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# The Coriolis Effect: Earth's Rotation and Its Effect on Weather
The Coriolis effect describes the pattern of deflection taken by objects not firmly connected to the ground as they travel long distances around the Earth.
3 - 12+
### Subjects
Earth Science, Meteorology, Geography, Physical Geography, Physics
Selected text level
Imagine you have amazing strength. You can throw a ball like Superman. You are standing at the Equator. The Equator is an imaginary line around the middle of Earth. You want to throw a ball to a friend, who is standing somewhere in North America. What's going to happen?
If you try throwing the ball straight to your friend, things won't go as planned. The ball will land slightly to your friend's right. The reason for this is the Coriolis effect.
The Coriolis effect is caused by the planet's rotation. Earth is constantly rotating, or spinning. It is spinning from west to east. This is left to right on a map. Every 24 hours, Earth makes a full spin. Not every point on Earth is moving at the same speed, though. Near the Equator, points are rotating faster than at the poles.
The Equator divides the planet into two halves. The northern half is called the Northern Hemisphere. The southern half is called the Southern Hemisphere. Earth is widest at the Equator. In order to make a full spin in 24 hours, points at the Equator have to go a longer distance. Points near the poles have to go a much shorter distance. That means points at the Equator are moving much faster than points near the poles. Regions near the Equator move at almost 1,600 kilometers (1,000 miles) an hour. Near the poles, Earth moves extremely slowly.
This is why the ball did not reach your friend in North America. At the Equator, you and the ball are moving east. Your friend is also moving east, but at a slower speed. When you throw the ball, it will move toward your friend at first. But it will also move east at a faster speed than him. So it will land to your friend's right.
Now let's pretend you're standing at the North Pole. When you throw the ball to your friend, it will again land to his right. This time, it's because he's moving faster than you; he has moved ahead of the ball.
In real life, the Coriolis effect has a large effect on the weather. It changes how winds blow. It makes winds bend to the right in the Northern Hemisphere. In the Southern Hemisphere, it makes them bend left. It also causes cyclones. These are large, rotating masses of air. In the Northern Hemisphere, cyclones rotate counterclockwise. In the Southern Hemisphere, they rotate clockwise.
The Coriolis Effect Closer to Home
Here's one last example of the Coriolis effect. It's something you can try yourself. Suppose you and a friend are throwing a ball back and forth. Imagine you're sitting on a merry-go-round. When the merry-go-round is still, playing catch is easy. Things are different when the merry-go-round is rotating. The ball won't reach your friend unless you throw it extra hard. If you throw it normally, the ball will curve to the right.
The ball is actually flying in a straight line. It is you and your friend who are moving out of the way. You are spinning because of the merry-go-round.
Fast Fact
Coriolis Force
The invisible force that appears to deflect the wind is the Coriolis force. The Coriolis force applies to movement on rotating objects. It is determined by the mass of the object and the object's rate of rotation. The Coriolis force is perpendicular to the object's axis. The Earth spins on its axis from west to east. The Coriolis force, therefore, acts in a north-south direction. The Coriolis force is zero at the Equator.
Though the Coriolis force is useful in mathematical equations, there is actually no physical force involved. Instead, it is just the ground moving at a different speed than an object in the air.
Fast Fact
Polar PowerThe Coriolis force is strongest near the poles, and absent at the Equator. Cyclones need the Coriolis force in order to circulate. For this reasons, hurricanes almost never occur in equatorial regions, and never cross the Equator itself.
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Jeannie Evers, Emdash Editing, Emdash Editing
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# A museum offers four video programs that run continuously
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A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?
A. 165 mins
B. 164 mins
C. 160 mins
D. 158 mins
E. 157 mins
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Dear all,
I have read the answer to this question but I am still confused. Does someone have an easier way to solve this problem?
Q: A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?
A.165 mins
B.164 mins
C.160 mins
D. 158 mins
E. 157 mins
First notice the minimum amount of time that needs to be taken: 15+10 + 30+10 + 45+10 + 40 = 160 mins
So D and E are out.
