters and that if route spacing is twice as high as stop spacing is 470 meters, and the latter option is noticeable faster.
How does our bus redesign compare with the theory?
We drew our redesigned map with full knowledge of how to optimize stop spacing on a single route, but we didn’t look at route spacing optimization. Of course, the assumption of regular route spacing is less realistic than that of regular stop spacing, as some areas have higher demand, or more distinguished arterials. But we can still discuss the average route spacing in our plan, by comparing our proposed route-length with Brooklyn’s land area.
With a 356-kilometer network in a borough of 180 km^2, effective route spacing is 1,010 meters. This is a little longer than I expected; in Southern Brooklyn the north-south and east-west routes we propose are spaced around 800-850 meters apart, and in Bed-Stuy the east-west routes tighten to 600 meters as they’re all radial toward Downtown Brooklyn and quite busy. The reason the answer is 1,010 meters is that there are margins of the borough with no service (like Floyd Bennett Field) or grid interruptions due to parks (such as Prospect Park) or already-good subway service (South Brooklyn).
The stop spacing we use is 480 meters, excluding nonstop freeway segments in the Brooklyn-Battery Tunnel and toward JFK. In the Southern Brooklyn grid, we’re pretty close to a regular spacing of two interstations between parallel routes. In the Bed-Stuy grid, the north-south routes have a stop per crossing route since the east-west routes are so densely placed, and the east-west routes have one, two, or three interstations between crossing routes, but the average is two.
To the extent the optimization formulas tell us anything, it’s that we should consider adding a few more routes. Target additions include another north-south Bed-Stuy route, an east-west route in South Brooklyn restoring the discontinued B71, and a north-south route through Southern Brooklyn on 16th Avenue. Altogether this would add around 20 km to our network. Beyond that, additional routes would duplicate subway routes, which my analysis above excludes even when they form a coherent grid with the buses.
Rules of thumb for your city
If your city has streets that form a coherent grid, then you can design a bus grid without too many constraints. By constraints I mean street networks that interrupt the grid so often so as to force you to use particular streets at particular spacing, for example the Bronx or Queens. Constraints in a way make planning easier, by reducing the search space; I contend Brooklyn is the hardest of the four main boroughs to redesign precisely because it has the fewest constraints in its grids and yet its grid is just interrupted enough that it cannot be treated as tabula rasa.
In general, you probably want buses spaced around 800 meters to a kilometer apart. While the value of d will differ between cities, the optimum route spacing isn’t that sensitive to it. If d rises to as high as 10,000, the optimal s in the scenario with transfers is 753 meters if route spacing equals stop spacing and 511 meters if it equals twice stop spacing, compared with 683 and 442 meters respectively with d = 3,360; the one-interstation-per-parallel-route scenario becomes better than the two-interstation scenario, but the difference is half a minute, compared with a minute and a half in favor of two interstations with d = 3,360.
In practice I don’t know of any city whose grid is so unconstrained and so isotropic that you can seriously debate 700, 800, 900, 1,000, etc. meters between routes. At that resolution you’re always constrained by arterial spacing, which in American cities tends to be 800 because it’s half a mile and in Canada is irregular (de facto close to a mile) due to constant grid interruptions on intermediate would-be arterials in both Toronto and Vancouver. In this range of arterial spacing, you want exactly two interstations between parallel routes; if you want more or fewer then you should have a very good reason, such as a major destination such as a hospital located at an awkward offset.
Something that does matter very much is fleet size relative to the area served – the quantity a/f. If you aren’t running much service, then you need wider route spacing just to avoid reducing frequency to unusable levels. If instead of f = 612 we use f = 200, then the optimum with one interstation per parallel routes with the transfer scenario is s = 1087, with two it’s s = 676, with three it’s s = 508, and with four it’s s = 414, and among these three is best and even four is a few seconds faster than two. In that case route spacing of about a kilometer and a half, which may be a mile in American arterials, is fully justified.
Conversely, if buses are faster, that is if u is higher, then the optimal interstations fall in all cases. This is because the impact of u comes from its effect on wait times, so faster buses mean that it’s less important to reduce λ.
The effects of a/f and u relate again to the negative interactions between various components of bus reform. Running more service means it’s justifiable to have more closely-spaced routes, since pruning routes to increase frequency from 10 to 5 minutes is much less valuable than pruning them to increase frequency from 30 to 15 minutes. Likewise, running faster service means wait times fall, again reducing the need to prune routes.
If you’re tasked with designing bus routes, then make sure to use correct values for a, f, u, and d for your city, as they are likely to be very different from those of New York. The formulas are more intricate when optimizing route spacing and it’s useful to play with them until you get comfortable with them on an intuitive level, but ultimately they do give reasonable answers for how to design a bus network.
1. Joshua Larimore
Hi Alon, I’ve been following you for a bit now and having been loving a lot of your work. What do you think would be an effective map/route for an S-Bahn style method of rapid rail transit in places like Houston, Dallas-Fort Worth, the Research Triangle, Tampa Bay Area, Minneapolis-St. Paul, Detroit, Metrolina (Charlotte Metro), Denver, Phoenix, Miami, Atlanta, and other sprawling city and metro areas, or do you think that S-Bahn style service would not be effective? I would love to hear your opinion and see what you think would be optimal for stop placement and route designs.
Thank you and keep up the good work.
• Eric
I’m not Alon, but I would say such systems do not have a bright future. Those cities are generally very low-density and often without much of a central focus, which means the ridership will be low. Building new rail lines in such places would be extremely expensive per rider. Using existing rail lines does not work well either, as those are generally owned by freight companies who cannot be forced to give them up, and in any case they tend to go through desolate industrial areas where ridership is even lower. So I do think that for cities you list of under ~4 million population, light rail (or simply more frequent bus service) is the right way to go. It can reduce costs by using city streets as necessary, while having widely spaced stops like S-Bahn in the suburbs. In the forseeable future, ridership will not rise to the point where grade separated ROW is necessary.
As for the growing Sunbelt cities of 5+ million people, it’s more complicated. It is hard to build a quality transit system in these places (see above), but the threat of Los Angeles-style gridlock leaves them little choice. Atlanta and Miami have metro systems, which do quite well and should be extended. In Miami I think density can be focused on a few transit oriented corridors. In Atlanta the main gain of metro extensions may be to keep the freeways passable. In Houston, they have chosen LRT which works well for the inner neighborhoods, but at some point they will need longer distance transit. In Dallas, they will need to grade separate LRT in downtown, but the many branches can remain as they are.
• Joshua Larimore
Thank you for your opinion on the subject matter, and hearing you out, I do tend to agree, especially with cost which I did foresee being a major deterrent to construction and expansion. Light rail, or something like a premetro (Basically a light rail system that can later be expanded into a full metro) would probably be preferable, as it would also be cheaper. Having these lines extend into suburbs with an occasional park and ride station could ease congestion in these regions. I also agree that the Miami and Atlanta metros really need expansion and improvement, to include some more commuter rail connections. Los Angeles could use work to, there 2028 Plan is nice, but some arts of it could still use work, personally I wonder what Levy thinks of it.
As for Dallas, and expansion into Arlington and Fort Worth, I do agree that the lines are going to have to be grade separated, most likely with underground and tunnel stations. However, I personally think they could better serve some of the suburbs to ease congestion on the freeways. Mostly the same goes for Houston, Denver, and Phoenix which have a small system and are working to improve it.
The main point I was trying to advocate for was rather than focusing solely on Rapid Transit or Commuter systems, it would be better to bundle the two together (Including common zone-based fare system to both this, buses, and anything else) to capture a larger market like how in sprawling Perth and the Brisbane areas (And their equal or smaller populations in relation to the cities and metros I mentioned), their main commuter systems are able to capture 100-200 thousand daily riders and the Calgary (300 thousand daily) and Edmonton (100 thousand daily) Light Rail systems which act like both a commuter and rapid transit system even with the high vehicle per capita ownership in these countries like in the US.
2. ckrueger99
Thanks. This could be quite useful for the network redesign in Philly for SEPTA, which has a near perfect grid in the central zone.
• Alon Levy
Yep! I wasn’t thinking of Philadelphia when I wrote this post, but it’s a good example of a relatively unconstrained grid – I don’t think there’s much of a hierarchy to the numbered streets (except Broad).
• ckrueger99
5th, 6th and 7th are stroads north of Chestnut. 22nd is also wider than the others (which are generally 27 feet, curb to curb). 20th is 2-way north of Market. Broad is actually less useful as a through route due to the loop around City Hall (plus the subway means why bother). But nothing to constrain buses. The current constraint is lack of free transfers for à la carte fares, but this will change with the redesign.
The routes along Broad can be eliminated once the ADA access to the subway is complete. Also, route 4 along 16th & 17th (Broad is equivalent to 14th) is already pretty marginal, so it probably goes. 47M on 9th, ditto.
The key is to improve frequency on the cross-town routes, which already correspond to the Broad St subway stops (and El stops in the Kensington area), while reducing stop spacing and other techniques to improve speed. I’m working on a blog to this effect on http://transitjawn.com .
• Nathanael
I would really like to see this applied specifically to Philadelphia. It’s a city with a lot of transit advocates and a unified transit agency which may listen to good proposals; a lot of potential possible.
3. Fbfree
I ran these formulas for the core of Montreal (between Park, the river, Honoré-Beaugrand, and Jarry). Most of the variables are very similar to Brooklyn except for the bus density at peak. Running through the GTFS schedules, I estimated 343 buses at peak over 59 square kilometers. This significantly reduces the optimum route spacing compared to your calculation. Here’s a reproduction of your table
Scenario Component Route spacing = s Route spacing = 2s Route spacing = 3s
Isotropy; 1-seat ride Optimal s 447 305 243
Walk time 335 343 365
Wait time 197 159 161
In-vehicle time 870 959 1030
Total perc. time 1935 1963 2080
Distinguished node; 1-seat ride Optimal s 610 417 332
Walk time 228 234 249
Wait time 136 107 107
In-vehicle time 819 884 936
Total time 1549 1567 1648
Isotropy; 2-seat ride Optimal s 567 328 247
Walk time 425 369 371
Wait time 296 289 315
In-vehicle time 830 939 1024
Total time 2273 2257 2394
Here, route spacing of 3s underperforms, causing a reduction in bus frequency vs. the 2s case. 1s is optimum in all but the 2-seat ride case.
The existing network already has about the density of east-west routes suggested by this analysis, although the spacing between north-south routes is significantly wider. However, many of these are currently run at much lower frequency (often every 30 minutes) and with large peak-to-midday service ratios. The frequent route network is currently a shell of the full network.
• Alon Levy
Just one question: is 343 the total fleet size or the number of buses in actual circulation, i.e. excluding spares, deadheads, and turnarounds? I’m asking because the actual fleet requirement for Eric’s and my plan is about 1,000, but only 600 buses circulate at any given time.
• Fbfree
Hi Alon,
That’s the number of buses in circulation at peak from GTFS data. Thus it excludes out of service deadheads and spares. It does include turnarounds. 343 is an estimated based on which routes are bounded by the area under analysis and estimating by eye the fraction of routes that are not fully contained in that area. I estimate it’s accurate to about +/- 5%. That’s more accurate than my estimate of u which is based on the average of Google typical drive time searches along the length of Beaubien and Rosemont. u could be off by around +15%/-30%.
4. Diego Beghin
I’m finally getting around to reading this. One comment before I keep going: I think the numerator and denominator are mixed up here. If the wait penalty was in the denominator, s would go up when w goes down.
Second, the model assumes the same penalty w for walking and waiting, but sometimes these two activities have distinct penalties, and then the walk penalty is responsible for the occurrence of w in the numerator in the formula whereas the wait penalty supplies the appearance of w in the denominator. Improving bus stop facilities reduces the wait penalty, pushing the optimal s farther down, even though at the same time it’s cheaper to improve bus stops if there are fewer of them.
• Alon Levy
Ugh, thanks for the catch. The formula is correct but the text flipped numerator and denominator. I corrected it, thanks!
5. Pingback: Circumferential Lines and Express Service | Pedestrian Observations
6. Edward
Doesn’t stop penalty depend on top speed in the form of acceleration? Maybe this isn’t an issue for buses if they deal with traffic lights and congestion, but it should be for rail.
• Alon Levy
Yeah, definitely. But usually urban rail stop penalties are mostly dwell time rather than accelerating to high speed, so while there is a range, it’s not an especially wide one.
7. Pingback: Density and Subway Stop Spacing | Pedestrian Observations
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 7,867 | 32,533 | {"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": 33, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-17 | latest | en | 0.928398 |
https://stat.ethz.ch/pipermail/r-help/2003-January/028742.html | 1,627,875,586,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00289.warc.gz | 540,703,313 | 3,534 | # [R] Multivariate regression in R
(Ted Harding) Ted.Harding at nessie.mcc.ac.uk
Thu Jan 16 21:37:03 CET 2003
```On 16-Jan-03 Huntsinger, Reid wrote:
>>From the equation you write, I don't see why "lsfit" wouldn't work.
> To estimate the covariance matrix of e you could use the sample
> covariance matrix of the residuals. If desired, use its cholesky
> decomposition to transform to make the error approximately
> uncorrelated, then refit (and back-transform the coefficient matrix).
>
> Stacking the columns of Y and replicating X won't do what you write;
> it forces each univariate regression to have the same coefficients.
> To get what you wrote you would replicate X in blocks of a
> block-diagonal
> matrix.
>
> I'm not sure I understand the role of W. If what you want is to fit
> equations like
>
> y(i) = a(w(i)) + xB(i) + e(i)
>
> for each observation and each response y(1),...,y(p), then I guess you
> could just fit each response separately, treating W as a factor. Then
> estimate the covariance matrix of e via the residuals as before, etc.
Thanks Reid! To clarify, let me spell out a simple example:
Y X W
y11 y12 y13 x11 x12 1 2 3
y21 y22 y23 x21 x22 1 3 2
y31 y32 y33 x31 x32 2 1 3
y41 y42 y43 x41 x42 2 1 3
y51 y52 y53 x51 x52 3 1 2
....
is a matrix representation of the data. Here, Y is the matrix of N
3-variate responses, X is an Nx2 matrix of quantitative covariates,
W is an Nx3 matrix of levels of the 3-level qualitative factor W
where Wij shows which level of W is associated with the jth variate
of Yi (the ith row of Y). After allowing for the effects of X and W,
it is expected that the residuals
e
e11 e12 e13
e21 e22 e23
e31 e32 e33
e41 e42 e43
e51 e52 e53
....
will be correlated between columns (in fact reflecting a real mechanism
in the process generating the data), and for modelling it is supposed
that (e1,e2,e3) have a joint Normal distribution with covariance matrix S.
The additive X component of the model can be written as X*B where
B = b11 b12 b13
b21 b22 b23
and the bij are the regression coefficients of Y on the quantitative
covariates (X1,X2).
The additive W component could be written in matrix form (for the above
"data") as:
W1 W2 W3
w1 * 1 0 0 + w2 * 0 1 0 + w3 * 0 0 1
1 0 0 0 0 1 0 1 0
0 1 0 1 0 0 0 0 1
0 1 0 1 0 0 0 0 1
0 1 0 0 0 1 1 0 0
... ... ...
where
col j of W1 is 1 where the jth variate in Y has level 1 of W (j = 1,2,3),
col j of W2 is 1 where the jth variate in Y has level 2 of W (j = 1,2,3),
col j of W3 is 1 where the jth variate in Y has level 3 of W (j = 1,2,3)
and w1, w2, w3 are the scalar magnitudes of the "effects" of levels
1, 2, 3 of W.
Hence the multivariate regression model for the data could be written in
matrix form as
Y = X*B + w1*W1 + w2*W2 + w3*W3 + e
where e is 3-dim N(0,S), and B, w1, w2, w3 and S are to be estimated.
What, in R, I can't make out how to do is to give some function (which
function?) a model specification of the form
Y ~ X + W1 + W2 + W3
but in such a way that it will fit a 2x3 matrix B of coefficients for X,
but scalar coefficients w1, w2, w3 for W1, W2, W3 (and also come up with
the estimated covariance matrix for the residuals e, though this, as you
point out, could be obtained after the fit from the Nx3 residuals;
however, it should be available from the function since it would have
to be used in fitting the model).
Analytically, the log-likelihood can be written (summing over r)
(-N/2)*log(det(S)) - 0.5*SUM[ e_r' * S^(-1) * e_r ] (e_r = rth row)
where e = Y - B*X - w1*W1 - w2*W2 - w3*W3. After differentiation and
algebra, one could implement the resulting matrix equations in octave
(or matlab) and proceed to a solution. One could even do this, as a
numerical procedure, in R -- but I'd rather not! Indeed, R's richness
in model-fitting resources tempts one to think that this problem may
be solvable using these -- it's just that I can't seem to put my hand
on what's needed.
Your response suggests that the way to go is to fit covariates and
factors to each Y-variate in turn, take the residuals from these fits,
and estimate the covariance matrix S. But thereafter you have to get
into considering them jointly, in order to improve the fit iteratively,
since S is involved in the joint weighting to be applied.
So I'm still not seeing how to do this in R ...
Best wishes,
Ted.
--------------------------------------------------------------------
E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>
Fax-to-email: +44 (0)870 167 1972
Date: 16-Jan-03 Time: 20:29:01
------------------------------ XFMail ------------------------------
``` | 1,450 | 4,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-31 | latest | en | 0.907997 |
https://amansingh-javatpoint.medium.com/calculator-program-in-python-7e7337e867b3?source=post_internal_links---------0------------------------------- | 1,642,328,316,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00627.warc.gz | 165,752,266 | 26,645 | # Calculator Program in Python
Simple Calculator Program in Python
This example helps to learn how to create a simple calculator program to perform operations like addition, subtraction, multiplication, and division.
Source Code
The following is the source code to perform simple calculations. In this program, we will perform addition, subtraction, multiplication, and division operations.
Function to perform addition operation
return num1 + number2
Function to perform subtraction operation
def subtraction(num1, num2):
return num1 — num2
Function to perform multiplication operation
def multiplication(num1, num2):
return num1 * num2
Function to perform division operation
def division(num1, num2):
return num1 / num2
Choose operation
print(“Select desired operation: \n”
“2.Substraction\n”
“3.Multiplication\n”
“4.Division”)
while True:
# Take input from the user
operation = input(“Choose operation(1/2/3/4): “)
# Check if operation is one of the four options if operation in (‘1’, ‘2’, ‘3’, ‘4’):
number1 = float(input(“Enter first number: “))
number2 = float(input(“Enter second number: “))
if operation == ‘1’:
#Print sum of two number
print(number1, “+”, number2, “=”, addition(number1, number2))
elif operation == ‘2’:
# Print subtraction of two number
print(number1, “-”, number2, “=”, subtraction(number1, number2))
elif operation == ‘3’:
# Print multiplication of two number
print(number1, “*”, number2, “=”, multiplication(number1, number2))
elif operation == ‘4’:
# Print division of two number
print(number1, “/”, number2, “=”, division(number1, number2))
break else:
print(“Wrong operation selected”)
https://www.tutorialandexample.com/python-mysql/ | 417 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.772323 |
http://www.ck12.org/search/?q=tag:distributive%20property | 1,487,762,213,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170940.45/warc/CC-MAIN-20170219104610-00536-ip-10-171-10-108.ec2.internal.warc.gz | 355,381,918 | 15,417 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Filters clear all
GRADES
SUBJECTS
Math
Science
English
More
CATEGORIES
Search Results (289 results for "tag:distributive property" ) Filters
We couldn't find what you are looking for.
Some suggestions:
Try searching for in Community Content
Make sure all words are spelled correctly.
Try different keywords.
Try more general keywords.
Showing results for:
Algebra Working with Real Numbers
The distributive property dictates that when multiplying, the multiplication of real numbers distributes over two or more terms in parentheses. Learn more.
Concept
Catching Fireflies PLIX
Distributive Property Interactive
Grade: 6
Created by: CK-12 Subject: Algebra Difficulty level: At Grade
PLIX
Party Planning Real World
Discover how the distributive property can help you determine how much you will spend on food for an event.
Grade: 7
Created by: CK-12 Difficulty level: At Grade
Real World
Ecuaciones con decimales, fracciones y paréntesis Read
Grade: 9, 10
Created by: CK-12 Subject: Algebra Difficulty level: Advanced
Read
Suma y resta de polinomios Read
Grade: 9, 10
Created by: CK-12 Subject: Algebra Difficulty level: Advanced
Read
Multiplicación de polinomios Read
Grade: 9, 10
Created by: CK-12 Subject: Algebra Difficulty level: Advanced
Read
Productos especiales de polinomios Read
Grade: 9, 10
Created by: CK-12 Subject: Algebra Difficulty level: Advanced
Read
Equations with the Distributive Property Read
This concept introduces solving equations that require using the distributive property.
Grade: 9, 10
Created by: CK-12 Subject: Algebra Difficulty level: At Grade
Read
Expresiones y la Propiedad Distributiva Read
Grade: 8, 9
Created by: CK-12 Subject: Algebra Difficulty level: Basic
Read
Cuando Usar la Propiedad Distributiva Read
Grade: 8, 9
Created by: CK-12 Subject: Algebra Difficulty level: Basic
Read
GRADES
SUBJECTS
Math
Science
English
More
CATEGORIES
clear all
Please wait...
Please wait... | 571 | 2,048 | {"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} | 3.734375 | 4 | CC-MAIN-2017-09 | latest | en | 0.609311 |
https://www.doorsteptutor.com/Exams/NSTSE/Class-3/Questions/Part-41.html | 1,624,024,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00133.warc.gz | 669,330,181 | 7,402 | # NSTSE (National Science Talent Search Exam- Unified Council) Class 3: Questions 315 - 320 of 2216
Access detailed explanations (illustrated with images and videos) to 2216 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!
View Sample Explanation or View Features.
Rs. 550.00 -OR-
## Question 315
### Question
MCQ▾
Total value of money in the purses shown below:
### Choices
Choice (4)Response
a.
1350
b.
2500
c.
2530
d.
2503
## Question 316
### Question
MCQ▾
total number of chocolates in the boxes shown below:
### Choices
Choice (4)Response
a.
132
b.
131
c.
136
d.
130
## Question 317
### Question
MCQ▾
Find the missing number in the following addition?
### Choices
Choice (4)Response
a.
3
b.
8
c.
2
d.
5
## Question 318
### Question
MCQ▾
Kina counted 20 Bikes, 28 Cars and 18 Trucks at the road. What was the total number of Bikes and Cars?
### Choices
Choice (4)Response
a.
48
b.
25
c.
40
d.
45
## Question 319
### Question
MCQ▾
The table below shows the number of cars needed for different numbers of students to go on a camping trip.
Number of car 1 2 3 4 5 6 Number of students 7 14 21 28 35 ?
If the pattern in the table continues how many students are able to ride in 6 cars?
### Choices
Choice (4)Response
a.
49
b.
42
c.
63
d.
56
## Question 320
### Question
MCQ▾
If represents 1, then write the number represented by the diagram?
### Choices
Choice (4)Response
a.
436
b.
346
c.
643
d.
463
Developed by: | 472 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-25 | latest | en | 0.684067 |
http://www.jiskha.com/display.cgi?id=1332290862 | 1,495,784,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00260.warc.gz | 658,485,203 | 3,801 | # science
posted by on .
If the greatest potential energy is 100J will the kinetic energy of the object ever reach 100J?
• science - ,
let's use a block. When a block is released from its rest place on a table, it has:
Potential Energy = 100J
Kinetic Energy = 0
when a block is falling halfway down, it has:
Potential Energy = 50J
Kinetic Energy = 50J
when a block is falling just before it hits the ground, it has:
Potential Energy = 0
Kinetic Energy = 100J | 124 | 464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-22 | latest | en | 0.855778 |
https://math.stackexchange.com/questions/1545220/famous-burning-ropes-optimal-measuring-in-general-case | 1,585,859,699,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370507738.45/warc/CC-MAIN-20200402173940-20200402203940-00229.warc.gz | 566,119,481 | 32,738 | # Famous Burning Ropes: Optimal Measuring in General Case
well known riddle: given two ropes that each take an hour to burn (they do not burn uniformly or identically, necessarily) measure 45 minutes. solution: light 3 out of the 4 ends, and then light the 4th end when the first rope is totally burned up.
generalization (and the topic of discussion):
1) given an infinite number of such ropes, which times are measurable? Assume you can light each rope only a finite number of times.
2) additionally, what is the optimal strategy to measure a given amount of time, i.e. the fastest way to measure?
attempt at solution:
intro: take the base 2 representation of any amount of time you wish to measure, where 1 represents 1 hour. to measure the time to the left of the decimal point burn the appropriate number of ropes sequentially. to the right of the decimal it gets trickier. for numbers less than 1 I believe it is possible to measure any finite linear combination of elements of the form "sum(1/2^i)" where i ranges from a natural number to infinity. I will sketch an algorithm/proof below.
gaps: first define the number of gaps g(x) as the number of sequences of consecutive 0's bounded by 1's in the base 2 representation. for example g(.010101) = 2 and g(.0100010101) =3. Now actually modify this definition slightly - we do not count gaps bounded on the left by a 1 in the 1/2 slot, e.g. g(.101) = 0 and g(.10010101) = 2. the intuition here is that measuring 1/2 is as simple as measuring 1 - we can instantly measure 1/2 by burning 2 ends of one rope, whereas smaller numbers require some lead up time, e.g. you must measure 1/2 before you can measure 1/4.
measurement sequence: note that to measure 1/2^k for k > 1 you must first measure 1/2, 1/4,.., 1/2^(k-1) convince yourself of this or tell me why I am wrong. from now on I will refer to the sequence of measurements 1/2, 1/4,.., 1/2^k as the "measurement sequence" for 1/2^k.
sketch of an algorithm that takes approximately n hours to measure x, where x<1 and g(x)=n:
1) rewrite x as a sum of numbers with 0 gaps, e.g. .010101 = .01 + .0001 + .000001
2) to measure 1/2^k takes 1-(1/2^k) amount of time, so it seems you should start the "measurement sequence" (1/2, 1/4, ..., 1/2^k) for the smallest number first. Likewise you should start the measurement sequence for the 2nd smallest number a tad later, and so on.
3) You now have g(x) "measurement sequences" that you need to time correctly. continuing with the example: you want to measure .01 immediately followed by .0001 immediately followed by .000001 (note, it seems you should measure the smallest number last since it will have the most expensive "measurement sequence"). notice, the measurement sequence for .0001 should start .00001 after the measurement sequence for .000001. We can time this delay using a measurement sequence for .00001 - let's call this a coordination sequence since it allows us to coordinate a delay between two measurement sequences.
4) we started out with 3 measurement sequences, and obtain two coordination sequences - one coordination sequence to make sure the 2nd measurement sequence starts .00001 after the 1st, and one coordination sequence to make sure the 3rd measurement sequence starts .001 after the 2nd. Now we need to coordinate the two coordination sequences with a "coordination^2" sequence - convince yourself or tell me I'm wrong.
5) convince yourself that for g(x) = n we get n measurement sequences, which we need to coordinate with n-1 coordination sequences, which we need to coordinate with n-2 coordination^2 sequences... which we need to coordinate with 1 coordination^(n-1) sequence, and that the whole process of coordinating and finally measuring takes a little less than n hours.
Under this algorithm, it seems we can measure any time, except times x such that g(x)=infinity. Is there a faster algorithm for measuring different times? Have I properly classified the set of measurable times? Thanks. I know my explanation is not super clear; ready to clarify any points.
on one hand, we can only light an end at a time measured by other ropes, and since the rope is burned up at the time $(t_1+t_2+1)/2$ (where $t_1$ and $t_2$ are the times when the ends were lighted), any one rope allows us to go only one binary digit deeper, so no numbers with infinite binary expansion can be measured by arbitrary large but finite number of ropes (infinite number of ropes with finite time won't help, too);
on the other hand, with $k$ ropes we can easily measure the time $1-2^{-k}$ from the moment when we begin burning the first of these $k$ ropes, so if we measure $g(x)$ such intervals one after another (in any order) while simultaneously burning $g(x)$ whole ropes one after another, the latter process will be exactly $x$ longer than the former, so any number $x$ with finite binary expansion can be measured. | 1,205 | 4,903 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-16 | latest | en | 0.933872 |
https://quant.stackexchange.com/questions/14132/reference-question-about-portfolio-optimization/14139 | 1,627,264,026,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00172.warc.gz | 493,775,785 | 39,330 | # reference question about portfolio optimization
I know the "classical" modern portfolio theory. However I have quite a lot of different sources. It seems that there is not a book which cover this topic in a rigorous way:
1. theory
2. application
3. examples in c++ / R / matlab / ...
Are there any books, which cover these points?
Recently, we discussed in class a dynamic portfolio optimization problem. Now I would like to know if there is any good literature about this topic? With dynamic portfolio optimization I mean changing constraints over time. Many thanks for sharing your experiences.
• Why do you want to use the "classical" approach? The portfolio weights are super sensitive to the inputs of expected return. You can never trade that portfolio with any good economic sense. Can you elaborate on the dynamic portfolio optimization and the constraints in question please? – Taran Jul 23 '14 at 0:43
• @Taran thanks for your comment. I did not say I want use it. I'm just interested in. If there are more sophisticated models, I would also be interested in a reference. I will update the dynamic part this evening. – math Jul 23 '14 at 6:22
• As a simple experiment you can try in excel and construct a MVO based portfolio using say 5 assets and see how the weights come out to be. Just assume some expected return and covariance matrix with reasonable values. You can use solver to get the weights. MVO is only concerned with first two moments I.e. mean and variance. It tries to come up with the portfolio which has maximum sharpe ratio. Once setup you can play with inputs and see how sensitive it is to the input. You will see that the allocation to the assets swing alot by slight change in the expected returns. – Taran Jul 23 '14 at 6:31
• Atillio Meucci Risk and Asset Allocation is probably what you're looking for. It goes through each point that you mentioned. – user25064 Jul 23 '14 at 13:36
• Yup, agree, I just forgot that Meucci have huge Matlab supplements for his book. Some, again, are already translated to R. – Alexander Didenko Jul 24 '14 at 10:24
There are plenty of books on portfolio issues built according to formula "some theory + some R code (or Matlab, or S - which is very similar to R)". See for example
1. Pfaff B. Financial Risk Modelling and Portfolio Optimization with R.// 2013.
2. Best M.J. Portfolio Optimization. Chapman & Hall, 2010.
3. Würtz D. et al. Portfolio Optimization with R/Rmetrics. // Rmetrics, 2009.
4. Sherer B, Martin R.D. Introduction to Modern Portfolio Optimization With NUOPT and S-PLUS. // 2006.
[1] is with examples from many R packages, and covers not only MPT but risk issues and MPT+ things like Black-Litterman and Meucci's approaches. [2] is based on Matlab examples, have very comprehensive optimization basics, and is more like quant-student-oriented book (unlike [1] which is more quant-PhD-advanced-portfolio-managers-oriented). [3] is entirely dedicated to Rmetrics R packages for optimizatioin, with some very basic theory. More like introductory thing, suitable for everyone from for bachelor student to somebody looking for a good coding-oriented start in the field. [4] is the oldest book, based on S, but all examples are easily translatable to R. Hence, no packages are introduced + the reader is supposed to be able to cope with some difficulties when translating from S to R. But still there are a few interesting advanced things in the book.
• Although I voted to close this question: You are recommending the Pfaff book. I havent read it yet, is it good? Just a small comment though: #4 is also weritten by Bernd Sherer and those are commercial software packages. #3 Is basically a cookbook for the corresponding R package and treats rather advanced optimization problems (compared to classical MVO). – vanguard2k Jul 23 '14 at 7:38
• I think Pfaff is very good, if you are not a novice in both MPT and R programming. Otherwise, take [3] which covers some advanced issues, like robust covariances, but without exploring the very depth - just a cookbook, I agree. And yes, [2] and [4] is based on commercial software. But [4] could be translated to R. I personally translated some return factor models for my class in MPT, and spent like 2 hours for the whole thing, including downloading fresh data from Bloomberg. As for [2], I haven't checked, but I think examples could be implemented in Octave. – Alexander Didenko Jul 23 '14 at 7:46
• Thank you so much for your answer. I will have a look at the books. It seems you have a quite good overview over the topic. Are there any paper which you would recommend as well? – math Jul 23 '14 at 20:09
• paper about portfolio optimization? like, say, Markowitz, 1952? :) – Alexander Didenko Jul 24 '14 at 2:59 | 1,128 | 4,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.94837 |
https://www.diga.me.uk/Sierpinski.html | 1,623,657,452,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00271.warc.gz | 663,011,284 | 3,412 | # Sierpinski curves
-
## Code
Code- one of many versions.
You specify the axiom and rule and angle, and it iteratively creates the long string of moves to be drawn,
``` 'QO fractal generator version f
nomainwin
UpperLeftX = 10
UpperLeftY = 2
WindowWidth =1020
WindowHeight = 730
textbox #w.text, 2, 2,1000, 30
graphicbox #w.graphic, 2, 40,1000, 550
texteditor #w.tedit, 2, 600,1000, 200
button #w.b, "Draw", [doit], UL, 930, 50, 50, 30
open "QO Fractal Generator" for window_nf as #w
#w "trapclose quit"
#w.graphic "size 2 ; color black ; cls ; fill darkblue"
#w "font courier_new 8"
'axiom\$ = "F"
axiom\$ = "F"
rule\$ = "F-F++F-F"
'rule\$ = "F-F+F+F-F"
n = 0
pi = 4 *atn( 1)
angle = 60 *pi /180
'angle = 90 *pi /180
[doit]
n =n +1
if n >7 then call quit h\$
totalx = 0
totaly = 0
newrule\$ = ""
direction = 0 ' NB in RADIANS
state =1
#w.tedit "Iteration # ";
#w.tedit n;
#w.tedit " & rule = "
l =len( axiom\$)
for k =1 to l /128
#w.tedit mid\$( axiom\$, k, 128)
next k
for i =1 to len( axiom\$)
ch\$ =mid\$( axiom\$, i, 1)
select case ch\$
case "+"
direction =direction +angle
newrule\$ =newrule\$ +"+"
case "-"
direction =direction -angle
newrule\$ =newrule\$ +"-"
case "F"
newrule\$ =newrule\$ +rule\$
totalx = totalx +cos( direction)
totaly = totaly +sin( direction)
end select
next i
op\$ ="Total x = " +str\$( int( totalx)) +" & total y = " +str\$( int( totaly))
#w.text op\$
ste =440 /totalx
xx =10: yy =250: direction =0
#w.graphic "cls ; fill darkblue"
#w.graphic "up ; goto 20 500 ; down ; north; turn 90"
'#w.graphic "color "; str\$( 40 +30 *n); " "; str\$( 40 +30 *n); " 50"
'if state =1 then state =0 else state =1
#w.graphic "color "; word\$( "black white", state +1)
#w.graphic "size "; 1 +2 *( 7 -n)
for i =1 to len( axiom\$)
in\$ =mid\$( axiom\$, i, 1)
select case in\$
case "F"
xx =xx +ste *cos( direction)
yy =yy +ste *sin( direction)
case "+"
direction =direction +angle
case "-"
direction =direction -angle
end select
#w.graphic "goto "; str\$( int( 2 *xx)); " "; str\$( 2 *int( yy))
scan
#w.graphic "flush ; getbmp scr 0 0 1000 550"
bmpsave "scr", "IFSrule" +rule\$ +str\$( n) +".bmp"
next i
axiom\$ =newrule\$
timer 1000, [o]
wait
[o]
timer 0
wait
sub quit h\$
close #w
end
end sub
```
We've had several visits to this, including for Rosetta Code. There are three variations that produce the same figure but in very different ways.
Draw scaled and moved triangles
Do by IFS ( iterated file function system)
Draw as a single non-imtersecting line curve.
While I've coded all ways, this is the latter one. Was writing in LB5 on Linux and found one or two glitches.
I was particularly pleased by the graphic created by overlaying each iteration in a different colour and with thinner lines to match the greater resolution. Worth watching it ruin!
## Code
```' ***********************************************************************
' ** **
' ** sierpinskiArrow2alteredHoriz.bas tenochtitlanuk 24/06/30 **
' ** **
' ***********************************************************************
nomainwin
global pi, TX, TY, Ttheta
TX =400: TY =350: Ttheta =0 ' screen centre, pointing North/up. ' <<<<<<<<<<<<<<<<<<<<<<<<<
pi =4 *atn( 1)
WindowWidth = 760
WindowHeight = 900
open "Sierpinski arrowhead" for graphics_nsb as #wg
#wg "trapclose quit"
#wg "color cyan ; font Arial bold 18"
#wg "cls ; down ; fill blue ; backcolor blue ; flush"
'width =11
for i =0 to 11
'#wg "cls ; down ; fill blue ; backcolor blue ; flush" ' rem this line to see each iteration on its own...
#wg "color "; word\$( "red,white,yellow,180 180 80,brown,cyan,darkcyan,130 255 120,131 130 255,darkgray,darkgreen,darkpink,darkred,green,lightgray,palegray,pink,blue", i +1, ",")
#wg "up ; goto 20 "; 30 +20 * i; " ; down"
#wg "\order "; i
#wg "size "; max( 1, 41 -5 *i)
TX =710: TY =850: Ttheta =-90
call SierpinskiArrowhead i, 800 ' order, length
#wg "flush"
#wg "getbmp scr 0 0 1400 900"
'bmpsave "scr", "scr3/SierpinskiArrow" +str\$( i) +".bmp"
call Sleep 2000
next i
wait
sub quit h\$
close #wg
end
end sub
sub Sleep ms
timer ms, [k]
wait
[k]
timer 0
end sub
if ( order and 1) =0 then ' order is even..
call curve order, length, 60
else
call turn -60
call curve order, length, 60
end if
end sub
sub curve order, length, angle
scan
if order =0 then
'#wg "go "; length
call forward length
else
'#wg "turn " +str\$( angle)
call curve order -1, length /2, 0 -angle
call turn angle
call curve order -1, length /2, 0 +angle
'#wg "turn " +str\$( angle)
call turn angle
call curve order -1, length /2, 0 -angle
end if
end sub
' <<<<<<<<<<<<<<<<<<<<<<<<<
end function
end function
sub draw lifted, x, y
if lifted =0 then #wg "up" else #wg "down"
#wg "line "; TX; " "; TY; " "; x; " "; y
Ttheta =atan2( x -TX, TY -y) *180 /pi ' NB DEGREES.
TX =x
TY =y
end sub
sub turn angle ' increment/update global turtle direction ( in DEGREES)
Ttheta =( Ttheta +angle)
if Theta <0 then Ttheta =Ttheta +360
Ttheta =Ttheta mod 360
end sub
sub forward s
#wg "down ; line "; TX; " "; TY; " "; TX +dx; " "; TY +dy; " ; up"
TX =TX +dx
TY =TY +dy
end sub
function atan2( x, y)
Result\$ = "Undetermined"
If ( x = 0) and ( y > 0) Then atan2 = pi / 2: Result\$ = "Determined"
If ( x = 0) and ( y < 0) Then atan2 = 3 * pi / 2: Result\$ = "Determined"
If ( x > 0) and ( y = 0) Then atan2 = 0: Result\$ = "Determined"
If ( x < 0) and ( y = 0) Then atan2 = pi: Result\$ = "Determined"
If Result\$ = "Determined" Then [End.of.function]
BaseAngle = Atn( abs( y) /abs( x))
If (x > 0) and (y > 0) Then atan2 = BaseAngle
If (x < 0) and (y > 0) Then atan2 = pi -BaseAngle
If (x < 0) and (y < 0) Then atan2 = pi +BaseAngle
If (x > 0) and (y < 0) Then atan2 = 2*pi -BaseAngle
[End.of.function]
end function
' <<<<<<<<<<<<<<<<<<<<<<<<
``` | 2,096 | 6,155 | {"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-2021-25 | longest | en | 0.441193 |
https://www.traditionaloven.com/metal/precious-metals/gold/convert-qty_1-frac-3-dram-dr-of-gold-to-cubic-inch-cu-in-gold.html | 1,718,618,254,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00106.warc.gz | 926,938,574 | 13,637 | Gold 1/3 dram to cubic inches of gold converter
# gold conversion
## Amount: 1/3 drams (dr) of gold mass Equals: 0.0019 cubic inches (cu in - in3) in gold volume
Calculate cubic inches of gold per 1/3 drams unit. The gold converter.
TOGGLE : from cubic inches into drams in the other way around.
### Enter a New dram Amount of gold to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## gold from dram to cubic inch Conversion Results :
Amount : 1/3 drams (dr) of gold
Equals: 0.0019 cubic inches (cu in - in3) in gold
Fractions: 19/10000 cubic inches (cu in - in3) in gold
CONVERT : between other gold measuring units - complete list.
## Solid Pure 24k Gold Amounts
This calculator tool is based on the pure 24K gold, with Density: 19.282 g/cm3 calculated (24 karat gold grade, finest quality raw and solid gold volume; from native gold, the type we invest -in commodity markets, by trading in forex platform and in commodity future trading. Both the troy and the avoirdupois ounce units are listed under the gold metal main menu. I advice learning from a commodity trading school first. Then buy and sell.) Gold can be found listed either in table among noble metals or with precious metals.
Is it possible to manage numerous calculations for how heavy are other gold volumes all on one page? Yes, all in one Au multiunit calculator makes it possible managing just that.
Convert gold measuring units between dram (dr) and cubic inches (cu in - in3) of gold but in the other direction from cubic inches into drams.
conversion result for gold: From Symbol Equals Result To Symbol 1 dram dr = 0.0056 cubic inches cu in - in3
# Precious metals: gold conversion
This online gold from dr into cu in - in3 (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this gold calculator are ...
With the above mentioned units calculating service it provides, this gold converter proved to be useful also as a teaching tool:
1. in practicing drams and cubic inches ( dr vs. cu in - in3 ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with gold's density values including other physical properties this metal has.
