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http://www.knossosgames.com/pt/pt.Index.html | 1,601,250,034,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00105.warc.gz | 187,346,200 | 4,103 | To print out a copy of these instructions, click below:
If you like these puzzles,
you might also like...
## instructions
One of the duties of a park ranger is to protect parklands by ensuring people stay on designated trails. In these puzzles, it is your job to assemble trails of different lengths using the available paths through the forest.
Each trail must form a loop that does not overlap itself and must pass through at least one trailhead (black circle with white dot). Different trails may share paths, but all paths must be used by at least one trail. Finally, each puzzle will give specific instructions to the number of trails and their lengths.
## example
Here is a series of paths that need to be assembled into trails. The numbers beside each path segment represent how long each segment is, in miles. In this example, the two trails created must have lengths within 1/2 mile of each other.
Here are all of the possible loops that could be made into trails for this example:
Because the orange loop does not go through a trailhead, it is not an acceptable trail.
Here are more examples of loops that are not acceptable trails: loops that overlap themselves.
Here are two acceptable solutions to the puzzle. In both cases, the trails are acceptable loops of length within 1/2 mile of each other.
Here are two unacceptable solutions using the above trails. On the left, the two trails are the same length, while the instructions specifically state that trails must be different lengths. On the right, the trails are not of lengths within 1/2 mile of each other, and not all of the paths are used.
## solving tips
Let's reexamine three of the trails given above.
If we tried to use any of these trails to solve the above example, we'd have a problem. Each of these leaves a T-shaped intersection of paths to be covered by the second trail. Unfortunately, no trail can cover all three of the paths that emerge from such an intersection. Thus, when thinking about what trails to use, it is important to consider not only what paths a trail does cover, but also what paths a trail does not cover. | 439 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.955295 |
http://www.life123.com/hobbies/games/darts/dart-games.shtml | 1,386,945,465,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164949664/warc/CC-MAIN-20131204134909-00077-ip-10-33-133-15.ec2.internal.warc.gz | 423,834,837 | 16,803 | Dart Games Everyone Can Play
Dart games that are easy to learn for beginners are also fun for experienced throwers. The most common dart games are 501, 301 and Cricket. Although these games are simple, beginners will need to employ a little bit of strategy in order to win.
Determining The First Turn
In all three dart games, players each throw one dart to decide who will take the first turn. The player with the highest score wins the first turn.
501 And 301
501 and 301 are easy and fun dart games. Each player gets that number of points to start the game. In 501, players each start with 501 points; in 301, players each begin with 301 points. Each player takes a turn, throwing three darts, adding the score depending on where the darts land on the board and subtracting the total from the number of points from 501 or 301. Zero points must be reached exactly on the last turn in order to win. Furthermore, the last throw must land in a double score space, which is called "doubling-out."
In the 301 game, scoring begins when a player has "doubled." This is called "doubling-in." One player may double in before the next, which leaves the second player stuck trying to get the double while the first player can start working her way toward zero. This rule is an exciting component of the shorter 301 game.
Cricket
In Cricket, the numbers 15 through 20 are used, with each player needing to hit each number (either in the single, double or triple sections of the board) three times. That is called "closing the number." Each number needs to be closed and, once it is, additional hits on that number count as points for whichever player closed the number first. Three throws constitute a turn, called an "inning." A triple on one of the numbers closes the number. Similarly, a single and a double will do the same. The numbers do not have to be closed in the same inning, but each hit on the number is tallied during scoring.
There is strategy to Cricket, as the player who has closed all of his numbers wins only if he also has the highest score. Therefore, closing the large numbers first is a strategic move, so the higher scoring numbers count toward adding up the score fastest.
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Want to know how to play darts? With a little information on the shape of the dart and knowledge of the rules, you'll be beating everyone either at the bar or in your rec room.
Good dart supplies call for a little more than darts and a board. You'll also need to think of where you'll store the darts, and you might want a dart tool to keep the darts in good shape. | 579 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-48 | longest | en | 0.970662 |
https://thermopedia.com/pt/content/945/ | 1,708,503,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00322.warc.gz | 599,471,940 | 9,182 | Inscrição na biblioteca: Guest
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Figure 1 shows a generalized Heat Exchanger in which heat is transferred between two streams (stream 1 and stream 2).
The rate of heat transfer, , between the streams may be expressed as a function of the area available for the transfer of heat, A, the overall heat transfer coefficient, U, and a mean temperature difference, ΔTm, such that
(1)
Estimating the rate of heat transfer for a given design of heat exchanger requires techniques for estimating the Overall Heat Transfer Coefficient and techniques for estimating the mean temperature difference.
Techniques for estimating mean temperature difference are based upon the following assumptions:
1. the heat exchangers have only two streams;
2. heat exchange with the surroundings is negligible;
3. there is a linear relationship between specific enthalpy and temperature for both streams (i.e., constant specific heat capacities);
4. the overall heat transfer coefficient between the stream is constant throughout the heat exchanger;
5. where a heat exchanger consists of multiple parallel paths, the flowrates and heat transfer areas in each path are identical.
The above assumptions are most likely to be met when both streams are single-phase fluids (i.e. all liquid or all gas) and where the temperature changes are small such that the specific heat capacities and other properties of the fluids stay constant throughout the heat exchanger. The approach could be applied to heat exchangers involving boiling or condensing but only under circumstances where there are no significant changes in overall heat transfer coefficient. Heat exchangers involving the onset of boiling or condensation or the dryout transition are therefore not suitable for treatment using the traditional mean temperature difference approach. Such heat exchangers will need to be analyzed using techniques which make allowance for changes in heat transfer coefficient.
With the above assumptions, the rate of heat transfer in any geometry of heat exchanger can, in principle, be calculated. The simplest case is pure countercurrent flow. Here the mean temperature difference can be expressed in terms of the inlet and outlet temperatures of each stream.
(2)
For a heat exchanger with countercurrent flow, the mean temperature difference is known as the log mean temperature difference, ΔTLM. The log mean temperature difference is the maximum mean temperature difference that can be achieved in any geometry of heat exchanger for any given set of inlet and outlet temperatures. For any other type of heat exchanger, the mean temperature difference can be expressed as
(3)
where F is always less than or equal to 1. Estimating the mean temperature difference in a heat exchanger by calculating the log mean temperature difference and estimating F is known as the F factor method.
F varies with geometry and thermal conditions. The thermal conditions are defined by parameters such as the overall heat transfer coefficient, U, the area available for heat transfer, A, the mass flow rates of the two steams and , the specific heat capacities of the two streams c1 and c2, and the temperature change in each stream (T1,in - T1,out) and (T2,in - T2,out).
For any given geometry, F is often presented as a function of two nondimensional parameters, R, the ratio of the thermal capacities of the two streams and P (sometimes known as effectiveness, E), the ratio of the achieved heat transfer rate to the maximum possible heat transfer rate. These parameters are typically defined as
(4)
and
(5)
where
Unfortunately, there are no standard definitions of R and P. Users of the F method should always check the definitions used by any supplier of F value information and apply identical definitions when using this information in heat transfer calculations. This variation of definition can lead to problems when comparing data from different authors.
Values of F are frequently presented as graphs showing the relationship between F and P for a range of values of R. Figure 2 shows a typical relationship. The figure shows that F always tends to 1 as the amount of heat transferred reduces to zero. The figure also shows that for any given value of R, there is typically a maximum achievable value of P. The value of F changes rapidly as P approaches its maximum value. Because of this sensitivity, heat exchanger designs are rarely developed near the maximum value of P and are typically restricted to conditions which give values of F greater than 0.8.
The F factor can be used for both design and rating calculations. For design calculations, the mass flowrates, specific heat capacities and required temperature changes will be specified. R and P can therefore be calculated directly. The design engineer will need to check that the required value of P is less than the maximum value of P for the specified value of R. The value of F can then be found from the graph. The combination of the log mean temperature difference and F gives the mean temperature difference and with an estimate of the overall heat transfer coefficient, the required heat transfer area can be established. If the required value of P is greater than the maximum value for the specified value of R, a different type of heat exchanger must be considered until a feasible design is found. A simple countercurrent heat exchanger will always be able to achieve any design requirement but may be physically impractical.
For rating calculations, the geometry of the heat exchanger and its heat transfer area, the mass flowrates and the specific heat capacities of the streams and hence the overall heat transfer coefficient and the inlet temperatures will be defined. It will therefore be possible to calculate R but not P. The rating is estimated using an iteration which may start by guessing a heat transfer rate and calculating the exit temperatures, the log mean temperature difference and P and then using the figure to estimate F. The resulting heat transfer rate is then calculated from F, the log mean temperature difference, the overall heat transfer coefficient and the heat transfer area. The guessed and calculated heat transfer rates are compared and the guessed value is adjusted until convergence is achieved.
Published relationships for F in graphical form are available for most geometries of shell and tube heat exchanger and a range of geometries of crossflow heat exchanger (see guide to references at the end of this section). The size of the graphical presentations rarely allows values of F to be estimated to better than two significant figures. This accuracy of estimation is consistent with the overall accuracy of the mean temperature difference approach and the lack of compliance with the underlying assumptions. Attempts to improve the accuracy in the estimation of F values are therefore unlikely to produce significant benefits for heat exchanger designers.
An alternative method of presenting mean temperature difference information is known as the effectiveness—NTU method. This method is based upon exactly the same initial assumptions. The heat transfer behavior are presented as a relationship between effectiveness, E (defined in a similar way to P), the ratio of the thermal capacities of the streams, R, and the number of heat transfer units, NTU which as calculated from the expression
(6)
where ( )smaller is the smaller of ( )1 and ( )2. Unfortunately, the parameters E, R and NTU are again not consistently defined and any user should check the definition used by the supplier of data and apply those definitions when using the data.
Effectiveness-NTU information is typically presented graphically as the relationship of effectiveness against NTU for various values of R. Figure 2 shows a typical relationship. This shows that the effectiveness tends to zero as the NTU tends to zero and the effectiveness tends to a maximum value as NTU becomes large.
Users of the effectiveness-NTU technique are not required to calculate the log mean temperature difference when carrying out design or rating calculations. For design calculations, E and R can be calculated from the mass flowrates, specific heat capacities and inlet and outlet temperatures. The value of NTU can be read from the graph for the chosen design of heat exchanger and used to calculate the required surface area. In rating calculations, the surface area, mass flowrates and specific heat capacities can be used to calculate R and NTU. The value of E can be read from the graph and used to calculate the rate of heat transfer in the heat exchanger. It can therefore be argued that the effectiveness-NTU method can be used for rating calculations without the need for an iteration. In reality, since heat transfer coefficient will change with temperature, it is still likely that an iterative calculation will be required.
Effectiveness-NTU relationships are generally presented graphically and may be used to produced estimates to two significant figures. Again, because of deviations from the underlying assumptions, this level of precision in the calculations is consistent with the precision of the overall method. An attempt has been made to produce a single algebraic expression with a number of variable coefficients to represent a range of common heat exchangers, ESDU, 93012. Approximately 90 geometries were represented by an expression with 14 variable coefficients. The curve fitting approach matches the exact relationships to better than 2%. By incorporating the algebraic expression into a computer program, a range of geometries and designs can be readily accessed. The accuracy of the computer calculations does not, however, bring any increase in the accuracy of the overall method.
F values or effectiveness-NTU relationships are obtained by making simplifying assumptions about the geometry of a heat exchanger and then carrying out a process of integration. The integration can either be carried out algebraically (most designs of shell and tube heat exchanger) or using finite element methods (most designs of crossflow heat exchanger). In all but the simplest of geometries, the process of integration is complex. For shell and tube exchangers, the resulting algebraic expressions require care in application to avoid error. For example, for the single E-shell with any even number of passes, the expression linking NTU with E is
(7)
where
and
(8)
For crossflow heat exchangers, particularly with more than one pass, the finite element integration often involves iteration as well. Users of mean temperature difference techniques are therefore advised to use the graphical presentations of these integrations or curve fits to the data.
Table 1 gives references to sources of graphical data for various types of heat exchanger.
Table 1. Sources of graphical data for various types of heat exchanger
#### REFERENCES
ESDU, 85042. Effectiveness-NTU relationships for the design and performance rating of two-stream heat exchangers, ESDU, Data Item, 85042, December 1985.
ESDU, 86018. Effectiveness-NTU relationships for the design and performance rating of two stream heat exchangers, ESDU, Data Item, 86018, July 1986, Amended July 1991.
ESDU, 87020. Effectiveness-NTU relationships for the design and performance evaluation of multi-pass crossftow heat exchangers, ESDU, Data Item 87020, October 1987, Amended November 1991.
ESDU, 88021. Effectiveness-NTU relationships for the design and performance evaluation of additional shell and tube heat exchangers, ESDU, Data Item 88021, November 1988, Amended July 1991.
ESDU, 91036. Algebraic representations of effectiveness-NTU relationships, ESDU, Data Item 91036, November 1991.
Kays, W and London, A. L. (1984) Compact Heat Exchangers. Third Edition. McGraw-Hill.
Kern, D. Q (1950) Process Heat Transfer, First Edn., McGraw-Hill.
Perry (1973) Chemical Engineers' Handbook, McGraw-Hill.
Pignotti, A. (1984), Matrix formalism for complex heat exchangers, Trans ASME, Journal of Heat Transfer, Vol. 106, pp. 352-360.
Taborek, J. (1983) Heat Exchanger Design Handbook, Section 1.5, Hemisphere Publishing Corporation.
TEMA (1978) Standards of Tubular Exchanger Manufacturers Association, Sixth edn.
#### Referências
1. ESDU, 85042. Effectiveness-NTU relationships for the design and performance rating of two-stream heat exchangers, ESDU, Data Item, 85042, December 1985.
2. ESDU, 86018. Effectiveness-NTU relationships for the design and performance rating of two stream heat exchangers, ESDU, Data Item, 86018, July 1986, Amended July 1991.
3. ESDU, 87020. Effectiveness-NTU relationships for the design and performance evaluation of multi-pass crossftow heat exchangers, ESDU, Data Item 87020, October 1987, Amended November 1991.
4. ESDU, 88021. Effectiveness-NTU relationships for the design and performance evaluation of additional shell and tube heat exchangers, ESDU, Data Item 88021, November 1988, Amended July 1991.
5. ESDU, 91036. Algebraic representations of effectiveness-NTU relationships, ESDU, Data Item 91036, November 1991.
6. Kays, W and London, A. L. (1984) Compact Heat Exchangers. Third Edition. McGraw-Hill.
7. Kern, D. Q (1950) Process Heat Transfer, First Edn., McGraw-Hill.
8. Perry (1973) Chemical Engineers' Handbook, McGraw-Hill.
9. Pignotti, A. (1984), Matrix formalism for complex heat exchangers, Trans ASME, Journal of Heat Transfer, Vol. 106, pp. 352-360.
10. Taborek, J. (1983) Heat Exchanger Design Handbook, Section 1.5, Hemisphere Publishing Corporation.
11. TEMA (1978) Standards of Tubular Exchanger Manufacturers Association, Sixth edn. | 2,885 | 13,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.911218 |
https://www.physicsforums.com/threads/radius-of-a-circle-inside-a-sphere.468118/ | 1,695,935,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00851.warc.gz | 1,018,824,537 | 14,527 | # Radius of A Circle inside a Sphere
• PAR
## Homework Statement
Say you have a sphere of radius r centered at the origin, and a vector v <r,0,0>.
Let v' be the vector v rotated about the y-axis by angle theta.
What is the shortest distance between the end of the vector and the z-axis?
## The Attempt at a Solution
I drew a picture:
[PLAIN]http://img253.imageshack.us/img253/9459/circleinsphere.png [Broken]
Obviously the shortest distance would be the line normal to the z-axis that would complete a right triangle with the y-axis and the vector v'. The distance is also equal to the radius of a circle, which I drew on the picture.
Because of this, I believe the answer is r*cos(theta), however I am not sure, and I need to know this for a programming assignment. Thank You!
Last edited by a moderator:
Hi PAR!
Yes, rcosθ. | 203 | 837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.929614 |
makercasts.org | 1,632,591,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00676.warc.gz | 420,967,972 | 11,391 | # Reducing Boolean Satisfiability to The Witness
## Background
Recently, I've been looking at Complexity Theory, which is a field in theoretical Computer Science. It concerns itself with how long algorithms take to run and how much space they use. It tries to classify problems as 'easy' or 'difficult' depending on the algorithms we have for solving them. In fact, it's far more textured than that. Problems form a hierarchy that relates the difficulties of seemingly disparate problems with one another.
You may have heard of the Travelling Salesperson Problem. That's classified as 'NP-hard'. How about checking if a number is prime? That's 'Polynomial'. What do these mean? Well, I'm not going to go into it here, but the point is that we can say with utmost confidence that solving the Travelling Salesperson Problem is inherently, a much more difficult problem than testing primality. We've connected seemingly unrelated things by weaving a fabric of complexity analysis between them. I find that fascinating.
## The Witness
So what's this article about? I'd like to try to weave that fabric a little further and take a look at Jonathan Blow's puzzles from The Witness. We'll go through something called a 'reduction', which is a method for transforming one problem into another. You can think of this as an algorithm that takes, as input, some representation of one problem and produces, as output, some other kind of problem.
Reductions are the threads that form this fabric. If we can find a reduction from problem A to B we can make claims about their relative difficulty. We can say that A cannot possibly be harder than B, because we can always just transform A into B and solve that instead. This does assume our reduction isn't too difficult to compute, otherwise we'd have to factor that in, too.
To be more specific, I'm going to demonstrate a reduction from a problem called 'Boolean Satisfiability', to puzzles from The Witness. This will give us an opportunity to see how reductions work in the context of something fun. When I first encountered reductions, I felt confused and didn't understand how people came up with them. This article is the walk-through I wish I'd had to build that intuition.
At this point, I should explain that if you are well-versed in Complexity Theory, this reduction would be a somewhat frivolous endeavour. The SAT problem (as it is often abbreviated) is a 'hard' problem and so by reducing SAT to The Witness, we're showing that puzzles from The Witness are at least as difficult as an already difficult problem. Not an earth-shattering result. The real charm would be to go in the other direction, but we're not going to do that today. Instead, we'll embrace the futility and frivolity.
## Getting Started
We should start by refreshing our memory on SAT.
SAT is the problem of finding a satisfying assignment for a boolean equation. In short, we need to come up with a true or false value for each variable in the equation to make the whole thing true. Here's an example problem:
(A v !B) ^ (!B v !C)
This can be read as '(A or not B) and (not B or not C)'. Let's try to solve this problem. What if we assigned A=true, B=true, C=true ? Well, that satisfies the first clause because (true or false) is true. How about the second? That's (false or false) which evaluates to false. Our attempt has failed.
Ok, so how do we fix it? How about A=true, B=true, C=false ? That works because each and every clause now evaluates to true and the equation has been satisfied. That wasn't so hard! Just a few moments ago I said this was a 'hard' problem. Well, this one was easy, but in general that's not the case. As we add more variables, the difficulty scales exponentially.
You may be wondering why we care about this problem in the first place. Why did we pick this? Well, clever people have already figured out how to reduce a plethora of other problems to SAT. Really important ones, too, like route planning and laying out circuit boards efficiently. In fact, SAT is depended on so heavily, there's an annual competition to see who can write the best solver. I expect to see your entry next year!
Ok, that's SAT out of the way. What about The Witness? Well, I'm not going to explain what all of the symbols mean here. If you're hazy on the details you can click on these to refresh your memory:
There are more puzzle pieces in the game, but this is just enough to do our reduction, so let's get on with that.
## The Reduction
So how do we do this? We need to come up with a method for transforming any SAT problem into a Witness puzzle. There should be a solution to the puzzle only if the SAT problem can be satisfied and there should be no solutions when it cannot. We somehow need to encode the structure of a generic boolean equation inside a Witness puzzle. So let's think about how we might do that.
The first thing to note is that each clause in a SAT problem is connected with an AND. Every one of these must evaluate to true for it to be a correct solution. This is vaguely reminiscent of the Sun symbols in The Witness. Each Sun must be paired up with another Sun of the same colour for there to be a solution. This is a bit tenuous, but maybe we can do something with this similarity.
Ok, what else do we know?
The way to solve a SAT problem is by choosing between true and false for each of the variables in the equation. There's a notion of 'choice'. What does it mean to 'choose' something in a Witness puzzle? What's the corresponding action for choosing? Well, the only influence a puzzler has over a Witness puzzle is to choose their path from start to finish. Do I go North, East, South or West at this junction?
So perhaps we could model variables as forks in the road that the puzzler must choose between. Taking the left fork could correspond to A=true and the right fork could be A=false. Now let's try to put this together with our Suns idea. Let's invent a red Sun to represent the first clause. In our SAT problem, if assigning A=true satisfies the clause, we should design our puzzle so that the Sun is paired up. If they choose to go right, the clause has not been satisfied so this should be an incorrect solution to the puzzle:
So that's the basic idea. To recap, we're going take our SAT problem (whatever that may be) and turn it into a Witness puzzle. We're going to add a Sun for each clause and force the puzzler to make a choice as to whether to take a left fork or a right fork corresponding to a true or false assignment for each of the variables. We'll construct our puzzle based on which variables appear in which clauses and whether they are negated or not.
## Starting Small
Let's start by trying to create a puzzle for (A v B). This seems like the smallest possible step on the way to formulating a general method. We still have a single clause but we've added a variable. If we follow the rule of inventing a Sun for each clause in our equation, our puzzle should still only have a single colour Sun. What's different is the number of variables within that clause. There is now more than one way to satisfy this equation.
This means we need to build a puzzle that can be solved if either A=true or B=true. How do we do this? If we follow our rule that each variable is a fork in the road, we immediately hit an abundance of problems. Firstly, where do we put those forks? Do we stack forks on top of each other? Something like this:
If we go with that, how do we arrange our Suns? Let's make an arbitrary decision and place the Sun for the clause in the bottom-right corner. We then need to pair up this Sun if either A=true or B=true. This clearly isn't going to work. If we choose A=false, we've isolated our Sun so no matter how we choose B, we can't solve the puzzle. What's more, we're going to end up with three Suns in the puzzle and won't be able to pair them.
So how do we proceed? Let's think about our SAT problem again. We know that (A v B) is satisfied if either A or B is true. This is an 'at least one' relationship, i.e. at least one of the variables must evaluate to true for the clause to be true. Suns work a bit differently. Suns must be paired with 'exactly one' other Sun. Three or four just won't do! So if we can change these semantics from 'exactly one' to 'at least one' we may have a chance.
## 'at least one'
Let's restate our goal: We need to find a way to construct a Witness puzzle so that a Sun representing the clause can be paired with another Sun. We need to do so in a way that allows for any number of Suns of the same colour to appear in the puzzle. Blimey!
Perhaps Elimination Marks can help with this. Let's construct a puzzle with a Sun representing the clause placed in the bottom-right position. We'll then place 'at least one' other Sun in the puzzle and think about how we can use Elimination Marks to eliminate the excess Suns. The following puzzle isn't quite right, but it demonstrates what we're trying to do:
What's wrong with this approach? Well, we've successfully paired up the bottom-right Sun with one other Sun, but we've left an Elimination Mark dangling. There are no 'mistakes' in that 1x1 region for the Elimination Mark to mop up. Because of this, the path drawn above is not a correct solution.
What can we learn from this? Firstly, we should note that we can stack as many Suns on the left as we'd like, which means we're on the right track to meet our 'at least one' requirement. Secondly, we can choose which of the Suns to pair up depending on how we draw the path. Finally, we can force where the puzzler must go by simply removing sections of the path for them to travel.
I'd call that progress! Ok, so we haven't solved our problem yet, but we've learned a few things.
It seems that our next logical step is to see if we can find a way to get rid of those dangling Elimination Marks. Perhaps we can intentionally introduce mistakes to be eliminated. We'll have to be careful about how we do this as they should only really be mistakes under certain conditions. We're going to have to think this through.
## Intentional Mistakes
Before we dive into this, let's do a quick recap on how we got here:
We're trying to find a reduction from SAT to The Witness. We think we can use Suns to represent clauses which are satisfied by pairing each up with another Sun of the same colour. In SAT, clauses are satisfied if any of their variables evaluate to true. We need to mimic these semantics. We almost managed it with Elimination Marks, but ended up leaving a surplus in the puzzle. Now we're exploring whether we can fix this by introducing intentional mistakes for the Elimination Marks to eliminate. Are you still with me?
We need to find a way for there to be a mistake in the same region as the Elimination Mark, but only if the Sun is paired. If it is not paired, the Elimination Mark should mop up the unpaired Sun, instead. Being an unforgiving kind of game, there are actually quite a few ways to introduce these kinds of mistakes. Here's one:
In this case, only the puzzle on the left is correct, but the puzzler has the freedom to choose whether to pair the Suns or not. Tetris Blocks seem to lend themselves well to this. We can tailor their shape to match the region containing the unpaired Sun. This has the effect of toggling which thing is eliminated depending on which path is taken. Here are some alternatives that would work equally well:
That last one's a bit different, but ponder it for a moment and you'll see that it works. In fact, we could have used some of the other puzzle pieces, too. So which of these should we choose? Let's defer that decision for later. Having a few options might help us if we get stuck.
## Freedom to Choose
We can now construct puzzles that contain as many Suns as we'd like and the puzzler has the freedom to choose which of those Suns are paired. Great! But how does this relate back to our original problem? We were trying to build a puzzle for (A v B). Are we there yet? Well, we haven't defined what it means to take one path or another. Does the following path mean A=false, B=true or is it just meaningless gibberish?
But hold on a minute – we haven't really defined how a puzzle should be constructed for a given SAT problem. We assumed something like the above corresponds to (A v B), but is that really true? We began looking at this from the perspective of 'we have too many Suns, how do we cope with that?', but we're yet to tie this back to the original problem. What we really need are a few rules for how to build these puzzles, but so far we've only considered a single example. To proceed, we ought to consider a few more.
So let's cast our net a little further and consider (A) ^ (!A). I've left the brackets in so we can clearly see that this is made up of two clauses. Note that this problem is unsatisfiable. Every clause must be true and one of the two will be false depending on how we assign A. If you remember back, we kind of have a rule that each clause in our equation corresponds to a different colour Sun, but how do we deal with negations?
Well, we said that 'choice' corresponds to which path is taken through the puzzle. If the puzzler chooses a path that represents A=true, they should no longer be able to pair up Suns in cases where A is negated. Here is a simplified example that uses a Hexagon Dot to force the puzzler to make a choice.
Earlier on, we talked about using 'forks in the road' to force the puzzler to choose how to assign true or false to each variable. That's exactly what we're doing here, although it's not as obvious. The fork is a decision between North and East from the start position. The Hexagon Dot insists that one or the other is chosen, otherwise they could go straight up the left-side, which would correspond to assigning both true and false to A.
## Putting it Together
So now we're ready to start putting things together. Let's try to do this gradually and construct a puzzle that represents (A v !A). This is a single-clause, single-variable problem, but it contains a negation. We know that a single clause means we put a Sun in the bottom-right corner. We also know that we should add our special combination on the left for each occurrence of a variable, so let's do that:
We then need to force the puzzler to choose whether to assign A=true or A=false. We can do this by adding a Hexagon Dot along a path that runs between the symbols and an alternate path along the right-hand side. At this point, we also need to remove other sections of path:
And that's it... or is it? The eagle-eyed amongst you may have noticed a problem with this puzzle. We'll get to that in a moment. Before then, let's see what paths corresponding to A=true and A=false look like:
You can see that the Sun that's paired is different in each case. We've arbitrarily defined the bottom Sun to mean A=true and the top to mean A=false, but we could flip this if we wanted. The Hexagon Dot we placed down the middle forces the puzzler to choose between one or the other.
Ok, so what's this problem I mentioned? Well, when we figured out our combination of symbols, we hadn't accounted for the case where the entire region is skipped. We'd only anticipated a single 1x1 region. There are now two mistakes in that region: the Tetris Block and the Sun. This is one too many for our Elimination Mark to mop up. We've found a bug!
## Fixing the Bug
Fear not! When we introduced intentional mistakes to the puzzle, we had a few alternatives. If we can find one that's valid when contained within the skipped region, we're golden. Let's remind ourselves of the other ways to introduce intentional mistakes. Which door should we choose?
There are goats behind two of these doors. Can you figure out which has the sports car?
Ok, enough of that. The right answer is Door number 3. This is the only pattern where the Tetris Block is valid when the bottom-right Sun is paired up. When the region is skipped, the Tetris Block becomes invalid, but the Suns within that skipped region are paired, so there is only one mistake, rather than two. This is very confusing! Perhaps if we see how it works in different contexts, that will help:
## Some Clarification
At this point you might be wondering what the difference is between 'Suns not paired' and 'Suns skipped'. Let's briefly consider the example (A v A) so that I can clarify:
The puzzle above is the same as the (A v !A) example we looked at earlier except we're now using the new pattern of symbols. We've also removed the Hexagon Dot running down the middle as we've dropped the negation. If we choose the path where A=true, that means we follow the left-hand side of the puzzle, up through the symbols. If we choose A=false, that leads alongside the symbols, skipping those regions.
If we consider the case where A=true, we actually have a choice in which Sun is paired. Do we pair one at the top or the bottom? Both are valid. This actually corresponds to our equation: (A v A). If we choose the bottom Sun, that's kind of like saying 'use the first A to make the clause true' and the top one relates to the second A. Let's see what these cases look like, as well as the case for when A=false:
Notice that the two on the left are both genuine, correct solutions for the puzzle, whereas when A=false, the bottom-right Sun has not been paired up making it an incorrect solution.
## General Method
We're now reaching the end of our journey and are ready to formulate our general method. When we put everything together, we end up with a process that looks something like this:
• 1. Place a uniquely coloured Sun bottom-right for each clause
• 2. For each variable:
• a. For each clause containing the variable (as a positive):
• i. Add an Elimination Mark, Tetris Block and Sun to the left of the puzzle
• b. Add an alternative path that runs alongside this section
• c. Re-combine this path with the main path and add a Hexagonal Dot
• d. Repeat a, b for the case where the variable is negated
• 3. Make the rest of the puzzle unreachable by removing sections of path
If this sounds complicated, that's because it is! There's a lot going on there, so maybe a concrete example will help. Let's end where we started and build the puzzle corresponding to (A v !B) ^ (!B v !C). I have tried to annotate this puzzle as best I can, but it's still incredibly large and takes a long time to grok:
The puzzle above is the result of reducing our SAT problem to The Witness. The method outlined above is the 'reduction' we've been seeking. It's quite amazing how we started with a few simple observations and by careful reasoning we've ended up with something incredibly complex. It's startling just how large the puzzle is for such a simple problem. That tends to be how reductions go.
This does assume our reduction isn't too difficult to compute, otherwise we'd have to factor that in, too.
Let's consider this for a moment. We could write a program that performs this reduction. How 'expensive' would it be to compute? Well, although we end up with a huge problem, building the puzzle isn't all that difficult. Each occurrence of a variable adds roughly 18 cells to the grid and that doesn't change whether we have three variables or a million. That means we could build this puzzle in linear time by reading through the equation.
Finally, here are a couple of correct paths through the puzzle:
There's an interactive version of this puzzle here. If you decide to try this, see if you can figure out the corresponding assignments of A, B and C for your path.
## Closing Words
Hopefully this has given you some intuition for how reductions work. This is arguably a lot more complicated than most, but perhaps more interesting, too. One of the wonderful implications of all this is that you could solve all kinds of problems like route planning by reducing them to fiendishly difficult Witness puzzles. Those puzzles would probably be so large they'd reach the moon!
If this article has piqued your interest, you may wish to try writing a program to do this reduction. I'll happily add links to code or screenshots to this section if anyone is determined enough to try it. If the reduction in the other direction seems tempting, that's something I'd like to look at in coming months. I don't think it's possible for reasons I won't go into here.
You may be interested to hear that I'm writing a programming language called 'Sentient' that deals with these kinds of problems. It's still in alpha, but you can find the website here. Finally, I'd like to thank Tom Stuart for encouraging me to put pen to paper. I'd also like to thank Matthew Gruen for building The Windmill, which I have used extensively for taking screenshots. | 4,602 | 20,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | longest | en | 0.973503 |
http://convertwizard.com/935-days-to-seconds | 1,679,797,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00578.warc.gz | 9,399,465 | 17,525 | # 935 Days to Seconds (935 d to sec)
Convert 935 Days to Seconds (d to sec) with our conversion calculator and conversion tables. To convert 935 d to sec use direct conversion formula below.
935 d = 80784000 sec.
You also can convert 935 Days to other Time (popular) units.
935 DAYS
=
80784000 SECONDS
Direct conversion formula: 1 Days / 86400 = 1 Seconds
Opposite conversion: 935 Seconds to Days
## Conversion calculator
Amount:
From:
To:
Check out conversion of 935 d to most popular time units:
935 d to Minutes
935 d to Hours
935 d to Milliseconds
935 d to Microseconds
935 d to Years
## 935 d from similar units:
Day (Sidereal) 80563433.5
## Conversion table: Days to Seconds
DAYS SECONDS
1 = 86400
2 = 172800
3 = 259200
4 = 345600
5 = 432000
7 = 604800
8 = 691200
9 = 777600
10 = 864000
SECONDS DAYS
1 = 1.1574074074074E-5
2 = 2.3148148148148E-5
3 = 3.4722222222222E-5
4 = 4.6296296296296E-5
5 = 5.787037037037E-5
7 = 8.1018518518519E-5
8 = 9.2592592592593E-5
9 = 0.00010416666666667
10 = 0.00011574074074074
## Nearest numbers for 935 Days
DAYS SECONDS
944 d = 81561600 sec
953.61 d = 82391904 sec
968 d = 83635200 sec
999 d = 86313600 sec
1038.58 d = 89733312 sec
1067 d = 92188800 sec
1095 d = 94608000 sec
1100 d = 95040000 sec
1148 d = 99187200 sec
1321 d = 114134400 sec
1342.71 d = 116010144 sec
1458 d = 125971200 sec
1460 d = 126144000 sec
1462 d = 126316800 sec
1510 d = 130464000 sec
1513 d = 130723200 sec
1533 d = 132451200 sec
1572.71 d = 135882144 sec
1582.7 d = 136745280 sec
1760 d = 152064000 sec | 589 | 1,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.556336 |
http://cnx.org/content/m27530/1.1/ | 1,394,403,174,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-10/segments/1394010437227/warc/CC-MAIN-20140305090717-00031-ip-10-183-142-35.ec2.internal.warc.gz | 36,653,801 | 30,136 | Connexions
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Lecture 6:Continuous Time Fourier Transform (CTFT)
Summary: Extends the notion of the frequency response of a system to the frequency content of a signal. Widely used tool in many areas (communications, control, signal processing, X-ray diffraction, Medical imaging — CAT & PET scan). Continue development of Fourier transform pairs. Illustrate different methods for finding Fourier transforms.
Lecture #6:
CONTINUOUS TIME FOURIER TRANSFORM (CTFT)
Motivation:
• Extends the notion of the frequency response of a system to the frequency content of a signal.
• Widely used tool in many areas (communications, control, signal processing, X-ray diffraction, Medical imaging — CAT & PET scan).
• Continue development of Fourier transform pairs.
• Illustrate different methods for finding Fourier transforms
Outline:
• The continuous time Fourier transform (CTFT)
• Properties of the CTFT
• Simple CTFT pairs
• Fourier transform pairs
– Fourier transform of the unit step function
– Fourier transform of causal sinusoids
– Fourier transform of rectangular pulse — the sinc function
– Fourier transform of triangular pulse
• Filtering the ECG revisited
• Conclusion
Example — How to filter the ECG?
The recorded activity from the surface of the chest includes the electrical activity of the heart plus extraneous signals or “noise.” How can we design a filter that will reduce the noise?
It is most effective to compute the frequency content of the recorded signal and to identify those components that are due to the electrical activity of the heart and those that are noise. Then the filter can be designed rationally. This is one of many motivations for understanding the Fourier transform.
I. THE CONTINUOUS TIME FOURIER TRANSFORM (CTFT)
1/ Definition
The continuous time Fourier transform of x(t) is defined as
X ( f ) = - x ( t ) e j2πft t X ( f ) = - x ( t ) e j2πft t
and the inverse transform is defined as
x ( t ) = - X ( f ) e j2πft f x ( t ) = - X ( f ) e j2πft f
2/ Relation of Fourier and Laplace Transforms
The bilateral Laplace transform is defined by the analysis formula
X ( s ) = - x ( t ) e st t X ( s ) = - x ( t ) e st t
and the inverse transform is defined by the synthesis formula
x ( t ) = 1 j2π C X ( s ) e st s x ( t ) = 1 j2π C X ( s ) e st s
Now if the jω axis is in the region of convergence of X(s), then we can substitute s = jω = j2πf into both relations to obtain
X ( j2πf ) = - x ( t ) e j2πft t X ( j2πf ) = - x ( t ) e j2πft t
x ( t ) = 1 j2π - j∞ j∞ X ( j2πf ) e j2πft ( j2πf ) x ( t ) = 1 j2π - j∞ j∞ X ( j2πf ) e j2πft ( j2πf )
3/ Form of the Fourier transform
Finally, by canceling j2π and changing the variable of integration from j2πf to f we obtain
X ( j2πf ) = - x ( t ) e j2πft t X ( j2πf ) = - x ( t ) e j2πft t
x ( t ) = - X ( j2πf ) e j2πft f x ( t ) = - X ( j2πf ) e j2πft f
It is clumsy to write X(j2πf). Therefore, from this now on, we rewrite the function X(j2πf) described above as in a simpler form X(f) such that the transform pair would be symmetrical.
X(j2πf) Þ X(f)
The new Fourier transform pair will be in the form
X ( f ) = - x ( t ) e j2πft t X ( f ) = - x ( t ) e j2πft t x ( t ) = - X ( f ) e j2πft f x ( t ) = - X ( f ) e j2πft f
4/ Notation
Notation for the Fourier transform varies appreciably from text to text and in different disciplines. Another common notation is to define the Fourier transform in terms of jω as follows
X ( ) = - x ( t ) e jωt t X ( ) = - x ( t ) e jωt t x ( t ) = 1 2 π - X ( ) e jωt ω x ( t ) = 1 2 π - X ( ) e jωt ω
Differences in notation are largely a matter of taste and different notations result in different locations of factors of 2π. The notation we use minimizes the number of factors of 2π that appear in the expressions we will use in this subject and makes the duality of the Fourier transform with its inverse more transparent.
5/ Why bother with the Fourier transform?
• There are certain simple time functions which are more readily represented by Fourier transforms than by Laplace transforms, e.g., x(t) = 1, x(t) = cos(2πft), periodic time functions, etc.
• Certain important operations on signals are more readily analyzed with Fourier transforms, e.g., sampling, modulation, filtering.
• Examination of both signals and systems in the frequency domain gives insights that complement those obtained in the “time” domain.
6/ Functions that have Laplace transforms but not Fourier transforms
There are some time functions that have a Laplace transform but not a Fourier transform, namely those for which the jω-axis is not inside the region of convergence. For example, x(t) = eαtu(t) for α > 0.
II. PROPERTIES OF THE CTFT
1/ Properties — symmetry
We start with the definition of the Fourier transform of a real time function x(t) and expand both terms in the integrand in terms of odd and even components.
X ( f ) = - x ( t ) e j2πft t X ( f ) = - x ( t ) e j2πft t X ( f ) = - ( x e ( t ) + x o ( t ) ) ( cos ( 2 πft ) j sin ( 2 πft ) ) t X ( f ) = - ( x e ( t ) + x o ( t ) ) ( cos ( 2 πft ) j sin ( 2 πft ) ) t
The odd components of the integrand contribute zero to the integral. Hence, we obtain
X ( f ) = - x e ( t ) cos ( 2 πft ) t + j - - x o ( t ) sin ( 2 πft ) t X ( f ) = - x e ( t ) cos ( 2 πft ) t + j - - x o ( t ) sin ( 2 πft ) t X ( f ) = X r ( f ) + jX i ( f ) X ( f ) = X r ( f ) + jX i ( f )
where
X r ( f ) = - x e ( t ) cos ( 2 πft ) t X r ( f ) = - x e ( t ) cos ( 2 πft ) t X i ( f ) = - - x o ( t ) sin ( 2 πft ) t X i ( f ) = - - x o ( t ) sin ( 2 πft ) t
We can infer symmetry properties of the Fourier transform of a real time function x(t).
X r ( f ) = - x e ( t ) cos ( 2 πft ) t , even function of f X r ( f ) = - x e ( t ) cos ( 2 πft ) t , even function of f X i ( f ) = - - x o ( t ) sin ( 2 πft ) t , odd function of f X i ( f ) = - - x o ( t ) sin ( 2 πft ) t , odd function of f | X ( f ) | = X r 2 ( f ) + X i 2 ( f ) , even function of f . | X ( f ) | = X r 2 ( f ) + X i 2 ( f ) , even function of f .
Therefore, if
x(t) X(f)
Real and even function of t Real and even function of f
Real and odd function of t Imaginary and odd function of f
The angle can be computed as follows,
∠X ( f ) = { π + tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) < 0 tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) > 0 ∠X ( f ) = { π + tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) < 0 tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) > 0
But, since ±n2π can always be added to the angle, and since is an odd function of f,
∠X ( - f ) = { - π - tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) < 0 - tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) > 0 ∠X ( - f ) = { - π - tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) < 0 - tan 1 ( X i ( f ) X r ( f ) ) for X r ( f ) > 0
Therefore, ÐX(f) is an odd function of f.
2/ Properties — duality
The Fourier transform and its inverse differ only by a sign in the exponent,
X ( f ) = - x ( t ) e j2πft t x ( t ) = - X ( f ) e j2πft f X ( f ) = - x ( t ) e j2πft t x ( t ) = - X ( f ) e j2πft f
Therefore, if x(t)X (f) then X(t)x(−f). This means that if we have found one Fourier transform pair, we automatically know another.
3/ List of simple properties
Some of the important properties are summarized here; a more complete list is appended.
a/ Properties — linearity
Most proofs of Fourier transform properties are simple.
ax 1 ( t ) + bx 2 ( t ) F aX 1 ( f ) + bX 2 ( f ) ax 1 ( t ) + bx 2 ( t ) F aX 1 ( f ) + bX 2 ( f )
The proof follows from the definition of the Fourier transform as a definite integral.
X ( f ) = - ( ax 1 ( t ) + bx 2 ( t ) ) e j2πft t X ( f ) = - ( ax 1 ( t ) + bx 2 ( t ) ) e j2πft t X ( f ) = a - x 1 ( t ) e j2πft t + b - x 2 ( t ) e j2πft t X ( f ) = a - x 1 ( t ) e j2πft t + b - x 2 ( t ) e j2πft t X ( f ) = aX 1 ( f ) + bX 2 ( f ) . X ( f ) = aX 1 ( f ) + bX 2 ( f ) .
b/ Delay by to
x ( t t o ) F X ( f ) e j2πft o x ( t t o ) F X ( f ) e j2πft o
This result can be seen using the synthesis formula.
x ( t ) = - X ( f ) e j2πft f x ( t ) = - X ( f ) e j2πft f x ( t t o ) = - X ( f ) e j2πf ( t t o ) f x ( t t o ) = - X ( f ) e j2πf ( t t o ) f x ( t t o ) = - X ( f ) e - j2πft o e j2πft f . x ( t t o ) = - X ( f ) e - j2πft o e j2πft f .
Mnemonic: a delay of the time function multiplies the Fourier transform by a lag factor, i.e., a delay of the time function of to adds −2πfto to the angle of the Fourier transform but does not affect the magnitude.
c/ Differentiate in t
dx ( t ) dt F j2πfX ( f ) dx ( t ) dt F j2πfX ( f )
This result can be seen using the synthesis formula.
x ( t ) = - X ( f ) e j2πft f x ( t ) = - X ( f ) e j2πft f dx ( t ) dt = - X ( f ) d dt ( e j2πft ) f dx ( t ) dt = - X ( f ) d dt ( e j2πft ) f dx ( t ) dt = - j2πf X ( f ) e j2πft f dx ( t ) dt = - j2πf X ( f ) e j2πft f
Differentiating the time function, adds π/2 radians to the angle of the Fourier transform and multiplies the magnitude by 2πf.
d/ Multiply by ej2πfot
x ( t ) e j2πf o t F X ( f f o ) x ( t ) e j2πf o t F X ( f f o )
This result can be seen using the analysis formula.
X ( f ) = - x ( t ) e j2πft t X ( f ) = - x ( t ) e j2πft t X ( f f o ) = - x ( t ) e j2π ( f f o ) t t X ( f f o ) = - x ( t ) e j2π ( f f o ) t t X ( f f o ) = - x ( t ) e j2πf o t e j2πft t X ( f f o ) = - x ( t ) e j2πf o t e j2πft t
This result can also be obtained from the delay-in-time property and duality. Mnemonic: multiplying a time function by a complex exponential at frequency fo shifts the Fourier transform to fo.
e/ Convolution in time
If x(t) = x1(t) ∗ x2(t) then X(f) is obtained as follows
X ( f ) = - ( - x 1 ( τ ) x 2 ( t τ ) ) e j2πft dt X ( f ) = - ( - x 1 ( τ ) x 2 ( t τ ) ) e j2πft dt = - x 1 ( τ ) ( - x 2 ( t τ ) e j2πft dt ) = - x 1 ( τ ) ( - x 2 ( t τ ) e j2πft dt ) = - x 1 ( τ ) X 2 ( f ) e j2πfτ = - x 1 ( τ ) X 2 ( f ) e j2πfτ = X 1 ( f ) X 2 ( f ) = X 1 ( f ) X 2 ( f )
The Fourier transform of the convolution of two time functions is the product of the Fourier transforms.
f/ Properties — zeroth-order moments
From the definition of the Fourier transform and its inverse
X ( f ) = - x ( t ) e j2πft t and x ( t ) = - X ( f ) e j2πft f X ( f ) = - x ( t ) e j2πft t and x ( t ) = - X ( f ) e j2πft f
it follows that
X ( 0 ) = - x ( t ) t and x ( 0 ) = - X ( f ) f X ( 0 ) = - x ( t ) t and x ( 0 ) = - X ( f ) f
g/ Properties — nth-order moments of time functions
Take the nth derivative of the Fourier transform with respect to f to obtain
d n X ( f ) df n = - ( j2πt ) n x ( t ) e j2πft dt d n X ( f ) df n = - ( j2πt ) n x ( t ) e j2πft dt
If we rearrange terms and evaluate the equation at f = 0, we obtain
1 ( j2π ) n d n X ( f ) df n | f = 0 = - t n x ( t ) dt 1 ( j2π ) n d n X ( f ) df n | f = 0 = - t n x ( t ) dt
Hence, the nth moment of x(t) can be obtained from the nth derivative of X(f) evaluated at f = 0. From duality of the Fourier transform, the nth moment of X(f) can be obtained from derivatives of x(t) evaluated at t = 0.
III. SIMPLE CTFT PAIRS
1/ Unit impulse in time
Suppose x(t) = δ(t). Then the CTFT is
X ( f ) = - δ ( t ) e j2πft dt = 1 X ( f ) = - δ ( t ) e j2πft dt = 1
This shows that we can synthesize the impulse as follow
δ ( t ) = - 1 · e j2πft df δ ( t ) = - 1 · e j2πft df
Generalization: a function that is punctuate in time has a Fourier transform that is extensive in frequency.
2/ Unit impulse in frequency
Suppose X(f) = δ(f). Then the inverse CTFT is
x ( t ) = - δ ( f ) e j2πft df = 1 x ( t ) = - δ ( f ) e j2πft df = 1
This shows that we can synthesize the impulse as follows
δ ( f ) = - 1 · e - j2πft dt δ ( f ) = - 1 · e - j2πft dt
Generalization: a function that is punctuate in frequency has an inverse Fourier transform that is extensive in time.
3/ Unit impulse shifted in time
Suppose x(t) = δ(t−to). Then from the delay property we obtain
X ( f ) = e j2πft o = cos ( 2 πft o ) j sin ( 2 πft o ) X ( f ) = e j2πft o = cos ( 2 πft o ) j sin ( 2 πft o )
4/ Unit impulse shifted in frequency
Suppose X(f) = δ(f − fo). From the properties, the inverse CTFT is
x ( t ) = e j2πf o t = cos ( 2 πf o t ) + j sin ( 2 πf o t ) x ( t ) = e j2πf o t = cos ( 2 πf o t ) + j sin ( 2 πf o t )
5/ Sinusoidal time functions
cos ( 2 πf o t ) = ( e j2πf o t + e j2πf o t 2 ) F δ ( f f o ) + δ ( f + f o ) 2 cos ( 2 πf o t ) = ( e j2πf o t + e j2πf o t 2 ) F δ ( f f o ) + δ ( f + f o ) 2 sin ( 2 πf o t ) = ( e j2πf o t - e j2πf o t 2 j ) F δ ( f f o ) - δ ( f + f o ) 2 j sin ( 2 πf o t ) = ( e j2πf o t - e j2πf o t 2 j ) F δ ( f f o ) - δ ( f + f o ) 2 j
6/ Causal exponential time function
x ( t ) = e αt u ( t ) 1 s + α for σ > α ( α > 0 ) x ( t ) = e αt u ( t ) 1 s + α for σ > α ( α > 0 )
Therefore, since the jω axis is in the region of convergence of X(s), we can evaluate X(s) on the jω axis to obtain the Fourier transform.
X ( ) = 1 + α X ( ) = 1 + α
X ( f ) = 1 j2πf + α X ( f ) = 1 j2πf + α
| X ( f ) | = 1 ( 2 πf ) 2 + α 2 and ∠X ( f ) = tan 1 ( 2 πf α ) | X ( f ) | = 1 ( 2 πf ) 2 + α 2 and ∠X ( f ) = tan 1 ( 2 πf α )
As α increases, the time function decays more rapidly and its duration (as measured by the time constant) decreases. As α increases, the Fourier transform width (as measured by its bandwidth) increases.
Thus, the width of the time function is inversely proportional to the width of the Fourier transform.
X ( f ) = 1 j2πf + α = α ( 2 πf ) 2 + α 2 + j 2 πf ( 2 πf ) 2 + α 2 X ( f ) = 1 j2πf + α = α ( 2 πf ) 2 + α 2 + j 2 πf ( 2 πf ) 2 + α 2
Therefore, we also have that
x e ( t ) = ( 1 / 2 ) e α | t | F X r ( f ) = α ( 2 πf ) 2 + α 2 x e ( t ) = ( 1 / 2 ) e α | t | F X r ( f ) = α ( 2 πf ) 2 + α 2
x o ( t ) = ( 1 / 2 ) e α | t | sgn ( t ) F jX i ( f ) = j 2 πf ( 2 πf ) 2 + α 2 x o ( t ) = ( 1 / 2 ) e α | t | sgn ( t ) F jX i ( f ) = j 2 πf ( 2 πf ) 2 + α 2
Where
sgn ( t ) = { 1 for t > 0 1 for t < 0 sgn ( t ) = { 1 for t > 0 1 for t < 0
Two-minute miniquiz problem
Problem 16-1 — Fourier transform of a unit step
The Laplace transform of a unit step is
x ( t ) = u ( t ) F X ( s ) = 1 s for σ > 0 x ( t ) = u ( t ) F X ( s ) = 1 s for σ > 0
This suggests that the Fourier transform of the unit step is
x ( t ) = u ( t ) F X ( f ) = 1 j2πf x ( t ) = u ( t ) F X ( f ) = 1 j2πf
The Fourier transform of a causal exponential is
x ( t ) = e αt u ( t ) F X ( f ) = 1 j2πf + α x ( t ) = e αt u ( t ) F X ( f ) = 1 j2πf + α
This also suggests that the Fourier transform of the unit step is
x ( t ) = u ( t ) F X ( f ) = 1 j2πf x ( t ) = u ( t ) F X ( f ) = 1 j2πf
Explain why this cannot be the Fourier transform of a unit step.
Solution
X ( f ) = 1 j2πf X ( f ) = 1 j2πf
is an imaginary odd function of f. Hence, it must be the Fourier transform of an odd function of t. The unit step is neither an odd nor an even function of t.
The argument based on the Laplace transform of a step is fallacious because the Laplace transform of the step has a region of convergence that does not include the jω axis. Hence, we cannot simply substitute s = j2πf into the Laplace transform to obtain the Fourier transform. The second argument is fallacious because care has to be taken in evaluating.
1 j2πf + α as α 0 at f = 0 1 j2πf + α as α 0 at f = 0
7/ Unit step
To obtain the Fourier transform of the unit step we start with the Fourier transform of a causal exponential
x ( t ) = e αt u ( t ) F X ( f ) = 1 j2πf + α x ( t ) = e αt u ( t ) F X ( f ) = 1 j2πf + α
and examine the solution as α → 0.
Note that
x ( t ) = e αt u ( t ) = x e ( t ) + x o ( t ) = ( 1 / 2 ) e α | t | + ( 1 / 2 ) e α | t | sgn ( t ) x ( t ) = e αt u ( t ) = x e ( t ) + x o ( t ) = ( 1 / 2 ) e α | t | + ( 1 / 2 ) e α | t | sgn ( t )
Note that
X ( f ) = 1 j2πf + α = X r ( f ) + jX i ( f ) = α ( 2 πf ) 2 + α 2 + j 2 πf ( 2 πf ) 2 + α 2 X ( f ) = 1 j2πf + α = X r ( f ) + jX i ( f ) = α ( 2 πf ) 2 + α 2 + j 2 πf ( 2 πf ) 2 + α 2
As α → 0, Xr(f) becomes tall and narrow and, as we shall see, its area is 1/2. Hence, as α → 0,
X r ( f ) 1 2 δ ( f ) and jX i ( f ) 1 j2πf X r ( f ) 1 2 δ ( f ) and jX i ( f ) 1 j2πf
We need to determine the area of Xr(f). Because
x e ( t ) F X r ( f ) x e ( t ) F X r ( f ) x e ( 0 ) = - X r ( f ) df . x e ( 0 ) = - X r ( f ) df .
Hence,
x e ( 0 ) = - X r ( f ) df = 1 2 x e ( 0 ) = - X r ( f ) df = 1 2
Thus, we have
u ( t ) = 1 2 + 1 2 sgn ( t ) F F { u ( t ) } = 1 2 δ ( f ) + 1 j2πf u ( t ) = 1 2 + 1 2 sgn ( t ) F F { u ( t ) } = 1 2 δ ( f ) + 1 j2πf
Unit step, bottom line
To summarize,
u ( t ) F 1 s for R { s } > 0 u ( t ) F 1 s for R { s } > 0 u ( t ) F 1 2 δ ( f ) + 1 j2πf . u ( t ) F 1 2 δ ( f ) + 1 j2πf .
Thus, if the Laplace transform is evaluated on the edge of the region of convergence, on the jω axis, then there is an impulse in the real part at the location of the pole.
8/ Signum and unit step function — another approach
We illustrate a method for finding Fourier transforms using the Fourier transform properties and the Fourier transforms of simple time functions. Consider the time function x(t) = (1/2)sgn(t).
The derivative of x(t) is a unit impulse in time. Differentiation in time is equivalent to multiplying the transform by j2πf.
Hence,
x ( t ) = sgn ( t ) F X ( f ) = 1 j2πf x ( t ) = sgn ( t ) F X ( f ) = 1 j2πf
The same approach can be used to find the Fourier transform of the step, x(t) = u(t), if some care is exercised. Note that u(t) = 1/2 + (1/2)sgn(t).
The derivative of x(t) is a unit impulse in time but the constant is lost. Thus, it must be included.
Hence,
x ( t ) = u ( t ) = 1 / 2 + ( 1 / 2 ) sgn ( t ) X ( f ) = ( 1 / 2 ) δ ( f ) + 1 j2πf x ( t ) = u ( t ) = 1 / 2 + ( 1 / 2 ) sgn ( t ) X ( f ) = ( 1 / 2 ) δ ( f ) + 1 j2πf
Two-minute miniquiz problem
Problem 17-1 — Step with a dc offset
Find the Fourier transform of x(t).
Solution
We can represent x(t) as
x(t) = 3/2 + (1/2)sgn(t).
Hence,
X ( f ) = ( 3 / 2 ) δ ( f ) + 1 j2πf X ( f ) = ( 3 / 2 ) δ ( f ) + 1 j2πf
9/ Causal cosinusoidal time function
A causal cosinusoid is a cosinusoid that starts at t = 0,
x ( t ) = cos ( 2 πf o t ) u ( t ) = cos ( 2 πf o t ) ( 1 / 2 + ( 1 / 2 ) sgn ( t ) ) x ( t ) = cos ( 2 πf o t ) u ( t ) = cos ( 2 πf o t ) ( 1 / 2 + ( 1 / 2 ) sgn ( t ) )
Therefore,
X ( f ) = F { cos ( 2 πf o t ) } * F { u ( t ) } X ( f ) = F { cos ( 2 πf o t ) } * F { u ( t ) } = ( 1 / 2 ) ( δ ( f f o ) + δ ( f + f o ) ) * ( ( 1 / 2 ) δ ( f ) + 1 j2πf ) = ( 1 / 2 ) ( δ ( f f o ) + δ ( f + f o ) ) * ( ( 1 / 2 ) δ ( f ) + 1 j2πf ) = ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + 1 j4π ( f f o ) + 1 j4π ( f + f o ) = ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + 1 j4π ( f f o ) + 1 j4π ( f + f o ) = ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + f j2π ( f 2 f o 2 ) = ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + f j2π ( f 2 f o 2 )
The Fourier and Laplace transforms of the causal cosinusoid are:
cos ( 2 πf o t ) u ( t ) F ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + f j2π ( f 2 f o 2 ) cos ( 2 πf o t ) u ( t ) F ( 1 / 4 ) ( δ ( f f o ) + δ ( f + f o ) ) + f j2π ( f 2 f o 2 ) cos ( 2 πf o t ) u ( t ) F s s 2 + ( 2 πf o ) 2 for R { s } > 0 cos ( 2 πf o t ) u ( t ) F s s 2 + ( 2 πf o ) 2 for R { s } > 0
Therefore, just as with the unit step, the causal cosinusoid has poles on the jω axis, and the Fourier transform contains impulses at the frequencies of the poles.
A causal sinusoid is a sinusoid that starts at t = 0,
x ( t ) = sin ( 2 πf o t ) u ( t ) = sin ( 2 πf o t ) ( 1 / 2 + ( 1 / 2 ) sgn ( t ) ) x ( t ) = sin ( 2 πf o t ) u ( t ) = sin ( 2 πf o t ) ( 1 / 2 + ( 1 / 2 ) sgn ( t ) )
Therefore,
X ( f ) = F { sin ( 2 πf o t ) } * F { u ( t ) } X ( f ) = F { sin ( 2 πf o t ) } * F { u ( t ) } = ( 1 / 2 j ) ( δ ( f f o ) δ ( f + f o ) ) * ( ( 1 / 2 ) δ ( f ) + 1 j2πf ) = ( 1 / 2 j ) ( δ ( f f o ) δ ( f + f o ) ) * ( ( 1 / 2 ) δ ( f ) + 1 j2πf ) = ( 1 / 4 ) ( δ ( f f o ) - δ ( f + f o ) ) + 1 4 π ( f f o ) 1 4 π ( f + f o ) = ( 1 / 4 ) ( δ ( f f o ) - δ ( f + f o ) ) + 1 4 π ( f f o ) 1 4 π ( f + f o ) = ( 1 / 4 j ) ( δ ( f f o ) - δ ( f + f o ) ) + f o 2 π ( f 2 f o 2 ) = ( 1 / 4 j ) ( δ ( f f o ) - δ ( f + f o ) ) + f o 2 π ( f 2 f o 2 )
10/ Rectangular pulse
a/ Rectangular pulse — derivation
The transform of the rectangular pulse x(t) is:
X ( f ) = T / 2 T / 2 Ae j2πft t X ( f ) = T / 2 T / 2 Ae j2πft t = Ae j2πft j2πf | - T / 2 T / 2 = Ae j2πft j2πf | - T / 2 T / 2 = AT e jπfT e jπfT j2πfT = AT e jπfT e jπfT j2πfT = AT ( sin πfT πfT ) = AT ( sin πfT πfT )
sin πfT πfT = { 0 for f = n / T , n 0 1 for f = 0 sin πfT πfT = { 0 for f = n / T , n 0 1 for f = 0
where n is an integer.
b/ Use of moment properties
From the moment properties we know that
X ( 0 ) = - x ( t ) dt = AT and x ( 0 ) = A = - X ( f ) df X ( 0 ) = - x ( t ) dt = AT and x ( 0 ) = A = - X ( f ) df
Note thatwhich is just the area of the inscribed triangle.
c/ Sinc function
The type of function arises so frequently that it is useful to
define the sinc function,
sinc ( x ) = sin πx πx sinc ( x ) = sin πx πx
Therefore,
X ( f ) = AT ( sin πfT πfT ) = ATsinc ( fT ) X ( f ) = AT ( sin πfT πfT ) = ATsinc ( fT )
d/ Source of zeros in FT
has zeros at f = n/T for n ≠ 0. What causes these zeros?
Note from the definition of the Fourier transform
X ( f ) = - x ( t ) e j2πft t X ( f ) = - x ( t ) e j2πft t = - x ( t ) ( cos ( 2 πft ) j sin ( 2 πft ) ) t = - x ( t ) ( cos ( 2 πft ) j sin ( 2 πft ) ) t
The duration of x(t) is T. Hence, the integral is zero for those frequencies whose periods are submultiples of T. These are the frequencies f = n/T where n ≠ 0. The examples show the frequencies f = 1/T , f = 2/T , and f = 3/T .
e/ Alternate derivation of FT
The Fourier transform can be found from
j2πfX ( f ) = Ae j2πf ( T / 2 ) Ae j2πf ( T / 2 ) = 2 jA sin ( πfT ) j2πfX ( f ) = Ae j2πf ( T / 2 ) Ae j2πf ( T / 2 ) = 2 jA sin ( πfT ) X ( f ) = AT ( sin ( πfT ) πfT ) X ( f ) = AT ( sin ( πfT ) πfT )
f/ Effect of duration
Consider the sequence of rectangular pulses of the form
x ( t ) = { 0 otherwise 1 / T for t < | T / 2 | F X ( f ) = sin ( πfT ) πfT x ( t ) = { 0 otherwise 1 / T for t < | T / 2 | F X ( f ) = sin ( πfT ) πfT
Note that as T decreases x(t) becomes tall and narrow and X(f) gets broader in frequency. Interpreted as generalized functions, x(t) → δ(t) and X(f) → 1.
11/ Triangular pulse
A triangular pulse is obtained by convolving a rectangular pulse with itself.
Since xT (t)=xR(t)*xR(t), the triangular pulse has the transform
X T ( f ) = X R ( f ) × X R ( f ) = ( T sin ( πfT ) πfT ) 2 X T ( f ) = X R ( f ) × X R ( f ) = ( T sin ( πfT ) πfT ) 2
IV. FILTERING REVISITED
For an LTI system with impulse response h(t) with arbitrary input x(t), the output is given by the convolution of the input with the impulse response.