Since all show timings are a multiple of 5 and the group needs 10 mins to regroup, there is no way we will get 164. Hence the only options that will work out are 165 or 160.
If 160 is the answer, we have no wriggle room. We must watch shows one after the other with only 10 mins in between.
Shows: 15 mins, 30 mins, 40 mins, 45 mins
If we start with 15 min show and then regroup in 10 mins, we need to start another show in 25 mins. But no show starts in 25 mins so let's not start with 15 mins. (Same logic can be used to show that we cannot watch 40 min and 45 min shows first if we want to spend a total of only 160 mins)
If we start with 30 min show and then regroup in 10 mins, we need to start another show in 40 mins. A show does start in 40 mins so let's watch that. Now 80 mins have passed. We take 10 mins to regroup and now 90 mins have passed. Both the other shows start now. We watch the 45 min show and regroup for 10 mins. Now 145 mins have passed. But the 15 min show will start in 5 mins so we will need to wait another 5 mins for it and then watch it.
Total time taken will be 165 mins.
We cannot watch all the shows in 160 mins.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 12 Oct 2012 Posts: 120 Followers: 1 Kudos [?]: 50 [0], given: 194 Re: A museum offers four video programs that run continuously [#permalink] ### Show Tags 24 Nov 2012, 22:57 VeritasPrepKarishma wrote: aditi2013 wrote: Dear all, I have read the answer to this question but I am still confused. Does someone have an easier way to solve this problem? Q: A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs? A.165 mins B.164 mins C.160 mins D. 158 mins E. 157 mins First notice the minimum amount of time that needs to be taken: 15+10 + 30+10 + 45+10 + 40 = 160 mins So D and E are out. Since all show timings are a multiple of 5 and the group needs 10 mins to regroup, there is no way we will get 164. Hence the only options that will work out are 165 or 160. If 160 is the answer, we have no wriggle room. We must watch shows one after the other with only 10 mins in between. Shows: 15 mins, 30 mins, 40 mins, 45 mins If we start with 15 min show and then regroup in 10 mins, we need to start another show in 25 mins. But no show starts in 25 mins so let's not start with 15 mins. (Same logic can be used to show that we cannot watch 40 min and 45 min shows first if we want to spend a total of only 160 mins) If we start with 30 min show and then regroup in 10 mins, we need to start another show in 40 mins. A show does start in 40 mins so let's watch that. Now 80 mins have passed. We take 10 mins to regroup and now 90 mins have passed. Both the other shows start now. We watch the 45 min show and regroup for 10 mins. Now 145 mins have passed. But the 15 min show will start in 5 mins so we will need to wait another 5 mins for it and then watch it. Total time taken will be 165 mins. We cannot watch all the shows in 160 mins. Answer A Karishma u are great..finally after breaking my head, i got it..is this a 700 level Q and can it be solved in 2 mins on the D-day? Manager Joined: 04 Oct 2011 Posts: 224 Location: India Concentration: Entrepreneurship, International Business GMAT 1: 440 Q33 V13 GMAT 2: 0 Q0 V0 GPA: 3 Followers: 0 Kudos [?]: 50 [0], given: 44 Re: A museum offers four video programs that run continuously [#permalink] ### Show Tags 25 Nov 2012, 16:40 VeritasPrepKarishma wrote: aditi2013 wrote: Dear all, I have read the answer to this question but I am still confused. Does someone have an easier way to solve this problem? Q: A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs? A.165 mins B.164 mins C.160 mins D. 158 mins E. 157 mins First notice the minimum amount of time that needs to be taken: 15+10 + 30+10 + 45+10 + 40 = 160 mins So D and E are out. Since all show timings are a multiple of 5 and the group needs 10 mins to regroup, there is no way we will get 164. Hence the only options that will work out are 165 or 160. If 160 is the answer, we have no wriggle room. We must watch shows one after the other with only 10 mins in between. Shows: 15 mins, 30 mins, 40 mins, 45 mins If we start with 15 min show and then regroup in 10 mins, we need to start another show in 25 mins. But no show starts in 25 mins so let's not start with 15 mins. (Same logic can be used to show that we cannot watch 40 min and 45 min shows first if we want