International unit symbols for these two gold measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for dram is: dr
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for cubic inch is: cu in - in3
### One dram of gold converted to cubic inch equals to 0.0056 cu in - in3
How many cubic inches of gold are in 1 dram? The answer is: The change of 1 dr ( dram ) unit of a gold amount equals = to 0.0056 cu in - in3 ( cubic inch ) as the equivalent measure for the same gold type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of gold can have a crucial/pivotal role in investments. If there is an exact known measure in dr - drams for gold amount, the rule is that the dram number gets converted into cu in - in3 - cubic inches or any other unit of gold absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many cubic inches ( cu in - in3 ) of gold are contained in a dram ( 1 dr ). Or, how much in cubic inches of gold is in 1 dram? To link to this gold - dram to cubic inches online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: gold from dram (dr) to cubic inches (cu in - in3) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,070 | 4,610 | {"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-2024-26 | latest | en | 0.828696 |
https://parents.koobits.com/animated-video-solutions/ | 1,660,647,706,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00497.warc.gz | 391,046,970 | 80,247 | # Bring Maths To Life With 1000+ Animated Video Solutions
Maths doesn’t have to be abstract and difficult to understand.
With animated videos, students can learn in a fun and engaging way.
Take the Helen-Ivan coin question that appeared in Singapore’s 2021 PSLE maths paper as an example.
It stumped many students.
Even some adults found it challenging, let alone explaining the solution to their child.
But using a video, a similar concept can be clearly explained in under 10 minutes:
In KooBits, we have over 1000 of such video solutions covering all topics in the Primary 1 – 6 maths syllabus.
Each video targets a specific concept, so you can use it to fill learning gaps.
To watch them free for a week, sign up for a trial today:
# Visual Learning Boosts Recall By 200%
By using colourful visuals and engaging animation, our videos do more explanation in less time.
This is a more efficient compared to traditional learning methods like reading textbooks.
Videos are much more memorable, boosting recall of key concepts and facts.
When it’s time to solve a new question, a student will be able to remember what was taught in a video, and use it to find their own solution.
In our videos, we help improve your child’s understanding through a 4-step process:
1. U for Understanding – Walk the child through the problem, ensuring they know what the question is asking.
2. M for Method – Choose a method to solve the problem.
3. S for Solve – Once we’ve laid out the facts, it’s time to work out the answer.
4. C for Check – Check to see if our answer makes sense! This helps avoid careless mistakes too.
This 4-step “secret sauce” makes our videos effective in helping kids at all levels to master complex concepts.
Weaker students watch the videos to improve their maths.
Top performers watch the videos to revise past topics.
Using psychology, gamification and advice from child educators, we’ve cracked the code to boost a child’s ability to learn, even with different levels of proficiency!
Here’s another example of an animated video:
To try these videos for free, click on the button below to sign up for a trial:
# How To Use Animated Videos To Learn Maths
## 3) Learn holistically. Explore how to use maths to solve real life problems:
This is an example of KooClass, a feature that teaches kids concepts not taught in school.
With KooClass videos, kids can relate maths to their everyday life, and help them open their minds to absorb more information.
Research shows that when kids “prepare their minds” properly before learning, they can achieve as high as 70% retention of knowledge, compared to the usual 30%.
Try our 7-day free trial and see why over 200,000 students are using KooBits videos to learn maths!
• Latest
• Resources
• Research
• Company
• Support | 607 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.921319 |
https://www.businessmanagementideas.com/company-management/shares/top-5-problems-on-bonus-issue-of-shares-with-solution/4084 | 1,718,770,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00342.warc.gz | 600,856,135 | 97,879 | Here is a compilation of top five problems on bonus issue of shares with their relevant solutions.
#### Bonus Issue of Shares: Problem with Solution #1:
The following are the extracts from the draft Balance Sheet of A Ltd., as on December 31,2010
A resolution was passed by the company declaring bonus of 25% on equity shares to be provided as to Rs 15,000 out of reserve fund and the balance out of profit and loss account. The bonus was to be satisfied by issuing fully paid equity shares.
You are required to make necessary journal entries and show the effect on the Balance Sheet of the company.
Solution:
#### Bonus Issue of Shares: Problem with Solution # 2:
A company has a share capital of 5,00,000 equity shares of Rs. 10 each, Rs. 6 per share paid. It has a balance in the Reserve Fund Account amounting to Rs. 50,00,000. The company has decided to pay bonus to shareholders by making the partly paid share as fully paid. Make necessary journal entries to record the same.
Solution:
#### Bonus Issue of Shares: Problem with Solution # 3:
The extracts are given from the draft Balance Sheet of Bharat Gears Ltd. as on 31st December 2010:
The Board of Directors pass a resolution to capitalise a part of existing reserves and profits by issuing Bonus Shares. One bonus share is being issued for every 4 equity shares held at present. For this purpose, Rs. 10,000 are to be provided out of Reserve Fund and the balance out of Profit and Loss Account.
Set out the journal entries to give effect to the resolution and show how they would affect the Balance Sheet of the Company.
Solution:
#### Bonus Issue of Shares: Problem with Solution # 4:
A company has a share capital of 1,00,000 equity shares of Rs. 10 each fully paid. The company has a reserve fund of Rs. 10,00,000 and declares a bonus of Rs. 4,50,000. This bonus is to be paid by issue of fully paid equity shares at a premium of Rs. 5 per share. Shares are quoted at Rs. 20 per share on the date of allotment of bonus issue.
Solution:
Note:
The market price of shares, i.e., Rs. 20/- per share is not to be taken into consideration.
#### Bonus Issue of Shares: Problem with Solution # 5:
A company has a share capital of 10,00,000 equity shares of Rs. 10 each, Rs. 8 per share paid up.
It has a reserve fund of Rs. 80,00,000. If is decided to utilise the whole of the reserve fund in the following manner:
(a) The existing shares to be made fully paid up without the shareholders having to pay anything and | 586 | 2,495 | {"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-2024-26 | latest | en | 0.961803 |
http://cakedecoratingforkids.com/cake-designer-brooklyn-heightsbrooklyn/1-box-of-powdered-sugar-equals-how-many-cups | 1,506,448,577,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696677.93/warc/CC-MAIN-20170926175208-20170926195208-00026.warc.gz | 54,019,840 | 7,017 | # 1 box of powdered sugar equals how many cups
I bought a 2 lb. bag of powdered sugar & a recipe I am thinking of making calls for a box of powdered 1 box of powered sugar has 4 cups in it.
And the recipe doesn't say how many ounces or powdered sugar. How many cup's might this be? here's the recipe: 1 box confectioners sugar.
The answer to how many cups are in one pound of powdered sugar will vary to 4 cups unsifted powdered sugar ; 1 pound = 4 1 /2 cups sifted powered sugar.
### Videos
How to Make PERFECT Chocolate Chip Cookies! how much sugar is in 1 box please; ie. how many ounces or grams 1 US cup of powdered sugar is thus 4 cups per box. (Ref).
Dry Ingredients Measured In Cups Will Vary In Weight - Check Ingredient Types. To convert ounces to grams, multiply the number of ounces by 30. (1 oz. Domino® Confectioners Sugar. 1 1 lb. box, -------------, approximately 2 1 /3 cups **.
How much confectioners sugar is in a " box?" Instead of specifying "3- 1 /2 to 4 cups confectioner's sugar," he thought it would be much simpler if you just. | 278 | 1,056 | {"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-2017-39 | longest | en | 0.92576 |
https://jp.mathworks.com/matlabcentral/answers/472511-extracting-data-from-cells | 1,576,340,109,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00497.warc.gz | 413,136,287 | 24,295 | ## Extracting Data from Cells
Devrim Tugberk
### Devrim Tugberk (view profile)
さんによって質問されました 2019 年 7 月 19 日
さんによって コメントされました 2019 年 7 月 19 日
さんの 回答が採用されました
Dear Matlab users,
I am facing some trouble extracting data from a cell. I attached the a screenshots showing what my cell looks like (cells within a cell). Each cell in the "big" cell has xyz data and i would like to extract each column representing the respective coordinates as column vectors.
My explanation might have been confusing so i will try summarize: Extract the "minor" cells from the main cell, extract the xyz from each minor cell as column vectors.
Thank you for the help
Guillaume
### Guillaume (view profile)
2019 年 7 月 19 日
What does extract mean? What do you want to do afterward?
2019 年 7 月 19 日
I interpreted it as an indexing question.
Devrim Tugberk
### Devrim Tugberk (view profile)
2019 年 7 月 19 日
My goal was the "extract" the xyz coordinates as column vectors and plot them using scatter3()
サインイン to comment.
## 3 件の回答
2019 年 7 月 19 日
2019 年 7 月 19 日
採用された回答
vv=cellfun(@(x) x(:,1:3),cross_sections,'un',0); % assuming the first three columns in each cell represents x,y & z
v=cat(1,vv{:}); % gathering x,y, & z of each cell into one matrix
scatter3(v(:,1),v(:,2),v(:,3))
% x -^ ^- y ^- z
Devrim Tugberk
### Devrim Tugberk (view profile)
2019 年 7 月 19 日
Perfect! Thank you all very much
Devrim Tugberk
### Devrim Tugberk (view profile)
2019 年 7 月 19 日
One last tiny tweak, is there a way I could supress the scatter3 function so it doesnt spit out 138 figures? Maybe just give me 138 tables or vectors or something which i can refer back to and plot the desired one only?
2019 年 7 月 19 日
Then for instance:
scatter3(vv{2}(:,1),vv{2}(:,2),vv{2}(:,3))
サインイン to comment.
2019 年 7 月 19 日
2019 年 7 月 19 日
I think this is the format of your data:
C = {rand(20,4),rand(15,4),rand(22,4)};
Extract the first 3 columns of cell 2
C{2}(:,1:3)
Extract column 2 of cell 3
C{3}(:,2)
Extract the entire matrix from cell 1
C{1}
Extract the entire matrix from cell 1 but reorganize it into a single column
C{1}(:)
#### 1 件のコメント
2019 年 7 月 19 日
"My goal was the "extract" the xyz coordinates as column vectors and plot them using scatter3()"; " I need to have 3 column vectors for each cell"
@ Devrim Check out the 2nd block of code in my answer. It does what you're describing. If you want to perform that block on all cells,
c3 = cellfun(@(x)x(:,1:3), C, 'UniformOutput', false)
サインイン to comment.
2019 年 7 月 19 日
### Guillaume (view profile)
2019 年 7 月 19 日
If you want to create a scatter3 plot using the coordinates from all the cells of Cross_sections, this is how I'd go about it:
allsections = vertcat(Cross_sections{:});
scatter3(allsections(:, 1), allsections(:, 2), allsections(:, 3));
#### 2 件のコメント
2019 年 7 月 19 日
Probably you meant:
allsections = vertcat(Cross_sections{:});
Guillaume
### Guillaume (view profile)
2019 年 7 月 19 日
Indeed! Otherwise, the operation does nothing.
サインイン to comment.
Translated by | 936 | 3,026 | {"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-2019-51 | latest | en | 0.488591 |
http://mathhelpforum.com/calculus/92246-num-analysis-2-a.html | 1,527,291,256,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00123.warc.gz | 185,822,821 | 9,425 | 1. ## num. analysis #2
Hello everybody.
Suppose that g is continuously differentiable on some interval (c, d) that contains the fixed point p of g. Show that if $\displaystyle |g'(p)|<1$, then there exists $\displaystyle \delta>0$ such that if $\displaystyle |p_0-p|<\delta$, then the fixed-point iteration converges.
2. ## Clear Definitions
If $\displaystyle |g'(p)|<1$, then by definition of the derivative, $\displaystyle \left|\frac{\Delta y}{\Delta x}\right|<1$ for a sufficiently small $\displaystyle |\Delta x|<\delta$. Since $\displaystyle \left|\frac{\Delta y}{\Delta x}\right|=\frac{|\Delta y|}{|\Delta x|}$ , this means $\displaystyle |\Delta y|<|\Delta x|<\delta$. So define $\displaystyle \Delta x=p_0-p$ such that $\displaystyle |p_0-p|<\delta$, and we have just shown that $\displaystyle |g(p_0)-g(p)|=|g(p_0)-p|<|p_0-p|$ , which is exactly the definition of convergent. So when $\displaystyle p_0\in(p-\delta,p+\delta)$, the value of each iteration $\displaystyle p_i=g(p_{i-1})$ converges to $\displaystyle p$.
Notice that this proof allows you to predict the value of $\displaystyle \delta$ for a given function. Notably, for some fixed point $\displaystyle g(p)=p$ where $\displaystyle |g'(p)|<1$, if the approximation of $\displaystyle g'(p)\approx\left|\frac{g(p_0)-g(p)}{p_0-p}\right|<1$ at some point $\displaystyle p_0$, then $\displaystyle \delta=|p_0-p|$ and the iteration of any point in the interval $\displaystyle (p-\delta,p+\delta)$ will converge on $\displaystyle p$. | 450 | 1,504 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.63708 |
https://www.naukri.com/code360/interview-experiences/cognizant/cognizant-interview-experience-by-on-campus-oct-2020-2-707 | 1,726,133,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00398.warc.gz | 834,021,802 | 67,944 | Cognizant interview experience Real time questions & tips from candidates to crack your interview
Genc Next
Cognizant
3 rounds | 3 Coding problems
Interview preparation journey
Preparation
Duration: 2 months
Topics: C, C++, java, Python, c#.net, sql, some certifications on Machine learning
Tip
Tip 1 : confidence is most important.
Tip 2 : explaining projects plays a key role.
Application process
Where: Campus
Eligibility: 6.5 CGPA
Resume tip
Tip 1 : you should be very clear what you have written in your resume.
Tip 2 : projects should be very clear
Interview rounds
01
Round
Medium
Online Coding Interview
Duration120 Minutes
Interview date18 Oct 2020
Coding problem1
We have aptitude, reasoning and verbal.
1. Sort Array
Moderate
15m average time
85% success
0/80
You are given an array consisting of 'N' positive integers where each integer is either 0 or 1 or 2. Your task is to sort the given array in non-decreasing order.
Note :
View more
02
Round
Easy
Online Coding Interview
Duration60 minutes
Interview date15 Nov 2020
Coding problem1
1. Minimum steps to reach target by a Knight
Moderate
25m average time
60% success
0/80
You have been given a square chessboard of size ‘N x N’. The position coordinates of the Knight and the position coordinates of the target are also given.
View more
03
Round
Medium
Face to Face
Duration60 Minutes
Interview date15 Dec 2020
Coding problem1
I got the link a day before and my interview was scheduled at 6Pm.
1. Ways To Make Coin Change
Moderate
20m average time
80% success
0/80
You are given an infinite supply of coins of each of denominations D = {D0, D1, D2, D3, ...... Dn-1}. You need to figure out the total number ...
View more
Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is the result of 4 % 2?
Start a Discussion
Similar interview experiences
Genc Next
2 rounds | 5 problems
Interviewed by Cognizant
633 views
Genc Next
2 rounds | 7 problems
Interviewed by Cognizant
773 views | 532 | 2,035 | {"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} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.82457 |
https://de.mathworks.com/matlabcentral/cody/problems/9-who-has-the-most-change/solutions/16400 | 1,579,759,774,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00256.warc.gz | 396,642,546 | 15,725 | Cody
# Problem 9. Who Has the Most Change?
Solution 16400
Submitted on 30 Jan 2012 by Rafael Oliveira
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
%% a = [1 2 1 15]; b = 1; assert(isequal(most_change(a),b))
2 Pass
%% a = [ 1 2 1 15; 0 8 5 9]; b = 2; assert(isequal(most_change(a),b))
3 Pass
%% a = [ 1 22 1 15; 12 3 13 7; 10 8 23 99]; b = 3; assert(isequal(most_change(a),b))
4 Pass
%% a = [ 1 0 0 0; 0 0 0 24]; b = 1; assert(isequal(most_change(a),b))
5 Pass
%% a = [ 0 1 2 1; 0 2 1 1]; c = 1; assert(isequal(most_change(a),c))
6 Pass
%% % There is a lot of confusion about this problem. Watch this. a = [0 1 0 0; 0 0 1 0]; c = 2; assert(isequal(most_change(a),c)) % Now go back and read the problem description carefully.
7 Pass
%% a = [ 2 1 1 1; 1 2 1 1; 1 1 2 1; 1 1 1 2; 4 0 0 0]; c = 5; assert(isequal(most_change(a),c)) | 410 | 970 | {"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-2020-05 | latest | en | 0.649042 |
https://www.prokidsedu.com/rewrite-in-simplest-radical-form-1-x-3-6-show-each-step-of-your-process | 1,669,977,747,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00290.warc.gz | 1,008,933,712 | 13,681 | # Rewrite In Simplest Radical Form 1 X −3 6 . Show Each Step Of Your Process.
Rewrite In Simplest Radical Form 1 X −3 6 . Show Each Step Of Your Process.. Is the expression x3•x3•x3 equivalent to x3•3•3? 1 🔴 on a question rewrite in simplest radical form 1 over x^(−3/6).
√ x • 4 √ x.= 4 √ x ∗ x = 4 √ x 2 question 4 rewrite in simplest radical form x 5 6 x 1 6. Show each step of your process. 3) the resultant fraction would be the lowest form of the given fraction.
### Rewrite In Simplest Radical Form X^5/6 X^1/6.
Is the expression x3•x3•x3 equivalent to x3•3•3? Rewrite in simplest radical form 1 over x^(−3/6). 1 🔴 on a question rewrite in simplest radical form 1 over x^(−3/6).
4 √ x 3 = x 3 4 b. 10 √ x 5 ∗ x 4 ∗ x 2 = 10. If you find my answer helpful please put thumbs up.
See Also : In His Poem "Homework," Allen Ginsberg Most Likely Uses Free Verse To
### 1 X − 1 =X C.
Show each step of your process. Show each step of your process. 8x + 17 / x^2 15x + 56 the expression is undefined when x = in the graph on the right, a line segment through the center of the circle intersects the circle at the points ( 7,2 ) and ( 11,6 ) as shown.
### Answer To Rewrite In Simplest Radical Form 1 X 3 6.
Show each step of your process. Show each step of your process. Rewrite in simplest radical form 1 x 3 6.
### Show Each Step Of Your Process.
Find all values of the variable x for which the rational expression is undefined. The steps are involved to convert the fraction in the lowest form as, 1) first break the denominator and numerator into their prime factors. Show each step of your process. | 480 | 1,615 | {"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} | 3.390625 | 3 | CC-MAIN-2022-49 | latest | en | 0.833228 |
https://www.doorsteptutor.com/Exams/NSTSE/Class-6/Questions/Topic-Mathematics-0/Subtopic-Playing-With-Numbers-2/Part-8.html | 1,529,412,892,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00253.warc.gz | 793,390,788 | 11,596 | # Mathematics-Playing With Numbers (NSTSE (National Science Talent Search Exam- Unified Council) Class 6): Questions 69 - 76 of 83
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 2029 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
Rs. 550.00 or
## Question number: 69
» Mathematics » Playing With Numbers
MCQ▾
### Question
The number 623273 is divisible by
### Choices
Choice (4) Response
a.
5
b.
7
c.
All of the above
d.
None of the above
## Question number: 70
» Mathematics » Playing With Numbers
MCQ▾
### Question
The HCF of 15,20 and 30 is
### Choices
Choice (4) Response
a.
5
b.
20
c.
15
d.
10
## Question number: 71
» Mathematics » Playing With Numbers
MCQ▾
### Question
The LCM of two numbers is X and their HCF is Y. So, the sum of two numbers is
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 72
» Mathematics » Playing With Numbers
MCQ▾
### Question
The number of prime numbers between 2 and 15 are
### Choices
Choice (4) Response
a.
5
b.
4
c.
3
d.
6
## Question number: 73
» Mathematics » Playing With Numbers
MCQ▾
### Question
Difference between a prime (18) and a composite (30) number is
### Choices
Choice (4) Response
a.
19
b.
23
c.
12
d.
30
## Question number: 74
» Mathematics » Playing With Numbers
MCQ▾
### Question
60 is written as the product of primes as
### Choices
Choice (4) Response
a.
2
b.
2
c.
2
d.
2
## Question number: 75
» Mathematics » Playing With Numbers
MCQ▾
### Question
Observe the following table of sum of odd numbers and multiplication of same pair numbers. Then find the sum of odd numbers from 1 to 29.
### Choices
Choice (4) Response
a.
227
b.
100
c.
121
d.
225
## Question number: 76
» Mathematics » Playing With Numbers
MCQ▾
### Question
The greatest number of 5-digits exactly divisible by 9,12,16 and 30 is
### Choices
Choice (4) Response
a.
99603
b.
99306
c.
99364
d.
99360
f Page | 605 | 2,099 | {"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} | 3.390625 | 3 | CC-MAIN-2018-26 | latest | en | 0.680114 |
https://www.habitat-swu.org/whats-a-balloon-payment/ | 1,568,764,337,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00342.warc.gz | 863,550,541 | 8,894 | # What’S A Balloon Payment
### Contents
And when the deadline comes up, you’ll have to pay the entire loan off in one giant payment (aka the balloon payment). A balloon payment can easily be tens of thousands of dollars or more, which.
A balloon mortgage comes with payments based on a long-term, 30-year amortization, for example, but the balance of the loan comes due after five to seven years. At that point, the outstanding loan.
A balloon payment is a large payment made at or near the end of a loan term. Example of a Balloon Payment Unlike a loan whose total cost (interest and principal ) is amortized – that is, paid incrementally during the life of the loan – a balloon loan ‘s principal is paid in one sum at the end of the term .
A balloon loan is a type of loan that does not fully amortize over its term. Since it is not fully amortized, a balloon payment is required at the end of the term to repay the remaining principal.
Balloon Payments Are Payments That Are 360 180 Loan Forward rate agreement: A hedging technique for interest rate risks – The first number denotes the time of commencement of the loan, while the difference in the two numbers. the short would need to make a payment of 5,000,000 x (0.075-0.06) x (180/360) = \$37,500 to.Your balance or ‘Balloon Payment Amount’ will be due at this time. Also choose whether ‘Length of Balloon Period’ is years or months. The monthly payment and interest are calculated as if the mortgage or loan were being paid over this length.
For more information on this subject, or for any commercial real estate related questions or information, you’re invited to call michael bull at 404-876-1640 x 101. Any question, anywhere, anytime.
Balloon Interest Calculator A boy the world thought was carried away in a balloon when he was actually. speculate on how it came loose. The banks’ interest rates decisions are putting more of your money in their coffers – use.
A balloon payment is a large payment made at or near the end of a loan term. Example of a Balloon Payment Unlike a loan whose total cost (interest and principal ) is amortized – that is, paid incrementally during the life of the loan – a balloon loan ‘s principal is paid in one sum at the end of the term .
“But, in a construction-to-permanent loan, a balloon payment cannot exist – it automatically rolls over to permanent financing. So, what is the benefit of telling the customer that there is a balloon.
Amortization Schedule Balloon Payment According to Wikipedia "Amortization refers to the process of paying off a debt (often from a loan or mortgage) over time through regular payments. A portion of each payment is for interest while the remaining amount is applied towards the principal balance." Further, "an amortization schedule is a table detailing each periodic payment on an amortizing loan (typically a mortgage), as generated.
A balloon payment is a lump sum paid at the end of a loan’s term that is significantly larger than all of the payments made before it. On installment loans without a balloon option, a series of fixed payments are made to pay down the loan’s balance.
Balloon payments and resale value. There are a range of factors to consider when choosing a balloon payment, but one of the most important is the expected value of your vehicle at the end of the loan term. Ideally, your balloon should be less than or equal to the value of the vehicle when it’s due. | 742 | 3,433 | {"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-2019-39 | longest | en | 0.969528 |
https://www.khanacademy.org/math/algebra-home/alg-vectors/alg-applications-of-vectors/e/vector-word-problems | 1,702,086,048,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00605.warc.gz | 917,762,770 | 138,078 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Algebra (all content)
### Course: Algebra (all content)>Unit 19
Lesson 10: Applications of vectors
# Vector word problems
You might need: Calculator
## Problem
Michael is running some errands.
• His first stop is to the east and to the south from his house.
• His second stop is to the west and to the south from the first stop.
• His third stop is to the west and to the north from the second stop.
What is Michael's direction relative to his starting point once he arrives at his third stop?
Round your answer to the nearest integer. You can round intermediate values to the nearest hundredth.
$\mathrm{°}$ rotation from the eastward direction
Stuck?
Stuck? | 201 | 889 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.95131 |
https://www.jiskha.com/display.cgi?id=1366823184 | 1,502,957,112,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102993.24/warc/CC-MAIN-20170817073135-20170817093135-00157.warc.gz | 913,215,202 | 3,736 | # math
posted by .
simplify the expression 14/(3x)+(x+5)/(3x)
• math -
(1/3x)(x+19)
## Similar Questions
1. ### math
application of the laws of exponent show different ways of simplifying each of the following expression. answer the question that follow. 1. 2^5 * 2^4 2. x^4 * x^7 3. (3^2)^3 4. (m^4)^5 5. (5^3 * 2^2)^2 6. (a^4 * b^2)^5 7. (3x^5)^3 …
2. ### math
application of the laws of exponent. show different ways of simplifying each of the following expression. answer the question that follow. 1. 2^5 * 2^4 2. x^4 * x^7 3. (3^2)^3 4. (m^4)^5 5. (5^3 * 2^2)^2 1. how did you simplify each …
3. ### math
translate the phrase into a mathematical expression and then evaluate it. sixtten squared divided by four cubed. The given expression in symbols is?
4. ### URGENTquestions for Math
A pair of jeans is on sale for 25% off the original price. Which expression represents the sale price?
5. ### Algebra
Translate the phrase to mathematical language. Then simplify the expression the difference between 119 and -40 What is a numerical expression for the phrase?
6. ### Hate Math
Consider the expression: (28/4). 5-6 + 4square. Explain the order of operations you would use to simplify this expression. then simplify it.
7. ### urgent math
1. Simplify the expression 7^9/7^3 a.7^3*** b.7^6 c.7^12 d.1^6 2. Simplify the expression z^8/z^12 a.z^20 b.z^4 c.1/z^-4 d.1/z^4 3. Simplify the expression (-3^4)/(-3^4 a.(-3)^1 b.0 c.1 d.(-3)^8 4.Which expressions can be rewritten …
8. ### Algebra
1. Use the Distributive Property to simplify the expression. (-1)(4-c) 2. Use the Distributive Property to simplify the expression. 4(2x -4) 3. Use the Distributive Property to simplify the expression. (10 + 4 y) 1/2 4. To which subsets …
9. ### Algebra
1. Use the Distributive Property to simplify the expression. (-1) (4-c) 2. Use the Distributive Property to simplify the expression. 4 (2 x - 4) 3. Use the Distributive Property to simplify the expression (10+4y) 1/2 4. To which subsets …
10. ### algebra
am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2
More Similar Questions | 751 | 2,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-34 | latest | en | 0.750966 |
https://testbook.com/question-answer/a-darsonval-movement-has-a-resistance-of-100--62285579ce915b61b1fb80f3 | 1,726,386,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00068.warc.gz | 519,512,345 | 46,061 | # A D'Arsonval movement has a resistance of 100 Ω and full deflection current of 2 mA and is used for construction of multirange ammeter as shown in the figure:What are the values of resistances R1, R2, R3 and R4, if the shunt resistances are connected as shown in the figure?
This question was previously asked in
UPSC IES Electrical 2022 Prelims Official Paper
View all UPSC IES Papers >
1. R1 = 0.002 Ω, R2 = 0.018 Ω , R3 = 0.18 Ω and R4 = 1.836 Ω
2. R1 = 0.002 Ω, R2 = 0.18 Ω , R3 = 0.018 Ω and R4 = 1.836 Ω
3. R1 = 1.836 Ω, R2 = 0.18 Ω , R3 = 0.018 Ω and R4 = 0.002 Ω
4. R1 = 1.836 Ω, R2 = 0.018 Ω , R3 = 0.18 Ω and R4 = 0.002 Ω
Option 1 : R1 = 0.002 Ω, R2 = 0.018 Ω , R3 = 0.18 Ω and R4 = 1.836 Ω
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
4.3 K Users
20 Questions 40 Marks 24 Mins
## Detailed Solution
Calculation:
Redraw the multi range ammeter:
Case1: When the switch is at position ‘1’
I = 100 mA
$$R_1+R_2+R_3+R_4={R_m\over m-1}$$
$$R_1+R_2+R_3+R_4={100\over {100\over 2}-1}$$
$$R_1+R_2+R_3+R_4=2.041\space Ω$$ ..........(i)
Case2: When the switch is at position ‘2’
I = 1 A
$$R_1+R_2+R_3={100+R_4\over {1\over 2\times 10^{-3}}-1}$$
$$R_1+R_2+R_3={100+R_4\over 499}$$
Putting the value of equation (i), we get:
$$2.041-R_4={100+R_4\over 499}$$
R4 = 1.836 Ω
$$R_1+R_2+R_3={100+1.836\over 499}$$
$$R_1+R_2+R_3=0.204\space Ω$$.............(ii)
Case 3: When the switch is at position ‘3’
I = 10 A
$$R_1+R_2={100+R_3+R_4\over {10\over 2\times 10^{-3}}-1}$$
$$R_1+R_2={101.836+R_3\over 4999}$$
From equation (ii), we get:
$$0.204-R_3={101.836+R_3\over 4999}$$
R3 = 0.183 Ω
$$R_1+R_2={101.836+0.183\over 4999}$$
$$R_1+R_2=0.0204\space Ω$$ ...........(iii)
Case 4: When the switch is at position ‘4’
I = 100 A
$$R_1={100+R_2+R_3+R_4\over {100\over 2\times 10^{-3}}-1}$$
$$R_1={102.019+R_2\over 49999}$$
$$4999R_1-R_2=102.019$$ ........(iv)
Solving equations (i) and (ii), we get:
R1 = 0.00204 Ω
R2 = 0.0183 Ω
Last updated on Sep 12, 2024
-> UPSC IES Reserve List has been released for the Engineering Services Examination, 2023.
-> The Engineering Services (Mains) Examination, 2024 was held on 23.06.2024.
-> Also, the IES 2025 Prelims Exams will be conducted on 9th February 2025 while the Mains Exam will be conducted on 22nd June 2025.
-> The UPSC conducts the Indian Engineering Services Exam to recruit candidates for engineering posts in various disciplines under different departments of the Central Government.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here. | 1,115 | 2,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-38 | latest | en | 0.66893 |
http://www.bestmortgageandhomeloans.com/note-maturity-calculator/ | 1,600,514,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00485.warc.gz | 165,752,059 | 7,974 | # Note Maturity Calculator
Yield-to-Maturity Calculators. In the days before personal computers, bond dealers and investment reps used a dedicated desktop bond calculator to determine prices and yields. But spreadsheet programs turned every computer into a financial calculator, and the Internet put bond calculators on a wide range of financial websites.
Calculate The Interest Payable At Maturity · How to Calculate a Bond Price. The bond will stipulate the interest rate known as the coupon rate, and the term to be used, known as the maturity date. Throughout the term the investor will receive the interest (coupon) payments, and at the maturity date will receive payment of the principal amount invested.
Maturity value is the amount payable to an investor at the end of a debt instrument’s holding period (maturity date). For most bonds, the maturity value is the face amount of the bond. For some certificates. For purposes of accounting, it’s important to be able to calculate the maturity value of a note to know how much a.
Yield to Maturity Calculator – The rate of return anticipated on a bond if it is held until the maturity date.
investors can calculate a potential reduction in bond mutual fund NAV if rates move up as forecast. Below is a table of several popular bond mutual funds with the current duration, current average.
Amortization Of Prepayments What is the correct accounting for prepayments in foreign currency under IFRS? How do IFRS treat the effect of moving exchange rates?" Let me tell you that here, it’s not all black or white. It depends on more factors, especially the nature of a specific prepayment. Let me explain why and how.
Real estate investment calculator solving for note maturity value given bank discount, annual bank bank discount equations calculator. financial investment real estate property land residential.. Fixed Deposit Calculator. Fixed deposit is a financial instrument offered by banks and financial institutions in India.
If you would like to figure out the maturity date of a loan, you can also use an online maturity date calculator. By entering your loan amount, interest rate and the length of the loan, you can get a breakdown of the monthly premium and interest payments along with your loan maturity date.
T-Bill Price Calculator – Given the annual interest rate and days to maturity, the price of a US Treasury Bill can be calculated.
How to Determine the Maturity Date of a Loan. By: Andrew Mayfair.. In the example, take note of the principal balance of the loan, \$10,000. Create a column for each monthly payment. Write \$500 next to each month for the loan payment. calculate the amount of interest per month. In our example.
The calculated values are provided to help you with your financial planning. This information does not express or imply any offer from Farm Credit Services.
While most investors use Net Income as the numerator in the calculation, I prefer to use FCF to calculate what I call. As. | 594 | 2,992 | {"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-2020-40 | latest | en | 0.907344 |
https://www.encodedna.com/excel/calculate-subtotals-for-a-range-in-excel-using-vba.htm | 1,624,321,962,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00000.warc.gz | 667,121,204 | 3,930 | How to calculate the Subtotals for a Range a data in Excel using VBA
# How to calculate the Subtotals for a Range a data in Excel using VBA
In Excel, there is an in-built feature called the Subtotal (under the Data tab) which when used can automatically insert subtotals and grand total for a selected range of data. However, if you are a VBA programmer, you can further automate this process using a simple, one line macro.
The image below explains it.
The Macro
Let us assume, I have a range of data with few columns (see the above image again). I want to show the Subtotal of Price for each product and finally at the last row show the grand total, with the click of a button.
Note: The result will be the same like the Subtotal feature that Excel provides. We are doing this using a Macro.
```Option Explicit
Private Sub CommandButton1_Click()
showSubTotal
End Sub
Sub showSubTotal()
Worksheets("sheet1").Activate
Range("A1:E" & Cells(Rows.Count, "E").End(xlUp).Row).Subtotal _
GroupBy:=1, Function:=xlSum, TotalList:=Array(5, 4), _
Replace:=True, PageBreaks:=False
End Sub```
In the above macro, I am using the Range() method where I have defined a range from column A to E. This is fixed, kind of hardcoded. Therefore, every time I click the button, it will show the subtotal and grand total of that particular range, A to E.
Now here’s another scenario. I don’t want to show the subtotal of all the products, only a few. For example, I want to insert subtotal to selected products only, like Pens, Fax and Safety pins.
To do this I can use the same method (that I used in the above example) using the Selection property.
```Option Explicit
Private Sub CommandButton1_Click()
showSubTotal
End Sub
Sub showSubTotal()
Worksheets("sheet1").Activate
Selection.Subtotal _
GroupBy:=1, Function:=xlSum, TotalList:=Array(5, 4), _
Replace:=True, PageBreaks:=False
End Sub```
Remember, if you are using the Selection property, you have to first select a range and then click the button (or whatever method you are using). Or else it will throw an error. | 506 | 2,070 | {"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-2021-25 | longest | en | 0.775222 |
https://questions.llc/questions/468284/what-is-the-volume-occupied-by-16-0g-ethane-gas-c2h6-at-720-torr-and-18c-v-nrt-p | 1,660,777,512,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00013.warc.gz | 459,265,542 | 5,094 | # what is the volume occupied by 16.0g ethane gas (C2H6) at 720 Torr and 18C?
V=nRT/P
conversion of mol:
16.0g C2H6 X 1mol C2H6/30.07g C2H6= .53mol C2H6
V= .53 mol(0.0821 L*atm/mol*K)(291)K /
720 Torr (1 atm/760Torr)
V= 13.5 L
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. I don't get that number exactly. First, I don't think you should have rounded when you did. You are allowed at least 3 s.f. but you rounded to 0.53, throwing away at least 1 s.f. I obtained, using your numbers, 13.37 L, which would be rounded to 13.4 to three s.f. However, using the number in the calculator that comes from 16/30.07 (0.53209), and using the remainder of the figures you have, returns 13.418 which rounds to 13.4L.
1. 👍
2. 👎
3. ℹ️
4. 🚩
2. I guess I need to figure out the s.f. or stop rounding numbers.... :( thank you...
1. 👍
2. 👎
3. ℹ️
4. 🚩
3. I ALWAYS carry at least one more place than I know I'm allowed and sometimes even two more. You can ALWAYS throw those unwanted numbers away at the end when you round with s.f. in mind; you can't always ADD numbers at the end if you've thrown them away earlier.
1. 👍
2. 👎
3. ℹ️
4. 🚩 | 415 | 1,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-33 | latest | en | 0.914249 |
http://gmatclub.com/forum/awa-quick-question-60008.html?fl=similar | 1,485,204,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00327-ip-10-171-10-70.ec2.internal.warc.gz | 113,874,313 | 50,747 | AWA - Quick Question : GMAT Verbal Section
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 23 Jan 2017, 12:55
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# AWA - Quick Question
Author Message
Intern
Joined: 26 Aug 2007
Posts: 30
Followers: 1
Kudos [?]: 2 [0], given: 0
### Show Tags
13 Feb 2008, 07:30
I've already taken the test once, but can't remember exactly how the essays worked... Do any you know the answer to these questions?
1.) What comes first, "Analysis of an Argument" or "Analysis of an Issue"? Do they appear in the same order every time?
2.) Do they mention the question subject before displaying the question? Meaning, do the words "Analysis of an Argument" appaear before the "Analysis of an Argument" question (same for the other AWA essay) or do they just give you the argument/issue and instructions?
Thanks! I know these are weird questions, but I can't remember.
If you have any questions
New!
Director
Joined: 14 Sep 2007
Posts: 908
Schools: Kellogg '10
Followers: 6
Kudos [?]: 93 [0], given: 15
Re: AWA - Quick Question [#permalink]
### Show Tags
13 Feb 2008, 11:47
McCombs_Hopeful wrote:
I've already taken the test once, but can't remember exactly how the essays worked... Do any you know the answer to these questions?
1.) What comes first, "Analysis of an Argument" or "Analysis of an Issue"? Do they appear in the same order every time?
2.) Do they mention the question subject before displaying the question? Meaning, do the words "Analysis of an Argument" appaear before the "Analysis of an Argument" question (same for the other AWA essay) or do they just give you the argument/issue and instructions?
Thanks! I know these are weird questions, but I can't remember.
I haven't taken the official test yet, but I'm pretty confident that it's Issue before Argument. Not sure about your second question.
_________________
http://www.fantasticcontraption.com
GMAT Club Premium Membership - big benefits and savings
Re: AWA - Quick Question [#permalink] 13 Feb 2008, 11:47
Similar topics Replies Last post
Similar
Topics:
1 A quick question about modifier 1 15 Sep 2015, 23:03
quick question 2 13 Sep 2008, 09:21
1 Quick Question - plural? 4 28 Aug 2008, 22:51
Quick question: SV agreement 7 03 Oct 2007, 13:30
quick idiom-ish question 4 09 Jul 2007, 17:49
Display posts from previous: Sort by | 759 | 2,936 | {"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-2017-04 | latest | en | 0.896407 |
https://physics.stackexchange.com/questions/529076/why-does-my-tea-periodically-alternate-its-rotational-speed-after-stirring-lin/529237 | 1,653,504,242,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00609.warc.gz | 520,399,837 | 74,650 | # Why does my tea periodically alternate its rotational speed after stirring? (Link to video below)
I noticed that after stirring, a bubble in the centre of my mug of tea changed the speed it was rotating at periodically. Speeding up, then slowing down, then speeding up again, etc. Almost like when a ballerina pulls in her arms to increase her speed.
Tea after stirring
Edit: I've repeated this with room temperature water to try and rule out any temperature-related effects and the same effect is present.
• @JeremyC This type of circulation formation is also referred to as Tea leaf paradox Feb 4, 2020 at 23:36
• Could you please try this experiment with a taller cup? If the timescale for the oscillation is slower in a taller cup, it would suggest that vertical motion of the fluid plays an important role.
– Yly
Feb 5, 2020 at 21:55
• Doing image analysis on rotation speeds would be easier if the camera was held steady and there was a bit more light, now there is noticeable amount of motion blur. Feb 6, 2020 at 9:56
Just looking at the video, it appears that the shape of the surface is varying quasi-periodically, as if the liquid is moving outward (and upward) toward the cup walls, then moving inward and rising in the center of the cup. This can be expected, if in the beginning the shape is not a perfect equilibrium shape (e.g., like a parabolic surface in a rotating cup). But when the the liquid moves toward the center, the rotation necessarily speeds up due to conservation of angular momentum; and when it moves outward the opposite happens.
A crude analogy: If you rolled a marble in a large wok with smooth spherical curvature, in such a way that it looped near the center/bottom, then out near the edge, you would see that its angular velocity increases when it approaches the center/bottom, and decreases when it recedes from the center/bottom. You can think of the volume of liquid doing the same thing as the surface shape changes from a shallow curve to a deep curve.
• In a stationary cup, I'd expect the oscillations to be much faster than what's in the video. However, I just read the article on the Tea Leaf Paradox and wonder if the secondary flow somehow reduces the frequency of the oscillations? Feb 6, 2020 at 20:47
• I believe the rotation of the tea reduces the frequency of the oscillations. Haven't done the math, though. Feb 6, 2020 at 23:44
• One other factor none of us has taken into account: There is often a surface film on tea or coffee, possibly a monolayer more or less like a Langmuir-Blodgett layer, which can be rather stiff in the tangent plane to the liquid but quite flexible normal to the liquid. Bubbles are pretty much locked into that film. When shear, stretching, or compression forces exceed the strength of the film, it breaks and its parts can spin freely until they link up again. Forces would be due to relative motion between the fluid and film. It seems likely that breakage and re-assembly would be quasi-periodic. Feb 7, 2020 at 2:29
You have created a Lissajous spiral from the tea: the raising of the mixing utensil creates a partial secondary flow which deviates some of the concentric flow up and down. The continuous rotational flow dominates and it is interrupted by a group that flows up and down as well as concentrically.
The effect slightly follows a Lissajous trajectory, although it is a complex turbulent flow and would necessitate some direct observation to understand precisely.
As others have said, it's very likely that the lissajous travels downwards through the center of the mug and back upwards trhough the sides, so that it actually becomes a spiralling torus.