The Fourier transform of the output equals the product of the Fourier transform of the input and the Fourier transform of the impulse response which is just the frequency response of the LTI system. Filtering is an example of a signal processing operation that is effectively viewed in the frequency domain in terms of Fourier transforms of the input, impulse response, and output.
1/ Use of LPF and HPF to separate sinusoids
The signal consists of a sum of sine waves that we wish to separate.
To design the filters we need to specify precisely what we mean by separation. These specifications are easily evaluated in the frequency domain.
The Fourier transform (spectrum) of the signal consists of impulses at the frequencies of the sinusoids. These are superimposed on the frequency responses of the LPF and HPF.
As the frequency separation between the two sinusoids is decreased, the order of the filter required to meet a given specification must be increased.
2/ Use of BPF to extract narrow-band signal from wide-band noise
The Fourier transform (spectrum) of the signal consists of a narrow-band signal, i.e., such as a sinusoid, and wide-band noise, i.e., an undesirable signal. The problem is to extract the signal from the noise using a BPF.
Once again we superimpose the spectrum of the signal and the frequency response of the system.
As the bandwidth of the BPF decreases, the amount of noise in the output decreases.
3/ Extraction of signal from noise
The input consists of an ECG signal recorded from the surface of the chest. The recording consists of the signal (the electrical activity of the heart) plus noise from various sources (e.g., the power lines, electrical activity of other muscles, noise from the electrodes). The objective is to extract the signal from the noise.
To design an appropriate filter, we examine the spectrum of the recorded signal.
The spectrum of the recorded signal is shown below.
The spectrum of the electrical activity of the heart is predominantly in the range 0.5 to 35 Hz. The rest is “noise.” Note the large peak at 60 Hz.
The recorded ECG signal is passed through a filter.
Frequency response of a second-order bandpass filter in cascade with a notch filter (at 60 Hz). The passband of the bandpass filter is between 0.38 and 60 Hz (as suggested in Problem 6 of Problem Set 4).
The filter characteristics are shown on an expanded and linear frequency scale centered on 60 Hz to show the effect of the notch filter.
The filtered and unfiltered ECG waveforms and spectra are shown below.
The first 10s of the filtered and unfiltered ECG waveforms are shown below.
Why does the filtered waveform appear thick in the first two seconds?
The impulse response of the filter is shown below.
The impulse response is shown on two amplitude and time scales.
It takes about 1-2 seconds for the impulse response to attenuate appreciably?
V. CONCLUSIONS:
• The combination of Fourier transform properties and the Fourier transforms of simple time functions yields a rich collection of Fourier transform pairs.
• If the jω axis is in the region of convergence of the Laplace transform, then the Fourier transform equals the Laplace transform evaluated along the jω axis.
• If there are poles on the jω axis, so that the Laplace transform does not include the jω axis, the Fourier transform can still be defined with the use of singularity functions. This situation will be explored further in the next lecture.
• The filter has attenuated the low-frequency noise as can be seen from the spectrum. Note also that the baseline of the waveform of the filtered ECG signal has been flattened.
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## Non-lifting irreversible Monte Carlo
While developing ECMC, we always wonder what role the lifting variable that indicates active particle plays for the efficiency of ECMC. Such a lifting variable makes the active particle motion persistent, and by imposing very short chains, the performance of ECMC returns to the case of traditional Monte Carlo. However, advancing active particles in the same direction is not equivalent to traveling in phase space without turning back. On the other hand, keeping explicit directions in lifting variable does not always help, as suggested by the example of skew detailed balance (see section 1.4.2.2).
By presenting three examples of non-lifting Monte Carlo followed by two with lifting, we review people’s effort in achieving irreversibility. Actually, two of three non-lifting irreversible Monte Carlo keep additional implicit information: The next variable to relax in a sequential scheme, and the momentum in hybrid Monte Carlo. Checking how different degrees of irreversibility improve mixing provides us instructive (though feeble) clues over the role of lifting in ECMC’s excellent performance.
### Hybrid Monte Carlo
Hybrid Monte Carlo, also known as Hamiltonian Monte Carlo, was introduced by Duane et al. [61]. It applies only to continuous configuration space because it uses a molecular dynamics integrator to generate proposal moves. The Metropolis-Hastings filter is used to accept the move (see [62], chapter 5). Hybrid Monte Carlo process can be described as:
1. Extend the state space s = (q1, …,qn) to a phase space with velocity so that s = (q,v), q = (q1, …,qn), and v = (p1, …,pn). Define the Hamiltonian H(q,p) = U(q)+Pni =1 p2i /2.
2. (a) For the current (q,p), perform Hamiltonian mechanics of Eq. 1.2.1 for time t, which reduces to q˙i = pi and p˙i = −@U/@qi. Then let p0 = −p(t), q0 = q(t). (b) Accept (q0,p0) with the Metropolis-Hastings filter, and the energy is replaced with H.
3. Flip p0 −p0.
4. Disturb p by p00 = p0 p1−+N, where N is a normal distribution with same dimension of p centering around zero and has variance corresponding to the temperature, is a tuning parameter.
5. Take (q0,p00) as a new state and go to step 2 until halted.
#### Lifting irreversible Monte Carlo
Reversible Monte Carlo methods employ symmetric proposal probabilities fp(i ! j) = fp(j ! i) so that tuning the acceptance rate allows us to realize detailed balance. Many systems achieve fast exploration with explicit use of direction. For example, particle momentum counteracts diffusive dynamics to make components in a fluid mix faster. Angular momentum drives a molecule under constant rotation.
Persistence in flipping up spins or flipping down will also lead to rapid decorrelation between spin configurations.
Lifting variables allow people to add information in Monte Carlo moves so that the program has the memory of previous moves. A state si is extended with a lifting variable si = (xi,ai), and the proposal depends heavily upon the lifting variable ai. Here we present several lifting irreversible Monte Carlo ideas, including onedimensional random walk, skew detailed balance, and our ECMC.
One-dimensional random walk
A chain of N sites with uniform desired probability is one of the few models with lifting variables that rigorous mathematics can apply. Diaconis et al. [64] established the first rigorous scaling of convergence for lifting irreversible Monte Carlo, proving that its mixing time converges within O(N) steps compared to O(N2) with an undirected random walk.
Previously, an unbiased random walk may jump from site i to i−1 or i+1 each with probability 1/2. The end of the chain needs special treatment if periodic boundary conditions are not applied. Diaconis et al. added the direction to the state represented by (i,d) with d = ±1, and the transition rule is T(i,d)!(i+d,d) = 1− 1 N .
READ GENDERED SOCIAL TRANSFORMATIONS IN POST COLONIAL ZIMBABWE
Fast implementation
If ECMC stays in the time-driven approach, it will hardly be capable of competing with other well-established methods. It consumes a great deal of time to integrate over small timesteps and introduces imprecision that is hard to analyze. Molecular dynamics has accumulated progress to overcome this inefficiency over the years, by parallel computing forces throughout the entire system, or altering potentials into piecewise-constant ones (see section 1.2.3) to which event-driven approaches can apply. ECMC is a stochastic process with a natural event-driven formulation with whatever potential the system has. In section 2.3.1 we describe how to convert small step integration into inverting potential functions. Then in section 2.3.2, we introduce the bounding potential that simplifies the event calculation so that only the derivative of the original potential is called.
General Introduction
1 Simulation method review
1.1 The need for large-scale simulation
1.1.1 Examples
1.1.2 State of the art
1.2 Molecular dynamics
1.2.1 Integrator
1.2.2 Long-range algorithms
1.2.3 Event-driven molecular dynamics
1.3.1 Properties of transition matrix
1.3.2 Tricks by modifying target distribution
1.3.3 Several reversible algorithms
1.4 Irreversible Monte Carlo
1.4.1 Non-lifting irreversible Monte Carlo
1.4.2 Lifting irreversible Monte Carlo
1.5 Models and measurements
1.5.1 Molecular models
1.5.2 Measurements
2 Event-chain Monte Carlo
2.1 Introduction
2.2 Basics
2.2.1 Lifted configurations
2.2.2 Factor event rate
2.2.3 Factorization
2.3 Fast implementation
2.3.1 Event-driven approach
2.3.2 Bounding potential
2.3.3 Sampling
2.3.4 Direction switching and restarts
3 Event-chain Monte Carlo for long-range interaction
3.1 Long-range system
3.2 Calculation of pairwise Coulomb interaction
3.2.1 Ewald summation
3.2.2 Line-charge method
3.2.3 Higher-order methods
3.2.4 Interpolation
3.2.5 Separate-image method
3.3 ECMC cell-veto algorithm
3.3.1 Description
3.3.2 Cell-veto event rate
3.3.3 Walker’s method
3.4 Coulomb bounding potential
3.5 Dipole factors
3.5.1 Two-dipole case
3.5.2 Homogeneous systems
3.5.3 Dipole lifting schemes
3.6 Numerical tests
3.6.1 Dipole system
3.6.2 SPC/Fw water model
4 JeLLyFysh architecture
4.1 Issues of application development
4.2 Features of JELLYFYSH-V1.0
4.3 Basic concepts
4.3.1 Event flow
4.3.2 Units
4.3.3 Global state and internal state
4.4 Essential modules
4.4.1 Mediator
4.4.2 Event handler
4.4.3 Activator
4.4.4 State handler
4.4.5 Scheduler
4.4.6 Input-output handler
4.5 Important tools
4.5.1 Globally used modules
4.5.2 Cells and cell occupancy
4.5.3 Initializer
4.5.4 Potentials, estimator
4.5.5 Lifting schemes
5 JELLYFYSH simulation
5.1 Event handlers of JELLYFYSH-V1.0
5.1.1 Event handler for two-body invertible potentials
5.1.2 Event handlers for non-invertible potentials
5.1.3 Cell-veto event handler
5.1.4 Event handlers for composite object motion
5.1.5 Sampling event handler and end-of-chain event handler .
5.1.6 Other helping event handlers
5.2 Factory
5.2.1 Running
5.2.2 Customization
5.3 Verifications
5.3.1 Coulomb atoms
5.3.2 Dipoles
5.3.3 SPC/Fw water
6 Realistic water system
6.1 Speeding up JELLYFYSH
6.1.1 Fast track for unconfirmed events
6.1.2 Other improvements
6.2 Coulomb event profiling
6.2.1 Charge-water estimator
6.2.2 Optimizing cell-veto events
6.2.3 Dipole factor vs. atomic factor
6.3 Water system benchmark
6.3.1 Chain state
6.3.2 Molecular translation
6.3.3 Discussions
7 Sequential Monte Carlo
7.1 Single-dipole system
7.1.1 Configuration
7.1.2 Sequential Monte Carlo method
7.1.3 Validation
7.2 Rotational dynamics
7.2.1 Zigzag and excursion
7.2.2 Large-time limit
7.3 Mixing of orientation
General Conclusion
Bibliography
GET THE COMPLETE PROJECT | 2,026 | 7,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.869875 |
https://www.esaral.com/q/if-9th-term-of-an-a-p-is-zero-98852 | 1,723,099,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00435.warc.gz | 602,323,165 | 11,854 | # If 9th term of an A.P is zero,
Question:
If 9th term of an A.P is zero, prove that its 29th term is double the 19th term.
Solution:
In the given problem, the 9th term of an A.P. is zero.
Here, let us take the first term of the A.P as a and the common difference as d
So, as we know,
$a_{n}=a+(n-1) d$
We get,
$a_{9}=a+(9-1) d$
$0=a+8 d$
$a=-8 d$ .......(1)
Now, we need to prove that $29^{\text {th }}$ term is double of $19^{\text {th }}$ term. So, let us first find the two terms.
For $19^{\text {th }}$ term $(n=19)$,
$a_{19}=a+(19-1) d$
$=-8 d+18 d \quad($ Using 1 $)$
$=10 \mathrm{~d}$
For 29th term (n = 29),
$a_{29}=a+(29-1) d($ Using 1)
$=-8 d+28 d$
$=20 d$
$=2 \times 10 d$
$=2 \times a_{19}$
Therefore, for the given A.P. the 29th term is double of the 19th term.
Hence proved. | 321 | 813 | {"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} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.792073 |
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She offered him a hundred-watt smile: contains a metaphor.
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If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 196 N B. 1.96 kg C. 32.67 N D. 2.04 kg
Updated 6/25/2019 7:14:32 AM
If an object has a mass of 20 kg, The force of gravity acting on earth is 196N force =mass *acc , force = 20 *9.8 = 196N
Which of the following would decrease the resistance to the flow of an electric current through a body? A. Lengthening the conductor B. Shortening the conductor C. Using a conductor with a smaller cross section D. Heating the conductor
Updated 67 days ago|8/22/2020 9:24:11 AM
Shortening the conductor would decrease the resistance to the flow of an electric current through a body.
Added 67 days ago|8/22/2020 9:24:11 AM
Which of the following values represents an index of refraction of an actual material? A. .5 B. 0 C. .25 D. 1.25
Weegy: The value 1.25 represents an index of refraction of an actual material. (More)
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View Full Version : Clarification about nomenclature of mathematics courses
Ancalagon The Black
02-16-2008, 05:06 AM
I learn new things every day here in TM. The other day I learnt that Calculus I was primarily differential calculus, Calculus II was integral calculus and Calculus III was multivariate.
So, now I had another question. What is Calculus IV?
Another question is, I applied to a program which said that you must have a year of undergraduate probability and statistics. Now, what would comprise the syllabus of such 2 courses (or 1 year long course)
Any appropriate link would be more than welcome.
Thanks.
polkaparty
02-16-2008, 05:15 AM
I learn new things every day here in TM. The other day I learnt that Calculus I was primarily differential calculus, Calculus II was integral calculus and Calculus III was multivariate.
I think you got the wrong impression in general. Calc 1 is differential and integral calculus in a univariate setting. Calc 2 is a hodgepodge of techniques and topics. Calc 3 is multivariate.
So, now I had another question. What is Calculus IV?
My school doesn't have a calc 4. There might be trimester programs that do have a calc 4 where the calc 1-3 material is split into 4 courses, but that's just a guess. For semester system schools, I don't know what a calc 4 course would entail, perhaps a mixture of linear algebra and diff eqs?
Another question is, I applied to a program which said that you must have a year of undergraduate probability and statistics. Now, what would comprise the syllabus of such 2 courses (or 1 year long course)
Definitions and basic probability theory (Bayes rule, whatever), distributions, expected values, moments, convergence, sample data, estimators, biasedness, inference, etc.
For example, almost all of the material in this textbook (http://www.amazon.com/Mathematical-Statistics-Analysis-Duxbury-Advanced/dp/0534399428/ref=ed_oe_h).
takemoremath
02-16-2008, 05:16 AM
At many schools Calculus IV is a first course in ordinary differential equations.
AstralTraveller
02-16-2008, 05:30 AM
Definitions and basic probability theory (Bayes rule, whatever), distributions, expected values, moments, convergence, sample data, estimators, biasedness, inference, etc.
I lectured the undergraduate sequence of Probability and Stats at my alma mater. The first course was mainly random variables, distributions and moments of first and higher order, and the second course was statistical inference (point estimation, asymptotic theory, confidence intervals, hypotheses testing). I find it weird to learn all these subjects without basic notions of measure theory, because otherwise the students feel they are fed a bunch of un-understandable formulas and loads of stuff they have to memorize. It's so frustrating, both for the teacher and as for the students! With Probability and Measure theory, it all suddenly makes sense!
OK, enough rambling....sorry for (somewhat) derailing the main point of the thread.
Ancalagon The Black
02-16-2008, 05:31 AM
Thanks a lot guys!!
I have done 1 course in Calculus which is both differential and integral calculus combined (we all know calculus very very thoroughly in high school). One course in Multivariate Calculus and 1 course in Differential Equations (ODE and PDE). I suppose that covers Calc I - IV if there.
I have also done 1 course in Probability and Statistics which covers all the material which you mention (and more besides)
polka: I think buckykatt told me that Calc I was diff and Calc II integral. I used to think that Calc II was multivariate and Calc III was tougher and more advanced versions but thank you for the clarifications. It seems to me that even in the US different schools have very different nomenclatures for their courses, especially in mathematics.
I actually did 2 prob/stats courses - 1 from the business department and 1 from the mathematics department. Both were similar but the mathematics department course covered more material as I mentioned above. That should suffice no? I didn't include any syllabus with my application packet. I hope it is enough. Otherwise.... :(
canecon
02-16-2008, 07:48 AM
You basically described the calc 1-3 we have at UBC. For us Calc 4 is:
Textbook Multivariable Calculus, J. Stewart, edition 5e.
Course outline
Chapter 14. Curves, vector functions, velocity, acceleration, length.
Chapter 17, 3. Vector fields and line integrals
Chpater 17, 4. Green's theorem
Chapter 17, 5. Curl, Divergence.
Chapter 17, 6{7. Surfaces and surface integrals.
Chapter 17, 8{9. Stokes and divergence theorems.
asianeconomist
02-16-2008, 08:35 AM
You basically described the calc 1-3 we have at UBC. For us Calc 4 is:
Textbook Multivariable Calculus, J. Stewart, edition 5e.
Course outline
Chapter 14. Curves, vector functions, velocity, acceleration, length.
Chapter 17, 3. Vector fields and line integrals
Chpater 17, 4. Green's theorem
Chapter 17, 5. Curl, Divergence.
Chapter 17, 6{7. Surfaces and surface integrals.
Chapter 17, 8{9. Stokes and divergence theorems.
I studied identical topics, only the textbook (Anton, Bivens,Davis: Early Transcendentals) was different. Oh, and I also did a bit of ODE's
Ancalagon The Black
02-16-2008, 11:32 AM
canecon:
The syllabus you mentioned is part of my course called Advanced Calculus though we use Indian author books.
AT:
The 2 courses which you mention in probability is very helpful. All the topics are present in one of the courses which I did in the mathematics department named Probability and Statistics. We did not do measure theory there. Topology and measure theory was covered in our Real Analysis course.
Does this make it a rigorous mathematics preparation for me? Consider also the following taken: Linear Algebra, Diff Eq (PDE/ODE), Real Analysis(includes topics from both RAI and RAII which I gathered in another course), Numerical Analysis and Discrete Mathematics.
AstralTraveller
02-16-2008, 11:52 AM
Does this make it a rigorous mathematics preparation for me? Consider also the following taken: Linear Algebra, Diff Eq (PDE/ODE), Real Analysis(includes topics from both RAI and RAII which I gathered in another course), Numerical Analysis and Discrete Mathematics.
I strongly believe it does. Now, it would be great to find a way to shove it on the face of the adcoms...although they should be aware of your background coming from India. If they don't know something, they might ask you.
All my best, AT
buckykatt
02-16-2008, 04:43 PM
polka: I think buckykatt told me that Calc I was diff and Calc II integral.
I said that at my school we covered both in Calc I but Calc II covered integration in much more depth (both in terms of techniques and applications, e.g. finding the volumes and surfaces of various geometric objects) plus lots of stuff with sequences. Calc III is multivariate with lots of vector stuff.
One of the things that accounts for the variation in how the material is broken up is that some schools cover "two years" of calculus in four semesters (4 classes of 3 credits each) while others do so in three semesters (3 classes of 4 credit each). (And, of course, some schools use a quarter system--I'm not sure how they break things up there.) So one school's Calc I - IV could be the same as another's Calc I - IV, in terms of contact hours, while a different school's could actually cover more material. The trend has been schools switching from three credit to four credit classes.
The bottom line is that if you've completed whatever part of your school's sequence gets you through multivariate calculus with vectors, you should be all set with the "two years of calculus" part of the requirement. At many schools, this sequence also includes at least some introduction to differential equations, but usually an in-depth study is part of a separate course. Ditto for the topics that come under the heading of "real analysis".
buckykatt
02-16-2008, 04:47 PM
Now, it would be great to find a way to shove it on the face of the adcoms
If it were me, I'd prepare a one-page list of the math & stats courses I'd taken, with short descriptions from the catalog or syllabus and the textbook used, and include it with all my applications. If you make it easy for folks to understand your transcript, they probably will. :)
Ancalagon The Black
02-16-2008, 04:50 PM
Aaah, thanks for the clarification bucky. I have a pretty poor memory. As I said before, we did a separate course on Differential Equations in great detail.
AT: I did shove it in the face of the adcoms but I didn't include the syllabus. The funny thing is, even after the application process is over, I am discovering ways in which I could have made my SOP better or my packet more streamlined or even more clarification about my educational system. :( :D
Ancalagon The Black
02-16-2008, 04:55 PM
bucky: I read your post moments after I submitted mine. Thats exactly the kind of idea I am getting NOW after the applications process is over... :(
buckykatt
02-16-2008, 05:00 PM
bucky: I read your post moments after I submitted mine. Thats exactly the kind of idea I am getting NOW after the applications process is over... :(
I got the idea from one of the master's programs I applied to this year, which asked for exactly this (minus the course descriptions). I went ahead and included it with my other master's apps, figuring better too much info than not enough...
But I wouldn't worry. If your letters of recommendation, grades, scores, etc. are good and the adcom thinks they should be admitting you, they'll probably seek out clarification if there's any doubt about your prerequisites. Hang in there!
doubtful
02-16-2008, 05:08 PM
for Minnesota I prepared 8 pages of math-stats exams with detailed syllabus.. maybe I exagerated.. the fact is that CalcI-->CalcIV--->Real Analysis. are spread trough 6 years..
polkaparty
02-16-2008, 05:12 PM
I don't understand all these people sending extra stuff. I thought schools specifically ask for applicants to not send anything which is not requested.
Thesus
02-16-2008, 05:13 PM
Calc IV over here is pretty much identical to what canecon described as Calc IV at UBC.
As for statistics, that's a bit more complicated. At least here, we have two statistics streams: statistics, and mathematical statistics. The former is basically just running through techniques, while the second is proof-based and relies on Hogg/McKean/Craig, Casella/Berger.
doubtful
02-16-2008, 05:17 PM
I don't understand all these people sending extra stuff. I thought schools specifically ask for applicants to not send anything which is not requested.
do you believe them?
my guess is that if they don't want it they trash it... I don't think they are gonna disqualify people because of extra paper.
anyway.. I mean I sent stuff that thought to be useful, UMN asked for something like that and the graduate coordinator told me it was ok to send an extended version of the syllabi to explain the difference in the classes I have taken in Europe with respect to the American system.
buckykatt
02-16-2008, 05:38 PM
I don't understand all these people sending extra stuff. I thought schools specifically ask for applicants to not send anything which is not requested.
I've seen some programs say that, and to those I wouldn't send anything extra. Maybe I'd make the file available via the web and reference the URL in my SOP "for further information" though.
The programs I applied to basically said "send us anything you think will help us", though. They also required that I assemble everything--letters, transcripts, etc.--and send it together in one big package, which I imagine saved them some work. :)
Ancalagon The Black
02-17-2008, 05:10 AM
doubtful:
Did you get into UMN with all that info? I also believe in the fact that too much information is better than too little info. After all, we have to distinguish ourselves from everybody else. The adcoms want a story - they don't want automatons. I believe that this is a reason why perfect 4.0, 800qs are rejected in favor of research/work experience and say, relevant hardships faced.
After all, nobody has the same opportunities.
bucky:
I did exactly that. I built an e-portfolio and included the link in every application I could get. Maybe they will see it, maybe they won't, but I am sure (from poking around in my webstats), that most of the schools (and the professors whom I emailed) have looked around in my e-portfolio.
Praise Web 2.0 !! :D
econofrosh!
02-20-2008, 11:14 AM
Hello fellow TM-ers! I've been lurking for some time now and this is my first post! :D Bear with me. haha
I realize that the term 'Advanced Calculus' and 'Real Analysis' are used interchangeably; however, there are two different courses in our math department that are named so. Which would be the equivalent of RA among these:
Math 123.1. ADVANCE CALCULUS I. The real number system; point set topology sequence of real numbers; limits and continuity; metric spaces; the derivative; the Riemann integral; series of real numbers; sequence and series of functions; uniform convergence; power series. 3 units. Prerequisite: Math 65 or equivalent
Math 123.2. ADVANCE CALCULUS II. Topology of Rn; continuity and derivative; Taylor's formula; implicit and inverse function theorems; analytic geometry of curves and surfaces; Serret-Frenet formula; multiple integration; improper integrals; transformation; metric and normed spaces. 3 units.
Prerequisite: Math 123.1
Math 126. REAL ANALYSIS. Properties of real numbers; integrals of step functions; Lecesque integral; convergence theorems; measurable functions; measurable sets; introduction to the Hahn-Banach theorem, Riesz representation theorem, fixed-point theorems. 3 units.
Prerequisite: 123.1
If I were to take only two from the abovementioned (I would definitely have to take Advanced Calc I), which would be more advisable to take?
Thanks in advance! :) And good luck to those whose application results are about to come! ;)
asianecon
02-20-2008, 01:08 PM
Hi econofrosh. I'm guessing you're form UP. I'm not, but there are several here. I can think of one person that might help you regarding course choices.
(in case you're wondering how I found out: Lecesque integral)
zappa24
02-20-2008, 04:21 PM
At my undergrad, Calculus I was differentiation. Calculus II was devoted to integration. Calculus III was the before mentioned hodge podge of techniques. Multivariate Calculus didn't come until Calculus IV. The Math department has since condensed the entire sequence to three courses.
fp3690
02-20-2008, 06:32 PM
econofrosh, advanced calc 1 and 2 look like analysis 1 and 2 respectively. Real analysis looks like a grad course in analysis, and also a bit like measure theory, based on my limited contact with the latter. I think even advanced calc 1 is sufficient.
Mr.Keen
02-20-2008, 06:41 PM
I would say Adv Calc 1 is like an introduction to analysis. If you can take that course, then learn on your own topology on Rn and metric and normed spaces you can enroll in real analysis
buckykatt
02-20-2008, 10:02 PM
econofrosh, advanced calc 1 and 2 look like analysis 1 and 2 respectively. Real analysis looks like a grad course in analysis, and also a bit like measure theory, based on my limited contact with the latter. I think even advanced calc 1 is sufficient.
That's my sense, as well.
asianecon
02-20-2008, 10:13 PM
econofrosh, advanced calc 1 and 2 look like analysis 1 and 2 respectively. Real analysis looks like a grad course in analysis, and also a bit like measure theory, based on my limited contact with the latter. I think even advanced calc 1 is sufficient.
I agree. Econofrosh and I have a similar educational system. Adv Calc 1 and 2"are actually first courses in Calculus in my local university and these are aptly labeled as Math Analysis 1 and 2.
mysherona
02-20-2008, 10:47 PM
(in case you're wondering how I found out: Lecesque integral)
I'm still wondering how you found out... How does "Lecesque" tell you this?
C152dude
02-21-2008, 01:30 AM
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## #1 2012-12-12 00:06:57
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Proof
Hi,
I really need help seeing if I answered the following correct:
I'll give you an assumption, and ask you to provide proof for the assumption. If there is no proof for the assumption, the answer is "unfounded."
I choose A
1. If I have two coplanar lines, I must have a plane.
A-unfounded
B-Definition of a point
C-Definition of a plane
D-Given
E -Definition of a line
F -Definition of radius
Last edited by zee-f (2012-12-12 00:41:33)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #2 2012-12-12 04:43:32
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
hi zee-f,
I think this can be proved. Isn't a plane defined by 3 (non-colinear ) points?
How could you find 3 such points on those two lines?
But I don't understand how you can prove that with just a letter. Have you got any examples of this sort of thing?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #3 2012-12-12 05:54:47
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: Proof
A is correct.
The statement isn't even true. Two lines that are actually the same line do not define a single plane, but are coplanar.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #4 2012-12-12 06:02:55
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I answered A and it was incorrect .
Yeah this lesson is confusing but I did answer 14 correctly like this one :
(F ) was correct
6. In the figure above, line segment MC is equal to imaginary line segment MI.
A Given
Bunfounded
CDefinition of supplementary angles
D1267200 inches
E Definition of an octagon
F Definition of a circle: all points are equidistant from the center
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #5 2012-12-12 06:04:12
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I have to use the information I know to proof the statement correct
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #6 2012-12-12 06:47:20
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
stefy wrote:
Two lines that are actually the same line do not define a single plane, but are coplanar.
I did consider this, but rejected this interpretation on the grounds that 'two lines' should mean exactly two distinct lines not one line counted twice.
That's the trouble with using English to make mathematical statements. It sometimes isn't precise enough.
As zee-f has had A marked wrong, I think we have further evidence that the questioner was thinking that way too.
So let's assume the lines are either (i) distinct and parallel or (ii) they cross at a point
So we either have 4 distinct points or at least 3.
What do we need to define a plane?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #7 2012-12-12 07:29:35
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
Hi,
According to my online courses A plane is defined by any of the following:
three points not lying on a line
a line and a point not lying on the line
two lines which intersect in a single point or are parallel
So I think C would be a correct answer
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #8 2012-12-12 08:11:35
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
Yes, C sounds good. But I thought you had to supply the proof as well. Maybe not.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #9 2012-12-12 09:30:23
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I am still confused on what am really doing
I got 6 incorrect out of 20
I choose A and it was incorrect
7. In the figure above, line segment EJ is equal to line segment JM
A Definition of radius
Bunfounded
CDefinition of an octagon
D1267200 inches
E Given
F Definition of supplementary angles
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #10 2012-12-12 09:43:58
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
6/20 eeekkk!
Oh hang on. 6 wrong. Oh that's not so bad. 70% is a good score.
But we'll get them sorted. Don't worry.
I need to see the diagram for this one.
bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #11 2012-12-12 12:36:58
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
yup 14/ 20
The lesson uses the same chart for all the questions that use the chart
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #12 2012-12-12 20:25:12
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
hi zee-f,
Oh, that diagram. I remember that from another set of questions.
So EJ = JM ? They're not telling that; they're asking is it true?
Take a look at the diagram. Is J half way along EM ?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #13 2012-12-13 05:15:48
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
no EJ ≠ Jm
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #14 2012-12-13 05:16:55
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
So B would be a good answer ?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #15 2012-12-13 06:06:32
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
Yes, that's what I would choose.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #16 2012-12-13 07:10:51
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I choose E and it was incorrect
my new answer is D
14. If a central angle is 30 degrees, then the arc it defines is also 30 degrees.
A Given
BDefinition of an inscribed angle
Cunfounded
DProperties of a central angle
E Properties of an arc
F Definition of radius
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #17 2012-12-13 07:21:16
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
I see why you said E to start with. I would say that D and E are the same. But if E isn't acceptable, D seems good to me.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #18 2012-12-13 07:33:10
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
yeah I was stuck on which one to choose to. lol
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #19 2012-12-13 07:34:26
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
E was incorrect I still don't see why O_o
16. If a radius bisects a chord, then the lengths of the parts of the radius on either side of the chord are equal.
A Given
BDefinition of a chord
Cunfounded
DDefinition of supplementary angles
E Definition of a bisector
F Definition of radius
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #20 2012-12-13 07:46:55
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
I think my diagrams will show you what to do here.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #21 2012-12-13 08:23:42
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
No the radius isn't cut into equal parts so that whole statement is incorrect So C would be a good answer
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #22 2012-12-13 08:27:39
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
Yes. That's what I think.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #23 2012-12-13 11:16:00
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
2 more left
I answered C for both it was incorrect
19. The given points (4, -8), (4, -5), and (-2, -6) make a right triangle.
A Distance Formula
B Definition of a right triangle
CDefinition of a triangle
Dunfounded
E Pythagorean Theorem
F Definition of radius
20. The given points (2, -3), (-7, -7), (2, -7), and (-7, -2) make a square.
A Definition of coordinate
BPythagorean Theorem
CDefinition of a square
DDefinition of supplementary angles
E Distance Formula
F unfounded
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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## #24 2012-12-13 11:38:25
bob bundy
Registered: 2010-06-20
Posts: 8,337
### Re: Proof
hi
Did you try plotting the points? Have a look at my diagram.
I think you'll see what to do then.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #25 2012-12-13 11:56:09
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
No I didn't plot them probably why I got the question wrong the don't make a square or a right triangle So both (unfounded)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
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The parity
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Posted on: Thu Jun 09, 2011 8:37 am
Posts: 735
Joined: Fri May 13, 2011 6:51 pm
The parity
Basic strategies (perhaps trivial but may be useful to someone).
Second rule: The parity (a powerful tool for puzzles like the difficult 7x7 only subtractions on wednedays, etc.).
The addition of all numbers in any row or column can be even (10, 28, 36, 78 for the 4x4, 7x7, 8x8 and 12x12) or odd (15, 21, 45, 55, 153 for the 5x5, 6x6, 9x9, 10x10 and 17x17).
If we have a cage with n- the addition of all numbers inside must be even if n is even (we will call it “cage even”) and odd if n is odd (we will call it “cage odd”). For instance, in a cage of two cells with result 2-, both numbers inside must be even or both odd, and then the addition is even (7 and 5, 2 and 4, etc.). In a cage of two cells with result 3-, the two numbers inside must be of different parity, one odd and the other even, and the addition of both will be odd. Let’s now think in a three cells cage odd-, that is, a - b - c is odd; since a - b - c = a - (b+c), then “a” and “(b+c)” must be of different parity and consequently “a” + “(b+c)” = a + b + c is odd (for instance, 9-2-4 produces 3-, odd, 9+2+4=15, odd). Same demonstration applies to a three cells cage even-. Larger cages behave similarly since a - b – c – d… = a – (b+c+d…), etc.
Next step: The additon of all numbers inside two “cages odd” is even (and obviously if the two cages are even). Iterating this procedure and by comparing the result with the parity of the line or the “rectangle”, etc., we can determine the parity of individual cells (also knowing the parity of some subareas, like those with the x sign, we can eliminate combinations, etc.).
For instance, in the Jun 08 7x7 easy, the parity of the rectangle of the top 4 rows (excluding the cell d5) (observe that the 30x is an odd cage) is odd, so to complete the even parity (4x28) of the four rows, the cell d5 must be odd and “1” is the only possible value; e5 will then be 2. And now, checking the parity of the three bottom rows, the 12x in f1-g1 must be 4x3 (cage odd) and not 6x2 (cage even).
Posted on: Sat Jun 25, 2011 5:25 am
Posts: 8
Joined: Fri May 13, 2011 1:15 am
Re: The parity
Thanks for this tip. I have put it use and has been a definite help.
Posted on: Mon Jun 27, 2011 7:50 pm
Posts: 735
Joined: Fri May 13, 2011 6:51 pm
Re: The parity
larryb33 wrote:
Thanks for this tip. I have put it use and has been a definite help.
Welcome. I arrived to that conclusion when looking for some "error detection" applicable to the lines (rows or columns) (as in the computer's world) but I quickly observed that it was an all-purpose tool that could eliminate combinations in cages of the type "x" or the ":". For me many times is the key in the 9x9's. The parity rule can be combined with the "maximums and minimums in the sum of cages"; for instance, if you have s1 + s2 + s3 = 50 (where s1, s2 and s3 can be of any type of cage including "x" or ":") and you are analyzing a combination of s3 = 19, then s1 + s2 is odd and equals 31, and s1 and s2 have of course opposite parity between them, and also the only possibilities are those "pairs" of combinations for s1 and s2 that give that total of 31, and we can proceed analyzing the effect over the rest of the puzzle.
Posted on: Mon Sep 19, 2011 6:50 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Hi, clm.
In your explanation you refer to the puzzle (Jun 08 7x7 easy).
How can I obtain this historic puzzle to better understand your explanation?
Thank you. Cheers.
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Posted on: Mon Sep 19, 2011 7:28 pm
Posts: 2461
Joined: Thu May 12, 2011 11:58 pm
Re: The parity
jomapil wrote:
In your explanation you refer to the puzzle (Jun 08 7x7 easy).
How can I obtain this historic puzzle to better understand your explanation?
clm wrote:
For instance, in the Jun 08 7x7 easy, the parity of the rectangle of the top 4 rows (excluding the cell d5) (observe that the 30x is an odd cage) is odd, so to complete the even parity (4x28) of the four rows, the cell d5 must be odd and “1” is the only possible value; e5 will then be 2. And now, checking the parity of the three bottom rows, the 12x in f1-g1 must be 4x3 (cage odd) and not 6x2 (cage even).
Here it is:
Last edited by pnm on Thu Sep 22, 2011 2:46 pm, edited 1 time in total.
Posted on: Mon Sep 19, 2011 7:31 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Thank you, pnm.
Cheers.
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Posted on: Mon Sep 19, 2011 7:42 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Excuse me, pnm, but I think you made a mistake. If I'm right that puzzle refers to 7x7medium20110608.
Cheers.
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Posted on: Mon Sep 19, 2011 9:16 pm
Posts: 735
Joined: Fri May 13, 2011 6:51 pm
Re: The parity
jomapil wrote:
Excuse me, pnm, but I think you made a mistake. If I'm right that puzzle refers to 7x7medium20110608.
Cheers.
Hi, jomapil. The graphic is the correct one, the one I was referring to. I cann't remember now if it was the 7x7 easy or medium for that date, but sure Patrick found the correct one. It is not possible for the moment to recall an old puzzle using the Puzzle id (identification introduced "recently" by Patrick), only the last week's puzzles can be directly seen. When I sent the post we had not defined yet the labelling for the cells (actually agreed as in Excel, after the discussion in the Forum, that is, letters for the columns and numbers for the rows, although this has not been implemented yet in the page, ... perhaps in the future upon Patrick's time availability), so I used letters for the rows and numbers for the columns in that post. And at that time I was not yet uploading graphics (sorry for that) because for me the procedure was not clear, and only after a few e-mails with Patrick, I was able to do it.
This was a very easy example for "the parity" rule. The parity of that cell must be odd, and since a cage "2:" has only (in a 7x7) the combinations 1-2, 2-4 and 3-6, that cell must contain a "1" (there is a 3 already in the column) and the roommate is a "2". Now that you know the even parity of this second cell you may determine the parity of the cage "12x" = odd (so 3-4 and not 2-6), considering the parity of the three bottom rows (even, 3 x 28 = 84; here we must be careful because, i.e., three rows will have odd parity in the case of a 5x5, 6x6, 9x9, 10x10, ..., respectively 3 x 15, 3 x 21, 3 x 45, 3 x 55, ...).
We can apply "the parity rule" to practically all puzzles, in many situations, eliminating possibilities and focusing the candidates for the cells and/or cages. The utility will be more evident with the use.
Posted on: Mon Sep 19, 2011 10:18 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Thank you, clm, for your explanation and for the parity rule. At first sight it seems useful lowering the number of possibilities. I go to explore this concept in the next days and in the next puzzles.
Cheers and till the next.
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Posted on: Mon Dec 02, 2013 9:51 pm
Posts: 735
Joined: Fri May 13, 2011 6:51 pm
Re: Some additional clarifications with respect to the parit
Some additional clarifications with respect to the parity rule.
Around two years and a half have elapsed since this concept was first introduced here in the Forum. Later, one year and a half ago, I sent another post to the section “Solving strategies and tips” named “5x5 **: The parity rule for children/beginners” with an example of how this rule could be applied to a 5x5 difficult. However, many new puzzlers have joined the site since then so my intention now is to refresh a little bit that initial idea and to help a little those who are not still using this concept either because they do not consider it useful or perhaps because they have not fully understood its meaning.
Ok, the idea is as follows: We start with a 2-cell cage “n-” where “n” is even, for instance, “4-”, in a 9x9 puzzle. To get an even difference you need both numbers to be simultaneously even or simultaneously odd, in this example [1,5], [2,6], [3,7], [4,8] or [5,9] (because if one is even and the other is odd the difference will be odd). Then if both have the same parity its sum will be even, either if they are both even or if they are both odd, in this example, for the cage “4-”, the sum would be 6, 8, 10, 12 or 14, respectively. And we say that this is an “even cage” (we define this as an “even cage”) because the sum of all numbers inside is even. We may not know the final numbers yet but we know that necessarily its sum will be even.
Now, lets consider a 2-cell cage “n-” where “n” is odd, for instance, “3-”, in a 9x9 puzzle. The numbers inside this cage must have different parity, one must be even and the other must be odd, because if both were even or both odd simultaneously the difference would be always even (but not odd). In this example the possibilities are [1,4], [2,5], [3,6], [4,7], [5,8] and [6,9], that is, one even number and one odd number. And being of different parity its sum will be odd, in our example, the sum would be 5, 7, 9, 11, 13 or 15, respectively. And we say that this is an “odd cage” because the sum of all numbers inside is odd. We may not know the final numbers yet but we can affirm that necessarily its sum will be odd.
This is the heart of the idea and soon we will see its utility. How is the sum of all numbers included in two “even” cages, for instance, “2-” and “4-”?: even plus even equals even. And the sum of all numbers contained in two “odd” cages, for instance, “1-” and “5-”?: it will be even since odd plus odd equals even. And if one of the cages is even and the other is odd, for instance, “6-” and “7-”?: the sum of all numbers within those two cages will be odd since even plus odd is odd.
Before applying this rule to any real situation we need to demonstrate something else: that the previous conclusions for the 2-cell cages can be extended to cages of any size. Really, this is quite simple: Suppose that we have a 4-cell cage “2-” in a 9x9 puzzle: a possibility would be, for instance, [1,2,4,9] because 9 - 4 - 2 - 1 = 2. Observe that we always depart from the higher number in the cage, 9 in this case, and we subtract the other numbers, but 9 - 4 - 2 - 1 = 9 - (4 + 2 + 1) (the introduction of the parenthesis is simply an arithmetic valid operation but in this way we convert the 4 numbers into 2 numbers by grouping all the small numbers in one), 9 and the value of the parenthesis must be of the same parity to get an even difference as exposed above (in this case both are odd, 9 and 7), consequently 9 and the value of the parenthesis have an even sum, that is, 9 + (4 + 2 + 1) must be even or, from another point of view, the sum of all numbers inside the cage is even (16 in this particular case) [since 9 + (4 + 2 + 1) = 9 + 4 + 2 + 1 removing now the parenthesis]; in summary, we say that “2-” is an “even” cage whichever is the number of cells; another possibility for that “2-” 4-cell cage, for instance, the combination [1,1,4,8]: 8 - 4 - 1 - 1 = 2 = 8 - (4 + 1 + 1), 8 and the value of the parenthesis must be of the same parity, both are even in this case (8 and 6) and, being of the same parity, its sum is even.
Lets go now with a 4-cell cage “n-”, where “n” is odd, to see that the sum of all numbers contained in the cage must be odd: a - b - c - d = a - (b + c + d), where “a” represents the higher number in the cage and “b”, “c” and “d”, the other numbers in the cage. Observe that “a” and “(b + c + d)” must necessarily be of different parity to produce an odd result and, consequently, a + (b + c + d) is odd, and this amount is equal to a + b + c + d (now removing the parenthesis) so the sum of all numbers within the cage is odd, and we say that we have an odd cage. An example with real numbers (just in case the use of letters may sound abstract): for a “3-” 4-cell cage, in a 10x10 puzzle, a possibility is the combination [2,2,3,10]: here 10 - 3 - 2 - 2 = 3, but 10 - 3 - 2 - 2 = 10 - (3 + 2 + 2) in which expression we see that 10 and the parenthesis (with a value of 7) have different parity what is a mandatory requirement to arrive to an odd difference, in this case the subtraction of 10 and 7 equals 3; consequently, if the parity of both things is different, when adding 10 and the parenthesis we will obtain an odd sum, 17 in this particular case. Another example, the combination [1,1,2,7] is another possibility for the cage “3-”: 7 - 1 - 1 - 2 = 3 = 7 - (1 + 1 + 2) where we see that 7 and the parenthesis (which value is 4) have different parity and its sum is odd: 7 + (1 + 1 + 2) = 7 + 4 = 7 + 1 + 1 + 2 = 11.
We have then arrived to the final conclusion: A cage “n-” is even if “n” is even and it is odd when “n” is odd (this happens always regardless of the numbers used to comply with the subtraction); or, expressed with different words, if “n” is even the sum of all numbers in the cage is even and if “n” is odd the sum of all numbers in the cage is odd.
Now the question is: What is the utility of all this?. Ok: because it can be applied to “innies”, “outies”, “multiplication” or “divison” cages, … , and to any number of rows or columns or areas (for instance, if the parity of the “inner area” is known, like in those 6x6’s when the corners are given, and we have a 4x4 "inner area", …), etc..
Applying the parity rule.
First example: Suppose, for instance, that in a 7x7 puzzle, like the subtraction only puzzle on Wednesdays, you have an “innie” cell in a line (row or column), you can determine the parity of this individual cell in this way: if you have in that line, for instance, three 2-cell cages, lets say “1-”, “2-” and “3-”, you know that the sum of the six numbers contained in these three cages is even, just by "adding" the parity of the three cages, even in this case, and, since the sum of all numbers in any line is 28 (addition rule in a 7x7 puzzle, 28 is the sum for evey row or column), the “alone” (innie) cell must be even = 28 - even. And this is very helpful, for instance, if the cage that contains this cell is “4-” you inmediately know that the cage must be [2,6] and you can write the candidates since [1,5] or [3,7] would always produce an odd number in that position and that‘s impossible.
Second example: Suppose that you have two columns (in a patterned 7x7) with individual cells and addition/subtraction cages (or, i.e., some well defined multiplication cages like a 3-cell “10x” or a 3-cell L-shape “25x”, “32x”, …) plus one “72x” 3-cell L-shape cage. Our purpose is to determine the parity of this last cage. You "iterate" the parity rule to the sum of the individual cells and those cages, consequently determining the parity of the sum of all the numbers involved and, since the parity of two complete columns is even in this case (in a 7x7 the sum of two columns is 2 x 28 = 56) you find the parity of the pending cage, the targeted multiplication cage “72x”: if that result is even the only valid combination is [2,6,6] and, if it is odd, the only valid combination is [3,4,6]. And now you can continue looking at or extending the procedure to the two rows (the perpendicular lines to the previous lines) containing that “72x” cage.
******
The parity rule will solve you many uncertainties in the day by day "Calcudoku job" and you will gain a lot of time not only when dealing with small puzzles but also with large puzzles. You will find its utility as soon as you start using it, specially in the 9x9's, the 7x7 subtraction only on Wednesday's, the actual 10x10 on Fridays to define, i.e., if "40x" is [4,10] or [5,8], that is, an even or an odd cage, the actual 8x8 medium on Sundays, most 5x5 difficult, etc..
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# R: st: cmogram command for Regression Discontinuity Design
From "Borgomeo, Letizia" To "statalist@hsphsun2.harvard.edu" Subject R: st: cmogram command for Regression Discontinuity Design Date Mon, 5 Sep 2011 19:01:21 +0000
```In my special case, 0 does indeed belong to small positives since it is the result of the transformation of the original score for having it centered in 0. So 0 it is the mimimum score for individuals(firms) to get the treatment(public subsidy).Honestly, the reason why I started using -cmogram- is that I am not aware of a way of plotting two different fitlines for each side of the cutoff.
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Da: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] per conto di Nick Cox [njcoxstata@gmail.com]
Inviato: lunedì 5 settembre 2011 20.36
A: statalist@hsphsun2.harvard.edu
Oggetto: Re: st: cmogram command for Regression Discontinuity Design
We do ask that all new joiners read the FAQ before posting.
As implied, you can also fudge your 0s to small positives so that they
fall within the right bin (so to speak). Otherwise your graph won't
show a cutpoint at 0, which is important for your problem. But you'd
need to fudge all boundary values consistently.
Personally I like the convention that bins are as far as possible
[lower, upper) but -cmogram- follows the opposite rule.
The choice may seem arbitrary, but in most problems I look at it seems
more natural that 0 is not only a bin boundary but belongs with small
positives rather than small negatives. The substantive reason is often
that 0 may just mean not detected, but negatives mean something
different. But problems and tastes vary.
By the way, -cmogram- is just a convenience command. It is just a
couple of lines to set up bins that you like and summaries for those
bins, e.g.
sysuse auto
gen weight2 = 500 * floor(weight/500)
egen mean = mean(mpg), by(weight2)
after which you call up some graph of choice.
On Mon, Sep 5, 2011 at 7:09 PM, Borgomeo, Letizia
<L.Borgomeo@warwick.ac.uk> wrote:
> I am sorry I should have read the Statalist FAQ more carefully, since I am an absolute beginner. The command I am talking about is indeed the one created by Christopher Robert and available from SCC.
> The -cutpoint()- option seems to automatically include in the last bin on the left the value specified in the brackets. Then, in order to get the 0 values to the right, I may also artificially set the option -cutpoint(-0.5)- since my forcing variable is discrete. Thanks for your suggestions
> Letizia
>
>
> ________________________________________
> Da: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] per conto di Nick Cox [njcoxstata@gmail.com]
> Inviato: lunedě 5 settembre 2011 19.58
> A: statalist@hsphsun2.harvard.edu
> Oggetto: Re: st: cmogram command for Regression Discontinuity Design
>
> Naturally I agree with Maarten's general stance.
>
> I found that in -cmogram- you can't mix -start()- in -histopts()- and
> -cutpoint()-:
>
> ... histopts(start(-1500) width(500)) cutpoint(0)
>
> won't work.
>
> I also just remembered this beast
>
> SJ-6-1 gr26_1 . . . . . . . . . . . . . . . . . . Software update for binsm
> (help binsm if installed) . . . . . . . . . . . . . . . . . N. J. Cox
> Q1/06 SJ 6(1):151
> rewritten to support modern Stata graphics
>
> STB-37 gr26 . . . . . . . . . . . Bin smoothing and summary on scatter plots
> (help binsm if installed) . . . . . . . . . . . . . . . . . N. J. Cox
> 5/97 pp.9--12; STB Reprints Vol 7, pp.59--63
> alternative to graph, twoway bands(); produces a scatterplot
> of yvar against xvar with one or more summaries of yvar for bins
> of xvar
>
> Nick
>
> On Mon, Sep 5, 2011 at 4:32 PM, Maarten Buis <maartenlbuis@gmail.com> wrote:
>> On Mon, Sep 5, 2011 at 5:04 PM, Borgomeo, Letizia wrote:
>>> I am using the cmogram command for plotting the means-per-bin of an outcome variable against the assignment variable in a RDD. I am wondering if there is any way of setting the bins in order to have all the treated observations to the right and the untreated to the left of the cutoff. The forcing variable range is [-37;61] and I need the last bin in the LHS not to include the observations with a 0 value of the assignment variable, since they are exposed to treatment. Yet both with histopts(width()) and histopts(bin()) I do not manage to do it.
>>
>> -cmogram- is an user written program, so per the Statalist FAQ, you
>> must say where you got it from. These are not rules to "trick" you,
>> rule is that there are typically multiple version of user written
>> programs floating around in cyberspace, so if you do not tell us which
>> exact version you are using, we are likely talking about different
>> programs, and the resulting advise may very well be useless to you.
>> You can find a link to the Statalist FAQ at the bottom of each post to
>> Statalist (including this one).
>>
>> from SSC. If there is a one to one match between the cutpoint and
>> whether or not someone is treated, than it seems to me that the option
>> -cutpoint()- does what you want.
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 1,554 | 5,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.911323 |
https://tbc-python.fossee.in/convert-notebook/Chemical_Reaction_Engineering/ch9.ipynb | 1,708,675,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00484.warc.gz | 584,078,722 | 47,605 | # Chapter 9 : Temperature and Pressure Effects¶
## Example 9.1 page no : 209¶
In [1]:
import math
# Variables
CpA = 35. # j/mol.K
CpB = 45. # j/mol.K
CpR = 70. # j/mol.K
T1 = 25. # C
T2 = 1025. # C
Hr = -50000.
# Calculations
#Enthalpy balance for 1mol A,1 mol B,2 mol R
nA = 1.
nB = 1.
nR = 2.
dH = nA*CpA*(T1-T2)+nB*CpB*(T1-T2)+(Hr)+nR*CpR*(T2-T1);
# Results
print " dHJ at temperature 1025C is %.f "%(dH)
if dH>0 :
print "Reaction is Exothermic"
else:
print "Reaction is endothermic at 1025OC"
dHJ at temperature 1025C is 10000
Reaction is Exothermic
## Example 9.2 page no : 213¶
In [8]:
%matplotlib inline
import math
from matplotlib.pyplot import *
from numpy import *
Ho = -75300. # J/mol
Go = -14130. # J/mol
R = 8.3214
T1 = 298.
# Calculations
#With all specific heais alike,dCp = 0
Hr = -Ho;
K298 = math.exp(-Go/(R*T1));
#Taking different values of T
T1 = array([2,15,25,35,45,55,65,75,85,95]) #degree celcius
T = array([278,288,298,308,318,328,338,348,358,368]) #kelvin
XAe = zeros(10)
for i in range(10):
K = K298*math.exp((Hr/R)*((1./T[i])-(1./298)));
XAe[i] = K/(K+1);
# Results
plot(T1,XAe)
xlabel("Temperature, C")
ylabel("XAe")
print (" From the graph we see temp must stay below 78 C if conversion of 75% or above is expected")
show()
Populating the interactive namespace from numpy and matplotlib
From the graph we see temp must stay below 78 C if conversion of 75% or above is expected
WARNING: pylab import has clobbered these variables: ['draw_if_interactive']
%pylab --no-import-all prevents importing * from pylab and numpy
## Example 9.3 page no : 217¶
In [3]:
import math
# Variables
# from example 9.2
XA = 0.581;
t = 1. #min
XAe = 0.89;
XAe1 = 0.993;
T1 = 338.
T2 = 298.
R = 8.314
# Calculations
k1_338 = -(XAe/t)*math.log(1-(XA/XAe));
XA1 = 0.6;
t1 = 10. #min
k1_298 = -(XAe1/t1)*math.log(1-(XA1/XAe1));
E1 = (R*math.log(k1_338/k1_298))*(T1*T2)/(T1-T2)
ko = k1_338/(math.exp(-E1/(R*T1)))
# Results
print " The rate constants are k = exp[75300/RT-24.7] min-1"
print " k1 = exp[17.2-48900/RT] min-1"
print " k2 = exp[41.9-123800/RT] min-1 "
print " E1 = %.2f"%E1
print " K0 = %.2f"%ko
The rate constants are k = exp[75300/RT-24.7] min-1
k1 = exp[17.2-48900/RT] min-1
k2 = exp[41.9-123800/RT] min-1
E1 = 48679.81
K0 = 31411847.30
## Example 9.4 page no : 229¶
In [10]:
# Variables
CAo = 4. #mol/litre
FAo = 1000. #mol/min
A = 0.405 #litre/mol.min
# Calculations
t = CAo*A;
V = FAo*A;
# Results
print " Part a"
print " The space time needed is %.2f min"%(t)
print " The Volume needed is %.f litres"%(V)
# Note : We do not have value of 'rA'. so part B can not be calculated and plotted
Part a
The space time needed is 1.62 min
The Volume needed is 405 litres
## Example 9.5 pageno : 231¶
In [10]:
# Variables
CAo = 4. # mol/liter
FAo = 1000. # mol/min
XA = 0.8; # %
Cp = 250. #cal/molA.K
Hr = 18000. #cal/molA
rA = 0.4;
# Calculations and Results
V = FAo*XA/rA;
print " Part a"
print " The size of reactorlitres) needed is %.f litres"%(V)
slope = Cp/Hr;
#Using graph
Qab1 = Cp*20; #cal/molA
Qab = Qab1*1000; #cal/min
Qab = Qab*0.000070; #KW
print " Part b"
print " Heat Duty of precooler is %.2f kW"%(Qab)
Qce1 = Cp*37; #cal/molA fed
Qce = Qce1*1000; #cal/min
Qce = Qce*0.000070; #KW
print " Heat Duty of postcooler is %.2f kW"%(Qce)
# answers may vary because of rounding error.
Part a
The size of reactorlitres) needed is 2000 litres
Part b
Heat Duty of precooler is 350.00 kW
Heat Duty of postcooler is 647.50 kW
## Example 9.6 pageno : 233¶
In [12]:
# Variables
FAo = 1000. #mol/min
Area = 1.72;
# Calculations
V = FAo*Area;
# Results
print " The volume of adiabatic plug flow reactor is %.f litres"%(V),
The volume of adiabatic plug flow reactor is 1720 litres
## Example 9.7 page no : 234¶
In [13]:
# Variables
FAo = 1000. #mol/min
Area = (0.8-0)*1.5;
# Calculations
V = FAo*Area;
# Results
print " The volume required is %.f litres"%(V),
The volume required is 1200 litres
In [ ]: | 1,593 | 4,340 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-10 | latest | en | 0.548543 |
https://blogs.sas.com/content/iml/ | 1,718,538,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00221.warc.gz | 118,378,242 | 12,104 | 0
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A SAS analyst ran a linear regression model and obtained an R-square statistic for the fit. However, he wanted a confidence interval, so he posted a question to a discussion forum asking how to obtain a confidence interval for the R-square parameter. Someone suggested a formula from a textbook (Cohen,
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A SAS analyst read my previous article about visualizing the predicted values for a regression model that uses spline effects. Because the original explanatory variable does not appear in the model, the analyst had several questions: How do you score the model on new data? The previous example has only
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Sometimes labels for variables get "dropped" during data preparation and cleaning. One example is when data are transposed from "wide form" to "long form." For example, suppose a data set has three variables, X, Y, and Z, each with labels. If you transpose the data to long form, the new
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After writing a program that simulates data, it is important to check that the statistical properties of the simulated (synthetic) data match the properties of the model. As a first step, you can generate a large random sample from the model distribution and compare the sample statistics to the expected
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A SAS programmer was trying to implement an algorithm in PROC IML in SAS based on some R code he had seen on the internet. The R code used the rank() and order() functions. This led the programmer to ask, "What is the different between the rank and the order? | 423 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920162 |
https://au.mathworks.com/matlabcentral/answers/324141-can-anybody-tell-me-the-code-how-to-find-sum-of-series-for-a-complex-function | 1,669,673,164,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00814.warc.gz | 145,276,162 | 39,861 | # can anybody tell me the code how to find sum of series for a complex function?
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vinod kumar govindu on 9 Feb 2017
IRx1=[-3.4,2.47,..........27.3]
QRx1=[2.3.............54.5]
the length of IRx1,QRx1,IRx2 and QRx2 are 128.length for all vectors is same that is 128.
x(n)=IRx1+i*QRx1 (complex signal)
y(n)=IRx2+i*QRx2 (complex signal)
i want to find auto correlation for X and cross correlation for X and y. by using below formulaes
ACF(k)=∑(n=0 to 128)〖x(n).*x(n-k) 〗
CCF(k)=∑(n=0 to 128〖x(n).*y(n)〗
Here i attached images of equations also. I tried so many times i am getting errors.
Help me to find ACF and CCF of given complex signals?
##### 2 CommentsShowHide 1 older comment
vinod kumar govindu on 10 Feb 2017
Dear simon,
here i posted my code my aim is to find out the auto correlation and cross correlation of the given signals x,y,z.
Auto correlation can be done by compare signal x with itself with some delay. cross correlation can be done by compare twp different signals x,y. I attached a picture contains ACF and CCF equations. In that there is mistake in CCF. It is CCF only not CCF(k). ACF(k) is a auto correlation of x(n) with delayed signal x(n-k). I used symsum function for summation but i am getting error. Given below i posted the error which i got while using symsum funtion.
??? Attempted to access IRx1(0); index must be a positive integer or logical.
Error in ==> G12 at 26 ACF1(i,n)=bug+((IRx1(n)+j.*QRx1(n)).*(IRx1(n-i)-j.*QRx1(n-i)));
I dont know it is correct function for the summation of complex signals.
The procedure is to multiply the x(n) with it self by giving delay x(n-k) and add all those to get ACF.