to spend a total of only 160 mins) If we start with 30 min show and then regroup in 10 mins, we need to start another show in 40 mins. A show does start in 40 mins so let's watch that. Now 80 mins have passed. We take 10 mins to regroup and now 90 mins have passed. Both the other shows start now. We watch the 45 min show and regroup for 10 mins. Now 145 mins have passed. But the 15 min show will start in 5 mins so we will need to wait another 5 mins for it and then watch it. Total time taken will be 165 mins. We cannot watch all the shows in 160 mins. Answer A Hey great Karishma!! I took 6 mins to solve... but still got it wrong as 160 mins... I wrote all possible combinations... Didnt even think i can eliminate D E and B so easily until i see ur solution _________________ GMAT - Practice, Patience, Persistence Kudos if u like Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7182 Location: Pune, India Followers: 2164 Kudos [?]: 14010 [0], given: 222 Re: A museum offers four video programs that run continuously [#permalink] ### Show Tags 05 Dec 2012, 22:20 aditi2013 wrote: Karishma u are great..finally after breaking my head, i got it..is this a 700 level Q and can it be solved in 2 mins on the D-day? The question isn't hard.. more like tedious. You need to work through the options to figure out that 160 is not possible. That could be time consuming. You can do it in 2 mins if you stay focused (which is hard to do under exam time constraints). So possibly you might end up spending more time on it. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: A museum offers four video programs that run continuously [#permalink]
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05 Dec 2012, 23:12
VeritasPrepKarishma wrote:
Karishma u are great..finally after breaking my head, i got it..is this a 700 level Q and can it be solved in 2 mins on the D-day?
The question isn't hard.. more like tedious. You need to work through the options to figure out that 160 is not possible. That could be time consuming. You can do it in 2 mins if you stay focused (which is hard to do under exam time constraints). So possibly you might end up spending more time on it.
Ya its hard in exam to stay focused on time...
hmm thanks for this question...
Looking forward more challenging questions to learn
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Re: A museum offers four video programs that run continuously [#permalink]
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23 Mar 2016, 18:21
A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10am and the last showing of each program ends at 4 pm. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?
A. 165 mins
B. 164 mins
C. 160 mins
D. 158 mins
E. 157 mins
oh damn..took some time to solve this one...got with the below best option:
start with the second movie, finish at 10:30. 10 mins pause, next movie starts at 10:40, the fourth one. 4th finishes at 11:20. another 10 mins pause. 11:30. at 11:30 starts the 3rd showing of the third movie. so at 12:15 finishes the 4th one. 10 mins pause. 12:25. well, we have to wait for another 5 mins while the last one (first) starts at 12:30. then more 15 mins to watch the first one, and finish at 12:45. total 2h45m. in minutes, 165. A
Re: A museum offers four video programs that run continuously [#permalink] 23 Mar 2016, 18:21
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Defining the basic types of matrices is essential to be able to build other much more complex types and methods.
The base is essential. And when we speak of base we are not referring to any mathematical concept. We are referring to the knowledge base. Matrices are one of the most important and widely used concepts in different fields of science.
In econometrics, in computer programming, in big data and in various fields in which it is a question of crossing data or working with a large amount of data.
## Square matrix
A square matrix satisfies that (m = n). In other words, it has the same number of rows and columns. So the dimension of the rows will be the same as the dimension of the columns.
The square matrix is very important because it is the basis for many matrix types and methods.
Example
Matrix dimension B = 2 x 2.
## Transposed Matrix
A transposed matrix consists of reordering the original matrix by changing the rows by columns and the columns by rows.