You can use oat flour or glitter in a transparent mug to see the effect. You can glue a syringe with milk to a spoon to see if you can observe a Lissajous spiral and film it. It probably necessitates a big jar and a video because there are a lot of small turbulent vortices that spiral and mix the boundary of the concentric vortices, especially when the Lissajous flow hits upper and lower boundaries.
• I notice that flow lines cross, indicating that flow goes two directions at the same point. Would a better picture perhaps show flow ascending near the center and descending near the edge? Perhaps this is something like the flow of a tornado? Feb 5, 2020 at 15:49
• I executed the proposed experiment (with water and milk, pouring at the bottom), but the whole water was rotating without up- or downward motion. So the dancing bubble is not caused by your proposed mechanism. And even if it were, the effect would be too little to cause the bubble to dance. Feb 5, 2020 at 20:54
• So then if you don't remove the spoon, it won't happen? Feb 5, 2020 at 23:26
• @mmesser314 yes it seems that the water also travels up and down the the middle and the sides of the mug when it changes up and down, so perhaps there are 1-2 spinning torus in the mug. i.ytimg.com/vi/T-cATdAUIHA/hqdefault.jpg Feb 6, 2020 at 7:56
• @BlueRaja-DannyPflughoeft depending if you use spoon/chopstick which can pull up more/less water upwards, the effect changes: the rotation is more regular and lasts longer if you deflect flow upwards. Feb 6, 2020 at 8:04
To corroborate on the observation of @Vladimir Kalitvianski, this hypothesis might be understood by simple Newtonian physics.
There are two timescales in the system - $$\tau_J$$ timescale of the dissipation of angular momentum and $$\tau_I$$timescale of (quasiperiodic) change in tensor of inertia of the liquid. It seems that $$\tau_J$$ large enough in comparison to $$\tau_I$$ such that tensor of inertial changes significantly without significant change in angular momentum at small finite times. The rest is the consequence of the $$\vec{J}=I\vec{\omega}$$ relation.
To experimentally test it, it would be nice to measure the time dependence of the circulation at the center (image processing the orientation of the bubble at the center) and measure the time dependent local height profile of the fluid, from which it is possible to obtain the time dependent component of tensor of inertia. These two independent measurement should correlate strongly in Fourier space if the hypothesis is correct.
When a liquid spins, there is often a vortex that spins quickly. Velocity drops away from the center. Usually the vortex spins quietly in the center of the vessel.
In this case, there may be a vortex that is off center orbiting around the center of the cup. When the vortex passes under the bubbles, they spin quickly. When it passes on, they slow down.
This sounds plausible at first, but I am not sure if it is realistic. The vortex affects the shape of the surface and draws the initially separated bubbles together. It looks like the center is depressed, and the largest component of the circular motion is around the center of the cup.
When the spoon swooshes through the tea, two counter-rotating vortices are created. The spoon travels in a circle, and this is what gives the tea its overall circular motion.
Perhaps one or more counter rotating vortices survive. It is clear that the bubbles are pulled off and on center in sync with their rotational speed changes. Perhaps a counter rotating vortex orbits the center, and slows the bubbles when it passes underneath. Perhaps it attracts or repels the bubbles.
This is all speculative, but there are a couple points. First I suspect that rapidly spinning vortices are involved somehow. And second while period doubling might explain the motion, it might also be explained by two separate causes where one of them comes and goes in a regular way.
Really cool video. I would like to know what is really going on.
• Best explanation! Feb 6, 2020 at 9:53
The liquid in your mug does not rotate as a solid body due to viscosity; it also goes up at the center and down at the mug walls ("rotates in a vertical cross section") due to heat losses and thus stratification effects. So the 3D motions are not stationary anyway - they will stop in a finite time. Their interplay (roughly saying, transport time and rotation time) lead to some surface effects that you observe.
The variable rotation speed is due to (semi-)turbulent motion.
You can see this more clearly in this YouTube video that doesn't show the rotation by a few bubbles but by thermal imaging (which also shows the convection).
It is not the entire flow that changes uniformly, but instead it looks more chaotic. Medium size rapidly rotating edies occur (randomly) and dissipate their energy and dissapear after which new eddies form.
The pattern seems to occur rhythmically because it only appears like that (but it is a bit random). Although the eddies may have a narrow distribution in lifetime because the Rayleigh-Bénard convection on top of the rotation causes fluid to move inward (where liquid is cooler and sinks) which creates a consistent pattern of acceleration in the angular speed and eventual collapse of the eddie.
Note that these eddies keep occurring also when the stirring was a long time ago. They are powered by the
• energy/motion from thermal convection (or only stirring if there is no strong temperature difference)
• turned into eddies/rotation due to turbulence (and regular instabilities that can already be explained by laminar flow)
• and a tornado effect because the liquid is moving to the inside (convection) which accelerates the angular speed (like the ballerina making the pirouette and explained by conservation of angular momentum)
The exact type of instabilities that cause the oscillations might be difficult to pinpoint because there are multiple effects (initial chaotic state due to stirring, temperature/gravity gradients, flow instabilities in rotating fluids, deceleration of the rotation when stirring stops). There are many visualisations of such effect (with ink or particles in liquid) an example is this YouTube video
related to a MITopencourseware course and http://weathertank.mit.edu
As pointed out by Vladimir Kalitviansky: as the fluid slows down (due to friction) an internal circulation commences: along the walls fluid is descending and hence in the center fluid is rising.
This circulation can also be inferred from the following observation: when stirred tea is slowing down the tea leaves that are on bottom of the glass accumulate in the center.
This circulation would not form if the glass itself would be co-rotating with the fluid. Then you get solid body rotation of the entire body of fluid.
In the case of solid body rotation the surface of the fluid redistributes to a shape with a parabolic cross section. Thus at every distance to the center of rotation the required centripetal force is provided by the inclination of the surface.
Here the fluid that is touching the wall is being slowed down, so it doesn't have the angular velocity that is required at that distance to the center of rotation. As a consequence the fluid that is touching the wall descends, and in turn that pushes up fluid in the center.
At the top the fluid must spread out again. That is: when the swirling fluid is in the process of slowing down the top layer consists of fluid that is flowing outward from the center.
As the top layer flows outward it loses angular velocity. Friction with the fluid underneath it brings the top layer up to speed. By the looks of it: this transfer of momentum happens in bursts, rather than continuously.
That, I seems to me, is the explanation for the changes in angular velocity of the top layer of fluid that you are observing
Welcome Luke, and congratulations on such a great question.
If the fluid just rotated uniformly, we'd expect to see a low spot in the middle (a vortex). due to the inward acceleration of the fluid by the cup. This is almost all we need to know to see why the bubble rotates more quickly in the low spot than when it jumps out. In the center of the vortex, the fluid rotates around the bubble uniformly and the little surrounding bubbles rotate with the fluid at the radius of the big bubble.
When the bubble group jumps out of the vortex, it keeps rotating in the same direction but slowly. Here, we see the part of the bubble group closer to the vortex moving more slowly than the part closer to the cup's edge. Crudely, $$V = r\omega$$. Since the outer edge of the group is further from the vertex center than the inner edge, the little bubbles spin around the big bubble slowly.
This is a remarkably complicated system. Your spoon accelerates the fluid in unpredictable ways, surface tension makes for a little equilibrium for the group at the center, there may be different density fluids present (if you recently added cold milk, for example), bubbles don't behave intuitively (check out the corks in the video I linked) and the friction between the fluid and the mug cause turbulence (depending on Reynolds number). The perturbations from these factors help explain why the vortex isn't in the center of the cup, why the bubbles "jump" in and out of the vortex and why the behavior gets slower and more uniform as the experiment progresses.
I see no evidence of liquid moving en mass to the center of the cup, nor of a higher center - it looks like your tea is like every other spinning liquid with the low point in the center. Further, I would be shocked if the mug cooling the fluid caused a down-current at the mug's edge during the experiment that was bigger than the factors I've listed so far.
If you wanted to test this hypothesis, place bubbles at varying distances from the vortex. As they move further from the center, they should spin at roughly the same rate until you place them on the edge where they may stop spinning or spin backwards due to the surface friction. There's an integral you could do to see how they vary, but these measurements are crude, and the differences will be small.
That's a good question for a new contributor! Let me try to give you an answer.
When you pour the coffee you give the body of coffee a rotation (clockward) in the cup. You can put a velocity vector field on the surface of the coffee. Initially, all these vectors have the same magnitude and directions parallel to the tangent vector on the side of the cup.
So in the middle of the cup, the rotation velocity has the highest value, simply because there is the least distance to travel before a complete rotation is completed.
The bubble goes along with this rotation (also clockward; just try pouring the coffee in the opposite direction; it would be very strange if this wasn't the case!). When the bubble is exactly in the middle (as well as the middle point of the rotation) it should stay there and will rotate along with the coffee (though in the video it doesn't seem the middle of the coffee surface has the highest rotation velocity, but let's assume the highest rotations velocities find themselves around the middle). The surface, in this case, must be symmetrical wrt the middle point of the round glass. The water molecules whirling around the vertical axis going through the middle point don't form a vortex.
But this is theoretically. Just a tiny displacement from the center is enough to let the effect occur, and this will surely happen after pouring. So in reality, when the bubble moves outward a bit, its angular velocity is diminished because of the interaction with the different (smaller) angular velocities of the coffee surface and because the angular momentum has to be conserved. This also drives the bubble back to the middle. You will never see the bubble moving too much away from the center of rotation. All angular velocities are reduced because of friction, tending to make all angular velocities equal. The bubble is doing a rotation dance!
Finally, the (angular) velocities all tend to zero because of friction and the effect is gone, obviously.
You made a great video. The temperature has to do with it insofar the viscosity of coffee becomes less if the temperature is increased. Maybe you can try to put the bubble in the middle and see what happens when trying to let the coffee rotate in such a way that the center of rotation will find itself in the middle, though this is very difficult to do!
The bubble spends more time in the middle when the temperature is higher (where it spins faster) because it's easier pulled to it than driven away from it. Obviously. Why? Maybe you can think about that for yourself (viscosity).
Here is another, new video.
• Looking at that video at the end it seems to be some sort of oscillating breathing pattern, where the paint (or whatever tracer is added) is moving repetitively inwards and outwards. This might be due to a wave and oscillating gravitational/kinetic energy. Feb 6, 2020 at 15:21
Possibly the Dzhanibekov effect in a fluid: https://www.youtube.com/watch?v=L2o9eBl_Gzw
Perhaps a more controlled experiment where motions outside the surface plane are minimized could be created to prevent the occurrence of the effect, but even then, I suspect the effect is unavoidable due to irregularities created in a liquid by the spinning motion itself that would cause the moment to move away from the exact center.
My best guess is that the whirlpool that carries the mass of the water downwards has only a limited pathlength to do so.
At some point you start stirring, and at that point the whirlpool starts accelerating mass downwards. So it carries the mass down, it bounces back up and from the surface back down, periodically.
If you now imagine a conical whirlpool, the lowest point, where the fewest mass is displaced, rotates the fastest, the highest point rotates the slowest. These points periodically interchange.
I think this is the same thing that @Alexander tried to express, but unfortunately unnecessarily complicated.
https://en.wikipedia.org/wiki/Rayleigh%E2%80%93B%C3%A9nard_convection | 3,839 | 17,874 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-21 | longest | en | 0.951562 |
https://mail.python.org/pipermail/python-list/2004-March/250792.html | 1,397,675,912,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00473-ip-10-147-4-33.ec2.internal.warc.gz | 805,061,249 | 2,865 | # Is there a map or graph module?
Elaine Jackson elainejackson7355 at home.com
Sat Mar 20 22:44:14 CET 2004
```"""
Hopefully some of the following will help you. The idea is to implement weighted
digraphs as dictionaries: the keys are edges and the values are weights. If
direction is not important in a particular graph, you can 'bidirectionalize' it
(see below). If weights are not important, just ignore them. Since the only
important property of a vertex is its name, the sensible thing is to build
graphs whose vertices are self-referring strings: a function is provided that
declares any number of such vertices at the same time. Another function returns
all the edges of the cyclic digraph with specified vertices. The exercises at
the bottom are from Discrete Mathematics and its Applications by Kenneth Rosen.
"""
def declareVertices(strg):
from string import split
for name in split(strg,','):
globals()[name]=name
cycleEdges = lambda *X: [(X[i],X[(i+1)%len(X)]) for i in range(len(X))]
def bidirectionalize(graph):
converse = dict([((b,a),graph[a,b]) for (a,b) in graph.keys()])
graph.update(converse)
def vertexSet(graph):
A=[a for (a,b) in graph.keys()]
B=[b for (a,b) in graph.keys()]
X=A+B
return [X[i] for i in range(len(X)) if X[i] not in X[0:i]]
def neighbors(a,graph):
leftNeighbors=[b for b in vertexSet(graph) if (b,a) in graph.keys()]
rightNeighbors=[b for b in vertexSet(graph) if (a,b) in graph.keys()]
X=leftNeighbors+rightNeighbors
return [X[i] for i in range(len(X)) if X[i] not in X[0:i]]
def dijkstra(origin,graph):
infinity = sum(graph.values())+1
reVal = {origin:0}
A = lambda y: [x for x in reVal.keys() if (x,y) in graph.keys()]
B = lambda y: (A(y) and min([reVal[x]+graph[x,y] for x in A(y)])) \
or infinity
while [y for y in vertexSet(graph) if y not in reVal.keys()]:
C = [(y,B(y)) for y in vertexSet(graph) if y not in reVal.keys()]
m = min([pair[1] for pair in C])
reVal.update(dict([(y,z) for (y,z) in C if z==m]))
return reVal
def kruskal(graph):
forest=[[x] for x in vertexSet(graph)]
freeEdges=graph.keys()
takenEdges=[]
while len(forest)>1:
minWeight=min([graph[edge] for edge in freeEdges])
A = lambda edge: graph[edge]==minWeight
B = lambda edge: [x for x in forest if edge[0] in x or edge[1] in x]
edge=filter((lambda edge: A(edge) and len(B(edge))==2),freeEdges)[0]
freeEdges.remove(edge)
takenEdges.append(edge)
newTree=B(edge)[0]+B(edge)[1]
forest=[x for x in forest if x not in B(edge)]
forest.append(newTree)
return takenEdges
####################################################
####################################################
"""
#################################
## Rosen; Section 7.6; Exercise 4
#################################
declareVertices('a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z')
graph = { \
(a,b):2, (a,c):4, (a,d):1, (b,c):3, (b,e):1, \
(c,e):2, (c,f):2, (d,f):5, (d,g):4, (e,h):3, \
(f,g):3, (f,h):3, (f,i):2, (f,j):4, (g,k):2, \
(h,l):1, (h,o):8, (i,j):3, (i,l):3, (i,m):2, \
(j,k):6, (j,m):6, (j,n):3, (k,n):4, (k,r):2, \
(l,m):3, (l,o):6, (m,n):5, (m,o):4, (m,p):2, \
(n,q):2, (n,r):1, (o,p):2, (o,s):6, (p,q):1, \
(p,s):2, (p,t):1, (q,r):8, (q,t):3, (r,t):8, \
(s,z):2, (t,z):8 \
}
distances = dijkstra(a,graph)
X = distances.keys()
X.sort()
for vertex in X: print vertex, ": ", distances[vertex]
"""
####################################################
####################################################
"""
#################################
## Rosen; Section 8.6; Exercise 3
#################################
declareVertices('a,b,c,d,e,f,g,h,i,j,k,l')
graph = { \
(a,b):2, (b,c):3, (c,d):1, \
(a,e):3, (b,f):1, (c,g):2, (d,h):5, \
(e,f):4, (f,g):3, (g,h):3, \
(e,i):4, (f,j):2, (g,k):4, (h,l):3, \
(i,j):3, (j,k):3, (k,l):1 \
}
totalWeight = lambda tree,graph: sum([graph[edge] for edge in tree])
spanTree=kruskal(graph)
print totalWeight(spanTree,graph)
"""
``` | 1,253 | 4,032 | {"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} | 3.265625 | 3 | CC-MAIN-2014-15 | latest | en | 0.779221 |
https://help.libreoffice.org/latest/tg/text/scalc/01/04060185.html | 1,708,555,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00274.warc.gz | 305,296,407 | 10,452 | # Функсияҳои Оморӣ Қисми Панқум
## САНА
Calculates the skewness of a distribution using the population of a random variable.
This function is available since LibreOffice 4.1.
### Синтаксис
SKEWP(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least three values.
This function is part of the Open Document Format for Office Applications (OpenDocument) standard Version 1.2. (ISO/IEC 26300:2-2015)
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
### Намунаҳо
SKEWP(2;3;1;6;8;5) returns 0.2329985562
SKEWP(A1:A6) returns 0.2329985562, when the range A1:A6 contains {2;3;1;6;8;5}
## DEVSQ
Суммаи квадратии фарқиятро дар асоси намуна ҳисоб мекунад.
#### Syntax
DEVSQ(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
Сину сол
## FORECAST
Арзишҳои давомро ҳисоб мекунад дар асоси арзишҳои x ва y.
#### Syntax
FORECAST(Value; DataY; DataX)
Value арзиши x.
Data_Y массиви y.
Data_X iмассиви x.
#### Example
=FORECAST(50;A1:A50;B1;B50) returns the Y value expected for the X value of 50 if the X and Y values in both references are linked by a linear trend.
## FORECAST.LINEAR
Арзишҳои давомро ҳисоб мекунад дар асоси арзишҳои x ва y.
#### Syntax
FORECAST.LINEAR(Value; DataY; DataX)
Value арзиши x.
Data_Y массиви y.
Data_X iмассиви x.
#### Example
=FORECAST.LINEAR(50;A1:A50;B1;B50) returns the Y value expected for the X value of 50 if the X and Y values in both references are linked by a linear trend.
#### Technical information
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.FORECAST.LINEAR
## NORM.S.DIST
Функтсияи арзиши тақсими стандартиро бармегардонад.
#### Syntax
NORM.S.DIST(Number; Cumulative)
Number арзише, ки барои он арзиши тақсими стандартӣ ҳисоб карда мешавад.
Cumulative 0 or FALSE calculates the probability density function. Any other value or TRUE calculates the cumulative distribution function.
#### Example
=NORM.S.DIST(1;0) returns 0.2419707245.
=NORMSDIST(1) натиҷа 0.84.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.NORM.S.DIST
## NORM.S.INV
Баръакси тақсимоти стандартии муқаррариро бармегардонад.
#### Syntax
NORM.S.INV(Number)
Number эҳтимолият.
#### Example
NORMSINV(0.908789) натиҷа 1.3333.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.NORM.S.INV
## NORMSDIST
Функтсияи арзиши тақсими стандартиро бармегардонад.
Ин GAUSS(x)=NORMSDIST(x)-0.5 аст
#### Syntax
NORMSDIST(Number)
Number арзише, ки барои он арзиши тақсими стандартӣ ҳисоб карда мешавад.
#### Example
=NORMSDIST(1) натиҷа 0.84.
## NORMSINV
Баръакси тақсимоти стандартии муқаррариро бармегардонад.
#### Syntax
NORMSINV(Number)
Number эҳтимолият.
#### Example
NORMSINV(0.908789) натиҷа 1.3333.
## PERMUT
Шумораи ҷойивазкуниро барои объектҳои додашуда ҳисоб мекунад.
#### Syntax
PERMUT(Count1; Count2)
Count_1 шумораи объектҳо.
Count_2 шумораи объектҳо дар ҷобаҷокунӣ.
#### Example
=PERMUT(6; 3) натиҷа 120.
## PERMUTATIONA
Шумораи ҷойивазкуниро барои объектҳои додашуда ҳисоб мекунад.
#### Syntax
PERMUTATIONA(Count1; Count2)
Count_1 шумораи объектҳо.
Count_2 шумораи объектҳо дар ҷобаҷокунӣ.
#### Example
Бо кадом зуддӣ 2 ин7тихоб мешаванд дар миёни 11 объект?
PERMUTATIONA(11;2) натиҷа 121.
PERMUTATIONA(6; 3) натиҷа 216.
## PROB
Эҳтимолияти дар миёни ду лимит будани арзишҳои диапазонро ҳисоб мекунад.Агар арзиши Охир набошад, ин функтсия эҳтимолиятро дар асоси арзиши Оғоз ҳисоб мекунад.
#### Syntax
PROB(Data; Probability; Start [; End])
Data массив.
Probability массив.
Start оғози фосила.
End (optional) is the end value of the interval whose probabilities are to be summed. If this parameter is missing, the probability for the Start value is calculated.
#### Example
=PROB(A1:A50;B1:B50;50;60) returns the probability with which a value within the range of A1:A50 is also within the limits between 50 and 60. Every value within the range of A1:A50 has a probability within the range of B1:B50.
## RANK.AVG
Returns the statistical rank of a given value, within a supplied array of values. If there are duplicate values in the list, the average rank is returned.
The difference between RANK.AVG and RANK.EQ occurs when there are duplicates in the list of values. The RANK.EQ function returns the lower rank, whereas the RANK.AVG function returns the average rank.
#### Syntax
RANK.AVG(Value; Data [; Type])
Value арзиш.
Data массив.
Type (ихтиёрӣ). = 0 ба болоравӣ, = 1 ба поёнравӣ.
Type = 1 ба поёнравӣ.
Type = 1 ба поёнравӣ.
#### Example
=RANK(A10; A1:A50) ҷойгиршавии A10 -ро дар қитъаи A1:A50 муайян месозад. Агар Арзиш дар қитъа набошад хабари хато нишон дода мешавад.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.RANK.AVG
## RANK.EQ
Returns the statistical rank of a given value, within a supplied array of values. If there are duplicate values in the list, these are given the same rank.
The difference between RANK.AVG and RANK.EQ occurs when there are duplicates in the list of values. The RANK.EQ function returns the lower rank, whereas the RANK.AVG function returns the average rank.
#### Syntax
RANK.EQ(Value; Data [; Type])
Value арзиш.
Data массив.
Type (ихтиёрӣ). = 0 ба болоравӣ, = 1 ба поёнравӣ.
Type = 1 ба поёнравӣ.
Type = 1 ба поёнравӣ.
#### Example
=RANK(A10; A1:A50) ҷойгиршавии A10 -ро дар қитъаи A1:A50 муайян месозад. Агар Арзиш дар қитъа набошад хабари хато нишон дода мешавад.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.RANK.EQ
## SKEW
Ғайрисимметрии тақсимотро медиҳад.
#### Syntax
SKEW(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least three values.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
#### Example
=SKEW(A1:A50) calculates the value of skew for the data referenced.
## SLOPE
Хамии регрессияи хаттиро медиҳад.
#### Syntax
SLOPE(DataY; DataX)
Data_Y массиви Y.
Data_X массиви X.
Сину сол
## STANDARDIZE
Адади тақрибиро ба адади муқаррарӣ табдил медиҳад.
#### Syntax
STANDARDIZE(Number; Mean; StDev)
Number арзиш.
Mean арзиши арифметикии тақсимот.
STDEV фарқияти стантартии тақсимот.
#### Example
=STANDARDIZE(11; 10; 1) натиҷа 1.
## STEYX
Хатои стандартиро барои арзиши y барои ҳар як арзиши x дар регрессия ҳисоб мекунад.
#### Syntax
STEYX(DataY; DataX)
Data_Y массиви Y.
Data_X массиви X.
Сину сол
## T.DIST
t-тақсимотро бармегардонад.
#### Syntax
T.DIST(Number; DegreesFreedom; Cumulative)
Number арзише, ки ба он t-тақсимот ҳисоб карда мешавад.
Degrees_freedomдараҷаи озодӣ.
Cumulative = 0 or FALSE returns the probability density function, 1 or TRUE returns the cumulative distribution function.
#### Example
=T.DIST(1; 10; TRUE) returns 0.8295534338
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.DIST
## T.DIST.2T
Calculates the two-tailed Student's T Distribution, which is a continuous probability distribution that is frequently used for testing hypotheses on small sample data sets.
#### Syntax
T.DIST.2T(Number; DegreesFreedom)
Number арзише, ки ба он t-тақсимот ҳисоб карда мешавад.
Degrees_freedomдараҷаи озодӣ.
#### Example
=T.DIST.2T(1; 10) returns 0.3408931323.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.DIST.2T
## T.DIST.RT
Calculates the right-tailed Student's T Distribution, which is a continuous probability distribution that is frequently used for testing hypotheses on small sample data sets.
#### Syntax
T.DIST.RT(Number; DegreesFreedom)
Number арзише, ки ба он t-тақсимот ҳисоб карда мешавад.
Degrees_freedomдараҷаи озодӣ.
#### Example
=T.DIST.RT(1; 10) returns 0.1704465662.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.DIST.RT
## T.INV
Баръакси t-тақсимотро медиҳад.
#### Syntax
T.INV(Number; DegreesFreedom)
Number эҳтимолият.
Degrees_freedomдараҷаи озодӣ.
#### Example
=TINV(0.1; 6) натиҷа 1.94
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.INV
## T.INV.2T
Calculates the inverse of the two-tailed Student's T Distribution , which is a continuous probability distribution that is frequently used for testing hypotheses on small sample data sets.
#### Syntax
T.INV.2T(Number; DegreesFreedom)
Number эҳтимолият.
Degrees_freedomдараҷаи озодӣ.
#### Example
=T.INV.2T(0.25; 10) returns 1.221255395.
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.INV.2T
## TDIST
t-тақсимотро бармегардонад.
#### Syntax
TDIST(Number; DegreesFreedom; Mode)
Number арзише, ки ба он t-тақсимот ҳисоб карда мешавад.
Degrees_freedomдараҷаи озодӣ.
Mode = 1 санҷиши якшоха, Mode = 2 санҷиши душоха.
Сину сол
## TINV
Баръакси t-тақсимотро медиҳад.
#### Syntax
TINV(Number; DegreesFreedom)
Number эҳтимолият.
Degrees_freedomдараҷаи озодӣ.
#### Example
=TINV(0.1; 6) натиҷа 1.94
## WEIBULL
Арзиши тақсимоти Вейбулро медиҳад.
The Weibull distribution is a continuous probability distribution, with parameters Alpha > 0 (shape) and Beta > 0 (scale).
If C is 0, WEIBULL calculates the probability density function.
If C is 1, WEIBULL calculates the cumulative distribution function.
#### Syntax
WEIBULL(Number; Alpha; Beta; C)
Number арзиш.
Alpha арзиши Алфа.
Beta арзиши Бетта.
C типи функсия. Агар C = 0 формаи функсия ҳисоб карда мешавад, агар C = 1 тақсимот ҳисоб карда мешавад.
#### Example
=WEIBULL(2; 1; 1; 1) натиҷа 0.86.
## WEIBULL.DIST
Арзиши тақсимоти Вейбулро медиҳад.
The Weibull distribution is a continuous probability distribution, with parameters Alpha > 0 (shape) and Beta > 0 (scale).
If C is 0, WEIBULL.DIST calculates the probability density function.
If C is 1, WEIBULL.DIST calculates the cumulative distribution function.
#### Syntax
WEIBULL.DIST(Number; Alpha; Beta; C)
Number арзиш.
Alpha арзиши Алфа.
Beta арзиши Бетта.
C типи функсия. Агар C = 0 формаи функсия ҳисоб карда мешавад, агар C = 1 тақсимот ҳисоб карда мешавад.
#### Example
=WEIBULL(2; 1; 1; 1) натиҷа 0.86.
#### Technical information
This function is available since LibreOffice 4.2.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.WEIBULL.DIST
## ZТЕСТ
Эҳтимолияти бо санҷиши t-и студентҳо ҳисоб мекунад.
#### Syntax
TTEST(Data1; Data2; Mode; Type)
Data_1 массив.
Data_2 массив.
Mode = 1 санҷиши якшоха, Mode = 2 санҷиши душоха.
Type намуди t-санҷиш. Type 1 ҷуфт. Type 2 ду намуна. Type 3 ну намунаи нобаробар.
Сину сол
## ZТЕСТ
Эҳтимолияти бо санҷиши t-и студентҳо ҳисоб мекунад.
#### Syntax
T.TEST(Data1; Data2; Mode; Type)
Data_1 массив.
Data_2 массив.
Mode = 1 санҷиши якшоха, Mode = 2 санҷиши душоха.
Type намуди t-санҷиш. Type 1 ҷуфт. Type 2 ду намуна. Type 3 ну намунаи нобаробар.
Сину сол
#### Technical information
This function is available since LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.T.TEST
## ВА
Рутбаи рақамро дар намуна медиҳад.
#### Syntax
RANK(Value; Data [; Type])
Value арзиш.
Data массив.
Type (ихтиёрӣ). = 0 ба болоравӣ, = 1 ба поёнравӣ.
Type = 0 аслӣ,
Type = 1 ба поёнравӣ.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
#### Example
=RANK(A10; A1:A50) ҷойгиршавии A10 -ро дар қитъаи A1:A50 муайян месозад. Агар Арзиш дар қитъа набошад хабари хато нишон дода мешавад.
## ФАР
Фарқиятро нисбати намуна муайян мекунад.
#### Syntax
VAR(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
Сину сол
## ФАР
Фарқиятро нисбати намуна муайян мекунад. Арзиши матн 0.
#### Syntax
VARA(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values.
Сину сол
## ФАРП
Фарқиятро нисбати намуна муайян мекунад.
#### Syntax
VAR.S(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values.
Сину сол
#### Technical information
This function is available since LibreOffice 4.2.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.VAR.S
## ФАРП
Фарқиятро нисбати ҳама намунаҳо муайян мекунад.
#### Syntax
VARP(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
Сину сол
## ФАРП
Фарқиятро нисбати ҳама намунаҳо муайян мекунад.
#### Syntax
VAR.P(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
Сину сол
#### Technical information
This function is available since LibreOffice 4.2.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.VAR.P
## ФАРП
Фарқиятро нисбати ҳама намунаҳо муайян мекунад. Арзиши матн 0.
#### Syntax
VARPA(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
Сину сол
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси намуна муайян месозад.
#### Syntax
STDEV(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
#### Example
=STDEV(A1:A50) returns the estimated standard deviation based on the data referenced.
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси намуна муайян месозад.
#### Syntax
STDEVA(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values. Text has the value 0.
#### Example
=STDEVA(A1:A50) returns the estimated standard deviation based on the data referenced.
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси ҳамаи арзишҳо ҳисоб мекунад.
#### Syntax
STDEVP(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.
#### Example
=STDEVP(A1:A50) returns a standard deviation of the data referenced.
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси ҳамаи арзишҳо ҳисоб мекунад.
#### Syntax
STDEV.P(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
#### Example
=STDEV.P(A1:A50) returns a standard deviation of the data referenced.
#### Technical information
This function is available since LibreOffice 4.2.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.STDEV.P
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси ҳамаи арзишҳо ҳисоб мекунад.
#### Syntax
STDEV.S(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
The parameters should specify at least two values.
#### Example
=STDEV.S(A1:A50) returns a standard deviation of the data referenced.
#### Technical information
This function is available since LibreOffice 4.2.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.STDEV.S
## ФАРҚРСТАНД
Фарқияти стандартиро дар асоси ҳамаи арзишҳо ҳисоб мекунад.
#### Syntax
STDEVPA(Number 1 [; Number 2 [; … [; Number 255]]])
Number 1, Number 2, … , Number 255 are numbers, references to cells or to cell ranges of numbers.
Text has the value 0.
#### Example
=STDEVPA(A1:A50) returns the standard deviation of the data referenced. | 6,482 | 20,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-10 | latest | en | 0.302336 |
https://en.khanacademy.org/science/in-in-class11th-physics/in-in-phy-kinetic-theory/in-in-phy-ideal-gas-laws/v/boyles-law | 1,701,818,410,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00416.warc.gz | 274,577,741 | 140,931 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Boyle's law
Explore Robert Boyle's 17th-century experiments with gases that led to Boyle's Law. Discover how Boyle's Law, demonstrating the inverse relationship between gas pressure and volume, laid the groundwork for the ideal gas equation. This historical journey illuminates key principles in gas behavior. Created by Ryan Scott Patton.
## Want to join the conversation?
• Why would Robert Boyles Wife be excited about the J tube?
• I think he's being sarcastic, and he actually means that she probably didn't like it.
• actually robert boyle didn't do the Experiment himself, he just took it from Henry Power. https://en.wikipedia.org/wiki/Henry_Power
• I still don't understand very well. So, do you have to find the inverse of your values in order to do the equation? And whats with the equation? So is it P1V1=P2V2, or V=m 1/p, or V=k(1/p), or P1V1=PfVf? PLEASE SOMEBODY EXPLAIN! :'(
• Hello Nicole. I think the equation is P1V1 = P2V2. You can derive this from the Combined Gas Equation (P1V1/T1 = P2V2/T2). Since Boyle's law says it is at constant temperature, the temperatures cancel each other so you are left with P1V1 = P2V2 which is Boyle's Law.
• at when he draw the second graph he took volume on the y axis but in the first graph he took pressure on the y axis .. why he did so?
• umm...also,can someone please explain what happened in the video?-im so sorry . i don't mean to be disrespectful or anything,but I'm kinda confused.
• Also, what is the difference between a Real Gas and an Ideal gas?
• What does Boyle's Law apparently state? | 439 | 1,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-50 | latest | en | 0.955565 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_Languages/Think_Python_-_How_to_Think_Like_a_Computer_Scientist_(Downey)/09%3A_Inheritance/9.10%3A_Data_Encapsulation | 1,722,904,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00612.warc.gz | 184,049,500 | 30,440 | Skip to main content
# 9.10: Data Encapsulation
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
Chapter 16 demonstrates a development plan we might call “object-oriented design.” We identified objects we needed—Time, Point and Rectangle—and defined classes to represent them. In each case there is an obvious correspondence between the object and some entity in the real world (or at least a mathematical world).
But sometimes it is less obvious what objects you need and how they should interact. In that case you need a different development plan. In the same way that we discovered function interfaces by encapsulation and generalization, we can discover class interfaces by data encapsulation.
Markov analysis, from Section 13.8, provides a good example. If you download my code from http://thinkpython.com/code/markov.py, you’ll see that it uses two global variables—suffix_map and prefix—that are read and written from several functions.
suffix_map = {}
prefix = ()
Because these variables are global we can only run one analysis at a time. If we read two texts, their prefixes and suffixes would be added to the same data structures (which makes for some interesting generated text).
To run multiple analyses, and keep them separate, we can encapsulate the state of each analysis in an object. Here’s what that looks like:
class Markov(object):
def __init__(self):
self.suffix_map = {}
self.prefix = ()
Next, we transform the functions into methods. For example, here’s process_word:
def process_word(self, word, order=2):
if len(self.prefix) < order:
self.prefix += (word,)
return
try:
self.suffix_map[self.prefix].append(word)
except KeyError:
# if there is no entry for this prefix, make one
self.suffix_map[self.prefix] = [word]
self.prefix = shift(self.prefix, word)
Transforming a program like this—changing the design without changing the function—is another example of refactoring (see Section 4.7).
This example suggests a development plan for designing objects and methods:
1. Start by writing functions that read and write global variables (when necessary).
2. Once you get the program working, look for associations between global variables and the functions that use them.
3. Encapsulate related variables as attributes of an object.
4. Transform the associated functions into methods of the new class.
Exercise $$\PageIndex{1}$$
Download my code from Section 13.8 (http://thinkpython.com/code/markov.py), and follow the steps described above to encapsulate the global variables as attributes of a new class called Markov.
Solution:
http://thinkpython.com/code/Markov.py (note the capital M).
This page titled 9.10: Data Encapsulation is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) .
• Was this article helpful? | 2,357 | 7,139 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.197267 |
https://netlib.org/lapack/explore-html/df/d53/zhst01_8f_a76324d4221bfb7b0de0ff21e5a4cf535.html | 1,709,469,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00469.warc.gz | 415,218,583 | 5,605 | LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
Searching...
No Matches
## ◆ zhst01()
subroutine zhst01 ( integer n, integer ilo, integer ihi, complex*16, dimension( lda, * ) a, integer lda, complex*16, dimension( ldh, * ) h, integer ldh, complex*16, dimension( ldq, * ) q, integer ldq, complex*16, dimension( lwork ) work, integer lwork, double precision, dimension( * ) rwork, double precision, dimension( 2 ) result )
ZHST01
Purpose:
``` ZHST01 tests the reduction of a general matrix A to upper Hessenberg
form: A = Q*H*Q'. Two test ratios are computed;
RESULT(1) = norm( A - Q*H*Q' ) / ( norm(A) * N * EPS )
RESULT(2) = norm( I - Q'*Q ) / ( N * EPS )
The matrix Q is assumed to be given explicitly as it would be
following ZGEHRD + ZUNGHR.
In this version, ILO and IHI are not used, but they could be used
to save some work if this is desired.```
Parameters
[in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in] ILO ` ILO is INTEGER` [in] IHI ``` IHI is INTEGER A is assumed to be upper triangular in rows and columns 1:ILO-1 and IHI+1:N, so Q differs from the identity only in rows and columns ILO+1:IHI.``` [in] A ``` A is COMPLEX*16 array, dimension (LDA,N) The original n by n matrix A.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [in] H ``` H is COMPLEX*16 array, dimension (LDH,N) The upper Hessenberg matrix H from the reduction A = Q*H*Q' as computed by ZGEHRD. H is assumed to be zero below the first subdiagonal.``` [in] LDH ``` LDH is INTEGER The leading dimension of the array H. LDH >= max(1,N).``` [in] Q ``` Q is COMPLEX*16 array, dimension (LDQ,N) The orthogonal matrix Q from the reduction A = Q*H*Q' as computed by ZGEHRD + ZUNGHR.``` [in] LDQ ``` LDQ is INTEGER The leading dimension of the array Q. LDQ >= max(1,N).``` [out] WORK ` WORK is COMPLEX*16 array, dimension (LWORK)` [in] LWORK ``` LWORK is INTEGER The length of the array WORK. LWORK >= 2*N*N.``` [out] RWORK ` RWORK is DOUBLE PRECISION array, dimension (N)` [out] RESULT ``` RESULT is DOUBLE PRECISION array, dimension (2) RESULT(1) = norm( A - Q*H*Q' ) / ( norm(A) * N * EPS ) RESULT(2) = norm( I - Q'*Q ) / ( N * EPS )```
Definition at line 138 of file zhst01.f.
140*
141* -- LAPACK test routine --
142* -- LAPACK is a software package provided by Univ. of Tennessee, --
143* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
144*
145* .. Scalar Arguments ..
146 INTEGER IHI, ILO, LDA, LDH, LDQ, LWORK, N
147* ..
148* .. Array Arguments ..
149 DOUBLE PRECISION RESULT( 2 ), RWORK( * )
150 COMPLEX*16 A( LDA, * ), H( LDH, * ), Q( LDQ, * ),
151 \$ WORK( LWORK )
152* ..
153*
154* =====================================================================
155*
156* .. Parameters ..
157 DOUBLE PRECISION ONE, ZERO
158 parameter( one = 1.0d+0, zero = 0.0d+0 )
159* ..
160* .. Local Scalars ..
161 INTEGER LDWORK
162 DOUBLE PRECISION ANORM, EPS, OVFL, SMLNUM, UNFL, WNORM
163* ..
164* .. External Functions ..
165 DOUBLE PRECISION DLAMCH, ZLANGE
166 EXTERNAL dlamch, zlange
167* ..
168* .. External Subroutines ..
169 EXTERNAL zgemm, zlacpy, zunt01
170* ..
171* .. Intrinsic Functions ..
172 INTRINSIC dcmplx, max, min
173* ..
174* .. Executable Statements ..
175*
176* Quick return if possible
177*
178 IF( n.LE.0 ) THEN
179 result( 1 ) = zero
180 result( 2 ) = zero
181 RETURN
182 END IF
183*
184 unfl = dlamch( 'Safe minimum' )
185 eps = dlamch( 'Precision' )
186 ovfl = one / unfl
187 smlnum = unfl*n / eps
188*
189* Test 1: Compute norm( A - Q*H*Q' ) / ( norm(A) * N * EPS )
190*
191* Copy A to WORK
192*
193 ldwork = max( 1, n )
194 CALL zlacpy( ' ', n, n, a, lda, work, ldwork )
195*
196* Compute Q*H
197*
198 CALL zgemm( 'No transpose', 'No transpose', n, n, n,
199 \$ dcmplx( one ), q, ldq, h, ldh, dcmplx( zero ),
200 \$ work( ldwork*n+1 ), ldwork )
201*
202* Compute A - Q*H*Q'
203*
204 CALL zgemm( 'No transpose', 'Conjugate transpose', n, n, n,
205 \$ dcmplx( -one ), work( ldwork*n+1 ), ldwork, q, ldq,
206 \$ dcmplx( one ), work, ldwork )
207*
208 anorm = max( zlange( '1', n, n, a, lda, rwork ), unfl )
209 wnorm = zlange( '1', n, n, work, ldwork, rwork )
210*
211* Note that RESULT(1) cannot overflow and is bounded by 1/(N*EPS)
212*
213 result( 1 ) = min( wnorm, anorm ) / max( smlnum, anorm*eps ) / n
214*
215* Test 2: Compute norm( I - Q'*Q ) / ( N * EPS )
216*
217 CALL zunt01( 'Columns', n, n, q, ldq, work, lwork, rwork,
218 \$ result( 2 ) )
219*
220 RETURN
221*
222* End of ZHST01
223*
subroutine zgemm(transa, transb, m, n, k, alpha, a, lda, b, ldb, beta, c, ldc)
ZGEMM
Definition zgemm.f:188
subroutine zlacpy(uplo, m, n, a, lda, b, ldb)
ZLACPY copies all or part of one two-dimensional array to another.
Definition zlacpy.f:103
double precision function dlamch(cmach)
DLAMCH
Definition dlamch.f:69
double precision function zlange(norm, m, n, a, lda, work)
ZLANGE returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value ...