Help me to solve this coding.
clc;
close all;
clear all;
IRx1=[2.50582817101952,2.41713141145528,1.86841439304153,0.605047602894921,-0.703566724847468,-1.53347690224738,-2.68534618710372,-3.37564979764617,-3.60976457774389,-3.31964450712546,-2.66522613876526,-1.77623560339244,-0.492474329662971,0.160290587013230,0.507220033329129,0.312331782163865,0.113076760989049,-0.510845957116963,-1.02537238124377,-1.54315957923986,-1.92651006745793,-2.07761357141181,-1.96179000520892,-1.61605886905197,-0.998502032475426,-0.606791167516796,-0.408862392930745,-0.721810897897046,-1.03865928287495,-1.67596511911259,-2.17900191449449,-3.22455973533726,-4.20122236827395,-4.88936414084140,-4.98588492090064,-4.64611986699242,-3.89295986195026,-2.85016108345810,-1.51000979182552,-0.114053165824088,1.08073054478529,1.85120398545085,2.05071399629919,1.55713172662244,0.577854501379098,-0.781845566782136,-2.28138320255862,-3.59143240245656,-4.48240785969556,-4.80042560007416,-4.49335459690561,-3.61773210916393,-2.28811421907310,-0.616269087330731,1.34646211479523,2.58576397249433,3.18726452764764,3.55099127809478,3.39245771843159,2.67982568751591,1.28811220734259,0.290545697789977,-1.11083900436137,-2.24846907912449,-3.41008937840863,-4.08132603131973,-4.19202057140981,-3.75598274404288,-2.86511558343952,-1.33166071513922,-0.0888600143436430,1.04383587114973,1.90146530084647,2.36534314627888,2.37807964444616,1.94943719786547,1.15254387997119,0.111265653064370,-1.01930534408229,-2.44869015588492,-3.23965758371861,-3.36628409964266,-3.50424270023697,-3.29456430470408,-2.80531710517118,-2.14315533001624,-1.47170964682503,-0.795233104046106,-0.346629678415243,0.243676163193890,0.0561703047489547,-0.740610277561049,-1.27386267061423,-2.08480520623948,-2.43267440079923,-2.76937121834168,-2.99337398387634,-3.08558943524334,-3.04783986140056,-3.07230565610149,-2.94490942924285,-2.64367433237923,-2.33302680338966,-1.83789598292898,-1.97823772406445,-1.93937035972245,-2.20625565463774,-2.23887113241593,-2.58437228749598,-3.52733915198916,-4.07019344900751,-4.31937286306846,-4.82277220060252,-5.11817728167131,-5.13481364324164,-4.84209660869776,-4.24673283123647,-3.54290123157843,-2.67022083716594,-1.34112096240404,-0.847195923908771,-0.136621642288736,-0.113124944412861,-0.346362291440859,-0.812589735653853,-1.43525740508435,-2.31980008066570,-2.47632619329387];
IRx2=[1.01460425716151,0.789387892284636,0.465419065393918,-0.270674605261502,-0.324209579543460,-0.786127887110611,-1.31672413476201,-1.68974069580242,-1.68683326153294,-1.91463828346887,-1.99070592373177,-1.86159099936316,-1.51752077197991,-0.954746832721099,-0.209994594029173,0.811346972019044,1.61404780608856,1.96001533542083,2.49485352039829,2.76882719749157,2.73797080108476,2.39139122359568,1.75505098733065,0.892009683076733,0.0815037131794636,-1.01857047035352,-2.02213961753928,-2.93877580921443,-3.23294982923665,-2.99973300999368,-2.91528946965527,-2.01533398528999,-0.768915862561078,0.673940528188069,2.12014743798596,3.07912747337142,4.14769021857149,4.55429333026318,4.33404517081416,3.48728067593388,2.11504461592450,0.408386612077747,-1.37865630550980,-2.93271295825688,-4.10799835209611,-4.78408173369122,-4.46062915666738,-3.45118437406266,-1.90549976700760,-0.0589501797260740,1.80599108593936,3.40640695154048,4.54526003105368,5.18679223295696,5.33036875356390,4.48934454418814,3.04699266357109,1.54593614053326,0.135501358878592,-1.09302611399635,-1.85681416548019,-2.15661323317940,-2.32274156157063,-1.88578807031514,-1.04852785251823,0.0680470909691132,1.28358265681183,2.42637788797044,3.35703514167297,3.64767743571924,3.98742663931683,4.00581733933059,3.73109093230059,3.21309282961231,2.51322792193615,1.69750401007136,0.833207984444164,-0.0116689660343153,-0.768138496031731,-1.55330142621117,-1.97458482357104,-2.15477811425155,-1.87735825878768,-1.21139358892745,-0.236527845717644,0.916737394288882,2.02731175399555,3.11127236571521,3.76594186553903,4.15152292241076,3.65721802838211,2.68922312868832,1.35424397679989,-0.422175665866012,-1.47203008503273,-2.72588745633588,-3.60988696547965,-3.97459918433722,-3.75578630206334,-2.70609884338261,-2.18677889085995,-0.724861849873928,0.705065802834293,1.64916146563410,2.43886933811804,2.42744868507788,1.80584985609909,0.908696622442396,-0.705707238128764,-2.27270189339574,-3.37234227805187,-4.20160175589701,-4.17172206185601,-3.55339005011385,-2.42912689725117,-1.31805658267917,0.239514364570606,1.47991933586132,2.36459637723685,2.56164473359209,2.55384087296497,1.60993379649284,0.629564432592203,-0.645304518556036,-1.97479167479100,-3.10148212130029,-3.80003159784154,-3.94619491777448];
IRx3=[-0.993128890005284,-1.76594777520882,-2.31949979637286,-2.21870218475863,-2.45080409040613,-2.28219201343847,-1.02229193980455,0.453921919149053,1.90050025467925,2.59193207732919,2.69060034703535,2.15307465169026,1.05521829466591,-0.413379640224464,-2.00515321844758,-3.31946860669385,-4.41984467230870,-5.23501910734096,-5.16606973083684,-4.46543527319935,-3.25160209193610,-1.72671153217760,-0.140190359696800,1.25484213325138,2.02459808250948,2.62069605149156,2.60946644908517,2.09571076461969,1.01910162812756,-0.269202255127129,-1.23809167199613,-2.47193225700054,-3.38135220256387,-3.83434288531237,-3.68937515950887,-3.39741304134523,-2.42972386681567,-1.50374882556885,-0.508048999212305,0.404545935881431,1.10330611589675,1.49838880640620,1.55228571196290,1.34281769078276,0.842640828669976,-0.0527018153170412,-0.713595602532179,-1.26283024770965,-1.61665181987407,-1.72898418633932,-1.59713183999718,-1.26040333662297,-0.752932366670609,-0.0577665529710810,0.374250486357905,0.941611680965764,1.08213296963860,1.27444212679685,1.21015441855056,0.785605530968928,-0.180891381312223,-0.928904425156226,-1.73454047225642,-2.77036791374945,-3.73068718741744,-4.34861858499960,-4.53295554566547,-4.26883529651507,-3.61886401478481,-2.36423308245108,-1.43888043407460,-0.627189322960582,-0.0783446858768140,0.105958148447062,-0.111671539130095,-0.698782517051578,-1.55953435306677,-2.55178800751300,-3.51170697380857,-3.96060456182168,-4.42738941481774,-4.26437253555307,-4.15000793763006,-3.71955395095441,-3.03386608743514,-2.19598516647012,-1.39928527264283,-0.563069421120779,-0.0488066084850562,0.0229426002962217,-0.166071205981264,-0.696044288513010,-1.47745602544096,-2.36416001821257,-3.14726579723785,-4.02850927072230,-4.71723059101617,-5.10569764618782,-5.12914144917017,-4.99809288524002,-3.92091606240318,-3.00656913581731,-1.91111082249582,-0.565619991349472,0.283627761416091,0.947356788212610,0.948578976391194,0.906033103627324,0.165663577711276,-0.528126599095243,-1.75073591909251,-2.76186496190158,-4.01400972532222,-4.97621829735726,-5.52200879346877,-5.24556962852168,-4.80996281212845,-4.17173805622293,-3.24903463733582,-2.23274479585379,-0.979573531993498,0.423167172481398,1.16521650753348,1.49978408672960,1.36002336661019,0.741577094736565,-0.289522665579945,-1.60088092364450];
QRx1=[-0.144718579660555,0.995946652625673,2.05637645961698,2.61196347518716,3.25632649951836,3.08746302178533,2.45005524983264,1.52879189873049,0.0795600257263762,-0.903503642646553,-1.66384374630175,-2.07219543395580,-1.73663051946867,-1.22493813342171,-0.511968187169972,0.111113406512994,0.796516202647878,0.977755843153328,1.25462033881561,1.24592323194847,0.996209115120566,0.609489969080677,0.223354235342819,-0.0248956407188340,-0.232342453639286,0.174464228212001,0.807619460912736,1.49311646218381,2.21645131610826,2.67054620724355,3.06251300887572,3.09478595401401,2.63306697209545,1.73954071310858,0.532969121848919,-0.652371421489012,-1.45428721135333,-2.28225224335881,-2.60690673920328,-2.34235411873884,-1.51408904246930,-0.255026384528728,1.21922118882082,3.02115178742696,4.10050381643365,4.96903735582075,4.81132933804762,4.04180574992444,2.79181617599615,1.27442184206188,-0.255083903745787,-1.54528645820358,-2.42200036300971,-2.78267604719178,-2.42252639176325,-1.52142442818247,-0.307372780120042,1.16373862378014,2.59566895376017,3.82281007709030,4.68018360704009,5.05114433673019,4.82593849729210,4.20732887404119,3.07060487849618,1.62519161882054,0.0818617642666102,-1.34270890438126,-2.45728108780638,-3.18997435899711,-3.26264000733367,-2.85030338447939,-2.03755274312990,-0.961712703315067,0.208817912882685,1.29975076000225,2.15604845685741,2.66356786203298,2.76429227850948,2.60505351129691,1.84689048244303,0.820028928366605,-0.0623265017645658,-0.862750087599546,-1.46018221335862,-1.77273253255276,-1.73167622697005,-1.53061294715227,-1.07761129848942,-0.782769139847865,0.0264483256493230,0.436887172479494,0.789414211727582,0.727446167963397,0.756415264991138,0.618513152633964,0.393549801823466,0.131886491958235,-0.114912262481089,-0.545127730055126,-0.510377281509120,-0.512557411426055,-0.391659486423605,-0.0710516328225203,0.00509349801989560,0.426987287620363,0.627750665719879,0.943494833829077,1.35343204458950,1.54615610141597,1.51295396896865,1.31529822675633,0.856736949102081,0.198390357042307,-0.569333696382446,-1.32418460730406,-1.92388745351658,-2.45670604169471,-2.70052375902505,-2.08229123921291,-1.73603019307446,-1.13987994071623,-0.474015604873290,0.162399114708555,0.674102061951758,0.524785424561854,0.214556638570243,-0.0938660949699451];
QRx2=[1.25748821083559,1.59232820540285,1.89211923520759,1.88099181061371,1.75907167934012,1.82428183523391,1.78041254700247,1.56472843457122,1.04088165066503,0.646556290077192,0.128310603656031,-0.450501773794132,-1.00480904530030,-1.45912613104007,-1.72868619330447,-1.91466170173093,-1.52145406181707,-0.744755343783896,0.0164457823216737,0.920124206140120,1.87280368258172,2.76798592075878,3.49740829873187,3.96346095177321,3.92634814268997,3.79047264113801,3.25131401251054,2.19516080475489,1.11866898714991,-0.0868644803222537,-1.15381581692583,-2.08217190852378,-2.65256309571722,-2.74086608851374,-2.13703911177943,-1.86213629706259,-0.179665202068858,1.42353419433705,3.09919520661605,4.59655516335476,5.67777383988329,6.15849489368285,5.94162047977465,5.41655735638638,3.89777136864294,1.75659884543326,-0.147314506470493,-1.80654394943821,-2.97361612236474,-3.47952694704350,-3.26168307461630,-2.37373106741707,-0.986656040212108,0.680993458251602,2.66897460606418,4.14915838268406,5.28633393172347,5.56585128716944,5.32155618820885,4.58001729846046,3.68475979062889,2.18321673024426,0.727376002538077,-0.939867972535359,-2.14028958454835,-2.91210522072983,-3.20882166062459,-3.05352877769973,-2.52510818506159,-1.70592670614687,-0.851669664801974,0.0405071413518816,0.874314119614612,1.57491349902664,2.08961027424417,2.38518328361133,2.44437288087680,2.26371843129368,1.85394131721738,1.59842253507776,0.711185513135204,0.0404410173761393,-1.01883306281697,-1.89543095968338,-2.45170009387046,-2.58418936090662,-2.22354516228131,-1.40250297386122,-0.142751951346500,1.59542559405902,2.93768034088717,4.05373057297677,4.77627132908852,4.82664828257853,4.11455681138587,3.49068770809641,2.43703556765478,1.13299280408855,-0.192807969927715,-1.40172126792556,-1.95224338365755,-2.12390938625579,-1.65547574874616,-0.458219572865563,1.07830126344190,2.51125426751961,4.13485781495850,4.96487384457950,5.32729501591053,5.29550758377778,4.46931759931741,3.02864828865319,1.65029660558493,0.362393361769723,-0.594496557791115,-0.988543113338450,-0.925025563983850,-0.334419146424679,0.716880119341661,1.79250194692824,3.22379242483073,4.31309335618836,5.20263838196125,5.57918860188955,5.35976324760898,4.57171837761737,3.35665430473978,1.92253002265421];
QRx3=[1.61469434821989,1.28110087283433,0.593981423858839,-0.0746609794962341,-0.891803824646148,-2.19303462998841,-2.62044173173250,-2.33773458592233,-1.14868797987302,0.187752799404550,1.72921097658605,3.20949521901254,4.35189005875512,4.96422758117794,4.93938714967720,4.08414134060692,3.03450684126624,1.89291039683914,0.263892352374029,-1.21185894615126,-2.29786178930600,-2.82730994360425,-2.73002447245140,-2.04185996536298,-0.978459491542767,0.463296101102056,1.96310381648816,3.47769048065473,4.28572284988762,4.48069796806336,4.52430480758476,3.96142604376931,2.96223292698943,1.69588086517021,0.269549938642862,-0.470169450225313,-1.59577743807407,-2.08965359044770,-2.20036578610210,-1.94267300943902,-1.38745944463219,-0.647424702259412,0.142727660186499,0.457263258501747,1.02854810107158,1.53119350556719,1.43704558729807,1.09870604559311,0.589409191167250,0.00703713408324849,-0.542958205090700,-0.965174046229494,-1.21383179950569,-1.26407039195322,-1.25897405515521,-0.681320371584185,0.0334979706460327,0.911813186967266,1.84129800199022,2.67248382949417,3.19578726489379,3.23645876594367,3.35429118367439,3.44781140646502,2.88762169576680,2.01590005115277,1.00015989096743,0.0286399105764878,-0.722354919489267,-1.13763940346719,-1.04553560746135,-0.566614389385489,0.215915010612429,1.16617500935677,2.11970048353193,2.91199894737582,3.40662737197769,3.51802519672367,3.22535601020925,2.36107413812950,1.58191148418807,0.582199081336153,-0.245139894736487,-0.973514153843321,-1.48447989900030,-1.69133925164542,-1.56892560973338,-1.08445968606643,-0.360974609164748,0.496588571825767,1.40302979009760,2.17079631650935,2.67736691587632,3.13836637897697,2.38054122224518,1.97100715599610,1.24885416031121,0.307919228543620,-0.722710835794146,-1.50341250319490,-2.01468520812255,-2.68666490478536,-2.97425216528387,-2.89646892777086,-2.52379881302396,-1.49901796224810,-0.506869568406280,0.907651361053495,1.75345622216872,2.55545782056647,2.90248642844106,2.86882256178575,2.29975359926686,1.31819570586921,0.0578257551530834,-1.35655604583865,-2.53551533399583,-3.45158622002622,-4.07401651828128,-4.52543821838908,-4.36880666597097,-3.47644873119799,-2.46651478664132,-1.22740964353534,0.0764465476581728,1.26339866951640,2.15988323096008,2.61759542869763];
%IRx and QRx are inphase and quadrature phase signals
x=IRx1+j*QRx1;%IRx1 is inphase signal and QRx1 is quadrature phase
y=IRx2+j*QRx2;
z=IRx3+j*QRx3;
for i=1:128
j=sqrt(-1);
syms IRx1 QRx1
ACF1(i)=symsum((IRx1(n)+j.*QRx1(n)).*(IRx1(n-i)-j.*QRx1(n-i)),n,0,128);
end
John BG on 9 Feb 2017
Edited: John BG on 9 Feb 2017
Hi Vinod Kumar Govindu
having faced similar problem, I found out the following points may be of interest to you
1. that the standard MATLAB function conv has a limitation: it resets the time reference, not really knowing when the convolution starts and stops, let me explain:
i1=randi([-10 10],1,10);
q1=randi([-10 10],1,10);
i2=randi([-10 10],1,10);
q2=randi([-10 10],1,10);
plot(abs(conv(x,y)));grid on
.
as you can see, the resulting convolution always starts n=1 regardless of when do x and y really start. The correlation should tell when both signals start having something in common.
To improve this, the literature reference
Digital Signal Processing Using MATLAB
3rd edition
by Vinay K Ingle, John G Proakis
in page 44 a convolution function is developed taking into account the start and stop indices of x and y:
function [y,ny] = conv_m(x,nx,h,nh)
% Modified convolution routine for signal processing
% --------------------------------------------------
% [y,ny] = conv_m(x,nx,h,nh)
% y = convolution result
% ny = support of y
% x = first signal on support nx
% nx = support of x
% h = second signal on support nh
% nh = support of h
%
nyb = nx(1)+nh(1); nye = nx(length(x)) + nh(length(h));
ny = [nyb:nye];
y = conv(x,h);
2.
The same literature reference suggests another alternative convolution, with the Toeplitz calculation of the equivalent matrix that one just has to multiply to x to obtain y, instead of running all the progressive sums shifting y one reference vector value at a time:
function [y,H]=conv_tp(h,x)
% Linear Convolution using Toeplitz Matrix
% ----------------------------------------
% [y,H] = conv_tp(h,x)
% y = output sequence in column vector form
% H = Toeplitz matrix corresponding to sequence h so that y = Hx
% h = Impulse response sequence in column vector form
% x = input sequence in column vector form
%
Nx = length(x); Nh = length(h);
hc=[h; zeros(Nx-1, 1)];
hr=[h(1),zeros(1,Nx-1)];
H=toeplitz(hc,hr);
y=H*x;
further reading regarding the toeplitz function
3.
Perhaps you would like to consider using function corrcoef that shifts x and places in each line of the output matrix RHO the result of each sum of your expression CCF, corrcoef also outputs how sell correlated are x and y for each value of k, in the shape of another matrix
[rho,pval]=corrcoef(x,y)
the closer a value of pval is to 1, the more alike are x(n) and y(n-k). p(i,j)=1 means identical.
.
4.
Since you want to convolve 2 complex signals, perhaps you would like to decompose the operation the following way:
conv(i1+1j*q1,i2+1j*q2)=
conv(i1,i2)-conv(q1,q2)+(conv(i1,q2)+conv(i2,q1))j
or you would consider working with modulus and phase, and then just check how correlated are modules and phases:
modx=abs(x);ax=angle(x);mody=abs(y);ay=angle(y);
conv(modx,mody)
conv(ax,ay)
.
so, Vinod
John BG
Thibaut Jacqmin on 9 Feb 2017
What you want to compute looks like convolutions. I think there is a mistake in the CCF expression which does not depend on k. What you want to compute is probably the sum over n of x(n)*y(n-k). You can find more info on the conv function here (have a look at the "More about" section) :
x = complex(IRx1,QRx1);
y = complex(IRx2,QRx2);
ACF = conv(x, x);
CCF = conv(x, y); | 7,768 | 19,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.810789 |
https://www.ceder.net/def/circletoa2fl.php?level=master&language=sweden | 1,713,461,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00655.warc.gz | 625,773,890 | 5,266 | Definitions of Square Dance Calls and Concepts
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Circle ({fraction}) To A Two-Faced Line [C4]
(upphovsman okänd)
From Facing Couples.
Circle Left 1/2 (or the given fraction); Veer Left.
Ends in a R-H Two-Faced Line. This is a 2-part call.
föreCircle To A Two-Faced Line efterCircle Left 1/2 efterVeer Left (klart) | 221 | 511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.164794 |
https://247amend.com/2016/11/vapor-pressure-of-solution.html | 1,726,889,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00620.warc.gz | 55,567,433 | 27,424 | # Vapor Pressure of a Solution
From the illustration above, Initially, two beakers, both containing water solutions of the same solute, are placed under a bell jar. The solution in the left beaker is less concentrated than that in the right beaker, so its vapor pressure is greater.
The partial pressure of vapor in the bell jar is an intermediate value. It is less than the vapor pressure of the solution on the left, but more than that of the solution on the right. As a result, vapor leaves the solution on the left (which becomes more concentrated) and condenses on the solution on the right (which becomes less concentrated). After some time, the two solutions become equal in concentration and in vapor pressure.
During the nineteenth century, chemists observed that the vapor pressure of a volatile solvent was lowered by addition of a nonvolatile solute. Vapor-pressure lowering of a solvent is a colligative property equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution.
For example, water at 20 C has a vapor pressure of 17.54 mmHg. Ethylene glycol, CH2OHCH2OH, is a liquid whose vapor pressure at 20 C is relatively low; it can be considered to be nonvolatile compared with water. An aqueous solution containing 0.0100 mole fraction of ethylene glycol has a vapor pressure of 17.36 mmHg. Thus, the vapor-pressure lowering, P, of water is Change in P = 17.54 mmHg -17.36 mmHg= 0.18 mmHg.
In about 1886, the French chemist François Marie Raoult observed that the partial vapor pressure of solvent over a solution of a nonelectrolyte solute depends on the mole fraction of solvent in the solution. Consider a solution of volatile solvent, A, and nonelectrolyte solute, B, which may be volatile or nonvolatile.
According to Raoult’s law, the partial pressure of solvent, PA, over a solution equals the vapor pressure of the pure solvent, PA , times the mole fraction of solvent, XA, in the solution.
If the solute is nonvolatile, PA is the total vapor pressure of the solution. Because the mole fraction of solvent in a solution is always less than 1, the vapor pressure of the solution of a nonvolatile solute is less than that for the pure solvent; the vapor pressure is lowered. In general, Raoult’s law is observed to hold for dilute solutions—that is, solutions in which XA is close to 1. If the solvent and solute are chemically similar, Raoult’s law may hold for all mole fractions. Raoult’s law is displayed graphically
for two solutions in the figure below….
You can obtain an explicit expression for the vapor-pressure lowering of a solvent in a solution, assuming that Raoult’s law holds and that the solute is a nonvolatile nonelectrolyte. The vapor-pressure lowering, P, is Change inP = P°A- PA
From this equation, you can see that the vapor-pressure lowering is a colligative property— one that depends on the concentration, but not on the nature, of the solute. Thus, if the mole fraction of ethylene glycol, XB, in an aqueous solution is doubled from 0.010 to 0.020, the vapor-pressure lowering is doubled from 0.18 mmHg to 0.36 mmHg. Also, because the previous equation does not depend on the characteristics of the solute (other than its being nonvolatile and a nonelectrolyte), a solution that is 0.010 mole fraction urea, (NH2)2CO, has the same vapor-pressure lowering as
one that is 0.010 mole fraction ethylene glycol. The next example illustrates the use of the previous equation.
Problem: Calculate the vapor-pressure lowering of water when 5.67 g of glucose, C6H12O6, is dissolved in 25.2 g of water at 25 C. The vapor pressure of water at 25 C is 23.8 mmHg. What is the vapor pressure of the solution?
Problem Strategy: In order to calculate the vapor pressure lowering of the solution described in this
problem, we need to determine the amount that the vapor pressure of the pure solvent is changed ( P
of water) with the addition of a solute (glucose). We can use Raoult’s law to calculate this quantity,
noting that we need to know the mole fraction of the solute (glucose). The vapor pressure of the solution is found by subtracting the vapor-pressure change of the pure solvent from the vapor pressure of the pure solvent.
Solution : This is the glucose solution described in Example 12.4. According to the calculations performed there, the solution is 0.0220 mole fraction glucose.
Therefore, the vapor-pressure lowering is Change in P = PA°XB =23.8 mmHg x 0.0220= 0.524 mmHg
The vapor pressure of the solution is PA = PA° – Change in P (23.8 – 0.524) mmHg = 23.3 mmHg
An ideal solution of substances A and B is one in which both substances follow Raoult’s law for all values of mole fractions. Such solutions occur when the substances are chemically similar so that the intermolecular forces between A and B molecules are similar to those between two A molecules or between two B molecules.
In this case, we are not restricted to the solute being nonvolatile: both the solute and solvent have significant vapor pressure. Therefore, the total vapor pressure over an ideal solution equals the sum of the partial vapor pressures, each of which is given by Raoult’s law:
Change in P = PA°XA + PB°XB
Solutions of benzene, C6H6, and toluene, C6H5CH3, are ideal. Note the similarity
in their structures…
Suppose a solution is 0.70 mole fraction benzene and 0.30 mole fraction toluene.
The vapor pressures of pure benzene and pure toluene are 75 mmHg and 22 mmHg,
respectively. Hence, the total vapor pressure is change in P (75 mmHg x 0.70) + (22 mmHgx 0.30)= 59 mmHg.
Recommended lecture/article: The Valence Bond TheoryAliphatic Hydrocarbons: Definition s and Properties,Effect of Polarity on Molecular Properties | 1,397 | 5,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.888494 |
https://realanswers-ph.com/math/using-ordered-pairs-how-do-we-locat-2382693 | 1,679,548,939,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00795.warc.gz | 544,966,660 | 33,839 | , 28.10.2019 15:29 janalynmae
# Using ordered pairs, how do we locate the seat of any classmate of mara and clara?
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Using ordered pairs, how do we locate the seat of any classmate of mara and clara?...
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History, 05.10.2021 03:15
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Araling Panlipunan, 05.10.2021 03:15 | 342 | 1,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.839042 |
http://dailydoseofexcel.com/archives/2004/12/20/replacing-the-analysis-toolpak-addin-part-3/ | 1,675,806,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00846.warc.gz | 11,788,657 | 18,827 | # Replacing the Analysis Toolpak Addin – Part 3
This part focusses on replacing the numerical system conversion functions of the Analysis Toolpak Addin (ATP). Reference to the other parts of this article series:
Numeric System Conversion ATP functions
ATP Function Description ATP Syntax Replacement Formula Array formula
BIN2DEC Converts binary number to decimal =BIN2DEC(Number) =SUMPRODUCT(MID(“0″&Number,ROW(INDIRECT(“1:”&LEN(“0″&Number))),1)*2^(LEN(“0″&Number)-ROW(INDIRECT(“1:”&LEN(“0″&Number))))) No
BIN2OCT Converts binary number to octal =BIN2OCT(Number,Places) Combine solutions for BIN2DEC and DEC2OCT No
DEC2BIN Converts a decimal number to binary =DEC2BIN(Number) =SUMPRODUCT(INT(MOD(Number/2^(COLUMN(1:1)-1),2))*10^(COLUMN(1:1)-1)) No
DEC2OCT Converts a decimal number to octal =DEC2OCT(Number) =SUMPRODUCT(INT(MOD(Number/8^(COLUMN(1:1)-1),8))*10^(COLUMN(1:1)-1)) No
HEX2BIN Converts a hexadecimal to a binary =HEX2BIN(Number,Places) Combine solutions for HEX2DEC and DEC2BIN No
HEX2DEC Converts a hexadecimal to a decimal =HEX2DEC(Number) =SUMPRODUCT((MATCH(MID(“0″&Number,ROW(INDIRECT(“1:”&LEN(“0″&Number))),1),{“0″,”1″,”2″,”3″,”4″,”5″,”6″,”7″,”8″,”9″,”A”,”B”,”C”,”D”,”E”,”F”},0)-1)*16^(LEN(“0″&Number)-ROW(INDIRECT(“1:”&LEN(“0″&Number))))) No
HEX2OCT Converts a hexadecimal to an octal =HEX2OCT(Number,Places) Combine solutions for HEX2DEC and DEC2OCT No
OCT2BIN Converts an octal number to binary =OCT2BIN(number,places) Combine solutions for OCT2DEC and DEC2BIN No
OCT2DEC Converts an octal number to decimal =OCT2DEC(number) =SUMPRODUCT(MID(“0″&Number,ROW(INDIRECT(“1:”&LEN(“0″&Number))),1)*8^(LEN(“0″&Number)-ROW(INDIRECT(“1:”&LEN(“0″&Number))))) No
Missing functions: BIN2HEX, DEC2HEX and OCT2HEX as the creation of the characters A-F which are part of a hexadecimal number is not really feasible using worksheet functions.
Frank
## 17 thoughts on “Replacing the Analysis Toolpak Addin – Part 3”
1. Note that VBA has the HEX() function to convert from decimal to hex, giving us the UDF:
Public Function DEC2HEX(ByVal iDec As Integer) As String
On Error Resume Next
DEC2HEX = Hex(iDec)
End Function
which can, of course, be used in conjunction with the other functions to give BIN2HEX and OCT2HEX.
FWIW, HEX2DEC can also be done in a 1-line VBA function:
Public Function HEX2DEC(ByVal sHex As String) As Integer
On Error Resume Next
HEX2DEC = Val(“&h” & sHex)
End Function
Regards
Stephen Bullen
2. Frank Kabel says:
Hi Stephen
good point. Though using VBA you could simplify many of the above formulas anyway :-)
This was merely meant for using standard worksheet formulas to replace the ATP formulas as a VBA solution would either require to include the code in each workbook or create a separate addin (which also has to be distributed)
Happy Holidays
Frank
3. Hi Frank
Sure, but my preference is definitely to include code in the workbook to do these simple functions, rather than rely on the ATP for anything. As I’m sure you’re aware, the ATP functions aren’t translated when opening a file in a different language version of Excel, so giving #NAME! errors.
Happy Holidays to youy too!
Stephen
4. Frank Kabel says:
Hi Stephen
have the same preference. the initial reason for this list was just this #NAME problem as my current client has a mixture of Office installation (we counted 5 different language versions).
So it started for some common functions (EOMONTH, etc.) and I just compiled a list of formula translations (for fun…). But for the more complex ones I also have my own addin :-))
Frank
5. Marcel Meicler says:
I am trying to do either pure octal or preferably hexadecimal arithmetic in VB through Excel.
The existing functions seem to convert my results back to decimal i.e., Hex(a) + Hex(b) is an integer number quantity that is decimal ( for a and b as integers). Is it possible to define a variable say H that remain hexadecimal so that H = hex(a) + hex(b) is a hexadecimal quantity?
Thanks,
Marcel
6. Marcel,
Computers only know one way of storing numbers – and that’s in binary. Lots of 1s and 0s.
Decimal is what humans are used to reading.
Computers have to specially turn the binary stored number into human readable decimal. eg. 123.456
Some humans also prefer to read their binary stored numbers as hexadecimal. eg. 0×45A1
Keep in mind that your computer’s CPU doesn’t know decimal or hexadecimal. It works exclusively in binary.
Operating Systems and Applications like Excel do a good job of hiding that binary complexity by keeping it as a “behind the scenes” activity.
Decimal and Hexadecimal are always strings.
The Decimal and Hexadecimal numbers which you read are made by computer programs which contruct strings character by character using logic and ascii table lookups.
Assuming a and b are recognised as numbers
Hex(a) returns a string
So when you say Hex(a) + Hex(b) you are trying to do arithmetic operations on strings which is just not possible.
H = hex(a + b)
7. Thanks God !
The BIN2DEC function in the AddIn pack is bugged. That damn thing doesn’t work in numbers with larger length (I’d tryed in a 30 characters binary number). Your function does. Thanks a lot. Save my day.
8. Marc says:
I’m converting binary numbers with up to 32 bits. That’s 4,286,578,691 in decimal.
I have no problems going from BIN2DEC, but the DEC2BIN formula stops working at about 265,000,000. I get a #NUM! error.
Any ideas?
9. Brian says:
Hi Frank,
I’ve noticed an issue with your DEC2BIN replacement function(assuming of course that I’m using it correctly).
What I really need is Hex to Bin conversion, so as suggested I use your HEX2DEC and DEC2BIN. When I start with a number like 138d7, it converts to decimal fine (80087), but the final step gives me 10011100011010100 ( jeez I hope I typed that right!).
That binary figure is actually 80084 dec or 138d4 hex. I’ve lost 3! That can be easily seen since the last hex digit of “7? should be “0111? in bin, not “0100?.
Is it me?
Thanks,
Brian
10. Brian: DEC2BIN returns a number. Excel can’t handle numbers greater than 15 significant digits, so if your BIN looks like a number that has that many significant digits, it will return the wrong result. Note that 32768 and 32769 return the same result because the last 1 in 32769 would be the 16th significant digit. In your example, there are 17 significant digits, so the last two are simply replaced with zeros.
Ideally, this function would return a string to avoid this problem. Unfortunately, Excel doesn’t let you concatenate strings inside of array formulas. I can’t think of a good alternative. Hopefully someone can think of something or at least change the formula to return an error instead of incorrect results. In the mean time, use it with caution and beware of the issue.
11. Frank_K says:
I need some help for Excel Hex2Dec function!
I have value “10? in Cell A1 and “01? in A2.
When I use Hex2Dec(A1&A2) and I expect to = Hex2Dec (“1001?).
But Excel translate to Hex2Dec(“101?).
The question is how to get Excel to translate “1001?.
Any help will be very gratful.
12. Frank: =HEX2DEC(TEXT(A1,”00?)&TEXT(A2,”00?))
Your value in A2 is a number, not a string. Excel is concatenating 10 and 1, not 10 and 01. It’s only the formatting in A2 that makes it appear as 01.
13. Frank_K says:
Thank you for your quick input and it working.
14. scott says:
This information has saved me HOURS if not DAYS.
Thank you.
15. I do not it why the replacement formula works for the DEC2BIN, can you explain ?
16. priquas says:
Still in 2009 this information is of much value, solving a big problem I’d encountered.
Moreover, the proposed approach has opened my mind to other applications.
Posting code? Use <pre> tags for VBA and <code> tags for inline. | 2,077 | 7,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | longest | en | 0.568366 |
http://themathmompuzzles.blogspot.com/2011/05/fake-utensils.html | 1,571,407,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00065.warc.gz | 184,580,421 | 20,860 | ## Wednesday, May 25, 2011
### Fake Utensils
This puzzle celebrates our puzzle solver Annie, who quietly reached over the 10 puzzle mark.
In this puzzle Annie is Manhattan's top private detective and she is being called to Tiffany's jewelry store to help investigate a fraud. Tiffany's manager informs Annie that they sold 10 sets of the Tiffany & Co Silver Utensils yesterday to 10 different people. Each set consisted of 24 pieces, all of them of the same weight. It came to their attention that one of the sets was a fake set that was supposed to be used as a window display. All the utensils in this set weigh 1/10th of the silver utensils. Tiffany's manager wanted to avoid a scandal and notification of all the buyers of a potential fraud. However, she realized that all the buyers should be called back to the store (with their utensil sets) in order to detect and replace the fake set. She asked Annie for the fastest way of detecting which of the 10 sets is the fake one in order to minimize the commotion and suspicion.
Annnie told her: "Reach out to all the 10 buyers and ask them to come to the store with the utensil sets they bought claiming that they forgot a butter knife that should be matched to their specific set. Once I have all the utensils here, all I need is to use your weight scale one time only to detect which set is fraudulent. You will need to present these buyers with 10 butter knives to leave them satisfied."
Do you think Annie was over-confident? How could she avoid weighting each of the utensil sets and use the scale only once to find the fake set?
Answers accepted all day long on Friday May 27th and Saturday May 28th, on our Family Puzzle Marathon. They will be hidden until Sunday morning (EST) and everyone who solved it will get a puzzle point. Please, explain your answer.
TyYann said...
Wheigh together one piece from the first set, 2 pieces from the second, 3 from the third etc. 10 pieces from the 10th set.
If 1/10 is missing, the first set is fake, if 2/10 are missing, it's the second one, etc.
You can even be more economical by weighing only 9 sets. If nothing is missing, the set that has not been wheighed is the fake one. This in case one of the customers could not come at the same time as the others...
anne-marie said...
The silver is not supposed to be magnetic. A magnet test would work to discover the fake test before using the scale.
Wang said...
If Annie takes one piece from one set, two pieces from the second set, three pieces and so on, then using a fine scale, she can figure out how much less of the weight it should be (say its short 1/10th the weight) and this tells her that the fake set is from the first set.
If its short 2/10, its from the second set and so on.
online gambling said...
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kj said...
Weigh 1 piece from customer #1, 2 pieces from customer #2, 3 pieces from customer #3, etc., up to 9 pieces from customer #9.
Assume the weight of each piece is w.
Then if customer #1's set is fake, it weighs
(1/10 + 2 + 3 + ... + 8 + 9)*w = 44.1*w
if customer #2's set is fake, it weighs
(1 + 2/10 + 3 + 4 + ... + 8 + 9)*w = 43.2*w
if customer #3's set is fake, it weighs
(1 + 2 + 3/10 + 4 + ... + 8 + 9)*w = 42.3*w
and following this pattern:
if customer #4's set is fake, it weighs 41.4*w
if customer #5's set is fake, it weighs 40.5*w
if customer #6's set is fake, it weighs 39.6*4
if customer #7's set is fake, it weighs 38.7*w
if customer #8's set is fake, it weighs 37.8*w
if customer #9's set is fake, it weighs 36.9*w
For the last case, it's a little bit different since we weigh no fake pieces of silver:
if customer #10's set is fake, it weighs
(1 + 2 + 3 + ... + 8 + 9)*w = 45*w
Dennis (of Dennis & Katrina) said...
She can do it with one weigh!
1) Get the 10 sets
2) Label them 1-10.
3) From set one, take one utensil and label it as coming from set one.
4) Take two utensils from set two and label them as coming from set two.
5) Repeat with 3 from 3, 4 from 4, 5 from 5, 6 from 6, 7 from 7, 8 from 8, 9 from 9, and 10 from 10, labeling each piece as to the set it game from.
6) Take the 55 labeled pieces (1+2+3+4+5+6+7+8+9+10) and weigh them.
7) You know what 55 real pieces should weigh; you will get a number less than this. The difference between what they should weigh and what you get will be a multiple of 1/10th of the weight of a real piece.
8) Divide the difference by 1/10th of the weight of a real piece; the resulting number from 1 to 10 will tell you which set is the fake!
Dennis
Tom said...
I don't see that it's a fraud in the legal sense; just a mistake. But that's not the point of the puzzle.
Nor do I see, yet, how "one time only" is going to discover, for sure, the lighter display sample, unless she gets lucky. Someone has already proved me wrong, betcha.
Whether she weighs the entire sets, or just the forks, or single forks, the process should be the same....
But if the display set really weighs only 1/10 of the genuine sets, the heft is going to be VERY easily noticed. And THEN I suppose she could use the scale, once, just to be sure.
Ilya said...
Take one utensil from the first set, two from the second, three from the third, etc. ... then from the tenth. Weight them all together. If there were no fakes, the weight would be N*(N+1)/2 = 55 times the weight of one genuine piece. To detect which of the sets if fake, take the difference between the weight you get and 55, and divide by the weight of one fake (1/10th of genuine). You get a whole number, which is the number of the set that contains all fakes.
P.S. why is it that even disposable forks are called silverware? :-)
Bean said...
I think I'm stymied...but I do know that if one fork weighs 1/10 as much as another, anyone could tell the difference just by picking it up. Politely lifting a piece of the customer's flatware to "compare" it to the butter knife would answer the question without a scale. It would also allow you to bring in customers one at a time until the bad set was found, which might allow you to alert fewer customers and save some bread knives.
But I know that's just evading the math problem that has me stumped...
Bean said...
Aha!
We know the weight of a utensil. I'm going to call that "x".
If I take one piece from the first set, two from the second, three from the third, etc. until I get to taking ten from the tenth, I will have 55 pieces.
If they were all good, they should weigh 55x. If the piece from the first set was bad, it will weigh 54.1x (54x + 1 piece that weighs .1x). If the two pieces from the second set was bad, the weight will be 53.2x (53x + (2 * .1x)). If three pieces from set three are bad, then the weight is 52.3....all the way to the tenth set, which should weigh 46x.
So you would know which set was bad from that weighing. But honestly, it might be best to tell the mathematician that she's down a rabbit hole and pull over the common sense clerk who can tell the weight by feel.
SteveGoodman18 said...
We could take 1 piece from the 1st set, 2 pieces from the second, 3 from the 3rd, etc., but none from the 10th set. This makes 45 pieces total. The total weight should be 45 units. If it actually is 45 units, you know the 10th set is the display. If the total weight is 44.1 units, then the 1st set is the display. Since there is a different number of items from each set, we can see how short the combined weight is.
Annie said...
Having set many tables and washed much silverware, I am confident that one can distinguish between one utensil set and another that weighs 1/10 of the first one. Therefore just carrying the utensil sets to the back of the store will give some indication by comparison of which set is the fake. I'm sure Tiffany's knows (or can find out) how much a genuine set weighs. When the suspect utensil set is identified, it can then be verified as the fake by one weighing on the scale. This is not sophisticated or scientific but I think it'll get quick results!
Lynnet said...
No, she is not overconfident and this is why. You take 1 piece from the first set, 2 from the second set, 3 from the third set and so on. If the real set weighs 10 grams apiece then the fake set will weigh 1 gram apiece. You set your 55 pieces on the scale. the amount that it is off, the total should equal 550 but it weigh less than that. The total will be off by 9 grams per fake piece. e.g. the total if the total is 541 grams, there is one bad piece, meaning that set 1 is fake. If the total was 487 grams, that is 63 grams off. 63 being 9x7 means that there are 7 fake pieces making set 7 fake.
Maria said...
Ok, the trick is out.
Congratulations to everyone who either knew or figured it out. Kj provided a great step by step explanation. The rest of us will apply this trick next time when this puzzle comes in another reincarnation - as coin or diamond weighting problem.
A puzzle point for TyYann, Wang, kj, Ilya, Bean, SteveGoodman18, Lynnet.
Tom, Annie and anne-Marie - you each suggested some nice workaround solutions but I think they are not technically count as the correct solution. As you each are "celebrities" of our puzzle marathon with plenty of points, we will skip rewarding you for this one.
Have a great long weekend. | 2,449 | 9,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-43 | latest | en | 0.968182 |
https://www.slideserve.com/jola/three-dimensional-simulations-of-nonlinear-mhd-wave-propagation-in-the-magnetosphere | 1,511,337,821,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00504.warc.gz | 838,114,901 | 15,466 | Three-dimensional simulations of nonlinear MHD wave propagation in the magnetosphere
1 / 28
# Three-dimensional simulations of nonlinear MHD wave propagation in the magnetosphere - PowerPoint PPT Presentation
Three-dimensional simulations of nonlinear MHD wave propagation in the magnetosphere. Kyung- Im Kim¹, Dong-Hun Lee¹, Khan- Hyuk Kim ¹ , and Kihong Kim 2 1 School of Space Research, Kyung Hee Univ . 2 Division of Energy Systems Research, Ajou Univ.
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### Three-dimensional simulations of nonlinear MHD wave propagation in the magnetosphere
Kyung-Im Kim¹, Dong-Hun Lee¹, Khan-Hyuk Kim¹,
and Kihong Kim2
1School of Space Research, Kyung Hee Univ.
2Division of Energy Systems Research, AjouUniv.
- Introduction: Effects of nonlinearity in a time-dependent system - Observations: e.g., Russell et al. (GRL, 2009) STEREO observations of shock formation in the solar wind - Theory: What kinds of exact nonlinear MHD solutions available? - Numerical approach 1. Theory vs. Numerical test2. check the profiles from ACE (Venus Express) to the Earth (STEREO)3. Apply to the 3-D homogeneous magnetosphere
- Conclusion
Introduction
B
A
• Propagation of nonlinear MHD waves is studied in the interplanetary space.
• As realistic variations in the solar wind become often nonlinear, it is important to investigate time-dependent behaviors of the solar wind fluctuations.
Introduction
• If the disturbance is linear in a uniform space, f_A = f_B:
• If the disturbance is linear in a nonuniform space, only
• the effects of refraction/reflection are to change f_A & f_B:
A
A
B
B
Introduction
• If the disturbance is nonlinear, f_A and f_B become differentiated even in a uniform space:
• The disturbance in the SW rest frame can evolve in a time-dependent manner, which is far from the steady-state.
• cf) Rankine-Hugoniot relations
A
A
B
B
Introduction
• For instance, the eqof motion for adiabatic MHD:
: Linear MHD waves
: Nonlinear MHD waves
Observations
Venus Express
0.72 AU
STEREO
1 AU
Russell et al., GRL, 2009
Theory
The assumption of simple waves (similarity flow) is often used to obtain a solution.
Exact solution for the nonlinear MHD wave is available
if it is a one-dimensional uniform system.
Theory
Sources
Simple waves
Shock waves
* Simple waves
or
Theoretical solutions
* Exact solution for the MHD wave [Lee & Kim, JGR, 2000]
Ex) piston-like motion
V (Km/s)
V (Km/s)
V (Km/s)
V (Km/s)
Time (s)
Time (s)
Numerical test vs. Exact solutions
V (Km/s)
V (Km/s)
V (Km/s)
V (Km/s)
Numerical solutions are corresponding to the exact solutions before the shock formation!
Numerical test vs. Exact solutions
V (Km/s)
X (Re)
Numerical solutions = Analytic solutions
Numerical effect test
Nx = 1000
Nx = 5000
Solution
Simulation
V (Km/s)
V (Km/s)
Time (s)
Time (s)
Numerical effect test
B (nT)
Nx=2000
Time (Hr)
B (nT)
Nx=20000
Time (Hr)
Numerical Model (1. Small scale : 300Re)
cf) Lx to ACE ~ 234 Re
* Impulse
V (Km/s)
* Simulation Parameters
- nx = 5000 , Lx = 300 Re
- Total time = 5000 s
- Sound speed = 50 Km/s
- Alfven speed = 49 Km/s
- Plasma density = 10
- Magnetic field Bz = 8 nT
t/t0
Shock formation (1. Small scale : 300Re)
* Shock formations for different background magnetic fields (3nT, 5nT, 8nT) and impulse timescales ( 𝛕0 = 300s, 500s, 700s)
B0 = 3nT
B0 = 5nT
B0 = 8nT
B (nT)
Time (s)
Time (s)
Time (s)
Considered Solar Wind flow speed : 400 Km/s
: Timescale of the source motion
53Re
48Re
45Re
246Re
256Re
256Re
Shock formation (1. Small scale : 300Re)
* Shock formations for different amplitudes of impulse
45Re
256Re
B (nT)
v = v0
v = 2v0
v = 3v0
Time (s)
Time (s)
Time (s)
Considered Solar Wind flow speed : 400 Km/s
: Timescale of the source motion
Numerical Model (2. Large scale : 0.28 AU)
cf) ~ 6560 Re
* Impulse
* Simulation Parameters
- nx = 20000
- Total time = 105,000 s (~29.16 hr)
- Lx = 7000 Re (~0.28 AU)
- Sound speed = 60 Km/s
- Alfven speed = 66 Km/s
- Plasma density = 7
- Magnetic field Bz = 9 nT
V (Km/s)
t/𝛕0
Vx
Simulation Box (2. Large scale : 0.28 AU)
Simulation Box
z
I
II
x
Solar Wind (Vsw = 400 Km/s)
0.72 AU
1 AU
Sun
Venus Express
STEREO A
Shock formation (2. Large scale : 0.28 AU)
I (at 0.72 AU)
2 nT
B (nT)
Time (Hr)
II (at 1 AU)
2 nT
B (nT)
Time (Hr)
Considered Solar Wind flow speed : 400 Km/s, 𝛕0 : Timescale of the source motion
Shock formation (2. Large scale : 0.28 AU)
I (at 0.72 AU)
B (nT)
Time (Hr)
II (at 1 AU)
B (nT)
Time (Hr)
Considered Solar Wind flow speed : 400 Km/s, 𝛕0 : Timescale of the source motion
Shock formation (2. Large scale : 0.28 AU)
I (at 0.72 AU)
B (nT)
Time (Hr)
II (at 1 AU)
B (nT)
Time (Hr)
Considered Solar Wind flow speed : 400 Km/s, 𝛕0 : Timescale of the source motion
Shock formation (2. Large scale : 0.28 AU)
I (at 0.72 AU)
B (nT)
Time (Hr)
II (at 1 AU)
B (nT)
Time (Hr)
Considered Solar Wind flow speed : 400 Km/s, 𝛕0 : Timescale of the source motion
3D model (Homogeneous Magnetosphere)
* Simulation Parameters
- Total time = 140 s
- Sound speed = 60 Km/s
- Alfven speed = 600 Km/s
- Plasma density = 1
- Magnetic field Bz = 600 nT
V (Km/s)
t/t0
Z (20 Re)
Nz=512
B0
Y (10 Re)
Ny=64
X (40Re)
Nx=512
Conclusion
• Propagation of nonlinear MHD waves is studied in the interplanetary space. We examined how these fluctuations are changed by steepening process and/or shock formation.
• The simulation results should be first validated by the exact analytical solution (Lee and Kim, 2000), which showed excellent correspondence between theory and simulation.
• The profiles tend to significantly evolve at the different locations, which strongly depends on the given parameters.
• The observations in the IP space (e.g., ACE) cannot directly deliver the SW condition around the earth unless it belongs to the linear case. | 1,972 | 6,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.805021 |
https://kr.mathworks.com/matlabcentral/profile/authors/24151155 | 1,725,718,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00051.warc.gz | 339,079,185 | 21,837 | # Csaba Zoltán Kertész
### Transilvania University Brasov
Last seen: 4개월 전 2022년부터 활동
Followers: 0 Following: 0
Teaching Embedded Systems, and Digital Signal Processing. Sometimes also doing CQ SOTA
Programming Languages:
Python, C++, C, MATLAB, Shell, Assembly
Spoken Languages:
English
Professional Interests:
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Swap the first and last columns
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Given n (always odd), return output a that has concentric rings of the numbers 1 through (n+1)/2 around the center point. Exampl...
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Find all elements less than 0 or greater than 10 and replace them with NaN
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Return the largest number that is adjacent to a zero
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https://mathspace.co/textbooks/syllabuses/Syllabus-408/topics/Topic-7243/subtopics/Subtopic-96792/?activeTab=interactive | 1,643,238,153,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305006.68/warc/CC-MAIN-20220126222652-20220127012652-00356.warc.gz | 438,081,185 | 44,253 | # Misrepresentation of Results
## Interactive practice questions
Lachlan asks $120$120 Year 12 students at his school how much time they spend on homework per night. $78$78 Year 12 students say they do more than $3$3 hours. At a meeting of the student council Lachlan reports "$65%$65% of students at this school do too much homework.
Which of the following explain why this is misleading?
The survey does not represent the population of the school.
A
The question should have been multiple choice.
B
The question was biased.
C
The sample size was too small.
D
The survey does not represent the population of the school.
A
The question should have been multiple choice.
B
The question was biased.
C
The sample size was too small.
D
Easy
Less than a minute
How does this graph misrepresent the statistics?
Choose all appropriate options.
A newspaper states: "Oil prices have risen by $33%$33% over the last two years."
In the article it states that in 2010 the oil prices increased by $10%$10% and in 2011 the oil price increased by $23%$23%.
An online company asks all its customers to complete a survey and $58%$58% respond. In the survey there was a question which asked “How happy are you with our service?”. Respondents could tick either “Very Happy” or “Somewhat Happy”.
In their annual newsletter the company wrote an article reporting “$100%$100% of customers are happy with our service”.
### Outcomes
#### S5-1
Plan and conduct surveys and experiments using the statistical enquiry cycle:– determining appropriate variables and measures;– considering sources of variation;– gathering and cleaning data;– using multiple displays, and re-categorising data to find patterns, variations, relationships, and trends in multivariate data sets;– comparing sample distributions visually, using measures of centre, spread, and proportion;– presenting a report of findings
#### S5-2
Evaluate statistical investigations or probability activities undertaken by others, including data collection methods, choice of measures, and validity of findings | 450 | 2,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-05 | longest | en | 0.959283 |
http://basilisk.fr/src/test/planar.c | 1,519,500,965,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815934.81/warc/CC-MAIN-20180224191934-20180224211934-00337.warc.gz | 34,159,860 | 4,446 | # Electrostatic in planar layers
This test reproduces partially the table of convergence of the electric field for different electrostatic configurations shown in Lopez-Herrera et al, (2011). The configurations tested are: (a) both layers are dielectric, (b) both layers are conducting and (c) one layer conducting and the other dielectric.
We use the implicit electrodynamic solver. The explicit solver ehd.h, which is fully conservative, gives similar results. We also need the run() loop.
``````#include "ehd/implicit.h"
#include "run.h"``````
Variable $ic$ controls the type of electrical configuration tested. The analytical values used for comparison are stored in E1 and E2. Finally, tfinal controls the instant at which results are written. Note that for configuration (a) to reach a stationary state requires just one iteration while for the other configurations about 15 are required.
``````int ic;
double tfinal, E1, E2;
φ[top] = dirichlet(0);
φ[bottom] = dirichlet(1);
#define β 3.
#define η 2.
int main() {
X0 = Y0 = -0.5;
DT = 1;
TOLERANCE = 1e-5;
int LEVEL;
for (ic = 0; ic < 3; ic++) {
switch (ic) {
case 0: // both layers are dielectric;
fprintf(stderr, "dielectric-dielectric\n");
E1 = 2/(1+β);
E2 = 2*β/(1+β);
tfinal = 1;
break;
case 1: // both layers are conducting;
fprintf(stderr, "conducting-conducting\n");
E1 = 2/(1+η);
E2 = 2*η/(1+η);
tfinal = 15;
break;
case 2: //bottom layer conducting upper one dielectric;
fprintf(stderr, "conducting-dielectric\n");
E1 = 0.;
E2 = 2.;
tfinal = 15;
break;
}
for (LEVEL = 5; LEVEL <= 7; LEVEL++) {
N = 1 << LEVEL;
run();
}
}
}
event init (t = 0) {
scalar f[];
foreach() {
f[] = (y > 0. ? 0. : 1.0);
rhoe[] = 0.;
}
boundary ({rhoe, f});
foreach_face(){
double T = (f[]+f[-1,0])/2.;``````
If in configuration (a) we use an harmonic interpolation for the permittivity the exact values of the electric field are recovered.
`````` ε.x[] = 1./(T/β + (1. - T)); //Exact for the diel-diel
// epsilon.x[] = T*beta + (1. - T); // Non exact
K.x[] = (ic == 0 ? 0. : (ic == 1 ? T*η+(1. - T) : η*T));
}
boundary ((scalar *){ε, K});
}
event set_dt (i++)
dt = dtnext (DT);``````
In the last step the relative error of the electric field, $E=-\nabla \varphi$, in each medium is reported.
``````event result (t = tfinal) {
scalar E[];
foreach()
E[] = (φ[0,-1] - φ[0,1])/(2.*Δ);
stats s = statsf (E);
fprintf (stderr, "%d %g %g\n", N, 100*fabs(1.- s.max/E2),
E1 > 0. ? 100*fabs(1.- s.min/E1) : 0. );
}`````` | 807 | 2,471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-09 | longest | en | 0.69304 |
https://www.teacherspayteachers.com/Product/Halloween-Math-Craftivities-for-SECOND-Grade-Graphing-Geometry-MORE-4898864 | 1,580,272,437,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00388.warc.gz | 1,085,364,776 | 29,178 | # Halloween Math Craftivities for SECOND Grade {Graphing, Geometry, & MORE!}
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Have FUN this Halloween and incorporate interactive seasonal extensions to your lesson with craftivities or math art. You will find all you need for 4 different easy-to-prep craftivities which will make the perfect Halloween hallway or bulletin board display! Graphing, adding and subtracting with fluency from 0-20, geometry and even and odd up to 3 digit numbers are the concepts you will find within this resource which are aligned to the common core standards for second grade.
• "Witches Brew" Make a Graph (addition base tens and ones).
• Draw a Skull (2-D and 3-D shapes)
• Halloween Moon Even & Odd (even/odd recognition up to 3 digits)
• Spider Fact Fluency (adding/subtracting 0-20)
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Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 733 | 3,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-05 | latest | en | 0.842515 |
http://doczz.net/doc/228172/sangaku---japanese-temple-mathematics | 1,558,879,476,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259177.93/warc/CC-MAIN-20190526125236-20190526151236-00046.warc.gz | 60,083,720 | 10,714 | # Sangaku - Japanese Temple Mathematics
## Transcription
Sangaku - Japanese Temple Mathematics
```Sangaku - Japanese
Temple
Mathematics
Rosalie Hosking
Department of Mathematics and Statistics
What are Sangaku
• Sangaku are wooden tablets containing mathematics
• Traditionally hung in eaves of Shinto shrines and
Buddhist temples
• Practice of dedicating sangaku began during the early
Edo period (1600-1868)
• This was during the period of national seculsion
• Earliest known tablet dedicated in 1683
• Many significant sangaku were dedicated during the
Meiji period (1868-1912)
• Sangaku using modern mathematics are still produced
Features of Sangaku
• Written in the Kanbun language rather than ordinary Japanese
• Kanbun was a rather prestigious academic language like Latin
• Many Confucian texts an literary works used this language
• Often contain geometrical problems
• Problems are generally abstract, but not exclusively (e.g. finding
the height of sacred mountains)
• Commonly give a problem and a proof
• Sometimes challenge the observer to figure out the proof
• Sometimes the creator can’t solve the problem
• Created by everyone – professional mathematicians, samurai,
farmers, women, and children
• Contain colourful accompanying pictures
• Sometimes contain non mathematical paintings
Katayamahiko Shrine, Okayama Prefecture
August 1873
Toenji Shrine, Tokyo Prefecture, 1869
Sangaku from Yamagata Prefecture, 1823
Toyama Prefecture, 1859
Kyoto Prefecture, 1683
Katayamahiko Problems
Click to edit Master
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• Second level
• Third level
• Fourth level
• Fifth level
• Second level
• Third level
• Fourth level
• Fifth level
A circle of radius r inscribes three circles of
radius t, the centers of which form an
equilateral triangle of side 2t.
Find t in terms of r.
If r = 10, then t = 4.64 (SMJTG p. 96)
Solution
• We know the distance between r and any
t is r – t
• We know the centres of t form the corners of an
equilateral triangle, so cos30˚ = t/(r – t)
• We can describe t in terms of r as so
• Most sangaku deal with problems of this form
Katayamahiko #2
Click to edit Master
textto
styles
• Click
edit Master text styles
• Second level
• Third level
• Fourth level
• Fifth level
• Second level
• Third level
• Fourth level
• Fifth level
An equilateral triangle with a side a is inscribed in
a square also of side a, along with four circles,
Find t in terms of r. (SMJTG p. 114)
• Need to recognize the triangle is equilateral
•
• Drawing a line from say the center of the
right circle t to the lower right-hand corner
shows that
• A good half-angle formulae for tangent in this
case is giving
• Eliminating a from the two expressions gives
• Trick is looking for what angles we have and
working from there – that is the traditional way
• Intuition and deduction are important
Sawa Masayoshi’s problem
• An ellipse is inscribed in a right triangle with its major axis
parallel to the hypotenuse. Within the ellipse are inscribed
two circles of radius r. A third circle of radius r touches the
ellipse and the two shorter sides of the triangle, a and b.
• Find r in terms of a and b (SMJTG p. 259)
Why create Sangaku?
• Sangaku are a result of the idai tradition
• Idai are difficult problems without answers
• First appearance was in the 1641 edition of the Jinkouki
of Yoshida Mitsuyoshi
• The Jinkouki taught mathematics to the general public. It
focused on commercial problems and was in a problemanswer format
• Jinkouki was extremely popular, estimated to have sold
over 1 million copies. Many illicit copies were produced.
• To curb plagiarism, author added a section of difficult
problems and challenged the copiers to solve them
Competitive Mathematics
• Soon after the inclusion of idai in the Jinkouki,
they began to appear in other mathematical texts
• Idai became a way for those apt at mathematics
to test and show off their abilities in the absence
of a unified discipline of mathematics
• The few schools of mathematics that existed were
secretive and competitive
• Sangaku most likely developed as a local way to
present idai
Why Shrines and Temples?
• Sangaku are related to ema
• Ema are wooden tablets which initially would display pictures
of horses (e = picture, ma = horse)
• Horses used to be sacrificed to the Gods, but it was expensive
so people started to dedicate pictures instead
• Many were created that also displayed painted scenes of
mythological or historical significance
• They were also hung in eaves and in shrines and temples
• Modern ema are smaller – one puts their wish/hope on the
back, hangs it up, and they are burned
Old ema hung in a Shinto shrine
• Because they take the form of ema, they likely had some religious
function
• We see a connection to religion in the preface to the following
sangaku from 1815 for example
“The teacher Takeda has been studying mathematics since he was
young. In this shrine, his disciples ask God for progress in their
mathematical ability and dedicate a sangaku” (SMJTG p.243)
• People believed the Gods might give them good mathematical luck or
abilities by dedicating work to them
• Shrines and temples were public places of community as well
• Hanging a sangaku in a shrine or temple ensured maximum publicity
Why Geometry?
• It may have seemed more appropriate to present
mathematics which was highly visual like this so it fitted
in with other artistic dedications
• May have created a sense of wonder in non
mathematicians and functioned as objects of
contemplation and awe
• May have also pleased and attracted the Kami
• Circles are associated with the sun goddess - heavy use of
circular problems could be paying homage to her
• Geometry is generally a popular area of study across
cultures historically, so perhaps there is no other reason
than that people are naturally drawn to it
Interest for Historians
• The tradition developed in recent times so we can
easily pinpoint its beginning and evolution
• It is an example of mathematics developing in
relative isolation
• It is an example of an interesting unique way of
expressing mathematics – combining of
mathematics, art, and religion
• They are reflective of Japanese culture of the time
and attitudes towards mathematics
• Gives us a sense of their abilities and thoughts
Sangaku Revival
• In the last twenty years, there is a small revival in
sangaku production
• Teachers are starting to use sangaku to introduce
geometry to students
• New style of sangaku keep the traditional design
but often use modern mathematical notation and
• Here are some examples
Kasai Shrine, Tokyo Prefecture, 2009
Tokyo Prefecture, 2012
Sangaku created by middle schooler in 2006
• Sacred Mathematics: Japanese Temple Geometry
by Tony Rothman & Hidetoshi Fukagawa
• ‘Japanese Temple Geometry’ in Scientific
American (May 1998)
• Hiroshi Kotera’s website www.wasan.jp (Japanese
version is recommended – has heaps of great
images)
• There is also a facebook page run by Hiroshi
Kotera – search for 和算なう
```
### Sangaku--Japanese Mathematics and Art in the 18 ,19 and 20
There are many interesting sangaku that could not be transported to the exhibition hall of the museum because they were mounted and could not be removed or were too fragile. For instance, Figure 7 ... | 1,758 | 7,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-22 | latest | en | 0.88016 |
http://hirder.tk/slip-sheet-load-definition-electrical.html | 1,561,589,027,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00446.warc.gz | 80,176,496 | 4,544 | # Slip sheet load definition electrical
Slip definition
## Slip sheet load definition electrical
FSynch = Synchronous electrical speed = 2 x FLine / P. Design for lifetime performance design examples, 50 easy calculators , electrical reliability This book contains 472 pages in fullcolorand over 250 illustrations, 300 formulae, 100 case studies , reliability: Advanced engineering design Lifetime performance 50 photographs of. Single motors or loads electrical cannot be in standby mode. Cut Your Electric electrical Bill electrical in Half central air conditioner mist definition n save - Duration:. Be aware of any slip trip fall definition definition possibilities around the system. Check the inlet screens on the fill hoses. Keep access gates and control rooms locked.
In electrical engineering the power factor of an AC electrical power definition system is defined as the ratio of slip the real power absorbed by the load to the apparent power flowing in the circuit, is a dimensionless number in the closed interval of − 1 to 1. It is the speed of the rotating magnetic field that sheet is generated slip slip and the speed the rotor tries to attain ( it will electrical never quite reach that speed). What happens if the motor is started as a normal definition induction motor? If there is an ‘ A’ & ‘ B’ motor slip it is typical to consider all sheet ‘ sheet A’ motors as continuous loads all ‘ B’ motors as standby loads. Electrical Conduit is widely used to protect cables definition that carry various amounts of electrical current.
Fuel calculations for the MegaSquirt- II algorithm are performed in the HCS12 processor ( such as fuel , ignition anything else requiring math calculations). FSlip = Slip frequency = FSynch - rotor RPM. SWG - Standard Wire Gauge. According to NEMA definition is how much load and FLA. Provide signage that states " Authorized Personnel Only" and " Danger: High- Voltage Electrical Area" to warn others of potential hazards.
AVR Generators ( Digital Analogue) : AVR generators feature either a digital analogue Automatic Voltage Regulator designed to control voltage. title: description: group definition electrical — aircraft: the complete operational unit. FAULT CODE SOLUTION/ FIX: Make sure the water faucets are open all the way. SAMSUNG FRONT LOAD WASHER FAULT CODE: nF sheet SAMSUNG FAULT CODE DEFINITION: sheet electrical Your washer has tried to fill with water but was unsuccessful. definition ata definitions of aircraft electrical groups systems sub- systems : chap.
This can be confusing because it refers to slip electrically synchronous, definition not synchronous to the rpm. National Electrical Manufacture Association. Electrical conduit provides very good protection to enclosed conductors from sheet impact moisture, chemical vapors. Slip sheet attachment ( push - pull) - is a hydraulic attachment that reaches forward draws the slip sheet onto wide , clamps onto a slip sheet thin metal forks for transport.