Generally, a transposed matrix is indicated with a superscript T or an apostrophe ('). To express it better, we opted for the superscript T.
Following the previous example it would be: BT.
Example
When the original matrix is a square matrix, as in our case, the dimension of the matrix remains the same because the number of rows and columns is the same.
Matrix dimension BT = 2 x2.
## Identity Matrix
The identity matrix is a square matrix in which all its elements are zeros except those that belong to its main diagonal. It is usually identified by the letter I.
The identity matrix can be quickly distinguished without doing any calculations.
We have assigned a 3 × 3 dimension in this case. However, this dimension can be larger or smaller. We only have to comply when the matrix remains square and meets the characteristic: all zeros except its main diagonal which must have ones.
Example
The identity matrix acts like the number 1 in common algebra. Let I be the identity matrix and B any matrix, the product of both has a neutral effect on matrix B. Therefore, matrix B is the same as IB.
## Triangular Matrix
A triangular matrix is a square matrix in which the elements below the main diagonal are zeros or the elements above the main diagonal are zeros.
The triangular matrix focuses on the location of triangles containing only zeros. Depending on its position with respect to the main diagonal, the triangular matrix will be called upper or lower.
Upper triangular matrix:
Lower triangular matrix (lower):
The triangular matrix participates in the Lower-Upper (LU) decomposition method, which is used to obtain the Cholesky decomposition. This method is widely used in quantitative finance to transform independent normal variables into correlated normal variables.
## Symmetric Matrix
A matrix is symmetric if it is a square matrix and coincides with its transpose (C = CT).
To find symmetric matrices in a simple way, we just have to look at the element triangles that are above and below the main diagonal.
Example
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A new approach suggests an extremely logical explanation for how and why the Earth first formed, and even how and why the Moon formed. A number of additional issues are also described in logical terms, such as why the inner planets have much higher densities than the outer ones.
There is an assumption that has always been made which had appeared to eliminate this possibility, but which is now seen to be incorrect. There are six "orbital elements" which complete define a planet's (or other revolving object) orbit. They are (1) average orbital radius or orbital semi-diameter; (2) orbital eccentricity; (3) inclination of the orbital plane; (4) direction if the perihelion or line of apsides; (5) direction of the tilt of the orbital plane or the longitude of the ascending node; and (6) the location of the planet or object along that orbit at some instant of time.
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In the study of Perturbations of planets on each other's orbits, it is fully accepted that all of these orbital elements EXCEPT THE FIRST can be affected and therefore changed. Even Laplace and LaGrange and all the early giants in astronomical calculations accepted that as absolutely true. Therefore, Perturbation Theory attempts to track the slow changes in the eccentricity of the orbit, and all the other parameters that can have such secular variations. For example, it is well established that the Earth is presently advancing toward a much more circular orbit. Currently, our eccentricity is around 0.016 (which means that we vary from a circular orbit by about one part in 60 in distance from the Sun, meaning that our average 93 million mile distance varies from 91.5 million (in early January) to 94.5 million (in early July). It is well established that in around 24,000 years, our eccentricity will have greatly reduced, to 0.003, which is extremely circular. After that, the eccentricity will increase for around 40,000 years until we are in an extremely eccentric orbit, (0.070 eccentricity). We can know these things due to a mathematical analysis of careful observations which is called a Fourier Analysis. The Earth's eccentricity is therefore able to vary in the range of 0.003 to 0.070 but that it will always stay within that range. The other orbital elements can equally be secularly perturbed in such ranges.
The single exception has always been the orbital semi-diameter. This orbital element has always been denied any possible variation! The reason seems to be obvious! Newton and others showed us that there is Conservationof Energy, which is also called the First Law of Thermodynamics, where energy can neither be created nor destroyed. He also showed us that there is Conservation of Angular Momentum, where that too can not be created nor destroyed (except due to the effect of an external action). When Laplace, Lagrange and others first carefully studied the motions of planets two hundred years ago, they applied these two Conservation laws, and that forced the assumption that the orbital semi-diameter cannot be perturbed.