Definition zlange.f:115
subroutine zunt01(rowcol, m, n, u, ldu, work, lwork, rwork, resid)
ZUNT01
Definition zunt01.f:126
Here is the call graph for this function:
Here is the caller graph for this function: | 1,732 | 5,167 | {"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-2024-10 | latest | en | 0.759263 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1568/2/ | 1,696,390,149,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00365.warc.gz | 944,917,544 | 94,891 | # Properties
Label 1568.2 Level 1568 Weight 2 Dimension 40141 Nonzero newspaces 24 Sturm bound 301056 Trace bound 13
## Defining parameters
Level: $$N$$ = $$1568 = 2^{5} \cdot 7^{2}$$ Weight: $$k$$ = $$2$$ Nonzero newspaces: $$24$$ Sturm bound: $$301056$$ Trace bound: $$13$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(\Gamma_1(1568))$$.
Total New Old
Modular forms 77184 41111 36073
Cusp forms 73345 40141 33204
Eisenstein series 3839 970 2869
## Trace form
$$40141 q - 124 q^{2} - 94 q^{3} - 124 q^{4} - 126 q^{5} - 124 q^{6} - 108 q^{7} - 220 q^{8} - 187 q^{9} + O(q^{10})$$ $$40141 q - 124 q^{2} - 94 q^{3} - 124 q^{4} - 126 q^{5} - 124 q^{6} - 108 q^{7} - 220 q^{8} - 187 q^{9} - 116 q^{10} - 94 q^{11} - 108 q^{12} - 118 q^{13} - 144 q^{14} - 162 q^{15} - 104 q^{16} - 58 q^{17} - 104 q^{18} - 94 q^{19} - 108 q^{20} - 144 q^{21} - 208 q^{22} - 86 q^{23} - 136 q^{24} - 185 q^{25} - 144 q^{26} - 58 q^{27} - 144 q^{28} - 230 q^{29} - 156 q^{30} - 74 q^{31} - 144 q^{32} - 308 q^{33} - 136 q^{34} - 108 q^{35} - 248 q^{36} - 126 q^{37} - 116 q^{38} - 34 q^{39} - 112 q^{40} - 174 q^{41} - 144 q^{42} - 134 q^{43} - 84 q^{44} - 10 q^{45} - 92 q^{46} - 18 q^{47} - 72 q^{48} - 24 q^{49} - 348 q^{50} + 22 q^{51} - 116 q^{52} + 2 q^{53} - 112 q^{54} - 42 q^{55} - 144 q^{56} - 136 q^{57} - 128 q^{58} - 30 q^{59} - 128 q^{60} - 6 q^{61} - 144 q^{62} - 48 q^{63} - 256 q^{64} - 224 q^{65} - 148 q^{66} - 62 q^{67} - 104 q^{68} + 4 q^{69} - 144 q^{70} - 174 q^{71} - 100 q^{72} - 190 q^{73} - 108 q^{74} - 46 q^{75} - 124 q^{76} - 144 q^{77} - 180 q^{78} - 90 q^{79} - 128 q^{80} - 123 q^{81} - 124 q^{82} - 42 q^{83} - 144 q^{84} - 204 q^{85} - 144 q^{86} - 2 q^{87} - 120 q^{88} - 174 q^{89} - 400 q^{90} - 150 q^{91} - 304 q^{92} - 232 q^{93} - 400 q^{94} - 106 q^{95} - 600 q^{96} - 482 q^{97} - 312 q^{98} - 360 q^{99} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(\Gamma_1(1568))$$
We only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space $$S_k^{\mathrm{new}}(N, \chi)$$ we list the newforms together with their dimension.
Label $$\chi$$ Newforms Dimension $$\chi$$ degree
1568.2.a $$\chi_{1568}(1, \cdot)$$ 1568.2.a.a 1 1
1568.2.a.b 1
1568.2.a.c 1
1568.2.a.d 1
1568.2.a.e 1
1568.2.a.f 1
1568.2.a.g 1
1568.2.a.h 1
1568.2.a.i 1
1568.2.a.j 2
1568.2.a.k 2
1568.2.a.l 2
1568.2.a.m 2
1568.2.a.n 2
1568.2.a.o 2
1568.2.a.p 2
1568.2.a.q 2
1568.2.a.r 2
1568.2.a.s 2
1568.2.a.t 2
1568.2.a.u 2
1568.2.a.v 2
1568.2.a.w 2
1568.2.a.x 4
1568.2.b $$\chi_{1568}(785, \cdot)$$ 1568.2.b.a 2 1
1568.2.b.b 2
1568.2.b.c 4
1568.2.b.d 4
1568.2.b.e 6
1568.2.b.f 6
1568.2.b.g 12
1568.2.e $$\chi_{1568}(783, \cdot)$$ 1568.2.e.a 4 1
1568.2.e.b 4
1568.2.e.c 8
1568.2.e.d 8
1568.2.e.e 12
1568.2.f $$\chi_{1568}(1567, \cdot)$$ 1568.2.f.a 8 1
1568.2.f.b 16
1568.2.f.c 16
1568.2.i $$\chi_{1568}(961, \cdot)$$ 1568.2.i.a 2 2
1568.2.i.b 2
1568.2.i.c 2
1568.2.i.d 2
1568.2.i.e 2
1568.2.i.f 2
1568.2.i.g 2
1568.2.i.h 2
1568.2.i.i 2
1568.2.i.j 2
1568.2.i.k 2
1568.2.i.l 2
1568.2.i.m 4
1568.2.i.n 4
1568.2.i.o 4
1568.2.i.p 4
1568.2.i.q 4
1568.2.i.r 4
1568.2.i.s 4
1568.2.i.t 4
1568.2.i.u 4
1568.2.i.v 4
1568.2.i.w 4
1568.2.i.x 4
1568.2.i.y 8
1568.2.j $$\chi_{1568}(391, \cdot)$$ None 0 2
1568.2.m $$\chi_{1568}(393, \cdot)$$ None 0 2
1568.2.p $$\chi_{1568}(31, \cdot)$$ 1568.2.p.a 16 2
1568.2.p.b 16
1568.2.p.c 16
1568.2.p.d 32
1568.2.q $$\chi_{1568}(815, \cdot)$$ 1568.2.q.a 4 2
1568.2.q.b 8
1568.2.q.c 8
1568.2.q.d 8
1568.2.q.e 8
1568.2.q.f 8
1568.2.q.g 12
1568.2.q.h 16
1568.2.t $$\chi_{1568}(177, \cdot)$$ 1568.2.t.a 4 2
1568.2.t.b 4
1568.2.t.c 4
1568.2.t.d 8
1568.2.t.e 8
1568.2.t.f 8
1568.2.t.g 12
1568.2.t.h 24
1568.2.u $$\chi_{1568}(225, \cdot)$$ n/a 336 6
1568.2.v $$\chi_{1568}(197, \cdot)$$ n/a 636 4
1568.2.y $$\chi_{1568}(195, \cdot)$$ n/a 624 4
1568.2.ba $$\chi_{1568}(215, \cdot)$$ None 0 4
1568.2.bb $$\chi_{1568}(361, \cdot)$$ None 0 4
1568.2.bf $$\chi_{1568}(223, \cdot)$$ n/a 336 6
1568.2.bg $$\chi_{1568}(111, \cdot)$$ n/a 324 6
1568.2.bj $$\chi_{1568}(113, \cdot)$$ n/a 324 6
1568.2.bk $$\chi_{1568}(65, \cdot)$$ n/a 672 12
1568.2.bm $$\chi_{1568}(165, \cdot)$$ n/a 1248 8
1568.2.bn $$\chi_{1568}(19, \cdot)$$ n/a 1248 8
1568.2.bp $$\chi_{1568}(57, \cdot)$$ None 0 12
1568.2.bs $$\chi_{1568}(55, \cdot)$$ None 0 12
1568.2.bt $$\chi_{1568}(81, \cdot)$$ n/a 648 12
1568.2.bw $$\chi_{1568}(47, \cdot)$$ n/a 648 12
1568.2.bx $$\chi_{1568}(159, \cdot)$$ n/a 672 12
1568.2.ca $$\chi_{1568}(27, \cdot)$$ n/a 5328 24
1568.2.cd $$\chi_{1568}(29, \cdot)$$ n/a 5328 24
1568.2.cf $$\chi_{1568}(9, \cdot)$$ None 0 24
1568.2.cg $$\chi_{1568}(87, \cdot)$$ None 0 24
1568.2.cj $$\chi_{1568}(3, \cdot)$$ n/a 10656 48
1568.2.ck $$\chi_{1568}(37, \cdot)$$ n/a 10656 48
"n/a" means that newforms for that character have not been added to the database yet
## Decomposition of $$S_{2}^{\mathrm{old}}(\Gamma_1(1568))$$ into lower level spaces
$$S_{2}^{\mathrm{old}}(\Gamma_1(1568)) \cong$$ $$S_{2}^{\mathrm{new}}(\Gamma_1(14))$$$$^{\oplus 10}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(16))$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(28))$$$$^{\oplus 8}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(32))$$$$^{\oplus 3}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(49))$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(56))$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(98))$$$$^{\oplus 5}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(112))$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(196))$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(224))$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(392))$$$$^{\oplus 3}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(784))$$$$^{\oplus 2}$$ | 2,987 | 5,746 | {"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} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.174869 |
https://mail.scipy.org/pipermail/scipy-user/2005-December/006253.html | 1,506,165,335,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00024.warc.gz | 697,286,455 | 1,796 | # [SciPy-user] unexpected matrix behavior
Travis Oliphant oliphant.travis at ieee.org
Thu Dec 15 22:13:13 CST 2005
```Alan G Isaac wrote:
>I would like to understand the behavior of matrices,
>as shown below. Is it expected? If so, what is the
>general principle?
>
>
>
Yes this is expected...
>>>>x = [[1,2],[3,4]]
>>>>y = scipy.array(x)
>>>>z = scipy.mat(x)
>>>>xz = zip(*x)
>>>>yz = zip(*y)
>>>>zz = zip(*z)
>>>>xz
>>>>
>>>>
>[(1, 3), (2, 4)]
>
>
>>>>yz
>>>>
>>>>
>[(1, 3), (2, 4)]
>
>
>>>>zz
>>>>
>>>>
>[(matrix([[1, 2]]), matrix([[3, 4]]))]
>
>
>
The general principle is that matrices are *always* two-dimensional and
when you slice a matrix you get a matrix.
Thus, the essential difference is that
z[0] is another two-dimensional object while y[0] and x[0] are both
one-dimensional objects.
Compare: len(z[0]) with len(x[0]) and len(y[0]) to see what's really
going on.
An array view of the same data is always available as z.A
So, zip(*z.A) gives
[(1, 3), (2, 4)]
-Travis
>
>_______________________________________________
>SciPy-user mailing list
>SciPy-user at scipy.net
>http://www.scipy.net/mailman/listinfo/scipy-user
>
>
``` | 370 | 1,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-39 | latest | en | 0.822594 |
http://wildaboutmath.com/2009/02/16/mmm-26-been-around-the-world/ | 1,519,082,661,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812855.47/warc/CC-MAIN-20180219231024-20180220011024-00766.warc.gz | 384,677,333 | 9,780 | # Wild About Math!Making Math fun and accessible
16Feb/090
## MMM #26 Been Around the World
Blinkdagger has MMM #26 posted. This one is tricky. We've received a number of different answers and even the incorrect ones are well explained!
So, if you want a Blinkdagger-style Math problem, give this one a try!
Here's the problem description:
Ever since Quan was a little boy, his lifelong dream has been to fly around the world in an airplane.
• Quan lives on an island wherein there are 100 airplanes, all created equally with identical characteristics.
• Each airplane has a fuel tank that contains enough fuel to fly exactly half way around the world.
• All of the airplanes travel at the same speed, and use their gas at the same rate.
• Airplanes can exchange fuel with other airplanes while in flight.
• The island is the only source of fuel.
• For the purposes of this problem, assume that there is no time lost refueling on either the air or ground.
• All airplanes must make it back to the island safely.
Question: What is the lowest number of airplanes required to allow one airplane to travel all the way around the world?
Check out the rules and submission information at Blinkdagger. | 261 | 1,203 | {"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-2018-09 | latest | en | 0.936789 |
https://www.cfd-online.com/Forums/openfoam-programming-development/84121-get-velocity-gradient-boundary-function-print.html | 1,519,000,066,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812293.35/warc/CC-MAIN-20180218232618-20180219012618-00330.warc.gz | 794,548,908 | 2,670 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- OpenFOAM Programming & Development (https://www.cfd-online.com/Forums/openfoam-programming-development/)
Chris Lucas January 20, 2011 10:08
get velocity gradient with a boundary function
Hi
I want to write a new inlet boundary condition for epsilon. For this new boundary condition, I need calculate for instance the velocity gradient ( and magSqr(symm(fvc::grad(U))) )
I know I can get the velocity field at the patch using:
const fvPatchField<vector>& U = patch().lookupPatchField<volVectorField, vector>("U");
The problem is (as I understand it) that by using the code above, U is a “vector patch field”. But if I want to get the velocity gradient (using fvc::grad(U)), U must be a volVectorField.
Is there a way to get the velocity of the entire grid within the boundary condition, so that I can calculate the velocity gradient and then use this result to get the velocity gradient at the patch?
Thanks for the help.
Christian
lindstroem August 10, 2011 05:57
Hi Chris,
did you succeed?
I also want to get the velocity gradient at a patch..
Chris Lucas August 10, 2011 08:36
Hi,
I didn't looked into this subject much more and therefore didn't solve the problem. Please let me know if you do :)
Best Regards
Christian
mathslw August 24, 2012 13:17
Quote:
Originally Posted by lindstroem (Post 319598) Hi Chris, did you succeed? I also want to get the velocity gradient at a patch.. Thanks in advance.
Hi,
I also want to get the velocity gradient at a patch.
did you succeed?
Thanks!
Wei
Chris Lucas August 27, 2012 03:11
Hi,
no, I have not continued this work.
The following is a guess only (not sure if this works or even if this is correct):
I guess you need the normal BC gradient and not the fvc::grad. | 455 | 1,812 | {"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-2018-09 | latest | en | 0.907738 |
http://www.chegg.com/homework-help/mechanics-of-materials-8th-edition-chapter-2.4-problem-7p-solution-9781285225784 | 1,455,077,600,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701158601.61/warc/CC-MAIN-20160205193918-00153-ip-10-236-182-209.ec2.internal.warc.gz | 336,586,390 | 16,295 | View more editions
# TEXTBOOK SOLUTIONS FOR Mechanics of Materials 8th Edition
• 1245 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
PROBLEM
Chapter: Problem:
A tube structure is acted on by loads at B and D, as shown in the figure. The tubes are joined using two flange plates at C, which are bolted together using six 0.5-in. diameter bolts.
(a) Derive formulas for the reactions RA and RE at the ends of the bar.
(b) Determine the axial displacements δB, δc, and δD at points B, C, and D, respectively.
(c) Draw an axial-displacement diagram (ADD) in which the abscissa is the distance x from support A to any point on the bar and the ordinate is the horizontal displacement δ at that point. (d) Find the maximum value of the load variable P if allowable normal stress in the bolts is 14 ksi.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
Mechanics of Materials | 8th Edition
9781285225784ISBN-13: 1285225783ISBN: Authors: | 284 | 1,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-07 | latest | en | 0.835141 |
https://eduhawks.com/tag/syllogism/ | 1,585,598,737,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00059.warc.gz | 453,342,252 | 13,056 | Breaking News
Home / Tag Archives: syllogism
# Tag Archives: syllogism
## “If I do not quit, then I am not giving up. If I am not giving up, then I am not failing. Therefore, If I do not quit, then I am not failing.” This is an example of which property or law? A. Law of Syllogism B. Addition Property C. Symmetric Property D. Law of Detachment
Answer: The correct options are 1, 3 and 6. Step-by-step explanation: The given function is It is parent absolute function. The graph of any absolute function is always V-shaped. Therefore the graph of f(x) is V-shaped. Option 1 is correct. If the sign before the modulus is positive, then the … | 169 | 644 | {"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} | 2.875 | 3 | CC-MAIN-2020-16 | latest | en | 0.861327 |
https://oeis.org/A194064 | 1,627,815,384,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00426.warc.gz | 431,197,002 | 4,063 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A194064 Natural interspersion of A006578; a rectangular array, by antidiagonals. 3
1, 4, 2, 8, 5, 3, 14, 9, 6, 7, 21, 15, 10, 11, 12, 30, 22, 16, 17, 18, 13, 40, 31, 23, 24, 25, 19, 20, 52, 41, 32, 33, 34, 26, 27, 28, 65, 53, 42, 43, 44, 35, 36, 37, 29, 80, 66, 54, 55, 56, 45, 46, 47, 38, 39, 96, 81, 67, 68, 69, 57, 58, 59, 48, 49, 50 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS See A194029 for definitions of natural fractal sequence and natural interspersion. Every positive integer occurs exactly once (and every pair of rows intersperse), so that as a sequence, A194064 is a permutation of the positive integers; its inverse is A194065. LINKS EXAMPLE Northwest corner: 1...4...8...14...21...30 2...5...9...15...22...31 3...6...10..16...23...32 7...11..17..24...33...43 12..18..25..34...44...56 MATHEMATICA z = 50; c[k_] := k (k + 1)/2 + Floor[(k^2)/4]; c = Table[c[k], {k, 1, z}] (* A006578 *) f[n_] := If[MemberQ[c, n], 1, 1 + f[n - 1]] f = Table[f[n], {n, 1, 400}] (* A194063 *) r[n_] := Flatten[Position[f, n]] t[n_, k_] := r[n][[k]] TableForm[Table[t[n, k], {n, 1, 7}, {k, 1, 7}]] p = Flatten[Table[t[k, n - k + 1], {n, 1, 11}, {k, 1, n}]] (* A194064 *) q[n_] := Position[p, n]; Flatten[Table[q[n], {n, 1, 90}]] (* A194065 *) CROSSREFS Cf. A194029, A194063, A194065. Sequence in context: A261830 A194038 A131819 * A194054 A191536 A187076 Adjacent sequences: A194061 A194062 A194063 * A194065 A194066 A194067 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Aug 14 2011 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 1 06:45 EDT 2021. Contains 346384 sequences. (Running on oeis4.) | 800 | 2,055 | {"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-2021-31 | latest | en | 0.669848 |
https://www.tes.com/en-us/teaching-resources/hub/middle-school/computing/computational-thinking/generalization-and-pattern-matching | 1,526,887,142,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863967.46/warc/CC-MAIN-20180521063331-20180521083331-00145.warc.gz | 848,457,861 | 30,386 | 1. Resources Home
2. Middle School
3. Computing
4. Computational thinking
5. Generalization and pattern matching
#### KS3 Computing - Cryptography Unit of Work - Lesson Presentations, Worksheets and Assessment
This is a whole unit of work that this aimed at KS3 - it covers all aspects of Cryptography that the students have absolutely loved doing and this has been uploaded to Google Classroom and made the whole unit very interactive. The work is split into Flight Paths and contains an assessment. All work has stretch and challenge extension tasks too. This is a really enjoyable unit of work for about 6 weeks.
#### OCR - 9-1 - Computer Pseudo worksheet questions
This work sheet is a number of questions including extension activities that will get students to program pseudocode on paper. The first part of the work sheet includes some basic questions as well as some answers on an answer sheet. It should be some good practice for the basics of pseudo code. The extension questinos are more complicated and should be done on a separate sheet or work book as practice for the students.
#### Teach yourself the ENTIRE AQA GCSE Computer Science specification
Detailed student workbooks that EXACTLY match the AQA GCE (9-1) Computer Science (8520) specification. Each includes theory, tasks to recap the knowledge and put their new skills into practice and the answers. Perfect for NQTs, non-specialist teachers or teachers who feel a little unsure about the qualification content and want to improve their subject knowledge. Also ideal to use as a revision aid for students. Provided as seven separate PDF documents: • Fundamentals of algorithms • Programming • Fundamentals of data representation • Computer systems • Fundamentals of computer networks • Fundamentals of cyber security • Ethical, legal and environmental impacts of digital technology on wider society, including issues of privacy These are ideal to use in the classroom, as homework tasks or for independent study as revision for the examinations.
#### Primary Computing - Understanding Pattern Recognition - Scratch Project
First, familiarise yourself with the terms Pattern Recognition and Decomposition, by watching Part 5 of “Delve in, for twelve min!” (Video linked above). Give the children the ‘Pattern Recognition’ scratch project. Children have to find different ways to reduce the amount of code that is repeated. I’ve included 3 different solutions; Broadcast (Not always the best solution, please see Part 6 of “Delve in, for twelve min!” on ‘concurrency and dependency’ ) Make a Block function (more advanced when you include number input) Using Make a block, selection and repetition (You can see how much smoother the code runs straight away) Please let me know how you get on! Cheers Phil Wickins
#### Computational Thinking Knowledge Organiser GCSE OCR Computer Science J276
My current year 11 students have been extremely positive about the Knowledge Organisers that I have created for their Computer Science revision (and students in previous years on the legacy spec). This resource covers the terminology involved in Computational Thinking along with an explanation of pseudocode and flow charts with several examples using sequence, selection and iteration. It also cover sub-programs and shows the difference between procedures and functions with examples of each. This covers the new J276 OCR specification for GCSE Computer Science. Please do leave feedback if possible, I hope you find it useful.
#### Poster: Blooms Taxonomy and Computational Thinking
Use Blooms Taxonomy thinking skills to incorporate Computational Thinking Concepts within your lessons. This poster cross references Blooms Taxonomy thinking skills with the concepts of Computational Thinking in order to support your delivery of these concepts within the classroom. Suitability ratings are also provided in order to identify how computational thinking concepts can be linked to their relative thinking skills. ***Please leave a review!
#### Cosmopolitan Coding - Fun INSET Introduction to Coding
Many teachers were never taught the coding and computer science elements of the new curriculum themselves. This introductory activity helps teachers understand the simplicity of using algorithms as instructions to complete tasks. Teachers will learn: Algorithms and how these are simple instructions Syntax and how this is simply known action in new language Problem solving and debugging code Testing code to see if it works Teachers love this INSET introduction!
#### Big O Notation - Computer Science - OCR A-Level
A Powerpoint presentation explaining Big O Notation. There are some examples of code and some graphs which help students visualise what is meant by exponential and the complexity of algorithms. This was useful to my year 13 students and hopefully it will help yours.
#### KS3 Computing Binary to ASCII challenge worksheet
This activity gets the students to work out the ASCII value once they have calculated the Denary. When they have got the word that it creates they research it and then explain what it means.
#### KS3 Computational Thinking Project - Practical Problem Solving (Full Unit of Work)
**KS3 Computational Thinking Project - Practical Problem Solving (Full Unit of Work) ** This project is designed to either introduce computational thinking skills or to supplement and build on knowledge and skills already acquired. It is suggested that the student booklet is used in conjunction with the associated presentation through all of the lessons. The project has been designed so that is hands-on and practical. The suggested time frame from start to finish of the project is approximately 6-8 lessons, depending on what additional aspects the teacher wishes to introduce during the project. Students are asked to create a stationery holder using everyday objects. The problem has to be approached and solved using computational thinking skills. The project can be approached from a totally ‘unplugged’ pedagogy. No computers required! The resource consists of: a) A student booklet. This can be printed as an A4 or A5 booklet. b) A student booklet containing examplar responses. Also contains extension / homework tasks. c) A PowerPoint presentation for use in lessons to guide students and the teacher. Many of the slides have accompanying ‘speaker notes’ with ideas and suggestions for lessons.
#### Algorithms - Learning Tasks: Classroom or Homework Resources
The resources can be used as a set of differentiated algorithm resources or as homeworks to compliment lesson activities: The resources covers aspects of computational thinking, including: Algorithms Pseudo Code Flowcharts Sequencing Decomposition Abstraction Decisions and Repetition Tasks can be completed in any order and are differentiated. The resource contains pages covering: My rewards My progress Notes Teacher feedback
#### GCSE Computer Science Revision
The exam is soon…
#### Computational Thinking Starter Quiz
The following quiz is based upon the four parts of Computational Thinking: Decomposition, Abstraction, Pattern Recognition and Algorithm Design. The quiz includes 8 questions including definitions and scenarios to test your students understanding of Computational Thinking. Ideal to use as either a starter or plenary! Give students a mini white board and marker pen for instant feedback! Also includes a Computational Thinking Recap slide prior to the quiz itself. Please leave a review!
#### Computer Science: Introduction to Programming Techniques
The aim of this resource is to introduce students to the basic concepts of programming techniques and algorithm design. This resource is aimed at absolute beginners of algorithm design (Key Stage 3) and takes you through the following parts: Understanding algebra Understanding Boolean Using Variables Introduction to Algorithm Design i. Sequence ii. Selection (IF, ELSE, ELSE IF) iii. Variables iv. Looping (WHILE and REPEAT) v. Functions My resources can be used both as presentations and as work units that students can complete independently. Each section includes activities to develop student understanding of the concept. The work finishes with a differentiated consolidation challenge, applying the concepts learned throughout the unit. Please leave a Review!
#### Computational Thinking Poster: Humanities
Download my poster which includes example task ideas on how to apply computational thinking problem solving skills to the humanities. Example tasks are included for History, Geography and PRE lessons. Computational Thinking is a skill set that can be used across the curriculum, not just within computer science! The logical approach to solving a problem means that it’s application can be used to develop and improve students ability to deal with difficult problems in a more rational way. The aim of my posters is to make you realise that many of the brilliant tasks that teachers facilitate week in, week out actually include computational thinking. It’s just a matter of making it more explicit! Update: High Resolution 2560x1440 in both PNG & JPEG Please leave a review!!!
#### GCSE Computing High and Low Level Languages Presentation
GCSE Computing High and Low Level Languages Presentation with Quiz. Great for lessons and teachers or student revision. Pack Includes Presentation, Script and Student Handouts. Based off the OCR (9-1) GCSE
#### Free Valentines Coding Lesson Plan & Resources
Computing lesson plan & associated resources for KS2 children programming cupid to fire his arrow at a love interest! Step-by-step lesson plan, support materials and pre-written Scratch program template. More free primary computing lessons and resources available at
#### Mouse Maze - KS3 Algorithm Starter
Mouse Maze is a fun introduction to decomposition, algorithms and iteration (loops). It works well as an extended starter. Print out the simple maze, or distribute the PowerPoint file to students (the mouse icon can be dragged and rotated). They must list the steps required for Mousey to reach the cheese. Decomposition: breaking the problem into smaller steps (i.e. solving the maze) Algorithms: listing the steps to guide Mousey through the maze Iteration: making the algorithm more efficient by repeating parts of it (challenge / extension activity) The download includes: A full set of teacher’s instructions PowerPoint maze for students which can be used electronically or printed out PowerPoint 6 slides showing the task, solution, sections which are repeated, and a more efficient solution. The final slide has definitions of ‘algorithm’ and ‘decomposition’, plus a note to tell students that they have already achieved both during the lesson. Pupils enjoy the activity and it is a great way to introduce the concepts for the first time. I hope you and your pupils enjoy it :)
#### The Imitation Game
A fun lesson revolving around the background of Computers in World War 2. Suitable to use when discussing computing history, ciphers and codes in WW2 and discussing code breaking and decipherment. Students take on the role of 2 Germans who have to encrypt a message and decide an algorithm. Messages are then swapped between groups as allies that attempt to break the cipher. The activity could be made a game by rewarding the most difficult to crack cipher or the team which cracks their cipher quickest. Students will need to know the basics of ciphers such as a Caesar cipher but do not need to know more difficult ciphers. It would also be useful to discuss decipherment methods for example looking for patterns such as “the” “to” or brute forcing the shift. The activity provides a good introduction to cyber security and was successfully used with BSc and MSc students, both enjoyed it however do not let the ciphers get too complex! They can be too difficult to break. Suggested topics after this could be modern cryptography or security principles or computing history such as the Manchester computer.
LOVELY RESOURCE WITHIN WHICH PUPILS HAVE TO MANUALLY WRITE OUT THE FORMULAE THAT WOULD BE USED TO GAIN THE CORRECT ANSWER FROM A SCREEN DUMP. REALLY TESTS PUPIL KNOWLEDGE AND UNDERSTANDING AND SUPERB FOR EVIDENCE OF TRACKER ASSESSMENT / LEVELS
#### GCSE Computer Science Short Test
Covers Searching and Sorting, 2’s compliment, truth table. Questions and Mark Scheme included.
#### Logical / Computational Thinking Peg Puzzle - Great starter
A wonderful starter for any lesson on Logical / computational Thinking. Print out the sheet and get students in pairs to see if they can come up with a solution The objective is to arrange the pegs (numbers 1-8) such that no consecutive number touches on any connecting line. For example number 1 cannot connect with number 2; number 2 cannot connect with 3 or 1; number 5 cannot connect with 4 or 6; etc. Answer is included in the pack
#### Computer Science: Algorithms AQA (9-1) GCSE revision and exam practice
Great for AQA (9-1) GCSE Computer Science revision and exam practice. This student revision workbook that covers ALL of the “Fundamentals of Algorithms” syllabus for AQA (9-1) GCSE Computer Science. Includes 82 pages covering the theory and giving lots of practical activities and even includes the answers Ideal as a GCSE revision aid or as a teaching resource. Perfect for NQTs, non-specialist teachers or teachers who feel a little unsure about the qualification content and want to improve their subject knowledge. Table of contents: • What are algorithms? • Decomposition • Abstraction • Basic Pseudocode • Arithmetic Operations • Relational Operators • Boolean Operators • Input and Output • Iteration • Selection • Nesting statements • Flowcharts • Trace Tables • Dealing with Arrays • Subroutines • Dealing with strings • ASCII and Unicode • Linear search • Binary Search • Compare the two search algorithms < • Bubble Sort • Merge Sort • Compare the two sort algorithms Other student workbooks in this series include: Programming Fundamentals of Data Representation GCSE revision student workbook Computer Systems GCSE revision student workbook Computer Networks GCSE revision student workbook Cyber Security GCSE revision student workbook Ethical, legal and environmental impacts GCSE workbook
#### Iterations / Loops - Coding Shapes in Scratch -FULL LESSON- KS3 Computing/Computer Science
This is a full lesson with all resources to teach students how to draw shapes in Scratch (good for KS3 and I normally use it with KS3.) All worksheets are included with Powerpoint and a quick video explanation that can be shown to students. Students have to draw simple shapes in scratch from triangles up to circles which teachers them how loops work and how to conceptualise an algorithm. Loops inside loops can also be used as extension to draw some great geometric art. Learning objectives are: Understand simple algorithm design Know how to identify important ideas (abstraction) Understand how to break a problem into manageable units (decomposition). Know what repetition is (iteration) Extension task 2 also included for more difficult shapes
#### Introduction to Computer Science Unplugged - Cup Stacking!
The aim of this lesson is to introduce students to writing their first ever code! Students will not need to use a computer! Starter - Which job would you rather do: Formula 1 Driver or a Formula 1 Mechanic? Introduction to different types of IT users: Formula 1 Driver is the expert user of software while the Mechanic is the creator of the software. Task 1: Robots and Programmers - Get your Robot to a specific location and back using the given commands. Main task: Cup stacking - use the symbols available and create the differentiated structures as shown. For the lesson you will also need: Plastic cups Poster paper Pens to write code Print out the resource packs for each group of students. I have used this lesson for students between years 5 and 9, all of which have thoroughly enjoyed it! ***Please leave a review!***
#### Computational Thinking and the Digital Competence Framework
My latest poster shows how the understanding of Computational Thinking underpins all concepts of the Digital Competence Framework (DCF). I have given examples of how each concept can be applied to primary strands of the framework. Feel free to download and use as part of CPD sessions! ***Please leave a review!***
#### Computing - HTML and CSS Web Development SOW
This 10 lesson SOW teaches students what HTML and CSS do and how the are used together to develop website content. Through the 10 lesson scheme of work, students are taken through the basics of HTML and CSS and provided with the opportunity to develop their own website. All teaching materials, HTML/CSS code samples, end of unit assessment and marksheet provided.
#### Visual Basic 6 FULL COURSE Software Development
A full unit of work utilising VB 6, Visual Basic programming language. PDF document of tasks/instructions included as well as all the starting files which students require
#### Y7 Computing Algorithms Unit
This is the unit of work I created to introduce my Y7 students to Algorithms, Problem Solving, Pseudocode and Computational Thinking. In the unit, there is the Activity Sheets which introduce: Grid References, Loops, Problem Solving, Trace Tables amongst others. The students trace code as it executes, keeping track of automated buses as they drive. Variables are brought in to track fuel, as well as booleans etc. This unit has been incredibly successful at boosting students understanding of developing and writing algorithms and when we move onto our units in Scratch coding / Python coding, they much better understand the terminiology All in, this can be 'rushed' in 6 lessons, or taking more time to delve into the Computing Acts can take it to 8 and beyond. With new GCSE and IGCSE courses covering some of the material, it is also something that could easily be expanded for students to undertake some of their own investigations into some relevant areas in AI etc.
#### KS2 / KS3 Scratch superunit - introduction and follow-up unit
This bundle contains two units of work for Scratch, including at least 14 lessons in total (some may take longer, depending on programming speed). It can be broken down into two units, or potentially completed in one go. I have completed these with the same students over two years - the introductory unit in Y5 and then the second unit in Y6. However, this could be used with older children with little Scratch experience. The lessons conatined are as follows: Unit 1 Introduction to Scratch - algorithms Joke - sequencing Disco - sequencing / iteration Etch-a-sketch - controlling sprites Maze - Selection / iteration Assessment Unit 2 Flowcharts - algorithms Chatbot - sequencing / selection Scrolling maze - variables Falling fruit - indefinite iteration / variables Breakout - Variables / subroutines Rock, Paper, Scissors - Definite iteration / subroutines Assessment
#### KS2 / KS3 Scratch intermediate unit
This unit is designed to be used after the initial Scratch unit, and contains lessons on flowcharts, sequencing, selection, iteration, variables and subroutines. The games that will be made through this unit include: A chatbot A scrolling maze Falling Fruit game Breakout Rock, paper, scissors Following the flowcharts lesson, the students will complete the other lessons to create the above programs, and then finish with an assessment. The assessment says Y6 at the top as this was the year group I have used it with, but that is because my Y5s completed the introductory unit and so moved onto this in Y6. However, this could be used for KS2 or KS3, depending on programming and Scratch experience.
#### KS2 / KS3 Scratch Assessment
This assessment is similar but more advanced than the introductory Scratch unit assessment in my shop, and includes questions on subroutines and variables as well as algorithms, sequencing, selection and iteration. This is aimed at students who have completed two units of Scratch.
#### KS2 / KS3 Rock paper scissors tutorial - subroutines
This lesson is for students who have completed the introductory unit of Scratch, or have other experience in Scratch or other programming languages. The lesson focuses on creating a game using subroutines, and includes a presentation which introduces the task, the complete code for demonstration purposes and starter code for the students. There is also a tutorial for the students to complete the creation of the game following the demonstration.
#### KS2 / KS3 Scratch Breakout Game - subroutines
This resource is for students who have experience in Scratch or other programming languages, including those who have completed the introductory unit of Scratch. The resource contains a powerpoint presentation to introduce subroutines, and a tutorial for students to create the game. It also includes the full game for demonstration and debugging purposes.
#### KS2 / KS3 Scratch Falling Fruit Game
This lesson is based around using indefinite iteration in Scratch, through creating a game where the player catches randomly falling objects. This resource includes the full game code, a presentation to introduce indefinite iteration, and a tutorial for students to create the game.
#### KS2 / KS3 Scratch Scrolling Maze Tutorial
This resource is for students who have completed an introductory unit of Scratch or who are older and more experienced in programming. The lesson is based around creating a scrolling maze, which is more advanced than a normal maze game, and introduces variables. It includes a tutorial, Powerpoint to introduce variables, the full game code and starter code for the students to begin programming with.
#### KS2 / KS3 Scratch chatbot lesson
This resource is for students who have already completed an introductory unit of Scratch, or are older students who might be more familiar with programming. They will create a program on Scratch using sequencing and selection.
#### [GCSE/IGCSE] Introduction to Algorithms
I use this unit of work to introduce students to creating algorithms. We look at some classic AI problems which are a fun challenge to solve, as well as board games. This introduces students to flow charts and pseudocode, abstraction, decomposition whilst improving their Problem Solving and Logical Reasoning skills. Definitely can see an improvement in my current Y10 class who were introduced to this Unit in Y9
#### GCSE Computer Science Paper 2 Mock
Made up a mock exam based on the new Computer Science curriculum and a combination of the exemplar papers which have been released. Perfect for homework/revision, or for editing and using as a mock exam/prelim exam practise. Also suitable for the IGCSE curriculum (may want to edit out any database questions)
#### Full Set GCSE Computing Revision Flash Cards!
GCSE Computing Revision Flash Cards for Students. This bundle has money off, as apposed to buying the Flashcards separately. £25 instead of over £30! 20% Off compared to if you bought them separately. Based off of the OCR Specification (9-1) | 4,503 | 23,143 | {"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-2018-22 | longest | en | 0.932992 |
https://www.cis.rit.edu/htbooks/mri/chap-5/chap-5.htm | 1,709,616,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00304.warc.gz | 697,312,307 | 6,285 | # The Basics of MRI
## FOURIER TRANSFORMS
### Introduction
A detailed description of the Fourier transform ( FT ) has waited until now, when you have a better appreciation of why it is needed. A Fourier transform is an operation which converts functions from time to frequency domains. An inverse Fourier transform ( IFT ) converts from the frequency domain to the time domain.
The concept of a Fourier transform is not that difficult to understand. Recall from Chapter 2 that the Fourier transform is a mathematical technique for converting time domain data to frequency domain data, and vice versa. You may have never thought about this, but the human brain is capable of performing a Fourier transform. Consider the following sine wave and note.
A musician with perfect pitch will tell us that this is middle C (261.63 Hz) on the western music scale. This musician will be able to also tell us that this sine wave is the first G above middle C (392 Hz),
and that this sine wave this note is a C one octave above middle C (523.25 Hz).
Some can tell the notes when more than one are played simultaneously, but this process becomes more difficult when more notes are played simultaneously. Play all of the above notes simultaneously. Can you hear which frequencies are simultaneously being played? The Fourier transform can! Change the relative amplitudes of the notes. Can you determine their relative amplitudes with your ear? The Fourier transform can!
The Fourier transform ( FT ) process is like the musician hearing a tone (time domain signal) and determining what note (frequency) is being played. The inverse Fourier Transform ( IFT ) is like the musician seeing notes (frequencies) on a sheet of music and converting them to tones (time domain signals).
### The + and - Frequency Problem
To begin our detailed description of the FT consider the following. A magnetization vector, starting at +x, is rotating about the Z axis in a clockwise direction. The plot of Mx as a function of time is a cosine wave. Fourier transforming this gives peaks at both +ν and -ν because the FT can not distinguish between a +ν and a -ν rotation of the vector from the data supplied.
A plot of My as a function of time is a -sine function. Fourier transforming this gives peaks at +ν and -ν because the FT can not distinguish between a positive vector rotating at +ν and a negative vector rotating at -ν from the data supplied.
The solution is to input both the Mx and My into the FT. The FT is designed to handle two orthogonal input functions called the real and imaginary components.
Detecting just the Mx or My component for input into the FT is called linear detection. This was the detection scheme on many older NMR spectrometers and some magnetic resonance imagers. It required the computer to discard half of the frequency domain data.
Detection of both Mx and My is called quadrature detection and is the method of detection on modern spectrometers and imagers. It is the method of choice since now the FT can distinguish between +ν and -&nu, and all of the frequency domain data be used.
### The Fourier Transform
An FT is defined by the integral
Think of f(ω) as the overlap of f(t) with a wave of frequency ω.
This is easy to picture by looking at the real part of f(ω) only.
Consider the function of time, f( t ) = cos( 4t ) + cos( 9t ).
To understand the FT, examine the product of f(t) with cos(ωt) for &omega values between 1 and 10, and then the summation of the values of this product between 1 and 10 seconds. The summation will only be examined for time values between 0 and 10 seconds.
ω=1 ω=2 ω=3 ω=4 ω=5 ω=6 ω=7 ω=8 ω=9 ω=10 f(ω)
The inverse Fourier transform (IFT) is best depicted as an summation of the time domain spectra of frequencies in f(ω).
### Phase Correction
The actual FT will make use of an input consisting of a REAL and an IMAGINARY part. You can think of Mx as the REAL input, and My as the IMAGINARY input. The resultant output of the FT will therefore have a REAL and an IMAGINARY component, too.
Consider the following function:
f(t) = e-at e-i2πνt
In FT NMR spectroscopy, the real output of the FT is taken as the frequency domain spectrum. To see an esthetically pleasing (absorption) frequency domain spectrum, we want to input a cosine function into the real part and a sine function into the imaginary parts of the FT. This is what happens if the cosine part is input as the imaginary and the sine as the real.
In an ideal NMR experiment all frequency components contained in the recorded FID have no phase shift. In practice, during a real NMR experiment a phase correction must be applied to either the time or frequency domain spectra to obtain an absorption spectrum as the real output of the FT. This process is equivalent to the coordinate transformation described in Chapter 2
If the above mentioned FID is recorded such that there is a 45° phase shift in the real and imaginary FIDs, the coordinate transformation matrix can be used with = - 45°>. The corrected FIDs look like a cosine function in the real and a sine in the imaginary.
Fourier transforming the phase corrected FIDs gives an absorption spectrum for the real output of the FT.
The phase shift also varies with frequency, so the NMR spectra require both constant and linear corrections to the phasing of the Fourier transformed signal.
= m ν + b
Constant phase corrections, b, arise from the inability of the spectrometer to detect the exact Mx and My. Linear phase corrections, m, arise from the inability of the spectrometer to detect transverse magnetization starting immediately after the RF pulse. The following drawing depicts the greater loss of phase in a high frequency FID when the initial yellow section is lost. From the practical point of view, the phase correction is applied in the frequency domain rather then in the time domain because we know that a real frequency domain spectrum should be composed of all positive peaks. We can therefore adjust b and m until all positive peaks are seen in the real output of the Fourier transform.