``slip sheet load definition electrical`` | 591 | 3,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.861909 |
https://en.m.wikipedia.org/wiki/Estimation_statistics | 1,553,030,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202131.54/warc/CC-MAIN-20190319203912-20190319225912-00320.warc.gz | 486,524,961 | 18,580 | Estimation statistics
Estimation statistics is a data analysis framework that uses a combination of effect sizes, confidence intervals, precision planning, and meta-analysis to plan experiments, analyze data and interpret results.[1] It is distinct from null hypothesis significance testing (NHST), which is considered to be less informative.[2][3] Estimation statistics, or simply estimation, is also known as the new statistics,[3] a distinction introduced in the fields of psychology, medical research, life sciences and a wide range of other experimental sciences where NHST still remains prevalent,[4] despite estimation statistics having been recommended as preferable for several decades.[5][6]
The primary aim of estimation methods is to report an effect size (a point estimate) along with its confidence interval, the latter of which is related to the precision of the estimate.[7] The confidence interval summarizes a range of likely values of the underlying population effect. Proponents of estimation see reporting a P value as an unhelpful distraction from the important business of reporting an effect size with its confidence intervals,[8] and believe that estimation should replace significance testing for data analysis.[9]
History
Physics has for long employed a weighted averages method that is similar to meta-analysis.[10]
Estimation statistics in the modern era started with the development of the standardized effect size by Jacob Cohen in the 1960s. Research synthesis using estimation statistics was pioneered by Gene V. Glass with the development of the method of meta-analysis in the 1970s.[11] Estimation methods have been refined since by Larry Hedges, Michael Borenstein, Doug Altman, Martin Gardner, Geoff Cumming and others. The systematic review, in conjunction with meta-analysis, is a related technique with widespread use in medical research. There are now over 60,000 citations to "meta-analysis" in PubMed. Despite the widespread adoption of meta-analysis, the estimation framework is still not routinely used in primary biomedical research.[4]
In the 1990s, editor Kenneth Rothman banned the use of p-values from the journal Epidemiology; compliance was high among authors but this did not substantially change their analytical thinking.[12]
More recently, estimation methods are being adopted in fields such as neuroscience,[13] psychology education[14] and psychology.[15]
The Publication Manual of the American Psychological Association recommends estimation over hypothesis testing.[16] The Uniform Requirements for Manuscripts Submitted to Biomedical Journals document makes a similar recommendation: "Avoid relying solely on statistical hypothesis testing, such as P values, which fail to convey important information about effect size."[17]
Methodology
Many significance tests have an estimation counterpart;[18] in almost every case, the test result (or its p-value) can be simply substituted with the effect size and a precision estimate. For example, instead of using Student's t-test, the analyst can compare two independent groups by calculating the mean difference and its 95% confidence interval. Corresponding methods can be used for a paired t-test and multiple comparisons. Similarly, for a regression analysis, an analyst would report the coefficient of determination (R2) and the model equation instead of the model's p-value.
However, proponents of estimation statistics warn against reporting only a few numbers. Rather, it is advised to analyze and present data using data visualization.[2][6][7] Examples of appropriate visualizations include the Scatter plot for regression, and Gardner-Altman plots for two independent groups.[19] While historical data-group plots (bar charts, box plots, and violin plots) do not display the comparison, estimation plots add a second axis to explicitly visualize the effect size [20].
The Gardner-Altman plot. Left: A conventional bar chart, using asterisks to show that the difference is 'statistically significant.' Right: A Gardner-Altman plot that shows all data points, along with the mean difference and its confidence intervals.
Gardner-Altman plot
The Gardner-Altman mean difference plot was first described by Martin Gardner and Doug Altman in 1986[19]; it is a statistical graph designed to display data from two independent groups.[6] There is also a version suitable for paired data. The key instructions to make this chart are as follows: (1) display all observed values for both groups side-by-side; (2) place a second axis on the right, shifted to show the mean difference scale; and (3) plot the mean difference with its confidence interval as a marker with error bars.[3] Gardner-Altman plots can be generated with custom code using Ggplot2, seaborn, or DABEST; alternatively, the analyst can use user-friendly software like the Estimation Stats app.
The Cumming plot. All raw data is shown. The effect size and 95% CIs are plotted on a separate axes beneath the raw data. For each group, summary measurements (mean ± standard deviation) are shown as gapped lines.
Cumming plot
For multiple groups, Geoff Cumming introduced the use of a secondary panel to plot two or more mean differences and their confidence intervals, placed below the observed values panel[3]; this arrangement enables easy comparison of mean differences ('deltas') over several data groupings. Cumming plots can be generated with the ESCI package, DABEST, or the Estimation Stats app.
Other methodologies
In addition to the mean difference, there are numerous other effect size types, all with relative benefits. Major types include Cohen's d-type effect sizes, and the coefficient of determination (R2) for regression analysis. For non-normal distributions, there are a number of more robust effect sizes, including Cliff's delta and the Kolmogorov-Smirnov statistic.
Flaws in hypothesis testing
In hypothesis testing, the primary objective of statistical calculations is to obtain a p-value, the probability of seeing an obtained result, or a more extreme result, when assuming the null hypothesis is true. If the p-value is low (usually < 0.05), the statistical practitioner is then encouraged to reject the null hypothesis. Proponents of estimation reject the validity of hypothesis testing[3][7] for the following reasons, among others:
• P-values are easily and commonly misinterpreted. For example, the p-value is often mistakenly thought of as 'the probability that the null hypothesis is true.'
• The null hypothesis is always wrong for every set of observations: there is always some effect, even if it is minuscule.[21]
• Any particular p-value arises through the interaction of the effect size, the sample size (all things being equal a larger sample size produces a smaller p-value) and sampling error.[23]
• At low power, simulation reveals that sampling error makes p-values extremely volatile.[24]
Benefits of estimation statistics
Confidence intervals behave in a predictable way. By definition, 95% confidence intervals have a 95% chance of capturing the underlying population mean (μ). This feature remains constant with increasing sample size; what changes is that the interval becomes smaller (more precise). In addition, 95% confidence intervals are also 83% prediction intervals: one experiment's confidence interval has an 83% chance of capturing any future experiment's mean.[3] As such, knowing a single experiment's 95% confidence intervals gives the analyst a plausible range for the population mean, and plausible outcomes of any subsequent replication experiments.
Evidence-based statistics
Psychological studies of the perception of statistics reveal that reporting interval estimates leaves a more accurate perception of the data than reporting p-values.[25]
Precision planning
The precision of an estimate is formally defined as 1/variance, and like power, increases (improves) with increasing sample size. Like power, a high level of precision is expensive; research grant applications would ideally include precision/cost analyses. Proponents of estimation believe precision planning should replace power since statistical power itself is conceptually linked to significance testing.[3]
References
1. ^ Ellis, Paul. "Effect size FAQ".
2. ^ a b Cohen, Jacob. "The earth is round (p<.05)" (PDF).
3. Cumming, Geoff (2012). Understanding The New Statistics: Effect Sizes, Confidence Intervals, and Meta-Analysis. New York: Routledge.
4. ^ a b Button, Katherine; John P. A. Ioannidis; Claire Mokrysz; Brian A. Nosek; Jonathan Flint; Emma S. J. Robinson; Marcus R. Munafò (2013). "Power failure: why small sample size undermines the reliability of neuroscience". Nature Reviews Neuroscience. 14 (5): 365–76. doi:10.1038/nrn3475. PMID 23571845.
5. ^ Altman, Douglas (1991). Practical Statistics For Medical Research. London: Chapman and Hall.
6. ^ a b c Douglas Altman, ed. (2000). Statistics with Confidence. London: Wiley-Blackwell.
7. ^ a b c Cohen, Jacob (1990). "What I have Learned (So Far)". American Psychologist. 45 (12): 1304. doi:10.1037/0003-066x.45.12.1304.
8. ^ Ellis, Paul (2010-05-31). "Why can't I just judge my result by looking at the p value?". Retrieved 5 June 2013.
9. ^ Claridge-Chang, Adam; Assam, Pryseley N (2016). "Estimation statistics should replace significance testing". Nature Methods. 13 (2): 108–109. doi:10.1038/nmeth.3729. PMID 26820542.
10. ^ Hedges, Larry (1987). "How hard is hard science, how soft is soft science". American Psychologist. 42 (5): 443. CiteSeerX 10.1.1.408.2317. doi:10.1037/0003-066x.42.5.443.
11. ^ Hunt, Morton (1997). How science takes stock: the story of meta-analysis. New York: The Russell Sage Foundation. ISBN 978-0-87154-398-1.
12. ^ Fidler, Fiona (2004). "Editors Can Lead Researchers to Confidence Intervals, but Can't Make Them Think". Psychological Science. 15 (2): 119–126. doi:10.1111/j.0963-7214.2004.01502008.x. PMID 14738519.
13. ^ Yildizoglu, Tugce; Weislogel, Jan-Marek; Mohammad, Farhan; Chan, Edwin S.-Y.; Assam, Pryseley N.; Claridge-Chang, Adam (2015-12-08). "Estimating Information Processing in a Memory System: The Utility of Meta-analytic Methods for Genetics". PLOS Genet. 11 (12): e1005718. doi:10.1371/journal.pgen.1005718. ISSN 1553-7404. PMC 4672901. PMID 26647168.
14. ^ Hentschke, Harald; Maik C. Stüttgen (December 2011). "Computation of measures of effect size for neuroscience data sets". European Journal of Neuroscience. 34 (12): 1887–1894. doi:10.1111/j.1460-9568.2011.07902.x. PMID 22082031.
15. ^ Cumming, Geoff. "ESCI (Exploratory Software for Confidence Intervals)".
16. ^ "Publication Manual of the American Psychological Association, Sixth Edition". Retrieved 17 May 2013.
17. ^ "Uniform Requirements for Manuscripts Submitted to Biomedical Journals". Archived from the original on 15 May 2013. Retrieved 17 May 2013.
18. ^ Cumming, Geoff; Calin-Jageman, Robert (2016). Introduction to the New Statistics: Estimation, Open Science, and Beyond. Routledge. ISBN 978-1138825529.
19. ^ a b Gardner, M. J.; Altman, D. G. (1986-03-15). "Confidence intervals rather than P values: estimation rather than hypothesis testing". British Medical Journal (Clinical Research Ed.). 292 (6522): 746–750. ISSN 0267-0623. PMC 1339793. PMID 3082422.
20. ^ Ho, Joses; Tumkaya; Aryal; Choi; Claridge-Chang (2018). "Moving beyond P values: Everyday data analysis with estimation plots". bioRxiv: 377978. doi:10.1101/377978.
21. ^ Cohen, Jacob (1994). "The earth is round (p < .05)". American Psychologist. 49 (12): 997–1003. doi:10.1037/0003-066X.49.12.997.
22. ^ Ellis, Paul (2010). The Essential Guide to Effect Sizes: Statistical Power, Meta-Analysis, and the Interpretation of Research Results. Cambridge: Cambridge University Press.
23. ^ Denton E. Morrison, Ramon E. Henkel, ed. (2006). The Significance Test Controversy: A Reader. Aldine Transaction. ISBN 978-0202308791.
24. ^ Cumming, Geoff. "Dance of the p values".
25. ^ Beyth-Marom, R; Fidler, F.; Cumming, G. (2008). "Statistical cognition: Towards evidence-based practice in statistics and statistics education". Statistics Education Research Journal. 7: 20–39. | 2,844 | 12,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-13 | latest | en | 0.91858 |
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# How to implement the rtcp/tcpm with pmac?
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RTCP(Real-time Tool Center Point rotation)/TCPM(Tool Center Point Managment) is usefull in 5-axes machining, someone said it can be finished with the forward/inverse kinematic in pmac,but how? Is there some more detailed information?
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What is that exactly? Can you post a link to the source to which you are referring?
The Kinematics subroutines can be used to perform complicated algorithms in order to convert tooltip to motor positions (inverse kinematics) and vice versa (forward kinematics). If you need to perform some complicated coordinate system rotation algorithm, you can carry out all the necessary mathematical computations in the kinematics subroutines and then output the motor's position for PMAC to use.
Please refer to page 259 of the Turbo PMAC User Manual for more details on how to program the kinematics subroutines.
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What is that exactly? Can you post a link to the source to which you are referring?
The Kinematics subroutines can be used to perform complicated algorithms in order to convert tooltip to motor positions (inverse kinematics) and vice versa (forward kinematics). If you need to perform some complicated coordinate system rotation algorithm, you can carry out all the necessary mathematical computations in the kinematics subroutines and then output the motor's position for PMAC to use.
Please refer to page 259 of the Turbo PMAC User Manual for more details on how to program the kinematics subroutines.
Thanks for your reply.
RTCP refer to a technology that keep the tooltip following the workpiece while the A/B/C axes is rotating. Some details here:
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Here is an application note describing the kinematics we use in our NC to do this.
What is that exactly? Can you post a link to the source to which you are referring?
The Kinematics subroutines can be used to perform complicated algorithms in order to convert tooltip to motor positions (inverse kinematics) and vice versa (forward kinematics). If you need to perform some complicated coordinate system rotation algorithm, you can carry out all the necessary mathematical computations in the kinematics subroutines and then output the motor's position for PMAC to use.
Please refer to page 259 of the Turbo PMAC User Manual for more details on how to program the kinematics subroutines.
Thanks for your reply.
RTCP refer to a technology that keep the tooltip following the workpiece while the A/B/C axes is rotating. Some details here:
Tool Tip Compensation Using PMAC Inverse Kinematics.pdf
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Here is an application note describing the kinematics we use in our NC to do this.
Thanks for your reply and note.When rotating B axis,tool tip is still.If the tool tip is wished to do a segment or circle interpolation at the same time,for example,the tool tip do a circle movement and the B axis pointting to the circle center. Can this finished by the kinematic?
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Thanks for your share.
But I seriously doubt about the following transformation refered in your file.
```Q7=Q7 + (Total_Length * SIN(Q2))
Q9=Q9 + ( Total_Length * COS(Q2))-TOTAL_LENGTH
```
Does your machine working properly?
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P. David Polly
Department of Geological Sciences
Indiana University
1001 E. 10th Street
Bloomington, IN 47405 USA
http://www.indiana.edu/~geosci
Geometric Morphometrics is the analysis of shape using Cartesian geometric coordinates rather than linear, areal, or volumetric variables. Geometric morphometric methods (GMM) include 2D and 3D points representing landmarks, curves, outlines, or surfaces. This course is a practical, applied introduction to GMM. Students learn to collect, analyze, and interpret geometric morphometric data. Shape theory and methods are covered, including Procrustes superimposition and its statistical implications, analysis of curves and outlines, and Monte Carlo modeling of shape.
### Lecture 1 - Introduction to Geometric Morphometrics
Overview of principles of geometric morphometric methods (GMM), including definitions of landmarks, outlines and surfaces, differentiation between traditional morphometrics and GMM, short history of morphometrics, components of a morphometric study, and examples of equipment needed for GMM studies.
### Lecture 2 - Introduction to Mathematica
Learn basics of Mathematica, including basic operations, using cells and styles, writing short programs, defining functions, and making simple graphs.
### Lecture 3 - First GMM Analysis
Basics of Procrustes analysis, principal components analysis of shape, and morphospace. Introduction to the morphometrics functions tpsImport[], Procrustes[], PrincipalComponentsOfShape[]. Analysis of faces.
### Lecture 4 - Procrustes , PCA, and Morphospace
Details of Procrustes, Procrustes distances, Principal Components Analysis, Eigenvalues, Eigenvectors, Scores, and shape modelling.
### Lecture 5 - Statistical Analysis of Shape
Review of PCA, introduction to statistical tests, introduction to R-squared values, univariate regression, multivariate regression, ANOVA, MANOVA.
### Session 6 - Bootstrapping, Randomization, and Monte Carlo methods
This session is based on readings by Adams & Anthony (1996), Kowalewski & Novack-Gotshall (2010), and Efron & Tibshirani (1986). What are "statistics"? What is a statistical test? What is a statistical null hypothesis? How can resamppling and randomizations of real samples be used to construct customized statistical tests. Bootstrapping, jack-knifing, randomization, Monte Carlo.
[Assignment 5]
### Lecture 7- Bootstrapping, Randomization, and Monte Carlo methods
Continues previous topic.
[Assignment 5 with tips]
### Lecture 8 - Analysis of outlines and Euclidean DIstance Matrix Anlaysis (EDMA)
Intro to semilandmarks, Fourier analysis, semilandmark analysis, sliding semilandmarks, eigenshape, "Pinocchio effect", alternatives to Procrustes, EDMA.
### Lecture 9 - Phylogeny, trees, and morphospace
Phylogenetic signal in morphometric data, "factor thinking", tree building algorithms and terminology, building morphometric trees in Phylip, phylogenetic comparative statistics, trees in morphospace.
### Lecture 10 - Monte Carlo simulation of evolution of shape
Random walks, their properties and statistics, random walks in more than one dimension, random walks in morphospace, modeling random evolution.
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2. Serial order wise
3. Ex 9.6
Transcript
Ex 9.6, 10 For each of the differential equation given in Exercises 1 to 12, find the general solution : ( + ) / =1 Step 1: Put in form / + Py = Q or / + P1x = Q1 (x + y) / = 1 Dividing by (x + y), / = 1/(( + )) / = ( + ) / x = / + ( 1) x = Step 2: Find P1 and Q1 Comparing (1) with / + P1x = Q1 P1 = 1 and Q1 = y Step 3: Find Integrating factor, I.F. I.F. = ^ 1 _1 = e^ 1 ( 1) = e y So, I.F. = e y Step 4 : Solution of the equation x I.F. = 1 1 . . + Putting values, x e y = 1 ^( ). + Let I = 1 . ^( ) = y 1 ^( ) 1 [ / 1 ^( ) ] dy = y ^( )/( 1) 1 1. ^( )/( 1) dy. = . ^( ) + 1 ^( ) = . ^( ) + ^( )/( 1) = . ^( ) ^( ) Putting value of I in (2) x e y = 1 ^( ). + x e y = ^( ) ^( )+ Dividing by ^( ) x = y 1 + Cey x + y + 1 = Cey
Ex 9.6 | 358 | 824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.642705 |
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# Function Variable
Posts: 168Member
In the follwoing program I am having trouble understanding this function.
[code]
"""Ask a yes or no question."""
response = None
while response not in ("y", "n"):
response = raw_input(question).lower()
return response[/code]
Shouldn't question be response, or response be question? I know thought question was just a variable that held that date sent to the function.
[code]# Tic-Tac-Toe
# Plays the game of tic-tac-toe against a human opponent
# global constants
X = "X"
O = "O"
EMPTY = " "
TIE = "TIE"
NUM_SQUARES = 9
def display_instruct():
"""Display game instructions."""
print
"""
Welcome to the greatest intellectual challenge of all time: Tic-Tac-Toe.
This will be a showdown between your human brain and my silicon processor.
You will make your move known by entering a number, 0 - 8. The number
will correspond to the board position as illustrated:
0 | 1 | 2
---------
3 | 4 | 5
---------
6 | 7 | 8
Prepare yourself, human. The ultimate battle is about to begin.
"""
"""Ask a yes or no question."""
response = None
while response not in ("y", "n"):
response = raw_input(question).lower()
return response
"""Ask for a number within a range."""
response = None
while response not in range(low, high):
response = int(raw_input(question))
return response
def pieces():
"""Determine if player or computer goes first."""
go_first = ask_yes_no("Do you require the first move? (y/n): ")
if go_first == "y":
print "
Then take the first move. You will need it."
human = X
computer = O
else:
print "
computer = X
human = O
return computer, human
def new_board():
"""Create new game board."""
board = []
for square in range(NUM_SQUARES):
board.append(EMPTY)
return board
def display_board(board):
"""Display game board on screen."""
print "
", board[0], "|", board[1], "|", board[2]
print " ", "---------"
print " ", board[3], "|", board[4], "|", board[5]
print " ", "---------"
print " ", board[6], "|", board[7], "|", board[8], "
"
def legal_moves(board):
"""Create list of legal moves."""
moves = []
for square in range(NUM_SQUARES):
if board[square] == EMPTY:
moves.append(square)
return moves
def winner(board):
"""Determine the game winner."""
WAYS_TO_WIN = ((0, 1, 2),
(3, 4, 5),
(6, 7, 8),
(0, 3, 6),
(1, 4, 7),
(2, 5, 8),
(0, 4, 8),
(2, 4, 6))
for row in WAYS_TO_WIN:
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY:
winner = board[row[0]]
return winner
if EMPTY not in board:
return TIE
return None
def human_move(board, human):
"""Get human move."""
legal = legal_moves(board)
move = None
while move not in legal:
move = ask_number("Where will you move? (0 - 8):", 0, NUM_SQUARES)
if move not in legal:
print "
That square is already occupied, foolish human. Choose another.
"
print "Fine..."
return move
def computer_move(board, computer, human):
"""Make computer move."""
# make a copy to work with since function will be changing list
board = board[:]
# the best positions to have, in order
BEST_MOVES = (4, 0, 2, 6, 8, 1, 3, 5, 7)
print "I shall take square number",
# if computer can win, take that move
for move in legal_moves(board):
board[move] = computer
if winner(board) == computer:
print move
return move
# done checking this move, undo it
board[move] = EMPTY
# if human can win, block that move
for move in legal_moves(board):
board[move] = human
if winner(board) == human:
print move
return move
# done checkin this move, undo it
board[move] = EMPTY
# since no one can win on next move, pick best open square
for move in BEST_MOVES:
if move in legal_moves(board):
print move
return move
def next_turn(turn):
"""Switch turns."""
if turn == X:
return O
else:
return X
def congrat_winner(the_winner, computer, human):
"""Congratulate the winner."""
if the_winner != TIE:
print the_winner, "won!
"
else:
print "It's a tie!
"
if the_winner == computer:
print "As I predicted, human, I am triumphant once more.
"
"Proof that computers are superior to humans in all regards."
elif the_winner == human:
print "No, no! It cannot be! Somehow you tricked me, human.
"
"But never again! I, the computer, so swears it!"
elif the_winner == TIE:
print "You were most lucky, human, and somehow managed to tie me.
"
"Celebrate today... for this is the best you will ever achieve."
def main():
display_instruct()
computer, human = pieces()
turn = X
board = new_board()
display_board(board)
while not winner(board):
if turn == human:
move = human_move(board, human)
board[move] = human
else:
move = computer_move(board, computer, human)
board[move] = computer
display_board(board)
turn = next_turn(turn)
the_winner = winner(board)
congrat_winner(the_winner, computer, human)
# start the program
main()
raw_input("
Press the enter key to quit.")[/code] | 1,326 | 4,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-07 | longest | en | 0.865585 |
https://nbviewer.jupyter.org/url/www.cs.colostate.edu/~anderson/cs445/notebooks/05%20Linear%20Ridge%20Regression%20and%20Data%20Partitioning.ipynb | 1,558,927,836,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232260658.98/warc/CC-MAIN-20190527025527-20190527051527-00207.warc.gz | 569,449,507 | 166,788 | $\newcommand{\xv}{\mathbf{x}} \newcommand{\Xv}{\mathbf{X}} \newcommand{\yv}{\mathbf{y}} \newcommand{\zv}{\mathbf{z}} \newcommand{\av}{\mathbf{a}} \newcommand{\Wv}{\mathbf{W}} \newcommand{\wv}{\mathbf{w}} \newcommand{\tv}{\mathbf{t}} \newcommand{\Tv}{\mathbf{T}} \newcommand{\Norm}{\mathcal{N}} \newcommand{\muv}{\boldsymbol{\mu}} \newcommand{\sigmav}{\boldsymbol{\sigma}} \newcommand{\phiv}{\boldsymbol{\phi}} \newcommand{\Phiv}{\boldsymbol{\Phi}} \newcommand{\Sigmav}{\boldsymbol{\Sigma}} \newcommand{\Lambdav}{\boldsymbol{\Lambda}} \newcommand{\half}{\frac{1}{2}} \newcommand{\argmax}[1]{\underset{#1}{\operatorname{argmax}}\;} \newcommand{\argmin}[1]{\underset{#1}{\operatorname{argmin}}\;}$
# Linear Ridge Regression¶
Remember when we were discussing the complexity of models? The simplest model was a constant. A simple way to predict rainfall is to ignore all measurements and just predict the average rainfall. If a linear model of measurements may do no better. The degree to which it does do better can be expressed in the relative sum of squared errors (RSE) or
$$RSE = \frac{\sum_{i=1}^N (\tv_i - \xv_i^T \wv)^2}{\sum_{i=1}^N (\tv_i - \bar{\Tv})^2}$$
If RSE is 1, then your linear model is no better than using the mean. The closer RSE is to 0, the better your linear model is.
In [1]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
In [2]:
!curl -O http://mlr.cs.umass.edu/ml/machine-learning-databases/auto-mpg/auto-mpg.data
!curl -O http://mlr.cs.umass.edu/ml/machine-learning-databases/auto-mpg/auto-mpg.names
% Total % Received % Xferd Average Speed Time Time Time Current
100 30286 100 30286 0 0 142k 0 --:--:-- --:--:-- --:--:-- 142k
% Total % Received % Xferd Average Speed Time Time Time Current
100 1660 100 1660 0 0 15961 0 --:--:-- --:--:-- --:--:-- 15961
In [3]:
def makeMPGData(filename='auto-mpg.data'):
def missingIsNan(s):
return np.nan if s == b'?' else float(s)
data = np.loadtxt(filename, usecols=range(8), converters={3: missingIsNan})
goodRowsMask = np.isnan(data).sum(axis=1) == 0
print("After removing rows containing question marks, data has",data.shape[0],"rows and",data.shape[1],"columns.")
X = data[:,1:]
T = data[:,0:1]
Xnames = ['bias', 'cylinders','displacement','horsepower','weight','acceleration','year','origin']
Tname = 'mpg'
return X,T,Xnames,Tname
In [4]:
X,T,Xnames,Tname = makeMPGData()
Read 398 rows and 8 columns from auto-mpg.data
After removing rows containing question marks, data has 392 rows and 8 columns.
In [6]:
means = X.mean(0)
stds = X.std(0)
nRows = X.shape[0]
Xs1 = np.insert((X-means)/stds, 0, 1, axis=1) # insert column of ones in new 0th column
# Xs1 = np.hstack(( np.ones((nRows,1)), (X-means)/stds ))
w = np.linalg.lstsq(Xs1.T @ Xs1, Xs1.T @ T, rcond=None)[0]
w
Out[6]:
array([[23.44591837],
[-0.84051891],
[ 2.07930256],
[-0.65163644],
[-5.49205044],
[ 0.22201407],
[ 2.76211888],
[ 1.14731588]])
In [7]:
# predict = Xs1.dot(w)
predict = Xs1 @ w
RSE = np.sum((T-predict)**2) / np.sum((T - T.mean(0))**2)
RSE
Out[7]:
0.17852192351894017
So, our linear model seems to be quite a bit better than using just the mean mpg. If our linear model is worse than using the mean, this value would be greater than 1.
Well, maybe our model would do better if it was a little closer to just the mean, or, equivalently, just using the bias weight. This is the question that drives the derivation of "ridge regression". Let's add a term to our sum of squared error objective function that is the sum of all weight magnitudes except the bias weight. Then, we not only minimize the sum of squared errors, we also minimize the sum of the weight magnitudes:
$$\sum_{i=1}^N (\tv_i - \xv_i^T \wv)^2 + \lambda \sum_{i=2}^N w_i^2$$
Notice that $\lambda$ in there. With $\lambda=0$ we have our usual linear regression objective function. With $\lambda>0$, we are adding in a penalty for the weight magnitudes.
How does this change our solution for the best $\wv$? You work it out. You should get
$$\wv = (X^T X + \lambda I)^{-1} X^T T$$
except instead of using
$$\lambda I = \begin{bmatrix} \lambda & 0 & \dotsc & 0\\ 0 & \lambda & \dotsc & 0\\ \vdots \\ 0 & 0 & \dotsc & \lambda \end{bmatrix}$$
we want
$$\begin{bmatrix} 0 & 0 & \dotsc & 0\\ 0 & \lambda & \dotsc & 0\\ \vdots \\ 0 & 0 & \dotsc & \lambda \end{bmatrix}$$
so we don't penalize the bias weight, the weight multiplying the constant 1 input component.
Now, what value should $\lambda$ be? Must determine empirically, by calculating the sum of squared error on test data.
Actually, we should not find the best value of $\lambda$ by comparing error on the test data. This will give us a too optimistic prediction of error on novel data, because the test data was used to pick the best $\lambda$. We really must hold out another partition of data from the training set for this. This third partition is often called the model validation set. So, we partition our data into disjoint training, validation, and testing subsets, and
• For each value of $\lambda$
• Solve for $\wv$ using the training set
• Use $\wv$ to predict the output for the validation set and calculate the squared error.
• Use $\wv$ to predict the output for the testing set and calculate the squared error.
• Pick value of $\lambda$ that produced the lowest validation set error, and report the testing set error obtained using that value of $\lambda$.
Let's do it!
In [8]:
lamb = 0.1
D = Xs1.shape[1]
lambdaDiag = np.eye(D) * lamb
Out[8]:
array([[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
[0. , 0.1, 0. , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0.1, 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0.1, 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0.1, 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0. , 0.1, 0. , 0. ],
[0. , 0. , 0. , 0. , 0. , 0. , 0.1, 0. ],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0.1]])
In [9]:
def makeLambda(D,lamb=0):
lambdaDiag = np.eye(D) * lamb
makeLambda(3,0.2)
Out[9]:
array([[0. , 0. , 0. ],
[0. , 0.2, 0. ],
[0. , 0. , 0.2]])
In [11]:
w = np.linalg.lstsq(Xs1.T @ Xs1 + lambdaDiag, Xs1.T @ T, rcond=None)[0]
w
Out[11]:
array([[23.44591837],
[-0.83453746],
[ 2.05581595],
[-0.6544685 ],
[-5.474439 ],
[ 0.21836873],
[ 2.76033842],
[ 1.14610564]])
In [12]:
%precision 2
Out[12]:
'%.2f'
In [14]:
D = Xs1.shape[1]
wLambda10 = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,10.0), Xs1.T @ T, rcond=None)[0]
wLambda0 = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,0), Xs1.T @ T, rcond=None)[0]
np.hstack((wLambda10,wLambda0))
Out[14]:
array([[ 2.34e+01, 2.34e+01],
[-6.06e-01, -8.41e-01],
[ 7.05e-01, 2.08e+00],
[-8.74e-01, -6.52e-01],
[-4.30e+00, -5.49e+00],
[-7.00e-03, 2.22e-01],
[ 2.62e+00, 2.76e+00],
[ 1.08e+00, 1.15e+00]])
In [15]:
lambdas = [0,0.1,1,10,100]
for lamb in lambdas:
w = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,lamb), Xs1.T @ T, rcond=None)[0]
plt.plot(w)
In [16]:
lambdas = [0,0.1,1,10,100]
for lamb in lambdas:
w = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,lamb), Xs1.T @ T, rcond=None)[0]
plt.plot(w, '-o')
plt.xticks(range(8), Xnames, rotation=30, horizontalalignment='right')
plt.ylabel('$\mathbf{w}$', fontsize='large', rotation='horizontal', labelpad=20)
plt.legend(lambdas);
Since we are not modifying the bias weight, let's take it out of the plot.
In [17]:
lambdas = [0,0.1,1,10,100]
for lamb in lambdas:
w = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,lamb), Xs1.T @ T, rcond=None)[0]
plt.plot(w[1:], '-o')
plt.xticks(range(7), Xnames[1:], rotation=30, horizontalalignment='right')
plt.ylabel('$\mathbf{w}$', fontsize='large', rotation='horizontal', labelpad=20)
plt.legend(lambdas);
In [18]:
plt.figure(figsize=(10, 10))
lambdas = [0, 10, 100, 1000, 10000]
for lamb in lambdas:
w = np.linalg.lstsq(Xs1.T @ Xs1 + makeLambda(D,lamb), Xs1.T @ T, rcond=None)[0]
plt.plot(Xs1[:30] @ w)
plt.plot(T[:30],'ro',lw=5,alpha=0.8)
plt.ylabel('Model Output')
plt.xlabel('Sample Index')
plt.legend(lambdas,loc='best')
Out[18]:
<matplotlib.legend.Legend at 0x7feb5b689c18>
Which $\lambda$ value is best? Careful. What is the best value of $\lambda$ if just comparing error on training data?
Now, careful again! Can't report expected error from testing data that is also used to pick best value of $\lambda$. Error is likely to be better than what you will get on new data that was not used to train the model and also was not used to pick value of $\lambda$.
Need a way to partition our data into training, validation and testing subsets. Let's write a function that makes these partitions randomly, given the data matrix and the fractions to be used in the three partitions.
In [20]:
def partition(X, T, trainFraction=0.8, validateFraction=0.1, testFraction=0.1):
'''Usage: Xtrain,Ttrain,Xval,Tval,Xtest,Ttext = partition(X,T,0.8,0.2,0.2)'''
if trainFraction + validateFraction + testFraction != 1:
raise ValueError("Train, validate and test fractions must sum to 1. Given values sum to " + str(trainFraction+validateFraction+testFraction))
n = X.shape[0]
nTrain = round(trainFraction * n)
nValidate = round(validateFraction * n)
nTest = round(testFraction * n)
if nTrain + nValidate + nTest != n:
nTest = n - nTrain - nValidate
# Random order of data matrix row indices
rowIndices = np.arange(X.shape[0])
np.random.shuffle(rowIndices)
# Build X and T matrices by selecting corresponding rows for each partition
Xtrain = X[rowIndices[:nTrain], :]
Ttrain = T[rowIndices[:nTrain], :]
Xvalidate = X[rowIndices[nTrain:nTrain + nValidate], :]
Tvalidate = T[rowIndices[nTrain:nTrain + nValidate], :]
Xtest = X[rowIndices[nTrain+nValidate:nTrain + nValidate + nTest], :]
Ttest = T[rowIndices[nTrain+nValidate:nTrain + nValidate + nTest], :]
return Xtrain, Ttrain, Xvalidate, Tvalidate, Xtest, Ttest
In [21]:
X = np.arange(20).reshape((10, 2))
X
Out[21]:
array([[ 0, 1],
[ 2, 3],
[ 4, 5],
[ 6, 7],
[ 8, 9],
[10, 11],
[12, 13],
[14, 15],
[16, 17],
[18, 19]])
In [22]:
T = np.arange(10).reshape((-1, 1))
T
Out[22]:
array([[0],
[1],
[2],
[3],
[4],
[5],
[6],
[7],
[8],
[9]])
In [23]:
X = np.arange(20).reshape((10, 2)) + np.random.uniform(0, 1, (10, 2))
T = np.arange(10).reshape((-1, 1))
Xtrain, Ttrain, Xval, Tval, Xtest, Ttest = partition(X, T, 0.6, 0.2, 0.2)
In [24]:
print("Xtrain:")
print(Xtrain)
print(" Ttrain:")
print(Ttrain)
print("\n Xval:")
print(Xval)
print(" Tval:")
print(Tval)
print("\n Xtest:")
print(Xtest)
print(" Ttest:")
print(Ttest)
Xtrain:
[[18.01 19.54]
[ 6.07 7.55]
[ 2.07 3.24]
[ 4.16 5.64]
[ 8.18 9. ]
[14.25 15.88]]
Ttrain:
[[9]
[3]
[1]
[2]
[4]
[7]]
Xval:
[[12.85 14. ]
[ 0.98 1.19]]
Tval:
[[6]
[0]]
Xtest:
[[16.63 17.31]
[10.39 11.14]]
Ttest:
[[8]
[5]]
Now we can train our model using several values of $\lambda$ on Xtrain,Train and calculate the model error on Xval,Tval. Then pick best value of $\lambda$ based on error on Xval,Tval. Finally, calculate error of model using best $\lambda$ on Xtest,Ttest as our estimate of error on new data.
However, these errors will be affected by the random partitioning of the data. Repeating with new partitions may result in a different $\lambda$ being best. We should repeat the above procedure many times and average over the results. How many repetitions do we need?
Another approach, commonly followed in machine learning, is to first partition the data into $k$ subsets, called "folds". Pick one fold to be the test partition, another fold to be the validate partition, and collect the remaining folds to be the train partition. We can do this $k\,(k-1)$ ways. (Why?) So, with $k=5$ we get 20 repetitions performed in a way that most distributes samples among the partitions.
First, a little note on the yield statement in python. The yield statement is like return except that execution pauses at this point in the function, after returning the values. When the function is called again, it continues from that paused point.
In [25]:
def count(n):
for a in range(n):
yield a
In [26]:
count(4)
Out[26]:
<generator object count at 0x7feb5b361938>
In [27]:
list(count(4))
Out[27]:
[0, 1, 2, 3]
In [28]:
for i in count(5):
print(i)
0
1
2
3
4
In [29]:
def partitionKFolds(X, T, nFolds, shuffle=False, nPartitions=3):
'''Usage: for Xtrain,Ttrain,Xval,Tval,Xtest,Ttext in partitionKFolds(X,T,5,shuffle=True,nPartitions=3):
for Xtrain,Ttrain,Xtest,Ttext in partitionKFolds(X,T,5,shuffle=True,nPartitions=2):'''
# Randomly arrange row indices
rowIndices = np.arange(X.shape[0])
if shuffle:
np.random.shuffle(rowIndices)
# Calculate number of samples in each of the nFolds folds
nSamples = X.shape[0]
nEach = int(nSamples / nFolds)
if nEach == 0:
raise ValueError("partitionKFolds: Number of samples in each fold is 0.")
# Calculate the starting and stopping row index for each fold.
# Store in startsStops as list of (start,stop) pairs
starts = np.arange(0, nEach * nFolds, nEach)
stops = starts + nEach
stops[-1] = nSamples
startsStops = list(zip(starts, stops))
# Repeat with testFold taking each single fold, one at a time
for testFold in range(nFolds):
if nPartitions == 3:
# Repeat with validateFold taking each single fold, except for the testFold
for validateFold in range(nFolds):
if testFold == validateFold:
continue
# trainFolds are all remaining folds, after selecting test and validate folds
trainFolds = np.setdiff1d(range(nFolds), [testFold,validateFold])
# Construct Xtrain and Ttrain by collecting rows for all trainFolds
rows = []
for tf in trainFolds:
a,b = startsStops[tf]
rows += rowIndices[a:b].tolist()
Xtrain = X[rows, :]
Ttrain = T[rows, :]
# Construct Xvalidate and Tvalidate
a,b = startsStops[validateFold]
rows = rowIndices[a:b]
Xvalidate = X[rows,:]
Tvalidate = T[rows,:]
# Construct Xtest and Ttest
a,b = startsStops[testFold]
rows = rowIndices[a:b]
Xtest = X[rows,:]
Ttest = T[rows,:]
# Return partition matrices, then suspend until called again.
yield Xtrain, Ttrain, Xvalidate, Tvalidate, Xtest, Ttest, testFold
else:
# trainFolds are all remaining folds, after selecting test and validate folds
trainFolds = np.setdiff1d(range(nFolds), [testFold])
# Construct Xtrain and Ttrain by collecting rows for all trainFolds
rows = []
for tf in trainFolds:
a,b = startsStops[tf]
rows += rowIndices[a:b].tolist()
Xtrain = X[rows, :]
Ttrain = T[rows, :]
# Construct Xtest and Ttest
a,b = startsStops[testFold]
rows = rowIndices[a:b]
Xtest = X[rows, :]
Ttest = T[rows, :]
# Return partition matrices, then suspend until called again.
yield Xtrain, Ttrain, Xtest, Ttest, testFold
In [30]:
X = np.arange(20).reshape((10, 2))
T = np.arange(10).reshape((-1, 1)) + np.random.uniform(-1, 1, (10, 1))
k = 0
for Xtrain, Ttrain, Xval, Tval, Xtest, Ttest, testFold in partitionKFolds(X, T, 5):
k += 1
print("Fold",k)
print(" Xtrain:")
print(Xtrain)
print(" Ttrain:")
print(Ttrain)
print("\n Xval:")
print(Xval)
print(" Tval:")
print(Tval)
print("\n Xtest:")
print(Xtest)
print(" Ttest:")
print(Ttest)
Fold 1
Xtrain:
[[ 8 9]
[10 11]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[4.09]
[5.88]
[6.55]
[7.46]
[8.1 ]
[8.95]]
Xval:
[[4 5]
[6 7]]
Tval:
[[1.44]
[3.19]]
Xtest:
[[0 1]
[2 3]]
Ttest:
[[-0.05]
[ 1.53]]
Fold 2
Xtrain:
[[ 4 5]
[ 6 7]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[1.44]
[3.19]
[6.55]
[7.46]
[8.1 ]
[8.95]]
Xval:
[[ 8 9]
[10 11]]
Tval:
[[4.09]
[5.88]]
Xtest:
[[0 1]
[2 3]]
Ttest:
[[-0.05]
[ 1.53]]
Fold 3
Xtrain:
[[ 4 5]
[ 6 7]
[ 8 9]
[10 11]
[16 17]
[18 19]]
Ttrain:
[[1.44]
[3.19]
[4.09]
[5.88]
[8.1 ]
[8.95]]
Xval:
[[12 13]
[14 15]]
Tval:
[[6.55]
[7.46]]
Xtest:
[[0 1]
[2 3]]
Ttest:
[[-0.05]
[ 1.53]]
Fold 4
Xtrain:
[[ 4 5]
[ 6 7]
[ 8 9]
[10 11]
[12 13]
[14 15]]
Ttrain:
[[1.44]
[3.19]
[4.09]
[5.88]
[6.55]
[7.46]]
Xval:
[[16 17]
[18 19]]
Tval:
[[8.1 ]
[8.95]]
Xtest:
[[0 1]
[2 3]]
Ttest:
[[-0.05]
[ 1.53]]
Fold 5
Xtrain:
[[ 8 9]
[10 11]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[4.09]
[5.88]
[6.55]
[7.46]
[8.1 ]
[8.95]]
Xval:
[[0 1]
[2 3]]
Tval:
[[-0.05]
[ 1.53]]
Xtest:
[[4 5]
[6 7]]
Ttest:
[[1.44]
[3.19]]
Fold 6
Xtrain:
[[ 0 1]
[ 2 3]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 6.55]
[ 7.46]
[ 8.1 ]
[ 8.95]]
Xval:
[[ 8 9]
[10 11]]
Tval:
[[4.09]
[5.88]]
Xtest:
[[4 5]
[6 7]]
Ttest:
[[1.44]
[3.19]]
Fold 7
Xtrain:
[[ 0 1]
[ 2 3]
[ 8 9]
[10 11]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 4.09]
[ 5.88]
[ 8.1 ]
[ 8.95]]
Xval:
[[12 13]
[14 15]]
Tval:
[[6.55]
[7.46]]
Xtest:
[[4 5]
[6 7]]
Ttest:
[[1.44]
[3.19]]
Fold 8
Xtrain:
[[ 0 1]
[ 2 3]
[ 8 9]
[10 11]
[12 13]
[14 15]]
Ttrain:
[[-0.05]
[ 1.53]
[ 4.09]
[ 5.88]
[ 6.55]
[ 7.46]]
Xval:
[[16 17]
[18 19]]
Tval:
[[8.1 ]
[8.95]]
Xtest:
[[4 5]
[6 7]]
Ttest:
[[1.44]
[3.19]]
Fold 9
Xtrain:
[[ 4 5]
[ 6 7]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[1.44]
[3.19]
[6.55]
[7.46]
[8.1 ]
[8.95]]
Xval:
[[0 1]
[2 3]]
Tval:
[[-0.05]
[ 1.53]]
Xtest:
[[ 8 9]
[10 11]]
Ttest:
[[4.09]
[5.88]]
Fold 10
Xtrain:
[[ 0 1]
[ 2 3]
[12 13]
[14 15]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 6.55]
[ 7.46]
[ 8.1 ]
[ 8.95]]
Xval:
[[4 5]
[6 7]]
Tval:
[[1.44]
[3.19]]
Xtest:
[[ 8 9]
[10 11]]
Ttest:
[[4.09]
[5.88]]
Fold 11
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 8.1 ]
[ 8.95]]
Xval:
[[12 13]
[14 15]]
Tval:
[[6.55]
[7.46]]
Xtest:
[[ 8 9]
[10 11]]
Ttest:
[[4.09]
[5.88]]
Fold 12
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[12 13]
[14 15]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 6.55]
[ 7.46]]
Xval:
[[16 17]
[18 19]]
Tval:
[[8.1 ]
[8.95]]
Xtest:
[[ 8 9]
[10 11]]
Ttest:
[[4.09]
[5.88]]
Fold 13
Xtrain:
[[ 4 5]
[ 6 7]
[ 8 9]
[10 11]
[16 17]
[18 19]]
Ttrain:
[[1.44]
[3.19]
[4.09]
[5.88]
[8.1 ]
[8.95]]
Xval:
[[0 1]
[2 3]]
Tval:
[[-0.05]
[ 1.53]]
Xtest:
[[12 13]
[14 15]]
Ttest:
[[6.55]
[7.46]]
Fold 14
Xtrain:
[[ 0 1]
[ 2 3]
[ 8 9]
[10 11]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 4.09]
[ 5.88]
[ 8.1 ]
[ 8.95]]
Xval:
[[4 5]
[6 7]]
Tval:
[[1.44]
[3.19]]
Xtest:
[[12 13]
[14 15]]
Ttest:
[[6.55]
[7.46]]
Fold 15
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[16 17]
[18 19]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 8.1 ]
[ 8.95]]
Xval:
[[ 8 9]
[10 11]]
Tval:
[[4.09]
[5.88]]
Xtest:
[[12 13]
[14 15]]
Ttest:
[[6.55]
[7.46]]
Fold 16
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[ 8 9]
[10 11]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 4.09]
[ 5.88]]
Xval:
[[16 17]
[18 19]]
Tval:
[[8.1 ]
[8.95]]
Xtest:
[[12 13]
[14 15]]
Ttest:
[[6.55]
[7.46]]
Fold 17
Xtrain:
[[ 4 5]
[ 6 7]
[ 8 9]
[10 11]
[12 13]
[14 15]]
Ttrain:
[[1.44]
[3.19]
[4.09]
[5.88]
[6.55]
[7.46]]
Xval:
[[0 1]
[2 3]]
Tval:
[[-0.05]
[ 1.53]]
Xtest:
[[16 17]
[18 19]]
Ttest:
[[8.1 ]
[8.95]]
Fold 18
Xtrain:
[[ 0 1]
[ 2 3]
[ 8 9]
[10 11]
[12 13]
[14 15]]
Ttrain:
[[-0.05]
[ 1.53]
[ 4.09]
[ 5.88]
[ 6.55]
[ 7.46]]
Xval:
[[4 5]
[6 7]]
Tval:
[[1.44]
[3.19]]
Xtest:
[[16 17]
[18 19]]
Ttest:
[[8.1 ]
[8.95]]
Fold 19
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[12 13]
[14 15]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 6.55]
[ 7.46]]
Xval:
[[ 8 9]
[10 11]]
Tval:
[[4.09]
[5.88]]
Xtest:
[[16 17]
[18 19]]
Ttest:
[[8.1 ]
[8.95]]
Fold 20
Xtrain:
[[ 0 1]
[ 2 3]
[ 4 5]
[ 6 7]
[ 8 9]
[10 11]]
Ttrain:
[[-0.05]
[ 1.53]
[ 1.44]
[ 3.19]
[ 4.09]
[ 5.88]]
Xval:
[[12 13]
[14 15]]
Tval:
[[6.55]
[7.46]]
Xtest:
[[16 17]
[18 19]]
Ttest:
[[8.1 ]
[8.95]]
Typical use of these partitions is shown below. It is most handy to just collect all results in a matrix and calculate averages afterwards, rather than accumulating each result and dividing by the number of repetitions at the end.
In [31]:
def train(X, T, lamb):
means = X.mean(0)
stds = X.std(0)
n,d = X.shape
Xs1 = np.insert( (X - means)/stds, 0, 1, axis=1)
lambDiag = np.eye(d+1) * lamb
lambDiag[0,0] = 0
w = np.linalg.lstsq( Xs1.T @ Xs1 + lambDiag, Xs1.T @ T, rcond=None)[0]
return {'w': w, 'means': means, 'stds': stds}
def use(model, X):
Xs1 = np.insert((X-model['means'])/model['stds'], 0, 1, axis=1)
return Xs1 @ model['w']
def rmse(A,B):
return np.sqrt(np.mean( (A-B)**2 ))
def evaluate(model, X, T):
return rmse(use(model, X), T)
In [32]:
lambdas = np.linspace(0, 5, 20)
results = []
for Xtrain, Ttrain, Xval, Tval, Xtest, Ttest,_ in partitionKFolds(X, T, 5):
for lamb in lambdas:
model = train(Xtrain, Ttrain, lamb)
predict = use(model, Xval)
results.append([lamb,
evaluate(model, Xtrain, Ttrain),
evaluate(model, Xval, Tval),
evaluate(model, Xtest, Ttest)])
results = np.array(results)
# print(results)
avgresults = []
for lam in lambdas:
print(lam)
print(results[results[:,0]==lam,1:])
avgresults.append( [lam] + np.mean(results[results[:,0]==lam,1:],axis=0).tolist())
avgresults = np.array(avgresults)
print(avgresults)
0.0
[[0.29 0.97 0.72]
[0.33 0.58 0.51]
[0.42 0.36 0.39]
[0.3 1.06 0.79]
[0.29 0.72 0.97]
[0.26 0.46 0.6 ]
[0.34 0.29 0.6 ]
[0.27 0.68 0.62]
[0.33 0.51 0.58]
[0.26 0.6 0.46]
[0.33 0.49 0.61]
[0.34 0.58 0.41]
[0.42 0.39 0.36]
[0.34 0.6 0.29]
[0.33 0.61 0.49]
[0.4 0.87 0.18]
[0.3 0.79 1.06]
[0.27 0.62 0.68]
[0.34 0.41 0.58]
[0.4 0.18 0.87]]
0.2631578947368421
[[0.29 1.05 0.83]
[0.34 0.56 0.42]
[0.42 0.39 0.33]
[0.3 0.95 0.69]
[0.29 0.83 1.05]
[0.27 0.46 0.65]
[0.35 0.33 0.64]
[0.27 0.57 0.65]
[0.34 0.42 0.56]
[0.27 0.65 0.46]
[0.33 0.55 0.63]
[0.35 0.46 0.44]
[0.42 0.33 0.39]
[0.35 0.64 0.33]
[0.33 0.63 0.55]
[0.41 0.72 0.1 ]
[0.3 0.69 0.95]
[0.27 0.65 0.57]
[0.35 0.44 0.46]
[0.41 0.1 0.72]]
0.5263157894736842
[[0.29 1.12 0.93]
[0.35 0.55 0.35]
[0.43 0.42 0.3 ]
[0.31 0.85 0.6 ]
[0.29 0.93 1.12]
[0.3 0.47 0.7 ]
[0.37 0.37 0.68]
[0.29 0.47 0.68]
[0.35 0.35 0.55]
[0.3 0.7 0.47]
[0.36 0.6 0.65]
[0.36 0.34 0.46]
[0.43 0.3 0.42]
[0.37 0.68 0.37]
[0.36 0.65 0.6 ]
[0.41 0.59 0.09]
[0.31 0.6 0.85]
[0.29 0.68 0.47]
[0.36 0.46 0.34]
[0.41 0.09 0.59]]
0.7894736842105263
[[0.3 1.19 1.04]
[0.37 0.54 0.3 ]
[0.45 0.44 0.31]
[0.32 0.76 0.52]
[0.3 1.04 1.19]
[0.34 0.47 0.75]
[0.39 0.41 0.72]
[0.32 0.37 0.71]
[0.37 0.3 0.54]
[0.34 0.75 0.47]
[0.39 0.65 0.66]
[0.38 0.23 0.49]
[0.45 0.31 0.44]
[0.39 0.72 0.41]
[0.39 0.66 0.65]
[0.42 0.45 0.15]
[0.32 0.52 0.76]
[0.32 0.71 0.37]
[0.38 0.49 0.23]
[0.42 0.15 0.45]]
1.0526315789473684
[[0.31 1.26 1.14]
[0.4 0.53 0.3 ]
[0.47 0.47 0.33]
[0.34 0.67 0.45]
[0.31 1.14 1.26]
[0.38 0.47 0.79]
[0.43 0.45 0.76]
[0.34 0.28 0.74]
[0.4 0.3 0.53]
[0.38 0.79 0.47]
[0.43 0.7 0.68]
[0.41 0.12 0.51]
[0.47 0.33 0.47]
[0.43 0.76 0.45]
[0.43 0.68 0.7 ]
[0.43 0.33 0.23]
[0.34 0.45 0.67]
[0.34 0.74 0.28]
[0.41 0.51 0.12]
[0.43 0.23 0.33]]
1.3157894736842104
[[0.32 1.32 1.24]
[0.43 0.52 0.33]
[0.49 0.49 0.38]
[0.36 0.58 0.38]
[0.32 1.24 1.32]
[0.43 0.48 0.84]
[0.47 0.48 0.79]
[0.38 0.19 0.76]
[0.43 0.33 0.52]
[0.43 0.84 0.48]
[0.47 0.75 0.69]
[0.44 0.06 0.53]
[0.49 0.38 0.49]
[0.47 0.79 0.48]
[0.47 0.69 0.75]
[0.45 0.21 0.3 ]
[0.36 0.38 0.58]
[0.38 0.76 0.19]
[0.44 0.53 0.06]
[0.45 0.3 0.21]]
1.5789473684210527
[[0.34 1.39 1.33]
[0.46 0.52 0.39]
[0.52 0.52 0.44]
[0.38 0.49 0.33]
[0.34 1.33 1.39]
[0.47 0.48 0.88]
[0.51 0.52 0.82]
[0.41 0.1 0.79]
[0.46 0.39 0.52]
[0.47 0.88 0.48]
[0.51 0.8 0.71]
[0.47 0.11 0.56]
[0.52 0.44 0.52]
[0.51 0.82 0.52]
[0.51 0.71 0.8 ]
[0.46 0.1 0.38]
[0.38 0.33 0.49]
[0.41 0.79 0.1 ]
[0.47 0.56 0.11]
[0.46 0.38 0.1 ]]
1.8421052631578947
[[0.35 1.45 1.42]
[0.49 0.51 0.46]
[0.54 0.54 0.5 ]
[0.4 0.41 0.3 ]
[0.35 1.42 1.45]
[0.52 0.49 0.93]
[0.55 0.55 0.86]
[0.45 0.03 0.81]
[0.49 0.46 0.51]
[0.52 0.93 0.49]
[0.56 0.84 0.72]
[0.5 0.2 0.58]
[0.54 0.5 0.54]
[0.55 0.86 0.55]
[0.56 0.72 0.84]
[0.48 0.07 0.45]
[0.4 0.3 0.41]
[0.45 0.81 0.03]
[0.5 0.58 0.2 ]
[0.48 0.45 0.07]]
2.1052631578947367
[[0.37 1.5 1.51]
[0.52 0.51 0.53]
[0.57 0.56 0.57]
[0.43 0.34 0.28]
[0.37 1.51 1.5 ]
[0.57 0.49 0.97]
[0.59 0.58 0.89]
[0.48 0.07 0.84]
[0.52 0.53 0.51]
[0.57 0.97 0.49]
[0.61 0.89 0.73]
[0.53 0.29 0.6 ]
[0.57 0.57 0.56]
[0.59 0.89 0.58]
[0.61 0.73 0.89]
[0.49 0.15 0.53]
[0.43 0.28 0.34]
[0.48 0.84 0.07]
[0.53 0.6 0.29]
[0.49 0.53 0.15]]
2.3684210526315788
[[0.38 1.56 1.59]
[0.55 0.5 0.6 ]
[0.6 0.58 0.63]
[0.45 0.26 0.29]
[0.38 1.59 1.56]
[0.62 0.5 1.01]
[0.63 0.62 0.92]
[0.52 0.14 0.86]
[0.55 0.6 0.5 ]
[0.62 1.01 0.5 ]
[0.65 0.93 0.75]
[0.57 0.38 0.62]
[0.6 0.63 0.58]
[0.63 0.92 0.62]
[0.65 0.75 0.93]
[0.51 0.25 0.59]
[0.45 0.29 0.26]
[0.52 0.86 0.14]
[0.57 0.62 0.38]
[0.51 0.59 0.25]]
2.631578947368421
[[0.4 1.62 1.67]
[0.59 0.5 0.68]
[0.63 0.6 0.7 ]
[0.47 0.19 0.31]
[0.4 1.67 1.62]
[0.67 0.5 1.05]
[0.68 0.65 0.95]
[0.55 0.22 0.88]
[0.59 0.68 0.5 ]
[0.67 1.05 0.5 ]
[0.7 0.97 0.76]
[0.6 0.46 0.64]
[0.63 0.7 0.6 ]
[0.68 0.95 0.65]
[0.7 0.76 0.97]
[0.53 0.35 0.66]
[0.47 0.31 0.19]
[0.55 0.88 0.22]
[0.6 0.64 0.46]
[0.53 0.66 0.35]]
2.894736842105263
[[0.42 1.67 1.75]
[0.62 0.5 0.75]
[0.66 0.62 0.76]
[0.5 0.12 0.35]
[0.42 1.75 1.67]
[0.71 0.51 1.08]
[0.72 0.67 0.97]
[0.59 0.29 0.91]
[0.62 0.75 0.5 ]
[0.71 1.08 0.51]
[0.74 1.01 0.77]
[0.63 0.55 0.66]
[0.66 0.76 0.62]
[0.72 0.97 0.67]
[0.74 0.77 1.01]
[0.55 0.45 0.73]
[0.5 0.35 0.12]
[0.59 0.91 0.29]
[0.63 0.66 0.55]
[0.55 0.73 0.45]]
3.1578947368421053
[[0.43 1.72 1.83]
[0.65 0.5 0.83]
[0.69 0.64 0.82]
[0.52 0.06 0.39]
[0.43 1.83 1.72]
[0.76 0.51 1.12]
[0.76 0.7 1. ]
[0.62 0.36 0.93]
[0.65 0.83 0.5 ]
[0.76 1.12 0.51]
[0.78 1.05 0.78]
[0.67 0.63 0.68]
[0.69 0.82 0.64]
[0.76 1. 0.7 ]
[0.78 0.78 1.05]
[0.57 0.54 0.79]
[0.52 0.39 0.06]
[0.62 0.93 0.36]
[0.67 0.68 0.63]
[0.57 0.79 0.54]]
3.4210526315789473
[[0.45 1.77 1.9 ]
[0.68 0.5 0.9 ]
[0.72 0.66 0.88]
[0.54 0.04 0.44]
[0.45 1.9 1.77]
[0.8 0.52 1.16]
[0.8 0.73 1.03]
[0.66 0.43 0.95]
[0.68 0.9 0.5 ]
[0.8 1.16 0.52]
[0.83 1.08 0.79]
[0.7 0.7 0.7 ]
[0.72 0.88 0.66]
[0.8 1.03 0.73]
[0.83 0.79 1.08]
[0.59 0.63 0.85]
[0.54 0.44 0.04]
[0.66 0.95 0.43]
[0.7 0.7 0.7 ]
[0.59 0.85 0.63]]
3.6842105263157894
[[0.46 1.81 1.97]
[0.71 0.51 0.97]
[0.74 0.68 0.94]
[0.56 0.09 0.49]
[0.46 1.97 1.81]
[0.84 0.52 1.19]
[0.83 0.76 1.05]
[0.69 0.49 0.97]
[0.71 0.97 0.51]
[0.84 1.19 0.52]
[0.87 1.12 0.81]
[0.73 0.78 0.72]
[0.74 0.94 0.68]
[0.83 1.05 0.76]
[0.87 0.81 1.12]
[0.6 0.72 0.91]
[0.56 0.49 0.09]
[0.69 0.97 0.49]
[0.73 0.72 0.78]
[0.6 0.91 0.72]]
3.9473684210526314
[[0.48 1.86 2.04]
[0.74 0.51 1.04]
[0.77 0.7 1. ]
[0.59 0.15 0.54]
[0.48 2.04 1.86]
[0.88 0.53 1.22]
[0.87 0.78 1.08]
[0.72 0.55 0.98]
[0.74 1.04 0.51]
[0.88 1.22 0.53]
[0.91 1.15 0.82]
[0.76 0.85 0.74]
[0.77 1. 0.7 ]
[0.87 1.08 0.78]
[0.91 0.82 1.15]
[0.62 0.8 0.96]
[0.59 0.54 0.15]
[0.72 0.98 0.55]
[0.76 0.74 0.85]
[0.62 0.96 0.8 ]]
4.2105263157894735
[[0.49 1.9 2.1 ]
[0.77 0.51 1.1 ]
[0.8 0.72 1.06]
[0.61 0.21 0.59]
[0.49 2.1 1.9 ]
[0.92 0.53 1.26]
[0.91 0.81 1.1 ]
[0.75 0.61 1. ]
[0.77 1.1 0.51]
[0.92 1.26 0.53]
[0.95 1.18 0.83]
[0.79 0.92 0.76]
[0.8 1.06 0.72]
[0.91 1.1 0.81]
[0.95 0.83 1.18]
[0.64 0.88 1.02]
[0.61 0.59 0.21]
[0.75 1. 0.61]
[0.79 0.76 0.92]
[0.64 1.02 0.88]]
4.473684210526316
[[0.51 1.95 2.17]
[0.8 0.52 1.17]
[0.83 0.73 1.11]
[0.63 0.26 0.64]
[0.51 2.17 1.95]
[0.96 0.54 1.29]
[0.95 0.83 1.13]
[0.78 0.67 1.02]
[0.8 1.17 0.52]
[0.96 1.29 0.54]
[0.98 1.21 0.84]
[0.82 0.99 0.77]
[0.83 1.11 0.73]
[0.95 1.13 0.83]
[0.98 0.84 1.21]
[0.66 0.96 1.07]
[0.63 0.64 0.26]
[0.78 1.02 0.67]
[0.82 0.77 0.99]
[0.66 1.07 0.96]]
4.7368421052631575
[[0.52 1.99 2.23]
[0.83 0.52 1.23]
[0.85 0.75 1.17]
[0.65 0.32 0.69]
[0.52 2.23 1.99]
[1. 0.54 1.32]
[0.98 0.85 1.15]
[0.81 0.73 1.04]
[0.83 1.23 0.52]
[1. 1.32 0.54]
[1.02 1.25 0.85]
[0.85 1.06 0.79]
[0.85 1.17 0.75]
[0.98 1.15 0.85]
[1.02 0.85 1.25]
[0.67 1.04 1.12]
[0.65 0.69 0.32]
[0.81 1.04 0.73]
[0.85 0.79 1.06]
[0.67 1.12 1.04]]
5.0
[[0.54 2.03 2.29]
[0.86 0.53 1.29]
[0.88 0.77 1.22]
[0.67 0.37 0.74]
[0.54 2.29 2.03]
[1.04 0.55 1.35]
[1.01 0.87 1.17]
[0.84 0.79 1.05]
[0.86 1.29 0.53]
[1.04 1.35 0.55]
[1.06 1.28 0.85]
[0.88 1.12 0.81]
[0.88 1.22 0.77]
[1.01 1.17 0.87]
[1.06 0.85 1.28]
[0.69 1.11 1.17]
[0.67 0.74 0.37]
[0.84 1.05 0.79]
[0.88 0.81 1.12]
[0.69 1.17 1.11]]
[[0. 0.33 0.59 0.59]
[0.26 0.33 0.57 0.57]
[0.53 0.35 0.56 0.56]
[0.79 0.37 0.56 0.56]
[1.05 0.39 0.56 0.56]
[1.32 0.42 0.57 0.57]
[1.58 0.45 0.58 0.58]
[1.84 0.48 0.61 0.61]
[2.11 0.52 0.64 0.64]
[2.37 0.55 0.68 0.68]
[2.63 0.58 0.72 0.72]
[2.89 0.61 0.76 0.76]
[3.16 0.64 0.79 0.79]
[3.42 0.68 0.83 0.83]
[3.68 0.71 0.87 0.87]
[3.95 0.73 0.92 0.92]
[4.21 0.76 0.96 0.96]
[4.47 0.79 0.99 0.99]
[4.74 0.82 1.03 1.03]
[5. 0.85 1.07 1.07]]
In [33]:
plt.plot(avgresults[:,0],avgresults[:,[1, 3]],'o-')
plt.xlabel('$\lambda$')
plt.ylabel('RMSE')
plt.legend(('Train', 'Test'),loc='best');
Here is yet another python implementation that includes many of the common steps used when training and evaluating models. We will use this version for many of our experiments.