The energy of an orbiting planet is in kinetic energy, which is described by 1/2 * M * V2. (In circular coordinates, since I = M * R2 and V = ω * R, this is more conveniently described as 1/2 * I * ω2) The (angular) momentum of a planet is described by (M * V) * R (Again, in circular coordinates, this is I * ω). Kepler (and then Newton) determined that the orbital semi-diameter is directly related to the orbital period, which means that the (average) velocity in the orbit is uniquely specified when the orbital semi-diameter is known.
The fact that one of these quantities depends on ω and the other on ω2 means that it is not possible to conserve both in any possible Perturbation transfer between two planets or other objects. As one of the objects would gain potential energy and the other would lose the exact same amount, there would necessarily be a violation regarding the changes of the angular momentums, therefore Laplace, Lagrange and all others have always concluded that no such perturbation of the orbital semi-diameter has been said possible.
Those statements ARE true, if only one plane of motion is considered. However, Euler expanded Newton's equations of motion for three dimensions, and a new possibility then arises.
The most obvious example of this is a high-quality child's gyroscope. Consider one where the support bearings are perfect, that is, there is no friction whatever, and it is operated in a total vacuum, such that the gyroscope rotor will spin forever and never slow down. Placed on the usual pedestal in a axle-horizontal position, we all know that the gyroscope will then precess around the pedestal. The question is "when the gyroscope is first released, it necessarily ACCELERATES up to the final precessional rate, so what is the source of that energy and angular momentum that appears during that acceleration?"
The gyroscope starts out with NO angular momentum around the precessional (Z) axis, but quickly develops the angular momentum due to the precession. According to the conventional description, this is a violation of the Conservation of Angular Momentum! The rotor did not slow down, so that was not the source of any angular momentum.
The answer is in the (third) Euler equation which describes the dynamics around the precessional axis. There is no external moment applied, so M = 0. The result is therefore that the angular acceleration of the precessional motion is due to (vertical) motion in a different plane! The support angle of the gyro body is very slightly lowered, which gives up some gravitational potential energy, which is then converted into the kinetic energy of the precessional motion. Conservation of Energy is maintained. The significant fact is that this transfer of Energy from one form (potential) to another (precessional kinetic) does NOT conserve Angular Momentum in the process!
Here is the (third) Euler equation.
M3 is the net external force (moment) acting around the Z-axis, which is zero. The first term on the right includes the acceleration of the precessional rotation rate and the other term involves the constant angular rotation of the spinning rotor and the angular rotation rate of falling (around axis 2). If we integrate this over time, we get:
We can see that we can now solve this for the relationship between ω3 (the final precessional angular velocity) and the total change in ω2 (which is the vertical angle through which the mass of the gyro has fallen, or θ2. The vertical angle of drop of that mass is therefore directly related to the increase in precessional angular velocity.
Further analysis of the energy content of the two show that Energy is Conserved, as they are the same quantity.
This suggests that the long-held assumption that Angular Momentum is always conserved is not really necessarily true when more than one plane of motion is considered, that gyroscopic precession certainly shows that flaw of reasoning.
Solar System objects move in various planes. This fact results in effects that are similar to the non-Conservation of Angular Momentum of the toy gyroscope. For example, the earth has an equatorial bulge that is rotating in a plane where the Sun and Moon nearly always act to gravitationally try to tilt that plane, which causes the Precession that the Earth experiences. Consider a "new earth" exactly like ours but not precessing. It would START to precess, in other words, the Precessional motion of the earth would ACCELERATE up to the rate it is now at. The energy that would supply that motion would seem to come from a slight Z-axis (Solar-System-vertical) relative movement of the Sun/Earth and Moon/Earth, (the earth would move through the potential energy field of the Sun) so Energy would be conserved with the "precessional acceleration up to the new precession rate". However, Angular Momentum in the Plane of the Ecliptic would NOT be conserved! New Angular Momentum would arise in that Plane.