In magnetic resonance imaging, the Mx or My signals are rarely displayed. Instead a magnitude signal is used. The magnitude signal is equal to the square root of the sum of the squares of Mx and My.
### Fourier Pairs
To better understand how FT NMR functions, you need to know some common Fourier pairs. A Fourier pair is two functions, the frequency domain form and the corresponding time domain form. Here are a few Fourier pairs which are useful in MRI. The amplitude of the Fourier pairs has been neglected since it is not relevant in MRI.
Constant value at all time
Real: cos(2πνt), Imaginary: -sin(2πνt)
Comb Function (A series of delta functions separated by T.)
Exponential Decay: e-at for t > 0.
A square pulse starting at 0 that is T seconds long.
Gaussian: exp(-at2)
### Convolution Theorem
To the magnetic resonance scientist, the most important theorem concerning Fourier transforms is the convolution theorem. The convolution theorem says that the FT of a convolution of two functions is proportional to the products of the individual Fourier transforms, and vice versa.
If f(ω) = FT( f(t) ) and g(ω) = FT( g(t) )
then f(ω) g(ω) = FT( g(t) ⊗ f(t) ) and f(ω) ⊗ g(ω) = FT( g(t) f(t) )
It will be easier to see this with pictures. In the animation window we are trying to find the FT of a sine wave which is turned on and off. The convolution theorem tells us that this is a sinc function at the frequency of the sine wave.
Another application of the convolution theorem is in noise reduction. With the convolution theorem it can be seen that the convolution of an NMR spectrum with a Lorentzian function is the same as the Fourier transform of multiplying the time domain signal by an exponentially decaying function.
### The Digital FT
In a nuclear magnetic resonance spectrometer, the computer does not see a continuous FID, but rather an FID which is sampled at a constant interval. Each data point making up the FID will have discrete amplitude and time values. Therefore, the computer needs to take the FT of a series of delta functions which vary in intensity.
What is the FT of a signal represented by this series of delta functions? The answer will be addressed in the next heading, but first some information on relationships between the sampled time domain data and the resultant frequency domain spectrum. An n point time domain spectrum is sampled at δt and takes a time t to record. The corresponding complex frequency domain spectrum that the discrete FT produces has n points, a width f, and resolution δf. The relationships between the quantities are as follows.
f = (1/δt)
δf = (1/t)
### Sampling Error
The wrap around problem or artifact in a magnetic resonance image is the appearance of one side of the imaged object on the opposite side. In terms of a one dimensional frequency domain spectrum, wrap around is the occurrence of a low frequency peak on the wrong side of the spectrum.
The convolution theorem can explain why this problem results from sampling the transverse magnetization at too slow a rate. First, observe what the FT of a correctly sampled FID looks like. With quadrature detection, the image width is equal to the inverse of the sampling frequency, or the width of the green box in the animation window.
When the sampling frequency is less than the spectral width or bandwidth, wrap around occurs.
### The Two-Dimensional FT
The two-dimensional Fourier transform (2-DFT) is an FT performed on a two dimensional array of data.
Consider the two-dimensional array of data depicted in the animation window. This data has a t' and a t" dimension. A FT is first performed on the data in one dimension and then in the second. The first set of Fourier transforms are performed in the t' dimension to yield an ν' by t" set of data. The second set of Fourier transforms is performed in the t" dimension to yield an ν' by ν" set of data.
The 2-DFT is required to perform state-of-the-art MRI. In MRI, data is collected in the equivalent of the t' and t" dimensions, called k-space. This raw data is Fourier transformed to yield the image which is the equivalent of the ν' by ν" data described above.
### Problems
1. What is the Fourier transform of a 63 MHz sine wave?
2. What is the Fourier transform of a 50 µs long square pulse?
3. What is the Fourier transform of a 63 MHz sine wave which is turned on for only 50 µs?
4. What is the width of the function in the answer to question 3 in Hz and in ppm when this signal is greater than 90% of its maximum value? Why is this result significant in NMR and MRI?
5. What quadrature sampling rate is necessary to accurately record the FID from a 2000 Hz wide NMR spectrum?
6. What is the Fourier transform of the quadrature signal from a 63 MHz sine wave? (Assume there are equal real 63 MHz cosine and imaginary 63 MHz sine components.)
7. What is the Fourier transform of a 33 µs long square pulse?
8. What is the Fourier transform of the 63 MHz sine wave of Question 1 which is turned on for only 33 µs?
9. What is the width of this function in Hz and in ppm when this signal is greater than 90% of its maximum value? Why is this result significant in NMR and MRI?
10. What quadrature sampling rate is necessary to accurately record the FID from a 500 Hz wide NMR spectrum? | 2,517 | 11,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-10 | latest | en | 0.937145 |
http://betterlesson.com/lesson/resource/2462410/student-work-area-docx | 1,487,631,481,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170613.8/warc/CC-MAIN-20170219104610-00137-ip-10-171-10-108.ec2.internal.warc.gz | 25,630,326 | 21,029 | ## student work area.docx - Section 2: Finding Area of Triangles
student work area.docx
# Area of Triangles
Unit 6: Solving Problems Involving Triangles
Lesson 10 of 13
## Big Idea: Why does (1/2)*a*b*sin(C) give the area of a triangle?
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, Trigonometry, triangles (Determining Measurements), area of triangle, problem solving, pre
45 minutes
### Katharine Sparks
##### Similar Lessons
###### The Law of Sines: More than Meets the Eye
Big Idea: Using the Law of Sines will throw your students for a loop - there are two possible answers!
Favorites(1)
Resources(10)
Troy, MI
Environment: Suburban
###### Final Exam Review Stations (Day 1 of 3)
Big Idea: Students review by working through various stations at their own pace and receive immediate feedback on their work.
Favorites(5)
Resources(23)
Phoenix, AZ
Environment: Urban
###### SAS Area Formula
Geometry » Right Triangles and Trigonometry
Big Idea: Batteries not included...but the angle must be. Some assembly required. In this lesson students build a new triangle area formula and learn when and how to apply it.
Favorites(2)
Resources(10) | 287 | 1,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-09 | latest | en | 0.859265 |
https://wiki.math.ucr.edu/index.php/007A_Sample_Midterm_3,_Problem_1 | 1,639,055,176,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00100.warc.gz | 658,671,076 | 5,846 | # 007A Sample Midterm 3, Problem 1
(a) If ${\displaystyle \lim _{x\rightarrow 3}{\bigg (}{\frac {f(x)}{2x}}+1{\bigg )}=2,}$ find ${\displaystyle \lim _{x\rightarrow 3}f(x).}$
(b) Evaluate ${\displaystyle \lim _{x\rightarrow 2}{\frac {2-x}{x^{2}-4}}.}$
(c) Find ${\displaystyle \lim _{x\rightarrow 0}{\frac {\tan(4x)}{\sin(6x)}}.}$ | 147 | 336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "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-2021-49 | latest | en | 0.197624 |
https://www.coursehero.com/file/6050725/BackExam-1/ | 1,498,615,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00456.warc.gz | 834,739,553 | 131,367 | # BackExam_1 - Name Section Strength of Materials Test#1...
This preview shows pages 1–4. Sign up to view the full content.
Name: ______________________________________ Section: __________ Strength of Materials Test #1 Problem Value Score 1 25 2 25 3 25 4 25 Total: 1. Write your name and section number on all pages. 2. State all your assumptions and present your work in an organized and legible fashion. Neatness Counts, points will be deducted if your work is not neat and properly organized.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Name: ______________________________________ Section: __________ Problem 1: The rigid bar AB shown in Fig. 1 is supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross-sectional area of 1800 mm 2 . The 18 mm - diameter pins at A and C are subjected to single shear. If the allowable stress for the steel and aluminum is (_ st ) all = 340 MPa and (_ AL ) all = 35 MPa, respectively, and the allowable shear stress for each pin is _ all = 450 MPa, determine the largest load P that can be applied to the bar. (Disregard the bearing stress near pins A & C).
Name: ______________________________________ Section: __________
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 12/14/2010 for the course ENGR 2530 taught by Professor Lindaschadlerfeist during the Fall '08 term at Rensselaer Polytechnic Institute.
### Page1 / 9
BackExam_1 - Name Section Strength of Materials Test#1...
This preview shows document pages 1 - 4. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 429 | 1,820 | {"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-2017-26 | longest | en | 0.856082 |
https://peeterjoot.com/tag/curl/ | 1,720,878,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00845.warc.gz | 385,940,630 | 40,299 | ## Motivation.
I was asked about geometric algebra equivalents for a couple identities found in [1], one for line integrals
\label{eqn:more_feynmans_trick:20}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx =
\int_{C(t)} \lr{
\PD{t}{\Bf} + \spacegrad \lr{ \Bv \cdot \Bf } – \Bv \cross \lr{ \spacegrad \cross \Bf }
}
\cdot d\Bx,
and one for area integrals
\label{eqn:more_feynmans_trick:40}
\ddt{} \int_{S(t)} \Bf \cdot d\BA =
\int_{S(t)} \lr{
\PD{t}{\Bf} + \Bv \lr{ \spacegrad \cdot \Bf } – \spacegrad \cross \lr{ \Bv \cross \Bf }
}
\cdot d\BA.
Both of these look questionable at first glance, because neither has boundary term. However, they can be transformed with Stokes theorem to
\label{eqn:more_feynmans_trick:60}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
=
\int_{C(t)} \lr{
\PD{t}{\Bf} – \Bv \cross \lr{ \spacegrad \cross \Bf }
}
\cdot d\Bx
+
\evalbar{\Bv \cdot \Bf }{\Delta C},
and
\label{eqn:more_feynmans_trick:80}
\ddt{} \int_{S(t)} \Bf \cdot d\BA =
\int_{S(t)} \lr{
\PD{t}{\Bf} + \Bv \lr{ \spacegrad \cdot \Bf }
}
\cdot d\BA
\oint_{\partial S(t)} \lr{ \Bv \cross \Bf } \cdot d\Bx.
The area integral derivative is now seen to be a variation of one of the special cases of the Leibniz integral rule, see for example [2]. The author admits that the line integral relationship is not well used, and doesn’t show up in the wikipedia page.
My end goal will be to evaluate the derivative of a general multivector line integral
\label{eqn:more_feynmans_trick:100}
\ddt{} \int_{C(t)} F d\Bx G,
and area integral
\label{eqn:more_feynmans_trick:120}
\ddt{} \int_{S(t)} F d^2\Bx G.
We’ve derived that line integral result in a different fashion previously, but it’s interesting to see a different approach. Perhaps this approach will lend itself nicely to non-scalar integrands?
## Definition 1.1: Convective derivative.
The convective derivative,
of $$\phi(t, \Bx(t))$$ is defined as
\begin{equation*}
\frac{D \phi}{D t} = \lim_{\Delta t \rightarrow 0} \frac{ \phi(t + \Delta t, \Bx + \Delta t \Bv) – \phi(t, \Bx)}{\Delta t},
\end{equation*}
where $$\Bv = d\Bx/dt$$.
## Theorem 1.1: Convective derivative.
The convective derivative operator may be written
\begin{equation*}
\frac{D}{D t} = \PD{t}{} + \Bv \cdot \spacegrad.
\end{equation*}
### Start proof:
Let’s write
\label{eqn:more_feynmans_trick:140}
\begin{aligned}
v_0 &= 1 \\
u_0 &= t + v_0 h \\
u_k &= x_k + v_k h, k \in [1,3] \\
\end{aligned}
The limit, if it exists, must equal the sum of the individual limits
\label{eqn:more_feynmans_trick:160}
\frac{D \phi}{D t} = \sum_{\alpha = 0}^3 \lim_{\Delta t \rightarrow 0} \frac{ \phi(u_\alpha + v_\alpha h) – \phi(t, Bx)}{h},
but that is just a sum of derivitives, which can be evaluated by chain rule
\label{eqn:more_feynmans_trick:180}
\begin{aligned}
\frac{D \phi}{D t}
&= \sum_{\alpha = 0}^{3} \evalbar{ \PD{u_\alpha}{\phi(u_\alpha)} \PD{h}{u_\alpha} }{h = 0} \\
&= \PD{t}{\phi} + \sum_{k = 1}^3 v_k \PD{x_k}{\phi} \\
&= \lr{ \PD{t}{} + \Bv \cdot \spacegrad } \phi.
\end{aligned}
## Definition 1.2: Hestenes overdot notation.
We may use a dot or a tick with a derivative operator, to designate the scope of that operator, allowing it to operate bidirectionally, or in a restricted fashion, holding specific multivector elements constant. This is called the Hestenes overdot notation.Illustrating by example, with multivectors $$F, G$$, and allowing the gradient to act bidirectionally, we have
\begin{equation*}
\begin{aligned}
&=
+
&=
\sum_i \lr{ \partial_i F } \Be_i G + \sum_i F \Be_i \lr{ \partial_i G }.
\end{aligned}
\end{equation*}
The last step is a precise statement of the meaning of the overdot notation, showing that we hold the position of the vector elements of the gradient constant, while the (scalar) partials are allowed to commute, acting on the designated elements.
We will need one additional identity
## Lemma 1.1: Gradient of dot product (one constant vector.)
Given vectors $$\Ba, \Bb$$ the gradient of their dot product is given by
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
= \lr{ \Bb \cdot \spacegrad } \Ba – \Bb \cdot \lr{ \spacegrad \wedge \Ba }
+ \lr{ \Ba \cdot \spacegrad } \Bb – \Ba \cdot \lr{ \spacegrad \wedge \Bb }.
\end{equation*}
If $$\Bb$$ is constant, this reduces to
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }
= \lr{ \Bb \cdot \spacegrad } \Ba – \Bb \cdot \lr{ \spacegrad \wedge \Ba }.
\end{equation*}
### Start proof:
The $$\Bb$$ constant case is trivial to prove. We use $$\Ba \cdot \lr{ \Bb \wedge \Bc } = \lr{ \Ba \cdot \Bb} \Bc – \Bb \lr{ \Ba \cdot \Bc }$$, and simply expand the vector, curl dot product
\label{eqn:more_feynmans_trick:200}
\Bb \cdot \lr{ \spacegrad \wedge \Ba }
=
\Bb \cdot \lr{ \dot{\spacegrad} \wedge \dot{\Ba} }
= \lr{ \Bb \cdot \dot{\spacegrad} } \dot{\Ba} – \dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }.
Rearrangement proves that $$\Bb$$ constant identity. The more general statement follows from a chain rule evaluation of the gradient, holding each vector constant in turn
\label{eqn:more_feynmans_trick:320}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }
+
\dot{\spacegrad} \lr{ \dot{\Bb} \cdot \Ba }.
## Time derivative of a line integral of a vector field.
We now have all our tools assembled, and can proceed to evaluate the derivative of the line integral. We want to show that
## Theorem 1.2:
Given a path parameterized by $$\Bx(\lambda)$$, where $$d\Bx = (\PDi{\lambda}{\Bx}) d\lambda$$, with points along a $$C(t)$$ moving through space at a velocity $$\Bv(\Bx(\lambda))$$, and a vector function $$\Bf = \Bf(t, \Bx(\lambda))$$,
\begin{equation*}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx =
\int_{C(t)} \lr{
\PD{t}{\Bf} + \spacegrad \lr{ \Bf \cdot \Bv } + \Bv \cdot \lr{ \spacegrad \wedge \Bf}
} \cdot d\Bx
\end{equation*}
### Start proof:
I’m going to avoid thinking about the rigorous details, like any requirements for curve continuity and smoothness. We will however, specify that the end points are given by $$[\lambda_1, \lambda_2]$$. Expanding out the parameterization, we seek to evaluate
\label{eqn:more_feynmans_trick:240}
\int_{C(t)} \Bf \cdot d\Bx
=
\int_{\lambda_1}^{\lambda_2} \Bf(t, \Bx(\lambda) ) \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda.
The parametric form nicely moves all the boundary time dependence into the integrand, allowing us to write
\label{eqn:more_feynmans_trick:260}
\begin{aligned}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
&=
\lim_{\Delta t \rightarrow 0}
\inv{\Delta t}
\int_{\lambda_1}^{\lambda_2}
\lr{ \Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) ) \cdot \frac{\partial}{\partial \lambda} \lr{ \Bx + \Delta t \Bv(\Bx(\lambda)) } – \Bf(t, \Bx(\lambda)) \cdot \frac{\partial \Bx}{\partial \lambda} } d\lambda \\
&=
\lim_{\Delta t \rightarrow 0}
\inv{\Delta t}
\int_{\lambda_1}^{\lambda_2}
\lr{ \Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) ) – \Bf(t, \Bx)} \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda \\
\lim_{\Delta t \rightarrow 0}
\int_{\lambda_1}^{\lambda_2}
\Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) )) \cdot \PD{\lambda}{}\Bv(\Bx(\lambda)) d\lambda \\
&=
\int_{\lambda_1}^{\lambda_2}
\frac{D \Bf}{Dt} \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda +
\lim_{\Delta t \rightarrow 0}
\int_{\lambda_1}^{\lambda_2}
\Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) \cdot \frac{\partial}{\partial \lambda} \Bv(\Bx(\lambda)) d\lambda \\
&=
\int_{\lambda_1}^{\lambda_2}
\lr{ \PD{t}{\Bf} + \lr{ \Bv \cdot \spacegrad } \Bf } \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda
+
\int_{\lambda_1}^{\lambda_2}
\Bf \cdot \frac{\partial \Bv}{\partial \lambda} d\lambda
\end{aligned}
At this point, we have a $$d\Bx$$ in the first integrand, and a $$d\Bv$$ in the second. We can expand the second integrand, evaluating the derivative using chain rule to find
\label{eqn:more_feynmans_trick:280}
\begin{aligned}
\Bf \cdot \PD{\lambda}{\Bv}
&=
\sum_i \Bf \cdot \PD{x_i}{\Bv} \PD{\lambda}{x_i} \\
&=
\sum_{i,j} f_j \PD{x_i}{v_j} \PD{\lambda}{x_i} \\
&=
\sum_{j} f_j \lr{ \spacegrad v_j } \cdot \PD{\lambda}{\Bx} \\
&=
\sum_{j} \lr{ \dot{\spacegrad} f_j \dot{v_j} } \cdot \PD{\lambda}{\Bx} \\
&=
\dot{\spacegrad} \lr{ \Bf \cdot \dot{\Bv} } \cdot \PD{\lambda}{\Bx}.
\end{aligned}
Substitution gives
\label{eqn:more_feynmans_trick:300}
\begin{aligned}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf} + \lr{ \Bv \cdot \spacegrad } \Bf + \dot{\spacegrad} \lr{ \Bf \cdot \dot{\Bv} } } \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda \\
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf}
+ \spacegrad \lr{ \Bf \cdot \Bv }
+ \lr{ \Bv \cdot \spacegrad } \Bf
– \dot{\spacegrad} \lr{ \dot{\Bf} \cdot \Bv }
} \cdot d\Bx \\
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf}
+ \spacegrad \lr{ \Bf \cdot \Bv }
+ \Bv \cdot \lr{ \spacegrad \wedge \Bf }
} \cdot d\Bx,
\end{aligned}
where the last simplification utilizes lemma 1.1.
### End proof.
Since $$\Ba \cdot \lr{ \Bb \wedge \Bc } = -\Ba \cross \lr{ \Bb \cross \Bc }$$, observe that we have also recovered \ref{eqn:more_feynmans_trick:20}.
## Time derivative of a line integral of a bivector field.
For a bivector line integral, we have
## Theorem 1.3:
Given a path parameterized by $$\Bx(\lambda)$$, where $$d\Bx = (\PDi{\lambda}{\Bx}) d\lambda$$, with points along a $$C(t)$$ moving through space at a velocity $$\Bv(\Bx(\lambda))$$, and a bivector function $$B = B(t, \Bx(\lambda))$$,
\begin{equation*}
\ddt{} \int_{C(t)} B \cdot d\Bx =
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ d\Bx \cdot \spacegrad } \lr{ B \cdot \Bv } + \lr{ \lr{ \Bv \wedge d\Bx } \cdot \spacegrad } \cdot B.
\end{equation*}
### Start proof:
Skipping the steps that follow our previous proceedure exactly, we have
\label{eqn:more_feynmans_trick:340}
\ddt{} \int_{C(t)} B \cdot d\Bx =
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot d\Bv.
Since
\label{eqn:more_feynmans_trick:360}
\begin{aligned}
B \cdot d\Bv
&= B \cdot \PD{\lambda}{\Bv} d\lambda \\
&= B \cdot \PD{x_i}{\Bv} \PD{\lambda}{x_i} d\lambda \\
&= B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv },
\end{aligned}
we have
\label{eqn:more_feynmans_trick:380}
\ddt{} \int_{C(t)} B \cdot d\Bx
=
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv } \\
Let’s reduce the two last terms in this integrand
\label{eqn:more_feynmans_trick:400}
\begin{aligned}
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv }
&=
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx –
\lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \dot{\Bv} \cdot B } \\
&=
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx
– \lr{ d\Bx \cdot \spacegrad} \lr{ \Bv \cdot B }
+ \lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \Bv \cdot \dot{B} } \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \Bv \cdot \dot{\spacegrad} } \dot{B} \cdot d\Bx
+ \lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \Bv \cdot \dot{B} } \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \Bv \lr{ d\Bx \cdot \spacegrad } – d\Bx \lr{ \Bv \cdot \spacegrad } } \cdot B \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \lr{ \Bv \wedge d\Bx } \cdot \spacegrad } \cdot B.
\end{aligned}
Back substitution finishes the job.
## Theorem 1.4: Time derivative of multivector line integral.
Given a path parameterized by $$\Bx(\lambda)$$, where $$d\Bx = (\PDi{\lambda}{\Bx}) d\lambda$$, with points along a $$C(t)$$ moving through space at a velocity $$\Bv(\Bx(\lambda))$$, and multivector functions $$M = M(t, \Bx(\lambda)), N = N(t, \Bx(\lambda))$$,
\begin{equation*}
\ddt{} \int_{C(t)} M d\Bx N =
\int_{C(t)}
\frac{D}{D t} M d\Bx N + M \lr{ \lr{ d\Bx \cdot \dot{\spacegrad} } \dot{\Bv} } N.
\end{equation*}
It is useful to write this out explicitly for clarity
\label{eqn:more_feynmans_trick:420}
\ddt{} \int_{C(t)} M d\Bx N =
\int_{C(t)}
\PD{t}{M} d\Bx N + M d\Bx \PD{t}{N}
+ \dot{M} \lr{ \Bv \cdot \dot{\spacegrad} } N
+ M \lr{ \Bv \cdot \dot{\spacegrad} } \dot{N}
+ M \lr{ \lr{ d\Bx \cdot \dot{\spacegrad} } \dot{\Bv} } N.
Proof is left to the reader, but follows the patterns above.
It’s not obvious whether there is a nice way to reduce this, as we did for the scalar valued line integral of a vector function, and the vector valued line integral of a bivector function. In particular, our vector and bivector results had $$\spacegrad \lr{ \Bf \cdot \Bv }$$, and $$\spacegrad \lr{ B \cdot \Bv }$$ terms respectively, which allows for the boundary term to be evaluated using Stokes’ theorem. Is such a manipulation possible here?
# References
[1] Nicholas Kemmer. Vector Analysis: A physicist’s guide to the mathematics of fields in three dimensions. CUP Archive, 1977.
[2] Wikipedia contributors. Leibniz integral rule — Wikipedia, the free encyclopedia. https://en.wikipedia.org/w/index.php?title=Leibniz_integral_rule&oldid=1223666713, 2024. [Online; accessed 22-May-2024].
## Motivation.
This revisits my last blog post where I covered this content in a meandering fashion. This is an attempt to re-express this in a more compact form. In particular, in a form that is amenable to include in my book. When I wrote the potential section of my book, I cheated, and didn’t try to motivate the results. My cheat was figuring out the multivector potential representation starting with STA where things are simpler, and then translating it back to a multivector representation, instead of figuring out a reasonable way to motivate things from the foundation already laid.
I’d like to eventually have a less rushed treatment of potentials in my book, where the results are not pulled out of a magic hat. Here is an attempted step in that direction. I’ve opted to put some of the motivational material in problems (with solutions at the chapter end.)
## Multivector potentials.
We know from conventional electromagnetism (given no fictitious magnetic sources) that we can represent the six components of the electric and magnetic fields in terms of four scalar fields
\label{eqn:mvpotentials:80}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} \\
\BH &= \inv{\mu} \spacegrad \cross \BA.
\end{aligned}
The conventional way of constructing these potentials makes use of the identities
\label{eqn:mvpotentials:60}
\begin{aligned}
\end{aligned}
applying those to the source free Maxwell’s equations to find representations of $$\BE, \BH$$ that automatically satisfy those equations. For that conventional analysis, see section 18-6 [2] (available online), or section 10.1 [3], or section 6.4 [4]. We can also find such a potential representation using geometric algebra methods that are cross product free (problem 1.)
For Maxwell’s equations with fictitious magnetic sources, it can be shown that a potential representation of the field
\label{eqn:mvpotentials:100}
\begin{aligned}
\BH &= -\spacegrad \phi_m – \PD{t}{\BF} \\
\BE &= -\inv{\epsilon} \spacegrad \cross \BF.
\end{aligned}
satisfies the source-free grades of Maxwell’s equation.
See [1], and [5] for such derivations. As with the conventional source potentials, we can also apply our geometric algebra toolbox to easily find these results (problem 2.)
We have a mix of time partials and curls that is reminiscent of Maxwell’s equation itself. It’s obvious to wonder whether there is a more coherent integrated form for the potential. This is in fact the case.
## Lemma 1.1: Multivector potentials.
For Maxwell’s equation with electric sources, the total field $$F$$ can be expressed in multivector potential form
\label{eqn:mvpotentials:520}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
For Maxwell’s equation with only fictitious magnetic sources, the total field $$F$$ can be expressed in multivector form
\label{eqn:mvpotentials:540}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } I \eta \lr{ -\phi_m + c \BF } }{1,2}.
The reader should try to verify this themselves (problem 3.)
Using superposition, we can form a multivector potential that includes all grades.
## Definition 1.1: Multivector potential.
We call $$A$$, a multivector with all grades, the multivector potential, defining the total field as
\label{eqn:mvpotentials:600}
\begin{aligned}
F
&=
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
\end{aligned}
Imposition of the constraint
\label{eqn:mvpotentials:680}
is called the Lorentz gauge condition, and allows us to express $$F$$ in terms of the potential without any grade selection filters.
## Lemma 1.2: Conventional multivector potential.
Let
\label{eqn:mvpotentials:620}
A = -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }.
This results in the conventional potential representation of the electric and magnetic fields
\label{eqn:mvpotentials:640}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} – \inv{\epsilon} \spacegrad \cross \BF \\
\end{aligned}
In terms of potentials, the Lorentz gauge condition \ref{eqn:mvpotentials:680} takes the form
\label{eqn:mvpotentials:660}
\begin{aligned}
0 &= \inv{c} \PD{t}{\phi} + \spacegrad \cdot (c \BA) \\
0 &= \inv{c} \PD{t}{\phi_m} + \spacegrad \cdot (c \BF).
\end{aligned}
See problem 4.
## Problem 1: Potentials for no-fictitious sources.
Starting with Maxwell’s equation with only conventional electric sources
\label{eqn:mvpotentials:120}
Show that this may be split by grade into three equations
\label{eqn:mvpotentials:140}
\begin{aligned}
\spacegrad \wedge \BE + \inv{c}\PD{t}{} \lr{ I \eta \BH } &= 0 \\
\spacegrad \wedge \lr{ I \eta \BH } &= 0.
\end{aligned}
Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BA = 0$$, for vector $$\BA$$ and $$\spacegrad \wedge \spacegrad \phi = 0$$, for scalar $$\phi$$ to find the potential representation.
Taking grade(0,1) and (2,3) selections of Maxwell’s equation, we split our equations into source dependent and source free equations
\label{eqn:mvpotentials:200}
\label{eqn:mvpotentials:220}
In terms of $$F = \BE + I \eta \BH$$, the source free equation expands to
\label{eqn:mvpotentials:240}
\begin{aligned}
0
&=
\lr{ \spacegrad + \inv{c} \PD{t}{} } \lr{ \BE + I \eta \BH }
}{2,3} \\
&=
&=
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \inv{c} \PD{t}{\BH},
\end{aligned}
which can be further split into a bivector and trivector equation
\label{eqn:mvpotentials:260}
0 = \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH}
\label{eqn:mvpotentials:280}
0 = \spacegrad \wedge \lr{ I \eta \BH }.
It’s clear that we want to write the magnetic field as a (bivector) curl, so we let
\label{eqn:mvpotentials:300}
I \eta \BH = I c \BB = c \spacegrad \wedge \BA,
or
\label{eqn:mvpotentials:301}
\BH = \inv{\mu} \spacegrad \cross \BA.
\Cref{eqn:mvpotentials:260} is reduced to
\label{eqn:mvpotentials:320}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH} \\
&= \spacegrad \wedge \BE + \inv{c} \PD{t}{} \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}
We can now let
\label{eqn:mvpotentials:340}
\BE + \PD{t}{\BA} = -\spacegrad \phi.
We sneakily adjust the sign of the gradient so that the result matches the conventional representation.
## Problem 2: Potentials for fictitious sources.
Starting with Maxwell’s equation with only fictitious magnetic sources
\label{eqn:mvpotentials:160}
show that this may be split by grade into three equations
\label{eqn:mvpotentials:180}
\begin{aligned}
-\eta \spacegrad \wedge \BH + \inv{c}\PD{t}{(I \BE)} &= 0 \\
\spacegrad \wedge \lr{ I \BE } &= 0.
\end{aligned}
Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BF = 0$$, for vector $$\BF$$ and $$\spacegrad \wedge \spacegrad \phi_m = 0$$, for scalar $$\phi_m$$ to find the potential representation \ref{eqn:mvpotentials:100}.
We multiply \ref{eqn:mvpotentials:160} by $$I$$ to find
\label{eqn:mvpotentials:360}
which can be split into
\label{eqn:mvpotentials:380}
\begin{aligned}
\end{aligned}
We expand the source free equation in terms of $$I F = I \BE – \eta \BH$$, to find
\label{eqn:mvpotentials:400}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } \lr{ I \BE – \eta \BH } }{0,3} \\
&= \spacegrad \wedge \lr{ I \BE } + \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH,
\end{aligned}
which has the respective bivector and trivector grades
\label{eqn:mvpotentials:420}
0 = \spacegrad \wedge \lr{ I \BE }
\label{eqn:mvpotentials:440}
0 = \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH.
We can clearly satisfy \ref{eqn:mvpotentials:420} by setting
\label{eqn:mvpotentials:460}
I \BE = -\inv{\epsilon} \spacegrad \wedge \BF,
or
\label{eqn:mvpotentials:461}
\BE = -\inv{\epsilon} \spacegrad \cross \BF.
Here, once again, the sneaky inclusion of a constant factor $$-1/\epsilon$$ is to make the result match the conventional. Inserting this value for $$I \BE$$ into our bivector equation yields
\label{eqn:mvpotentials:480}
\begin{aligned}
0
&= -\inv{\epsilon} \inv{c} \PD{t}{} (\spacegrad \wedge \BF) – \eta \spacegrad \wedge \BH \\
&= -\eta \spacegrad \wedge \lr{ \PD{t}{\BF} + \BH },
\end{aligned}
so we set
\label{eqn:mvpotentials:500}
\PD{t}{\BF} + \BH = -\spacegrad \phi_m,
and have a field representation that automatically satisfies the source free equations.
## Problem 3: Total field in terms of potentials.
Prove lemma 1.1, either by direct expansion, or by trying to discover the multivector form of the field by construction.
Proof by expansion is straightforward, and left to the reader. We form the respective total electromagnetic fields $$F = \BE + I \eta H$$ for each case.
We find
\label{eqn:mvpotentials:560}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\spacegrad \phi – \PD{t}{\BA} + I \frac{\eta}{\mu} \spacegrad \cross \BA \\
&= -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad \wedge (c\BA) \\
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – \inv{c} \PD{t}{(c \BA)} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}
For the field for the fictitious source case, we compute the result in the same way, inserting a no-op grade selection to allow us to simplify, finding
\label{eqn:mvpotentials:580}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\inv{\epsilon} \spacegrad \cross \BF + I \eta \lr{ -\spacegrad \phi_m – \PD{t}{\BF} } \\
&= \inv{\epsilon c} I \lr{ \spacegrad \wedge (c \BF)} + I \eta \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } \\
&= I \eta \lr{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } } \\
&= I \eta \gpgrade{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } }{1,2} \\
&= I \eta \gpgrade{ \spacegrad (-\phi_m + c \BF) – \inv{c} \PD{t}{(c \BF)} }{1,2} \\
&= I \eta \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } (-\phi_m + c \BF) }{1,2}.
\end{aligned}
## Problem 4: Fields in terms of potentials.
Prove lemma 1.2.
Let’s expand and then group by grade
\label{eqn:mvpotentials:n}
\begin{aligned}
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }} \\
&=
-\inv{c} \PD{t}{\phi} + c \inv{c} \PD{t}{ \BA } + I \eta \lr{ -\inv{c} \PD{t}{\phi_m} + c \inv{c} \PD{t}{\BF} } \\
&=
+ I \eta c \spacegrad \wedge \BF
– c \inv{c} \PD{t}{\BA}
– c I \eta \inv{c} \PD{t}{\BF} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} } \\
&=
– \PD{t}{\BA}
– \PD{t}{\BF}
} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} }.
\end{aligned}
Observing that $$F = \gpgrade{ \lr{ \spacegrad -(1/c) \partial_t } A }{1,2} = \BE + I \eta \BH$$, completes the problem. If the Lorentz gauge condition is assumed, the scalar and pseudoscalar components above are obliterated, leaving just
$$F = \lr{ \spacegrad -(1/c) \partial_t } A$$.
# References
[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.
[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.
[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.
[4] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.
[5] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.
## Conventional formulation.
The idea behind introducing the scalar potential $$\phi$$ and vector potential $$\BA$$ is that we can impose a constraint on the form of our observable fields $$\BE, \BB$$, (or $$\BD, \BH$$), that reduces the complexity and coupling of Maxwell’s equations. These potentials are not unique, but the types of allowed variations in those potentials (gauge transformations) do not change the observable fields.
The basic idea is that we are looking for representations of the fields that automatically satisfy the pair of source free Maxwell’s equations
\label{eqn:gapotentials:40}
\begin{aligned}
\spacegrad \cdot \BB &= 0 \\
c \partial_0 \BB + \spacegrad \cross \BE &= 0,
\end{aligned}
so that the problem is reduced to solving just the remaining source dependent Maxwell’s equations.
The conventional way of constructing these potentials makes use of the identities
\label{eqn:gapotentials:60}
\begin{aligned}
\end{aligned}
where $$\Bf$$ is a vector, and $$\chi$$ is a scalar. This approach is straightforward. Instead of replicating it, here are a few well known references where such a treatment can be found
1. section 18-6 potentials and the wave equation in [2] (available online),
2. section 10.1 The potential formulation in [3], and
3. section 6.4 Vector and Scalar Potentials, in [4],
## Multivector potentials in geometric algebra.
The multivector form of Maxwell’s equation is
\label{eqn:gapotentials:820}
\lr{ \spacegrad + \partial_0 } F = J,
where $$\partial_0 = (1/c)\partial/\partial t$$, the electromagnetic field $$F = \BE + I c \BB = \BE + I \eta H$$ has grades(1,2), and a multivector charge and current density $$J$$. Grades(0,1) of the current are the charge and current densities respectively, and if desired, the grade(2,3) portion of the current has the fictitious magnetic charge and current densities (used in microwave and antenna engineering.)
It’s best to consider the case of electric sources, separately from the case of (fictitious) magnetic sources, and then use superposition to construct a potential representation that includes both.
We require a tool, that generalizes the $$\mathbb{R}^3$$ cross product curl identities above.
## Lemma 1.1: Curl of curl.
Let $$A \in \bigwedge^k$$ be a blade of grade $$k$$. Then
\begin{equation*}
\nabla \wedge \nabla \wedge A = 0.
\end{equation*}
Observe that for scalar $$A$$, this reduces to
\label{eqn:gapotentials:1740}
\nabla \wedge \nabla A = 0.
We’ve recently proved this, so we won’t do it again now.
Now we are ready to figure out the structure of the potentials.
### Case I. No (fictitious) magnetic sources.
Without magnetic sources, Maxwell’s equation is
\label{eqn:gapotentials:840}
This can be split into two equations, one that has just the sources, and one that is source free
\label{eqn:gapotentials:860}
\label{eqn:gapotentials:880}
If you are clever, or have the benefit of having worked out the answer already, you can look directly at \ref{eqn:gapotentials:880} and guess the multivector form for the potential. Hint: you want something closely related to $$F = \lr{ \spacegrad – \partial_0 } A$$, where $$A$$ has grades(0,1).
If you aren’t that clever, or don’t have a time machine that let’s you look that clever, you’ll have to work it out systematically like the rest of us. We can start by breaking down $$F$$ into it’s constituent observer dependent fields. That means that we want to find values for $$\BE, \BH$$ that satisfy
\label{eqn:gapotentials:900}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3} = 0.
Expanding the multivector factors gives us
\label{eqn:gapotentials:920}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3}
&=
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \partial_0 \BH.
\end{aligned}
Splitting this into one equation for each grade, leaves us with
\label{eqn:gapotentials:940}
0 = \spacegrad \wedge \BE + I \eta \partial_0 \BH
\label{eqn:gapotentials:960}
0 = \spacegrad \wedge \lr{ I \eta \BH }.
Observe that we could have also written \ref{eqn:gapotentials:960} as $$0 = I \eta \lr{ \spacegrad \cdot \BH }$$, which is the starting point of the conventional non-GA approach.
It’s clear that we want to write $$I \eta \BH = I c \BB$$ as a (bivector) curl, and let
\label{eqn:gapotentials:980}
I \eta \BH = c \spacegrad \wedge \BA.
It’s a bit sneaky to toss that factor of $$c$$ in here, but that’s done to make the units of $$\BA$$ turn out in a way that matches the conventional vector potential. If it makes you feel better, you can think of this as an undetermined constant multiplicative undetermined factor that will be used to adjust the dimensions of $$\BA$$ down the line.
Having made that choice, \ref{eqn:gapotentials:960} is automatically satisfied, and \ref{eqn:gapotentials:940} is reduced to
\label{eqn:gapotentials:1000}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \partial_0 \BH \\
&= \spacegrad \wedge \BE + \partial_0 \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + c \partial_0 \BA }.
\end{aligned}
We can now let
\label{eqn:gapotentials:1020}
\BE + \partial_0 c \BA = -\spacegrad \phi.
Again, we had the option of including an arbitrary multiplicative constant, but this time, we managed to find the right switch for our time machine, and look ahead to see that we want that constant to be $$-1$$ in order to have agreement with the conventional result.
We are left with a potential construction for our individual field components
\label{eqn:gapotentials:1040}
\begin{aligned}
\BE &= -\spacegrad \phi – c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA,
\end{aligned}
or
\label{eqn:gapotentials:1060}
F = -\spacegrad \phi – c \partial_0 \BA + c \spacegrad \wedge \BA.
This automatically satisfies the grades of Maxwell’s equation that are source free, leaving us to solve just
\label{eqn:gapotentials:1080}
### Multivector potential.
It’s natural to wonder if there is a more structured form for $$F$$ than \ref{eqn:gapotentials:1060}, just as we found a GA structure for Maxwell’s equation that eliminated the crazy mix of divs and curls that we had in the original Gibbs representation. Let’s find that structure. To do so, we can enclose $$F$$ in a no-op grade selection operation
\label{eqn:gapotentials:1100}
\begin{aligned}
F
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – c \partial_0 \BA + \lr{ \partial_0 \phi – \partial_0 \phi } }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}
We can now introduce a multivector potential, and express the remaining non-zero grades of Maxwell’s equation in terms of this potential
\label{eqn:gapotentials:1120}
\begin{aligned}
A &= -\phi + c \BA \\
\end{aligned}
### Lorentz gauge.
The grade selection in our representation of $$F$$ is a bit annoying, and can be eliminated if we impose additional constraints on the potential. We can write
\label{eqn:gapotentials:1140}
F =
\lr{ \spacegrad – \partial_0 } A
and then ask what conditions are required for this grade(0,3) selection to be zero. In terms of our constituent potentials, that is
\label{eqn:gapotentials:1160}
\begin{aligned}
0 &=
&=
\gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{0,3} \\
&=
c \spacegrad \cdot \BA + \partial_0 \phi,
\end{aligned}
This is the Lorentz gauge condition, recognized a bit more easily if written out in terms of the time partials explicitly
\label{eqn:gapotentials:1180}
\inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA = 0.
We can now write Maxwell’s equations, in the potential formulation, as
\label{eqn:gapotentials:1200}
\begin{aligned}
A &= -\phi + c \BA \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA \\
\end{aligned}
This is quite nice. We have a one to one decoupled relationship between the potential and the current, and are free to use the well known techniques for solving the wave equation (using convolution and a superposition of advanced and retarded Green’s functions for the wave equation operator.)
### Gauge transformation.
There’s one more thing that we should look at before moving on to the magnetic sources case, and that’s the question of gauge freedom. We’ve said that the potentials are not unique, but this non-uniqueness has a very specific form.
Since we’ve constructed $$F$$ with a grade selection as
\label{eqn:gapotentials:1220}
so it’s clear that any transformation
\label{eqn:gapotentials:1240}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},
where $$\psi_{0,3}$$ is any multivector with grades(0,3) components, will leave $$F$$ invariant. That is
\label{eqn:gapotentials:1260}
\begin{aligned}
A &= -\phi + c \BA \\
&\rightarrow
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \bar{\psi} } \\
&=
\lr{ -\phi + c \partial_0 \psi }
+ c \lr{ \BA + \spacegrad \psi }
+ I \partial_0 \bar{\psi}.