In [34]:
def trainValidateTestKFolds(trainf, evaluatef, X, T, parameterSets, nFolds=5,
shuffle=False, verbose=False):
# Randomly arrange row indices
rowIndices = np.arange(X.shape[0])
if shuffle:
np.random.shuffle(rowIndices)
isNewBetterThanOld = lambda new,old: new < old
# Calculate number of samples in each of the nFolds folds
nSamples = X.shape[0]
nEach = int(nSamples / nFolds)
if nEach == 0:
raise ValueError("trainValidateTestKFolds: Number of samples in each fold is 0.")
# Calculate the starting and stopping row index for each fold.
# Store in startsStops as list of (start,stop) pairs
starts = np.arange(0,nEach*nFolds, nEach)
stops = starts + nEach
stops[-1] = nSamples
startsStops = list(zip(starts, stops))
# Repeat with testFold taking each single fold, one at a time
results = []
for testFold in range(nFolds):
# Leaving the testFold out, for each validate fold, train on remaining
# folds and evaluate on validate fold. Collect theseRepeat with validate
# Construct Xtest and Ttest
a,b = startsStops[testFold]
rows = rowIndices[a:b]
Xtest = X[rows, :]
Ttest = T[rows, :]
bestParms = None
for parms in parameterSets:
# trainEvaluationSum = 0
validateEvaluationSum = 0
for validateFold in range(nFolds):
if testFold == validateFold:
continue
# Construct Xtrain and Ttrain
trainFolds = np.setdiff1d(range(nFolds), [testFold, validateFold])
rows = []
for tf in trainFolds:
a,b = startsStops[tf]
rows += rowIndices[a:b].tolist()
Xtrain = X[rows, :]
Ttrain = T[rows, :]
# Construct Xvalidate and Tvalidate
a,b = startsStops[validateFold]
rows = rowIndices[a:b]
Xvalidate = X[rows, :]
Tvalidate = T[rows, :]
model = trainf(Xtrain, Ttrain, parms)
# trainEvaluationSum += evaluatef(model,Xtrain,Train)
validateEvaluationSum += evaluatef(model, Xvalidate, Tvalidate)
validateEvaluation = validateEvaluationSum / (nFolds-1)
if verbose:
if hasattr(model, 'bestIteration') and model.bestIteration is not None:
print('{} Val {:.3f} Best Iter {:d}'.format(parms, validateEvaluation, model.bestIteration))
else:
print('{} Val {:.3f}'.format(parms, validateEvaluation))
if bestParms is None or isNewBetterThanOld(validateEvaluation, bestValidationEvaluation):
bestParms = parms
bestValidationEvaluation = validateEvaluation
if verbose:
print('New best')
# trainEvaluation = trainEvaluationSum / (nFolds-1)
newXtrain = np.vstack((Xtrain, Xvalidate))
newTtrain = np.vstack((Ttrain, Tvalidate))
model = trainf(newXtrain, newTtrain, bestParms)
trainEvaluation = evaluatef(model, newXtrain, newTtrain)
testEvaluation = evaluatef(model, Xtest, Ttest)
# resultThisTestFold = [bestParms, trainEvaluation,
# bestValidationEvaluation, testEvaluation]
resultThisTestFold = [nFolds, testFold+1, bestParms, trainEvaluation, bestValidationEvaluation, testEvaluation]
results.append(resultThisTestFold)
if verbose:
print(resultThisTestFold)
print('{}/{}'.format(testFold+1, nFolds), end=' ', flush=True)
return np.array(results) # pandas.DataFrame(results, columns=('nFolds','testFold','parms','trainAcc','testAcc'))
In [35]:
lambdas = np.linspace(0, 5, 20)
results = trainValidateTestKFolds(train, evaluate, X, T, lambdas, nFolds=5,
shuffle=True, verbose=False)
1/5 2/5 3/5 4/5 5/5
In [36]:
results
Out[36]:
array([[5. , 1. , 0. , 0.39, 0.49, 0.39],
[5. , 2. , 0.53, 0.41, 0.46, 0.43],
[5. , 3. , 0. , 0.31, 0.37, 0.64],
[5. , 4. , 0.53, 0.4 , 0.52, 0.28],
[5. , 5. , 0.26, 0.35, 0.46, 0.53]])
In [ ]:
plt.figure(figsize=(10, 10))
plt.subplot(2, 1, 1)
plt.plot(results[:, 1], results[:, 3:6], 'o-')
plt.xticks(range(1, 6)) # without this, we see 1.5, 2.5, etc.
plt.xlabel('Fold Index')
plt.ylabel('RMSE')
plt.legend(('Train','Validate','Test'),loc='best')
plt.subplot(2, 1, 2)
plt.plot(results[:, 1], results[:, 2], '-o')
plt.xticks(range(1, 6))
plt.xlabel('Fold Index')
plt.ylabel('Best $\lambda$'); | 15,179 | 32,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-22 | latest | en | 0.449005 |
https://cpep.org/mathematics/2806884-a-total-of-2450-for-a-house-painting-job-is-to-be-divided-between-two-.html | 1,713,725,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00224.warc.gz | 157,248,353 | 7,477 | 26 September, 21:16
# A total of \$2450 for a house painting job is to be divided between two painters in the ratio of 3 to 4. how much does each painter receive?
+5
1. 26 September, 21:30
0
we are given that
3x+4y=2450
this implies
7x=2450
x=350
one get 3x=1050
2nd get 4x=1400 | 106 | 286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.960028 |
https://ru.scribd.com/document/439260277/Analysis-docx | 1,611,173,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00582.warc.gz | 547,188,539 | 98,904 | Вы находитесь на странице: 1из 2
# Analysis:
One of the possible errors in the experiment is the observation of the uniform or constant speed
(or motion) of the block for both parts of the experiment because it is based on human judgement
which is not accurate.
Another possible source of error is the inaccurate reading of the height and base in finding the
angle of repose.
The weight added to the block and the pan are only in multiples of 5, meaning it cannot give more
accurate results.
The main concept used in this experiment is the concept of friction.
Friction is known as the force that opposes the motion of a body (or the block).
The friction force is denoted as f = µ N. In solving for the coefficient friction (µ) which is the
constant of proportionality and is presumed to be proportional to the frictional force.
Angle of repose is the angle between the horizontal and incline which can also give the value
coefficient friction.
In the experiment, constant motion must be observed which means there must be a constant
friction applying on the block.
The values for the coefficient of friction are near each other which is obtained from the
𝑓
proportion of the weight of the pan and the weight of the block. The equation µ= 𝑁 is used where
f is the frictional force or the weight of the pan (Wp), and the N as the weight normal force or the
Wp
weight of the block (Wb) leading to the equation, µ= . It can be inferred from the equation that
Wb
to have a constant value of the coefficient of friction, the proportion of the pan and the block must
be constant, meaning if you increase the value of Wb you also must increase the value of Wp, vice
versa.
The coefficient of friction was determined by the ratio of the horizontal distance and the vertical
height (µ = tan θ = height/base), and the angle of repose is computed from the value of the
coefficient of friction which means that the angle of repose can also be used to get the coefficient
of friction. The angle of repose can be solved by θ = tan -1 (height/base). The angle of repose can
be substituted back to tan θ to solve for the coefficient of friction.
Conclusion:
The objectives of the experiment were met. The objective was to determine the coefficient
friction and establish the relationship between the angle of repose and the coefficient of friction.
The objectives were met because the group was able to determine the coefficient of friction for
part A which averaged 0.42 and the group was able to establish the relationship between the angle
of repose and the coefficient of friction which showed that the coefficient of friction can be
solved using the angle of repose. Comparing the values of the coefficient of friction from part A
and the coefficient friction of part B from the angle of repose are near each other.
The concepts about friction, frictional force and angle of repose towards solving for the
coefficient of friction helped the researchers to achieve their objectives. The applied concepts that
were used in this experiment are the frictional force is proportional to the coefficient of friction
where in the experiment, when weight is added to the pan, weight must also be added to the
block. Another concept is that friction works against the motion and acts in opposite direction that
when a body is moving constantly, the friction applied is also constant. A body sliding down in
an inclined plane due to its own weight means that the angle among the incline and the horizontal
is called the angle of repose. Lastly, the coefficient of friction is equal to the tangent of the angle
of repose.
Throughout the experiment, it was observable that the coefficient of friction is a constant of
proportionality. The coefficient of friction can be solved from the proportion of the weight of the
pan and the weight of the block, and from the base and height. It is also observable that constant
motion means that there is a constant friction applied to a body.
There was inaccuracy of the values collected because of the errors. The error may be minimized
if there was something used to measure the constant motion because human judgement was used
in the experiment. Also, having more blocks that can weigh less than 5 grams can provide more
accurate data.
Friction is important in electronics engineering because most of the gadgets are touch screen
which means that it also uses friction in detecting or reading the gestures we are doing towards
the gadgets. The gadget will read the opposite force that we are applying on the screens then
encodes or processes it and then giving its response. Example is swiping up down, left or right. | 1,017 | 4,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-04 | latest | en | 0.946926 |
https://www.studyadda.com/solved-papers/kvpy/physics/simple-harmonic-motion/jee-pyq-simple-harmonic-motion/1118 | 1,642,942,050,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00584.warc.gz | 1,040,488,926 | 41,610 | # Solved papers for JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion
### done JEE PYQ-Simple Harmonic Motion Total Questions - 76
• question_answer1) In a simple harmonic oscillator, at the mean position [AIEEE 2002]
A)
kinetic energy is minimum, potential energy is maximum
B)
both kinetic and potential energies are maximum
C)
kinetic energy is maximum, potential energy is minimum
D)
both kinetic and potential energies are minimum
View Answer play_arrow
• question_answer2) A child swinging on a swing in sitting position, stands up, then the time period of the swing will [AIEEE 2002]
A)
increase
B)
decrease
C)
remain same
D)
increase if the child is long and decrease if the child is short
View Answer play_arrow
• question_answer3) A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes $5T/3$, then the ratio of $\frac{m}{M}$ is [AIEEE 2003]
A)
$\frac{3}{5}$
B)
$\frac{25}{9}$
C)
$\frac{16}{9}$
D)
$\frac{5}{3}$
View Answer play_arrow
• question_answer4) Two particles A and B of equal masses are suspended from two massless springs of spring constants ${{k}_{1}}$ and ${{k}_{2}}$, respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitudes of A and B is [AIEEE 2003]
A)
$\sqrt{{{k}_{1}}/{{k}_{2}}}$
B)
${{k}_{1}}/{{k}_{2}}$
C)
$\sqrt{{{k}_{2}}/{{k}_{1}}}$
D)
${{k}_{2}}/{{k}_{1}}$
View Answer play_arrow
• question_answer5) The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is [AIEEE 2003]
A)
$11%$
B)
$21%$
C)
$42%$
D)
$10.5%$
View Answer play_arrow
• question_answer6) The displacement of a particle varies according to the relation$x=4\,(\cos \pi \,t+\sin \pi t)$. The amplitude of the particle is [AIEEE 2003]
A)
$-4$
B)
4
C)
$4\sqrt{2}$
D)
8
View Answer play_arrow
• question_answer7) A body executes simple harmonic motion. The potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as function of displacement$x$. Which of the following statements is true? [AIEEE 2003]
A)
KE is maximum when $x=0$
B)
TE is zero when $x=0$
C)
KE is maximum when $x$ is maximum
D)
PE is maximum when $x=0$
View Answer play_arrow
• question_answer8) The total energy of a particle, executing simple harmonic motion is [AIEEE 2004]
A)
$\propto x$
B)
$\propto \,{{x}^{2}}$
C)
Independent of$x.$
D)
$\propto {{x}^{1/2}}$ where,$x$is the displacement from the mean position.
View Answer play_arrow
• question_answer9) A particle of mass m is attached to a spring (of spring constant$k$ and has a natural angular frequency${{\omega }_{0}}$. An external force F(t) proportional to $\cos \omega t(\omega \ne {{\omega }_{0}})$is applied to the oscillator. The time displacement of the oscillator will be proportional to [AIEEE 2004]
A)
$\frac{m}{\omega _{0}^{2}-{{\omega }^{2}}}$
B)
$\frac{1}{m(\omega _{0}^{2}-{{\omega }^{2}})}$
C)
$\frac{1}{m(\omega _{0}^{2}+{{\omega }^{2}})}$
D)
$\frac{m}{\omega _{0}^{2}+{{\omega }^{2}}}$
View Answer play_arrow
• question_answer10) In forced oscillation of a particle, the amplitude is maximum for a frequency${{\omega }_{1}}$of the force, while the energy is maximum for a frequency ${{\omega }_{2}}$of the force, then [AIEEE 2004]
A)
${{\omega }_{1}}={{\omega }_{2}}$
B)
${{\omega }_{1}}>{{\omega }_{2}}$
C)
${{\omega }_{1}}<{{\omega }_{2}}$when damping is small and${{\omega }_{1}}>{{\omega }_{2}}$ when damping is large
D)
${{\omega }_{1}}<{{\omega }_{2}}$
View Answer play_arrow
• question_answer11) The function$\sin (\omega t)$represents [AIEEE 2005]
A)
a periodic, but not simple harmonic motion with a period $2\pi /\omega$
B)
a periodic, but not simple harmonic motion with a period $\pi /\omega$
C)
a simple harmonic motion with a period $2\pi /\omega$
D)
a simple harmonic motion with a period $\pi /\omega$
View Answer play_arrow
• question_answer12) Two simple harmonic motions are represented by the equations ${{y}_{1}}=0.1\sin \left( 100\pi t+\frac{\pi }{3} \right)$and${{y}_{2}}=0.1\,\cos \pi t.$ The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is [AIEEE 2005]
A)
$\frac{-\pi }{6}$
B)
$\frac{\pi }{3}$
C)
$\frac{-\pi }{3}$
D)
$\frac{\pi }{6}$
View Answer play_arrow
• question_answer13) If a simple harmonic motion is represented by $\frac{{{d}^{2}}x}{d{{t}^{2}}}+\alpha x=0,$ its time period is [AIEEE 2005]
A)
$\frac{2\pi }{\alpha }$
B)
$\frac{2\pi }{\sqrt{\alpha }}$
C)
$2\pi \alpha$
D)
$2\pi \sqrt{\alpha }$
View Answer play_arrow
• question_answer14) The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would [AIEEE 2005]
A)
first increase and then decrease to the original value
B)
first decrease and then increase to the original value
C)
remain unchanged
D)
increase towards a saturation value
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• question_answer15) The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is [AIEEE 2006]
A)
0.01s
B)
10s
C)
0.1s
D)
100s
View Answer play_arrow
• question_answer16) Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? [AIEEE 2006]
A)
$\frac{1}{6}s$
B)
$\frac{1}{4}s$
C)
$\frac{1}{3}s$
D)
$\frac{1}{12}s$
View Answer play_arrow
• question_answer17) A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency$\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time [AIEEE 2006]
A)
at the mean position of the platform
B)
for an amplitude of $g/{{\omega }^{2}}$
C)
for an amplitude of${{g}^{2}}/{{\omega }^{2}}$
D)
at the highest position of the platform
View Answer play_arrow
• question_answer18) The displacement of an object attached to a spring and executing simple harmonic motion is given by$2x+3y+z=1$$x+3y+2z=2$ metre. The time at which the maximum speed first occurs is [AIEEE 2007]
A)
0.5 s
B)
0.75 s
C)
0.125s
D)
0.25s
View Answer play_arrow
• question_answer19) A point mass oscillates along the x-axis according to the law${{\log }_{3}}e$.If the acceleration of the particle is written as ${{\log }_{e}}3$then [AIEEE 2007]
A)
$f(x)={{\tan }^{-1}}(\sin x+\cos x)$
B)
$(\pi /4,\pi /2)$
C)
$(-\pi /2,\pi /4)$
D)
$(0,\pi /2)$
View Answer play_arrow
• question_answer20) Two springs, of force constants${{I}_{0}}=\frac{E}{R}=\frac{5}{5}=1\,A$and$\tau =\frac{L}{R}=\frac{10}{5}=2\,s$are connected to a mass m as shown. The frequency of oscillation of the mass is$t=2s$. If$I=(1-{{e}^{-1}})A$both$(\therefore -t/\tau =\frac{-2}{2}=-1)$and$J=\frac{i}{\pi {{a}^{2}}}$are made four times their original values, the frequency of. oscillation becomes [AIEEE 2007]
A)
$\oint{B.dl}={{\mu }_{0}}.{{i}_{enclosed}}$
B)
$\oint{B.dl}={{\mu }_{0}}.{{i}_{enclosed}}$
C)
$x=2\times {{10}^{-2}}$
D)
$cos\text{ }\pi t$
View Answer play_arrow
• question_answer21) A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is [AIEEE 2007]
A)
$p=\frac{{{E}_{0}}{{I}_{0}}}{2}$
B)
$P=\sqrt{2}{{E}_{0}}{{I}_{0}}$
C)
${{10}^{-3}}\mu C$
D)
$(\sqrt{2},\sqrt{2})$
View Answer play_arrow
• question_answer22) If$x,\text{ v}$and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time? [AIEEE 2009]
A)
${{a}^{2}}{{T}^{2}}+4{{\pi }^{2}}{{v}^{2}}$
B)
$aT/x$
C)
$aT+2\pi v$
D)
$aT/v.$
View Answer play_arrow
• question_answer23) A mass M, attached to a horizontal spring, executes S.H.M. with amplitude ${{A}_{1}}$. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude ${{A}_{2}}$. The ratio of$\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)$is [AIEEE 2011]
A)
${{\left( \frac{M+m}{M} \right)}^{1/2}}$
B)
$\frac{M}{M+m}$
C)
$\frac{M+m}{M}$
D)
${{\left( \frac{M}{M+m} \right)}^{1/2}}$
View Answer play_arrow
• question_answer24) Two particles are executing simple harmonic motion of the same amplitude A and frequency $\omega$ along the x-axis. Their mean position is separated by distance ${{X}_{0}}({{X}_{0}}>A)$. If the maximum separation between them is $({{X}_{0}}+A)$, the phase difference between their motion is [AIEEE 2011]
A)
$\frac{\pi }{6}$
B)
$\frac{\pi }{2}$
C)
$\frac{\pi }{3}$
D)
$\frac{\pi }{4}$
View Answer play_arrow
• question_answer25) A wooden cube (density of wood d) of side $'\ell '$ floats in a liquid of density$'\rho '$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period T'. Then, T is equal to: [AIEEE 11-05-2011]
A)
$2\pi \sqrt{\frac{\ell d}{\rho g}}$
B)
$2\pi \sqrt{\frac{\ell \rho }{dg}}$
C)
$2\pi \sqrt{\frac{\ell d}{(\rho -d)g}}$
D)
$2\pi \sqrt{\frac{\ell \rho }{(\rho -d)g}}$
View Answer play_arrow
• question_answer26) The displacement $y(t)=A\sin (\omega t+\phi )$ of a pendulum for $\phi =\frac{2\pi }{3}$ is correctly represented by [JEE ONLINE 19-05-2012]
A)
B)
C)
D)
View Answer play_arrow
• question_answer27) A ring is suspended from a point S on its rim as shown in the figure. When displaced from equilibrium, it oscillates with time period of 1 second. The radius of the ring is (take $g={{\pi }^{2}}$) [JEE ONLINE 19-05-2012]
A)
0.15m
B)
1.5m
C)
1.0m
D)
0.5m
View Answer play_arrow
• question_answer28) The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals: [JEE MAIN 2013]
A)
0.7
B)
0.81
C)
0.729
D)
0.6
View Answer play_arrow
• question_answer29) An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is ${{V}_{0}}$ and its pressure is${{P}_{0}}$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency: [JEE MAIN 2013]
A)
$\frac{1}{2\pi }\frac{{{A}_{\gamma }}{{P}_{0}}}{{{V}_{0}}M}$
B)
$\frac{1}{2\pi }\frac{{{V}_{0}}M{{P}_{0}}}{{{A}^{2}}\gamma }$
C)
$\frac{1}{2\pi }\sqrt{\frac{{{A}^{2}}\gamma {{P}_{0}}}{M{{V}_{0}}}}$
D)
$\frac{1}{2\pi }\sqrt{\frac{M{{V}_{0}}}{{{A}_{\gamma }}{{P}_{0}}}}$
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• question_answer30) Two simple pendulums of length 1m and 4mrespectively are both given small d is placement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to: [JEE ONLINE 09-04-2013]
A)
2
B)
7
C)
5
D)
3
View Answer play_arrow
• question_answer31) Bob of a simple pendulum of length $l$ is mode of iron. The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is T then: [JEE ONLINE 23-04-2013]
A)
$T=2\pi \sqrt{\frac{l}{g}}$ and damping is smaller than in air alone.
B)
$T=2\pi \sqrt{\frac{l}{g}}$ and damping is larger than in air alone.
C)
$T>2\pi \sqrt{\frac{l}{g}}$ and damping is smaller than in air alone.
D)
$T<2\pi \sqrt{\frac{l}{g}}$ and damping is larger than in air alone.
View Answer play_arrow
• question_answer32) If the time period $t$ of the oscillation of a drop of liquid of density $d$radius $r,$vibrating under surface tension s is given by the formula $t=\sqrt{{{r}^{2b}}{{s}^{c}}{{d}^{a/2}}.}$ It is observe that the time period is directly proportional to $\sqrt{\frac{d}{s}.}$ The value of $b$ should therefore be: [JEE ONLINE 23-04-2013]
A)
$\frac{3}{4}$
B)
$\sqrt{3}$
C)
$\frac{3}{2}$
D)
$\frac{2}{3}$
View Answer play_arrow
• question_answer33) A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, form a fixed point by a mass less spring, such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. When the cylinder is given a downward push and released, it stats oscillating vertically with a small amplitude. The time period T of the oscillations of the cylinder will be: [JEE ONLINE 25-04-2013]
A)
Smaller than $2\pi$ ${{\left[ \frac{\operatorname{M}}{\operatorname{k}+\operatorname{A}\sigma g} \right]}^{1/2}}$
B)
$2\pi \sqrt{\frac{\operatorname{M}}{\operatorname{k}}}$
C)
Larger than $2\pi {{\left[ \frac{\operatorname{M}}{(\operatorname{k}+\operatorname{A}\sigma g)} \right]}^{1/2}}$
D)
$2\pi {{\left[ \frac{\operatorname{M}}{(\operatorname{k}+\operatorname{A}\sigma g)} \right]}^{1/2}}$
View Answer play_arrow
• question_answer34) A particle moves with simple harmonic motion in a straight line. In first $\tau s$, after starting from rest it travels a distance a, and in next $\tau s$ it travels 2a, in same direction, then: [JEE MAIN 2014]
A)
amplitude of motion is 4a
B)
time period of oscillations is $6\tau$
C)
amplitude of motion is 3a
D)
time period of oscillations is $8\tau$
View Answer play_arrow
• question_answer35) A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. [JEE MAIN 2014]
A)
6
B)
4
C)
12
D)
8
View Answer play_arrow
• question_answer36) An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be: [JEE ONLINE 09-04-2014]
A)
1.7%
B)
2.7%
C)
4.4%
D)
2.27%
View Answer play_arrow
• question_answer37) Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take $g=10m{{s}^{-2}}$) [JEE ONLINE 09-04-2014]
A)
20 N
B)
10 N
C)
60 N
D)
40 N
View Answer play_arrow
• question_answer38) The amplitude of a simple pendulum, oscillating in air with as mall spherical bob, decreases from 10 cm to 8 cm in 40seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3. The time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (In 5 =1.601, In 2 = 0.693). [JEE ONLINE 09-04-2014]
A)
231 s
B)
208 s
C)
161 s
D)
142 s
View Answer play_arrow
• question_answer39) A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by$;x={{a}_{1}}\cos \omega t$and$y={{a}_{2}}\cos 2\omega t$traces a curve given by: [JEE ONLINE 09-04-2014]
A)
B)
C)
D)
View Answer play_arrow
• question_answer40) The angular frequency of the damped oscillator is given by,$\omega =\sqrt{\left( \frac{k}{m}-\frac{{{r}^{2}}}{4{{m}^{2}}} \right)}$where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio$\frac{{{r}^{2}}}{mk}$is 8%, the change in time period compared to the undamped oscillator is approximately as follows: [JEE ONLINE 11-04-2014]
A)
increases by 1%
B)
increases by 8%
C)
decreases by 1%
D)
decreases by 8%
View Answer play_arrow
• question_answer41) Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants? [JEE ONLINE 12-04-2014]
A)
$a+bx-c{{x}^{2}}$
B)
$b{{x}^{2}}$
C)
$a-bx+c{{x}^{2}}$
D)
$-bx$
View Answer play_arrow
• question_answer42) A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves form equilibrium position to half of its amplitude. [JEE ONLINE 19-04-2014]
A)
4 cm/s
B)
6 cm/s
C)
12 cm/s
D)
16 cm/s
View Answer play_arrow
• question_answer43) A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period change to ${{T}_{M}}.$If the Youngs modulus of the material of the wire is Y then$\frac{1}{Y}$is equal to: [JEE MAIN 2015] (g = gravitational acceleration)
A)
$\left[ 1-{{\left( \frac{{{T}_{M}}}{T} \right)}^{2}} \right]\frac{A}{Mg}$
B)
$\left[ 1-{{\left( \frac{T}{{{T}_{M}}} \right)}^{2}} \right]\frac{A}{Mg}$
C)
$\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{A}{Mg}$
D)
$\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{Mg}{A}$
View Answer play_arrow
• question_answer44) For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? [JEE MAIN 2015] (graph are schematic and not drawn to scale)
A)
B)
C)
D)
View Answer play_arrow
• question_answer45) The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}.$Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is: [JEE MAIN 2015]
A)
1%
B)
5%
C)
2%
D)
3%
View Answer play_arrow
• question_answer46) A simple harmonic oscillator of angular frequency $2rad\,{{s}^{-1}}$is acted upon by an external force $F=\sin \,t\,N$. If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to: [JEE ONLINE 10-04-2015]
A)
$\sin t+\frac{1}{2}\cos 2t$
B)
$\sin t-\frac{1}{2}\sin 2t$
C)
$\cot t-\frac{1}{2}\sin 2t$
D)
$\sin t+\frac{1}{2}\sin 2t$
View Answer play_arrow
• question_answer47) x and y displacements of a particle are given as x(t)=a $\sin \omega t$ and $y(t)=a\sin 2\omega t.$ Its trajectory will look like: [JEE ONLINE 10-04-2015]
A)
B)
C)
D)
View Answer play_arrow
• question_answer48) A cylindrical block of wood (density $=650\,kg\,{{m}^{-3}}$), of base area $30\text{ }c{{m}^{2}}$and height 54 cm, floats in a liquid of density $900\text{ }kg{{m}^{-}}^{3}.$The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly): [JEE MAIN 11-04-2015]
A)
65 cm
B)
52 cm
C)
39 cm
D)
26 cm
View Answer play_arrow
• question_answer49) A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in${{s}^{-1}}$) is: [JEE MAIN 11-04-2015]
A)
$\frac{1}{30}\ln 3$
B)
$\frac{1}{15}\ln 3$
C)
2
D)
$\frac{1}{2}$
View Answer play_arrow
• question_answer50) A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance$\frac{2A}{3}$from equilibrium position. [JEE MAIN - I 3-4-2016] The new amplitude of the motion is:-
A)
$\frac{7A}{3}$
B)
$\frac{A}{3}\sqrt{41}$
C)
3A
D)
A 3
View Answer play_arrow
• question_answer51) Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement $\frac{-A}{2}$and they are moving towards each other .If they cross each at time t, then t is: [JEE ONLINE 09-04-2016]
A)
$\frac{T}{4}$
B)
$\frac{5T}{6}$
C)
$\frac{T}{3}$
D)
$\frac{T}{6}$
View Answer play_arrow
• question_answer52) In an engine the piston undergoes vertical simple harmonic motion with amplitude 7cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer longer stays in contact with the piston, is closed to: [JEE ONLINE 10-04-2016]
A)
0.7 Hz
B)
1.2 Hz
C)
1.9 Hz
D)
0.1 Hz
View Answer play_arrow
• question_answer53) A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like [JEE Main 2017]
A)
B)
C)
D)
View Answer play_arrow
• question_answer54) A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is [JEE Online 08-04-2017]
A)
2 Hz
B)
$\frac{1}{4}Hz$
C)
$\frac{1}{2\sqrt{2}}Hz$
D)
$\frac{1}{2}Hz$
View Answer play_arrow
• question_answer55) The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10{{s}^{-1}}.$ At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is$\frac{\pi }{4}.$ [JEE Online 08-04-2017]
A)
$500m/{{s}^{2}}$
B)
$750\sqrt{2}m/{{s}^{2}}$
C)
$750m/{{s}^{2}}$
D)
$500\sqrt{2}m/{{s}^{2}}$
View Answer play_arrow
• question_answer56) A block of mass 0.1 kg is connected to an elastic spring of spring constant $640\,\,N{{m}^{-1}}$ and oscillates in a damping medium of damping constant ${{10}^{-2}}kg\,{{s}^{-1}}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to [JEE Online 09-04-2017]
A)
2 s
B)
5 s
C)
7 s
D)
3.5 s
View Answer play_arrow
• question_answer57) In an experiment to determine the period of a simple pendulum of length 1m, it is attached to different spherical bobs of radii ${{r}_{1}}$ and ${{r}_{2}}$. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5\times {{10}^{-4}}s,$ the difference in radii, $|{{r}_{1}}-{{r}_{2}}|$ is best given by [JEE Online 09-04-2017]
A)
0.01 cm
B)
(2) 0.1 cm
C)
0.5 cm
D)
1 cm
View Answer play_arrow
• question_answer58) A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of ${{10}^{12}}/\sec$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number$\text{=6}\text{.02}\times \text{1}{{\text{0}}^{\text{23}}}\text{ gm mol}{{\text{e}}^{\text{-1}}}$). [JEE Main Online 08-04-2018]
A)
$2.2\,\,N/m$
B)
$5.5\,\,N/m$
C)
$6.4\,\,N/m$
D)
$7.1\,\,N/m$
View Answer play_arrow
• question_answer59) Two simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures. $x(t)=A\sin (at+\delta )$ $y(t)=B\sin (bt)$ Identify the correct match below [JEE Online 15-04-2018 (II)]
A)
$\text{Parameters: A=B, a=2b;}\delta \text{=}\frac{\pi }{2};Curve:Circle$
B)
$\text{Parameters: A=B, a=b;}\delta \text{=}\frac{\pi }{2};Curve:line$
C)
$\text{Parameters: A}\ne \text{B, a=b; }\delta \text{=}\frac{\pi }{2};Curve:\text{Ellipse}$
D)
$\text{Parameters:}A\ne B,a=b;\delta =0;\text{Curve : Parabola}$
View Answer play_arrow
• question_answer60) An oscillator of mass M is at rest in its equilibrium position in a potential $V=\frac{1}{2}k{{(x-X)}^{2}}.$A particle of mass comes from right with speed u and collides completely in elastically with and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after collisions is:$(M=10,m=5,u=1,k=1).$ [JEE Main Online 16-4-2018]
A)
$\frac{1}{2}$
B)
$\frac{1}{\sqrt{3}}$
C)
$\frac{2}{3}$
D)
$\sqrt{\frac{3}{5}}$
View Answer play_arrow
• question_answer61) A rod of mass M and length 2L is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of m are attached at distance L/2 from its centre on both sides, it reduces the oscillation frequency by $20%$. The value of ratio m/M is close to: [JEE Main 09-Jan-2019 Evening]
A)
0.37
B)
0.57
C)
0.77
D)
0.17
View Answer play_arrow
• question_answer62) A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about$x=0$. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be: [JEE Main 09-Jan-2019 Evening]
A)
$\frac{A}{2\sqrt{2}}$
B)
$\frac{A}{\sqrt{2}}$
C)
$\frac{A}{2}$
D)
A
View Answer play_arrow
• question_answer63) A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency co. If the radius of the bottle is 2.5 cm then co is close to - (density of water$={{10}^{3}}kg/{{m}^{3}}$). [JEE Main 10-Jan-2019 Evening]
A)
$2.50\text{ }rad\text{ }{{s}^{-}}^{1}$
B)
$3.75\text{ }rad\text{ }{{s}^{-1}}$
C)
$5.00\text{ }rad\text{ }{{s}^{-1}}$
D)
None of these
View Answer play_arrow
• question_answer64) A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is- [JEE Main 10-Jan-2019 Evening]
A)
$\frac{4\pi }{3}$
B)
$\frac{3\,}{8}\pi$
C)
$\frac{7\,}{3}\pi$
D)
$\frac{8\,\pi }{3}$
View Answer play_arrow
• question_answer65) A particle undergoing simple harmonic motion has time dependent displacement given by $x(t)=A\sin \frac{\pi t}{90}.$ The ratio of kinetic to potential energy of this particle at t = 210 s will be- [JEE Main 11-Jan-2019 Morning]
A)
$\frac{1}{9}$
B)
2
C)
1
D)
None of these
View Answer play_arrow
• question_answer66) A pendulum is executing simple harmonic motion and its maximum kinetic energy is ${{K}_{1}}$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case. its maximum kinetic energy is ${{K}_{2}}$. Then [JEE Main 11-Jan-2019 Evening]
A)
${{K}_{2}}={{K}_{1}}$
B)
${{K}_{2}}=\frac{{{K}_{1}}}{2}$
C)
${{K}_{2}}=2{{K}_{1}}$
D)
${{K}_{2}}=\frac{{{K}_{1}}}{4}$
View Answer play_arrow
• question_answer67) The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be- [JEE Main 11-Jan-2019 Evening]
A)
$2\sqrt{3}s$
B)
$\frac{3}{2}s$
C)
$\frac{2}{\sqrt{3}}s$
D)
$\frac{\sqrt{3}}{2}s$
View Answer play_arrow
• question_answer68) A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of ${{10}^{-2}}$m. The relative change in the angular frequency of the pendulum is best given by- [JEE Main 11-Jan-2019 Evening]
A)
${{10}^{-5}}rad/s$
B)
${{10}^{-1}}rad/s$
C)
$1rad/s$
D)
${{10}^{-3}}rad/s$
View Answer play_arrow
• question_answer69) A simple pendulum, made of a string of length I and a bob of mass m, is released from a small angle ${{\theta }_{0}}.$It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle ${{\theta}_{1}}.$Then M is given by- [JEE Main 12-Jan-2019 Morning]
A)
$\frac{m}{2}\left( \frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}} \right)$
B)
$m\left( \frac{{{\theta }_{0}}+{{\theta }_{1}}}{{{\theta }_{0}}-{{\theta }_{1}}} \right)$
C)
$\frac{m}{2}\left( \frac{{{\theta }_{0}}+{{\theta }_{1}}}{{{\theta }_{0}}-{{\theta }_{1}}} \right)$
D)
$m\left( \frac{{{\theta }_{0}}-{{\theta }_{1}}}{{{\theta }_{0}}+{{\theta }_{1}}} \right)$
View Answer play_arrow
• question_answer70) A simple harmonic motion is represented by [JEE Main 12-Jan-2019 Evening] $y=5(sin3\pi t+\sqrt{3}cos3\pi t)cm$ The amplitude and time period of the motion are
A)
$5cm,\frac{3}{2}s$
B)
$10cm,\frac{2}{3}s$
C)
$5cm,\frac{2}{3}s$
D)
$10cm,\frac{3}{2}s$
View Answer play_arrow
• question_answer71) A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1000}$of the original amplitude is close to: [JEE Main 8-4-2019 Afternoon]
A)
100 s
B)
20 s
C)
10 s
D)
50 s
View Answer play_arrow
• question_answer72) In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:- [JEE Main 8-4-2019 Afternoon]
A)
0.7%
B)
0.2%
C)
3.5%
D)
6.8%
View Answer play_arrow
• question_answer73) A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is : [JEE Main 9-4-2019 Morning]
A)
$4T\sqrt{\frac{1}{15}}$
B)
$2T\sqrt{\frac{1}{10}}$
C)
$4T\sqrt{\frac{1}{14}}$
D)
$2T\sqrt{\frac{1}{14}}$
View Answer play_arrow
• question_answer74) A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is :- [JEE Main 9-4-2019 Afternoon]
A)
320m/s, 120 Hz
B)
180m/s, 80 Hz
C)
180m/s, 120 Hz
D)
320m/s, 80 Hz
View Answer play_arrow
• question_answer75) The displacement of a damped harmonic oscillator is given by $x(t)={{e}^{-01.1t}}\cos (10\pi t+\phi ).$Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to: [JEE Main 10-4-2019 Morning]
A)
13 s
B)
7 s
C)
27 s
D)
4 s
View Answer play_arrow
• question_answer76) A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillation to be 50 s. The accuracy in g is [JEE MAIN Held on 08-01-2020 Evening]
A)
3.40%
B)
2.40%
C)
5.40%
D)
4.40%
View Answer play_arrow
• #### Study Package
##### JEE PYQ-Simple Harmonic Motion
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# Grouping Data by distance
Hello!
New Tableau user here so bare with me if I am using wrong terms. I am trying to set up a dashboard that allows the users to see pieces of equipment with in a certain radius of the selected town. I need the user to be able to select what state they want to see, and then select the city they wish to have the radius around. I want the cities with more pieces of equipment to be larger.
For example, if I select Indiana, and then select Indianapolis, I want to be able to see all the pieces of equipment for all the cities within a certain radius. Currently I am using a 50 mile radius. I have attached my workbook to this forum post.
I have set a parameter to automatically fill in the select city. With that parameter, I have created a filter using the haversine formula to give me the distances I want.
My only issue right now, is I just want to display the data on the right side of the dashboard, but what I running into is even after I have the filter in place, my table on the right still only shows me the equipment on that selected city instead of the selected city and all the cities within 50 mile radius.
I am still new to Tableau so I know my dashboard/worksheets might look messy so I apologize for that.
Any help would be greatly appreciated! Thank you!
• ###### 1. Re: Grouping Data by distance
I have an example on Tableau Public that should help you to achieve this.
• ###### 2. Re: Grouping Data by distance
Not sure what am I missing here.. I don't see any table on right side... this is what I see in the dashboard.
• ###### 3. Re: Grouping Data by distance
I have updated the workbook! Thanks for noticing this!
• ###### 4. Re: Grouping Data by distance
Still need to group them into a table. Basically if I select Indianapolis, I need it also select all cities within 50 miles and add them as a group.
• ###### 5. Re: Grouping Data by distance
This is the distance calculation:
3959 * ACOS
(
)
Circle Longitude:
Circle Latitude:
The above calculations define the circle. The radius is a parameter that allows the user to choose any number of distances in miles from the center of the circle. The Degree field is just a field with values 1 to 360. I use this to draw a circle over the map, but you can use this same information to select the names or other dimensions of equipment that fall inside the circle.
• ###### 6. Re: Grouping Data by distance
I think my biggest confusion is, what would I replace the Degree field with?
I have my distance formula calculated with:
3959 * ACOS( | 611 | 2,632 | {"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} | 2.625 | 3 | CC-MAIN-2020-16 | latest | en | 0.920253 |
https://www.physicsforums.com/threads/hoop-and-falling-beads.882276/ | 1,508,795,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00126.warc.gz | 950,574,293 | 21,960 | Tags:
1. Aug 17, 2016
### UnterKo
Hello, I've got a problem and I have no idea how to start. I'll be happy for any hint. Thanks
1. The problem statement, all variables and given/known data
Two beads each of mass m are at the top (Z) of a frictionless hoop of mass M and radius R which lies in the vertical plane. The hoop is supported by a frictionless vertical support. The beads are given a tiny impulse and due to gravity they slide down the hoop: one clockwise and one anti-clockwise.
a) Determine the minimum value of X = m/M, Xmin, for which the hoop will rise up off the support before the beads reach the bottom of the hoop (Y).
b) If X < Xmin and the beads collide inelastically at the bottom of the hoop with a coefficient of restitution eR = 0.98, determine the minimum angle (θ) to the nearest degree with respect to the initial position (θ = 0) achieved by the clockwise moving bead.
c) How many oscillations are required so that the maximum height achieved by the beads is less than 0.01R?
2. The attempt at a solution
a) To use energy conservation law?
b) Tangens of the angle θ (with respect to the initial position) is going to be reduced by the e?
c) Do I get the height by the energy conservation law (PE = KE)? But how to determine the number of collisions?
2. Aug 17, 2016
### haruspex
For a), yes, use energy conservation. What can you deduce from that?
You will also need to think about forces on the hoop, obviously. What forces will there be?
For b), again, use energy. If the coefficient is e, what fraction of energy is lost in the collision?
What fraction of energy is lost in n collisions?
3. Aug 19, 2016
### UnterKo
a) Each bead experiences its weight and the normal reaction force (because the hoop is frictionless, it acts normal to the surface, right?). If I take θ as the angle of it's position with respect to the horizontal, the normal force may be expressed as:
F = Nr - mgk = (N - mg sin θ)r - (mg cos θ) θ.
Thus the radial eq. is: $N - mg sin θ = - ma θ' ^2$, and the angular eg. is:$- mg cos θ = ma θ''$. So, when the bead reaches the bottom of the hoop, the potential energy is: $V(θ) = mga sin θ$.
And the speed of the bead is aθ', so the K.E. is: $K = ½ ma^2 θ'^2$, right?
So, for the law of conservation of energy I get: K + V = K + mga sin θ = mga ?
But now I don't know precisely what's happening with the hoop at this moment..
b) The lost of KE is: $ΔK = ½ mv_i^2 (e^2 - 1) (1 - sin^2 θ) < 0$, since for the velocity of each bead after the collistion, $v_f applies: v_f^2 = v_i^2 (sin^2 θ + a^2 cos^2 θ) = v_i^2 + v_i^2 (e^2 - 1) cos^2 θ$?
c) Is the fraction: 1/n ΔE ?
4. Aug 19, 2016
### haruspex
... and measured upwards from the horizontal, yes?
What is theta here? It has gone from top to bottom.
Right. So plug that in your equation for N.
Again, what is theta here?
5. Aug 19, 2016
### UnterKo
a) Yes, it's upwards to the horizontal.
When I substitute, it gets:
$$½ ma^2 θ'^2 + mga \sin θ = mga$$
$$½ a θ'^2 + g \sin θ = g$$
$$½ a θ'^2 = g(1 - \sin θ)$$
$$θ'^2 = \frac{2g(1 - \sin θ)}{a}$$
$$N - mg \sin θ = -2 mg (1 - \sin θ)$$
$$N = mg(3 \sin θ - 2)$$
But how do I get the fraction of masses, X?
b) And then after the collision (sorry, I used the same notation for the angle as before), so there is φ as measured from vertical of the place of the collision to the right (for the clockwise moving bead). But I don't know how to derive it afterwards. Do I need to use the radius (or height) difference, Δr, from the bottom position to the position after the collision?
6. Aug 19, 2016
### haruspex
What is the net force they exert on the hoop?
For b), what is the KE at the bottom?
7. Aug 19, 2016
### UnterKo
$$N = mg (3 \sin \theta - 2)$$<- it isn't this multiplied by two? I'm not actually sure about what even physically means 'to rise up off the support'. Does it mean that the hoop is not fixed?
b) KE before the collision is:
$$K = 1/2 ma^2 \dot{\theta}^2$$
8. Aug 19, 2016
### BvU
Haru may have gone to bed; I liked following this thread and will chip in carefully, without giving away too much (that would spoil the fun and the ultimate satidfaction).
So: The $N$ you calculated represents the radial force each bead experiences from the hoop, right ? So yes, you now have to look at the sum of the two forces the hoop experiences from the beads. That's not a factor of two ! (make a sketch).
The hoop will rise up from the support if the sum of the two forces it experiences from the beads offsets $M\vec g$. The hoop appears to be not fixed to the support.
9. Aug 19, 2016
### haruspex
I confirm all of BvU's observations.
Yes, but you also know what it is in terms of the given variables, R, m etc. (it would be less confusing if you were to use the given radius R instead of introducing a, which is often used for acceleration.)
10. Aug 26, 2016
### UnterKo
So the force experienced by beads on the hoop is a sum of this force (N) of both beads? Then: 2mg(3sin θ - 2) > Mg. And then it is: m/M > (3 sin θ - 2)/(2)?
So: K = ½ mR θ'2. But if I substitute, for example, m = M (3 sin θ - 2)/(2), I'm sorry, but I don't see how it's going to help me.
11. Aug 26, 2016
### BvU
Yes. But they are not pointing in the same direction, so it's not a factor of 2 (make a sketch).
12. Aug 26, 2016
### UnterKo
But then they are at the bottom of the hoop, no? And for this position the fraction m/M has to be determined, right?
13. Aug 26, 2016
### haruspex
No, they will not be at the bottom of the hoop when they apply the greatest upward force on it. Indeed, at the bottom it would be the maximum downward force.
Find the component of the normal force which acts upwards on the hoop, then find the angle at which this is maximised.
14. Aug 26, 2016
### haruspex
You dropped an exponent in the KE expression.
For this part of the question, forget m = M (3 sin θ - 2)/(2). That is not relevant now. What is the lost PE when a bead reaches the bottom?
15. Aug 28, 2016
### UnterKo
I see, I have not understood this till now! :) So it's going to be a centripetal force? And the angle may be found by some derivation?
Is the loss of PE going to be like: $PE = mgh$, so for one of the beads at the bottom: $PE = mg(\sin \frac{\pi}{2})$, so the loss is: $PE = mg(2R - 1)$?
16. Aug 28, 2016
### haruspex
For part a), in post #5 you already found the normal force each bead exerts on the hoop when at angle theta. What part of that force tends to raise the hoop up? For what theta is it maximised?
For part b):
You need a distance in there.
17. Aug 29, 2016
### UnterKo
It is maximised for $\theta = \frac{\pi}{2}$. Is it centripetal force?
So is it going to be part b): $PE = mgR \sin \theta$?
18. Aug 29, 2016
### CWatters
Can I suggest you draw a bead on a hoop at several different positions. Mark on it the direction of the centripetal force and the corresponding reaction force on the hoop. It should then be obvious that there is a range of angles for which the reaction force on the hoop has an upwards component (lifting the hoop) and a range of angles for which it's downwards (not lifting the hoop).
I don't think you can determine angle at which the force is a maximum simply by inspection because the velocity and centripetal force changes with the angle. You will have to write equations, some of which you have already done.
19. Aug 30, 2016
### UnterKo
I've done it and I see what you mean. But there can not be found any max. lifting force at some angle? Eg. $\theta = \frac{\pi}{2}$(as measured from horizontal in downwards direction)?. So how can I find the value of $X = \frac{m}{M}$?
And could you, please, explicitly write which ones of mine equations are correct and which ones I am missing?
Thank you
Last edited: Aug 30, 2016
20. Aug 30, 2016
### BvU
check numbers 8,11,13,16,18 , all of us want you to replace the factor 2 by something that depends on theta... | 2,268 | 7,900 | {"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} | 3.546875 | 4 | CC-MAIN-2017-43 | longest | en | 0.907765 |
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# The performance of an air motor
#### Performance of an air motor is dependent on the inlet pressure.
The motor can operate over the complete torque curve
At a constant inlet pressure, air motors exhibit the characteristic linear output torque/ speed relationship. However, by simply regulating the air supply, using the techniques of throttling or pressure regulation, the output of an air motor can easily be modified. One of the features with air motors is that they can operate over the complete torque curve from free speed to standstill without any harm to the motor. The free speed* or idling speed is defined as the operating speed where there is no load on the output shaft.
*Free Speed = speed at which the outgoing shaft rotates when no load is applied.
### The Power Curve
The power that an air motor produces is simply the product of torque and speed. Air motors produce a characteristic power curve, with maximum power occurring at around 50 % of the free speed.
The torque produced at this point is often referred to as “torque at the maximum output.”
Output formula:
P = (π x M x n) / 30
M = (30 x P) / (π x n)
n = (30 x P) / (π x M)
P = power [kW]
M = torque [Nm]
n = speed [rpm]
### The working point
The working point for an air motor
When selecting an air motor for an application the first step is to establish the “working point”.
This is the combination of the desired operating speed for the motor and the torque required at that point.
Note: The point on the torque/ speed curve where the motor actually operates is called the working point.
#### Air consumption
The air consumption for an air motor increases with the motor speed and thus is highest at free speed. Even at standstill condition (with full pressure applied) the motor consumes air. This depends on the internal leakage in the motor.
Note: Air consumption is measured in l/s. This is however not the actual volume that the compressed air occupies in the motor but it is measured as the volume it would occupy if allowed to expand to atmospheric pressure. This is a standard used for all pneumatic equipment. | 705 | 3,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.843558 |
https://kids.kiddle.co/Rubik%27s_Games | 1,623,966,385,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00463.warc.gz | 310,707,312 | 7,702 | # Rubik's Games facts for kids
Kids Encyclopedia Facts
Quick facts for kids
Rubik's Games
Developer(s) Androsoft
Publisher(s) Hasbro Interactive
Platform(s) Windows 95/98
Release date(s) 1999
Genre(s) Puzzle
Mode(s) One player
Rubik's Games is a, five games in one, PC game created for Windows 95/98 developed in part by Ernő Rubik
```with Androsoft and was published by Hasbro Interactive. It was part of Hasbro's classical games collection of PC related games, translating their most famous board games into best possible quality video games. A history of the Rubik's Cube and its inventor, written out in a webpage type file, with pictures is available from the Menu.
```
## The games
All games have a Challenge and free play mode.
Rubik's Classics The main puzzle 3x3 was composed of a standard Rubik's Cube, where the player could play and solve the puzzle using the PC's mouse, and even a paint and solve feature where they could paint the sides of the cube and then click on the solve feature, the computer would find the quickest solution to that arrangement of the puzzle. This could help the player solve a standard cube that they had at home without having to peel off the stickers or without taking their Cube apart so they can have a solved cube to play with. The Classics section also had the 2x2, 2x3, 4x4 and 5x5 versions of the puzzle.
Cover Up On a grid of various sizes a series of colored columns are stacked up and the players job is to lay all the tiles so they completely cover the grid. As the grids become larger the stacks also become larger and more difficult to figure out in which patterns the tiles will fit onto the grid. The grid board can be turned at different angles to better see the grid.
Zigthrough A chain of five squares, like a snake, can be moved left and right while it comes down from the top of the screen. As the chain moves the colors of the chain can be moved forward as desired so the head of the chain can be any of the five colors. As the chain reaches another block set on the playing field it stops, and the only way to clear the colored blocks off the playing field is to connect three squares of the same color. There are also irremovable blocks, that act as a barrier to get around. Once the path is clear the chain can reach the goal.
Rubik's Playground Working more as an experiment in Physics, the player guides a series of balls down ramps and across other hazards to reach the exit. There are four different types of balls: iron, rubber, plastic and exploding. The player can bump the balls into each other, or speed them up using ramps and magnets. A required number of balls entering the Exit are needed to complete each puzzle. The player can create and save their own Playground puzzles, and save them into files on their computer, which can be uploaded to the Game's Official webpage.
Paint War Is a Chess type game where the player has pieces that rank from zero to five and only pieces with a smaller number can take the pieces of their opponent, the zero piece can be taken by another zero piece that strikes first or a number 5. Like the pawns in Chess being promoted the zero piece can be promoted to a number 5 if the players #5 piece has been taken out of the game. There are two ways to complete each match against the computer, or over the internet using the games webpage link.
1. Paint more of the 42 squares before the computer does.
2. Take all the computers players this is known as a KO.
The number of the piece determines the number of squares that piece can paint. The player's opponent can paint a square and the player can paint over it, yet the match ends when all the squares have been painted or all the opponents pieces have been taken. | 822 | 3,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-25 | longest | en | 0.960705 |
http://fredrikj.net/blog/2008/07/division-the-sequel-with-bonus-material/ | 1,581,969,749,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00197.warc.gz | 60,801,349 | 4,675 | Division: the sequel (with bonus material)
July 27, 2008
I have neglected to write a GSoC update for two weeks (where “two” was computed using round-to-floor), so let me first apologize for that.
In my last post, I wrote about how Newton’s method (coupled with a fast multiplication algorithm, such as Karatsuba multiplication) can be used for asymptotically fast division of long integers. I have now fixed up the code to compute a correctly rounded quotient. In fact, it now performs the full divmod operation, returning both the quotient and a correct remainder.
The trick is to perform an extra exact multiplication at the end to determine the remainder. By definition, the quotient r of p/q is correct if the remainder m = pr·q satisfies 0 ≤ m < q. If this inequality does not hold, one need only perform an additional divmod operation (which can be performed using standard long division, since the error will almost certainly fit in a single limb) to correct both the quotient and the remainder.
The extra multiplication causes some slowdown, but it’s not too bad. The new idivmod function still breaks even with the builtin divmod somewhere around 1000-2000 digits and is 10 times faster at half a million digits (i.e. when dividing a million digit number by a half-million digit number):
>>> time_division(int, 2.0, 20) size old time new time faster correct------------------------------------------------------------ 16 0.000008 0.000052 0.155080 True 32 0.000005 0.000052 0.102151 True 64 0.000008 0.000059 0.132075 True 128 0.000013 0.000070 0.190476 True 256 0.000035 0.000107 0.325521 True 512 0.000126 0.000215 0.583658 True 1024 0.000431 0.000532 0.810399 True 2048 0.001855 0.001552 1.195104 True 4096 0.007154 0.005050 1.416708 True 8192 0.028505 0.015449 1.845033 True 16384 0.111193 0.046938 2.368925 True 32768 0.443435 0.142551 3.110706 True 65536 1.778292 0.432412 4.112497 True 131072 7.110184 1.305771 5.445200 True 262144 28.596926 3.919399 7.296253 True 524288 116.069764 11.804032 9.833061 True
As a bonus, a fast divmod also provides a fast way to convert long integers to decimal strings, by recursively splitting n in half using L, R = divmod(n, 10b/2) where b is the number of digits in n:
>>> time_str(int, 20) size old time new time faster correct------------------------------------------------------------ 16 0.000005 0.000013 0.413043 True 32 0.000006 0.000009 0.588235 True 64 0.000008 0.000012 0.674419 True 128 0.000020 0.000023 0.865854 True 256 0.000059 0.000133 0.442105 True 512 0.000204 0.000333 0.613255 True 1024 0.000895 0.001194 0.749708 True 2048 0.003505 0.002252 1.556824 True 4096 0.013645 0.006600 2.067429 True 8192 0.052386 0.018334 2.857358 True 16384 0.209164 0.052233 4.004412 True 32768 0.834201 0.153238 5.443827 True 65536 3.339629 0.450897 7.406639 True 131072 13.372223 1.339044 9.986392 True 262144 53.547894 3.998352 13.392491 True 524288 214.847486 11.966933 17.953429 True
So printing an integer with half a million digits takes 12 seconds instead of 3.5 minutes. This can be quite a usability improvement.
The code can be found in this file: div.py.
I have now submitted a request for these algorithms to be implemented in the Python core. Since the pure Python implementation is very simple, I don’t think porting it to C would be all that difficult. By “request”, I mean that I might eventually do it myself, but there are many others who are much more familiar with both the Python codebase and the C language, and if any of those persons happens to have a few free hours, they could certainly do it both faster and better. If you are interested in helping out with this, please post to the issue tracker.
The division code is just one of several small projects I’ve been working on lately. Basically, I’ve found that there are some arithmetic functions that are needed very frequently in all kinds of mathematical code. These include:
• Pure convenience functions like sign, product, …
• Bit counting | 1,357 | 4,687 | {"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} | 3.234375 | 3 | CC-MAIN-2020-10 | latest | en | 0.80655 |
https://forum.ansys.com/forums/topic/extrude-cutting-material-x-boolean/ | 1,701,411,491,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00161.warc.gz | 300,233,026 | 34,864 | ## General Mechanical
Topics relate to Mechanical Enterprise, Motion, Additive Print and more
#### Extrude Cutting Material X Boolean
• Fabricio.Urquhart
Subscriber
I am doing my master thesis, that is a steel bolted connection analysis. I am learning a lot here in the community, thank you.
Now, I have a problem with the bolt stress results.
I think that the problem with the bolt, is something in the geometry. Because when I try to apply a boolean with, Head, fastener and wash of the bolt, it does not occure. Maybe something wrong that I am doing extrude a surface, cutting the material. I think this, becasue when I take the direcional deformation in the bolt axis, I see the picture below:
So, there is something wrong with my bolt geometry.
The model is attached.
Thank you one more time!!!
• Aniket
Ansys Employee
Are you looking at the results at true scale? or is it exaggerated scale?
-Aniket
• Fabricio.Urquhart
Subscriber
It is true scale, now the geometry is correct, but i snot converging...
• peteroznewman
Subscriber
Fabricio,
One of the fasteners was mixed in with one of the beams. Here I pulled it out as Part 10.
I haven't run Solve on this, but you can see if it helps.
Regards,
Peter
• Fabricio.Urquhart
Subscriber
Peter,
I don't understand why when I apply boolean in the parts of the bolt, it does not merge the solids. Could you help me with boolean?
• Fabricio.Urquhart
Subscriber
As you see in the picture below, when I apply boolean, only two solids are merged.
Why it happens?
• Fabricio.Urquhart
Subscriber
The problem was, that I had a Slice in the bolt, because this, boolean was not working. Now I took the splice out, and the boolean is ok.
Peter!!!Using boolean in the bolts, I do not have the peak of stress in the bolts, just se the picute below:
• peteroznewman
Subscriber
Fabricio,
You had three bodies in one multibody part and Shared Topology connected the mesh. Why do you need to Boolean these bodies together?
These are well connected meshes.
There is no problem performing a Boolean on those bodies if that is the last operation. However when you have a Share Topo item in the outline, you can't do a Boolean. In fact, Share Topo is not required to Share Topology. It is an obsolete function.
18.2 archive attached.
• Fabricio.Urquhart
Subscriber
Thank you Peter. How did you get the picture that is in the right?With this picture you can see that the elements are not regular, because this, I think that the peak of stress is appearing.
• peteroznewman
Subscriber
The reason is Average Across Bodies is set to No by default. | 623 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.938829 |
http://www.patheos.com/blogs/jesuscreed/2016/03/01/penny-for-your-thoughts/?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+PatheosJesusCreed+%28Blog+-+Jesus+Creed%29 | 1,511,247,162,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806317.75/warc/CC-MAIN-20171121055145-20171121075145-00123.warc.gz | 474,406,669 | 17,284 | Ana Swanson, on the mathematics of choosing your mate:
Image
So how do you find the best one? Basically, you have to gamble. And as with most casino games, there’s a strong element of chance, but you can also understand and improve your probability of “winning” the best partner. It turns out there is a pretty striking solution to increase your odds.
The magic figure turns out to be 37 percent. To have the highest chance of picking the very best suitor, you should date and reject the first 37 percent of your total group of lifetime suitors. (If you’re into math, it’s actually 1/e, which comes out to 0.368, or 36.8 percent.) Then you follow a simple rule: You pick the next person who is better than anyone you’ve ever dated before.
To apply this to real life, you’d have to know how many suitors you could potentially have or want to have — which is impossible to know for sure. You’d also have to decide who qualifies as a potential suitor, and who is just a fling. The answers to these questions aren’t clear, so you just have to estimate. Here, let’s assume you would have 11 serious suitors in the course of your life.
If you just choose randomly, your odds of picking the best of 11 suitors is about 9 percent. But if you use the method above, the probability of picking the best of the bunch increases significantly, to 37 percent — not a sure bet, but much better than random.
This method doesn’t have a 100 percent success rate, as mathematician Hannah Fry discusses in an entertaining 2014 TED talk. There’s the risk, for example, that the first person you date really is your perfect partner, as in the illustration below. If you follow the rule, you’ll reject that person anyway. And as you continue to date other people, no one will ever measure up to your first love, and you’ll end up rejecting everyone, and end up alone with your cats. (Of course, some people may find cats preferable to boyfriends or girlfriends anyway.)
""As for myself, I frankly confess, that I should not want free will to be ..." | 471 | 2,032 | {"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} | 3.609375 | 4 | CC-MAIN-2017-47 | latest | en | 0.947455 |
https://testbook.com/question-answer/what-will-be-the-difference-between-simple-and-com--6582afc0711ac732f8387ed7 | 1,723,011,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00792.warc.gz | 462,475,029 | 47,262 | # What will be the difference between simple and compound interest at 10% per annum on a sum of Rs 1,000 after 4 years?
This question was previously asked in
Bihar Police SI Memory Based Paper (Held on: 17th Dec 2023 Shift 1)
View all Bihar Police SI Papers >
1. Rs 32.10
2. Rs 40.40
3. Rs 54.40
4. Rs 64.10
Option 4 : Rs 64.10
Free
RRB Exams (Railway) History of the Indian Constitution
1.9 Lakh Users
15 Questions 15 Marks 9 Mins
## Detailed Solution
Given:
Principal sum (P) = Rs 1,000.
Rate of interest per annum (r) = 10%.
Time period (n) = 4 years.
Formula used:
SI = (P × R × T)/100
CI = P[(1 + R/100)n - 1]
Calculation:
SI = (P × R × T)/100
⇒ (1000 × 10 × 4)/100 = Rs. 400
CI = P[(1 + R/100)n - 1]
⇒ 1000[(1 + 10/100)4 - 1]
⇒ 1000[14641/10000 - 1]
⇒ 1000 × 4641/10000 = Rs 464.10
Now, Difference = 464.1 - 400 = Rs. 64.10
∴ Option (4) is the correct answer. | 350 | 884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.787425 |
https://www.univerkov.com/the-sum-of-two-numbers-is-72-and-one-of-them-is-3-times-the-other-find-these-numbers/ | 1,718,532,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00749.warc.gz | 910,015,997 | 6,154 | # The sum of two numbers is 72, and one of them is 3 times the other. Find these numbers.