The effect described here is very small, and very slow. In all practical situations, Conservation of Angular Momentum will be seen to appear true. It is only where Euler equation transfers from one plane to another can occur that any variances with that Conservation can occur. Conservation of Energy appears to still always be true.
For that "new Earth" that is initially not precessing, we can easily calculate how much kinetic energy there is in our precession. It is 1/2 * I * ω2. We know that the rotational inertia (I) of the earth is 8.07 * 1037 kg-meters2. We know that ω is one precessional revolution in 25,800 years or one radian in 1.296 * 1011 seconds. Therefore, the kinetic energy the Earth has in precessing is around 2.4 * 1015 joules. In planetary dynamics, that is not very much, but it still is kinetic energy that did not used to exist, and must have been converted from some other previous form of energy!
Several years of research into the Physics and Engineering of "forced vibrations" and resonance as related to the gravitational interactions between planets and satellites has suggested a number of potential consequences of this phenomenon that seem never to have been previously discussed.
Asteroids in mutual perturbation with the planet Jupiter would have several slow, long-term effects. The semi-major axis dimension of individual asteroids would be able to change slowly, to come into commensurable, or actually NEAR commensurable, orbits with Jupiter. This could explain the presence and the origin of the Kirkwood gaps in the asteroid belt. The same effect would also apply to the four Galilean satellites of Jupiter, to cause the very distinctive near commensurable relationships to have developed over long periods of time (less than a million years in their case). The extreme thin shape and the gaps in the rings of Saturn, and the complex interrelationships between those ring particles and the various satellites of Saturn might also show the same reasoning. There are many other similar relationships. Interestingly, this seems to suggest that our common reasoning might have been backwards regarding the rings of Saturn; that instead of being the remains of previous satellites ripped apart by going within the Roche limit, they may instead represent particles that are gradually collecting to form future satellites.
That reasoning is this. By a particle being able to have its semi-major axis altered by the perturbative effect of an existing satellite, the particles might tend to be "channeled" into preferred orbital distances. There appear to be similar effects that might reduce eccentricity and orbital inclination. Among other things, this might explain the extremely thin thickness of the rings of Saturn and the other planets. This reasoning would suggest that a long-term study of Saturn's rings might show certain dynamic changes where some areas might be becoming more dense, while other areas of the rings might tend to clear out. This seems to show that the orbital inclination and eccentricity of the individual particle orbits have already been greatly affected, but that there might eventually be generated a rather narrow and fairly circular well-defined, higher density ring of material circling the planet. The reasoning would continue in suggesting that certain locations along that ring might have incrementally greater density, and therefore greater gravitational attraction, suggesting that the ring might gradually "clump up" into significant chunks, and that eventually, those co-revolving chunks would collect into a planet.
This line of reasoning seems to suggest an entirely new premise regarding the origin of the Earth! It has always been assumed that the Earth formed, in its present place, pretty much due exclusively to its own gravitational actions among its component parts. This must certainly have been true, of course, but a critically important external effect may also have been necessary, or at least important, primarily involving the planet Jupiter.
The previously unconsidered conjectured gravitational effect discussed here may have tended to "funnel" random materials of the Solar System into an area near what is now the orbital path of the Earth. We referred to that above as a narrow and compact ring. This would greatly have increased the local density of material that could eventually have accreted to become our Earth, temporarily creating essentially a wide and then very narrow ring-like structure around the Sun. This premise suggests that such a ring would then have been a precursor to relatively conventional view of accreting planetary formation, and not the result of the destruction of an earlier planet.
It should be noted that this premise does not involve any "mechanisms" such as gravity waves or electromagnetic phenomena. Rather, this premise is simply the long term result of repetitive patterns of the positions of objects orbiting the Sun, and the perturbations and resonance effects that would therefore exist. Nothing more than standard Newtonian gravitation is involved. The Engineering field of Forced Vibration becomes valuable in this exploration.