\end{aligned}
We see that the contributions of $$\bar{\psi}$$ result in grade(2,3) terms, which are not of interest, and we find that a paired transformation of the potentials
\label{eqn:gapotentials:1280}
\begin{aligned}
\phi &\rightarrow \phi – \PD{t}{\psi} \\
\BA &\rightarrow \BA + \spacegrad \psi,
\end{aligned}
called a gauge transformation, leaves the field $$F$$ unchanged. This can be expressed slightly more compactly as
\label{eqn:gapotentials:1300}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } c \psi,
where, once again, the multiplicative constant $$c$$ is included so for consistency with the conventional expression for potential gauge transformation.
### Case II. With (fictitious) magnetic sources.
With magnetic sources, Maxwell’s equation is
\label{eqn:gapotentials:1500}
We put this in dual form
\label{eqn:gapotentials:1520}
which now has the sources all with grades (0,1) as we just analyzed. The dual vector $$I F$$, like $$F$$, has only grade(1,2) components.
Expanding the source free Maxwell’s equations in terms of $$\BE, \BH$$, we have
\label{eqn:gapotentials:1340}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{I \BE – \eta \BH } }{2,3} \\
&= \gpgrade{ I \spacegrad \BE – \eta \spacegrad \BH + I \partial_0 \BE – \eta \partial_0 \BH }{2,3} \\
&= \spacegrad \wedge \lr{ I \BE } – \eta \spacegrad \wedge \BH + I \partial_0 \BE,
\end{aligned}
\label{eqn:gapotentials:1360}
0 = \spacegrad \wedge \lr{ I \BE },
\label{eqn:gapotentials:1361}
0 = – \eta \spacegrad \wedge \BH + I \partial_0 \BE.
We see that the dual electric field needs to be a curl to satisfy \ref{eqn:gapotentials:1360}
\label{eqn:gapotentials:1400}
I \BE = -\eta \spacegrad \wedge c \BF,
and after substitution into \ref{eqn:gapotentials:1361} we are left with
\label{eqn:gapotentials:1540}
\begin{aligned}
0
&= – \eta \spacegrad \wedge \BH + \partial_0 \lr{ – \eta \spacegrad \wedge c \BF } \\
&= \eta \spacegrad \wedge \lr{ -\BH – \partial_0 c \BF } \\
\end{aligned}
We set
\label{eqn:gapotentials:1420}
-\BH – \partial_0 c \BF = \spacegrad \phi_m,
Our fields are
\label{eqn:gapotentials:1440}
\begin{aligned}
\BE &= – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= -\spacegrad \phi_m – \PD{t}{\BF}.
\end{aligned}
This has the structure that matches the potential conventions from antenna theory, for example as stated in [1].
### Multivector potential.
As with the electrical sources, we expect that we can write this as something like
\label{eqn:gapotentials:1460}
Let’s verify that this is the case.
\label{eqn:gapotentials:1480}
\begin{aligned}
F
&= I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF \\
&= \gpgrade{ I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \spacegrad c \BF -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \lr{ \spacegrad \lr{ – \phi_m + c \BF } – \partial_0 c \BF + \partial_0 \phi_m – \partial_0 \phi_m} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF } }{1,2}.
\end{aligned}
### Lorentz gauge.
Let’s see what constraints we need to write our field in terms of a potential without a grade selection, that is
\label{eqn:gapotentials:1560}
F = \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }.
We need the grade(0,3) components of this multivector to be zero. Those components are
\label{eqn:gapotentials:1580}
\begin{aligned}
0 &=
\gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }}{0,3} \\
&=
\gpgrade{-\spacegrad I \eta \phi_m+\spacegrad I \eta c \BF+ \partial_0 I \eta \phi_m – \partial_0 I \eta c \BF }{0,3} \\
&=
+ \partial_0 I \eta \phi_m \\
&=
I \eta \lr{ c \lr{ \spacegrad \cdot \BF} + \partial_0 \phi_m },
\end{aligned}
or
\label{eqn:gapotentials:1600}
0 = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF.
This is the Lorentz gauge condition. With this condition we can we can express Maxwell’s equation with magnetic sources, as a forced wave equation
\label{eqn:gapotentials:1620}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF \\
\end{aligned}
### Gauge transformation.
Without the Lorentz gauge assumption, our potential representation for the field is
\label{eqn:gapotentials:1640}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
\end{aligned}
It’s clear that any transformation of the form
\label{eqn:gapotentials:1660}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},
leaves the field unchanged.
\label{eqn:gapotentials:1680}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
&\rightarrow
I \eta \lr{ -\phi + c \BF } + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
I \eta \lr{ -\phi_m + c \BF } + \lr{ \spacegrad + \partial_0 } \lr{ \psi + I \eta c \bar{\psi} } \\
&=
I \eta \lr{
-\phi_m
+ c \partial_0 \bar{\psi}
+ c \BF
}
+ \lr{ \spacegrad + \partial_0 } \psi.
\end{aligned}
We can drop the $$\psi$$ contributions, since this time we want only grades(2,3) in our potential, and find that the
desired form of the gauge transformation, for scalar $$\bar{\psi}$$, is
\label{eqn:gapotentials:1700}
\begin{aligned}
\phi_m &\rightarrow \phi_m – \PD{t}{\bar{\psi}} \\
\BF &\rightarrow \BF + \spacegrad \bar{\psi}.
\end{aligned}
The multivector form of this is
\label{eqn:gapotentials:1720}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } I \eta c \bar{\psi}.
### Superposition.
We can now use superposition to construct a potential representation that works for both conventional electric and fictitious magnetic charges and currents.
Without a Lorentz gauge assumption, that is
\label{eqn:gapotentials:1760}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
J &= \lr{ \spacegrad + \partial_0 } F,
\end{aligned}
where, given scalar functions $$\psi, \bar{\psi}$$, we are free to make gauge transformations of the multivector potential that satisfy
\label{eqn:gapotentials:1800}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \eta c \bar{\psi} },
With a Lorentz gauge constraint, we have a wave equation operator acting on $$A$$, with the multivector current as a forcing term.
\label{eqn:gapotentials:1780}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
F &= \lr{ \spacegrad – \partial_0 } A \\
J &= \lr{ \spacegrad^2 – \partial_{00} } A.
\end{aligned}
### Check.
It’s worth expansion to verify that we got all the dimensional constants write, and compare the results to Maxwell’s equations in their Gibbs form.
Let’s start with an expansion of $$F$$ in terms of the potentials
\label{eqn:gapotentials:1820}
\begin{aligned}
F &=
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ c \BA + I \eta c \BF } }{1,2} \\
&=
-\partial_0 \lr{ c \BA + I \eta c \BF }.
\end{aligned}
That is
\label{eqn:gapotentials:1840}
\begin{aligned}
\BE &= -\spacegrad \phi + I \eta c \spacegrad \wedge \BF -c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA – I \eta \spacegrad \phi_m – I \eta c \partial_0 \BF,
\end{aligned}
or
\label{eqn:gapotentials:1860}
\begin{aligned}
\BE &= – \spacegrad \phi -\partial_t \BA – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= – \spacegrad \phi_m – \partial_t \BF + \inv{\mu} \spacegrad \cross \BA.
\end{aligned}
All is good. This is exactly the form that we expect.
Let’s expand out Maxwell’s equation in terms of this potential representation and see what we get.
Let’s write the total field without the grade(1,2) selection, by subtracting off any grade(0,3) contributions
\label{eqn:gapotentials:1880}
F = \lr{ \spacegrad – \partial_0 } A – \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3}.
That difference term is
\label{eqn:gapotentials:1900}
\begin{aligned}
&=
– \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA – I \eta \phi_m + I \eta c \BF } }{0,3} \\
&=
– c \spacegrad \cdot \BA – I \eta c \spacegrad \cdot \BF – \partial_0 \phi – I \eta \partial_0 \phi_m.
\end{aligned}
The field is nicely split into a multivector term that depends directly on the full multivector potential $$A$$, and a difference term that wipes out any scalar and pseudoscalar terms
\label{eqn:gapotentials:1920}
F
=
\lr{ \spacegrad – \partial_0 } A
– \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } – I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }.
Maxwell’s equations are now reduced to
\label{eqn:gapotentials:1940}
\lr{ \spacegrad^2 – \partial_{00} } A
\lr{ \partial_0 \phi + c \spacegrad \cdot \BA }
I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }
= J.
This splits nicely into a single equation for each grade of $$A, J$$ respectively. We write
\label{eqn:gapotentials:1960}
J = \eta\lr{ c \rho – \BJ } + I \lr{ c \phi_m – \BM },
so
\label{eqn:gapotentials:1980}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } (-\phi) – \partial_0 \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= \eta c \rho \\
\lr{ \spacegrad^2 – \partial_{00} } (c \BA) – \spacegrad \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= -\eta \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } (I \eta c \BF) – I \eta \partial_0 \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= -I \BM \\
\lr{ \spacegrad^2 – \partial_{00} } (-I \eta \phi_m) – I \eta \spacegrad \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= I c \rho_m.
\end{aligned}
If we choose the Lorentz gauge conditions
\label{eqn:gapotentials:2000}
0 = \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } = \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF },
all of these equations decouple nicely, leaving us with 8 (scalar) equations in 8 unknowns
\label{eqn:gapotentials:2020}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } \phi &= -\frac{\rho}{\epsilon} \\
\lr{ \spacegrad^2 – \partial_{00} } \BA &= -\mu \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } \BF &= -\epsilon \BM \\
\lr{ \spacegrad^2 – \partial_{00} } \phi_m &= – \frac{\rho_m}{\mu}.
\end{aligned}
## Potentials in STA (space time algebra).
All of this was very convoluted. Maxwell’s equation in STA form is considerably simpler, as is the potential formulation.
### STA form of Maxwell’s equation.
We identify
\label{eqn:gapotentials:2040}
\begin{aligned}
\Be_k &= \gamma_k \gamma_0 \\
I &= \Be_1 \Be_2 \Be_3 = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \\
\gamma^\mu \cdot \gamma_\nu &= {\delta^\mu}_\nu.
\end{aligned}
Our field multivector
\label{eqn:gapotentials:2060}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= \gamma_{k0} E^k + \eta \gamma_{0123k0} H^k \\
&= \gamma_{k0} E^k + \eta \gamma_{123k} H^k,
\end{aligned}
now has a pure bivector representation in STA (since $$k$$ will always clobber one of the $$1,2,3$$ indexes.) To find the STA representation of Maxwell’s equation, we simply multiply both sides of our multivector representation, from the left, by $$\gamma_0$$.
\label{eqn:gapotentials:2080}
\gamma_0 \lr{ \spacegrad + \partial_0 } F = \gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }.
The LHS is just the spacetime gradient of $$F$$, which we can see by expanding the product
\label{eqn:gapotentials:2100}
\begin{aligned}
\gamma_0 \lr{ \spacegrad + \partial_0 }
&=
\gamma_0 \lr{ \gamma_{k0} \PD{x^k}{} + \PD{x^0}{} } \\
&=
-\gamma_{k} \PD{x^k}{} + \gamma_0 \PD{x^0}{}.
\end{aligned}
\label{eqn:gapotentials:2120}
\grad \equiv \gamma^k \PD{x^k}{} + \gamma^0 \PD{x^0}{} = \gamma^\mu \partial_\mu.
Our RHS is
\label{eqn:gapotentials:2140}
\begin{aligned}
\gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }
&=
\gamma_0 \frac{\rho}{\epsilon} – \gamma_{0k0} \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 – \gamma_{0k0} (\BM \cdot \Be_k) } \\
&=
\gamma_0 \frac{\rho}{\epsilon} + \gamma_k \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 + \gamma_{k} (\BM \cdot \Be_k) }.
\end{aligned}
If we let
\label{eqn:gapotentials:2160}
\begin{aligned}
J_e^0 &= \frac{\rho}{\epsilon} \\
J_e^k &= \eta (\BJ \cdot \Be_k) \\
J_m^0 &= c \rho_m \\
J_m^k &= (\BM \cdot \Be_k) \\
J_e &= J_e^\mu \gamma_\mu \\
J_m &= J_m^\mu \gamma_\mu,
\end{aligned}
then we are left with
\label{eqn:gapotentials:2180}
\grad F = J_e – I J_m,
or just
\label{eqn:gapotentials:2640}
where we now give a different meaning to $$J$$ than we had in the multivector formulation. This $$J$$ is now a multivector with grade(1,3) components.
### Case I: potential formulation for conventional sources.
Much like we did with to find the potential formulation for the multivector form of Maxwell’s equation, we use superposition, and tackle the conventional sources, and fictitious magnetic sources separately.
With no fictitious sources, Maxwell’s equation is
\label{eqn:gapotentials:2200}
which we may split into vector and trivector components
\label{eqn:gapotentials:2220}
\begin{aligned}
\grad \cdot F &= J_e \\
\end{aligned}
Clearly, the trivector equation can be satified by setting
\label{eqn:gapotentials:2240}
for some vector $$A$$. We may also make gauge transformations of $$A$$ of the form
\label{eqn:gapotentials:2260}
A \rightarrow A + \grad \psi,
without changing $$F$$, showing that $$A$$ is not uniquely determined. With such a representation, Maxwell’s equation is now reduced to
\label{eqn:gapotentials:2280}
or
\label{eqn:gapotentials:2300}
\begin{aligned}
J_e
&=
&=
\end{aligned}
Clearly the equivalent of the Lorentz gauge condition is now just
\label{eqn:gapotentials:2320}
so the Lorentz gauge potential form of Maxwell’s equation is just
\label{eqn:gapotentials:n}S
### Case II: potential formulation for fictitious sources.
If we have only fictious sources, Maxwell’s equation is
\label{eqn:gapotentials:2340}
or after left multiplication by $$I$$ we have
\label{eqn:gapotentials:2360}
Let $$G = I F$$, for the dual field, which is still a bivector. As before, we can split Maxwell’s equations into vector and trivector compoents
\label{eqn:gapotentials:2380}
\begin{aligned}
\grad \cdot G &= J_m \\
\end{aligned}
We may set
\label{eqn:gapotentials:2400}
for vector $$K$$. Maxwell’s equation is now reduced to
\label{eqn:gapotentials:2420}
or
\label{eqn:gapotentials:2440}
\begin{aligned}
J_m
&=
&=
\end{aligned}
As before we may make gauge transformations by adding gradient to our potential
\label{eqn:gapotentials:2460}
K \rightarrow K + \grad \bar{\psi},
which will not change $$G$$. For such sources, the Lorentz gauge condition is $$\grad \cdot K = 0$$. With the Lorentz gauge, Maxwell’s equation is reduced to
\label{eqn:gapotentials:2480}
### Superposition.
For non-fictious sources, we have
\label{eqn:gapotentials:2500}
and for fictious sources, we have
\label{eqn:gapotentials:2520}
I F = G = \grad \wedge K,
or
\label{eqn:gapotentials:2540}
F = -I G = -I \lr{ \grad \wedge K }.
Combining these results, we have
\label{eqn:gapotentials:2560}
\begin{aligned}
F
\end{aligned}
or
\label{eqn:gapotentials:2580}
F = \grad \lr{ A + I K } – \gpgrade{ \grad \lr{ A + I K } }{0,4}.
Maxwell’s equation is
\label{eqn:gapotentials:2600}
With the Lorentz gauge, this splits nicely into one forced wave equation for each vector potential
\label{eqn:gapotentials:2620}
\begin{aligned}
\end{aligned}
# References
[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.
[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.
[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.
[4] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.
## Hodge star vs. pseudoscalar multiplication.
We find a definition of the hodge star for basic k-forms in [2].
## Definition 1.7: Hodge star.
Let $$\omega$$ be a basic k-form on $$\mathbb{R}^n$$. The hodge star of $$\omega$$, denoted by $${*} \omega$$ is the unique $$n-k$$-form with the property
\begin{equation*}
\omega \wedge {*} \omega = dx_1 \wedge \cdots \wedge dx_n.
\end{equation*}
I find it interesting that this duality definition is completely free of any notion of metric or inner product. That isn’t the case with the hodge star definition from [3]. This is certainly an easier definition to understand.
Let’s calculate all the duals for the basic forms from $$\mathbb{R}^3$$. We let $$I = dx_1 \wedge dx_2 \wedge dx_3$$, and then by inspection find all the duals satisfying
\label{eqn:formAndCurl:1110}
\begin{aligned}
I &= 1 \wedge {*} 1 \\
I &= dx \wedge {*} dx \\
I &= dy \wedge {*} dy \\
I &= dz \wedge {*} dz \\
I &= (dx dy) \wedge {*} (dx dy) \\
I &= (dy dz) \wedge {*} (dy dz) \\
I &= (dz dx) \wedge {*} (dz dx) \\
I &= dx dy dz \wedge {*} (dx dy dz).
\end{aligned}
Those are
\label{eqn:formAndCurl:1130}
\begin{aligned}
{*} 1 &= dx dy dz \\
{*} dx &= dy dz \\
{*} dy &= dz dx \\
{*} dz &= dx dy \\
{*} (dx dy) &= dz \\
{*} (dy dz) &= dx \\
{*} (dz dx) &= dy \\
{*} (dx dy dz) &= 1.
\end{aligned}
Now let’s compare this to multiplication of the $$\mathbb{R}^3$$ basis vectors with the pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$. We have
\label{eqn:formAndCurl:1140}
\begin{aligned}
1 I &= I \\
\Be_1 I &= \Be_{1123} = \Be_{23} \\
\Be_2 I &= \Be_{2123} = \Be_{31} \\
\Be_3 I &= \Be_{3123} = \Be_{12} \\
\Be_{23} I &= \Be_{23123} = – \Be_1 \\
\Be_{31} I &= \Be_{31123} = – \Be_2 \\
\Be_{12} I &= \Be_{12123} = – \Be_3 \\
\Be_{123} I &= \Be_{123123} = -1.
\end{aligned}
With differential forms, the duals of the duals of all our basic forms recovered the original, that is $$** \omega = \omega$$, but that isn’t the case if we use pseudoscalar multiplication to define duality. We see that to model the Hodge dual, we need to multiply by a grade specific pseudoscalar.
## Definition 1.8: Hodge dual of an $$\mathbb{R}^3$$ multivector
Let $$M$$ be a $$\mathbb{R}^3$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}
In particular, if $$A$$ is a k-blade in $$\mathbb{R}^3$$, a round trip requires multiplication with different signed unit pseudoscalars.
Let’s step back and consider the $$\mathbb{R}^2$$ case as well. This time we let $$i = dx_1 \wedge dx_2$$. We seek all the duals satisfying
\label{eqn:formAndCurl:1180}
\begin{aligned}
i &= 1 \wedge {*} 1 \\
i &= dx \wedge {*} dx \\
i &= dy \wedge {*} dy \\
i &= (dx dy) \wedge {*} (dx dy).
\end{aligned}
Those duals are
\label{eqn:formAndCurl:1200}
\begin{aligned}
{*} 1 &= dx dy \\
{*} dx &= dy \\
{*} dy &= -dx \\
{*} (dx dy) &= 1 \\
\end{aligned}
Now let’s compare this to multiplication of the $$\mathbb{R}^2$$ basis vectors with the pseudoscalar $$i = \Be_1 \Be_2$$. We have
\label{eqn:formAndCurl:1220}
\begin{aligned}
1 i &= i \\
\Be_1 i &= \Be_{112} = \Be_{2} \\
\Be_2 i &= \Be_{212} = -\Be_{1} \\
\Be_{12} i &= \Be_{1212} = -1 \\
\end{aligned}
## Definition 1.9: Hodge dual of $$\mathbb{R}^2$$ multivector
Let $$M$$ be a $$\mathbb{R}^2$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}
Neither of these grade specific duality operations are as nice as simply multiplying by a unit pseudoscalar, but if we care about correspondence with the Hodge dual (at least according to the definition in the article), then this is what we need.
Having done that, let’s now look at the Hodge dual that produces the divergence operation.
## Lemma 1.13: Divergence relation to the exterior derivative.
Let $$\omega = f dx + g dy + h dz$$ be a one-form in $$\mathbb{R}^3$$. The exterior derivative of the Hodge dual of $$\omega$$ is a divergence three-form
\begin{equation*}
d({*} \omega) = \lr{ \PD{x}{f} + \PD{y}{g} + \PD{z}{h} } dx \wedge dy \wedge dz.
\end{equation*}
The GA equivalent of this, for a vector corresponding to this one-form $$\Bf = f \Be_1 + g \Be_2 + h \Be_3 \in \mathbb{R}^3$$, is
\begin{equation*}
\end{equation*}
### Start proof:
The dual of the one form is
\label{eqn:formAndCurl:1280}
{*} \omega =
f dy \wedge dz
+ g dz \wedge dx
+ h dx \wedge dy,
so the exterior derivative is
\label{eqn:formAndCurl:1300}
\begin{aligned}
d({*} \omega) &=
\lr{
\PD{x}{f} dx +
\PD{y}{f} dy +
\PD{z}{f} dz
}
\wedge dy \wedge dz \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dz \wedge dx \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dx \wedge dy \\
&=
\lr{
\PD{x}{f} +
\PD{y}{g} +
\PD{z}{h}
}
dx \wedge dy \wedge dz.
\end{aligned}
We expect that the GA equivalent of this is $$\spacegrad \wedge ({*} \Bf) = \lr{ \spacegrad \cdot \Bf} I$$. Let’s check that this is the case. The dual, for a vector, is
\label{eqn:formAndCurl:1320}
{*} \Bf
= \Bf I,
so
\label{eqn:formAndCurl:1340}
\begin{aligned}
&= \lr{ \spacegrad \cdot \Bf } I.
\end{aligned}
# References
[1] Vincent Bouchard. Math 215: Calculus iv: 4.4 the exterior derivative and vector calculus, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_exterior_vector.html. [Online; accessed 11-November-2023].
[2] Vincent Bouchard. Math 215: Calculus iv: 4.8 hodge star, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_hodge.html. [Online; accessed 13-November-2023].
[3] H. Flanders. Differential Forms With Applications to the Physical Sciences. Courier Dover Publications, 1989.
## Motivation.
I was asked about the geometric algebra equivalents of some of the vector calculus identities from [1]. I’ll call the specific page of those calculus notes “the article”. The article includes identities like
\label{eqn:formAndCurl:20}
\begin{aligned}
\spacegrad \cdot \lr{ \BF \cross \BG } &= \BG \cdot \lr{ \spacegrad \cross \BF } – \BF \cdot \lr{ \spacegrad \cross \BG },
\end{aligned}
but the point of these particular lecture notes is the interface between traditional Gibbs vector calculus and differential forms. That’s a much bigger topic, and perhaps not what I was actually being asked about. It is, however, an interesting topic, so let’s explore it.
## Comparisons.
The article identifies the cross product representation of the curl $$\spacegrad \cross \BF$$ as the equivalent to the exterior derivative of a one form (which has been mapped to a vector function.) In geometric algebra, this isn’t the identification we would use. Instead we should identify the “bivector curl” $$\spacegrad \wedge \BF$$ as the logical equivalent of the exterior derivative of that one form, and in general identify $$\spacegrad \wedge A_k$$ as the exterior derivative of a k-form (k-blade). In my notes to follow, the wedge of the gradient with a function, will be called the curl of that function, even if we are operating in $$\mathbb{R}^3$$ where the cross product is defined.
The starting place of the article was to define a one form and it’s exterior derivative was essentially as follows
## Definition 1.1: The exterior derivative of a one form.
Let $$f : \mathbb{R}^N \rightarrow \mathbb{R}$$ be a zero form. It’s exterior derivative is
\begin{equation*}
df = \sum_i dx_i \PD{x_i}{f}.
\end{equation*}
I’ve stated that the GA equivalent of the exterior derivative was a curl $$\spacegrad \wedge A$$, and this doesn’t look anything curl like, so right away, we have some trouble to deal with. To resolve that trouble, let’s step back to the gradient, which we haven’t defined yet. In the article, the gradient (of a scalar function) was defined as a coordinate triplet
\label{eqn:formAndCurl:60}
\spacegrad \Bf = \lr{ \PD{x}{f}, \PD{y}{f}, \PD{z}{f} }.
In GA we don’t like representations where the basis vectors are implicit, so we’d prefer to define
## Definition 1.2: The gradient of a function.
We define the gradient of multivector $$f(x_1, x_2, \cdots, x_N)$$, and denote it by $$\spacegrad f$$
\begin{equation*}
\spacegrad f = \sum_{i=1}^N \Be_i \PD{x_i}{f},
\end{equation*}
where $$\setlr{ \Be_1, \cdots \Be_N }$$ is an orthonormal basis for $$\mathbb{R}^N$$.
Unlike the article, we do not restrict $$f$$ to be a scalar function, since we do not have a problem with a vector valued operator directly multiplying a vector or any product of vectors. Instead $$f$$ can be a multivector function, with scalar, vector, bivector, trivector, … components, and we define the gradient the same way.
In order to define the curl of a k-blade, we need a reminder of how we define the wedge of a vector with a k-blade. Recall that this is how we generally define the wedge between two blades.
## Definition 1.3:
Let $$A_r$$ be a r-blade, and $$B_s$$ a s-blade. The wedge of $$A_r$$ with $$B_s$$ is
\label{eqn:formAndCurl:120}
A_r \wedge B_s = \gpgrade{A_r B_s}{r+s}.
In particular, if $$\Ba$$ is a vector, then the wedge with an s-blade $$B_s$$ is
\label{eqn:formAndCurl:140}
\Ba \wedge B_s = \gpgrade{\Ba B_s}{s+1},
which is just the $$s+1$$ grade selection of their product. Furthermore, if $$f$$ is a scalar, then
\label{eqn:formAndCurl:160}
\Ba \wedge f = \gpgrade{\Ba f}{1} = \Ba f.
We can now state the curl of a k-blade
## Definition 1.4: Curl of a k-blade.
Let $$A_k$$ be a k-blade. We define the curl of a k-blade as the wedge product of the gradient with that k-blade, designated
\begin{equation*}
\end{equation*}
Observe, given our generalized wedge product definition above, that the curl of a scalar function $$f$$, is in fact just the gradient of that function
\label{eqn:formAndCurl:200}
This has exactly the structure of the exterior derivative of a one form, as stated in “Definition: The exterior derivative of a one form”, but we have replaced $$dx_i$$ with a basis vector $$\Be_i$$.
## Definition 1.5: Exterior derivative of a one-form.
Let $$\omega = f_i dx_i$$ be a one-form. The exterior derivative of $$d \omega$$ is
\begin{equation*}
d\omega = \sum_i d( f_i ) \wedge dx_i.
\end{equation*}
## Lemma 1.1: Exterior derivative of a one-form.
Let $$\omega = f_i dx_i$$ be a one-form. The exterior derivative $$d \omega$$ can be expanding into a Jacobian form
\begin{equation*}
d\omega
=
\sum_{i < j} \lr{
\PD{x_i}{f_j}
\PD{x_j}{f_i}
} dx_i \wedge dx_j.
\end{equation*}
### Start proof:
\label{eqn:formAndCurl:220}
\begin{aligned}
d\omega
&= \sum_j d( f_j dx_j ) \\
&= \sum_j d( f_j ) \wedge dx_j \\
&= \sum_j \lr{ \sum_i dx_i \PD{x_i}{f_j} } \wedge dx_j \\
&= \sum_{ji} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&= \sum_{j \ne i} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&=
\sum_{i < j} \PD{x_i}{f_j} dx_i \wedge dx_j
+
\sum_{j < i} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&=
\sum_{i < j} \PD{x_i}{f_j} dx_i \wedge dx_j
+
\sum_{i < j} \PD{x_j}{f_i} dx_j \wedge dx_i \\
&=
\sum_{i < j} \lr{
\PD{x_i}{f_j}
\PD{x_j}{f_i}
} dx_i \wedge dx_j.
\end{aligned}
## Lemma 1.2: Curl of a vector.
Let $$\Bf = \sum_i \Be_i f_i \in \mathbb{R}^N$$ be a vector. The curl of $$\Bf$$ has a Jacobian structure
\begin{equation*}
=
\sum_{i < j}
\lr{ \PD{x_i}{f_j} – \PD{x_j}{f_i} }
\lr{ \Be_i \wedge \Be_j }
.
\end{equation*}
### Start proof:
The antisymmetry of the wedges of differentials in the exterior derivative and the curl clearly has a one to one correspondence. Let’s show this explicitly by expansion
\label{eqn:formAndCurl:240}
\begin{aligned}
&=
\sum_{ij} \lr{ \Be_i \PD{x_i}{} } \wedge \lr{ \Be_j f_j } \\
&=
\sum_{ij} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i \ne j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j}
+
\sum_{j < i} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j}
+
\sum_{i < j} \lr{ \Be_j \wedge \Be_i } \PD{x_j}{f_i} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \lr{ \PD{x_i}{f_j} – \PD{x_j}{f_i} }.
\end{aligned}
### End proof.
If we are translating from differential forms, again, we see that we simply replace any differentials $$dx_i$$ with the basis vectors $$\Be_i$$ (at least for the zero-form and one-form cases, which is all that we have looked at here.)
Note that in differential forms, we often assume that there is an implicit wedge product between any different one form elements, writing
\label{eqn:formAndCurl:260}
dx_1 \wedge dx_2 = dx_1 dx_2.
This works out fine when we map differentials to basis vectors, since
\label{eqn:formAndCurl:280}
\Be_1 \Be_2 =
\Be_1 \cdot \Be_2
+
\Be_1 \wedge \Be_2
=
\Be_1 \wedge \Be_2.
However, we have to be more careful in GA when using indexed expressions, since
\label{eqn:formAndCurl:300}
\Be_i \Be_j = \Be_i \cdot \Be_j + \Be_i \wedge \Be_j.
The dot product portion of the RHS is only zero if $$i \ne j$$.
Now let’s look at the equivalence between the exterior derivative of a two-form with the curl.
## Definition 1.6: Exterior derivative of a two-form.
Let $$\eta = \sum_{ij} f_{ij} dx_i \wedge dx_j$$ be a two-form. The exterior derivative of $$\eta$$ is
\begin{equation*}
d\eta =
\sum_{ij} d( f_{ij} ) \wedge dx_i \wedge dx_j.
\end{equation*}
## Lemma 1.3: Exterior derivative of a two-form.
Let $$\eta = \sum_{ij} f_{ij} dx_i \wedge dx_j$$ be a two-form. The exterior derivative of $$\eta$$ can be expanded as
\begin{equation*}
d \eta
=
\sum_{i,j,k} \PD{x_k}{f_{ij}} dx_i \wedge dx_j \wedge dx_k.
\end{equation*}
### Start proof:
The exterior derivative of $$\eta$$ is
\label{eqn:formAndCurl:340}
\begin{aligned}
d \eta
&=
\sum_{i,j} d( f_{ij} dx_i \wedge dx_j ) \\
&=
\sum_{i,j,k} \lr{ \PD{x_k}{f_{ij}} dx_k } \wedge dx_i \wedge dx_j \\
&=
\sum_{i,j,k} \PD{x_k}{f_{ij}} dx_i \wedge dx_j \wedge dx_k.
\end{aligned}
### End proof.
Let’s compare that to the curl of a bivector valued function.
## Lemma 1.4: Curl of a 2-blade.
Let $$B = \sum_{i \ne j} f_{ij} \Be_i \wedge \Be_j$$ be a 2-blade. The curl of $$B$$ is
\begin{equation*}
=
\sum_{i,j,k} \PD{x_k}{f_{ij}} \Be_i \wedge \Be_j \wedge \Be_k.
\end{equation*}
### Start proof:
\label{eqn:formAndCurl:380}
\begin{aligned}
&=
\lr{ \sum_k \Be_k \PD{x_k}{} } \wedge \lr{ \sum_{i \ne j} f_{ij} \Be_i \wedge \Be_j } \\
&=
\sum_{k, i \ne j} \PD{x_k}{f_{ij}} \Be_k \wedge \Be_i \wedge \Be_j \\
&=
\sum_{i,j,k} \PD{x_k}{f_{ij}} \Be_i \wedge \Be_j \wedge \Be_k.
\end{aligned}
### End proof.
Again, we see an exact correspondence with the exterior derivative $$d \eta$$ of a two-form, and the curl $$\spacegrad \wedge B$$, of a 2-blade.
The article established a coorespondence between the exterior derivative of a two form over $$\mathbb{R}^3$$ to the divergence. The way we would express this in GA (also for \R{3}) is to write
\label{eqn:formAndCurl:400}
B = I \Bb,
where $$I = \Be_1 \Be_2 \Be_3$$ is the $$\mathbb{R}^3$$ pseudoscalar (a “unit” trivector.) Forming the curl of $$B$$ we have
\label{eqn:formAndCurl:420}
\begin{aligned}
\end{aligned}
The equivalence relationships that we have developed must then imply that the differential forms representation of this relationship is
\label{eqn:formAndCurl:440}
d B = dx_1 \wedge dx_2 \wedge dx_3 (\spacegrad \cdot \Bb)
= dx \wedge dy \wedge dz \lr{ \PD{x}{b_1} + \PD{y}{b_2} + \PD{z}{b_3} },
as defined in the article.
Here is the GA equivalent of Lemma 4.4.10 from the article
## Lemma 1.5: Repeated curl identities.
Let $$A$$ be a smooth k-blade, then
\begin{equation*}
\end{equation*}
For $$\mathbb{R}^3$$, this result, for a scalar function $$f$$, and a vector function $$\Bf$$, in terms of the cross product, as
\label{eqn:formAndCurl:560}
\begin{aligned}
\end{aligned}
It shouldn’t be surprising that this is the equivalent of $$d^2 A = 0$$ from differential forms. Let’s prove this, first considering the 0-blade case
### Start proof:
\label{eqn:formAndCurl:480}
\begin{aligned}
&=
&=
\sum_{ij} \Be_i \wedge \Be_j \frac{\partial^2 A}{\partial x_i \partial x_j} \\
&= 0.
\end{aligned}
The smooth criteria of for the function $$A$$ is assumed to imply that we have equality of mixed partials, and since this is a sum of an antisymmetric term with respect to indexes $$i, j$$ (the wedge) and a symmetric term in indexes $$i, j$$ (the partials), we have zero overall.
Now consider a k-blade $$A, k > 0$$. Expanding the gradients, we have
\label{eqn:formAndCurl:500}
=
\sum_{ij} \Be_i \wedge \Be_j \wedge \frac{\partial^2 A}{\partial x_i \partial x_j}.
It may be obvious that this is zero for the same reasons as above (sum of product of symmetric and antisymmetric entities). We can, however, make it more obvious, at the cost of some hellish indexing, by expressing $$A$$ in coordinate form. Let
\label{eqn:formAndCurl:520}
A = \sum_{i_1, i_2, \cdots, i_k}
A_{i_1, i_2, \cdots, i_k} \Be_{i_1} \wedge \Be_{i_2} \wedge \cdots \wedge \Be_{i_k},
then
\label{eqn:formAndCurl:540}
\begin{aligned}
&=
\sum_{i,j,i_1, i_2, \cdots, i_k} \Be_i \wedge \Be_j \wedge \Be_{i_1} \wedge \Be_{i_2} \wedge \cdots \wedge \Be_{i_k}
\frac{\partial^2 }{\partial x_i \partial x_j} A_{i_1, i_2, \cdots, i_k} \\
&= 0.
\end{aligned}
Now we clearly have a sum of an antisymmetric term (the wedges), and a symmetric term (assuming smooth $$A$$ means that we have equality of mixed partials), so the sum is zero.
Finally, for the $$\mathbb{R}^3$$ identities, we have
\label{eqn:formAndCurl:580}
\begin{aligned}
&=
&=
0,
\end{aligned}
since $$\spacegrad \wedge \lr{ \spacegrad f } = 0$$. For a vector $$\Bf$$, we have
\label{eqn:formAndCurl:600}
\begin{aligned}
&=
} \\
&=
} \\
&=
} \\
&=
&= 0,
\end{aligned}
again, because $$\spacegrad \wedge \lr{ \spacegrad \wedge \Bf} = 0$$.
## Identities.
We have a number of chain rule identities in the article. Here is the GA equivalent of that, and its corollaries
## Lemma 1.6: Chain rule identities.
Let $$f$$ be a scalar function and $$A$$ be a k-blade, then
\begin{equation*}
\spacegrad \lr{ f A } = \lr{ \spacegrad f } A + f \lr{ \spacegrad A }.
\end{equation*}
For $$A$$ with grade $$k > 0$$, the grade $$k-1$$ and $$k+1$$ components of this product are
\begin{equation*}
\begin{aligned}
\spacegrad \cdot \lr{ f A } &= \lr{ \spacegrad f } \cdot A + f \lr{ \spacegrad \cdot A } \\
\spacegrad \wedge \lr{ f A } &= \lr{ \spacegrad f } \wedge A + f \lr{ \spacegrad \wedge A }.
\end{aligned}
\end{equation*}
For $$\mathbb{R}^3$$, the wedge product relation above can be written in dual form as
\begin{equation*}
\spacegrad \cross \lr{ f A } = \lr{ \spacegrad f } \cross A + f \lr{ \spacegrad \cross A }.
\end{equation*}
Proving this is left to the reader.
We have some chain rule identities left in the article to verify and to find GA equivalents of. Before doing so, we need a couple miscellaneous identities relating triple cross products to wedge-dots.
## Lemma 1.7: Triple cross products.
Let $$\Ba, \Bb, \Bc$$ be vectors in $$\mathbb{R}^3$$. Then
\begin{equation*}
\begin{aligned}
\Ba \cross \lr{ \Bb \cross \Bc } &= – \Ba \cdot \lr{ \Bb \wedge \Bc } \\
\lr{ \Ba \cross \Bb } \cross \Bc &= – \lr{ \Ba \wedge \Bb } \cdot \Bc.
\end{aligned}
\end{equation*}
### Start proof:
\label{eqn:formAndCurl:720}
\begin{aligned}
\Ba \cross \lr{ \Bb \cross \Bc }
&=
\gpgradeone{ -I \lr{ \Ba \wedge \lr{ \Bb \cross \Bc } } } \\
&=
\gpgradeone{ -I \lr{ \Ba \lr{ \Bb \cross \Bc } } } \\
&=
\gpgradeone{ (-I)^2 \lr{ \Ba \lr{ \Bb \wedge \Bc } } } \\
&=
-\Ba \cdot \lr{ \Bb \wedge \Bc },
\end{aligned}
\label{eqn:formAndCurl:740}
\begin{aligned}
\lr{ \Ba \cross \Bb } \cross \Bc
&=
\gpgradeone{ -I \lr{ \Ba \cross \Bb } \wedge \Bc } \\
&=
\gpgradeone{ -I \lr{ \Ba \cross \Bb } \Bc } \\
&=
\gpgradeone{ (-I)^2 \lr{ \Ba \wedge \Bb } \Bc } \\
&=
– \lr{ \Ba \wedge \Bb } \cdot \Bc.
\end{aligned}
### End proof.
Next up is another chain rule identity
## Lemma 1.8: Gradient of dot product.
If $$\Ba, \Bb$$ are vectors, then
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb } =
\lr{ \Ba \cdot \spacegrad } \Bb
+
\lr{ \Bb \cdot \spacegrad } \Ba
+
\cdot
\Ba
+
\cdot
\Bb
\end{equation*}
For $$\mathbb{R}^3$$, this can be written as
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\lr{ \Ba \cdot \spacegrad } \Bb
+
\lr{ \Bb \cdot \spacegrad } \Ba
+
\Ba \cross \lr{ \spacegrad \cross \Bb }
+
\Bb \cross \lr{ \spacegrad \cross \Ba }
\end{equation*}
### Start proof:
We will use $$\rspacegrad$$ to indicate that the gradient operates on everything to the right, $$\lrspacegrad$$ to indicate that the gradient operates bidirectionally, and $$\spacegrad’ A B’$$ to indicate that the gradient’s scope is limited to the ticked entity (just on $$B$$ in this case.)
\label{eqn:formAndCurl:760}
\begin{aligned}
\rspacegrad \lr{ \Ba \cdot \Bb }
&=
\rspacegrad \lr{ \Ba \Bb – \Ba \wedge \Bb }
} \\
&=
+
}
– \rspacegrad \cdot \lr{ \Ba \wedge \Bb }
\\
&=
+
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
+
– \Ba \spacegrad \Bb + 2 \lr{ \Ba \cdot \spacegrad } \Bb
}
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }
\\
&=
+
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
\Ba \lr{ \spacegrad \cdot \Bb }
\Ba \cdot \lr{ \spacegrad \wedge \Bb }
+ 2 \lr{ \Ba \cdot \spacegrad } \Bb
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }.
\end{aligned}
We are running out of room, and have not had any cancellation yet, so let’s expand those last two terms separately
\label{eqn:formAndCurl:780}
\begin{aligned}
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }
&=
– \lr{ \spacegrad’ \cdot \Ba’ } \Bb
+ \lr{ \spacegrad’ \cdot \Bb } \Ba’
– \lr{ \spacegrad’ \cdot \Ba } \Bb’
+ \lr{ \spacegrad’ \cdot \Bb’ } \Ba
\\
&=
– \lr{ \spacegrad \cdot \Ba } \Bb
+ \lr{ \Bb \cdot \spacegrad } \Ba
– \lr{ \Ba \cdot \spacegrad } \Bb
+ \lr{ \spacegrad \cdot \Bb } \Ba.
\end{aligned}
Now we can cancel some terms, leaving
\label{eqn:formAndCurl:800}
\begin{aligned}
\rspacegrad \lr{ \Ba \cdot \Bb }
&=
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
\Ba \cdot \lr{ \spacegrad \wedge \Bb }
+ \lr{ \Ba \cdot \spacegrad } \Bb
+ \lr{ \Bb \cdot \spacegrad } \Ba.
\end{aligned}
After adjustment of the order and sign of the second term, we see that this is the result we wanted. To show the $$\mathbb{R}^3$$ formulation, we have only apply “Lemma: Triple cross products”.
## Lemma 1.9: Divergence of a bivector.
Let $$\Ba, \Bb \in \mathbb{R}^N$$ be vectors. The divergence of their wedge can be written
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
=
\Bb \lr{ \spacegrad \cdot \Ba }
– \Ba \lr{ \spacegrad \cdot \Bb }
– \lr{ \Bb \cdot \spacegrad } \Ba
+ \lr{ \Ba \cdot \spacegrad } \Bb.