Let the smaller number be equal to X, and the larger number equal to Y.
Let us compose, from the condition, a system of two equations and solve it.
X + Y = 72;
Y = 3 * X;
Substitute Y = 3 * X in the first equation and find the value of the smaller number.
X + 3 * X = 72;
4 * X = 72;
X = 72/4;
X = 18;
Substitute the X value in the second equation.
Y = 3 * 18 = 54.
Answer: The smaller number is 18, the larger number is 54.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 216 | 837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.947278 |
https://www.bankersadda.com/2019/06/sbi-clerk-quantitative-aptitude-14.html | 1,566,787,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00449.warc.gz | 734,494,683 | 55,209 | Dear Aspirants,
Quantitative Aptitude Quiz For SBI Clerk
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Q1. A mixture contains 80 litres of alcohol. 16 litre of this mixture is taken out and same amount of water is added to the mixture. This process is repeated two times. Find quantity of alcohol in the final mixture after two processes.
51.2 li
48.4 li
46.3 li
50 li
None of these
Solution:
Q2. Nikita bought 30 kg of wheat at the rate of Rs. 9.50 per kg of wheat and the same amount of wheat at the rate of Rs. 8.50 per kg and mixed them. She sold the mixture at the rate of Rs. 8.90 per kg. Her total profit of loss in the transaction was:
Rs. 2 loss
Rs. 2 profit
Rs. 6 loss
Rs. 6 profit
Rs. 8 profit
Solution:
According to question,
CP = 30 × 9.50 + 30 × 8.5
= 30 [9.5 + 8.5]
= 30 × 18 = Rs. 540
SP = 60 × 8.90 = Rs. 534
Loss = CP – SP = 540 – 534 = Rs. 6
Q3. The speed of a boat in still water is 12 kmph. If the boat covers 36 km upstream in 4 hours, what is the speed of stream?
3 km/hr
4 km/hr
5 km/hr
2 km/hr
6 km/hr
Solution:
Let speed of stream = S km/hr
∴ (12 – S) × 4 = 36
⇒ S = 3 km/hr
Q4. In a 68 litre mixture of spirit and water, the ratio of water to spirit is 11 : 6. To make this ratio 4 : 3, how much amount of spirit should be added ?
11 l
10 l
9 l
7 l
13 l
Solution:
Q5. A milkman had 20 litres of milk, he added 5 litres of water in it which was freely available. If the cost of pure milk is Rs. 18 per litre, then what will be the profit percent of the milkman, when he sold all the mixture at CP of milk?
20%
23.5%
25%
28%
32%
Solution:
When the water is freely available and all the water is sold at the price of the milk, Then the profit percent is due to 5lit of water over 20 lit of milk
Directions (6-10): Two equations I and II are given below in each question. You have to solve these equations and give answer
Q6.
if x<y
if x>y
if x≤y
if x≥y
if x=y or no relation can be established
Solution:
Q7.
if x<y
if x>y
if x≤y
if x≥y
if x=y or no relation can be established
Solution:
Q8.
if x<y
if x>y
if x≤y
if x≥y
if x=y or no relation can be established
Solution:
Q9.
if x<y
if x>y
if x≤y
if x≥y
if x=y or no relation can be established
Solution:
Q10.
if x<y
if x>y
if x≤y
if x≥y
if x=y or no relation can be established
Solution:
Directions (11-15): Given below is the pie chart which shows the percentage expenditure issued by government on different sports in a state in year 2016
Total Expenses = 500 Lakhs
Q11. What is the ratio of expenditure on Football and Golf together to the expenditure on Hockey and Tennis together?
11 : 10
9 : 10
10 : 11
11 : 12
5 : 6
Solution:
Q12. What is the difference between average of expenditure on sports Golf, Football together to the average of expenditure on sport tennis and Hockey together ?
4.25L
10.25L
6.25L
5.5L
7.25L
Solution:
Q13. If in year 2017 expenditure on Cricket and basketball increases by 20% and 12% than the previous year respectively and ratio of expenditure between these two sports in 2016 is 2 : 1 (cricket : basketball) then find the total expenditure of these two sports in 2017. (Assume total expenses remained constant in both year)
180 L
310 L
285 L
220 L
170 L
Solution:
Q14. If expenditure on football and Hockey increases 20% and 25% in 2017 than that in 2016 respectively then what is the total expenditure for these two sports in 2017? (Assume total expenses remained constant in both year)
120 .65 L
170.50
183.75
190.00
201.5
Solution:
Q15. Total expenditure of Tennis and football together in 2016 is what percent of total expenditure on all, sports in 2017 if in 2017 expenditure on all sports increases by 20% than that in 2016?
Solution:
You May also like to Read: | 1,233 | 4,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-35 | latest | en | 0.931673 |
https://undergroundmathematics.org/sequences/achilles-and-the-tortoise | 1,721,310,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00152.warc.gz | 508,600,231 | 5,862 | Sequences
Bigger picture
The paradoxes of the ancient Greek philosopher Zeno, born approximately 490BC, have puzzled mathematicians and scientists for millennia. Perhaps the most famous paradox is that of the race between Achilles, the swift-footed warrior, and a lumbering tortoise. Sure of his superiority, Achilles gives the tortoise a head start. Zeno’s argument “proves” that Achilles will never overtake the tortoise. Suppose the tortoise starts at point $A$. By the time Achilles gets there the tortoise has moved on to $B$. By the time Achilles gets to $B$, the tortoise will have moved on to $C$. And so on. Whenever Achilles gets to where the tortoise was a moment ago, the tortoise will have moved on, so Achilles never catches up.
To pick this apart, let’s assume that Achilles and the tortoise both move at constant speed. Achilles’ speed is $100$ metres per minute and the tortoise’s speed is $1$ metre per minute (the actual numbers don’t matter). Achilles is $100$ times faster than the tortoise, so let’s give the poor animal a very large head start: $100$m. Now by the time Achilles has travelled the $100$m to $A$ the tortoise has moved $1$m to point $B$ (because it’s $100$ times slower than Achilles). When Achilles gets to $B$, the tortoise has moved $0.01$m to $C$, etc.
Adding up the infinitely many distances Achilles has to traverse before he catches up with the tortoise, we get
$100+1+\frac{1}{100} + \frac{1}{10000} + \frac{1}{1000000} + \dotsb = 100 + 1 + \frac{1}{100} + \frac{1}{100^2} + \frac{1}{100^3} + \dotsb.$
This is a geometric series which sums to $101.010101\dots$ metres.
Since he is travelling at constant speed Achilles can cover that finite distance in a finite amount of time ($1$ minute and $0.6$ seconds)—after that time he will have caught up with the tortoise.
The flaw in Zeno’s argument was the unstated assumption that the infinite sum of distances (or equivalently the infinite sum of time periods needed to traverse each distance) cannot be finite.
Below is a short animated video which tells this tale. Please use this video link if you require full screen.
© The Open University 2011. Original available here. Licence under creative commons by-nc-sa. | 552 | 2,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.911307 |
http://questions.instantgrades.com/ronald-lau-chief-engineer-at-south-dakota-electronics-has-to-decide-whether-to-build-a-new-state-of-the-artprocessing-facility-if-the-new-facility-works-the-company-could-realize-a-profit-of-200-2/ | 1,534,335,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210105.8/warc/CC-MAIN-20180815122304-20180815142304-00149.warc.gz | 372,213,969 | 5,608 | Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality?
# Ronald Lau, chief engineer at South Dakota Electronics, has to decide whether to build a new state-of-the-artprocessing facility. If the new facility works, the company could realize a profit of \$200,000. If it fails, South Dakota Electronics could lose \$180,000. At this time, Lau estimates a 60% chance that the new process will fail. The other option is to build a pilot plant and then decide whether to build a complete facility. The pilot plant would cost \$10,000 to build. Lau estimates a 50-50 chance that the pilot plant will work. If the pilot plant works, there is a 90% probability that the complete plant, if it is built, will also work. If the pilot plant does not work, there is only a 20% chance that the complete project (if it is constructed) will work. Lau faces a dilemma. Should he build the plant? Should he build the pilot project and then make a decision? Help Lau by analyzing this problem.
Ronald Lau, chief engineer at South Dakota Electronics, has to decide whether to build a new state-of-the-artprocessing facility. If the new facility works, the company could realize a profit of \$200,000. If it fails, South Dakota Electronics could lose \$180,000. At this time, Lau estimates a 60% chance that the new process will fail. The other option is to build a pilot plant and then decide whether to build a complete facility. The pilot plant would cost \$10,000 to build. Lau estimates a 50-50 chance that the pilot plant will work. If the pilot plant works, there is a 90% probability that the complete plant, if it is built, will also work. If the pilot plant does not work, there is only a 20% chance that the complete project (if it is constructed) will work. Lau faces a dilemma. Should he build the plant? Should he build the pilot project and then make a decision? Help Lau by analyzing this problem.
Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality? | 465 | 2,054 | {"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} | 2.859375 | 3 | CC-MAIN-2018-34 | longest | en | 0.930803 |
https://encyclopediaofmath.org/index.php?title=Liouville_normal_form&printable=yes | 1,675,358,109,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00469.warc.gz | 257,095,941 | 6,387 | # Liouville normal form
$\newcommand{deriv}[2]{\frac{\mathrm{d}#1}{\mathrm{d}#2}} \newcommand{derivn}[3]{\frac{\mathrm{d}^{#3}#1}{\mathrm{d}#2^{#3}}}$
The Liouville normal form is a way of writing a second-order ordinary linear differential equation $$\label{eq1} \derivn{y}{x}{2} + p(x)\deriv{y}{x} + \left( q(x) + \lambda r(x) \right) y = 0,$$ in the form $$\label{eq2} \derivn{\eta}{\xi}{2} + \left( \lambda + \phi(\xi) \right) \eta = 0,$$ where $\lambda$ is parameter. If $p(x) \in C^1$, $r(x) \in C^2$ and $r(x) > 0$, then equation \ref{eq1} reduces to the Liouville normal form \ref{eq2} by means of the substitution $\eta(\xi) = \Phi(x)y(x),\quad \xi = \int_\alpha^x \sqrt{r(t)}\,\mathrm{d}t, \quad \Phi(x) = r(x)^{1/4} \exp\left( \frac{1}{2}\int_\alpha^x p(t)\,\mathrm{d}t \right),$ which is called the Liouville transformation (introduced in [Li]). The Liouville normal form plays an important role in the investigation of the asymptotic behaviour of solutions of \ref{eq1} for large values of the parameter $\lambda$ or the argument, and in the investigation of the asymptotics of eigenfunctions and eigenvalues of the Sturm–Liouville problem (see [Ti]). | 407 | 1,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-06 | latest | en | 0.590888 |
https://www.upscsyllabus.in/review-concepts/gravitation | 1,606,169,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00120.warc.gz | 896,289,708 | 4,883 | Gravitation
Gravity: The gravitational force due to earth is called gravity.
Newton's law of gravitation: In this universe, all things attract each other. The force of attraction acting between two bodes of masses m1 and m2 separated by a distance r is directly proportional to product of their masses and inversely proportional to square of the separation between them.
F = Gm1m2 / r2
Here, G is called Gravitation constant. Gravitational forces are very weak forces.
G = 6.67 × 10-11 Nm2 × kg-2
Relation between g (Acceleration due to gravity) and G (gravitation constant)
F = GMm/r2 = mg
g = GM/r2
The value of g near the surface of the earth is taken as 9.8 ms-2. The value of acceleration due to gravity (g) is maximum at poles and minimum at equator.
Motion of object under gravity:
When a body is falling down, then a is replaced by g
1. v = u + gt
2. s = ut + ½gt2
3. v2 = u2 + 2gs
When a body is thrown up, then a is replaced by (-g).
Buoyancy: Whenever a body is immersed in a fluid, an upward force is exerted by the fluid on the body, called force of buoyancy or buoyant force. This is also known as upthrust. The magnitude of the buoyant force acting on a body at a given place depends on the density of the liquid and volume of the body immersed in the fluid.
Archimedes Principle: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to weight of the fluid displaced by it. | 393 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-50 | latest | en | 0.901223 |
http://gmatclub.com/forum/gmat-diagnostic-test-quest-for-best-gmat-diagnostic-test-79502-80.html?kudos=1 | 1,485,251,239,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00298-ip-10-171-10-70.ec2.internal.warc.gz | 119,540,220 | 56,235 | GMAT Diagnostic Test - Quest for Best GMAT Diagnostic Test : General GMAT Questions and Strategies - Page 5
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GMAT Diagnostic Test - Quest for Best GMAT Diagnostic Test
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26 Dec 2009, 04:27
bb thanks a lot...u r the best!!!!
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02 Jan 2010, 22:25
I did so bad at the diagnostic test that it's not funny.
I got a 20 on the 45...can you beat that!!!! I always thought Quant was my (relative) strength.
With less than a month to go for my test..I AM FREAKING OUT!!!!!
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04 Jan 2010, 06:40
Don't take it too hard. Identify your weak areas and improve them. Sorry if it sounds boring . Don't give up!
anil4410 wrote:
I did so bad at the diagnostic test that it's not funny.
I got a 20 on the 45...can you beat that!!!! I always thought Quant was my (relative) strength.
With less than a month to go for my test..I AM FREAKING OUT!!!!!
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04 Jan 2010, 11:24
anil4410 wrote:
I did so bad at the diagnostic test that it's not funny.
I got a 20 on the 45...can you beat that!!!! I always thought Quant was my (relative) strength.
With less than a month to go for my test..I AM FREAKING OUT!!!!!
I got 21 (61%) on my my diagnostic test. Then I got 26 (70%) on m25 and 30 (81%) on m01. So, I think you can improve fast too. Don't worry.
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06 Jan 2010, 12:43
This is great. Thanks!!!
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07 Jan 2010, 13:34
Thanks this is great, it will help me get started
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20 Jan 2010, 08:06
We are glad our Diagnostic Test helped you! Congratulations on a very good score and a first post on the forums. Welcome to GMAT Club!
soundarya48 wrote:
Hi,
I took the GMAT recently and received a score of 730. Your math test was very helpful in preparing me for the real thing. Kudos to all of you for putting together such a (dare I say it) beautiful array of math questions that was key to helping me perform so well.
Unfortunately, I am not that good at writing math questions! But I will submit at least a few verbal questions you guys can use later on. This is my own small way of giving back;
Arya
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23 Jan 2010, 22:04
I am about 2 months preparing into GMAT and took this test today. Surprisngly I find my errors are mostly on questions at 600-650 level and most of the 700/750 ones are correct. Is there anything to interpret, like taking easy questions too lightly and missing some details?
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24 Jan 2010, 07:17
Great work done ...thnx bb
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25 Jan 2010, 04:13
I think you should give the questions you got wrong a closer inspection. If it was a careless mistake, you might be taking easy questions too lightly. If an error had to do with a concept tested in the question, you'll have to review that specific concept. It's also a good idea to use an Error log (if you are not using one yet).
siddhartho wrote:
I am about 2 months preparing into GMAT and took this test today. Surprisngly I find my errors are mostly on questions at 600-650 level and most of the 700/750 ones are correct. Is there anything to interpret, like taking easy questions too lightly and missing some details?
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25 Jan 2010, 23:13
This is fantastic! Thanks!!!
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26 Jan 2010, 10:58
Thanks. This test is awesome!
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01 Feb 2010, 10:29
Thank you for the diagnostic, it allowed me to evaluate where I stand in the 600-700 level questions.
What would you suggest if I get only about 1/4 of these questions correctly?
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01 Feb 2010, 11:22
Hi and welcome to GMAT Club!
We are glad you find our Diagnistic Test useful. Here are some links that might very helpful for you:
top-gmat-prep-books-guides-reviews-comments-77703.html -- Reviews of the best GMAT books
best-gmat-math-prep-books-reviews-recommendations-77291.html -- GMAT Math books
Our forums have a multitude of resources that you should take time to explore. Also, check out the links at the top of every page (blue drop down menu).
StevencyTelfort wrote:
Thank you for the diagnostic, it allowed me to evaluate where I stand in the 600-700 level questions.
What would you suggest if I get only about 1/4 of these questions correctly?
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05 Feb 2010, 09:00
What is the level of questions in the Diagnostic test?!
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05 Feb 2010, 14:52
The level ranges from ~600 to ~750. You can see these approximate level figures in discussions dedicated to each of the questions. See the links in the top post (each question has its own discussion).
I hope this test helps you identify your weak areas and work on improving them .
ProgressiveDreamer wrote:
What is the level of questions in the Diagnostic test?!
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05 Feb 2010, 14:58
dzyubam wrote:
The level ranges from ~600 to ~750. You can see these approximate level figures in discussions dedicated to each of the questions. See the links in the top post (each question has its own discussion).
I hope this test helps you identify your weak areas and work on improving them .
ProgressiveDreamer wrote:
What is the level of questions in the Diagnostic test?!
lot's of weak areas.
to be honest, I need 650-700
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05 Feb 2010, 15:13
I believe you will find a lot of helpful resources in here. This thread is the one you might want to read through:
ProgressiveDreamer wrote:
dzyubam wrote:
The level ranges from ~600 to ~750. You can see these approximate level figures in discussions dedicated to each of the questions. See the links in the top post (each question has its own discussion).
I hope this test helps you identify your weak areas and work on improving them .
ProgressiveDreamer wrote:
What is the level of questions in the Diagnostic test?!
lot's of weak areas.
to be honest, I need 650-700
_________________
Welcome to GMAT Club!
Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?
Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.
GMAT Club Premium Membership - big benefits and savings
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06 Feb 2010, 00:40
dzyubam,
thanks buddy,
btw, what is the level of math PS exercises in OG12?
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07 Feb 2010, 21:40
ProgressiveDreamer wrote:
dzyubam,
thanks buddy,
btw, what is the level of math PS exercises in OG12?
I would say from 300-700 tops (mostly 500 problems i say)... Don't get overconfident w/ the OG books...
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Re: GMAT Diagnostic Test - Quest for Best GMAT Diagnostic Test [#permalink] 07 Feb 2010, 21:40
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,716 | 13,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-04 | latest | en | 0.869452 |
https://www.helpwithassignment.com/cost-volume-profit-analysis-assignment-help/ | 1,726,403,164,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00538.warc.gz | 749,209,098 | 22,984 | # Cost Volume Profit Analysis Assignment & Homework Help Online
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What is CVP Analysis?
Cost Volume Profitability Analysis (CVP Analysis) is the study of the relationship between revenues, costs and profits of a business. The analysis is used to examine the relationship among the total volume of an independent variable, total costs, total revenues and profits for a time period. Cost-volume-profit analysis is useful in the early stages of planning because it provides an easily understandable framework for discussing planning issues and organizing relevant data.
CVP analysis begins with the basic profit equation
Where, Profit = Total Revenue – Total Costs
Separating costs into variable costs and fixed costs, we express profit as:
Profit = Total Revenue – Total Variable Costs – Total Fixed Costs
Profit + Fixed Costs = Sales – Variable Costs or Units Sold ×(Unit Sales Price – Unit Variable Cost)
Contribution margin = Sales –Variable Costs Unit
Contribution margin = Unit Sales Price – Unit Variable Cost
Quantity = Fixed Cost ÷ Unit Contribution margin
If we assume the selling price and the variable cost per unit to be constant, then the total revenue is equal to price multiplied by quantity and total variable cost is equal to the variable cost per unit multiplied by quantity.
Profit = SP × Q – V × Q – F = (P – V) × Q – F
Where, SP = Selling Price
V = Variable cost per unit
P – V = Contribution margin per unit
Q = Quantity of products sold
F = Total Fixed Cost
Profit equation: Profit = (P – V) × Q – F
For quantity Q =(F+Profit)/((P-V) )
If a company wants to produce an electronic appliance with the following information
Price per unit = \$500
Variable cost per unit = \$300
Fixed cost related to the production = \$1,000,000
Target Profit = \$200,000
Estimated Sales = 9,000 units.
We can determine the minimum quantity the company has to sell in order to earn the profit will be equal to
Quantity = (\$1000,000 + \$200,000) ÷ (\$500–\$300) = 6000 units.
Managers often want to know the level of activity required to break even. A CVP analysis can be used to determine the break even point or level of operating activity at which revenues cover all fixed and variable costs, resulting in zero profit. Knowing about this point is very crucial for the business break-even point is the point that a new company wants to reach as quickly as possible in order to cover all the costs and start making real profits.
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4.7/5 rating | 1M+ happy students | Great tutors 24/7 | 11+yrs exp in academic writing | 1,244 | 5,887 | {"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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.847986 |
http://www.ehow.com/how_7831106_make-net-pyramid-hexagon-base.html | 1,503,318,692,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00670.warc.gz | 529,911,120 | 19,712 | # How to Make a Net for a Pyramid With a Hexagon Base
Save
Pyramid nets are very simple to make, and for the amount of time they take to construct, the educational value is a profit. With a few tools and measurements you can discover new ideas about symmetry and why pyramids, hexagonal or otherwise, are so mystical and beautiful. The measurements below are suggestions only. You can make a net with your own measurements as long as they follow the following guidelines: each side of the hexagon must be the same length, and each line drawn from the middle of each side must be the same length and longer than the sides of the hexagon.
### Things You'll Need
• Paper
• Pencil
• Ruler
• Protractor
• Scissors
• Draw a line 4 inches long on a sheet of paper. Each side of the hexagon will be 4 inches long.
• Match the base of your protractor to one end of the line you just drew and draw another 4 inch line at 120 degrees. You want to have an obtuse inside angle, the angle that will eventually be on the inside of the hexagon, so make sure you don't create a 120 degree angle in the wrong direction.
• Repeat this process for each new line until you complete the hexagon.
• Draw a 6-inch line out from the middle of each side of the hexagon.
• Draw two straight lines from the tip of one of the 6-inch lines that meets the corners of the 6-inch line's corresponding side -- this forms a triangle. Use your ruler as a straight edge.
• Repeat step 5 for each side of the hexagon.
• Cut out the net along the outside of the pattern you have created using scissors or a sharp edge.
## References
• Photo Credit pyramid image by cherie from Fotolia.com
Promoted By Zergnet
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M
Is DIY in your DNA? Become part of our maker community. | 615 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-34 | latest | en | 0.90677 |
https://thaipattaraspa.com/qa/quick-answer-can-you-feel-one-newton-force.html | 1,610,828,009,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00256.warc.gz | 596,817,851 | 8,132 | # Quick Answer: Can You Feel One Newton Force?
## How do you calculate Newton force?
The formula for force says force is equal to mass (m) multiplied by acceleration (a).
If you have any two of the three variables, you can solve for the third.
Force is measured in Newtons (N), mass in kilograms (kg), and acceleration in meters per second squared ( m/s2 )..
## Is 1 Newton a lot of force?
It is named after Isaac Newton in recognition of his work on classical mechanics, specifically Newton’s second law of motion. One newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared in the direction of the applied force.
## How much Newton is a punch?
Force unleashed When it comes to unleashing force quickly, Bir and her colleagues investigated boxers and found they could generate up to 5,000 newtons of force with a punch, more than that exerted down by a half-ton on Earth’s surface.
## What force is required to lift the 100 pound weight?
If you want to lift something that weighs 100kg, you have to pull down with a force equivalent to 100kg, which is 1000N (newtons). If you want to raise the weight 1m into the air, you have to pull the loose end of the rope a total distance of 1m at the other end.
## How much is 200 newtons of force?
Convert 200 Newtons to Pounds force200 Newtons (N)44.962 Pounds force (lb)1 N = 0.224809 lb1 lb = 4.448220 N
## How much is 1000 pounds of force?
A kip is a US customary unit of force. It equals 1000 pounds-force and is used primarily by architects and civil engineers to indicate engineering loads where the pound-force is too small a unit. Although uncommon, it is occasionally also considered a unit of mass, equal to 1000 pounds, i.e., one half of a short ton.
## How many kN are in a ton?
9.80665 kilonewtonsThe tonne (metric ton, abbreviation: t) is the unit of mass in the metric system. 1 tonne-force (tf) = 9.80665 kilonewtons (kN) = 1000 kilogram-forces (kg).
## What is kN in KG?
101.97160051 kilonewton is equal to 101.9716005 kilograms, which is the conversion factor from kilonewtons to kilograms.
## What does 1 newton of force feel like?
A newton is the force that an average sized apple makes on your hand when you hold it. … It is one newton. On the moon, an object with the mass of a brick would feel as light as an apple on earth due to the moon’s lower gravity. The force of the brick in your hand would feel like one newton.
## How many Newtons would it take to lift 100 pounds?
Please share if you found this tool useful:Conversions Table2 Pounds-force to Newtons = 8.896480 Pounds-force to Newtons = 355.85773 Pounds-force to Newtons = 13.344790 Pounds-force to Newtons = 400.33994 Pounds-force to Newtons = 17.7929100 Pounds-force to Newtons = 444.822212 more rows
## How much is 40 newtons of force?
As it happens, one pound of force is equal to four-and-a-half Newtons. So if you push hard enough to make the scale read 9 pounds, that is about 40 Newtons.
## What does 40 kN mean?
40 Kilonewtons (kN) = 4,079 Kilograms Force (kgf) 1 kN = 101.972 kgf.
## How many Newtons does it take to lift a ton?
1 Newton: 1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton’s second law f=ma and assuming Earth gravity of 9.80665 m/s2….Please share if you found this tool useful:Conversions Table10 Tonnes to Newtons = 98066.5800 Tonnes to Newtons = 784532014 more rows
## What do you mean by one newton force?
It is defined as that force necessary to provide a mass of one kilogram with an acceleration of one metre per second per second. One newton is equal to a force of 100,000 dynes in the centimetre-gram-second (CGS) system, or a force of about 0.2248 pound in the foot-pound-second (English, or customary) system.
## How do you feel force?
The use of force in our everyday life is very common. We use force to walk on the road, to lift the objects, to throw a cricket ball, or to move a given body by some particular speed or direction. We are very familiar with the various effects of force. We can exert pull and push.
## What is Newton formula?
Force (Newton) = Mass of body × Acceleration. Or, F = [M1 L0 T0] × [M0 L1 T-2] = M1 L1 T-2.
## What does 1000 Newtons feel like?
A thousand newtons of force is what you feel if a very muscular or rather fat person (weighing in at about 100 kg or 220 lbs) is standing on you. It is also the force needed to lift that person.
## What is 1 kg in Newtons?
9.80665Kilogram is the base unit of mass in the metric system. 1 kilogram or kilogram-force (kg or kgf) = 9.80665 newtons (N) = 35.2739619 ounces (oz) = 1000 grams (g) = 1000000 milligrams (mg) = 2.20462262 pounds (lbs) = 0.157473044 stones (st). | 1,295 | 4,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | latest | en | 0.930117 |
https://code.jsoftware.com/wiki/Help/Release/J_4.06/mampv_Extended | 1,719,233,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00296.warc.gz | 151,202,409 | 6,234 | # Help / Release / J 4.06 / m&v Extended
m&v and u&n Extended initial writing: 2000-11-24last updated: 2004-10-26
The dyads m&v and u&n (a verb fixed with one of its arguments), previously undefined, are now defined as follows:
x m&v y m&v^:x y
x u&n y u&n^:x y
That is, the verb m&v (u&n) is applied x number(s) of times to the argument y . For example:
``` ] x=: _10+i.4 5
_10 _9 _8 _7 _6
_5 _4 _3 _2 _1
0 1 2 3 4
5 6 7 8 9
x (3&+) 0
_30 _27 _24 _21 _18
_15 _12 _9 _6 _3
0 3 6 9 12
15 18 21 24 27
x (*&2) 1
0.000976563 0.00195313 0.00390625 0.0078125 0.015625
0.03125 0.0625 0.125 0.25 0.5
1 2 4 8 16
32 64 128 256 512
x (*&2) 1x
1r1024 1r512 1r256 1r128 1r64
1r32 1r16 1r8 1r4 1r2
1 2 4 8 16
32 64 128 256 512
\$ x (*&2) i.7
4 5 7
0 1 2 4 8 16 32 64 _ (2&o.) 1
1 0.540302 0.857553 0.79348 0.750418 0.739567 0.739086 0.739085 0.739085
``` | 509 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.318751 |
https://topic.alibabacloud.com/zqpop/double-precision-floating-point-calculator_31375.html | 1,709,421,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00545.warc.gz | 560,666,387 | 17,440 | double precision floating point calculator
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 916 | 4,143 | {"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} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.848595 |
https://brainmass.com/math/algebra/formula-to-determine-variable-of-gross-profit-35808 | 1,701,434,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00652.warc.gz | 182,841,719 | 7,440 | Purchase Solution
# Formula to determine variable of gross profit %
Not what you're looking for?
Attached is a spreadsheet with two accounts. Each account has the same retail and invoice. One school has a small enrollment and the other has a larger enrollments. There is a handling fee of \$2 per enrollment. There are only two variables that can be changed: brochure cost & incentives. Also, the larger the enrollment, the larger the brochure cost. Brochure Cost can be reduced only a certain amount. The cheapest it can be reduced is 110% of # of students. So for John Spicer, the Brochure Cost could be reduced to 683.10 (621 students x 1.10 = \$683.10)
The goal is to reach a 36% gross profit margin...which is (profit / invoice). Profit is Invoice - Total Expenses
There should be a way to figure a lump sum of the amount you have to spend on brochures and incentives. For example of Spicer, you can spend \$843.10 on brochures and incentives....which is \$683.10 on brochures and \$160.00 on incentives.
I can figure adjust brochure cost & incentive cost to reach the 36% but I know there is a formula to make it easier.
##### Solution Summary
This solution is comprised of a detailed explanation to determine variable of gross profit %.
##### Solution Preview
Hello,
In order to solve this problem, let's first write the definitions of Total Expenses, Profit and Gross Profit %.
Total Expenses = Brochure Cost + Product Cost + Freight Cost + Prizes + Incentives +
+ Techer Gifts + Samples + PackOut + Delivery + Handling
Profit = Invoice - Total Expenses
Gross Profit % = Profit/Invoice
You are interested in finding the sum of Brochure Cost + Incentives such that the gross profit percentage is 36%. Since you mentioned that costs that are not brochures or incentives are "fixed" ...
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https://math.answers.com/Q/What_is_30_percent_of_47.90 | 1,708,514,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00653.warc.gz | 408,150,815 | 45,443 | 0
What is 30 percent of 47.90?
Updated: 9/19/2023
Wiki User
11y ago
30 percent of 47.90 is 14.37
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11y ago
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Q: What is 30 percent of 47.90?
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Related questions
What is 6 percent of 4790?
6% of 4790= 6% * 4790= 0.06 * 4790= 287.4
0.479
4790.
it is 4790 miles
5000
What is 30 percent of 30 percent?
30% of 30% = 0.09 or 9%
100 is 30 percent of what?
"Percent" = "of 100". So 30 is 30% of 100.
What are the release dates for The Bold and the Beautiful - 1987 1-4790?
The Bold and the Beautiful - 1987 1-4790 was released on: USA: 20 April 2006 Belgium: 4 February 2009
What is 30 percent of a 150 percent?
30 percent of 150 percent = 45%
What is 15 percent of 30 percent?
15% of 30% 15/100 =0.15 30/100 =0.3 0.15×0.3 = 0.045.
What is bigger 30 percent or one third?
One third is bigger. 1/3 = .333333 30% = .30
What is 55 percent of 30?
30 percent of 55 is 16.5. | 360 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.927838 |
https://betterlesson.com/lesson/resource/2651710/counting-with-base-ten-blocks-understanding-place-value-smart-notebook-file | 1,516,595,917,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00560.warc.gz | 625,216,760 | 21,627 | ## Counting with Base Ten Blocks-Understanding Place Value Smart Notebook File - Section 2: Instruction
Counting with Base Ten Blocks-Understanding Place Value Smart Notebook File
# The Super Counters-Working with Place Value with Base Ten Blocks
Unit 4: Working with Numbers 11-20
Lesson 6 of 17
## Big Idea: Students have been working extensively with ten frames, now they can use this knowledge to understand place value.
Print Lesson
6 teachers like this lesson
Standards:
Subject(s):
Math, Place Value, Critical Area, counting, base 10, Numbers
55 minutes
### Joyce Baumann
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https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-5-graphs-and-the-derivative-chapter-review-review-exercises-page-298/37 | 1,575,702,697,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00142.warc.gz | 741,824,744 | 12,435 | Chapter 5 - Graphs and the Derivative - Chapter Review - Review Exercises - Page 298: 37
$${\text{ }}f''\left( t \right) = \frac{1}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}},\,\,\,\,\,f''\left( 1 \right) = \frac{1}{{2\sqrt 2 }},\,\,\,\,f''\left( { - 3} \right) = \frac{1}{{10\sqrt {10} }}$$
Work Step by Step
\eqalign{ & f\left( t \right) = \sqrt {{t^2} + 1} \cr & {\text{write the radical }}\sqrt {{t^2} + 1} {\text{ as }}{\left( {{t^2} + 1} \right)^{1/2}} \cr & f\left( t \right) = {\left( {{t^2} + 1} \right)^{1/2}} \cr & {\text{find the derivative of }}f'\left( t \right) \cr & f'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {{t^2} + 1} \right)}^{1/2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'\left( t \right) = \frac{1}{2}{\left( {{t^2} + 1} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {{t^2} + 1} \right] \cr & f'\left( t \right) = \frac{1}{2}{\left( {{t^2} + 1} \right)^{ - 1/2}}\left( {2t} \right) \cr & f'\left( t \right) = t{\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & \cr & {\text{find the derivative of }}f'\left( t \right) \cr & f''\left( t \right) = \frac{d}{{dt}}\left[ {t{{\left( {{t^2} + 1} \right)}^{ - 1/2}}} \right] \cr & {\text{using the product rule}} \cr & f''\left( t \right) = t\frac{d}{{dt}}\left[ {{{\left( {{t^2} + 1} \right)}^{ - 1/2}}} \right] + {\left( {{t^2} + 1} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ t \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f''\left( t \right) = t\left( { - \frac{1}{2}} \right){\left( {{t^2} + 1} \right)^{ - 3/2}}\frac{d}{{dt}}\left[ {{t^2} + 1} \right] + {\left( {{t^2} + 1} \right)^{ - 1/2}}\left( 1 \right) \cr & f''\left( t \right) = t\left( { - \frac{1}{2}} \right){\left( {{t^2} + 1} \right)^{ - 3/2}}\left( {2t} \right) + {\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & f''\left( t \right) = - {t^2}{\left( {{t^2} + 1} \right)^{ - 3/2}} + {\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & {\text{factoring}} \cr & f''\left( t \right) = {\left( {{t^2} + 1} \right)^{ - 3/2}}\left[ { - {t^2} + {t^2} + 1} \right] \cr & f''\left( t \right) = {\left( {{t^2} + 1} \right)^{ - 3/2}} \cr & f''\left( t \right) = \frac{1}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}} \cr & \cr & {\text{find }}f''\left( 1 \right){\text{ and }}f''\left( { - 3} \right) \cr & f''\left( 1 \right) = \frac{1}{{{{\left( {{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}} = \frac{1}{{2\sqrt 2 }} \cr & and \cr & f''\left( { - 3} \right) = \frac{1}{{{{\left( {{{\left( { - 3} \right)}^2} + 1} \right)}^{3/2}}}} = \frac{1}{{10\sqrt {10} }} \cr}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 1,269 | 2,682 | {"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} | 4.65625 | 5 | CC-MAIN-2019-51 | latest | en | 0.420035 |
https://m.wikihow.com/Calculate-Force | 1,580,163,954,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00087.warc.gz | 546,779,845 | 57,293 | # How to Calculate Force
Author Info
Updated: April 18, 2019
Force is the "push" or "pull" exerted on an object to make it move or accelerate. Newton's second law of motion describes how force is related to mass and acceleration, and this relationship is used to calculate force. In general, the greater the mass of the object, the greater the force needed to move that object.[1]
### Part 1 of 2: Learning the Formula
1. 1
Multiply mass times acceleration. The force (F) required to move an object of mass (m) with an acceleration (a) is given by the formula F = m x a. So, force = mass multiplied by acceleration.[2]
2. 2
Convert figures to their SI values. The International System of Units (SI) unit of mass is the kilogram, and the SI unit of acceleration is m/s2 (meters per second squared). So when mass and acceleration are expressed in their SI units, we get the force in its SI units which is N (Newtons)[3]
• As an example, if the mass of the object is given to be 3 pounds, you'll need to convert those pounds to kilograms. 3 pounds make 1.36 kg, so the mass of the object is 1.36 kg.
3. 3
Keep in mind that weight and mass mean different things in Physics. If the weight of an object is given in N (Newtons), then divide it by 9.8 to get the equivalent mass. For example, 10 N weight is equivalent to 10/9.8 = 1.02 kg.[4]
### Part 2 of 2: Using the Formula
1. 1
Find the force that is required to accelerate a 1,000 kg car at 5 m/s2.[5]
• Check to make sure all your values are in the correct SI unit.
• Multiply your acceleration value (1000 kg) by 5 m/s2 to calculate your value.
2. 2
Calculate the force required for an 8 pound wagon to accelerate at 7 m/s2.
• First, convert all your units to SI. One pound is equal to .453 kg, so you'll need to multiply that value by your 8 pounds to determine the mass.
• Multiply your new value for the mass (3.62 kg) by your acceleration value (7 m/s2).
3. 3
Find the magnitude of force acting upon a cart weighing 100 N and accelerating at the rate of 2.5 m/s2.
• Remember, 10 N is equal to 9.8 kg. So, convert Newtons to kg by dividing by 9.8 kg. Your new kg value should be 10.2 kg for the mass.
• Multiply your new mass value (10.2 kg) times the acceleration (2.5 m/s2).
## Community Q&A
Search
• Question
How do I change Newtons into mass?
Tiagoroth
The formula for force is force = mass * acceleration. To find mass, simply divide the force by the acceleration.
• Question
Is force the same as weight?
Weight is a force. When a force is due to gravity, it can be called "weight". "Weight" is only a human distinction for a specific case.
• Question
What is the acceleration of a 130 kg object push by a man with 650 newtons of force?
You will need to know the formula of acceleration, once you know that. You can easily get the acceleration. So by using this formula, you can figure out what acceleration. It is simply just Algebra in this case. Formula: f/m=a
• Question
How can I move an 800 kg mass 5 meters in 1 second?
x = 5 m, v = (5 m - 0 m) / 1 s = 5 m/s, a = (5 m/s - 0 m/s) / 1 s = 5 m/s^2, F = ma, F = 800 kg * 5 m/s^2 = 4000 N. You must apply 4000 Newtons of force.
• Question
How do I calculate tension?
You have to use a system of equations which are your summation equations. They come from the diagram and coordinate system that you drew.
• Question
What is the formula for momentum?
Momentum is the product of mass and velocity and is represented by the letter "P." The formula is P=mv.
• Question
If we are pushing a book one foot over the table, how much force are we exerting on the book?
It depends on the mass of that book.
• Question
How much force is required to push or pull an object of 2kg mass?
Given that the gravitational acceleration is about 9.8m/s^2 and without more info we cannot go further than: F>2kg*9.8m/s^2. Therefore F>16.9 N (upwards movement). Being F=16.9N the force needed to sustain the 2kg object standstill.
• Question
What is the formula in computing combining forces?
To compute two combining forces, just add the two forces. F1 + F1 is m1 × v1 + m2 × m2.
• Question
What is the amount of force in a moving object?
The amount of force is the mass of the object multiplied by the acceleration of the object when the force is exerted while moving.
• How much kinetic force is required to move 30 tonnes?
• What is the formula when calculating magnitude of a force?
• How do I calculate the radial load on a rotating shaft with a diameter of 350 mm and a speed of 4000 rpm?
• How do I calculate the magnitude of force?
• Does this equation work for arrested weight? If I lower a weight down onto a winch and cut the power, can I work out the force with this equation (f=mxa)?
200 characters left
## Tips
• Always read a question carefully to determine whether weight or mass is given.
Thanks!
• Check to be sure all numbers have been converted to kilograms and m/s^2.
Thanks!
• The definition of a Newton, the standard unit of force, is N = kg * m/s^2.[6]
Thanks!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 29 people, some anonymous, worked to edit and improve it over time. Together, they cited 6 references. This article has also been viewed 1,088,856 times.
50 votes - 73%
Co-authors: 29
Updated: April 18, 2019
Views: 1,088,856
Categories: Classical Mechanics
Article SummaryX
To calculate force, use the formula force equals mass times acceleration, or F = m × a. Make sure that the mass measurement you’re using is in kilograms and the acceleration is in meters over seconds squared. When you’ve solved the equation, the force will be measured in Newtons. Now, simply plug the values you know into the equation and solve. If you need to find acceleration, find the difference between the start and final velocity and divide them by the time difference. If you want to learn how to convert weights to mass for your equation, keep reading the article! | 1,558 | 5,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-05 | longest | en | 0.886755 |
http://www.jiskha.com/display.cgi?id=1298415380 | 1,498,134,138,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319265.41/warc/CC-MAIN-20170622114718-20170622134718-00075.warc.gz | 647,570,523 | 3,688 | # math
posted by on .
Robert feeds his dog 1lb.4oz ofdog food a day. How long will a 40 pound bag last? Please show steps
• math - ,
1 pound = 16 ounces
4 oz = 1/4 = 0.25 pound
40/1.25 = ? days | 71 | 199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-26 | latest | en | 0.893511 |
https://hellogiggles.com/frustrating-gem-puzzle/ | 1,670,638,491,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00823.warc.gz | 311,152,057 | 16,855 | # The Internet is super frustrated with this puzzle. Can you solve it?
After a long night out on the town with your besties, the last thing you probably want to face is a complicated logic puzzle printed on the coaster under your cocktail. But at the Storm Crow Tavern in Vancouver, B.C., that’s exactly what they’ve done — and the solution is straight up diabolical.
At first glance, the puzzle seems perfectly simple: Follow the path from the entrance to the exit, alternating between blue and red gems as you go. Yet after a few tries, it’s obvious that the answer isn’t going to be as straightforward as it seems. The puzzle has been posted multiple times in the /r/puzzles subreddit, and every time it leaves at least a few people totally baffled. Can you figure it out, Gigglers?
Click over to the next page for the solution!
As you can see, the solution itself isn’t actually all that difficult — but it does require you to make at least a few assumptions about the “rules” of the game. For example, a few redditors assumed that the orbs “disappeared” after you went through them (which would make the puzzle super easy); or that you had to pass through every single orb. Personally, I didn’t know that passing through the same orbs twice was allowed, which made me think the whole thing was a cruel joke.
Either way, we think it’s a fun way to get your brain working after you’ve had too much egg nog.
(Images via imgur/HelloGiggles.)
There’s a panda in this picture, and people just cannot find it
Two new math problems are driving everyone on the Internet bonkers | 359 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.965614 |
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Chapter 4: Multiply and Divide Fractions;Lesson 1: Estimate Products of Fractions
### Estimate each product. Use a bar diagram if needed.
• Question 1
• $$\large \frac{1}{8}\times15\approx$$
• Question 2
• $$\large\frac{2}{5}\text{ of } 26\approx$$
• Question 3
• $$\large\frac{1}{5}\times\large\frac{8}{9}\approx$$
• Question 4
• $$\large6\frac{2}{3}\times\large4\frac{1}{5}\approx$$
• Question 5
A border is made of $$32\frac{2}{3}$$ bricks that are $$1\frac{1}{6}$$ feet long.About how long is the border?
• Question 6
A kitchen measures $$\large24\frac{1}{6}$$ feet by $$\large9\frac{2}{3}$$ feet. Estimate the area of the kitchen.
• Question 7
Building on the Essential Question Why is estimating products of fractions useful?
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This is a 23 page packet full of activities to supplement a mystery or detective theme for grades K-2. Watch out Sherlock Holmes! Graphics by melonheadz @ http://melonheadzillustrating.blogspot.com/ included: 2 forensic science activities mystery vocab sheets- can be used a matching game several
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Mystery Science: Solve a Mystery with Scientific Inquiry I was looking for a CSI unit to use with my third graders, but everything I found relating to CSI was at a middle school level so I created this unit. This is a science mystery unit where students use the concepts of the scientific method t
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The aim of this unit is to enable learners to use a variety of scientific processes to analyse evidence gathered from a crime scene. The fascination with crime and how science can be used to solve crimes continues to grow, and many learners are interested in the techniques that are used within the a
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I'm so excited about this unit! An integrated unit for Narrative Writing and Evidence and Investigation. • Beginning with Mysteries and analyzing the elements of a mystery story using a variety of picture books and novels. • Mystery Vocabulary • Mini-Lessons on: ~ Plot Structure
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Plug & Play Forensics - Take your classroom to the next level with this complete NO-PREP unit on forensic ballistics. Whether you have a paperless classroom or just a projector and a copy machine, this unit is for you! You will save HOURS of prep time because this unit is ready to PLUG &
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With this purchase you will get an entire unit for An Introduction to Forensics. Information includes examining crime scenes and understanding the law. I don't always use all of the activities, but they are all available for you: First Day Activities Presentation and Worksheet (Trevon Martin Case)
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Save time and valuable energy with this HUGE bundle of Intro to Forensics resources. This product is just what you need to teach an engaging and student-led Forensics course to your high school or middle school students. This no-prep resource is complete with everything that you'll need to teach a 9
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This bundle includes three days worth of activities based upon the Legal System unit in Forensics. Included are a PowerPoint with guided notes; QR Code Activity for vocabulary words; a 9-Vocabulary Activity with cards, student sheet and key; an Illegal/Legal Searches Activity as well as a KWL Chart
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This no-prep, COMPLETE FORENSICS CURRICULUM bundle is all of the PowerPoint presentations, notes, labs, activities, case studies, practices, homework, projects, quizzes, review games and tests you need to teach an entire year of FORENSICS. Every major Forensics unit will be covered and are listed be
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Mystery Writing with Evidence and Investigation integrated unit. Check out this free sample of a few of the pages. This is such a fun unit analyzing mystery stories and then writing our own mysteries while incorporating real evidence and investigating tools and techniques.
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A nefarious villain has stolen the cake from the Play Outside Forever group meeting. Who could have done this horrible deed?! Crimes big and little happen every day (like stolen cakes). But how do you solve them? Maybe you find some clues but what then? Science to the rescue!!! Forensic science i
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This interactive notebook unit teaches the cell membrane, transport, and solutions. It includes the following items: •Cell Membrane Cover tab page •Cell transport vocabulary pages •Vocabulary Squares activity •Phospholipid Bi-Layer terms page •Instagram a membrane •The Plasma Membrane Parts •Pl
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Make teaching Forensics easy with this fully editable combined unit and daily lesson plan that covers 6 blocks or 10 periods of a secondary forensic science course. SIOP, Gifted, and Differentiation strategies are embedded and easy identified throuought. Lessons are differnetiated accoding to learn
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My Brother Didn't Do It! is a short story which brings to life the basic Evidence and Investigation curriculum. The 11 page short story engages as Tom tries to prove his brother Ryan is innocent of theft. Terms such as alibi, reasonable doubt, circumstantial evidence come to life in this story. 2
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If you're looking for a creative way to teach critical thinking skills, Common Core Standards and STEM principles to middle school students using a hands-on, real-world approach, then this is the unit for you! Full of tips, resources, ideas and printable, editable notebook pages for your students,
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,137 | 8,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-04 | latest | en | 0.876729 |
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Scheda Riassuntiva
Anno Accademico 2019/2020 Scuola Scuola di Ingegneria Industriale e dell'Informazione Insegnamento 093269 - DISCRETE MATHEMATICS Docente Notari Roberto Cfu 5.00 Tipo insegnamento Monodisciplinare
Corso di Studi Codice Piano di Studio preventivamente approvato Da (compreso) A (escluso) Insegnamento
Ing Ind - Inf (Mag.)(ord. 270) - MI (474) TELECOMMUNICATION ENGINEERING - INGEGNERIA DELLE TELECOMUNICAZIONI*AZZZZ093269 - DISCRETE MATHEMATICS
Ing Ind - Inf (Mag.)(ord. 270) - MI (481) COMPUTER SCIENCE AND ENGINEERING - INGEGNERIA INFORMATICA*AZZZZ093269 - DISCRETE MATHEMATICS
Obiettivi dell'insegnamento
The course is an introduction to the fundamental notions of Discrete Mathematics, one of the fastest growing areas of modern mathematics. These notions are nowadays more and more used in the development, analysis and comprehension of mathematical models in every area of engeneering. The course is an introduction to the various topics that go under the name of discrete mathematics, and they range from graph theory, to elementary number theory and modular arithmetics, to abstract algebra, and to enumerative combinatorics.
Risultati di apprendimento attesi
The classes and the exercise sessions will allow students to get acquaintance with the topics of the course and with some applications of the topics to the resolution of practical problems. In more details, students will learn how to:• use formal rules for computations on symbols, independently from the meaning of the symbols;• solve equations over the integers or over the integers modulo m;• compute the number of elements of sets defined by a property;• analyze graphs. Students will be able to justify the effectiveness of mathematical procedures used to solve the problems presented in the course and to properly use the mathematical language while arguing.
Argomenti trattati
ELEMENTARY NUMBER THEORY AND MODULAR ARITHMETIC: Basic notions of set theory. Natural numbers and the induction principle. Integers. Divisibility and prime numbers. Euler's function. Congruences. Integers modulo m. Euler's theorem. Fermat's little theorem. Elements of group, rings and fields. Finite fields. Chinese remainder theorem. Primality test: Wilson's theorem and its converse. ENUMERATIVE COMBINATORICS: Cardinality of a set. Basic counting principles. Binomial numbers, Stiefel's formula and other identities on binomial numbers. Permutations and selections from a set. Binomial theorem. Formal power series. Partial fractions. Generating functions. Linear homogeneous recursions. Fibonacci and Lucas numbers. Closed form of the Fibonacci numbers. Stirling numbers of the second kind. Partitions of a natural number. Ferrers diagrams. The Tower of Hanoi problem. Principle of inclusion and exclusion. Formulas of Sylvester and Da Silva. Derangements. GRAPHS: Basic definitions. Planarity of graphs. Euler's formula. Bipartite graphs. Eulerian graphs. Hamiltonian graphs. Trees. Binary trees. Spanning trees. Vertex colorings. Chromatic number. Edge colorings. Chromatic index. K\"{o}nig's theorem. Matchings in bipartite graphs. Hall's theorem. Adjacency matrix, incidence matrix with respect to an orientation, Laplacian matrix. Theorem of Poincare'. Spectrum of a graph.
Prerequisiti
Students are required to have basic knowledge of mathematical analysis and linear algebra.
Modalità di valutazione
The final evaluation consists of an oral exam on the topics presented in the course. The aim of the exam is to verify if the student has got the following expertises: knowledge of the definitions and results on the topics of the course and ability both of applying that knowledge to solve exercises and of proving mathematical results presented in the course.
Bibliografia
N. Biggs, Discrete Mathematics, Editore: Clarendon Press, Oxford, Anno edizione: 1985 R. Graham, D. Knuth, O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Editore: Addison-Wesley, Anno edizione: 1989 F. S. Roberts, Applied Combinatorics, Editore: Prentice-Hall, Anno edizione: 1984
Software utilizzato
Nessun software richiesto
Forme didattiche
Tipo Forma Didattica Ore di attività svolte in aula
(hh:mm)
Ore di studio autonome
(hh:mm)
Lezione
37:30
56:15
Esercitazione
12:30
18:45
Laboratorio Informatico
0:00
0:00
Laboratorio Sperimentale
0:00
0:00
Laboratorio Di Progetto
0:00
0:00
Totale 50:00 75:00
Informazioni in lingua inglese a supporto dell'internazionalizzazione
Insegnamento erogato in lingua Inglese Disponibilità di materiale didattico/slides in lingua inglese Disponibilità di libri di testo/bibliografia in lingua inglese Possibilità di sostenere l'esame in lingua inglese Disponibilità di supporto didattico in lingua inglese
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T102 CHAP 9 REVIEW
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Unformatted text preview: likely to occur as letter A. a) Sketch such a spinner. b) Find the simple event probabilities for the experiment of spinning the spinner once. 8) The probability that a first proposal passes is 3/7. a) What are the odds in favor of the first proposal passing? b) What are the odds against the first proposal passing? 9) The probability of winning a certain raffle is 1/600. Find the odds against winning. T102 Chapter 9 Review Page 1 10) a) b) If you flipped a fair coin 9 times and got 9 heads, what would be the probability of getting a head on the next toss? If you rolled a fair die 5 times and got the numbers 1, 2, 3, 4, and 5, what would be the probability of rolling a 6 on the next turn? 11) Suppose there are 3 red, 4 blue, and 5 green chips in a box and you win by selecting either a red or green chip. If you get a red chip you win \$4, a green chip pays you \$3. What is the expected average winnings of this game if it costs you \$3.50 to play? Is it a fair game? Why? 12) An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. a) Find the probability that the sum is either a 2, 3, or 12. b) Find the probability that the sum is a 7 given that the sum is less than 8. 13) A consumer buys a package of 5 light bulbs not knowing that 1 of the bulbs is bad. List the outcomes in the sample space if 2 bulbs are selected randomly and then find the probabilities of the following events. a) Both bulbs are good b) At least one of the bulbs is bad 14) A game consists of the following: a single fair die is rolled and you are paid back the number of dollars corresponding to the number of dots facing up. a) What is the expected value of this game? b) If you pay \$4.00 to play the game, is it a fair game? 15) In a horse race, how many different finishes are possible among the first 3 places (win - place- show) if 10 horses are running? (exclude ties) 16) Eight distinct points are selected on the circumference of a circle. How many triangles can be drawn...
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{[ snackBarMessage ]} | 609 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-17 | latest | en | 0.913433 |
https://rdrr.io/cran/clue/man/lattice.html | 1,670,071,160,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00872.warc.gz | 502,926,520 | 9,474 | # lattice: Cluster Lattices In clue: Cluster Ensembles
lattice R Documentation
## Cluster Lattices
### Description
Computations on the lattice of all (hard) partitions, or the lattice of all dendrograms, or the meet semilattice of all hierarchies (n-trees) of/on a set of objects: meet, join, and comparisons.
### Usage
```cl_meet(x, y)
cl_join(x, y)
```
### Arguments
`x` an ensemble of partitions or dendrograms or hierarchies, or an R object representing a partition or dendrogram or hierarchy. `y` an R object representing a partition or dendrogram or hierarchy. Ignored if `x` is an ensemble.
### Details
For a given finite set of objects X, the set H(X) of all (hard) partitions of X can be partially ordered by defining a partition P to be “finer” than a partition Q, i.e., P ≤ Q, if each class of P is contained in some class of Q. With this partial order, H(X) becomes a bounded lattice, with intersection and union of two elements given by their greatest lower bound (meet) and their least upper bound (join), respectively.
Specifically, the meet of two partitions computed by `cl_meet` is the partition obtained by intersecting the classes of the partitions; the classes of the join computed by `cl_join` are obtained by joining all elements in the same class in at least one of the partitions. Obviously, the least and greatest elements of the partition lattice are the partitions where each object is in a single class (sometimes referred to as the “splitter” partition) or in the same class (the “lumper” partition), respectively. Meet and join of an arbitrary number of partitions can be defined recursively.
In addition to computing the meet and join, the comparison operations corresponding to the above partial order as well as `min`, `max`, and `range` are available at least for R objects representing partitions inheriting from `"cl_partition"`. The summary methods give the meet and join of the given partitions (for `min` and `max`), or a partition ensemble with the meet and join (for `range`).
If the partitions specified by `x` and `y` are soft partitions, the corresponding nearest hard partitions are used. Future versions may optionally provide suitable “soft” (fuzzy) extensions for computing meets and joins.
The set of all dendrograms on X can be ordered using pointwise inequality of the associated ultrametric dissimilarities: i.e., if D and E are the dendrograms with ultrametrics u and v, respectively, then D ≤ E if u_{ij} ≤ v_{ij} for all pairs (i, j) of objects. This again yields a lattice (of dendrograms). The join of D and E is the dendrogram with ultrametrics given by \max(u_{ij}, v_{ij}) (as this gives an ultrametric); the meet is the dendrogram with the maximal ultrametric dominated by \min(u_{ij}, v_{ij}), and can be obtained by applying single linkage hierarchical clustering to the minima.
The set of all hierarchies on X can be ordered by set-wise inclusion of the classes: i.e., if H and G are two hierarchies, then H ≤ G if all classes of H are also classes of G. This yields a meet semilattice, with meet given by the classes contained in both hierarchies. The join only exists if the union of the classes is a hierarchy.
In each case, a modular semilattice is obtained, which allows for a natural metrization via least element (semi)lattice move distances, see Barthélémy, Leclerc and Monjardet (1981). These latticial metrics are given by the BA/C (partitions), Manhattan (dendrograms), and symdiff (hierarchies) dissimilarities, respectively (see `cl_dissimilarity`).
### Value
For `cl_meet` and `cl_join`, an object of class `"cl_partition"` or `"cl_dendrogram"` with the class ids or ultrametric dissimilarities of the meet and join of the partitions or dendrograms, respectively.
### References
J.-P. Barthélémy, B. Leclerc and B. Monjardet (1981). On the use of ordered sets in problems of comparison and consensus of classification. Journal of Classification, 3, 187–224. doi: 10.1007/BF01894188.
### Examples
```## Two simple partitions of 7 objects.
A <- as.cl_partition(c(1, 1, 2, 3, 3, 5, 5))
B <- as.cl_partition(c(1, 2, 2, 3, 4, 5, 5))
## These disagree on objects 1-3, A splits objects 4 and 5 into
## separate classes. Objects 6 and 7 are always in the same class.
(A <= B) || (B <= A)
## (Neither partition is finer than the other.)
cl_meet(A, B)
cl_join(A, B)
## Meeting with the lumper (greatest) or joining with the splitter
## (least) partition does not make a difference:
C_lumper <- as.cl_partition(rep(1, n_of_objects(A)))
cl_meet(cl_ensemble(A, B, C_lumper))
C_splitter <- as.cl_partition(seq_len(n_of_objects(A)))
cl_join(cl_ensemble(A, B, C_splitter))
## Another way of computing the join:
range(A, B, C_splitter)\$max
```
clue documentation built on Nov. 19, 2022, 5:05 p.m. | 1,230 | 4,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.893236 |
http://mathhelpforum.com/advanced-algebra/229582-help-laplace-transform-dirac-comb.html | 1,524,709,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948047.85/warc/CC-MAIN-20180426012045-20180426032045-00268.warc.gz | 194,061,575 | 12,030 | # Thread: Help Laplace transform of Dirac comb
1. ## Help Laplace transform of Dirac comb
Please I need some help to solve this problem the Laplace transform of :
2. ## Re: Help Laplace transform of Dirac comb
the summation over what?
3. ## Re: Help Laplace transform of Dirac comb
from 0 to +inf
4. ## Re: Help Laplace transform of Dirac comb
Originally Posted by ameva
from 0 to +inf
there is no summation index in the expression.... is it a?
Yes it's a.
6. ## Re: Help Laplace transform of Dirac comb
Originally Posted by ameva
Yes it's a.
take a single element of the series
$f(t)=\delta(t-k T)$
I used a sampling time $T$ in the expression. You can set this so 1 if you like.
The Laplace transform of $\delta(t)$ is $1$
The Laplace transform of $g(t-kT)$ is $e^{-skT}G(s)$
so
$\large \mathscr{L} \{ \delta(t-kT) \} = e^{-s k T} * 1 = e^{-s k T}$
The Laplace transform is linear so the Laplace transform of the sum is the sum of the transforms, i.e.
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} f_k(t) \} = \displaystyle{\sum_{k=0}^\infty} F_k(s)$ where $F_k(s) = \mathscr{L}\{ f_k(t) \}$
thus
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} \delta(t - kT) \} = \displaystyle{\sum_{k=0}^\infty} e^{-s k T}$
This last term is a geometric series in $k$
$\large \displaystyle{\sum_{k=0}^\infty} e^{-s k T} = \dfrac 1 {1 - e^{-s T}}$
so
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} \delta(t - kT) \} = \dfrac 1 {1 - e^{-s T}}$
7. ## Re: Help Laplace transform of Dirac comb
that is what I looking for! thanks!
8. ## Re: Help Laplace transform of Dirac comb
just a question, did you change "a" by "kT"?
9. ## Re: Help Laplace transform of Dirac comb
Originally Posted by ameva
just a question, did you change "a" by "kT"?
yes. In the final answer just set $T=1$
i.e.
$\dfrac 1 {1 - e^{-s}}$
Thanks!
11. ## Re: Help Laplace transform of Dirac comb
I've just put the summation in a calculator program and it returned me this. Is it correct? Why is that?
12. ## Re: Help Laplace transform of Dirac comb
PD: On the paper I get the same as you
13. ## Re: Help Laplace transform of Dirac comb
Originally Posted by ameva
PD: On the paper I get the same as you
divide that second solution top and bottom by $\large e^s$. They are the same answer.
,
,
,
### dirac comb function uppercase
Click on a term to search for related topics. | 771 | 2,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-17 | latest | en | 0.758418 |
https://discusstest.codechef.com/t/how-many-games-spoj/8615 | 1,628,026,135,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00036.warc.gz | 221,894,541 | 5,772 | How many Games SPOJ
Problem: http://www.spoj.com/problems/GAMES/ My solution is giving right answer for the test cases given in the question and the cases which i tried but i m still getting wrong answer.May be there is some problem related to precision but i m not able to find the bug. Please help.
#include < iostream>
#include < cmath>
using namespace std;
include < stdio.h>
bool inline isInt(long double avg)
{
long double eps=0.0001;
if((avg-floor(avg))<eps)
return true;
``````return false; }
``````
`````` int main()
{
long int t,i=1;
long double avg,dup;
scanf("%ld",&t);
while(t--)
{
i=1;
cin>>avg;
dup=avg;
while(isInt(dup)==false)
{
dup=avg;
i=i+1;
dup=dup*i;
}
cout<<i<<"\n";
}
return 0; }``````
you have a precision error here because you are working on floats ,
here is the correct approach :-
write the average score x as a reduced fraction x=p/q. This means that p and q are integers, that q is positive and that q is minimal (or, equivalently, that p and q have no nontrivial common factor). Then the player can have played any multiple of q games hence the minimum number of games the player should have played is q.
When x=−30.25, note that −30.25=−121/4 and −121 and 4 have no common factors except +1 and −1, hence the minimum number of games is indeed 4 .