The full reasoning of the Forced Vibration aspects shows that it would be extremely unstable of two planets were precisely commensurate, as Laplace and LaGrange and the others regularly noted. Forced vibration analysis shows that the amplitude of perturbations would be extreme for exact commensurable orbiting objects.
In the field of Engineering, the mechanically destructive effects of resonant vibrations due to forced vibrations are analyzed. The differential equations of motion of an object having a natural frequency of ωn while being forced by an exterior force acting at a frequency defined by ω, can be written in the form of:
and
The solution to these differential equations can be written in the form of:
And
where e represents a variable generally called eccentricity (but which has a different meaning than the astronomical meaning of the term).
One can see that if the forcing frequency were exactly the same as the natural frequency, the denominator goes to zero and the amplitude of the oscillatory motion therefore goes to infinity. In mechanical systems, this is akin to the situation when a device disintegrates due to unexpected vibrations. These equations are for the situation for a system which has no damping factor, which cannot actually occur in any real mechanical device, but which is essentially true for systems of planets orbiting the Sun.
The claim here is that this Engineering approach of forced vibration can be applied for the situation of one planet perturbing another. The equations are more complex than these simple ones because the strength of the perturbing force constantly varies with the distance between the two planets per the inverse square rule, and they are each travelling in elliptic orbits. So the actual mathematics of this is more complex, but the reasoning is as indicated here.
Therefore, this shows that if the perturbing (forcing) frequency were commensurate with the natural frequency (inverse of period) of a planet, the perturbing effect would be extremely unstable and essentially catastrophic. This shows the well known effect of this fact that exact commensurability cannot exist among planets or satellites or asteroids with Jupiter.
There are therefore two adjacent orbital radii of near commensurability, one on each side of the exact commensurable orbital radius, where a meta-stable relationship can and will occur. By inserting the actual parameters for any specific two planets in the differential equations above, it should be possible to solve them to calculate the actual meta-stable solutions. Such solutions contain the above and also factors that depend on the cosine or sine of the motion of the perturbed planet, so the perturbative effects vary by both orbital motions. It is my belief that this should then indicate/calculate the slight differences of the Galilean satellites from exact commensurability with each other, and the other slight differences which exist in such relationships.
The Earth happens to be near the outer of the two which would represent the perturbations of Jupiter (at a harmonic factor of 12:1). Roughly 1.35 million miles inward of where we now are, there should be another meta-stable location, where debris should collect, as a response to Jupiter's perturbations. It seems reasonable to speculate that if a smaller amount of material initially went into that other ring, and it also coalesced into an object, we might have a very practical explanation for the Moon. It can even include why the Moon is made of different material than the Earth. The two might have been able to revolve in extremely similar orbits for a long time, but the great mass of both, and the very near passes they would make every century or so, and the rather small differential velocity that would exist between them, might easily explain how they later captured each other and became our double planet of the Earth and Moon.
For several centuries, it has been noted that there are an assortment of "interesting patterns" involving some Solar System objects. Titius-Bode's Law presented a surprisingly accurate simple formula regarding the relative distances of the planets from the Sun. No one has ever found a theoretical explanation for it. The four large Galilean Moons of Jupiter have orbits that are fairly close (but not exactly!) to having periods in the ratio of 1:2:4:8. Jupiter and Saturn have orbital periods around the Sun that are very close (but not exactly!) in the ratio of 2:5. Among the many thousands of asteroids circling the Sun, none seem to have orbital periods that are exactly simple fractions of the period of Jupiter (Kirkwood gaps), but many are fairly close to one or another of such fractions.
I am suggesting that there is a possibility that the violation of Conservation of Angular Momentum due to the precession-gyroscope effect might enable slow alteration of the semi-major axes of celestial objects, and the effects of Engineering's forced vibration analysis might clarify why exact commensurability cannot occur, and that this combination might provide some answers to some previously unanswered questions.