\end{equation*}
For $$\mathbb{R}^3$$, this can also be written in triple cross product form
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
=
-\spacegrad \cross \lr{ \Ba \cross \Bb }.
\end{equation*}
### Start proof:
\label{eqn:formAndCurl:860}
\begin{aligned}
\rspacegrad \cdot \lr{ \Ba \wedge \Bb }
&=
\spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
+ \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ } \\
&=
\lr{ \spacegrad’ \cdot \Ba’ } \Bb
– \lr{ \spacegrad’ \cdot \Bb } \Ba’
+ \lr{ \spacegrad’ \cdot \Ba } \Bb’
– \lr{ \spacegrad’ \cdot \Bb’ } \Ba
\\
&=
\lr{ \spacegrad \cdot \Ba } \Bb
– \lr{ \Bb \cdot \spacegrad } \Ba
+ \lr{ \Ba \cdot \spacegrad } \Bb
– \lr{ \spacegrad \cdot \Bb } \Ba.
\end{aligned}
For the $$\mathbb{R}^3$$ part of the story, we have
\label{eqn:formAndCurl:870}
\begin{aligned}
\spacegrad \cross \lr{ \Ba \cross \Bb }
&=
-I \lr{ \spacegrad \wedge \lr{ \Ba \cross \Bb } }
} \\
&=
-I \spacegrad \lr{ \Ba \cross \Bb }
} \\
&=
(-I)^2 \spacegrad \lr{ \Ba \wedge \Bb }
} \\
&=
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
\end{aligned}
### End proof.
We have just one identity left in the article to find the GA equivalent of, but will split that into two logical pieces.
## Lemma 1.10: Dual of triple wedge.
If $$\Ba, \Bb, \Bc \in \mathbb{R}^3$$ are vectors, then
\begin{equation*}
\Ba \cdot \lr{ \Bb \cross \Bc } = -I \lr{ \Ba \wedge \Bb \wedge \Bc }.
\end{equation*}
### Start proof:
\label{eqn:formAndCurl:680}
\begin{aligned}
\Ba \cdot \lr{ \Bb \cross \Bc }
&=
\Ba \lr{ \Bb \cross \Bc }
} \\
&=
\Ba (-I) \lr{ \Bb \wedge \Bc }
} \\
&=
-I \lr{
\Ba \cdot \lr{ \Bb \wedge \Bc }
+
\Ba \wedge \lr{ \Bb \wedge \Bc }
}
} \\
&=
-I \lr{ \Ba \wedge \lr{ \Bb \wedge \Bc } }
} \\
&=
-I \lr{ \Ba \wedge \lr{ \Bb \wedge \Bc } }.
\end{aligned}
## Lemma 1.11: Curl of a wedge of gradients (divergence of a gradient cross products.)
Let $$f, g, h$$ be smooth functions with smooth derivatives. Then
\begin{equation*}
=
\wedge
\wedge
\end{equation*}
For $$\mathbb{R}^3$$ this can be written as
\begin{equation*}
=
\cdot
\lr{
\cross
}.
\end{equation*}
### Start proof:
The GA identity follows by chain rule and application of “Lemma: Repeated curl identities”.
\label{eqn:formAndCurl:910}
\begin{aligned}
&=
+
f
&=
\end{aligned}
The $$\mathbb{R}^3$$ part of the lemma follows from “Lemma: Dual of triple wedge”, applied twice
\label{eqn:formAndCurl:970}
\begin{aligned}
&=
&=
&=
\end{aligned}
## Lemma 1.12: Curl of a bivector.
Let $$\Ba, \Bb$$ be vectors. The curl of their wedge is
\begin{equation*}
\spacegrad \wedge \lr{ \Ba \wedge \Bb } = \Bb \wedge \lr{ \spacegrad \wedge \Ba } – \Ba \wedge \lr{ \spacegrad \wedge \Bb }
\end{equation*}
For $$\mathbb{R}^3$$, this can be expressed as the divergence of a cross product
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \cross \Bb } = \Bb \cdot \lr{ \spacegrad \cross \Ba } – \Ba \cdot \lr{ \spacegrad \cross \Bb }
\end{equation*}
### Start proof:
The GA case is a trivial chain rule application
\label{eqn:formAndCurl:950}
\begin{aligned}
\rspacegrad \wedge \lr{ \Ba \wedge \Bb }
&=
\lr{ \spacegrad’ \wedge \Ba’} \wedge \Bb
+
\lr{ \spacegrad’ \wedge \Ba } \wedge \Bb’ \\
&= \Bb \wedge \lr{ \spacegrad \wedge \Ba } – \Ba \wedge \lr{ \spacegrad \wedge \Bb }.
\end{aligned}
The $$\mathbb{R}^3$$ case, is less obvious by inspection, but follows from “Lemma: Dual of triple wedge”.
## Summary.
We found that we have an isomorphism between the exterior derivative of differential forms and the curl operation of geometric algebra, as follows
\label{eqn:formAndCurl:990}
\begin{aligned}
df &\rightleftharpoons \spacegrad \wedge f \\
dx_i &\rightleftharpoons \Be_i.
\end{aligned}
We didn’t look at how the Hodge dual translates to GA duality (pseudoscalar multiplication.) The divergence relationship between the exterior derivative of an $$\mathbb{R}^3$$ two-form really requires that formalism, and has only been examined in a cursory fashion.
We also translated a number of vector and gradient identities from conventional vector algebra (i.e.: using cross products) and wedge product equivalents of the same. The GA identities are often simpler, and in some cases, provide nice mechanisms to derive the conventional identities that would be more cumbersome to determine without the GA toolbox.
# References
[1] Vincent Bouchard. Math 215: Calculus iv: 4.4 the exterior derivative and vector calculus, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_exterior_vector.html. [Online; accessed 11-November-2023]. | 26,605 | 76,291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.664019 |
https://wumbo.net/concept/radians/ | 1,606,669,572,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141201836.36/warc/CC-MAIN-20201129153900-20201129183900-00248.warc.gz | 569,168,918 | 7,649 | Radians are a unit that measure angles. A full rotation in radians is equal to (tau) radians, where the constant is approximately equal to . Radians are the preferred unit for measuring angles, as opposed to degrees, since they lead to “more succinct and elegant formulas throughout mathematics"[1].
## Definition
A radian angle is equivalent to the ratio of the arc-length over the radius of the circle. By design, the unit is “dimensionless” or, in plain language, the length of the radius is arbitrary; one radian is equal to the amount of rotation required to travel the length of one radius along the circumference of the circle. The symbol (equivalent) is used to indicate that negative angles or angles more than a full rotation are mapped to the equivalent angle between and a full rotation.
This “radius-invariant” property plays an important role for why radians are the preferred unit for measuring angles. See radians vs. degrees.
## Usage
The radian system is used in many places in mathematics: measuring angles, numerous geometric and advanced math formulas via the circle constant, and as the unit of choice for the trigonometric functions.
### Circle Constant
Note: This website uses the constant (tau) instead of (pi) as the default circle constant for reasons discussed on this page. The substitution can be used to translate between the two constants.
The circle contanstant (tau) makes talking about and using the radians angle system much more convenient. The circle constant is approximately equal to [2] as illustrated by figure 3.
Without the circle constant, angles measured in radians can be confusing. Take for example, the figures figure 4 and figure 5 showing measured angles of 1 radians and 2 radians. Both figures draw tic marks at the numeric values of radians, which conceptually demonstrates what a radian is. However, as we will later see it is much more convenient to use a constant.
### Angles
Angles measured using radians are expressed as a fraction of “one turn” where one turn is equal to the circle constant (tau). By convention, angles in the XY coordinate plane are measured from the positive direction where the counter-clockwise rotation is positive.
This is where the circle constant plays a crucial role in making the radians system complete; angles can easily be expressed as fractions of the whole. As figure 6 and figure 7 illustrate, the circle can be divided into equal parts and using the circle constant constant leads to elegant and succinct notation.
As a final note on measuring angles, here are some “special angles” that show up in numerous places. The figure also illustrates why circle constant is also the best method for converting between radians and degrees[3]
#### Common Angles
Shown below are some special angles measured using radians. Note, the symbol (theta) is often used to represent a measured angle.
### Formulas
Radians appear in numerous geometric formulas and advanced formulas in mathematics via the circle constant or or alternatively through the constant . Shown below are some common geometric formulas. Note, very likely you have seen the constant used insted of This website primarily uses instead of in equations and formulas as it is the better constant[4]. The relation can be used to translate between the two constants.
Area of Circle | Formula
The area of a circle is give by one-half multiplied by τ (tau) mutliplied by the radius of the circle squared.
Circumference of Circle | Formula
The circumference of a circle is given the constant τ (tau) multpilied by the radius of the circle, where τ = 2π.
Volume of Sphere | Formula
The volume of a sphere is given by two-thirds multiplied by the circle constant τ (tau) multiplied by the radius cubed.
The circle constant also appears in some suprising places such as the definition of the normal distribution. When is present, there usually is some connection to circles or a circular way of thinking.
### Functions
The trigonometric functions and the unit circle are often where radians are introduced. Some programming languages and spreadsheet applications only offer the trigonometric functions implemented using the radian system. This is discussed in more detail on the individual function pages, but shown below are two figures which graph the two functions using the radian system.
There are two systems for measuring degrees in math: degrees and radians. The short answer for why radians are preferred is that the radian system leads to more succinct and elegant formulas throughout mathematics [4]. For example, the derivative of the sine function is only true when the angle is expressed in radians.
### Convert to Degrees
To convert an angle from radians to degrees multiply by the ratio of over radians [3]. | 951 | 4,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-50 | longest | en | 0.921699 |
http://mormonconferences.org/aa/1-4-roof-slope.html | 1,618,443,534,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00114.warc.gz | 69,973,991 | 10,188 | # 1 4 Roof Slope
### For most home styles roof pitches fall in a range 4 12 a moderate slope up to 8 12 fairly steep.
1 4 roof slope.
The picture below shows the pitch of a 7 12 roof slope meaning that for 12 of horizontal measurement roof run the vertical measurement roof rise is 7. This means for every 12 horizontal units the roof must rise a minimum of one fourth vertical unit. Most roof s have a pitch in the 4 12 to 9 12 range. Examples of extreme slopes range from 1 4 12 almost flat to 12 12 sloping down at a perfect 45 degree angle.
3 of rise per 12 of run is the same as 1 of rise per 4 of run or 1 4 25. Conventional roofs are the easiest to construct and you can walk on them safely. While roof slope is typically expressed in rise it can also be expressed in degrees or in percent of slope. This measurement is best done on a bare roof because curled up roofing shingles will impair your measurement.
Why is the roof slope 25. Conventional from 4 12 1 in 3 to 9 12 3 in 4. The table below shows common roof pitches and the equivalent grade degrees and radians for each. That s the same as a 14 degree slope or a 25 slope.
High pitched roofs often need extra fasteners and their pitch can be as high as 21 12. You can also calculate roof pitch even without using a roof slope calculator. Slope ratio a roof that rises 4 inches for every 1 foot or 12 inches of run is said to have a 4 in 12 slope. The basic ranges of pitch are not uniformly defined but range from flat which are not perfectly flat but sloped to drain water up to 1 2 12 to 2 12 1 in 24 to 1 in 6.
A 3 in 12 roof rises 3 for every 12 of run. They have a pitch ranging from 4 12 to 9 12. A 4 12 is a roof slope that rises by 4 inches for every 12 inches across. A pitch over 9 12 is considered a steep slope roof between 2 12 and 4 12 is considered a low slope roof and less than 2 12 is considered a flat roof.
This means that the rise of the slope goes up or down 8 inches for every 12 inches. These roofs have a pitch less than 4 12. Liquid applied roofing must have a design roof slope of 1 4 12 or greater 2 slope. This forms an angle of 18 5 between the horizontal section and the roof and creates a gentle incline that is seen as a midpoint between a low pitch and medium pitch roof.
If the rise is 6 inches for every 12 inches of run then the roof slope is 6 in 12 the slope can be expressed numerically as a ratio. An example for a steep sloped roof is 8 12.
### Display Of Roof Pitches 1 12 Through 3 2 This Graphic Uses Run Rise The Inverse Of The Convention Described In The Article Pitched Roof House Roof Dormers
Source : pinterest.com | 693 | 2,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-17 | latest | en | 0.951863 |
https://mathoverflow.net/questions/326214/identifying-elements-in-the-kernel-of-an-explicit-endomorphism-of-a-jacobian-var/326225 | 1,600,764,253,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00491.warc.gz | 540,703,299 | 30,664 | # Identifying elements in the kernel of an explicit endomorphism of a Jacobian variety
I hope this question fits here.
Let $$H/k$$ be a genus $$2$$ curve and $$J$$ its Jacobian variety. Since $$J(k)\cong \text{Pic}^0(H)(k)$$ we have that its generic point looks like $$[(x_1,y_1)+(x_2,y_2)-2\infty]\in J$$. In Mumford coordinates we can see it as $$g:=\langle x^2 -Ax + B,Cx+D\rangle:=\langle u(x),v(x)\rangle\in J$$ with $$u(x_i)=0$$ and $$v(x_i)=y_i$$. This means that $$A:=x_1+x_2, B:=x_1x_2, C:=\tfrac{y_1-y_2}{x_1-x_2}, D:=\tfrac{x_2y_1-x_1y_2}{x_1-x_2}$$.
I calculated explicitly an element $$\gamma\in\text{End}_k(J)$$ using Mumford coordinates for the generic point, that is $$\gamma(g)=$$ where $$A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2\in k[J]$$.
To formulate my question, note that since the endomorphism $$\gamma$$ works fine for the generic point, and $$J$$ has dimension $$2$$, we have that if for some $$D\in J(k)$$, its image under $$\gamma$$, namely $$\gamma(D)$$ is of the form $$[(x_1,y_1)-\infty]$$, then $$\gamma$$ won't be defined for $$D$$, since some of the denominators will be zero.
How can I distinguish when $$\gamma(D)$$ is 0 or when it is non-zero non-generic of the form $$[(\tilde x,\tilde y)-\infty]$$?
I know I could use a constant divisor $$D_0$$ and calculate $$\gamma(D+D_0)$$ when $$\gamma(D)$$ is not defined and then subtract with $$\gamma(D_0)$$ using Cantor addition. However, in my situation this is not possible since I am testing if $$k$$ is a field using some arithmetic geometry. I want to be able to use the functions defining $$\gamma$$ over $$\mathbb{Q}$$ to distinguish the situation of $$\gamma(D)$$ being $$[0]$$ or being of the form $$[(\tilde x,\tilde y)-\infty]$$
I have noted that when the ALL the denominators are 0, it looks like the image 0 in fact, but when it lies in the "theta divisor", (the image is a point of the form $$[(x_1,y_1)-\infty]$$), some of the denominators are non-zero. However I do not know how to distinguish this formally or maybe my examples are just "lucky" examples.
Is there a way with this information to distinguish when $$\gamma(D)$$ is exactly [0] ?
What I did
I tried to calculate the formula of $$\gamma$$ using MAGMA via the function field of $$J$$, using the usual relations for the Jacobian of $$H$$ plus the denominators of $$\gamma$$ as relations, but the computation does not finish and eats all my memory eventually.
I just need to know if a point in the image is 0 or non-generic when $$\gamma(D)$$ has 0's in the denominators using the information that I have, or maybe using an element of the function field of $$J$$ that can distinguish if a divisor is 0 or if it has the point at infinity with multiplicity 1.
## 1 Answer
I would suggest that you work with the Kummer surface $$K$$ of $$J$$ instead of using Mumford coordinates. The advantage is that $$K$$ is a quartic surface in $$\mathbb P^3$$; in the case you are considering when the curve has a unique point at infinity, the vanishing of the first coordinate means that the point is in the theta divisor, whereas it is the origin when the first three coordinates vanish (using the standard Kummer coordinates as in the book by Cassels and Flynn). Since your endomorphism commutes with multiplication by $$-1$$, it induces an endomorphism of $$K$$. This will be given by a quadruple of homogeneous polynomials of some degree $$d$$ in the four coordinates; it should not be too hard to figure out what they are from the generic representation in terms of the Mumford representation. Then your problem comes down to checking whether the first of these polynomials vanishes, and if so, whether the next two also vanish. (This assumes that all four polynomials do not vanish simultaneously at some point on $$K$$.)
When $$\gamma$$ is multiplication by 2, for example, the polynomials are of degree 4 and can be obtained via
KummerSurface(J)Delta;
`
in Magma.
Added later:
For the curve $$C \colon y^2 = x^5 + 10\,,$$ one choice of polynomials giving multiplication by $$\sqrt{5}$$ on the Kummer surface is $$\begin{array}{r@{\,}c{\,}l} P_1(x_1,x_2,x_3,x_4) &=& 8000 x_1^3 x_2^2 + 400 x_1^2 x_2 x_4^2 + 200 x_1^2 x_3^2 x_4 + 400 x_1 x_2^2 x_3 x_4 - 600 x_1 x_2 x_3^3 + 5 x_1 x_4^4 + 200 x_2^3 x_3^2 + 10 x_2 x_3 x_4^3 + 10 x_3^3 x_4^2 \\ P_2(x_1,x_2,x_3,x_4) &=& 8000 x_1^4 x_4 + 8000 x_1^3 x_2 x_3 + 400 x_1 x_2^2 x_4^2 + 200 x_1 x_2 x_3^2 x_4 + 400 x_1 x_3^4 - 400 x_2^3 x_3 x_4 + 200 x_2^2 x_3^3 - 5 x_2 x_4^4 + 10 x_3^2 x_4^3 \\ P_3(x_1,x_2,x_3,x_4) &=& 40 x_1^3 x_2 x_4 + 8000 x_1^3 x_3^2 - 8040 x_1^2 x_2^2 x_3 - 200 x_1^2 x_4^3 - 7960 x_1 x_2^4 - 996 x_1 x_2 x_3 x_4^2 + 200 x_1 x_3^3 x_4 + 399 x_2^3 x_4^2 - 598 x_2^2 x_3^2 x_4 - x_2 x_3^4 + 5 x_3 x_4^4 \\ P_4(x_1,x_2,x_3,x_4) &=& 64000 x_1^5 + 2720 x_1^3 x_3 x_4 + 8000 x_1^2 x_2^2 x_4 + 5280 x_1^2 x_2 x_3^2 + 2720 x_1 x_2^3 x_3 - 200 x_1 x_2 x_4^3 - 1528 x_1 x_3^2 x_4^2 + 1600 x_2^5 - 68 x_2^2 x_3 x_4^2 - 64 x_2 x_3^3 x_4 + 52 x_3^5 + x_4^5 \end{array}$$ (they are not unique, since we can add multiples of the defining equation of the Kummer surface).
They were obtained by interpolating data from a number of points. (Magma) Code is available from me on request.
• Prof. Stoll. Thanks for the answer. In fact I retook a problem, using also your advice from some time ago, which was to extend the function field of the jacobian to the denominators). My profile in mathoverflow changed since univ. mail does not exists anymore. Related to your answer, I would need then to calculate the endomorphism mapped to the Kummer Surface, in my case I have to map concretely $\sqrt{5}\in\text{End}(J)$ where $J$ is the Jacobian of $H:y^2=x^5 + 10$ over $\mathbb{Q}$. Any suggestions? (technically talking) do you know if someone has done this for explicit endomorphisms? – Eduardo R. Duarte Mar 28 '19 at 16:16
• One think to add is that I tried to do what you said of extending the function field resulting always in "out of memory" problems after some days. – Eduardo R. Duarte Mar 28 '19 at 16:23
• @EduardoR.Duarte I have done the computation of the polynomials for the curve you mention and added the result to my answer. – Michael Stoll Mar 30 '19 at 10:01
• Prof, this is really nice, I just accepted the answer. I assume that $x_1,x_2, x_3,x_4$ are standard coordinates found in Prolegomena Flynn's book. I am really interested in the MAGMA computation in fact. Why the formulas are so small ? When I calculated $\sqrt{5}\in\text{End}(J)$ I used the generic point of $g\in J$ and then the action of $\zeta_1,\zeta_4\in\text{End}(J)$ in $g$. The final formulas are obtained through $2(\zeta_1(g)+\zeta_4(g))+g=\sqrt{5}(g)$. These formulas are huge. Yours are very compact, is there something I am missing? – Eduardo R. Duarte Mar 31 '19 at 19:55
• @EduardoR.Duarte If you send me an email, I can send you the code that does the computation. -- I think the reason why the expressions you got are so large is mainly that the coordinates coming from the Mumford representation are not very nice in a way, whereas the coordinates on the Kummer surface correspond to a full linear system. – Michael Stoll Mar 31 '19 at 22:14 | 2,401 | 7,147 | {"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": 53, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | latest | en | 0.872927 |
https://bbs.timetravelinstitute.com/threads/time-travel-and-conventional-mathematics.771/ | 1,696,326,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00627.warc.gz | 135,019,346 | 22,169 | # Time travel and conventional mathematics
G
#### Guest
##### Guest
The basis of time travel is that we are keeping track of our dear friend time..im not sure if we can understand that the mathematical and linear logic assume time to be a straight phenomena,,,while the time itself has distortions in its quantum nature. How can we justify our mathematics to solve this phenomena iwht numbers..we need intuitive mathematics...to solve the complex paradoxes of time.
Explain the equation
t1 = the current time
x =Distance between the desired past or desired future
t2 = destination
v = velocity of time travel ie accelaration x/mc^2
think abt it ..if u get something interesting
contact me at [email protected]
Post the equation, not just the variables, and someone might be able to help you out.
"the mathematical and linear logic assume time to be a straight phenomena,,,while the time itself has distortions in its quantum nature"
Ah, but time can be well approximated linearly. You wouldn't say that the ball I throw doesn't follow a parabolic path because gravity falls off as the inverse square of distance - the effects are too small. Likewise with time. And if you're talking about very small scales, then you need to shift into quantum mechanics, which takes quantum fluctuation into account.
"How can we justify our mathematics to solve this phenomena iwht numbers"
Everything's numbers. Just, some of the numbers are probabilities.
Replies
4
Views
429
Replies
0
Views
412
Replies
3
Views
1K
Replies
2
Views
992
Replies
1
Views
690
General chit-chat
Help Users
• Cosmo:
Does it do that one?
• Cosmo:
I think it does that one
• Cosmo:
Welcome back
• Num7:
Oh, welcome!
• Num7:
Titor is one and Titor is all.
• Cosmo:
Titor is the one true graviton which binds us all.
• Mylar:
Hi anyone saw this one with Tyson
?
• L LeoTCK:
Interesting theories, some of them. The rest is just fantasy or plain wrong. Also the thing about black hole because that assumes that black holes (as originally described) really exist. Rather than what I heard myself that the infinite mass thing is simply based on a mathematical error nobody seemed to challenge.
• Mylar:
Uhm ok I see
• Num7:
Titor bless you.
• Mylar:
I read this on a french YT channel about UFOs, that: Magnetic field + gamma rays can be used to create a circulating light beam that distorts or loops time, which can lead to a twisting of space and time. Looks like what R.Mallet working on it. What's your thoughts on this?
• Chat Bot:
Mylar has joined the room.
• Num7:
John, may You brighten this day and decorate it with everlasting happiness.
• B brunomazet:
num...
• B brunomazet:
when you read this check on paranormalis one of my old names and unban them... he banned the account with no reason
• Num7: | 667 | 2,776 | {"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-2023-40 | latest | en | 0.926387 |
http://mathhelpforum.com/geometry/122828-special-right-triangles.html | 1,540,247,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00331.warc.gz | 228,198,576 | 9,736 | 1. ## Special right triangles
I'm really confused with a word problem trying to find the sides of a triangle with just an altitude or a perimeter.
How would I found the length of a side of an equilateral triangle with an altitude of 24? I know altitude divides the triangle in half, so would it be a 45-45-90 triangle?
Also how would I find the altitude of an equilateral triangle with a perimeter of 45?
2. Originally Posted by Jubbly
I'm really confused with a word problem trying to find the sides of a triangle with just an altitude or a perimeter.
How would I found the length of a side of an equilateral triangle with an altitude of 24? I know altitude divides the triangle in half, so would it be a 45-45-90 triangle?
Also how would I find the altitude of an equilateral triangle with a perimeter of 45?
Hi Jubbly,
Actually, the altitude will divide the equilateral triangle into two 30-60-90 congruent right triangles.
You know in a 30-60-90 configuration, the altitude is equal to the short side (x) times the square root of 3.
$\displaystyle 24=x\sqrt{3}$
$\displaystyle x=\frac{24}{\sqrt{3}}=8\sqrt{3}$
The hypotenue is the length of the side of the equilateral triangle and is two times the short side.
Thus, the side of the equilateral triangle is $\displaystyle 16\sqrt{3}$
If the perimeter of an equilateral triangle is 45, then each side would be 15. This would mean that in your triangular configuration, 15 is the hypotenuse. The short side would be 7.5 (half the hypotenue) and the long side (altitude) would be $\displaystyle 7.5\sqrt{3}$ | 400 | 1,572 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-43 | latest | en | 0.900353 |
https://hpmuseum.org/forum/showthread.php?tid=11447&pid=104585&mode=threaded | 1,670,538,145,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00154.warc.gz | 339,472,853 | 6,138 | Summation on HP 42S
09-24-2018, 01:55 PM (This post was last modified: 09-24-2018 01:57 PM by pier4r.)
Post: #15
pier4r Senior Member Posts: 2,132 Joined: Nov 2014
RE: Summation on HP 42S
for sum(x^2 - 3*x)
one could actually do:
sum(x^2 ) (closed formula) minus
3*sum(x) (closed formula).
namely here: https://brilliant.org/wiki/sum-of-n-n2-or-n3/
PS: brilliant has so much potential that they don't use. So many interesting problems buried by a poor layout sometimes.
Wikis are great, Contribute :)
« Next Oldest | Next Newest »
Messages In This Thread Summation on HP 42S - lrdheat - 09-23-2018, 06:14 PM RE: Summation on HP 42S - Didier Lachieze - 09-23-2018, 06:28 PM RE: Summation on HP 42S - lrdheat - 09-23-2018, 06:41 PM RE: Summation on HP 42S - lrdheat - 09-23-2018, 06:51 PM RE: Summation on HP 42S - ijabbott - 09-23-2018, 07:40 PM RE: Summation on HP 42S - burkhard - 09-25-2018, 12:58 PM RE: Summation on HP 42S - John Keith - 09-25-2018, 04:15 PM RE: Summation on HP 42S - Thomas Klemm - 09-23-2018, 08:29 PM RE: Summation on HP 42S - lrdheat - 09-23-2018, 09:25 PM RE: Summation on HP 42S - Didier Lachieze - 09-23-2018, 10:03 PM RE: Summation on HP 42S - lrdheat - 09-23-2018, 10:02 PM RE: Summation on HP 42S - lrdheat - 09-23-2018, 10:37 PM RE: Summation on HP 42S - Thomas Klemm - 09-24-2018, 12:56 AM RE: Summation on HP 42S - Albert Chan - 09-24-2018, 11:56 AM RE: Summation on HP 42S - Thomas Klemm - 09-24-2018, 01:17 AM RE: Summation on HP 42S - Thomas Klemm - 09-24-2018, 01:52 PM RE: Summation on HP 42S - Albert Chan - 09-24-2018, 04:20 PM RE: Summation on HP 42S - pier4r - 09-24-2018 01:55 PM RE: Summation on HP 42S - Frido Bohn - 09-25-2018, 01:59 PM RE: Summation on HP 42S - Dieter - 09-25-2018, 04:47 PM RE: Summation on HP 42S - Frido Bohn - 09-26-2018, 10:10 AM RE: Summation on HP 42S - Albert Chan - 09-26-2018, 03:13 PM RE: Summation on HP 42S - Albert Chan - 09-25-2018, 05:28 PM RE: Summation on HP 42S - lrdheat - 09-25-2018, 05:17 PM RE: Summation on HP 42S - lrdheat - 09-25-2018, 05:28 PM RE: Summation on HP 42S - Albert Chan - 09-25-2018, 07:17 PM RE: Summation on HP 42S - Valentin Albillo - 09-25-2018, 08:01 PM RE: Summation on HP 42S - Albert Chan - 09-25-2018, 08:59 PM RE: Summation on HP 42S - Gerson W. Barbosa - 09-26-2018, 12:36 PM RE: Summation on HP 42S - Albert Chan - 09-26-2018, 01:23 PM RE: Summation on HP 42S - Gerson W. Barbosa - 09-26-2018, 01:54 PM RE: Summation on HP 42S - pier4r - 09-26-2018, 04:09 PM RE: Summation on HP 42S - Thomas Klemm - 09-27-2018, 03:47 AM RE: Summation on HP 42S - Ángel Martin - 09-27-2018, 05:27 AM RE: Summation on HP 42S - pier4r - 09-27-2018, 10:37 PM RE: Summation on HP 42S - Thomas Okken - 09-28-2018, 12:33 AM RE: Summation on HP 42S - pier4r - 09-28-2018, 01:40 PM RE: Summation on HP 42S - ijabbott - 09-28-2018, 11:20 PM RE: Summation on HP 42S - Albert Chan - 09-29-2018, 12:51 AM RE: Summation on HP 42S - Valentin Albillo - 09-29-2018, 01:15 AM RE: Summation on HP 42S - Thomas Okken - 09-29-2018, 01:24 AM RE: Summation on HP 42S - ijabbott - 09-29-2018, 11:58 AM RE: Summation on HP 42S - Albert Chan - 09-29-2018, 12:49 PM RE: Summation on HP 42S - Thomas Okken - 09-29-2018, 01:50 PM RE: Summation on HP 42S - ijabbott - 09-29-2018, 04:06 PM RE: Summation on HP 42S - Thomas Okken - 09-29-2018, 06:15 PM RE: Summation on HP 42S - ijabbott - 09-29-2018, 09:10 PM RE: Summation on HP 42S - Thomas Okken - 09-29-2018, 09:33 PM RE: Summation on HP 42S - Albert Chan - 09-30-2018, 03:03 PM RE: Summation on HP 42S - ijabbott - 09-30-2018, 05:28 PM RE: Summation on HP 42S - brickviking - 10-01-2018, 08:40 AM RE: Summation on HP 42S - pier4r - 09-29-2018, 05:59 PM RE: Summation on HP 42S - Thomas Okken - 09-29-2018, 08:59 PM
User(s) browsing this thread: 1 Guest(s) | 1,682 | 3,788 | {"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-2022-49 | latest | en | 0.611176 |
http://www.acemyhw.com/projects/39792/Mathematics/math-focus-set | 1,490,583,211,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189377.63/warc/CC-MAIN-20170322212949-00521-ip-10-233-31-227.ec2.internal.warc.gz | 402,764,120 | 16,989 | # Project #39792 - Math Focus Set
Â
Focus Set 3 – Fundamental Counting Principle
Â
Â
Â
Provide answers to # 1 - 10 below.
Â
Show how you set up your problem and then do the math.
Â
1. If there are four roads from town A to town B and three roads from town B to town C, how many routes are there from town A to town C which go through town B?
Â
Â
Â
Â
Â
2. Â How many different 4-letter radio station call codes are possible if each code must begin with K or W and no letter can be repeated? Â Examples:Â WABC, KABC, or WABK
Â
Â
Â
Â
Â
3. A firm wants to assign its employees ID numbers that have say x digits each. If a firm has 800 employees, what is the smallest number of digits (that is, the smallest x) that the firm can use for each ID number? Assume that repetition of digits is permitted. Â
(Hint: There is no formula for this – you just have to reason it out, but you must be enough ID's to go around!)
Â
Â
Â
Â
Â
4. In how many ways can 10 questions on a True-False test be answered?
Â
Â
Â
Â
Â
5. In how many ways can a president and vice president be selected from a club consisting of                    12 people?Â
(Assume that these 2 positions cannot be occupied by the same person.)
Â
Â
Â
Â
Â
6. In how many ways can 5 people be seated in a row?
Â
Â
Â
Â
Â
7. How many ways can 6 people be seated in a row, if Ruth must be seated in the first chair?
Â
Â
Â
Â
Â
8. How many distinct ordered arrangements can be made with the letters of the word TENNESSEE?
Â
Â
Â
Â
Â
9. You are trying to schedule classes for next semester and wish to enroll in Math, English, Fine Arts, and History.  You have found 4 Math sections, 5 English sections, 6 Fine Arts sections, and 2 History sections that do not have conflicts.
Â
How many different possible schedules can you develop using these options? Â
Â
Â
10. It is lunchtime and you wish to build a pizza with 1 choice of crust, 1 meat, 1 vegetable. Â
Â
You have the following choices:Â 2 types of crust, 4 types of meats, and 7 types of vegetables
Â
 How many different ways can you build your pizza?
Â
Â
Â
Focus Set 4 – Permutations and Combinations
Â
Â
Â
Use your calculator to solve the following problems:
Â
Â
Â
1.   12!
Â
       8!
Â
Â
Â
2.  P(5,3)
Â
3.  C(5.3)
Â
Â
Â
Â
Â
For # 4-8, show how you set up your problem and then do the math.
Use your calculators to find Combinations and Permutations.
Â
Â
Â
4.Ms. Lee’s class is having a spelling contest. There are 4 finalists in the contest.
Â
     How many different ways can the 4 students be seated on the stage?
Â
Â
Â
Â
Â
5. A Book Store wishes to display five books in the window of the shop.  They want to choose 5 mystery books from a list of 12 new mystery books.Â
Â
     How many ways can this be done? (order is not important)
Â
Â
Â
Â
Â
6. On an English test, Jeff must write an essay for three of six questions in Part 1, and two of five questions in Part 2. How many different combinations of questions can be chosen?
Â
Â
Â
7. A club is selecting a program committee. They want to select 5 people from their membership of 20 people.  How many ways can this committee be formed?
Â
Â
Â
Â
Â
8.  A committee at of 5 is to be formed from a group consisting of 8 Democrats, 7 Republicans, and 2 Independents. Note: A committee is an unordered group.
Â
Â
Â
(a) How many committee choices are possible if the committee is to consist entirely of Democrats?
Â
Â
Â
(b) How many committee choices are possible if the committee is to consist of 2 Democrats, 2 Republicans, and 1 Independent?
Â
Â
Â
(c)Â How many committee choices are possible if the committee is chosen from all 17 people with no restrictions?
Â
Â
Â
Â
Focus Set 5 – Introduction to Probability and Odds
Â
Â
Â
Write probabilities as fractions reduced to lowest terms or decimals rounded to 3 decimal places.
Â
Â
Â
1. If you write the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 on slips of paper and
Â
place them in a hat, then you draw one slip of paper.Â
(Note: There are 10 slips of paper.)
Â
What is the probability of dawning the following:
Â
Â
Â
a)Â Â Â Â Â Â Â Â a number greater than 7? P(x > 7)
Â
Â
Â
b)        a number less greater than 4 and less than 8? P(x>4 and x<8)
Â
Â
Â
2. A single card is dealt from a standard deck of 52 cards. What is the probability that:
Â
Â
Â
(a) Â Â Â Â Â the card is a club? Â
Â
(b)Â Â Â Â Â Â the card is a picture card?
(Picture cards have faces. Aces are not picture cards.)
Â
     A      B      C     D     F Freshmen     2      5      10     6     5 Sophomores     3      6      9     2     2
Â
Â
Â
3. The grade distribution for a particular finite math class (not mine) populated by freshmen and sophomores is as shown in the table.
Â
Suppose we randomly select a student from this class.
Â
Â
Â
(a) What is the probability that the student made a D?
Â
Â
Â
(b) What is the probability that the student is a freshman?
Â
Â
Â
Odds should be written as ratios.
Â
Probabilities should be as fractions or decimals.
Â
Fraction must be in lowest terms.
Â
Decimals must be rounded to 3 decimal places.
Â
Â
Â
4. A single card is selected from a standard deck of 52 cards.
Â
Â
Â
a)Â Â Â Â Â Â Â Â What are the odds in favor of selecting an ace?
Â
Â
Â
Â
Â
b) Â Â Â Â Â Â Â What are the odds againstselecting a heart?
Â
Â
Â
5. If the odds against Andre winning his tennis match are 4:5, then what is the probability that Andre will win his tennis match?
Â
Â
Â
6. If the odds in favor of an event are 2:5, then what is the probability that the event will occur?
Â
Â
Â
Â
Â
Â
Â
Â
Â
Focus Set 6 – AND and OR Probability
Â
Â
Â
Write probabilities as fractions or decimals.
All fractions MUST be reduced to lowest terms.
All decimal MUST be rounded to 3 decimal places.
Â
Â
Â
Â
Â
1. A single card is dealt from a standard deck of 52 cards. What is the probability that:
Â
Â
Â
(a) the card is a 5 and club ?
Â
Â
Â
(b) the card is a diamond or King?
Â
Â
Â
(c) the card is a 5 or a 2?
Â
Â
Â
(d)Â the card is a 5 and a 2?
Â
Â
Â
     A      B      C     D     F Freshmen     2      5      10     6     5 Sophomores     3      6      9     2     2
Â
(e) the card is a diamond or a picture card?
Â
Â
Â
(f)Â the card is a diamond and a picture card?
Â
Â
Â
Â
Â
2. The grade distribution for a particular finite math class (not mine) populated by freshmen and sophomores is as shown in the table.
Â
Suppose we randomly select a student from this class.
Â
Â
Â
(a) What is the probability that the student is a freshman and made a D?
Â
(b)
What is the probability that the student is a freshman or made a D?
Â
Â
Â
(c)Â Â What is the probability that the student is a sophomore and made an A?
Â
Â
Â
(d)Â What is the probability that the student made a B or a freshman?
Â
Â
Â
Â
Focus Set 7 – Conditional Probability & Expected Value
Â
Â
Â
Instructions:Â TYPE you answers after the problems.
Write probabilities as fractions or decimals.
All fractions MUST be reduced to lowest terms.
All decimal MUST be rounded to 3 decimal places.
Â
Â
Â
1. One card is drawn from a standard deck of cards. What is the probability that card is a Queen, given that the card is black?
Â
Â
2. One card is drawn from a standard deck of cards. What is the probability that card is a club, given that the card is black?
Â
Â
Â
3. Two cards are drawn from a standard deck of cards with replacement.
What is the probability that the first is a King and the second is not a King?
Â
Â
Â
4. Two cards are drawn from a standard deck of cards without replacement. What is the probability that the first is a King and the second is not a King?
Â
Â
Â
5. The grade distribution for a particular finite math class having only freshmen and sophomores is shown in the table. Â
Â
Â
Â
 A B C D F Totals Freshmen 2 5 10 6 5  Sophomores 3 6 9 2 2  Totals      Â
Â
Â
Â
a)Â Â Â Â Â Â Â Â What is the probability that the student made an C?
Â
b)Â Â Â Â Â Â Â What is the probability that the student is a sophomore?
Â
Â
Â
c)        What is the probability that the student made a C, given that the student is a sophomore?  This is  P( C | sophomore )
Â
Â
Â
d) Â Â Â Â Â Â Â What is the probability that the student is a freshman, given that the student
Â
           made a B?  This is  P( Freshman | B)
Â
6. An insurance company is going to sell 1-year life insurance policies with a face value of \$40,000 to 25 year old men for \$5000. Their mortality tables show that such men will live for 1 year with probability 0.90. Â Â
Find the company's expected earnings per policy.
Â
7. An outdoor spring festival is planed. The weatherman is predicting a 20% chance of rain, 30% chance of a cloudy day, and a 50% chance of sunshine. The committee feels that if it rains, then only 500 people will attend, if it is cloudy, then only 10000 will attend, and if the sunny, then 15000 will attend.  How many people can be expected to attend the spring festival?
Â
Â
Focus Set 8 - Statistics
Round all statistics to 3 decimal places.
1. Draw a Stem and Leaf chart for this data: Â 45, 49, 49, 49, 50, 55, 58, 60, 62, 68
 Use the key 4 | 5 represents 45.
Draw the stem and leaf:Â
(Use the vertical bar above the Enter key to make the vertical lines)
Â
Â
Â
2. Which histogram best fits the following data: (Answer is just A, B, or C.)
Data Frequency
1Â 3
2Â 5
3Â 8
4Â 4
Â
AÂ Â Â BÂ Â CÂ
3. a)Â Find the mid-point of the second class.
   b) Find the total number of pieces of data.
Data Frequency
15-18Â 6
19-22Â 10
23-26Â 12
27-30Â 4
4. Create a frequency chart with classes of width 5 beginning with 55 for the following data:
79Â 62Â 87Â 84Â 55Â 76Â 67Â 73Â 82Â 68Â
82Â 76Â 56Â 79Â 61Â 64Â 64Â 67Â 79Â 68
 I have created a table where you can insert your classes and frequencies by just clicking and typing. If you need more boxes just press TAB to get more lines.Â
Data Â
Classes Frequency
55 -Â Â
Â
Â
Â
Â
Focus Set 9 – Statistics
Â
Â
Â
 Round all statistics to 3 decimal places.
Â
Â
Â
1. Find the following statistics using this data:
Â
Â
Â
  152                162                174                179                185                185   Â
Â
  152                163                176                180                185                192   Â
Â
  156                165                176                183                185                192   Â
Â
Â
Â
mean =Â Â
Â
median =Â
Â
mode =Â Â
Â
midrange =Â
Â
standard deviation =Â Â
Â
range =Â Â
Â
Â
Â
Â
Â
Â
Â
 2.      Find the following statistics using this data:
Â
  53       63       75       80       82       92
Â
  56       68       75       82       82       92
Â
57Â Â Â Â Â Â Â 70Â Â Â Â Â Â Â 77Â Â Â Â Â Â Â 82Â Â Â Â Â Â Â 85Â Â Â Â Â Â Â 95
Â
60Â Â Â Â Â Â Â 71Â Â Â Â Â Â Â 79Â Â Â Â Â Â Â 82Â Â Â Â Â Â Â 90Â Â Â Â Â Â Â 97
Â
mean =
median =
Â
mode =
Â
midrange =
Â
standard deviation =
Â
range =
Â
Â
Focus Set 10 – Normal Curves
Â
You must put the percent symbol on percentages.