Basically we have to convert the avg into fraction part p/q then calculate the gcd(p,q)
here is the implementation of idea :-
2 Likes
I too tried to solve it using double and i got TLE. Try solving it using char array or strings. It’s more easy to implement.
i tried to sovle it using the same method but iam still getting wrong answer!!!1
here is my source code and for 1.0009 there is some weird thing happening …
include
int gcd (long int a ,long int b)
{
int temp=0;
if(a<b){temp=a;a=b;b=temp;}
if(b==0)return a;
else return gcd(b,a%b);
}
int main()
{
int t=0;double avg=0;long int a ,b;double atemp;
scanf("%d",&t);
while(t–)
{
scanf("%lf",&avg);
if(avg==1.0009){printf(“10000”);continue;}
atemp=avg*10000;b=10000;
a=(int)atemp;
// printf(“avg is :%lf ,a is :%ld ,atemp is :%lf ,b is :%ld ,gcd(a,b) is: %d\n”,avg,a,atemp,b,gcd(a,b));
// printf(“minimum no’ of games :%ld\n”,b/gcd(a,b));
printf("%ld\n",b/(gcd(a,b)));
}
return 0;
}
Can any body help me also??
my code passes all the test case I could come up with…but still SPOJ says wrong answer.
Try changing
if((avg-floor(avg))<eps) return true;
to
if((avg-floor(avg))<eps || (ceil(avg)-avg)<eps) return true;
Hope it may give you right answer.
I used the some method but added this one too.It gives correct answer to me.
1 Like
What is wrong with this solution? I multiply the decimal part till an integer is obtained, still spoj shows wrong answer.
#include<bits/stdc++.h>
using namespace std;
int main() {
int t,int_part,num;
double n,dec_part;
cin>>t;
while(t–) {
cin>>n;
num=1;
int_part=int(n10000)/10000;
dec_part=(n-int_part);
if(dec_part!=0) {
while(1) {
float temp=dec_part
num;
int int_part_temp=int(temp*10000)/10000;
double dec_part_temp=(temp-int_part_temp);
if(dec_part_temp==0)
break;
else
num++;
}
}
cout<<num<<endl;
}
}
// | 907 | 3,099 | {"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} | 3.296875 | 3 | CC-MAIN-2021-31 | latest | en | 0.84536 |
https://testbook.com/question-answer/in-a-uniform-circular-motion--5f5b674e90bea74971ecc478 | 1,708,813,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00349.warc.gz | 552,610,354 | 47,404 | # In a uniform circular motion:
This question was previously asked in
MP Patwari (15th March 2023 Shift 2) Memory Based Full Test
View all MP Patwari Papers >
1. Velocity is constant
2. Distance is constant
3. Displacement is constant
4. Speed is constant
Option 4 : Speed is constant
Free
50 K Users
10 Questions 10 Marks 7 Mins
## Detailed Solution
The correct answer is Speed is constant.
Key Points
• Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path.
• The distance and displacement are different at every instance at any particular point of time.
Confusion Points
• While the speed is a scalar quantity, in uniform motion, it remains constant.
• As, Velocity is a vector quantity, and the direction of movement changes continuously, velocity changes continuously.
• Due to this, the object is in accelerated motion.
Important Points | 203 | 910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-10 | latest | en | 0.915912 |
https://forum.bebac.at/mix_entry.php?id=17264&order=time&view=mix | 1,642,796,730,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00472.warc.gz | 314,366,519 | 14,782 | Yura
★
Belarus,
2017-04-25 15:14
(1732 d 07:10 ago)
Posting: # 17264
Views: 11,156
## question of adjustment [RSABE / ABEL]
Dear all,
Correction of alpha is carried out by CVWR and CVWT-R (nR and nT-R, GRMR and
GRMT-R, respectively, for TRR / RTR / RRT) or only by CVWR? Since it is necessary to calculate CIT-R.
Regards
Helmut
★★★
Vienna, Austria,
2017-04-26 14:17
(1731 d 08:08 ago)
@ Yura
Posting: # 17267
Views: 10,244
## TIE depends on CVwR (and n)
Hi Yura,
can you please explain what you mean by the abbreviations you used?
» Correction of alpha is carried out by CVWR and CVWT-R (nR and nT-R, GRMR and
» GRMT-R, respectively, for TRR / RTR / RRT) or only by CVWR? Since it is necessary to calculate CIT-R.
The inflation of the Type I Error depends on CVwR (and to a minor extent on the sample size). CVwT – which is indeed nice to know – is not accessible in the partial replicate design.
Examples of the TIE and adjusting α for an assumed true ratio of 0.9:
1. n = 48, balanced sequences
library(PowerTOST) scABEL.ad(CV=0.35, n=48, design="2x3x3") +++++++++++ scaled (widened) ABEL ++++++++++++ iteratively adjusted alpha (simulations based on ANOVA evaluation) ---------------------------------------------- Study design: 2x3x3 (TRR|RTR|RRT) log-transformed data (multiplicative model) 1,000,000 studies in each iteration simulated. CVwR 0.35, n(i) 16|16|16 (N 48) Nominal alpha : 0.05 True ratio : 0.9000 Regulatory settings : EMA (ABEL) Switching CVwR : 0.3 Regulatory constant : 0.76 Expanded limits : 0.7723 ... 1.2948 Upper scaling cap : CVwR > 0.5 PE constraints : 0.8000 ... 1.2500 Empiric TIE for alpha 0.0500 : 0.05663 Power for theta0 0.9000 : 0.801 Iteratively adjusted alpha : 0.04405 Empiric TIE for adjusted alpha: 0.05000 Power for theta0 0.9000 : 0.785
2. Three dropouts
scABEL.ad(CV=0.35, n=c(16, 14, 15), design="2x3x3") +++++++++++ scaled (widened) ABEL ++++++++++++ iteratively adjusted alpha (simulations based on ANOVA evaluation) ---------------------------------------------- Study design: 2x3x3 (TRR|RTR|RRT) log-transformed data (multiplicative model) 1,000,000 studies in each iteration simulated. CVwR 0.35, n(i) 16|14|15 (N 45) Nominal alpha : 0.05 True ratio : 0.9000 Regulatory settings : EMA (ABEL) Switching CVwR : 0.3 Regulatory constant : 0.76 Expanded limits : 0.7723 ... 1.2948 Upper scaling cap : CVwR > 0.5 PE constraints : 0.8000 ... 1.2500 Empiric TIE for alpha 0.0500 : 0.05634 Power for theta0 0.9000 : 0.779 Iteratively adjusted alpha : 0.04420 Empiric TIE for adjusted alpha: 0.05000 Power for theta0 0.9000 : 0.762
Due to the smaller sample size in #2 the TIE is less inflated (0.05634 < 0.05663) and less adjustment is required (0.04420 > 0.4405) to preserve the patient’s risk. On the other hand the narrower CI (91.16% < 91.19%) cannot outweigh the loss in power (76.2% < 78.5%).
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Helmut Schütz
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Yura
★
Belarus,
2017-04-26 17:28
(1731 d 04:56 ago)
@ Helmut
Posting: # 17269
Views: 10,063
## TIE depends on CVwR (and n)
Hi, Helmut
Yes of course
Adjusted alpha is used to construct confidence intervals of the pharmacokinetic parameters for the index T-P?
Regards
Helmut
★★★
Vienna, Austria,
2017-04-26 18:00
(1731 d 04:25 ago)
@ Yura
Posting: # 17270
Views: 10,135
## TIE depends on CVwR (and n)
Hi Yura,
» Adjusted alpha is used to construct confidence intervals of the pharmacokinetic parameters for the index T-P?
What do you mean by “the index T-P”? For the EMA use the ‘Method A’ or ‘Method B’ as given in the Q&A-document. But instead of using the nominal α of 0.05 (i.e., the 100(1–2α) = 90% CI) apply the respective adjusted α. If the TIE would be inflated with α 0.05, the adjusted CI is always wider (i.e., conservative).
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Helmut Schütz
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Yura
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Belarus,
2017-04-26 18:55
(1731 d 03:30 ago)
@ Helmut
Posting: # 17272
Views: 10,081
## TIE depends on CVwR (and n)
Hi, Helmut
Yes of course
We evaluate the difference between T-R, and also R-R - for expansion, if necessary. Therefore, CV T-R and CV R-R are obtained. Other CV, therefore, different alpha. What use alpha to build confidence interval difference T-R?
Regards
Yura
★
Belarus,
2017-04-28 11:13
(1729 d 11:11 ago)
@ Yura
Posting: # 17275
Views: 10,004
## TIE depends on CVwR (and n)
Hi, Helmut
Did I understand correctly, when constructing a confidence interval for T-R differences, use the adjusted alpha for R-R?
Regards
Helmut
★★★
Vienna, Austria,
2017-04-28 19:16
(1729 d 03:08 ago)
@ Yura
Posting: # 17277
Views: 10,182
## TIE = p(BE) at expanded limits
Hi Yura,
» Did I understand correctly, when constructing a confidence interval for T-R differences, use the adjusted alpha for R-R?
Exactly – now you got it!
In a nutshell, the Type I Error is the probability of falsely claiming BE. The TIE can be estimated by setting theta0 to one of the limits of the acceptance range. Easy for ABE (since an explicit solution exists).
library(PowerTOST) CV <- 0.3 des <- "2x2x4" n <- 34 U <- 1.25 power.TOST(CV=CV, n=n, theta0=U, design=des) # [1] 0.05
Alternatively you can perform simulations.
power.TOST.sim(CV=CV, n=n, theta0=U, design=des, nsims=1e7) # [1] 0.0500178
The convergence is lousy. If you have a lot of time increase the number of simulations.
When it comes to reference-scaling no explicit formula for power exists. Hence, we need simulations. There is a complication: ABEL is a framework of decisions where the Null-hypothesis is constructed in face of the data. In other words we don’t know the expanded limits until we have calculated CVwR. Unlike in ABE the limits are random variables themselves.
CVwR <- 0.3 reg <- "EMA" U <- scABEL(CV=CVwR, regulator=reg)[["upper"]] power.scABEL(CV=CVwR, n=n, theta0=U, design=des, regulator=reg, nsims=1e6) # [1] 0.081626
If you don’t like (or trust in) simulating the underlying statistics, with the latest version of PowerTOST you can simulate subject data as well.
power.scABEL.sdsims(CV=CVwR, n=n, theta0=U, design=des, regulator=reg, nsims=1e6) # [1] 0.081602
Note that in the final evaluation (i.e., the PE and its CI) the EMA’s model assumes that CVwT = CVwR which is strong meat. To get the adjusted α, and the TIEs for nominal and adjusted α:
res <- scABEL.ad(CV=CVwR, n=n, design=des, regulator=reg, print=FALSE) TIE0 <- res[["TIE.unadj"]] adj <- res[["alpha.adj"]] TIE1 <- res[["TIE.adj"]] cat(TIE0, adj, TIE1, "\n") # 0.081626 0.028572 0.05
Now imagine a study which passed with the nominal α 0.05 (90% CI):
round(100*CI.BE(alpha=0.05, pe=0.9, CV=CV, n=n, design=des), 2) # lower upper # 82.78 97.85
The study would still pass with the adjusted α 0.028572 (94.2856% CI) but it is a more close shave:
round(100*CI.BE(alpha=adj, pe=0.9, CV=CV, n=n, design=des), 2) # lower upper # 81.69 99.16
If you want to avoid surprises increase the sample size.
res <- sampleN.scABEL.ad(CV=CVwR, theta0=0.9, design=des, regulator=reg, print=FALSE) adj <- res[["alpha.adj"]] n.a <- res[["Sample size"]] cat(adj, n.a, "\n") # 0.028311 42
Note that the TIE depends also on the sample size. Therefore, slightly more adjustment is needed (n=34: 0.028572, n=42: 0.028311). With 42 subjects you should be on the safe side.
round(100*CI.BE(alpha=adj, pe=0.9, CV=CV, n=n.a, design=des), 2) # lower upper # 82.49 98.20
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Helmut Schütz
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Yura
★
Belarus,
2017-04-29 13:01
(1728 d 09:23 ago)
@ Helmut
Posting: # 17279
Views: 9,886
## TIE = p(BE) at expanded limits
Dear Helmut
As always, you are on top
Regards
pjs
★
India,
2018-02-28 14:33
(1423 d 06:51 ago)
@ Helmut
Posting: # 18483
Views: 8,599
## TIE depends on CVwR (and n)
Dear All,
Request you to share your thoughts for the requirement for Type 1 error estimation and adjustment of alpha for below different scenarios.
» The inflation of the Type I Error depends on CVwR (and to a minor extent on the sample size).
study is conducted as partial replicate design with 60 subjects. Now CVwr has turned out to be 31% in the study. In the sample size T/R ratio was assumed to be 0.90. In the actual conducted study T/R ratio had come to 100% (product essentially similar to reference product-Hypothetical scenario). Now study is passing the SCABE criteria. Study would have also passed incase limits would not have been scaled. Essentially there would not have been difference in study conclusion if the Scaling approach would have applied or not applied. As per my understanding Type 1 error would arise incase there is difference in study conclusion when there is uncertainty in the ISCV and due to that difference in study conclusion incase different method for study conclusion is utilized (like study passing in SCABE but failing in ABE, borderline case). DO any such case require adjustment of alpha although there is borderline high variability (incases where there is maximum probability of type 1 error).
Regards
Pjs
Helmut
★★★
Vienna, Austria,
2018-02-28 14:48
(1423 d 06:37 ago)
@ pjs
Posting: # 18485
Views: 8,646
## TIE depends on CVwR (and n)
Dear Pjs,
» […] study is passing the SCABE criteria.
If you are referring to the FDA’s RSABE approach, there is not problem with inflation of the Type I Error if (the true) CVwR ≥30%. Only with CVwR <30% problems might be massive (see this presentation).
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Helmut Schütz
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pjs
★
India,
2018-03-01 07:35
(1422 d 13:49 ago)
@ Helmut
Posting: # 18488
Views: 8,507
## TIE depends on CVwR (and n)
Dear Helmut,
Thanks for the feedback.
Just need some clarification.
For FDA limits are scaled from 0.25 but applied at 0.294 unlike EU in which limits are scaled from 0.294.
Hence at any Swr limits would be more wider for FDA compared to EU. Hence chances of establishing BE would be more for USFDA compared to EU in the nearby CV from cutoff limit of 30%. Considering the same There is more possibility for difference in BE results when applying two different methods Scaling and ABE incase of FDA for study conclusion. Hence alpha inflation should be more for FDA instead of EU. Please correct me if i have misunderstood the concept.
Also incase of EU scaling is applicable for Cmax parameter only and not for AUC parameter in contrary to FDA for which scaling is applicable for all the primary metric. This could also play certain role in the calculation of alpha inflation.
Regards
Pjs
Helmut
★★★
Vienna, Austria,
2018-03-01 13:32
(1422 d 07:52 ago)
@ pjs
Posting: # 18489
Views: 8,693
## Comparing methods for (S)ABE
Hi Pjs,
your considerations are essentially correct. OK, a little bit theoretical because in all jurisdictions we need the respective region’s reference product. But yes, if we consider the same data set, the conclusions might differ if we apply different reference-scaling methods – especially in borderline cases.
What do we have now?
1. US-FDA, CFDA
RSABE for both AUC and Cmax, no clinical justification required, fixed effects model for the partial replicate design and mixed effects model for full replicate design. GMR restriction 80.00–125.00%. If swR <0.294, mixed effects model irrespective of the design.
2. EEA, Russia, EEU, Egypt, ANVISA
ABEL for Cmax (MR products additionally: Cmin, pAUCs), upper cap on scaling at CVwR 50%, clinical justification required, fixed effects model. GMR restriction 80.00–125.00%.
3. WHO
Like #3. Pilot phase for AUC. 4-period full replicate design mandatory, comparison of swT with swR (though no conditions for passing given so far).
ABEL for AUC, upper cap on scaling at CVwR 57.4%, clinical justification required, mixed effects model. GMR restriction 80.0–125.0%.
GMR of Cmax within 80.0–125.0% (no CI needed).
5. GCC member states
Widening of the acceptance limits for Cmax only (fixed and pre-specified to 75–133%), clinical justification required, CVwR >30% demonstrated in a full replicate design, GMR restriction 80.00–125.00%.
In all methods (except #4: Cmax and #5: assessed by ABE) inflation of the Type I Error is possible.
At the 2nd International Conference of the Global Bioequivalence Harmonization Initiative (Rockville, Sep 2016) an entire session was devoted to reference-scaling. No consensus reached. On the contrary. Each agency defended its concept as if it is an eternal truth. Disappointing.
BTW, we have lacking harmonization even in ABE. For NTIDs the EMA’s acceptance range is 90.00–111.11%, whereas for Health Canada it is 90.0–112.0%.
library(PowerTOST) round(100*CI.BE(pe=1.05, CV=0.12, n=24), 2) lower upper 98.96 111.41
The same study would fail for the EMA but pass for HC.
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Helmut Schütz
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pjs
★
India,
2018-03-05 14:50
(1418 d 06:35 ago)
@ Helmut
Posting: # 18496
Views: 8,476
## Comparing methods for (S)ABE
Hi Helmut,
Yes agree with you, harmonization should be there for evaluating BE criteria incase of HVDP or NTI drugs or any such evaluation criteria.
Have gone through some of the post related to alpha correction incase of SABE approach in the forum. In the simulations done most extreme GMR of 1.25 is assumed for the simulation and calculation of possible alpha inflation.
I would like to understand as the actual alpha adjustment would be based on actual Swr observed in the study and number of subjects, what could be the rationale of doing adjustment of alpha based on the calculation which is done on extreme GMR while for the actual study conducted T/R ratio could be very much close to unity (let's say 0.95 or 0.97). I do understand the most deviated GMR would lead to maximum probability for the alpha inflation but applying this extreme case and alpha adjustment in each and every study is required?
Regards
Pjs
Helmut
★★★
Vienna, Austria,
2018-03-05 17:40
(1418 d 03:44 ago)
@ pjs
Posting: # 18497
Views: 8,509
## Simulating the Null
Hi Pjs,
» Have gone through some of the post related to alpha correction incase of SABE approach in the forum. In the simulations done most extreme GMR of 1.25 is assumed for the simulation and calculation of possible alpha inflation.
Not exactly. In order to simulate the type I error we assume that the Null Hypothesis (of bioinequivalence) is true. In SCABE (ABEL: expanded limits, RSABE: implied limits) only if CVwR ≤30% the GMR of the Null is at 1.25 (or 0.8). Therefore, with higher CVs we have to simulate at higher GMRs as well.
Note that in ABE the Null is always at the borders of the acceptance range (0.8 and 1.25). Hence, here we don’t need simulations but can directly calculate the power (i.e., chance of passing BE = falsely rejecting the Null):
library(PowerTOST) power.TOST(CV=0.3, theta0=1.25, design="2x2x4", n=40) # [1] 0.05
Of course, we can simulate the TIE as well:
power.TOST.sim(CV=0.3, theta0=1.25, design="2x2x4", n=40, nsims=1e6) # [1] 0.05004
TOST (or the CI inclusion method) is not the most powerful test. If the sample size is very low, the TIE will be ≤5%:
power.TOST(CV=0.3, theta0=1.25, design="2x2x4", n=12) # [1] 0.04977286
On the other hand, a very high sample size would still preserve the nominal level:
power.TOST(CV=0.3, theta0=1.25, design="2x2x4", n=1200) # [1] 0.05
That’s a comforting property.
In SCABE the scaled limits (and hence the GMR which we use in simulating the Null) depend on the CVwR. Example for CVwR 40%:
scABEL(CV=0.4, regulator="EMA") # lower upper # 0.746177 1.340165 scABEL(CV=0.4, regulator="FDA") # lower upper # 0.7090232 1.4103911
» I would like to understand as the actual alpha adjustment would be based on actual Swr observed in the study and number of subjects, what could be the rationale of doing adjustment of alpha based on the calculation which is done on extreme GMR while for the actual study conducted T/R ratio could be very much close to unity (let's say 0.95 or 0.97).
The observed GMR is only an estimate. We don’t know where the population’s true GMR lies. Think about the 90% CI. There is a 5% chance at each CL that the true GMR is outside. Imagine in ABE you observe a GMR of 1 with a 90% CI of 0.8–1.25. What is the TIE?
» I do understand the most deviated GMR would lead to maximum probability for the alpha inflation …
Correct.
» … but applying this extreme case and alpha adjustment in each and every study is required?
That’s maybe the best we have so far. OK, we could go even further (suggested by Molins et al.)* Simulating the Null at the scaled limits still assumes that the CVwR estimated in the study is the true value – which might not be correct. If one wants to get the most conservative adjusted α one should simulate at CVwR 30% – irrespective of what we observe (since this is the location of the maximum TIE both in ABEL and RSABE). However, there is no free lunch. Either power will be compromised or 20–25% more subjects are needed to preserve the desired power.
• Molins E, Cobo E, Ocaña J. Two‐stage designs versus European scaled average designs in bioequivalence studies for highly variable drugs: Which to choose? Stat Med. 2017;36(30):4777–88. doi:10.1002/sim.7452.
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Helmut Schütz
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Helmut
★★★
Vienna, Austria,
2018-03-07 16:21
(1416 d 05:03 ago)
@ pjs
Posting: # 18504
Views: 8,254
Hi Pjs,
extending the end of my post. No R-code yet because it requires the development-version of PowerTOST. Essentially we have three options:
1. Assuming that the observed CVwR is the true one.
2. Since the true CVwR is unknown, assume the worst, i.e., always adjust for a CVwR of 30%.
3. Calculate a conservative CI of the observed CVwR. If the CI includes 30%, adjust for CVwR=30%. If not adjust for the CL which is closer to 30%.
Sample sizes were estimated for a GMR of 0.9 and target power 0.8, 4-period full replicate design, and the EMA’s ABEL. NA denotes cases where no adjustment is necessary (since the Type I Error with the nominal α is ≤0.05).
Adjustment based on observed CVwR (Labes and Schütz 2016): reg design n CVwR TIE.nom alpha.adj TIE.adj pwr.des pwr.act EMA 2x2x4 14 0.175 0.04988 NA NA 0.8007 0.8007 EMA 2x2x4 18 0.200 0.04998 NA NA 0.8007 0.8007 EMA 2x2x4 24 0.225 0.04958 NA NA 0.8229 0.8229 EMA 2x2x4 28 0.250 0.05180 0.04825 0.05 0.8116 0.8069 EMA 2x2x4 32 0.275 0.05980 0.04148 0.05 0.8082 0.7824 EMA 2x2x4 34 0.300 0.08163 0.02857 0.05 0.8028 0.7251 EMA 2x2x4 34 0.325 0.06971 0.03418 0.05 0.8010 0.7492 EMA 2x2x4 34 0.350 0.06557 0.03630 0.05 0.8118 0.7728 EMA 2x2x4 32 0.375 0.06271 0.03782 0.05 0.8110 0.7760 EMA 2x2x4 30 0.400 0.05912 0.04024 0.05 0.8066 0.7800 EMA 2x2x4 30 0.425 0.05451 0.04454 0.05 0.8219 0.8094 EMA 2x2x4 28 0.450 0.04889 NA NA 0.8112 0.8112 EMA 2x2x4 28 0.475 0.04114 NA NA 0.8162 0.8162 EMA 2x2x4 28 0.500 0.03317 NA NA 0.8143 0.8143 EMA 2x2x4 28 0.525 0.03787 NA NA 0.8073 0.8073 EMA 2x2x4 30 0.550 0.04165 NA NA 0.8211 0.8211 EMA 2x2x4 30 0.575 0.04420 NA NA 0.8047 0.8047 EMA 2x2x4 32 0.600 0.04630 NA NA 0.8101 0.8101 EMA 2x2x4 34 0.625 0.04779 NA NA 0.8128 0.8128 EMA 2x2x4 36 0.650 0.04856 NA NA 0.8127 0.8127 'Worst case' adjustment based on CVwR=30% (Molins et al. 2017): reg design n CVwR CV.adj TIE.nom alpha.adj TIE.adj pwr.des pwr.act EMA 2x2x4 14 0.175 0.3 0.07904 0.03022 0.05 0.8007 0.4073 EMA 2x2x4 18 0.200 0.3 0.07928 0.02988 0.05 0.8007 0.4942 EMA 2x2x4 24 0.225 0.3 0.08064 0.02927 0.05 0.8229 0.5951 EMA 2x2x4 28 0.250 0.3 0.08057 0.02910 0.05 0.8116 0.6517 EMA 2x2x4 32 0.275 0.3 0.08075 0.02904 0.05 0.8082 0.7055 EMA 2x2x4 34 0.300 0.3 0.08160 0.02864 0.05 0.8028 0.7239 EMA 2x2x4 34 0.325 0.3 0.08160 0.02864 0.05 0.8010 0.7239 EMA 2x2x4 34 0.350 0.3 0.08160 0.02864 0.05 0.8118 0.7239 EMA 2x2x4 32 0.375 0.3 0.08075 0.02904 0.05 0.8110 0.7055 EMA 2x2x4 30 0.400 0.3 0.08123 0.02888 0.05 0.8066 0.6753 EMA 2x2x4 30 0.425 0.3 0.08123 0.02888 0.05 0.8219 0.6753 EMA 2x2x4 28 0.450 0.3 0.08057 0.02910 0.05 0.8112 0.6517 EMA 2x2x4 28 0.475 0.3 0.08057 0.02910 0.05 0.8162 0.6517 EMA 2x2x4 28 0.500 0.3 0.08057 0.02910 0.05 0.8143 0.6517 EMA 2x2x4 28 0.525 0.3 0.08057 0.02910 0.05 0.8073 0.6517 EMA 2x2x4 30 0.550 0.3 0.08123 0.02888 0.05 0.8211 0.6753 EMA 2x2x4 30 0.575 0.3 0.08123 0.02888 0.05 0.8047 0.6753 EMA 2x2x4 32 0.600 0.3 0.08075 0.02904 0.05 0.8101 0.7055 EMA 2x2x4 34 0.625 0.3 0.08160 0.02864 0.05 0.8128 0.7239 EMA 2x2x4 36 0.650 0.3 0.08173 0.02852 0.05 0.8127 0.7454 Conservative adjustment based on 99.9% CI of observed CVwR: reg design n CVwR lower CL upper CL CV.adj TIE.nom alpha.adj TIE.adj pwr.des pwr.act EMA 2x2x4 14 0.175 0.1022 0.4536 0.3000 0.07818 0.03072 0.05 0.8007 0.4096 EMA 2x2x4 18 0.200 0.1237 0.4407 0.3000 0.07944 0.03005 0.05 0.8007 0.4935 EMA 2x2x4 24 0.225 0.1475 0.4300 0.3000 0.08040 0.02933 0.05 0.8229 0.5952 EMA 2x2x4 28 0.250 0.1683 0.4503 0.3000 0.08122 0.02886 0.05 0.8116 0.6513 EMA 2x2x4 32 0.275 0.1892 0.4737 0.3000 0.08130 0.02860 0.05 0.8082 0.7018 EMA 2x2x4 34 0.300 0.2082 0.5088 0.3000 0.08163 0.02857 0.05 0.8028 0.7251 EMA 2x2x4 34 0.325 0.2251 0.5547 0.3000 0.08163 0.02857 0.05 0.8010 0.7251 EMA 2x2x4 34 0.350 0.2420 0.6014 0.3000 0.08163 0.02857 0.05 0.8118 0.7251 EMA 2x2x4 32 0.375 0.2560 0.6641 0.3000 0.08130 0.02860 0.05 0.8110 0.7018 EMA 2x2x4 30 0.400 0.2694 0.7334 0.3000 0.08102 0.02888 0.05 0.8066 0.6777 EMA 2x2x4 30 0.425 0.2856 0.7864 0.3000 0.08102 0.02888 0.05 0.8219 0.6777 EMA 2x2x4 28 0.450 0.2979 0.8680 0.3000 0.08122 0.02886 0.05 0.8112 0.6513 EMA 2x2x4 28 0.475 0.3136 0.9263 0.3136 0.07365 0.03238 0.05 0.8162 0.6668 EMA 2x2x4 28 0.500 0.3291 0.9864 0.3291 0.06893 0.03485 0.05 0.8143 0.6814 EMA 2x2x4 28 0.525 0.3446 1.0480 0.3446 0.06639 0.03625 0.05 0.8073 0.6965 EMA 2x2x4 30 0.550 0.3646 1.0730 0.3646 0.06362 0.03748 0.05 0.8211 0.7421 EMA 2x2x4 30 0.575 0.3800 1.1360 0.3800 0.06183 0.03838 0.05 0.8047 0.7580 EMA 2x2x4 32 0.600 0.4000 1.1610 0.4000 0.05905 0.04008 0.05 0.8101 0.8035 EMA 2x2x4 34 0.625 0.4198 1.1890 0.4198 0.05430 0.04416 0.05 0.8128 0.8464 EMA 2x2x4 36 0.650 0.4397 1.2170 0.4397 0.04819 NA NA 0.8127 0.8127
As you can see, power is compromised (pwr.des = achieved power in sample size estimation, pwr.act = actual power if the study is evaluated with the adjusted α). IMHO, power <0.7 is not desirable.
Molin’s approach is extremely conservative. Imagine an observed CVwR of 100%. According to ABEL we will employ the maximum expansion (69.84–143.19%) of the BE limits and the decision will practically lead by the GMR-restriction (80.00–125.00%). But how likely is a true CVwR of 30%? Less than 10–15! The adjusted α (4-period full replicate, n 68) will be 0.02748 and power 0.7161. Borderline.
Will it help to use a less strict CI of CVwR?
Conservative adjustment based on 99.0% CI of observed CVwR: reg design n CVwR lower CL upper CL CV.adj TIE.nom alpha.adj TIE.adj pwr.des pwr.act EMA 2x2x4 14 0.175 0.1135 0.3535 0.3000 0.07818 0.03072 0.05 0.8007 0.4096 EMA 2x2x4 18 0.200 0.1359 0.3603 0.3000 0.07944 0.03005 0.05 0.8007 0.4935 EMA 2x2x4 24 0.225 0.1604 0.3660 0.3000 0.08040 0.02933 0.05 0.8229 0.5952 EMA 2x2x4 28 0.250 0.1822 0.3895 0.3000 0.08122 0.02886 0.05 0.8116 0.6513 EMA 2x2x4 32 0.275 0.2039 0.4146 0.3000 0.08130 0.02860 0.05 0.8082 0.7018 EMA 2x2x4 34 0.300 0.2240 0.4471 0.3000 0.08163 0.02857 0.05 0.8028 0.7251 EMA 2x2x4 34 0.325 0.2423 0.4863 0.3000 0.08163 0.02857 0.05 0.8010 0.7251 EMA 2x2x4 34 0.350 0.2605 0.5261 0.3000 0.08163 0.02857 0.05 0.8118 0.7251 EMA 2x2x4 32 0.375 0.2763 0.5758 0.3000 0.08130 0.02860 0.05 0.8110 0.7018 EMA 2x2x4 30 0.400 0.2914 0.6292 0.3000 0.08102 0.02888 0.05 0.8066 0.6777 EMA 2x2x4 30 0.425 0.3090 0.6725 0.3090 0.07533 0.03144 0.05 0.8219 0.6875 EMA 2x2x4 28 0.450 0.3231 0.7326 0.3231 0.07043 0.03402 0.05 0.8112 0.6755 EMA 2x2x4 28 0.475 0.3402 0.7787 0.3402 0.06698 0.03590 0.05 0.8162 0.6924 EMA 2x2x4 28 0.500 0.3573 0.8257 0.3573 0.06493 0.03691 0.05 0.8143 0.7083 EMA 2x2x4 28 0.525 0.3742 0.8736 0.3742 0.06322 0.03784 0.05 0.8073 0.7255 EMA 2x2x4 30 0.550 0.3952 0.9005 0.3952 0.05981 0.03969 0.05 0.8211 0.7744 EMA 2x2x4 30 0.575 0.4121 0.9487 0.4121 0.05704 0.04200 0.05 0.8047 0.7946 EMA 2x2x4 32 0.600 0.4330 0.9759 0.4330 0.05217 0.04695 0.05 0.8101 0.8410 EMA 2x2x4 34 0.625 0.4539 1.0040 0.4539 0.04565 NA NA 0.8128 0.8128 EMA 2x2x4 36 0.650 0.4747 1.0330 0.4747 0.03821 NA NA 0.8127 0.8127
Only a little.
Dif-tor heh smusma 🖖
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes | 10,288 | 26,327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.541878 |
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# 7-Atmospheric Motion-v2011-v2 - Atmospheric Motion Leila M...
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Atmospheric Motion Leila M. V. Carvalho
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What is wind? Well, wind is air in movement. We can feel the wind and we notice its presence, we can fear, and we can even use its energy to move and generate power
Isaac Newton (1642-1727) first law of motion states that an object at rest will remain at rest and object in motion will remain in motion (and travel at a constant velocity along a straight line) as long as no force is exerted in the object” So, if wind is air in movement, a force must exist to move air from rest to movement. Let’s examine the forces that act upon the atmosphere Imagine a volume with molecules at a given temperature. They are randomly vibrating and colliding with each other, depending on the temperature Horizontal Wind Wind is therefore the movement of a given volume of air. F = ma (mass x accelaration) , and F=F1+ F2 +…+ Fn
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Low Pressure High Pressure In this condition, what should be the direction of the wind? Here, we observe an horizontal pressure change or horizontal pressure gradient. Gradient measures the spatial variation of a scalar (pressure, temperature, density, etc). Pressure gradient = difference in pressure/distance, PG=Δp/d
According with Newton second law, a force must act upon an object to change its speed (from rest to a speed V). This force, in this case is the Gradient Pressure Force Low Pressure p2 High Pressure p1 wind direction Pressure Gradient REMEMBER THAT A GRADIENT” ALWAYS POINT TOWARD THE HIGHEST MAGNITUDES OF THE SCALAR . THE PRESSURE GRADIENT FORCE POINTS FROM HIGH TOWARDS LOW PRESSURE Pressure gradient = difference in pressure/distance, PG=Δp/d The Pressure Gradient Force per unity of mass (PGF) is proportional to - Δp/d, Where Δp=p2-p1 In mathematical terms: , where ρ is the atmosphere density at a given level Pressure Gradient Force E W s p Fg - = ρ 1
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Let’s recall the definition of air pressure Air pressure is simply the weight of mass of air above a given level, the force per unit area exerted against a surface by the weight of the air molecules above that surface. P2 P1 P1>P2
Pressure measurements (pg. 100 text book) Inches or millimeters of mercury Pressure is measured based on the compression of chambers Aneroid barometer Barograph Cylinder moves and pressure is registered along the days Mercury barometer Barometer: literally means measure ‘Bar’, which is one unity of pressure. Bar is a relatively large unity, and because surface pressure changes are normally small, the common unity in Meteorology: millibar (mb), 1mb=1/1000 bar (1mb = hecto Pascal (100 Pa), where Pa=N/m2) STANDARD PRESSURE AT SEA LEVEL: 1013.25mb = 29.92 in Hg = 76cm
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We also learned that pressure and density decrease exponentially with height : rapidly at first and slowly at higher altitudes P2 P1 P1>P2
4,322m 1000m If you are climbing a mountain from the sea level to about 1000m, you will notice that pressure will decrease about 100mb/km = 10mb/100m (10mb/328 ft) Δp/ Δz = 100mb/km
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## This note was uploaded on 12/27/2011 for the course GEOG 110 taught by Professor Leila during the Fall '09 term at UCSB.
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7-Atmospheric Motion-v2011-v2 - Atmospheric Motion Leila M...
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Discussion of Problem 1792. Hamming Code
My solution
Posted by Pearl 4 Oct 2018 04:27
First, let give the bits some names, I put them inside the parentheses:
Petal 1 (p1) = II (c2) + III (c3) + IV (c4)
Petal 2 (p2) = I (c1) + III (c3) + IV (c4)
Petal 3 (p3) = I (c1) + II (c2) + IV (c4)
As p1, p2 and p3 are created from c1, c2, c3, c4, we check each change in c1, c2, c3, c4:
When c1 change: p2 and p3 won't match the value calculated using the formular above (i.e: c1 + c3 + c4 != p2 and c1 + c2 + c4 != p3)
When c2 change: p1 and p3 will go wrong
When c3 change: p1 and p2 will go wrong
As c4 contribute to the value of all three petals, if it is changed then p1, p2 and p3 will all go wrong.
In short:
Calculate ep1, ep2 and ep3 from the first 4 bits (the e in ep stand for expected), then compare them with the given p1, p2 and p3:
All three pairs do not match: c4 is changed
p1 and p2 do not match: c3 is changed
p1 and p3 do not match: c2 is changed
p2 and p3 do not match: c1 is changed
Only p1 not match: p1 is changed
Only p2 not match: p2 is changed
Only p3 not match: p3 is changed | 463 | 1,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-51 | latest | en | 0.848511 |
https://app.jove.com/science-education/v/15601/bulk-modulus?trialstart=1 | 1,726,500,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00235.warc.gz | 87,274,293 | 23,041 | Faculty Resource Center
The bulk modulus is a scientific term used to describe a material's resistance to uniform compression. It is the proportionality constant that links a change in pressure to the resulting relative volume change.
This concept becomes clearer when an isotropic material element is visualized as a cube of unit volume. When this cube is subjected to normal stresses, it undergoes deformation, changing its shape into a rectangular parallelepiped with a different volume. The discrepancy between this new volume and the original one is termed the dilatation of the material. Dilatation can be computed as the cumulative sum of the strains in the three spatial directions. When the body is under uniform hydrostatic pressure, each stress component equals the negative of this pressure. Inserting these values into the dilatation formula gives an expression that introduces the bulk modulus.
This modulus has the same units as the modulus of elasticity. Under hydrostatic pressure, stable materials reduce in volume, rendering the dilatation negative and the bulk modulus positive. An ideal material with a zero Poisson's ratio could stretch in one direction without lateral contraction. On the other hand, a material with a Poisson's ratio of 0.5 would be perfectly incompressible.
Privacy | 248 | 1,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.888897 |
http://stackoverflow.com/questions/7618374/bresenhams-circle-algorithm | 1,386,921,212,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164919525/warc/CC-MAIN-20131204134839-00027-ip-10-33-133-15.ec2.internal.warc.gz | 172,041,658 | 14,781 | # Bresenham's circle algorithm
I have the following code for drawing a circle :
``````#include<stdio.h>
#include<conio.h>
#include<graphics.h>
#include<math.h>
void main()
{
int xc, yc, x, y, p[100], r, k;
int gdriver=DETECT, gmode, errorcode;
printf("\nEnter the center point(xc,yc): ");
scanf("%d%d", &xc, &yc);
scanf("%d", &r);
printf("\nPlotting...\n");
sleep(5);
clrscr();
initgraph(&gdriver, &gmode, "");
p[0]=1-r;
x=0;
y=r;
for(k=0;k<=y;k++)
{
putpixel(xc+x, yc+y, 9);
putpixel(xc-x, yc-y, 9);
putpixel(xc+x, yc-y, 9);
putpixel(xc-x, yc+y, 9);
putpixel(xc+y, yc+x, 9);
putpixel(xc-y, yc-x, 9);
putpixel(xc+y, yc-x, 9);
putpixel(xc-y, yc+x, 9);
if(p[k]>0)
{
p[k+1]= p[k]+ 2*(x+1)+1-2*(y+1);
x++;
y--;
}
else
{
p[k+1]=p[k]+2*(x+1)+1;
x++;
}
}
getch();
}
``````
This part of code :
``````putpixel(xc+x, yc+y, 9);
putpixel(xc-x, yc-y, 9);
putpixel(xc+x, yc-y, 9);
putpixel(xc-x, yc+y, 9);
putpixel(xc+y, yc+x, 9);
putpixel(xc-y, yc-x, 9);
putpixel(xc+y, yc-x, 9);
putpixel(xc-y, yc+x, 9);
``````
Is mainly for plotting the points with respect to the circle, and it works because of the symmetric property of circle.
But I couldn't figure out what this part of code is exactly doing ;
``````if(p[k]>0)
{
p[k+1]= p[k]+ 2*(x+1)+1-2*(y+1);
x++;
y--;
}
else
{
p[k+1]=p[k]+2*(x+1)+1;
x++;
}
``````
Can anyone explain me what it does? Thanks in advance.
-
The update formulae look a little weird, and I will give what I think are the correct steps below:
You are starting from the topmost point in the circle and rotating clockwise until the angle reaches 45 degrees.
Now, the points on the circle roughly satisfy (x^2 + y^2 = r^2).
The idea is to draw one pixel at a time, moving in the positive `x` direction. If you find that the next point (without shifting down) is too far from the center of the circle, then that point should be drawn one unit lower. For example, if you look at pixellated circles, you will see that they can be essentially broken down into a series of horizontal lines and pixels. Each end of the horizontal line marks a point where extending the line would be too far from the circle, and hence you see a drop.
Note that there is some element of discretion here regarding which points you choose. There are 3 circle drawing disciplines:
1. Inner circle: Choose points such that no point is drawn outside of the circle (so that `x^2 + y^2 < (r+1)^2 for each point r` -- note that its `r+1` here and not `r`)
2. Outer circle: Choose points such that no point is drawn inside of the circle (so that `x^2 + y^2 > (r-1)^2 for each point r` -- note that its `r-1` here and not `r`)
3. Middle circle: Choose points that minimize `abs(x^2 + y^2 - r^2)`.
You can choose any of these disciplines in the algorithm. The methods are identical except for that code block (and the changes there are minor).
In each case, you have to calculate how far each point deviates from the circle. This requires knowing `x^2 + (y-1)^2 - r^2`. Let's call that sequence `p[k]`. If `x^2 + (y-1)^2 - r^2 <= 0`, then moving down would show the point too close to the center of the circle, so the next point should be `(x+1, y)`. In that circumstance, then the next deviation will be:
``````p[k+1] = (x+1)^2 + (y-1)^2 - r^2 = x^2 + (y-1)^2 - r^2 + 2x + 1 = p[k] + 2*(x + 1) - 1
``````
If `x^2 + y^2 - r^2 > 0`, then the next point should be `(x+1,y-1)`, so that
``````p[k+1] = (x+1)^2 + (y-2)^2 - r^2 = x^2 + (y-1)^2 - r^2 + 2x + 1 - 2y + 3 = q[k] + 2*(x + 1) - 2*(y - 1) = p[k] + 2*(x+1) - 2 * (y + 1)
``````
These formulae change based on whether you are interested in finding the outer circle (pixels are never too close), inner circle (pixels are never too far), or center circle (roughly in line), but this is the essential idea.
-
The lines aren't clear to me ```The trick here is basically to increment x each time, shifting down a row every time the distance is too far. There is some element of discretion here. If you want to stay outside the theoretical circle, which is what happens here, then you should check if moving down the y scale would not make the deviation negative.``` Can you be bit more clearer? – sriram Oct 1 '11 at 8:56
@GroovyUser I tried to make it a little clearer. I will go back and update the wiki article later on, but before that, let me know if there are any other points that are unclear (or anything else that should be added) – Foo Bah Oct 1 '11 at 13:12
@Ant's In each step, you are moving horizontally by one column. You are going to draw one point in that column. There are two possible `y` values (since you are drawing it in the `0-45 degree` octant): the current `y` value, or one step lower. Thus, you can "guess" that one choice is correct (in our case, we guess that moving down a row is correct) and check the error. If you are wrong, you just move horizontally, If you're right, you also move down a row. As a physical analogy, when you solve a maze, you look ahead to figure out the next step (I hope you dont actually draw the lines) – Foo Bah Oct 1 '11 at 13:46
@GroovyUser Like i tried to explain in the response for Ant's, when you move horizontally, you can either stay in the same row (dont change `y`) or move down a row (`y-1`). You are actually checking if you should shift downward (which is why you see `y-1` and `y-2` in the error formulae). If you wanted to check if moving horizontally (without going down) is correct, you would have a different loop. Why dont you try working it out :) – Foo Bah Oct 1 '11 at 13:49
@GroovyUser After you get past the `C` language kerfluffle, it's really just math :) I also didnt know there was a wikipedia article on the matter until I saw Miguel's answer. To make sure you understand what is going on, why don't you try to write the explanation in the wiki article (and then I'll go back and clean up any points of confusion :) – Foo Bah Oct 1 '11 at 13:56 | 1,807 | 5,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.749731 |
https://marketsplash.com/tutorials/matlab/meshgrid-matlab/ | 1,685,577,628,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00551.warc.gz | 452,219,779 | 59,358 | Imagine you're a captain navigating the vast seas of data analysis, and you're looking for a reliable compass. In the ocean of MATLAB, that compass is the Meshgrid function. It's the unsung hero of the high seas of 3D graphics, and in this article, we're going to make it sing. Buckle up, it's going to be a fun ride.
• The Basics Of Meshgrid In MATLAB
• The Power Of Meshgrid For 3D Visualization
• Meshgrid Vs. Other MATLAB Functions
• Common Pitfalls And Troubleshooting With Meshgrid
• Leveraging Meshgrid For Increased Website Traffic
• Important disclosure: we're proud affiliates of some tools mentioned in this guide. If you click an affiliate link and subsequently make a purchase, we will earn a small commission at no additional cost to you (you pay nothing extra). For more information, read our affiliate disclosure.
The Basics Of Meshgrid In MATLAB
Imagine you're an artist about to sketch a 3D landscape. What's the first thing you do? You lay out the gridlines on your canvas. That's essentially what Meshgrid does. It's the quiet hero that turns your 2D vectors into a 3D grid, the starting point for any 3D plot. The syntax, my friends, is simpler than ordering a coffee: `[X, Y] = meshgrid(x, y)`.
Now let's add some real-life spice to this. Suppose we're tracking the flight path of a super-intelligent pigeon (don't ask why). We have its longitude `x = 1:5` and latitude `y = 1:5`. With Meshgrid, we can easily visualize this flight pattern as a 3D grid.
The Power Of Meshgrid For 3D Visualization
Do you remember those magic eye posters from the 90s? The ones where you squinted until you saw a 3D image? Well, Meshgrid is like the magic eye of MATLAB. It can turn a boring 2D dataset into an exciting 3D visualization.
Think about our pigeon friend from earlier. With Meshgrid, we can create a 3D plot that shows us exactly where our feathered friend has flown. This is like having a GPS tracker on the pigeon, but without annoying the local wildlife authorities.
Meshgrid isn't just for tracking super-intelligent pigeons. It's got more uses than a Swiss army knife at a camping convention. From generating computational models to simulating complex physics phenomena, Meshgrid is your go-to function.
For instance, let's say you're a weather forecaster (or a very enthusiastic hobbyist). You've got temperature and pressure data for different points on a map. Using Meshgrid, you can create a 3D model that visualizes how these two factors interact, making your weather predictions more accurate than your local news station!
Meshgrid Vs. Other MATLAB Functions
MATLAB has more functions than Hollywood has reboots. And like choosing between Spiderman 1, 2, or 3, it can be hard to know when to use Meshgrid versus other similar functions like `ndgrid` or `griddata`. Let's break it down:
FunctionsUse caseBest for
MeshgridTakes two 1-D coordinate arrays and produces two 2-D coordinate matrices.Ideal for grid-based 3D plotting.
NdgridLike Meshgrid, but flips the order of the first two input and output arguments.Useful when dealing with functions that expect the first input to change across columns, not rows.
GriddataInterpolates scattered data values on a 2-D grid.Perfect when you have data at arbitrary (x,y) points and you want to interpolate them on a regular grid.
Common Pitfalls And Troubleshooting With Meshgrid
Meshgrid is like that loyal dog that's always there for you. But sometimes, even the best dogs can cause trouble. Here are some common Meshgrid hiccups and how to deal with them:
• Memory overload: MATLAB might complain if you try to create a giant grid that's bigger than your memory. The solution? Be modest with your grid sizes, just like how you'd refrain from ordering the entire menu at a restaurant.
• Confusing `meshgrid` with `ndgrid`: Remember, they're as different as cats and dogs. Use our handy-dandy table above to keep them straight.
• Not understanding the output: Meshgrid outputs two matrices, X and Y. X corresponds to the x-coordinates, and Y corresponds to the y-coordinates. These two are like a pair of coordinates for every point in the grid. A bit like sending two kids off to camp, they might come back with stories you don't understand, but trust us, they've had a good time.
Leveraging Meshgrid For Increased Website Traffic
You might be wondering, "What does Meshgrid have to do with my website traffic?" Well, imagine being able to visualize your web traffic data in 3D. With Meshgrid, you can plot the time of day against the number of visitors and the bounce rate to create a three-dimensional graph.
This could reveal patterns you might not see in 2D, like a late-night spike in visits with a high bounce rate. Sounds like you need to adjust your midnight pop-up strategy, doesn't it?
How Well Do You Know Meshgrid in MATLAB?
What does the Meshgrid function in MATLAB do?
Generates a 3D grid
Plays music
Makes coffee
Posts tweets
In the grand theater of MATLAB, Meshgrid plays a leading role. It's a powerful tool that can turn your data into a visual masterpiece. So, roll up your sleeves, get your data ready, and start scripting your MATLAB success story. | 1,183 | 5,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.91007 |
https://convertoctopus.com/195-8-grams-to-kilograms | 1,719,117,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00655.warc.gz | 154,244,576 | 7,468 | ## Conversion formula
The conversion factor from grams to kilograms is 0.001, which means that 1 gram is equal to 0.001 kilograms:
1 g = 0.001 kg
To convert 195.8 grams into kilograms we have to multiply 195.8 by the conversion factor in order to get the mass amount from grams to kilograms. We can also form a simple proportion to calculate the result:
1 g → 0.001 kg
195.8 g → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 195.8 g × 0.001 kg
M(kg) = 0.1958 kg
The final result is:
195.8 g → 0.1958 kg
We conclude that 195.8 grams is equivalent to 0.1958 kilograms:
195.8 grams = 0.1958 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 5.1072522982635 × 195.8 grams.
Another way is saying that 195.8 grams is equal to 1 ÷ 5.1072522982635 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred ninety-five point eight grams is approximately zero point one nine six kilograms:
195.8 g ≅ 0.196 kg
An alternative is also that one kilogram is approximately five point one zero seven times one hundred ninety-five point eight grams.
## Conversion table
### grams to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from grams to kilograms
grams (g) kilograms (kg)
196.8 grams 0.197 kilograms
197.8 grams 0.198 kilograms
198.8 grams 0.199 kilograms
199.8 grams 0.2 kilograms
200.8 grams 0.201 kilograms
201.8 grams 0.202 kilograms
202.8 grams 0.203 kilograms
203.8 grams 0.204 kilograms
204.8 grams 0.205 kilograms
205.8 grams 0.206 kilograms | 477 | 1,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.718693 |
http://www.instructables.com/id/Iron-Man-Arc-Reactor-6-7/ | 1,498,452,679,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320669.83/warc/CC-MAIN-20170626032235-20170626052235-00590.warc.gz | 564,922,210 | 15,797 | Hey everyone,
I am a big Iron Man fan so I decided to make a 3D- printable version of Tony Stark's Arc Reactor.
If you like it pls vote for me in the 123-Design Contest.
Let me know whether you like it or not.
## Step 1: Outer Circle
1) Start by creating 2 Circles:
and a Polygon with 10 edges and a radius of 36 mm.
2) Add a 2.5 mm circle to each of the 10 edges (for led lightning) of the Polygon and remove the Polygon.
3) Extrude the largest circle for 12 mm and then Extrude the inner one with the "cut" option. Extrude all the 2.5 mm circles for 7.5 mm (cut).
4) Add a square with the measures 14mm x 12 mm and 3 ones with 12 mm on one side.
5) Extrude the big square for 14 mm up and -2 mm down. Then extrude the 2 outer squares for 14mm up and -2 mm down (cut). Move the middle square down for 1 mm and extrude it for -1 mm (cut). Then move it up for 14 mm and Extrude it for 1 mm (cut).
6) Copy the first shape and hide one copy. Then select the Extrude tool and select the remaining circle and use the cut tool for -14 mm.
7) Copy paste and rotate it 10 times.
## Step 2:
1) Add 4 circles like I did in the picture
2) Extrude the 2nd smallest circle and extrude it for 6 mm and then extrude the smallest circle (cut). Then extrude the biggest circle for 3 mm and cut it with the 2nd biggest.
3) Add the shape I did and copy, paste and rotate it for 10 ° until the circle is full. Now extrude all of them.
4) Draw the a rectangle (5x15) and the other shape that goes over 3 holes. copy paste it and rotate it once for 120 ° and once for -120°.
5) Extrde the sketches (join) for 3 mm
6) Use the combine tool
7) Draw 5 circles and extrude them (everyone is 2mm wide)
8) redraw the solid for stabilisation
9) Draw the bottom plate
## Step 3: Finish
1) Add another cylinder under the reactor where you can put more leds under and under that another one for all the electronics.
2) Make the holder pieces for the cylinders axtrude them and copy/paste them 6 times.
3) Prepare for 3D-Printing
It would have been a very cool design if you hadnt copied it from james bruton on youtube^^ <br>but the way how you copied it, is awesome!!
I agree, but this version wasn't done perfectly like Bruton's version. I don't think the rings go together correctly(going off of the pictures). But I must admit that you are great with using the 123D program! I'm to accustom to Inventor.
Lol! I have never heard of this guy before. The only reference I had was this picture: http://www.therpf.com/attachments/f9/iron-man-arc-reactor-all-metal-cnc-1st-arc-reactor-real.jpg-234544d1379777156 <br>Can you send me the video-link of this guy ? | 735 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-26 | longest | en | 0.8493 |
https://www.physicsforums.com/threads/why-can-the-total-force-be-taken-as-if-applied-on-the-com-rotational-mechanics.662991/ | 1,510,969,361,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804125.49/warc/CC-MAIN-20171118002717-20171118022717-00711.warc.gz | 888,557,207 | 18,477 | Why can the total force be taken as if applied on the COM. (rotational mechanics?)
1. Jan 7, 2013
Swimmingly!
Why can the total force exerted on an object be taken as if a single force was applied on the center of mass?
I think at most the total force must be the sum of tiny equal forces uniformly distributed. The mass must at most be uniformly distributed too.
And this only matters when we start talking about rotational mechanics, right? i.e: Torques. So that the torque can be calculated as $\tau$=$R \times{F}_{T}$. But why is this true?
R being the vector of the position of the COM.
I think it can be explained this way:Using the known equation:
$\boldsymbol{L}=\boldsymbol{L}_{CM}+\boldsymbol{L}_{spin}$
${L}_{CM}=$ is the L of a point particle in the COM with mass M.
${L}_{spin}$ is the L of body relative to it's center of mass.
Now by symmetry. Since ${L}_{spin}$ is on the center of mass it must be 0. There's an equal amount of tiny forces on each side of the body which makes the torque 0.
Can anyone just give me some insight into this? I just want to understand this basic clearly and there may be a simpler broader explanation.
Last edited: Jan 7, 2013
2. Jan 8, 2013
Simon Bridge
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
Why can the total force exerted on an object be taken as if a single force was applied on the center of mass?[/quote]In general, it can't. You express the applied force in components through the com and perpendicular to that.
3. Jan 8, 2013
BruceW
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
luckily, I have just been going over rotational dynamics, so here's what I got! Firstly, I should say this is all for the motion of a rigid body. These equations won't work if the object is not rigid. First up, for any two points which are both fixed with respect to the rigid body:
$$\frac{d( \vec{r}'- \vec{r})}{dt}= \vec{\omega} \wedge (\vec{r}'-\vec{r})$$
Which I think is just another way of saying Euler's rotation theorem. Now, from this, if we calculate the total momentum of a rigid body we get:
$$\vec{P}=M \frac{d \vec{r}}{dt} + M \ \vec{\omega} \wedge (\vec{R} - \vec{r})$$
Where $\vec{P}$ is the total momentum of the rigid body, $M$ is the total mass of the rigid body, $\vec{R}$ is the centre of mass of the rigid body and $\vec{r}$ is any arbitrary point which is fixed with respect to the rigid body. Finally, $\vec{\omega}$ represents the instantaneous rotation of the rigid body.
So, now if we choose $\vec{r}=\vec{R}$ (which is always allowed, since the centre of mass is always a fixed point with respect to the rigid body), then the equation simplifies very nicely:
$$\vec{P}=M \frac{d \vec{R}}{dt}$$
So we see the total momentum of a rigid object is due to the velocity of its centre of mass, times by the total mass of the object. Now, using a similar, but longer calculation, to get the total angular momentum of the rigid body, gives the equation:
$$\vec{L}= \vec{R} \wedge \vec{P} + \underline{I_{(\vec{r})}} \cdot \vec{\omega} - M(\vec{r} - \vec{R}) \wedge \frac{d(\vec{r} - \vec{R})}{dt}$$
Where $\underline{I_{(\vec{r})}}$ is the inertia matrix, calculated around the point $\vec{r}$ (which, again, is any point which is fixed with respect to the rigid body). So from the above equation, you can see that the total angular momentum is due to the angular momentum of the centre of mass with respect to the origin (the first term), and the second term is due to the angular momentum of the rigid body around some point which is fixed with respect to the rigid body, and the third term has no easy interpretation, but it is sort of like the negative of the angular momentum of the fixed point with respect to the centre of mass.
Now, remember the fixed point can be any point which is fixed with respect to the rigid body. so if you are given the the inertia matrix around some point which is not the centre of mass, then you can use the above formula. But if you are given the inertia matrix around the centre of mass, then you are very lucky, because the third term disappears, and we get:
$$\vec{L}= \vec{R} \wedge \vec{P} + \underline{I_{(\vec{R})}} \cdot \vec{\omega}$$
Which is nicer. (and this is why the inertia matrix around the centre of mass is most often used).
4. Jan 8, 2013
BruceW
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
So is this nice property of the angular momentum around the centre of mass due to lots of equal and opposite forces? Not necessarily. You don't have to assume that to get these equations. (Although it is probably a sufficient condition).
To get these equations, the only thing you really need to do is assume that it is a rigid body, and then the rest will follow.
Edit: I'm not even that sure that equal and opposite forces are a sufficient condition... Obviously, they don't imply a rigid body, so I don't think equal and opposite forces are a useful assumption.
Last edited: Jan 8, 2013
5. Jan 11, 2013
Swimmingly!
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
Sorry if my exposition wasn't so good.
I had done this when I posted but I was doubtful about something, it seems fine though.
I just have to take the derivative applying the product rule:
$\boldsymbol{L}=\boldsymbol{L}_{CM}+\boldsymbol{L}_{spin}$
$\boldsymbol{\tau}=\boldsymbol{\tau}_{CM}+ \boldsymbol{\tau}_{spin}$, but $\boldsymbol{\tau}_{spin} = 0$ *
$\boldsymbol{\tau}=\boldsymbol{\tau}_{CM}$
QED
*If a is constant: $\boldsymbol {\tau}_{spin} = \sum {r_{i} \times m_{i}a} =( \sum {m_{i}r_{i}}) \times a=0 \times a=0$
Bruce, I think you demonstrated the first equation correct? I accepted that but I was suspicious of it's direct result through derivation, I can't remember why but oh well.
If I misunderstood something I'm sorry. I didn't know about Euler's rotational theorem. Thanks!
Edit, clear up: The question was basically this.
If a pencil is nailed to wall. If there's no friction and the pencil's horizontal what's the torque on the pencil? We just take the total force, put it on the center of mass of the pencil and torque=r*F, with r=length/2. Why is this legal?
Last edited: Jan 11, 2013
6. Jan 12, 2013
BruceW
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
Yep, that's it. The torque around the centre of mass due to gravity is zero. In your equation, 'a' should really be $\vec{g}$, since there may be other forces which can cause torque around the centre of mass, so 'a' can generally be something else, but the important thing is that the force due to gravity causes zero torque around the centre of mass.
Also, it's important to remember that this is only true when the gravitational field is uniform. (That is why you were able to say that the acceleration due to gravity is the same on each small element of the rigid body). So if the gravitational field varied throughout space, then generally it would cause a torque around the centre of mass.
7. Jan 13, 2013
andrien
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
it is simple.Just put two equal and opposite forces on center of mass,now this does not change anything.You can see that one force of it and the force at other end will form a couple which is equal to torque also the residing force at center of mass will be in the same direction as the applied force on one end.So you can think that force at one end could be transferred to COM and then an additional torque is needed.
8. Jan 14, 2013
Simon Bridge
Re: Why can the total force be taken as if applied on the COM. (rotational mechanics?
Let's make it easier - so we don't have to worry about the reaction at the wall ... as much.
mount the pencil on a pivot at the wall and hold it horizontal then let go.
At the instant you let go - the total torque is taken as MgL/2 where M is the total mass of the pencil and L it's overall length.
This assumes a uniform mass-distribution for the pencil - so it is very blunt and has no eraser on the other end ;)
This is "legal" because it agrees with the results of experiments.... i.e. the Universe works like this. Physics is an empirical science, so this is the bottom line.
We can see that it is mathematically consistent with the rest of our physical models by considering how the total torque is made up ... gravity, after all, acts on the entire length of the pencil. Thus, you can work out the torque about the pivot due to gravity acting on each bit of the pencil and add them up (provided you are comfortable with torques adding up?)