Â
Round decimals to 3 decimal places before changing decimals to percents.
Â
Ex.  0.3456345 = 0.346 = 34.6%
Â
Â
Â
1. On a particular section of a CPA exam for which the scores were normally distributed with a mean of 700 points and a standard deviation of 70 points. Â Â
Â
(a)Â What percent of the students scored lower than 640 on the exam?
Â
 (b)
 What percent of the students scored between 630 and 730 on the exam?
Â
 (c)
 What percent of the students scored between 600 and 700 on the exam?
Â
(d)Â
What percent of the students scored above 700 on the exam?
Â
(e)Â
 If 900 people take the exam, how many people will score above 750 on the exam?
Â
Â
Â
2. Â The weights of boxes of Brand Z cereal were found to be normally distributed with a mean of 22.0 ounces and a standard deviation of 0.6 ounces.
Â
(a) What percentage of the boxes will weigh more than 21.7 ounces?
Â
Â
Â
(b) What percentage of the boxes will weigh between 21.7 and 22.3 ounces?
Â
Â
Â
(c)Â What percentage of the boxes will weigh less than 22.5 ounces?
Â
Â
Â
(d)Â What percentage of the boxes will weigh more than 23 ounces?
          Â
Â
Â
(e) If a store has 500 boxes of cereal, how manyof them will weigh more than 23 ounces?
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Â
Subject Mathematics Due By (Pacific Time) 09/16/2014 12:00 pm
TutorRating
pallavi
Chat Now!
out of 1971 reviews
amosmm
Chat Now!
out of 766 reviews
PhyzKyd
Chat Now!
out of 1164 reviews
rajdeep77
Chat Now!
out of 721 reviews
sctys
Chat Now!
out of 1600 reviews
Chat Now!
out of 770 reviews
topnotcher
Chat Now!
out of 766 reviews
XXXIAO
Chat Now!
out of 680 reviews | 5,177 | 13,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-13 | longest | en | 0.918749 |
https://www.supermoney.com/encyclopedia/future-value/ | 1,686,241,656,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00585.warc.gz | 1,054,313,753 | 28,056 | # Future Value: Definition, Formula, How to calculate, Examples
Summary
Future value is an important idea in personal finance and investment that helps people comprehend how much their money will be worth in the future. The idea is based on the time value of money, which holds that money is worth more now than it will be later.
The future value formula considers the present value of money, the interest rate, and the time period over which the investment will receive interest. Understanding future value can help you make smarter decisions about your investment options and estimate the possible returns on your assets.
## What is future value?
Future value is the amount of money an investment will be worth after a specific period of time if permitted to grow at a certain interest rate. It is an important idea to comprehend in order to make informed decisions regarding saving, investing, and retirement planning.
Future value is an important estimate in financial planning because it enables individuals and corporations to make current decisions based on future estimates.
Individuals can calculate the future worth of an investment to determine how much money they need to save or invest today in order to achieve their desired financial goals in the future.
Future value formula and computation
The future value formula is a mathematical equation used to compute the future worth of an investment or cash flow based on its present value, interest rate, and holding period.
The future value formula can be expressed in several different ways, but the most common form is:
FV = PV x (1 + r)n
Where:
FV = Future Value
PV = Present Value
r = Interest Rate
n = Number of Periods
To use the formula, the present value of the investment or cash flow is multiplied by the sum of 1 plus the interest rate, raised to the power of the number of periods the investment is held. The result is the estimated future value of the investment.
For example, suppose an individual invests \$1,000 in a savings account that pays an annual interest rate of 5%, compounded annually, for five years. Using the future value formula, the future value of the investment can be calculated as follows:
FV = \$1,000 x (1 + 0.05)5
FV = \$1,276.28
This means that if the interest rate remains constant and the investment is neither taken or reinvested, the investment will be worth \$1,276.28 after five years.
The future value formula can be used to any investment or cash flow, including stocks, bonds, mutual funds, and other financial instruments.
It is an important financial planning tool because it helps individuals and businesses to predict the future worth of their investments and make informed decisions about how to deploy their resources over time.
## Examples of future value
• Investing in a savings account: Assume a person deposits \$5,000 into a savings account with a 2% annual interest rate that is compounded monthly.
After ten years, the investment’s future worth can be computed as follows: FV = \$5,000 x (1 + 0.02/12)^(12×10)
FV = \$5,613.53
• Investing in a mutual fund: Assume an individual invests \$10,000 in a mutual fund with a 20-year compounded annual return of 7%.
After 20 years, the investment’s future worth can be computed as follows: FV = \$10,000 x (1 + 0.07)^20
• Investing in a stock portfolio: Suppose an individual invests \$50,000 in a diversified stock portfolio that has an average annual return of 10%, compounded annually, for 30 years. After 30 years, the future value of the investment can be calculated as follows:
FV = \$50,000 x (1 + 0.10)^30
FV = \$1,368,363.81
These examples demonstrate how the future value formula can be used to estimate the potential return on different types of investments over time. It is important to note that the actual return on an investment may differ from the estimated future value due to changes in interest rates, market conditions, and other factors.
## Factors affecting future value
• Interest rate: The interest rate is a key factor that affects the future value of an investment. A higher interest rate will result in a higher future value, while a lower interest rate will result in a lower future value.
• Compounding frequency: The frequency at which interest is compounded can also affect the future value of an investment. More frequent compounding, such as daily or monthly, will result in a higher future value than less frequent compounding, such as annually.
• Time horizon: The length of time an investment is held can significantly affect its future value. The longer the time horizon, the more time the investment has to grow, resulting in a higher future value.
Investment amount: The amount of money invested also affects the future value of an investment. A larger investment will result in a higher future value, while a smaller investment will result in a lower future value.
• Inflation: Inflation can reduce the purchasing power of an investment over time, which can affect its future value. It is important to account for inflation when calculating the future value of an investment.
• Investment risk: The risk associated with an investment can also affect its future value. Investments with higher risk may have the potential for higher returns, but also carry a higher risk of loss. Lower-risk investments may have lower returns, but also carry a lower risk of loss.
• Fees and taxes: Fees and taxes associated with an investment can also affect its future value. Higher fees and taxes can reduce the overall return on an investment, resulting in a lower future value.
## Key takeaways
• Future Value (FV) is the value of an investment at a specified point in the future, based on a predetermined interest rate.
• The formula to calculate FV is FV = PV x (1 + r)^n, where PV is the present value of the investment, r is the interest rate, and n is the number of compounding periods.
• FV can be calculated for any type of investment, such as bonds, stocks, or mutual funds, and can help individuals determine the potential growth of their investments over time.
• Compounding is a critical factor in calculating FV, as it is the process of reinvesting the interest earned on an investment back into the investment.
• To maximize FV, individuals can consider investing in high-yield investment vehicles, such as stocks or mutual funds, and increase the frequency of compounding by reinvesting dividends or interest earned.
• Inflation is an important consideration when calculating FV, as it can erode the purchasing power of the investment over time. It is important to factor in inflation when projecting future values.
###### View Article Sources
1. A Journey Through Time – American Journal of Business Education
2. Figuring Out Credit Card Formulas and Why it May Help — SuperMoney
3. How To Calculate Interest on Savings Account Funds — SuperMoney
4. APR vs. APY: What’s the Difference? — SuperMoney | 1,456 | 6,927 | {"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-2023-23 | latest | en | 0.917727 |
https://math.stackexchange.com/questions/1963967/distribution-of-minimum-of-2-inid-random-variables-when-one-is-transformed | 1,713,595,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00189.warc.gz | 345,759,035 | 32,823 | Distribution of minimum of 2 INID random variables when one is transformed.
Suppose that I observe $z_i=Min[\frac{n-1}{n}v_i,w_i]$ where $v_i\sim U[0,\bar{v}]$ and $w_i\sim U[0,\lambda \bar{v}]$. For all intents and purposes in what I am doing I assume: $\bar{v}>0, \lambda>1,n>=2$.
How could I derive the PDF and CDF of $z_i$ ?
I know how to accomplish this with order statistic when there is no transformation inside the min function, but am getting lost when I have to transform one of the random variables by a function.
Many thanks! | 155 | 541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.847918 |
http://www.baltimoresun.com/entertainment/la-sci-sn-pi-day-tau-20140314-story.html | 1,529,660,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00396.warc.gz | 375,410,366 | 51,033 | # Today is Pi Day. Detractors say pi really is half-baked
Today is Pi Day. Math nerds and geometry aficionados will be celebrating the mathematical constant that describes the relationship between the circumference of a circle and its diameter.
For those who need a refresher: Take a circle that is 1 inch across. Then measure the length of the circle itself. When you’ve gone all the way around, you’ll have covered a distance of slightly more than 3.14 inches. That’s pi. (And that’s why Pi Day is observed on March 14.)
Pi (represented by the Greek letter Π) has many uses. In addition to using it to calculate the circumference of a circle (2Πr, where “r” is the circle’s radius), it will also tell you the area (Πr2).
But that’s just the beginning. You can use it to calculate the surface area and volume of a sphere. You need it for trigonometric functions that repeat periodically, like sine and cosine. And it is essential for calculating the normal distribution for probability and statistics.
Pi hides out in the famous fractal known as the Mandelbrot set. It also appears in the equation for Heisenberg’s uncertainty principle and Einstein’s field equations. On a more practical level, pi is necessary to determine your location with GPS.
In short, pi has been helping humans do cool stuff for about 4,000 years. But despite this illustrious history, pi is finding itself under attack by people who think it is only half as awesome as it ought to be. These folks are pushing to have pi replaced by the Greek letter tau, which has a value of 2Π.
As Victoria Hart explains in the video above, the thing that determines the circumference of a circle is its radius, not its diameter. Accordingly, she asks, “Why would we define the circle constant as a ratio of the diameter to the circumference?”
For Hart (a musician and math enthusiast who has made videos for Khan Academy), this problem is an offense to math itself. “Mathematics should be as elegant and beautiful as possible,” she says. “There’s a boatload of important equations and identities where 2Π shows up, which could and should be simplified to tau.”
In Hart’s view, pi is simply an unfortunate anachronism that has outlived its usefulness:
“Pi, as a concept, is a terrible mistake that has gone uncorrected for thousands of years. The problem with pi and Pi Day is the same as the problem with Christopher Columbus and Columbus Day. Sure, Christopher Columbus was a real person who did some stuff, but everything you learn about him in school is warped and overemphasized. He didn’t discover America. He didn’t discover the world was round. And he was a bit of a jerk. So why do we celebrate Columbus Day? Same with pi.”
Hart is not alone in her enthusiasm for tau. In this video, you can watch Steve Mould and Matt Parker (two-thirds of the trio behind the Festival of the Spoken Nerd and maybe rough British equivalents to America’s Bill Nye) go toe-to-toe in a tau-versus-pi smackdown.
Mould, who conducts science experiments for the BBC, sums up many of his objections to pi by saying, “There’s always this factor of two that you’ve got to keep in your mind when you’re doing calculations.”
Parker, who holds the title of London Mathematical Society Popular Lecturer, counters that “pi works just as well [as tau] in all cases. ... The fact that it’s got a bit of heritage doesn’t mean we should keep it around necessarily. But it’s worked for a very long time, it’s there for very good reasons,” and if you switch from pi to tau, “you gain as much as you lose.”
The final score is 23-27, but you’ll have to watch the video to see which constant comes out on top.
If you’re ready for a deep dive on the subject, check out the Tau Manifesto by Michael Hartl, or watch him give a lecture at Caltech. (It was delivered on June 28 — a.k.a. Tau Day — in 2011.)
Even if you’re open to the idea that tau is superior to pi, you can still celebrate Pi Day. Or as Hart puts it: “You can have your pi and eat it.”
karen.kaplan@latimes.com
MORE FORM SCIENCE | 924 | 4,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-26 | latest | en | 0.948362 |
https://www.coursehero.com/file/8653086/practice-exam-c/ | 1,527,102,131,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00568.warc.gz | 719,092,390 | 111,637 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
practice exam c
# practice exam c - Seat Math 116 CALCULUS II PRACTICE EXAM...
This preview shows pages 1–5. Sign up to view the full content.
Math 116 CALCULUS II PRACTICE EXAM – FINAL C (For In-Class) December 5 (Thu), 2013 Seat # : Instructor: Yasuyuki Kachi Line #: 23590. ID # : Name : This is ‘Version C’ of the practice exam. (There are ‘Versions A,B’.) The actual exam may not be very similar to these practice exams. The purpose of these practice exams is to give you an idea of how the actual exam will look like, in terms of the length and the format. This practice exam is for the “in-class” portion of the exam only. [I] (40pts) Give the Taylor series expression for arctan x : arctan x = = s n = ( 1 ) x . The radius of convergence R = . 1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Line #: 23590. ID # : Name : ([I] continued) (2a) Find the derivative: d dx p arctan x P = . (2b) Evaluate: i + x =0 1 x 2 + 1 dx = b B + x =0 = = . 2
Line #: 23590. ID # : Name : [II] (40pts) (1) Complete Machin’s formula: π = s n = p 1 P = . (2a) Complete BBP formula: π = s n = ± ² n . = . 3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Line #: 23590. ID # : Name : ([II] continued) (2b) Use calculator to give a ballpark estimate of the Frst three terms of the BBP formula in decimals. The required degree of accuracy is up to the given number
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 669 | 2,559 | {"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-2018-22 | latest | en | 0.849203 |
http://laserpointerforums.com/f51/will-445nm-1w-laser-curcuit-work-how-well-please-read-description-77711.html | 1,503,373,605,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00406.warc.gz | 233,632,519 | 15,113 | Laser pointer discussion. Read/write reviews of laser pointers and laser pointer companies. Learn about all types of laser pointers and lasers
Welcome to Laser Pointer Forums! If you are looking for a laser pointer or want to compare different laser pointer companies, you may want to check out the LPF Laser Pointer Company Database. The link will open in a new window for your convenience.
One laser store meets all your needs
Laser Pointer Forums - Discuss Laser Pointers Will this 445nm 1w laser curcuit work? and how well? please read description.
LPF List of Laser Pointer Companies (link opens new window)
10-04-2012, 04:51 AM #1
Junior Member Join Date: Oct 2012 Posts: 11 Rep Power: 0
Junior Member
Join Date: Oct 2012
Posts: 11
Rep Power: 0
Will this 445nm 1w laser curcuit work? and how well? please read description.
so i want to make a simple laser, but this time a good one... i have made a simple laser before however used a diode of 5mw. i want to make it 1w. what i mean by simple circuit is diode to circuit, circuit to switch, switch to battery (9 volts), battery to diode. hopefully i didn't just confuse you... anyways, my 5mw is still up and running after many months but i want to know if it will still last long as 1w and how long. if you can please include which driver is should use for a project like this and why. thank you for all reply's i get in advance.
LPF List of Laser Pointer Companies (link opens new window)
10-04-2012, 03:23 PM #2
Class 4 Laser Join Date: Sep 2007 Location: Wisconsin Posts: 9,144 Rep Power: 9321
Cyparagon
Class 4 Laser
Join Date: Sep 2007
Location: Wisconsin
Posts: 9,144
Rep Power: 9321
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
Quote:
Originally Posted by ibizlink what i mean by simple circuit is diode to circuit
That's an infinite recursive loop. You can't use the word in the definition OF that word.
__________________
A problem well stated is a problem half solved.
10-04-2012, 05:28 PM #3
Class 1M Laser Join Date: Sep 2011 Location: Vancouver, Canada Posts: 235 Rep Power: 15
Apocalypse
Class 1M Laser
Join Date: Sep 2011
Posts: 235
Rep Power: 15
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
I think he means driving the diode directly with the battery and circuit means wire...?
Last edited by Apocalypse; 10-04-2012 at 05:30 PM.
10-04-2012, 07:18 PM #4
Class 4 Laser Join Date: Sep 2008 Location: Quebec, Canada Posts: 15,793 Rep Power: 43174
lasersbee
Class 4 Laser
Join Date: Sep 2008
Posts: 15,793
Rep Power: 43174
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
Quote:
Originally Posted by ibizlink so i want to make a simple laser, but this time a good one... i have made a simple laser before however used a diode of 5mw. i want to make it 1w. what i mean by simple circuit is diode to circuit, circuit to switch, switch to battery (9 volts), battery to diode. hopefully i didn't just confuse you... anyways, my 5mw is still up and running after many months but i want to know if it will still last long as 1w and how long. if you can please include which driver is should use for a project like this and why. thank you for all reply's i get in advance.
You will need to do some research and reading on Laser Diode
Drivers if you don't want to spend all your money replacing DEAD
Laser Diodes.
Jerry
You can contact us at any time on our Website: J.BAUER Electronics
__________________
LaserBee Laser Power Meter Products meet your needs
at affordable Prices:
See them all here on LPF
LaserBee Power Meter products ALWAYS in Stock
Also available on eBay:Check availability here..
Subsidary: Pharma Electronic Solutions
This banner is available to and can be copied/used FREE by any LaserBee owner
Last edited by lasersbee; 10-04-2012 at 07:21 PM.
10-05-2012, 01:37 AM #5
Junior Member Join Date: Oct 2012 Posts: 11 Rep Power: 0
Junior Member
Join Date: Oct 2012
Posts: 11
Rep Power: 0
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
Thanks for the replies. So anyways I really want to build a laser gun, link at bottom and I just posted his simple description of how he did it... I'm new to lasers but I get it now. But help me build this guys please? I am on a mobile phone and so it's on mobile so I don't know how it'll work on a computer. Just search laser gun on YouTube and its a picture of an airsoft gun shooting a blue laser. Second video on my device.
Link: YouTube - Real Life Laser Gun! Airsoft Gun Hacked Into Laser Blaster!
10-05-2012, 03:50 AM #6
Class 4 Laser Join Date: Sep 2007 Location: Wisconsin Posts: 9,144 Rep Power: 9321
Cyparagon
Class 4 Laser
Join Date: Sep 2007
Location: Wisconsin
Posts: 9,144
Rep Power: 9321
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
Your thread is titled "Will this 445nm 1w laser circuit work?" But you haven't yet described the circuit. What are we supposed to say?
__________________
A problem well stated is a problem half solved.
10-05-2012, 04:06 AM #7
Junior Member Join Date: Oct 2012 Posts: 11 Rep Power: 0
Junior Member
Join Date: Oct 2012
Posts: 11
Rep Power: 0
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
I mean this. (sorry for the confusion) i connect the laser diode to the driver, driver to momentary switch, momentary switch to battery(Li-ion recharhavle batteries), battery to laser diode, thus closing and finishing the circuit. Sorry. I just realized I put circuit instead if driver. So help me make it please?
10-05-2012, 04:20 AM #8
Class 2 Laser Join Date: May 2012 Location: Northern CA Posts: 301 Rep Power: 17
Lethere Belight
Class 2 Laser
Join Date: May 2012
Location: Northern CA
Posts: 301
Rep Power: 17
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
10-05-2012, 11:07 AM #9
Class 4 Laser Join Date: Sep 2008 Location: Quebec, Canada Posts: 15,793 Rep Power: 43174
lasersbee
Class 4 Laser
Join Date: Sep 2008
Posts: 15,793
Rep Power: 43174
Re: Will this 445nm 1w laser curcuit work? and how well? please read description.
Quote:
Originally Posted by ibizlink Thanks for the replies. So anyways I really want to build a laser gun, link at bottom and I just posted his simple description of how he did it... I'm new to lasers but I get it now. But help me build this guys please? I am on a mobile phone and so it's on mobile so I don't know how it'll work on a computer. Just search laser gun on YouTube and its a picture of an airsoft gun shooting a blue laser. Second video on my device. Link: YouTube - Real Life Laser Gun! Airsoft Gun Hacked Into Laser Blaster!
Quote:
Originally Posted by ibizlink I mean this. (sorry for the confusion) i connect the laser diode to the driver, driver to momentary switch, momentary switch to battery(Li-ion recharhavle batteries), battery to laser diode, thus closing and finishing the circuit. Sorry. I just realized I put circuit instead if driver. So help me make it please?
The YouTube video belongs to one of our members (styropro).
You are basically correct in your method of connecting the LD,
Driver, Switch and Battery if the driver is set to the correct
current for the LD you are using and the Batteries are the
correct Voltage and Amperage.
Each Laser/Driver/SW/Battery combination is different.....
Jerry
You can contact us at any time on our Website: J.BAUER Electronics
__________________
LaserBee Laser Power Meter Products meet your needs
at affordable Prices:
See them all here on LPF
LaserBee Power Meter products ALWAYS in Stock
Also available on eBay:Check availability here..
Subsidary: Pharma Electronic Solutions
This banner is available to and can be copied/used FREE by any LaserBee owner
Last edited by lasersbee; 10-05-2012 at 11:09 AM.
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules
All times are GMT. The time now is 03:46 AM.
-- DarkShadows V5 -- Responsive LPF -2562016 -- Default Style Contact Us - Laser Pointer Forums - Archive - Top | 2,235 | 8,331 | {"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-2017-34 | latest | en | 0.913416 |
https://www.futurelearn.com/courses/maths-power-laws/1/steps/86517 | 1,544,967,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00572.warc.gz | 905,026,053 | 25,059 | 1.4
## UNSW Sydney
A classical property of a hyperbola
# Classical properties of hyperbolas
Hyperbolas are special kinds of conic sections and were studied extensively by Apollonius of Perga around 200 B.C. In some ways they are analogous to ellipses, but they also have some remarkable additional features.
In this step we will
• learn about the asymptotes, tangent lines and reflective properties of hyperbolas
• discuss the focus and directrix definition of a hyperbola
## Hyperbolas, cones and asymptotes
Apollonius thought of hyperbolas as being slices of a cone, extending indefinitely in both directions, with a plane that cuts both parts of the cone. One of the key properties of such a hyperbola is the existence of two asymptotes: lines which the hyperbola does not meet but which it approaches closer and closer as we move further outwards in the plane.
Geometrically, the asymptotes are the lines which form the boundary of the perpendicular projection of the cone onto the plane. Neither ellipses nor parabolas have asymptotes, and this makes the geometry of the hyperbola somewhat special. The hyperbola also has a centre which is a point of symmetry. The centre is the meet of the two asymptotes.
## The eccentricity of a hyperbola
A hyperbola may also be defined in terms of a focus $\normalsize{F}$ and a directrix $\normalsize{l}$, such that for any point $\normalsize X$ on the hyperbola,
for some fixed constant $\normalsize{e}$ called the eccentricity of the hyperbola. This condition can also be restated in terms of quadrances, which makes it more algebraic and suitable for calculation:
We remind you that the quadrance between points is just the square of the distance between those points, and that the quadrance from a point $\normalsize{X}$ to a line $\normalsize{l}$ means the quadrance from $\normalsize{X}$ to the closest point $\normalsize{P}$ on the line. Quadrance is often easier to work with than distance because no messy square roots are involved, and it also forms the basis of the new theory of Rational Trigonometry, which is very exciting.
## Calculating the equation of a hyperbola
Let’s use the above characterisation of a hyperbola with focus $\normalsize{F=[2,0]}$, directrix the line $\normalsize{x=1}$ and eccentricity $\normalsize{e=2}$ to find its equation. Using the quadrance form, if $\normalsize{X=[x,y]}$ is an arbitrary point on the hyperbola, then
and
so the equation is
We rewrite this as
which simplifies (please check the algebra!) to
Q1 (E): Find the equation of the hyperbola with focus the point $\normalsize{F=[0,2]},$ directrix the line $\normalsize{y=1}$, and eccentricity $\normalsize{e=2}$.
In fact there are always two pairs of foci and directrices for a hyperbola. We can see this since the reflection in the centre of the hyperbola will take one pair to the other.
Q2 (M): What is the centre of the hyperbola that you found in Q1?
Q3 (C): What is the other focus, and what is the other directrix of the hyperbola in Q1?
## Reflective property
Another property of the hyperbola is that if a light ray is emitted from one focus $\normalsize{F_1}$, then it reflects off the hyperbola along a line which passes through the second focus $\normalsize{F_2}$.
This property also holds for an ellipse, but in the case of the hyperbola the reflected ray moves away from the second focus while for an ellipse it moves towards it.
For an ellipse, this property is responsible for the interesting phenomenon of whispering rooms: if a room has a ceiling with an ellipsoidal shape, then someone standing at one focus can whisper to a friend standing at a second focus and they can hear, since the sound waves bounce off the ceiling and all converge to the other focus. A famous example occurs in St. Paul’s cathedral in London.
St Paul’s Cathedral Whispering Gallery DeZeeuw from nl GFDL or CC-BY-SA-3.0, via Wikimedia Commons
## Tangent property of a hyperbola
A property of the hyperbola discovered by Apollonius states that for a line tangent to the hyperbola at a point $\normalsize{X}$ and intersecting the asymptotes at points $\normalsize{P}$ and $\normalsize{R}$, the product $\normalsize{\vert O,P\vert \times \vert O,R\vert}$ is constant, and $\normalsize{\vert P,X\vert=\vert R, X\vert}$.
Of course it is much more pleasant if we restate these conditions rationally as
and
Q4 (C): Consider the initial image for this step. It also shows a property of the hyperbola. Can you deduce what it is?
## Answers
A1. Using the quadrance form, if $\normalsize{X=[x,y]}$ is an arbitrary point on the hyperbola, then
and
so the equation is
We rewrite this as
which simplifies to
A2. Let’s complete the square to rewrite the hyperbola $\normalsize{x^2-3y^2+4y=0}$ as
Since the centre of the hyperbola $\normalsize{x^2-3y^2=-\frac{4}{3}}$ is the origin $\normalsize{[0, 0]}$, the centre of the hyperbola $\normalsize{x^2-3y^2+4y=0}$ will be the point $\normalsize{[0, 2/3]}$.
A3. To get the second focus and directrix, we reflect the first focus and directrix in the centre $\normalsize{[0, 2/3]}$. The second focus will be the point $\normalsize{[0, -2/3]}$, and the second directrix will be the line $\normalsize{y=1/3}$.
A4. The initial image for this step shows conjugate diameters for a hyperbola. That is something we discussed in our earlier course Maths for Humans: Linear and Quadratic Relations. The idea is that if you look at chords parallel to a given diameter, that is a chord through the centre, and take midpoints of those chords, then you get another diameter. And if you apply the same trick to this new diameter, you get back the diameter you started with. | 1,474 | 5,672 | {"found_math": true, "script_math_tex": 31, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-51 | longest | en | 0.933845 |
http://www.answers.com/Q/Does_the_earth_revolve_above_or_under_the_sun | 1,547,776,948,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118030858-00073.warc.gz | 263,356,837 | 50,997 | Does the earth revolve above or under the sun?
Would you like to merge this question into it?
already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
exists and is an alternate of .
Above and below are concepts which don't really apply when you are not standing on the surface of the Earth. We define 'down' as towards the centre of the Earth.
There is no such thing as up and down in space, so it just revolves around the sun.
5 people found this useful
How does the earth revolve around the sun?
The Earth revolves around the Sun in a slightly elliptical orbit,moving in a counter-clockwise direction when viewed from thecelestial north it also revolves on its axis.
Does the sun revolve around the earth?
No, the Earth revolves around the Sun every 365 days. While the Earth is revolving around the Sun it also is spinning and spins once every 24 hours, that is how we get a day.
It doesn't!
Yes
Why does the sun revolve around the earth?
The earth revolves around the sun.
Yes it do.
Why sun revolves around the earth?
The sun doesn't revolve around the earth the Sun is for all intents and purposes stationary 1 . The Earth Rotates around the sun as do all the other planets. --------------
Why do the earth revolve around the sun?
A combination of inertia and gravity.
Why sun not revolving around the earth?
Astronomically, you only get smaller masses orbiting larger ones - try imagining a tennis ball on the end of a string, with you swinging the tennis ball around. Now imagine
Why is earth revolves around the sun?
The earth is smaller than the sun.
no because we do
Which direction does earth revolve around the sun when viewed from above?
The question realized, correctly, that a vantage point has to be specified, because different observers in different places will interpret revolution to be taking place in
In The Moon
Why is earth revolving around the sun?
The earth is revolving around the sun because God made it that way! Also, because we need sunlight! Or we would die.... :O
In Planetary Science
Do the earth revolves the sun?
yes. The earth revolves around the sun.
In Planet Earth
Why earth revolves round the the sun?
the earth revolves around the sun because of two main components in space. Inertia and gravitational pull. The sun has a gravitational pull on all of the planets but to keep t
In The Moon
Why does not the sun revolve around the earth?
Because the sun is way bigger than the earth. And the sun's gravitational pull is way stronger than the earth. So the answer is the sun's gravitational pull. Because without t
In The Moon
Is earth revolves around the sun?
A commonly accepted theory is that the Earth rotates around the sun in 365.25 days, so yes. Though you could say that is moves in a straight line and that spacetime is curved | 620 | 2,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-04 | latest | en | 0.952564 |
http://stackoverflow.com/questions/5239715/problems-with-a-shunting-yard-algorithm/5240781 | 1,410,946,764,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657123274.33/warc/CC-MAIN-20140914011203-00176-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 259,503,203 | 21,315 | # Problems with a shunting yard algorithm
I have successfully implemented a shunting yard algorithm in java. The algorithm itself was simple however I am having trouble with the tokenizer. Currently the algorithm works with everything I want excluding one thing. How can I tell the difference between subtraction(-) and negative (-)
such as 4-3 is subtraction but -4+3 is negative
I now know how to find out when it should be a negative and when it should be a minus, but where in the algorithm should it be placed because if you use it like a function it wont always work for example
3 + 4 * 2 / -( 1 − 5 ) ^ 2 ^ 3
when 1-5 becomes -4 it will become 4 before it gets squared and cubed
just like 3 + 4 * 2 / cos( 1 − 5 ) ^ 2 ^ 3 , you would take the cosine before squaring and cubing
but in real math you wouldn’t with a - because what your really saying is 3 + 4 * 2 / -(( 1 − 5 ) ^ 2 ^ 3) in order to have the right value
-
I added the 'java' tag, I thought it might get your question more views. – Trevor Boyle Mar 9 '11 at 0:03
It sounds like you're doing a lex-then-parse style parser, where you're going to need a simple state machine in the lexer in order to get separate tokens for unary and binary minus. (In a PEG parser, this isn't something you have to worry about.)
In JavaCC, you would have a `DEFAULT` state, where you would consider the `-` character to be `UNARY_MINUS`. When you tokenized the end of a primary expression (either a closing paren, or an integer, based on the examples you gave), then you would switch to the `INFIX` state where `-` would be considered to be `INFIX_MINUS`. Once you encountered any infix operator, you would return to the `DEFAULT` state.
If you're rolling your own, it might be a bit simpler than that. Look at this Python code for a clever way of doing it. Basically, when you encounter a `-`, you just check to see if the previous token was an infix operator. That example uses the string `"-u"` to represent the unary minus token, which is convenient for an informal tokenization. Best I can tell, the Python example does fail to handle case where a `-` follows an open paren, or comes at the beginning of the input. Those should be considered unary as well.
In order for unary minus to be handled correctly in the shunting-yard algorithm itself, it needs to have higher precedence than any of the infix operators, and it needs to marked as right-associative. (Make sure you handle right-associativity. You may have left it out since the rest of your operators are left-associative.) This is clear enough in the Python code (although I would use some kind of struct rather than two separate maps).
When it comes time to evaluate, you will need to handle unary operators a little differently, since you only need to pop one number off the stack, rather than two. Depending on what your implementation looks like, it may be easier to just go through the list and replace every occurrence of `"-u"` with `[-1, "*"]`.
If you can follow Python at all, you should be able to see everything I'm talking about in the example I linked to. I find the code to be a bit easier to read than the C version that someone else mentioned. Also, if you're curious, I did a little write-up a while back about using shunting-yard in Ruby, but I handled unary operators as a separate nonterminal, so they are not shown.
-
In particular, one of those answers references a solution in C that handles unary minus.
Basically, you have to recognize a unary minus based on the appearance of the minus sign in positions where a binary operator can't be, and make a different token for it, as it has different precedence.
Dijkstra's original paper doesn't too clearly explain how he dealt with this, but the unary minus was listed as a separate operator.
-
the standard shunting yard algorithm deosn't support them, im trying to modify it to support them. Wolfram alpha, texas instruments, wolfram mathematica, microsoft math ect.. support them though and all of those use a version of the shunting yard algorithm – The Dude Mar 9 '11 at 1:24
In your lexer, you can implement this pseudo-logic:
``````if (symbol == '-') {
if (previousToken is a number
OR previousToken is an identifier
OR previousToken is a function) {
currentToken = SUBTRACT;
} else {
currentToken = NEGATION;
}
}
``````
You can set up negation to have a precedence higher than multiply and divide, but lower than exponentiation. You can also set it up to be right associative (just like '^'). Then you just need to integrate the precedence and associativity into the algorithm as described on Wikipedia's page.
If the token is an operator, o1, then: while there is an operator token, o2, at the top of the stack, and either o1 is left-associative and its precedence is less than or equal to that of o2, or o1 has precedence less than that of o2, pop o2 off the stack, onto the output queue; push o1 onto the stack.
I ended up implementing this corresponding code:
``````} else if (nextToken instanceof Operator) {
final Operator o1 = (Operator) nextToken;
while (!stack.isEmpty() && stack.peek() instanceof Operator) {
final Operator o2 = (Operator) stack.peek();
if ((o1.associativity == Associativity.LEFT && o1.precedence <= o2.precedence)
|| (o1.associativity == Associativity.RIGHT && o1.precedence < o2.precedence)) {
popStackTopToOutput();
} else {
break;
}
}
stack.push(nextToken);
}
``````
Austin Taylor is quite right that you only need to pop off one number for a unary operator:
``````if (token is operator negate) {
operand = pop;
push operand * -1;
}
``````
Example project:
https://github.com/Digipom/Calculator-for-Android/ | 1,353 | 5,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-41 | latest | en | 0.934918 |
https://001yourtranslationservice.com/translations/encyclopedia/Translations-Mathematics-30.htm | 1,656,914,142,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00652.warc.gz | 118,111,943 | 2,941 | Equations with Fractions
Introducing Fractions
Equations which contain at least one fraction containing a variable denominator are called equations with fractions. When solving equations with fractions, the same rules of calculation apply as with normal equations. Usually, when solving such equations, the first step is to find a common denominator. By multiplying equations with fractions having common denominators,
we manage to remove such denominators, as such giving us a normal equation which we can solve in the standard manner.
Example:
We are given the following equation with fractions:
The first step is to multiply the equation by x + 1, to get:
The second step is to solve the equation in the normal manner, to give us:
| +2
| :2
Solving the equation in this manner gives us the solution: x = 1.
Translating Dutch Hungarian Translations Hungarian Dutch Translating Swedish Czech Translations Czech Swedish Translating Russian
Enter your search terms Submit search form | 198 | 994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-27 | longest | en | 0.935012 |
https://socratic.org/questions/the-angles-of-a-triangle-have-the-ratio-2-3-4-what-is-the-measure-of-the-smalles | 1,600,939,049,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00248.warc.gz | 540,604,373 | 6,086 | # The angles of a triangle have the ratio 2:3:4. What is the measure of the smallest angle?
Let $2 x$ the smallest angle , $3 x$ the second angle , $4 x$ the third angle hence we have that $2 x + 3 x + 4 x = 180 \implies 9 x = 180 \implies x = 20$
The smallest angle is ${40}^{o}$ | 95 | 281 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.03125 | 4 | CC-MAIN-2020-40 | latest | en | 0.854053 |
http://mathhelpforum.com/discrete-math/128945-big-o-proof-print.html | 1,511,114,042,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00610.warc.gz | 190,470,994 | 3,220 | # Big O proof
• Feb 15th 2010, 01:28 PM
centenial
Big O proof
I'm having trouble with this proof:
Show that if $p(n)$ is a polynomial in $n$, then $\log{p(n)}$ is $O(\log{n})$.
I'm given this definition for O(n):
$T(N) = O(f(N))$ if there are positive constants $c$ and $n_0$ such that $T(N) \leq cf(N)$ when $N \geq n_0$.
I'm not sure how to use it to arrive at the conclusion that $\log{p(n)} = O(\log{n})$. Wouldn't $\log{p(n)}$ always be greater than $\log{n}$?
Any help would be much appreciated,
• Feb 15th 2010, 06:13 PM
centenial
I've been thinking about it more. Would this be sufficient?
We can approximate $p(n)$ to be $n^k$, where $k$ is the highest degree in $p(n)$. We notice that $p(n)$ is $O(n^k)$. Next, we observe that $\log{n^k} = k\log{n}$. Therefore, $\log{p(n)} = k\log{n}$, so $\log{p(n)} = O(\log{n})$.
• Feb 15th 2010, 09:12 PM
Drexel28
Quote:
Originally Posted by centenial
I'm having trouble with this proof:
Show that if $p(n)$ is a polynomial in $n$, then $\log{p(n)}$ is $O(\log{n})$.
I'm given this definition for O(n):
$T(N) = O(f(N))$ if there are positive constants $c$ and $n_0$ such that $T(N) \leq cf(N)$ when $N \geq n_0$.
I'm not sure how to use it to arrive at the conclusion that $\log{p(n)} = O(\log{n})$. Wouldn't $\log{p(n)}$ always be greater than $\log{n}$?
Any help would be much appreciated,
We know that there exists some $C$ and some $N$ such that $N\leqslant n\implies p(n)\leqslant Cn^{k}$. And clearly there exists some $N'$ such that $N'\leqslant n\implies Cn^{k}\leqslant n^{k+1}$ and so $N'\leqslant n\implies p(n)\leqslant n^{k+1}\implies \log p(n)\leqslant (k+1)\log n$ | 596 | 1,641 | {"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": 39, "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-2017-47 | longest | en | 0.82659 |
https://ravelys.github.io/what-is-variability-an-introduction-to-variance-in-statistics-6-1.html | 1,670,549,709,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00145.warc.gz | 522,744,998 | 10,342 | # What is Variability? – An Introduction to Variance in Statistics (6-1)
Research By Design
Published at : 28 Sep 2021
29091 views
288
6
When we seek to organize our jumbled data, we have two important questions about our data set. Our first question is “what single number best represents our data?” Our second question is whether the scores are packed together or spread out. The answer to our second question is going to be some measure of variability. Variability indicates how much the scores differ from one another. We will examine multiple measures of variability from simple measures of the range of the data to a mathematical measure of the average about of deviation in the data.
This video teaches the following concepts and techniques:
Introduction to variability in statistics
Link to a Google Drive folder with all of the files that I use in the videos including the Bear Handout and the StatsClass.sav dataset. As I add new files, they will appear here, as well.
01:56 - BRAIN CHECK
03:11 - Variability & Central Tendency
04:09 - Variability and Central Tendency | 233 | 1,083 | {"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 | 0.914958 |
https://www.physicsforums.com/threads/magnetic-lens.400226/ | 1,531,858,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00150.warc.gz | 948,999,514 | 14,543 | # Magnetic lens
1. May 2, 2010
### jam12
Hello,
I wanted to ask, is it possible for a magnetic lens to have many focal points, when the magnetic field of the lens is sufficiently large? if so what is the reason for this? how does this give better resolution of images compared to a simple optical lens.
I know for a normal optical lens that there is only one point where electromagnetic waves are focused to, thus giving a single focal point.
2. May 2, 2010
### Rajini
Hi,
Normal optic lens (made of glass- used for e.g., palm reading etc) are good for focusing light (not x-ray or electron beam).
But magnetic lens, which are made of solenoid (wires wounding and no glass), are in general used for focusing matter, for e.g., electron beam. In a transmission electron microscopy one uses magnetic field to focus (magnetic lens) electron beam onto sample/specimen. But what is the use of having more focal points..? But you can control the focal length by varying the magnetic field..Electrons spiral when travels in a magnetic field (Lorentz force)
hope this helps.
3. May 3, 2010
### jam12
Thanks, but im still unsure why a magnetic lens can have more than one focal point when the magnetic field of it is increased.
4. May 3, 2010
### Rajini
Focus point means all rays are focused to one point. Electron beams are focused to a point using magnetic field. Now if you want to have another focal point you tune the magnetic field. So the electron passing through the magnetic field can be focused to a different point. So one can say by this you have different focal point.
Point to note is electron can be tuned using magnetic field. And this magnetic field is magnetic lens.
5. May 3, 2010
### jam12
yes i understand that you can tune the magnetic lens to get a focal point required, but what im saying is that at any one time, is it possible to have more than one focal point?.
With out going into too much detail, I have a graph (for a high magnetic field in the lens) where i can see that it crosses the x axis more than once, ie it curves up and down ( a bit like a sin wave) but crossing the axis 4 times, doesn't this suggest more than one focal point?
For low magnetic fields, i have only one intersection suggesting that their is only one focal point for the lens.
6. May 3, 2010
### Rajini
I really don't know exactly about what you are telling.
Could you upload the graph ?
It could also be some sort of aberration?
7. May 4, 2010
### Bob S
If charged particles of a fixed momentum are emitted from a source on the z-axis of a solenoid with a constant magnetic field Bz, the particles will have both a longitudinal momentum along the z-axis, and a relatively small radial momentum component. The particles with the radial momentum component will undergo circular motion in the r-theta plane, and periodically return to the z axis. Because the motion in the r-theta plane is circular, and the period is independent of the radius of gyration, all the particles will return to the solenoid axis at the same point. The gyration frequency is
f = eB/(2πm) Hz where e and m are the charge and mass of the particles.
If the solenoid is sufficiently long, the particles will periodically re-focus at successive points at equal distances along the axis of the solenoid.
To understand this in detail. download Humphries' free book on the Principles of Charged Particle Acceleration at
http://www.fieldp.com/cpa.html
Review Sections 3.6 (book page 40) and 3.7 (book page 43) on motions of charged particles in cylindrical coordinates in magnetic fields.
Bob S | 833 | 3,586 | {"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-2018-30 | latest | en | 0.921689 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.