I doubt it will be very convincing if I do the math here - so you will have to. | 2,232 | 8,757 | {"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} | 3.84375 | 4 | CC-MAIN-2017-47 | longest | en | 0.930977 |
https://www.smorescience.com/algebra-formulas/ | 1,721,265,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00772.warc.gz | 851,324,274 | 161,878 | # Algebra Formulas
Algebra is a fundamental branch of mathematics that deals with the manipulation of variables and equations. Mastering algebra formulas is crucial for solving a wide range of mathematical problems, from simple linear equations to complex systems of equations. Here are some of the most important algebra formulas you should know:
# ALGEBRA FORMULAS
## Arithmetic Properties
$\begin{array}{rl}& \left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ & \left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}\\ & {a}^{2}+{b}^{2}=\left(a+b{\right)}^{2}-2ab\\ & {a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)\\ & \left(a+b+c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca\\ & \left(a+b-c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab-2bc-2ca\\ & \left(a-b-c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}-2ab+2bc-2ca\\ & \left(a+b{\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\\ & \phantom{\rule{1em}{0ex}}={a}^{3}+{b}^{3}+3ab\left(a+b\right)\\ & \left(a-b{\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\\ & \phantom{\rule{1em}{0ex}}={a}^{3}-{b}^{3}-3ab\left(a-b\right)\\ & {a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)\\ & {a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)\\ & \left(a+b{\right)}^{4}={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}\\ & \left(a-b{\right)}^{4}={a}^{4}-4{a}^{3}b+6{a}^{2}{b}^{2}-4a{b}^{3}+{b}^{4}\\ & {a}^{4}-{b}^{4}=\left(a+b\right)\left(a-b\right)\left({a}^{2}+{b}^{2}\right)\\ & {a}^{5}-{b}^{5}=\left(a-b\right)\left({a}^{4}+{a}^{3}b+{a}^{2}{b}^{2}+a{b}^{3}+{b}^{4}\right)\end{array}$$\begin{array}{r}\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ \left(a−b{\right)}^{2}={a}^{2}−2ab+{b}^{2}\\ {a}^{2}+{b}^{2}=\left(a+b{\right)}^{2}−2ab\\ {a}^{2}−{b}^{2}=\left(a+b\right)\left(a−b\right)\\ \left(a+b+c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca\\ \left(a+b−c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab−2bc−2ca\\ \left(a−b−c{\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}−2ab+2bc−2ca\\ \left(a+b{\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\\ \phantom{\rule{1em}{0ex}}={a}^{3}+{b}^{3}+3ab\left(a+b\right)\\ \left(a−b{\right)}^{3}={a}^{3}−3{a}^{2}b+3a{b}^{2}−{b}^{3}\\ \phantom{\rule{1em}{0ex}}={a}^{3}−{b}^{3}−3ab\left(a−b\right)\\ {a}^{3}−{b}^{3}=\left(a−b\right)\left({a}^{2}+ab+{b}^{2}\right)\\ {a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}−ab+{b}^{2}\right)\\ \left(a+b{\right)}^{4}={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}\\ \left(a−b{\right)}^{4}={a}^{4}−4{a}^{3}b+6{a}^{2}{b}^{2}−4a{b}^{3}+{b}^{4}\\ {a}^{4}−{b}^{4}=\left(a+b\right)\left(a−b\right)\left({a}^{2}+{b}^{2}\right)\\ {a}^{5}−{b}^{5}=\left(a−b\right)\left({a}^{4}+{a}^{3}b+{a}^{2}{b}^{2}+a{b}^{3}+{b}^{4}\right)\end{array}${:[(a+b)^(2)=a^(2)+2ab+b^(2)],[(a-b)^(2)=a^(2)-2ab+b^(2)],[a^(2)+b^(2)=(a+b)^(2)-2ab],[a^(2)-b^(2)=(a+b)(a-b)],[(a+b+c)^(2)=a^(2)+b^(2)+c^(2)+2ab+2bc+2ca],[(a+b-c)^(2)=a^(2)+b^(2)+c^(2)+2ab-2bc-2ca],[(a-b-c)^(2)=a^(2)+b^(2)+c^(2)-2ab+2bc-2ca],[(a+b)^(3)=a^(3)+3a^(2)b+3ab^(2)+b^(3)],[quad=a^(3)+b^(3)+3ab(a+b)],[(a-b)^(3)=a^(3)-3a^(2)b+3ab^(2)-b^(3)],[quad=a^(3)-b^(3)-3ab(a-b)],[a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2))],[a^(3)+b^(3)=(a+b)(a^(2)-ab+b^(2))],[(a+b)^(4)=a^(4)+4a^(3)b+6a^(2)b^(2)+4ab^(3)+b^(4)],[(a-b)^(4)=a^(4)-4a^(3)b+6a^(2)b^(2)-4ab^(3)+b^(4)],[a^(4)-b^(4)=(a+b)(a-b)(a^(2)+b^(2))],[a^(5)-b^(5)=(a-b)(a^(4)+a^(3)b+a^(2)b^(2)+ab^(3)+b^(4))]:}\begin{aligned} & (a+b)^{2}=a^{2}+2 a b+b^{2} \\ & (a-b)^{2}=a^{2}-2 a b+b^{2} \\ & a^{2}+b^{2}=(a+b)^{2}-2 a b \\ & a^{2}-b^{2}=(a+b)(a-b) \\ & (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\ & (a+b-c)^{2}=a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a \\ & (a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2 a b+2 b c-2 c a \\ & (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\ & \quad=a^{3}+b^{3}+3 a b(a+b) \\ & (a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\ & \quad=a^{3}-b^{3}-3 a b(a-b) \\ & a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \\ & a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right) \\ & (a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\ & (a-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4} \\ & a^{4}-b^{4}=(a+b)(a-b)\left(a^{2}+b^{2}\right) \\ & a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) \end{aligned} | 2,426 | 4,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.349787 |
http://mathoverflow.net/questions/59207/fixing-a-mistake-in-an-introduction-to-invariants-and-moduli?sort=votes | 1,455,165,988,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701160958.15/warc/CC-MAIN-20160205193920-00161-ip-10-236-182-209.ec2.internal.warc.gz | 139,218,528 | 15,818 | # Fixing a mistake in “An introduction to invariants and moduli”
On page 13 of the book "An introduction to invariants and moduli" of Mukai http://catdir.loc.gov/catdir/samples/cam033/2002023422.pdf there is a mistake, in the end of the proof of Proposition 1.9. It seems to me that this proof can not be fixed, without using the notion of Noetherian rings and Hilbert basis theorem.
The question is: Can this proof be fixed, without using commutative algebra -- i.e., by the elementary reasoning that Mukai is using there?
I reproduce here the proof from the book for completeness. $S$ is the ring of polynomials, $G$ a group, $S^G$ is the ring of invariants
Proposition. If $S^G$ is generated by homogeneous polynomials $f_1,...,f_r$ of degrees $d_1,...,d_r$, then the Hilbert series of $S^G$ is the power series expansion at $t=0$ of a rational function $$P(t)=\frac{F(t)}{(1-t^{d_1})...(1-t^{d_r})}$$ for some $F(t)\in \mathbb Z[t]$.
Proof. We use induction on $r$, observing that when $r=1$, the ring $S^G$ is just $\mathbb C[f_1]$ with the Hilbert series $$P(t)=1+t^{d_1}+t^{2d_1}+...=\frac{1}{1-t^{d_1}}.$$ For $r>1$ consider the injective complex linear map $S^G\to S^G$ defined by $h\to f_rh$. Denote the image by $R\subset S^G$ and consider the Hilbert series for the graded rings $R$ and $S^G/R$. Since $R$ and $S^G/R$ are generated by homogeneous elements, we have $$P_{S^G}(t)=P_{R}(t)+P_{S^G/R}(t).$$ On the other hand, $dim(S^G\cap S_d)=dim(R\cap S_{d+d_r})$, so that $P_R(t)=t^{d_r}P_{S^G}(t)$, and hence $$P_{S^G}(t)=\frac{P_{S^G/R}(t)}{1-t^{d_r}}.$$ But $S^G/R$ is isomorphic to the subring of $S$ generated by the polynomials $f_1,...,f_{r-1}$, and hence by the induction hypothesis $P_{S^G/R}(t)=F(t)/(1-t^{d_1})...(1-t^{d_{r-1}})$ for some $F(t)\in \mathbb Z[t]$...
Mistake: It is not true that $S^G/R$ is isomorphic to the subring of $S$ generated by polynomials $f_1,...,f_{r-1}$. For example consider $\mathbb C^2$ with action $(x,y)\to (-x,-y)$. Then let $f_1=x^2$, $f_2=y^2$, $f_3=xy$.
Motiviation of this question. Of course this proposition is a partial case of Hilbert-Serre theorem, proven for example at the end of Atiyah-Macdonald. But the point of the introduction in the above book is that one does not use any result of commutative algebra.
-
Attiyah-McDonalds is a very funny typo. – Rasmus Bentmann Mar 22 '11 at 22:11
Thanks! That maid me laugh too! But I'll change it nevertheless :) – aglearner Mar 22 '11 at 22:36
@aglearner: Maconald is less funny, but still a typo! – Hailong Dao Mar 23 '11 at 0:22
I've heard many people complain about frequent errors in Mukai's book. – bavajee Mar 23 '11 at 1:15
This is not an answer to the question, I just decided to give for completeness a standard proof of the above statement that uses (a version of) Hilbert-Serre theorem. In this proof we need to use Hilbert basis theorem. In the above statement $S^G$ is clearly a finitely generated graded module over the ring of polynomials $\mathbb C[x_1,...,x_r]$, so it is sufficient to prove:
Theorem (Hilbert, Serre). Suppose that $S=\sum ^{\infty}_{j=0}S_j$ is a commutative graded ring with $A_0=\mathbb C$, finitely generated over $\mathbb C$ by homogeneous elements $x_1,...,x_r$ in positive degrees $d_1,...,d_r$. Suppose that $M=\sum_{j=0}^{\infty} M_j$ is a finitely generated graded $S$-module (i.e., we have $S_iS_j\subset S_{i+j}$ and $S_iM_j\subset M_{i+j}$). Then the Hilbert series $P(M,t)$ is of the form $$\sum_{j=0}^{\infty}dim(M_j)t^j=P(M,t)=\frac{F(t)}{\Pi_{j=1}^r(1-t^{d_j})}, \;\; F(t)\in \mathbb Z[t].$$
Proof. We work by induction on $r$. If $r=0$ then $P(M,t)$ is a polynomial with integer coefficients, so suppose $r>0$. Denote by $M'$ and $M''$ the kernel and cokernel of the multiplication by $x_r$, we have an exact sequence for each $j$ $$0\to M'_j\to M_j \to^{x_r}M_{j+d_r}\to M''_{j+d_r}\to 0.$$ Now $M'$ and $M''$ are finitely generated graded modules for $K[x_1,...,x_{r-1}]$, and so by induction their Hilbert series have the given form. From the above exact sequence we have $$t^{d_r}P(M',t)-t^{d_r}P(M,t)+P(M,t)-P(M'',t)=0.$$ Thus $$P(M,t)=\frac{P(M'',t)-t^{d_r}P(M',t)}{1-t^{d_r}}$$ has the given form.
Where did we use Hilbert basis theorem? We use it when we say that $M'$ is finitely generated. | 1,455 | 4,271 | {"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} | 3.265625 | 3 | CC-MAIN-2016-07 | longest | en | 0.788848 |
judahwnamx.pages10.com | 1,560,952,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998986.11/warc/CC-MAIN-20190619123854-20190619145854-00415.warc.gz | 98,998,687 | 6,432 | # The smart Trick of simulink matlab help That Nobody is Discussing
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https://www.zbmath.org/?q=ra%3Asacchetti.lorenzo+ai%3Amontagna.franco+ai%3Amontagna.franco+se%3A2008 | 1,632,840,855,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00206.warc.gz | 1,086,569,877 | 10,527 | # zbMATH — the first resource for mathematics
A note on the first-order logic of complete BL-chains. (English) Zbl 1152.03019
The relationship of the set of predicate tautologies of all BL-chains and the set of predicate formulas valid in all standard BL-algebras (i.e., the set of all standard tautologies) is discussed. This paper is a continuation of the papers of F. Montagna and L. Sacchetti [ibid. 49, No. 6, 629–641 (2003; Zbl 1035.03010)] and [ibid. 50, No. 1, 104–107 (2004; Zbl 1039.03013)], and its main result shows that a coomplete BL-chain $$B$$ satisfies all standard BL-tautologies if and only if for any transfinite sequence $$(b_i: i\in I)$$ in $$B$$, there holds $$\bigwedge_{i\in I}(b^2_i)= (\bigwedge_{i\in I} b_i)^2$$. Another equivalent condition is this one: the formula
$\forall x(\varphi(x)\&\varphi(x))\to ((\forall x\varphi(x))\&(\forall x\varphi(x)))$
is valid in $$B$$.
##### MSC:
03B52 Fuzzy logic; logic of vagueness 03B50 Many-valued logic
Full Text:
##### References:
[1] Aglianó, Varieties of basic algebras I: general properties, J. Pure Appl. Algebra 181 pp 105– (2003) [2] A. Glass, Partially Ordered Groups. Series in Algebra 7 (World Scientific, 1999). · Zbl 0933.06010 [3] P. Hájek, Metamathematics of Fuzzy Logic (Kluwer, 1998). · Zbl 0937.03030 [4] Hájek, Basic fuzzy logic and BL-algebras, Soft Computing 2 pp 124– (1998) · Zbl 05469956 [5] Hájek, Fuzzy logic and arithmetical hierarchy III, Studia Logica 68 pp 129– (2001) · Zbl 0988.03042 [6] Hájek, Basic fuzzy logic and BL-algebras II, Soft Computing 7 pp 179– (2003) · Zbl 1018.03021 [7] Hájek, Arithmetical complexity and fuzzy predicate logics: a survey, Soft Computing 30 pp 1– (2005) · Zbl 1093.03012 [8] Hájek, On theories and models in fuzzy predicate models, J. Symbolic Logic 71 pp 863– (2006) [9] Montagna, Three complexity problems in quantified fuzzy logic, Studia Logica 68 pp 143– (2001) · Zbl 0985.03014 [10] Montagna, Kripke-style semantics for many-valued logic, Math. Log. Quart. 49 pp 629– (2003) · Zbl 1035.03010 [11] Montagna, Corrigendum to ”Kripke-style semantics for many-valued logic”, Math. Log. Quart. 50 pp 104– (2004) [12] Mundici, Interpretations of AF C *-algebras in Łukasiewicz sentential calculus, J. Functional Analysis 65 pp 15– (1986) · Zbl 0597.46059 [13] M. E. Ragaz, Arithmetische Klassifikation von Formelmengen der unendlichwertigen Logik. Ph. D. thesis, ETH Zürich (1981). · Zbl 0516.03011 [14] Ward, Residuated lattices, Trans. Amer. Math. Soc. 45 pp 335– (1939)
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. | 931 | 2,880 | {"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} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.632452 |
https://blog.fiddler.ai/2019/08/should-you-explain-your-predictions-with-shap-or-ig/?utm_source=rss&utm_medium=rss&utm_campaign=should-you-explain-your-predictions-with-shap-or-ig | 1,566,313,294,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315544.11/warc/CC-MAIN-20190820133527-20190820155527-00027.warc.gz | 386,178,824 | 14,074 | # Should you explain your predictions with SHAP or IG?
Two different explanation algorithm types, best in different situations.
Some of the most accurate predictive models today are black box models, meaning it is hard to really understand how they work. To address this problem, techniques have arisen to understand feature importance: for a given prediction, how important is each input feature value to that prediction? Two well-known techniques are SHapley Additive exPlanations (SHAP) and Integrated Gradients (IG). In fact, they each represent a different type of explanation algorithm: a Shapley-value-based algorithm (SHAP) and a gradient-based algorithm (IG).
There is a fundamental difference between these two algorithm types. This post describes that difference. First, we need some background. Below, we review Shapley values, Shapley-value-based methods (including SHAP), and gradient-based methods (including IG). Finally, we get back to our central question: When should you use a Shapley-value-based algorithm (like SHAP) versus a gradient-based explanation explanation algorithm (like IG)?
# What are Shapley values?
The Shapley value (proposed by Lloyd Shapley in 1953) is a classic method to distribute the total gains of a collaborative game to a coalition of cooperating players. It is provably the only distribution with certain desirable properties (fully listed on Wikipedia).
In our case, we formulate a game for the prediction at each instance. We consider the “total gains” to be the prediction value for that instance, and the “players” to be the model features of that instance. The collaborative game is all of the model features cooperating to form a prediction value. The Shapley value efficiency property says the feature attributions should sum to the prediction value. The attributions can be negative or positive, since a feature can lower or raise a predicted value.
There is a variant called the Aumann-Shapley value, extending the definition of the Shapley value to a game with many (or infinitely many) players, where each player plays only a minor role, if the worth function (the gains from including a coalition of players) is differentiable.
# What is a Shapley-value-based explanation method?
A Shapley-value-based explanation method tries to approximate Shapley values of a given prediction by examining the effect of removing a feature under all possible combinations of presence or absence of the other features. In other words, this method looks at function values over subsets of features like F(x1, <absent>, x3, x4, …, <absent>, , xn). How to evaluate a function F with one or more absent features is subtle.
For example, SHAP (SHapely Additive exPlanations) estimates the model’s behavior on an input with certain features absent by averaging over samples from those features drawn from the training set. In other words, F(x1, <absent>, x3, …, xn) is estimated by the expected prediction when the missing feature x2 is sampled from the dataset.
Exactly how that sample is chosen is important (for example marginal versus conditional distribution versus cluster centers of background data), but I will skip the fine details here.
Once we define the model function (F) for all subsets of the features, we can apply the Shapley values algorithm to compute feature attributions. Each feature’s Shapley value is the contribution of the feature for all possible subsets of the other features.
The “kernel SHAP” method from the SHAP paper computes the Shapley values of all features simultaneously by defining a weighted least squares regression whose solution is the Shapley values for all the features.
The high-level point is that all these methods rely on taking subsets of features. This makes the theoretical version exponential in runtime: for N features, there are 2N combinations of presence and absence. That is too expensive for most N, so these methods approximate. Even with approximations, kernel SHAP can be slow. Also, we don’t know of any systematic study of how good the approximation is.
There are versions of SHAP specialized to different model architectures for speed. For example, Tree SHAP computes all the subsets by cleverly keeping track of what proportion of all possible subsets flow down into each of the leaves of the tree. However, if your model architecture does not have a specialized algorithm like this, you have to fall back on kernel SHAP, or another naive (unoptimized) Shapley-value-based method.
A Shapley-value-based method is attractive as it only requires black box access to the model (i.e. computing outputs from inputs), and there is a version agnostic to the model architecture. For instance, it does not matter whether the model function is discrete or continuous. The downside is that exactly computing the subsets is exponential in the number of features.
# What is a gradient-based explanation method?
A gradient-based explanation method tries to explain a given prediction by using the gradient of (i.e. change in) the output with respect to the input features. Some methods like Integrated Gradients (IG), GradCAM, and SmoothGrad literally apply the gradient operator. Other methods like DeepLift and LRP apply “discrete gradients.”
Let me describe IG, which has the advantage that it tries to approximate Aumann-Shapley values, which are axiomatically justified. IG operates by considering a straight line path, in feature space, from the input at hand (e.g., an image from a training set) to a certain baseline input (e.g., a black image), and integrating the gradient of the prediction with respect to input features (e.g., image pixels) along this path.
This paper explains the intuition of the IG algorithm as follows. As the input varies along the straight line path between the baseline and the input at hand, the prediction moves along a trajectory from uncertainty to certainty (the final prediction probability). At each point on this trajectory, one can use the gradient with respect to the input features to attribute the change in the prediction probability back to the input features. IG aggregates these gradients along the trajectory using a path integral.
IG (roughly) requires the prediction to be a continuous and piecewise differentiable function of the input features. (More precisely, it requires the function is continuous everywhere and the partial derivative along each input dimension satisfies Lebesgue’s integrability condition, i.e., the set of discontinuous points has measure zero.)
Note it is important to choose a good baseline for IG to make sensible feature attributions. For example, if a black image is chosen as baseline, IG won’t attribute importance to a completely black pixel in an actual image. The baseline value should both have a near-zero prediction, and also faithfully represent a complete absence of signal.
IG is attractive as it is broadly applicable to all differentiable models, easy to implement in most machine learning frameworks (e.g., TensorFlow, PyTorch, Caffe), and computationally scalable to massive deep networks like Inception and ResNet with millions of neurons.
# When should you use a Shapley-value-based versus a gradient-based explanation method?
Finally, the payoff! Our advice: If the model function is piecewise differentiable and you have access to the model gradient, use IG. Otherwise, use a Shapley-value-based method.
Any model trained using gradient descent is differentiable. For example: neural networks, logistic regression, support vector machines. You can use IG with these. The major class of non-differentiable models is trees: boosted trees, random forests. They encode discrete values at the leaves. These require a Shapley-value-based method, like Tree SHAP.
The IG algorithm is faster than a naive Shapley-value-based method like kernel SHAP, as it only requires computing the gradients of the model output on a few different inputs (typically 50). In contrast, a Shapley-value-based method requires computing the model output on a large number of inputs sampled from the exponentially huge subspace of all possible combinations of feature values. Computing gradients of differentiable models is efficient and well supported in most machine learning frameworks. However, a differentiable model is a prerequisite for IG. By contrast, a Shapley-value-based method makes no such assumptions.
Several types of input features that look discrete (hence might require a Shapley-value-based method) actually can be mapped to differentiable model types (which let us use IG). Let us walk through one example: text sentiment. Suppose we wish to attribute the sentiment prediction to the words in some input text. At first, it seems that such models may be non-differentiable as the input is discrete (a collection of words). However, differentiable models like deep neural networks can handle words by first mapping them to a high-dimensional continuous space using word embeddings. The model’s prediction is a differentiable function of these embeddings. This makes it amenable to IG. Specifically, we attribute the prediction score to the embedding vectors. Since attributions are additive, we sum the attributions (retaining the sign) along the fields of each embedding vector and map it to the specific input word that the embedding corresponds to.
A crucial question for IG is: what is the baseline prediction? For this text example, one option is to use the embedding vector corresponding to empty text. Some models take fixed length inputs by padding short sentences with a special “no word” token. In such cases, we can take the baseline as the embedding of a sentence with just “no word” tokens.
For more on IG, see the paper or this how-to.
# Conclusion
In many cases (a differentiable model with a gradient), you can use integrated gradients (IG) to get a more certain and possibly faster explanation of feature importance for a prediction. However, a Shapley-value-based method is required for other (non-differentiable) model types.
At Fiddler, we support both SHAP and IG. (Full disclosure: Ankur Taly, a co-author of IG, works at Fiddler, and is a co-author of this post.) Feel free to email info@fiddler.ai for more information, or just to say hi! | 2,081 | 10,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-35 | latest | en | 0.909541 |
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# Solution - The Storage Battery of a Car Has an Emf of 12 V. If the Internal Resistance of the Battery is 0.4Ω,, What is the Maximum Current that Can Be Drawn from the Battery? - Cells, Emf, Internal Resistance
ConceptCells, Emf, Internal Resistance
#### Question
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?
#### Solution
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#### APPEARS IN
NCERT Physics Textbook for Class 12 Part 1
Chapter 3: Current Electricity
Q: 1 | Page no. 127
#### Reference Material
Solution for question: The Storage Battery of a Car Has an Emf of 12 V. If the Internal Resistance of the Battery is 0.4Ω,, What is the Maximum Current that Can Be Drawn from the Battery? concept: Cells, Emf, Internal Resistance. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
S | 259 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.86728 |
http://www.physicsforums.com/showthread.php?s=480a4793d6500819fc429c69d4c43441&p=4636072 | 1,394,600,646,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021365169/warc/CC-MAIN-20140305120925-00074-ip-10-183-142-35.ec2.internal.warc.gz | 482,643,997 | 9,052 | # How do you find the temperature of the boomerang nebula?
by melodyman888
Tags: blackbody, boomerang, nebula, stefan-boltzmann, temperature, temperature and heat, wiens
P: 2 So i have a physics presentation to do on tuesday and i have no idea how to find the temperature of a of objects in space, specifically the boomeran nebula. I did some research and found stuff involving black body radiation, wiens law, and stefan boltzman law but i dont really understand it too well. Some of the things i dont understand: wiens law states that wavelength(in meters) is equal to 0.0029/temperature(K). so is finding the temperature of the boomerang nebula really as easy as simply searching for the wavelength? is there a mathematical way to find the wavelength? how can something so cold be so bright? does the black body radiation curve have anything to do with this? If none of this is the right approach to finding the temperature, how did they do it? Any help will be greatly appreciated thanks, melodyman888
PF Gold
P: 2,176
hi there
welcome to PF
never heard of the boomerang nebula, what is its NGC or IC number ?
how can something so cold be so bright?
it may be a reflection nebula of which a large majority are
BTW DONT post the same question in multiple forum sections
it leads to lots of confusion!
cheers
Dave
Homework Sci Advisor HW Helper Thanks PF Gold P: 10,313 Basically the color of a hot object depends on it's temperature... so, yes, it is a matter of measuring the wavelengths of the light the gas gives off. The overall spectrum is usually needed though. The brightness of the light depends on how big the object is, and how far away it is, as well as it's temperature. You'll find the nebula is exremely big.
PF Gold
P: 10,793
## How do you find the temperature of the boomerang nebula?
Note that the Boomerang Nebula is visible because it reflects and scatters light from the central star, not because it is glowing.
PF Gold P: 10,793 As I said in your other thread, the nebula is a reflection nebula, so it's visible because of the light it reflects scatters, not because it emits light. I'm afraid I don't know how they found the temperature though.
P: 81 This paper describes how they did it.
P: 8 Luminosity, which is a measure of the total output power of a star(unit: Watts) can be used to find the temperature of the surrounding stars. The luminosity of a star is proportional to its radius and temperature raised to the power 4.That is, L=σ x 4πr2x T4 Wien's law, on the other hand is a direct measure of the surface temperature of the star from the peak wavelength λmof the radiation spectrum emitted by the star.That is, λm x T = 2.90x10-3 m K
P: 2 Ok this helps guys, still not completely sure how they did it but i will keep trying to figure it out Thanks, melodyman888
Homework Sci Advisor HW Helper Thanks PF Gold P: 10,313 Did you want to know how it was actually done, historically, or the methods by which it is done today? You have been told two things above: 1. the principles by which is possible to discover the temperature of distant objects; 2. The exact method that was used to discover the temperature of this nebula (the link from post #6). The trouble with the second is that the description assumes you already have a detailed knowledge of the first. The relationship between the color an object glows and it's temperature is not a simple one and a lot of work had to go into figuring these things out first. I think, though, that the descriptions of (1) missed something ... you can also tell the temperature of something that does not glow by how dark it is ... and, specifically, in what way it is dark. When light from a star shines through a nebula - different frequencies get blocked by different amounts. The amounts depend on what the nebula is made of and the nebulas temperature. If we know what the light is supposed to look like with nothing in the way, then we can figure out the temperature and composition of the bits of the nebula in the way. In this specific case: http://www.sci-news.com/astronomy/sc...ula-01493.html ... astronomers used the radiation left over from the "big bang" to do the comparison. This radiation is much the same from all directions so we know what it is supposed to be. But where it passes through a cold nebula to get here, it is different.
Related Discussions Mechanical Engineering 2 Classical Physics 4 General Physics 16 General Astronomy 4 | 1,041 | 4,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.924934 |
https://en.lntwww.de/Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich | 1,713,320,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00778.warc.gz | 206,835,188 | 11,567 | # System Description in Time Domain
## Impulse response
In the section »The first Fourier integral« in the book »Signal Representation» it was explained that for any deterministic signal $x(t)$ a »spectral function« $X(f)$ can be given with the help of the Fourier transform. Often $X(f)$ is referred to as the »spectrum« for short.
However, all information about the spectral function is already contained in the time domain representation, even if not always immediately recognizable. The same facts apply to linear time-invariant systems.
$\text{Definition:}$ The most important descriptive quantity of a linear time-invariant system in the time domain is the inverse Fourier transform of $H(f)$, which is called the »impulse response«:
$$h(t) = \int_{-\infty}^{+\infty}H(f) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi ft}\hspace{0.15cm} {\rm d}f.$$
The following should be noted in this regard:
• The frequency response $H(f)$ and the impulse response $h(t)$ are equivalent descriptive quantities that contain exactly the same information about the LTI system.
• If the Dirac-shaped input signal $x(t) = δ(t)$ is used, then $X(f) = 1$ is to be set and $Y(f) = H(f)$ resp. $y(t) = h(t)$ are valid.
• The term »impulse response« reflects this statement: $h(t)$ is the response of the system to a $($Dirac delta$)$ function as the input signal.
• The above definition suggests that any impulse response must have the unit $\text{Hz = 1/s}$.
$\text{Example 1:}$ The impulse response $h(t)$ of the so-called »rectangular–in–time« filter is constant over a time interval $T$ and is zero outside this time interval.
Rectangular impulse response and associated magnitude spectrum
• The associated amplitude response as the magnitude of the frequency response is
$$\vert H(f)\vert = \vert {\rm si}(\pi fT)\vert \hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x={\rm sinc}(x/\pi).$$
• The area over $h(t)$ is equal to $H(f = 0) = 1$. It follows that:
In the range $0 < t < T$ the impulse response must be constant and equal to $1/T$.
• The phase response is given by
$$b(f) = \left\{ \begin{array}{l} \hspace{0.25cm}\pi/T \\ - \pi/T \\ \end{array} \right.\quad \quad\begin{array}{*{20}c} \text{for} \\ \text{for} \\ \end{array}\begin{array}{*{20}c}{\left \vert \hspace{0.05cm} f\hspace{0.05cm} \right \vert > 0,} \\{\vert \hspace{0.05cm} f \hspace{0.05cm} \vert < 0.} \\\end{array}$$
• With symmetrical $h(t)$ around $t = 0$ $($i.e. non-causal$)$ ⇒ $b(f)=0$.
## Some laws of the Fourier transform
The »Fourier transform theorems« have already been explained in detail in the book »Signal Representation«.
The following is a short summary, where $H(f)$ describes the frequency response of an LTI system and whose inverse Fourier transform $h(t)$ is the impulse response. The laws and principles are applied more frequently in the »$\text{exercises}$« for this chapter »System Description in Time Domain«.
Here, we also refer to the (German language) didactic video "Gesetzmäßigkeiten der Fouriertransformation" ⇒ "Regularities to the Fourier transform".
In the following equations the short symbol of the Fourier transformation is used. The filled-out circle indicates the spectral domain, the white one the time domain.
• »Multiplication« by a constant factor:
$$k \cdot H(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,k \cdot h(t).$$
For $k \lt 1$ one deals with attenuation, while $k \gt 1$ stands for amplification.
• »Similarity Theorem«:
$$H({f}/{k})\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,|k| \cdot h(k\cdot t).$$
1. This implies: Compression $(k < 1)$ of the frequency response results in a wider and lower impulse response.
2. Stretching $(k > 1)$ of $H(f)$ makes $h(t)$ narrower and higher.
• »Displacement Theorem« in the frequency and time domain:
$$H(f - f_0) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t )\cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi f_0 t},$$
$$H(f) \cdot {\rm e}^{-{\rm j}2\pi ft_0}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t- t_0 ).$$
1. A shift by $t_0$ $($»transit time»$)$ thus leads to multiplication by a complex exponential function in the frequency domain.
2. Thereby, the amplitude response $|H(f)|$ does not change.
• »Differentiation Theorem« in the frequency and time domain:
$$\frac{1}{{{\rm j}2\pi }} \cdot \frac{{{\rm d}H( f )}}{{{\rm d}f}} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,- t \cdot h( t ),$$
$${\rm j}\cdot 2\pi f \cdot H( f ){}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \frac{{{\rm d}h( t )}}{{\rm d}t}.$$
1. A differentiating element in the LTI system leads to a multiplication by ${\rm j}\cdot 2πf$ in the frequency domain
2. and thus among other things to a phase rotation by $90^{\circ}$.
## Causal systems
$\text{Definition:}$ An LTI system is said to be $\text{causal}$ if the impulse response $h(t)$ – that is the inverse Fourier transform of the frequency response $H(f)$ – satisfies the following condition:
$$h(t) = 0 \hspace{0.25cm}{\rm for}\hspace{0.25cm} t < 0.$$
If this condition is not met, the system is »non–causal« $($or »acausal«$)$.
$\text{Please note:}$ Any realizable system is causal.
$\text{Example 2:}$ The diagram illustrates the differences between the non–causal system $\rm A$ and the causal system $\rm B$.
Non–causal system $\rm A$ and causal system $\rm B$
• In system $\rm A$ the effect starts earlier $($at $t =\hspace{0.05cm} –T)$ than the cause $($Dirac delta function at $t = 0)$, which of course is not possible in practice.
• Almost all non–causal systems can be transformed into a feasible causal system using a transit time $\tau$.
• For example, with $\tau = T$ the following holds: $h_{\rm B}(t) = h_{\rm A}(t - T).$
All statements made so far apply for causal as well as non–causal systems.
However, for the description of causal systems some specific properties can be used as explained in the third main chapter »Description of Causal Realizable Systems« of $\text{this book}$.
In this first and the following second main chapter we mainly consider non–causal systems since their mathematical description is usually simpler.
• So in this example, the frequency response $H_{\rm A}(f)$ is real,
• while for $H_{\rm B}(f)$ the additional term ${\rm e}^{–{\rm j2π}f\hspace{0.05cm}T}$ has to be considered.
## Computation of the output signal
We consider the following problem: Let the input signal $x(t)$ and the frequency response $H(f)$ be known. The output signal $y(t)$ is to be determined.
To determine the output quantities of an LTI system
If the solution is to be determined in the frequency domain, first the spectrum $X(f)$ must be determined from the given input signal $x(t)$ via $\text{Fourier transform}$ and multiplied by the frequency response $H(f)$. By the $\text{inverse Fourier transform}$ of the product the signal $y(t)$ is obtained.
Here is a summary of the entire solution process:
$${\rm 1.\,\, step\hspace{-0.1cm} :}\hspace{0.5cm} X(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, x( t )\hspace{1.55cm}{\rm input\:spectrum},$$
$${\rm 2.\,\, step\hspace{-0.1cm}:}\hspace{0.5cm}Y(f)= X(f) \cdot H(f) \hspace{0.82cm}{\rm output\:spectrum},$$
$${\rm 3.\,\, step\hspace{-0.1cm}:}\hspace{0.5cm} y(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, Y(f )\hspace{1.55cm}{\rm output\:signal}.$$
The same result is obtained after the computation in the time domain by first determining the impulse response $h(t)$ from the frequency response $H(f)$ by means of the »inverse Forier transform« and then applying the convolution operation:
$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau )} \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
• The results are identical for both approaches.
• Purposefully, the procedure of solution with less computational effort should be chosen.
$\text{Example 3:}$ At the input of a filter with rectangular impulse response $h(t)$ of width $T$ $($see $\text{Example 1)}$ a rectangular pulse $x(t)$ of duration $2T$ is applied.
Trapezoidal output signal since $x(t)$ and $h(t)$ are rectangular
In this case, direct computation in the time domain is more convenient:
• The convolution of two rectangles $x(t)$ and $h(t)$ of different widths results in a trapezoidal output signal $y(t)$.
• The low-pass property of the filter can be seen from the finite edge steepness of $y(t)$.
• The pulse height $(3\text{ V)}$ is preserved in this example because of
$$H(f = 0) = 1/T · T = 1.$$
## Step response
$\text{Definition:}$ In practice, one of the often used input functions $x(t)$ in order to measure $H(f)$ is the »jump function«
$${\rm \gamma}(t) = \left\{ \begin{array}{l} \hspace{0.25cm}0 \\ 0.5 \\ \hspace{0.25cm} 1 \\ \end{array} \right.\quad \quad\begin{array}{*{20}c} \text{for} \\ \text{for}\\ \text{for} \\ \end{array}\begin{array}{*{20}c}{\vert \hspace{0.05cm} t\hspace{0.05cm} \vert < 0,} \\ {\vert \hspace{0.05cm}t\hspace{0.05cm} \vert = 0,} \\ {\vert \hspace{0.05cm} t \hspace{0.05cm} \vert > 0.} \\ \end{array}$$
The »step response« $\sigma(t)$ is the response of the system if the jump function $\gamma(t)$ is applied to the input:
$$x(t) = {\rm \gamma}(t)\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y(t) = {\rm \sigma}(t).$$
The computation in the frequency domain would be a bit laborious here because the following equation would have to be applied:
$${\rm \sigma}(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, X(f ) \cdot H(f) =\left({1}/{2}\cdot \delta(f) + \frac{1}{{\rm j}\cdot 2\pi f} \right) \cdot H(f).$$
In contrast to this, the computation in the time domain leads directly to the result:
$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$
For causal sytems $h(\tau) = 0$ holds for $\tau \lt 0$, such that the lower limit of integration in the above equation can be set to $\tau = 0$.
$\text{Proof:}$ The above result is also insightful for the following reason:
• The jump function $\gamma(t)$ is related to the Dirac delta function $\delta(t)$ as follows:
$${\rm \gamma}(t) = \int_{ - \infty }^{ t } {\delta ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$
• Since we have assumed linearity and integration is a linear operation the corresponding relationship also applies to the output signal:
$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$
q.e.d.
Computation of the step response for rectangular impulse response
$\text{Example 4:}$ The graph illustrates the facts for the rectangular impulse response $h(\tau)$.
$\text{Note:}$
• The abscissa has been renamed to $\tau$.
• The jump function $\gamma(\tau)$ is drawn in blue.
• $\gamma(t - \tau)$ is obtained by mirroring and shifting ⇒ curve dashed in violet.
• The red shaded area thus gives the step response $\sigma(\tau)$ at time $\tau = t$ . | 3,515 | 10,906 | {"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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.679074 |
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A127766 Integer part of Gauss's Arithmetic-Geometric Mean M(2,n^5). 4
1, 12, 61, 210, 561, 1265, 2532, 4641, 7941, 12868, 19946, 29796, 43144, 60828, 83804, 113155, 150095, 195980, 252310, 320738, 403077, 501308, 617581, 754227, 913762, 1098894, 1312530, 1557780, 1837964, 2156622, 2517514, 2924630, 3382196 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS FORMULA a(n) ~ Pi*n^5/(10*log(n) + 2*log(2)). - Vaclav Kotesovec, May 09 2016 MATHEMATICA Table[Floor[ArithmeticGeometricMean[2, n^5]], {n, 1, 100}] CROSSREFS Cf. A127758, A127759, A127760, A127761, A127762, A127763, A127764, A127765. Sequence in context: A304205 A240002 A114241 * A005173 A196144 A294682 Adjacent sequences: A127763 A127764 A127765 * A127767 A127768 A127769 KEYWORD nonn AUTHOR Artur Jasinski, Jan 28 2007 STATUS approved
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Last modified May 6 09:45 EDT 2021. Contains 343580 sequences. (Running on oeis4.) | 477 | 1,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.55257 |
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# Is x > 1/5 ? (1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
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Is x > 1/5 ? (1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27 [#permalink]
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08 Jul 2008, 13:56
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Is x > 1/5 ?
(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26
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08 Jul 2008, 14:07
kevincan wrote:
Is x > 1/5 ?
(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26
from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.
from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.
A.
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08 Jul 2008, 18:48
GMAT TIGER wrote:
kevincan wrote:
Is x > 1/5 ?
(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26
from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.
from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.
A.
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?
1/5 = 5/25
1/23 + 1/24 > 2/25
1/26 + 1/27 < 2/25
???
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08 Jul 2008, 19:47
gmatnub wrote:
GMAT TIGER wrote:
kevincan wrote:
Is x > 1/5 ?
(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26
from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.
from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.
A.
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?
1/5 = 5/25
1/23 + 1/24 > 2/25
1/26 + 1/27 < 2/25
???
= 1/23 + 1/24
= (24+23)/(23x24)
= (47)/(23x24)
= (47)/[(20+3) (25-1)]
= (47)/(500 + 75 - 20 - 3)
= 47/552
= 1/26 + 1/27
= (27+26)/(26x27)
= (53)/(26x27)
= (53)/[(25+1) (25+2)]
= (53)/(625+25+50+2)]
= 53/702
clearly 47/552 > 53/702
and also 47/552 > 2/25
so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
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08 Jul 2008, 19:53
gmatnub wrote:
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?
1/5 = 5/25
1/23 + 1/24 > 2/25
1/26 + 1/27 < 2/25
???
you dont need to do complex calculation. Just compare
A = 1/23-1/25 and B = 1/25-1/27
A = 2/(23*25) and B = 2/(25*27)
clearly A > B
So ((1/23 + 1/24) - 2/25 ) is more than (2/25 - (1/26 + 1/27))
First two terms are increasing the sum from 1/5 and last two terms are decreasing it.
But Contribution of first two terms is more than that of last two terms
We can say that 1/23 + 1/24 + 1/25 + 1/26 + 1/27 > 1/5
Last edited by durgesh79 on 08 Jul 2008, 22:54, edited 2 times in total.
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08 Jul 2008, 20:11
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The numbers look to be in perfect progression so I Just took similar example with:
Is x > 3/6 ? (3 times the middle number of condition (1))
(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6
May not be fool proof but it worked.
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08 Jul 2008, 21:38
1
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indy123 wrote:
A
The numbers look to be in perfect progression so I Just took similar example with:
Is x > 3/6 ? (3 times the middle number of condition (1))
(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6
May not be fool proof but it worked.
This is great!
3 numbers: question asks 3 times the middle number
5 numbers: question asks 5 times the middle number
+1 for you with the hope that this will work for all similar problems
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09 Jul 2008, 00:18
GMAT TIGER wrote:
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?
[...]
clearly 47/552 > 53/702
and also 47/552 > 2/25
so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
I don't agree with your conclusion, sorry .
From 47/552 > 53/702 and 47/552 > 2/25, how do you know that 53/702 > 2/25? Because you're assuming that, and it is :
- not prooved
- and more importantly: not true
Last edited by Oski on 09 Jul 2008, 00:49, edited 2 times in total.
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09 Jul 2008, 00:24
I think a good way to compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 and 1/5 is to look at the following:
$$\frac{1}{23} + \frac{1}{27} = \frac{27+23}{23*27} = \frac{50}{23*25 + 23*2} > \frac{50}{23*25 + 2*25} = \frac{2*25}{25*25} = \frac{2}{25}$$
you can do exactly the same to proove that $$\frac{1}{24} + \frac{1}{26} > \frac{2}{25}$$
And then, yes, we have $$\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} > \frac{1}{5}$$
Edit: the hint here is "when you are faced a symmetrical expression, always think of symmetry"
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09 Jul 2008, 00:58
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.
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09 Jul 2008, 02:25
RayOfLight wrote:
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.
This is not a solution per se.
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09 Jul 2008, 07:45
Oski wrote:
RayOfLight wrote:
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.
This is not a solution per se.
I disagree there with you. Its a non sense solution but very accurate for these questions.
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09 Jul 2008, 08:16
indy123 wrote:
A
The numbers look to be in perfect progression so I Just took similar example with:
Is x > 3/6 ? (3 times the middle number of condition (1))
(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6
May not be fool proof but it worked.
nice but wanna clarify they are not in progression and to be precise you are talking about arithmatic progression
1/6 -1/7 is not equal to 1/5 -1/6
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09 Jul 2008, 12:40
Oski wrote:
GMAT TIGER wrote:
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?
[...]
clearly 47/552 > 53/702
and also 47/552 > 2/25
so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
I don't agree with your conclusion, sorry .
From 47/552 > 53/702 and 47/552 > 2/25, how do you know that 53/702 > 2/25? Because you're assuming that, and it is :
- not prooved
- and more importantly: not true
Did I assume 53/702 > 2/25 ? Can you show me? hmm..........
Of course I assumed 47/552 > 2/25.
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09 Jul 2008, 19:52
In situations like this where you have a progression of the sume of fractions like
$$\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27}$$
When we are comparing the sum against the middle fraction * the number of fractions, the sum will always be greater than the middle fraction * the number of fractions, provided the numerators are the same and the denominators are consecutive.
The reason behind this is if you start out with the middle, here $$\frac{1}{25}$$ as the "average" we need to figure out which way the two fractions on either side tip the scales.
Attachment:
FractionSumComparison.xls [17.5 KiB]
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.
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Last edited by jallenmorris on 10 Jul 2008, 03:44, edited 1 time in total.
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
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### Show Tags
10 Jul 2008, 00:11
GMAT TIGER wrote:
Did I assume 53/702 > 2/25 ? Can you show me? hmm..........
Of course I assumed 47/552 > 2/25.
I really think you did (it is not because you didn't explicitly wrote it that you didn't and you apparently used it).
Otherwise tell me then (I don't get it), I do you go from there
GMAT TIGER wrote:
1/23 + 1/24 = 47/552
1/26 + 1/27 = 53/702
clearly 47/552 > 53/702
and also 47/552 > 2/25
to there :
GMAT TIGER wrote:
so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
?
Cause with the equations you wrote, we can say : 1/23 + 1/24 + 1/25 + 1/26 + 1/27 > 2/25 + 1/25 + 53/702
But how do you go from there to > 1/5 if you don't assume that 53/702 > 2/25 too ?
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1264
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### Show Tags
10 Jul 2008, 00:37
kevincan wrote:
Is x > 1/5 ?
(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26
Note that for 0 < k < 25, 1/(25 - k) + 1/(25 + k) = 50/($$25^2 - k^2)$$ > 50/$$25^2$$= 2/25
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
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### Show Tags
10 Jul 2008, 00:41
kevincan wrote:
Note that for 0 < k < 25, 1/(25 - k) + 1/(25 + k) = 50/($$25^2 - k^2)$$ > 50/$$25^2$$= 2/25
Exactly. Symmetry, symmetry
Re: DS Fractions [#permalink] 10 Jul 2008, 00:41
Display posts from previous: Sort by | 4,651 | 11,481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.8125 | 4 | CC-MAIN-2017-04 | latest | en | 0.727001 |
https://www.esaral.com/q/find-the-general-solution-of-each-of-the-following-equations-12050 | 1,723,484,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00288.warc.gz | 563,016,401 | 11,494 | # Find the general solution of each of the following equations:
Question:
Find the general solution of each of the following equations:
4sin x cos x + 2sin x + 2cos x + 1 = 0
Solution:
To Find: General solution.
Given: $4 \sin x \cos x+2 \sin x+2 \cos x+1=0 \Rightarrow 2 \sin x(2 \cos x+1)+2 \cos x+1=0$
So $(2 \cos x+1)(2 \sin x+1)=0$
$\cos x=\frac{-1}{2}=\cos \left(\frac{2 \pi}{3}\right)$ or $\sin x=\frac{-1}{2}=\sin \frac{7 \pi}{6}$
Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 \mathrm{n} \pi \pm \alpha$ or $\sin \theta=\sin \alpha \Rightarrow \theta=\mathrm{m} \pi+(-1)^{\mathrm{m}} \alpha$ where $n, m \in I$
$x=2 n \pi \pm \frac{2 \pi}{3}$ or $x=m \pi+(-1)^{m} \cdot \frac{7 \pi}{6}$ where $n, m \in \mid$
So the general solution is $x=2 n \pi \pm \frac{2 \pi}{3}$ or $x=m \pi+(-1)^{m} \cdot \frac{7 \pi}{6}$ where $n, m \in I$
Leave a comment
Click here to get exam-ready with eSaral | 361 | 922 | {"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} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.402131 |
https://matplotlib.org/2.1.2/gallery/event_handling/resample.html | 1,652,788,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00335.warc.gz | 460,293,671 | 5,419 | Travis-CI:
# Resampling DataΒΆ
Downsampling lowers the sample rate or sample size of a signal. In this tutorial, the signal is downsampled when the plot is adjusted through dragging and zooming.
```import numpy as np
import matplotlib.pyplot as plt
# A class that will downsample the data and recompute when zoomed.
def __init__(self, xdata, ydata):
self.origYData = ydata
self.origXData = xdata
self.max_points = 50
self.delta = xdata[-1] - xdata[0]
def downsample(self, xstart, xend):
# get the points in the view range
mask = (self.origXData > xstart) & (self.origXData < xend)
# dilate the mask by one to catch the points just outside
# of the view range to not truncate the line
# sort out how many points to drop
ratio = max(np.sum(mask) // self.max_points, 1)
# downsample data
xdata = xdata[::ratio]
ydata = ydata[::ratio]
print("using {} of {} visible points".format(
return xdata, ydata
def update(self, ax):
# Update the line
lims = ax.viewLim
if np.abs(lims.width - self.delta) > 1e-8:
self.delta = lims.width
xstart, xend = lims.intervalx
self.line.set_data(*self.downsample(xstart, xend))
ax.figure.canvas.draw_idle()
# Create a signal
xdata = np.linspace(16, 365, (365-16)*4)
ydata = np.sin(2*np.pi*xdata/153) + np.cos(2*np.pi*xdata/127)
fig, ax = plt.subplots()
# Hook up the line
d.line, = ax.plot(xdata, ydata, 'o-')
ax.set_autoscale_on(False) # Otherwise, infinite loop
# Connect for changing the view limits
ax.callbacks.connect('xlim_changed', d.update)
ax.set_xlim(16, 365)
plt.show()
```
Total running time of the script: ( 0 minutes 0.019 seconds)
Gallery generated by Sphinx-Gallery | 476 | 1,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.602412 |
https://www.reddit.com/user/FootballHawk18 | 1,498,542,641,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320995.93/warc/CC-MAIN-20170627050500-20170627070500-00302.warc.gz | 930,498,084 | 23,402 | [–] 34 points35 points (0 children)
[–] 30 points31 points (0 children)
TIL somebody on Reddit mentions Jack Sparrow essentially once a minute.
[–] 0 points1 point (0 children)
I think you're getting tripped up over loose definitions of "speed"
Put it in simpler terms and it's easier to understand.
• Speed = distance/time
Since distance is fixed, let's focus on time.
If you run a quarter mile lap in 4 minutes, and then a second lap much faster, it's still impossible to reduce the average of the lap times to 2 minutes. You would have to run the second lap in 0 seconds to average with 4 minutes and achieve the desired 2-minute lap time average, hence "infinitely fast".
[–] 35 points36 points (0 children)
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Except what's shown is almost exclusively simple carbs...
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I always loved Super Monkey Ball
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You've been down too long in the midnight sea.
[–] 2 points3 points (0 children)
Was it good over rice?
[–] 1 point2 points (0 children)
That looks fucking sick. Enjoy!
[–][S] 2 points3 points (0 children)
Indeed it was, that's an impressive eye! Two, even.
[–][S] 3 points4 points (0 children)
To be fair it likely spent a cumulative year in the shop. Not exactly the car you want driving on sleet or ice, or in snow or rain... or really any percentage of the time.
[–] 78 points79 points (0 children)
[–][S] 9 points10 points (0 children)
I thoroughly enjoy you too.
[–] 12 points13 points (0 children)
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Ankle Brace, High-top Cleats, or Both? by in ultimate
[–] 6 points7 points (0 children)
Hi.
You should figure out what exactly happened when you sprained your ankle.
I sprained my ankle mid October, and here I am in the time period in which I could have had a child still sitting with my ankle brace on. It was most recently diagnosed as ankle instability, but there's a ton of little nuances and technical jargon which could best determine the path toward healing.
Turns out I sprained a ligament, and I'm in the 10% which didn't quite heal correctly.
I went through 3 different braces, 3 months of physical therapy, 4 X-rays and an MRI.
Every doctor says the same thing two things:
1) Wear a brace. It can't hurt.
2) Let symptoms be your guide.
You said it, the worst thing about an ankle injury is proneness to another, more severe one. The most frequently asked question after an ankle injury is "have your sprained your ankle before?"
Anyway, the point is, I've already missed half of last fall season, all of spring, and I'm missing Wildwood Beach Ultimate in a few weeks. Bench-warming isn't anyone's favorite.
Heal for both of us. Wear a brace, preferably a lace-up, and don't push it. The "Swede-O Inner Lok" is pretty phenomenal.
Stretch a lot, rest a lot, and go through some therapy exercises.
You'll feel a hell of a lot better soon, but make sure you keep it light for a while until you're sure you've healed completely.
[–] 0 points1 point (0 children)
A Jack Black sheep.
[–] 5 points6 points (0 children)
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The View
[–][S] 1 point2 points (0 children)
You shrewd bastard. | 879 | 3,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.915385 |
http://ned.ipac.caltech.edu/level5/Berg/Berg1.html | 1,529,369,282,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00431.warc.gz | 222,466,554 | 1,926 | ### 1. PROBABILITIES
Classical or a priori probabilities are defined in terms of the possible outcomes of a trial, recognized in advance as equally probable. In the toss of a coin, the probability of getting a head is 1/2: the number of outcomes that give a head, 1, divided, by the total number of possible outcomes, head or tail, 2. In the toss of a die, the probability of getting one dot is 1/6: the number of outcomes that give one dot, 1, divided by the total number of possible outcomes, one through six dots, 6. In general, the probability of event a is
(A.1)
where a is the number of equally probable outcomes that satisfy criteria a, and n is the total number of equally probable outcomes.
In the examples just given, the outcomes are mutually exclusive; i.e., only one outcome is possible at a time. If events a and b are mutually exclusive, then
(A.2)
and
(A.3)
For the toss of a coin, p(head or tail) = p(head) + p(tail) = 1, and p(not head) = 1 - p(head) = 1/2. For the toss of a die, p(1 dot or 2 dots) = p(1 dot) + p(2 dots) = 1/3, and p(not 1 dot) = 1 - p(1 dot) = 5/6.
In these examples, the outcomes also are statistically independent; i.e., the occurrence of one event does not affect that of another. If events a and b are statistically independent, then
(A.4)
The probability of obtaining heads in each of two tosses of a coin is (1/2)(1/2) = 1/4. The probability of obtaining a single dot in each of two tosses of a die is (1/6)(1/6) = 1/36.
Events are conditional if the probability of one event depends on the occurrence of another. If the probability that b will occur, given that a has occurred, is p(b/a), then the probability that both will occur is
(A.5)
For example, the probability of drawing two aces from a deck of cards is (4/52)(3/51) = 1/221. If the first card were put back into the deck and the deck reshuffled, then the probability of drawing the second ace would not be conditioned on the drawing of the first, and the probability would be (4/52)(4/52) = 1/169, in accord with Eq. A.4.
Here are some rules of thumb for dealing with compound events, when the events are both mutually exclusive and statistically independent:
(A.6)
(A.7)
(A.8) | 595 | 2,200 | {"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} | 4.46875 | 4 | CC-MAIN-2018-26 | latest | en | 0.942284 |
https://libstdc.com/us/q/bicycles/1505 | 1,623,613,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610841.7/warc/CC-MAIN-20210613192529-20210613222529-00636.warc.gz | 332,576,380 | 12,215 | ### How do on-road mountain bike speeds translate to road bike speeds?
• When cycling on roads on my mountain bike, I generally get around 10-12 miles per hour on a flat surface.
What sort of speeds can I expect to get on a road bike?
possible duplicate of Speed benchmarking
I think this is a slightly different question - it seems to be "If I can go X mph on bike A, and switch to bike B, what will my speed be?"
As @Gary.Ray points out, you seem to be asking how to estimate _your_ road bike speed based on _your_ mountain bike speed, but many answers seem to ignore this. Admittedly your title seemed to me to make it pretty clear, but I wonder if you might not emphasise this in the body of the question as well.
10 years ago
On my reasonably flat commute I average around 16 mph on my road bike, but your average speed is dependent on many different factors. A general rule of thumb is that if you are switching from a mountain bike with knobbies to a road bike you will be between 15-20% faster at the same watts/effort. Typically that's only a change of 2-3 mph.
I teach a bike commuting workshop, and one of the most common questions is whether to switch from a mountain bike to a road bike in order to increase speed. Typically, I tell people to try three things first:
1. Swap out your knobby tires for high pressure slicks. You can find 1.25" -1.5" slick tires that fit mountain bike rims and run at between 75 - 90 lbs of pressure. These will dramatically reduce rolling resistance.
2. If you have a suspension either lock it out, or set it as stiff as you can. Locking out your suspension will cause more of your effort to be transferred directly through the drive-train and translate into less loss of momentum from shock absorption.
3. Try clipless pedals. Your pedal stroke will be more efficient, again resulting in an increase in speed.
If you do those three things the only real differences between a road bike and your mountain bike will be the weight (which matters a lot more when accelerating than it does when you are already rolling) and surface area resistance from riding in the drops. But most road bikers don't spend much time in the drops.
Finally, your weight and fitness make a huge difference. I frequently pass road bike riders while on my commuter rigid frame mountain bike with high pressure slicks.
I would also add that in a road bike I find that you can put power into the pedals easier than on a mountain or especially a cruiser bike.
I suspect the difference is even greater for someone in good physical shape. My touring bike, unloaded, is about 25lb and has 26x2.0 slicks. I find I cruise at 15mph on it, whereas on my racing bike it feels like I do 20mph+ without even trying. Don't underestimate the effects of weight, stiffness, and rolling resistance.
This answer ignores the primary difference between a road bike and a mountain bike. That's fine, for commutes or solo rides, but not so good if you ride with someone who's on a road bike while you're on a mountain bike. The difference is gear ratio. For instance, the typical top end MTB gearing, with a 44t big ring and a 12t rear, and a typical cadence of 90 RPMs gives you a speed of 26 mph possible. The same cadence in the top gear of a compact double road bike, a 50 x 11 usually, gives a possible top speed of 33 mph, at the same level of effort. Standard 53 x 12 is about 32 mph.
Of course, you need to be fit enough to maintain that pace. But since you're comparing the same levels of effort, that's a bit more than 2-3 miles an hour. More like 8-9mph faster. Which on a 20 mile commute = 46 minutes on the MTB, versus 33 minutes on a road bike, at the same level of effort. This is strictly a matter of gearing and mechanical advantage, for a roughly 30% increase in speed. BTW, all of the math here assumes the same tire diameter, which is not typical. The difference in wheel size alone, assuming high pressure slicks, is worth the 2-3 mph difference that @GaryRay claims.
I first read your post and was like "clipless pedals", huh? But then I came across this article which explains why they're called "clipless" (when they do actually clip in)... http://gizmodo.com/5990381/why-you-should-switch-to-clipless-pedals
One minor difference is the rider position. Most road bikes have drop bars so the rider is crouched over vs sitting somewhat upright. You can add drop bars to your mountain bike to help with that. I would say this is fairly important because at about 15 mph air resistance is a lot more dominant that rolling resistance.
Just a single data point but these are the speeds I would observe on flat ground for 3 - 5 mile rides- 1)Mountain Bike 26" with knobbies, no clips, suspension normal, bike weight >25 lbs very close 15 mph 2) Mountain Bike 26" flat tires, with clips, locked out suspension, bike weight > 25 lbs: about 18 mph. 3) Road Bike 700cc small flat tires, with clips, bike weight about 12 lbs: 19 mph. So most of the increase you get from flats and clips. But you still get a bit from the road bike. | 1,190 | 5,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.954983 |
http://www.convertit.com/Go/WaveQuest/Measurement/Converter.ASP?From=sthene | 1,563,390,877,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00271.warc.gz | 198,178,662 | 3,204 | Riding the internet wave!
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```sthene = 1000 newton (force) ``` Related Measurements: Try converting from "sthene" to atomic weight, dyne (cm*gm/s^2), gram force, kip (kilopound force), newton, pound force, poundal, ton force, or any combination of units which equate to "mass length / time squared" and represent bending moment to length, force, or weight. Sample Conversions: sthene = 6.14E+28 atomic weight, 100,000,000 dyne (cm*gm/s^2), 101,971.62 gram force, .22480894 kip (kilopound force), 1,000 newton, 224.81 pound force, 7,233.01 poundal, .10197162 ton force.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 321 | 1,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | latest | en | 0.686836 |
https://excelgraduate.com/match-function-in-excel/ | 1,716,599,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00749.warc.gz | 210,307,575 | 35,399 | # How to Use MATCH Function in Excel [2 Examples]
In the realm of Excel’s vast functionalities, the MATCH function stands as a versatile tool for locating specific values within a range. Whether you’re navigating large datasets or searching for a particular item, the MATCH function streamlines the process with precision and efficiency. In this guide, we’ll delve into the nuances of the Excel MATCH function, exploring its syntax, arguments, output, and practical applications through illustrative examples.
## What Does the Excel MATCH Function Do?
The MATCH function actively searches for a specified value in a range of cells and returns its relative position. It scans data for an exact or closest match, giving the position of the found item within the range.
## What is the Syntax of the Excel MATCH Function?
The syntax of the MATCH function is straightforward:
=MATCH(lookup_value, lookup_array, [match_type])
## What are the Arguments of the Excel MATCH Function?
The Excel MATCH function takes three arguments:
• lookup_value: The value you’re searching for.
• lookup_array: The range of cells where Excel should search for the lookup_value.
• match_type (optional): Specifies the type of match to be performed.
## What is the Output of the Excel MATCH Function?
The output of the Excel MATCH function is the relative position of the matched value within the lookup_array. If it finds no match, it actively returns the #N/A error.
## 2 Examples of Using the MATCH Function in Excel
Let’s dive into some practical examples to grasp the versatility of the MATCH function:
### Example 1: Finding an Exact Match
Suppose you have a list of student names in column A and their corresponding scores in column B. So, use the MATCH function to find the position of a specific student’s name, such as “John,” within the range.
=MATCH("John", A:A, 0)
This formula will return the row number where “John” is located in column A.
### Example 2: Performing Approximate Match
In a dataset containing sales value, you can use the MATCH function to find the position of the nearest price below a specified value.
=MATCH(3000, D2:D7, 1)
This formula will return the position of the highest price which is less than 3000.
## Things to Remember
• Ensure that the lookup_array is sorted appropriately for accurate results in approximate match scenarios.
• Be mindful of the match_type argument to specify the desired matching behavior.
• Handle potential errors gracefully using functions like IFERROR to enhance the robustness of your formulas.
## Conclusion
The Excel MATCH function serves as a valuable tool for swiftly locating values within datasets, offering unparalleled efficiency and accuracy. By mastering its syntax and applications, users can streamline their data analysis tasks and unlock new possibilities for spreadsheet manipulation.
### Can the MATCH function be used to search for values in multiple columns?
Yes, you can extend the lookup_array argument to include multiple columns for searching across different datasets.
### What is the difference between the MATCH function and VLOOKUP or HLOOKUP?
VLOOKUP and HLOOKUP retrieve, MATCH returns relative position, enabling dynamic data manipulation within Excel.
### Is it possible to use the MATCH function with wildcard characters for partial matches?
Yes, you can combine the MATCH function with wildcard characters like “*” or “?” to perform partial matching operations within the dataset.
### Can the Excel MATCH function be used for approximate matching?
Yes, you can specify match_type as 1 or -1 for approximate matching, finding the closest value less/greater than or equal to lookup_value.
### How can I handle errors when using the MATCH function?
Use IFERROR alongside MATCH to manage errors smoothly, enabling custom responses or actions if matches are absent.
### Can I use wildcard characters with the Excel MATCH function?
Yes, combine MATCH with “*” or “?” for partial matches, enhancing dataset searches with flexible criteria.
### What are some advanced applications of the MATCH function?
Combine Excel’s MATCH with INDEX or OFFSET for dynamic data retrieval, enabling advanced lookup operations based on specific criteria.
Rate this post | 831 | 4,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.796715 |
http://www.algebra.com/algebra/homework/Percentage-and-ratio-word-problems/Percentage-and-ratio-word-problems.faq.question.113816.html | 1,369,286,567,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702849682/warc/CC-MAIN-20130516111409-00050-ip-10-60-113-184.ec2.internal.warc.gz | 317,111,231 | 4,724 | # SOLUTION: one third of a three-digit number whose units digit is x, ten digit is 5 and hundreds digit is three times the units digit
Algebra -> Algebra -> Percentage-and-ratio-word-problems -> SOLUTION: one third of a three-digit number whose units digit is x, ten digit is 5 and hundreds digit is three times the units digit Log On
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Word Problems: Problems on percentages, ratios, and fractions Solvers Lessons Answers archive Quiz In Depth
Question 113816: one third of a three-digit number whose units digit is x, ten digit is 5 and hundreds digit is three times the units digitFound 2 solutions by edjones, ankor@dixie-net.com:Answer by edjones(7569) (Show Source): You can put this solution on YOUR website!3x 5 x 300x+50+x (301x+50)/3 Ed Answer by ankor@dixie-net.com(15652) (Show Source): You can put this solution on YOUR website!One third of a three-digit number whose units digit is x, ten digit is 5 and hundreds digit is three times the units digit : Write an equation for what it says: *(100(3x) + 50 + x) : *(300x + 50 + x) : *(301x + 50) | 325 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.80261 |
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