Jupiter is, by far, the most massive planet in our Solar System, and so it is universally accepted that it gravitationally affects all of the other objects, but deep analysis of the very-long-term consequences of these effects seems never to have been pursued. My research of the past several years has involved attempting to apply commonly accepted standard Engineering/Physics concepts regarding gyroscopes and "forced vibration" to such gravitational systems.
Another effect of the combination of these effects seems to be a general reduction of the orbital eccentricity (over very extended time), but with an instability for perfectly circular orbits.
Yet another effect seems to be a general reduction of the orbital inclination (again over very extended time), but again with an instability for perfectly co-planar orbits.
## Planetary Formation
Over extremely long periods of time, stray atoms, molecules and gas pockets and dust grains, would have been affected by Jupiter's gravity, either to be eventually pulled into Jupiter itself or to have orbital elements (around the Sun) altered as mentioned above. These materials might have been left-over components from the original formation of the Solar System OR they may have been deep-space materials that just happened to be in the path of the Sun and its family on its way through the Galaxy. The objects would (virtually) all be very small and easily perturbed in the ways described.
Over an extended period of time, a substantial amount of this material would initially have orbits that were relatively random in orbital radius, in orbital eccentricity, and in orbital inclination. The effects discussed here would gradually cause these various small particles to have orbits which tended toward having one of several specific preferred orbital radii and shapes and locations. Many would gradually tend to all have relatively circular orbits around the Sun, in a plane that is relatively the same as that of Jupiter's orbit.
Essentially, the Sun (with an axis of rotation completely independent of this, around 7° different in inclination, would develop a substantial ring system, possibly all within the orbit of Jupiter. One (pair of) favored radii might have been at a radius equivalent to an orbital period of around 1/12 that of Jupiter. This would then have created a relatively circular, relatively narrow ring around the Sun, at the orbital radius of where the Earth is now, or a pair of such rings around 1.35 million miles different in semi-major axis.
This would then provide the environment where the conventional view of an Earth then beginning to gravitationally self-coalesce, from the appropriate ring or pair of rings, with the same effect happening to create Venus, Mercury and Mars. The present asteroid belt may not then have existed, or it might have been an unsuccessful accretion into a planet for some reason, or it may still be an ongoing attempt at creating another planet. The former seems like a more likely scenario, suggesting that the objects in the present asteroid belt may be extremely old but the collecting them into a "belt" may have been a more recent process, which is still proceeding.
This premise suggests several interesting consequences:
• The asteroids might then NOT be remnants of some previously destroyed larger object, but a future planet in the early process of forming, due to Jupiter's gravitational effect. This might even be confirmable, by collectively doing Fourier Analysis the orbital elements of many the asteroids over time, to determine whether they are becoming more random or more organized.
• All of the four inner planets might then still be in the process of "growing" as confirmed by the many tons of meteoritic dust that falls to earth every day. The assumption has always been that our daily collection of meteoroids is from random space, where this might suggest that it is actually from a previously collected (by the effect of Jupiter) very faint ring of material currently sharing our orbit. There is a simple way to either prove or disprove this. If our spacecraft that have been sent to other portions of the solar system detect greater or lesser concentrations of micrometeoroids than we accumulate each day, then there might be such differences. If the detected quantities far from Earth are similar, then this suggestion is wrong. In the very distant past, those four planets might have been significantly smaller and less massive. Mercury would logically have been smaller, for having swept out a smaller volume of space and for having been farthest from Jupiter's gravitational forced-vibration effects.
• The elemental composition of the four inner planets might then better be explainable, and even the variations between them might be a result of just what available particles and materials had been available at roughly those orbital distances from the Sun.
• This premise suggests the critical requirement of a Jupiter-like planet, to permit the efficient accumulation of materials in a restricted ring-like spaces. This might imply that virtually all stars that have a Jupiter-sized planet orbiting them (except double-stars), would eventually accumulate small inner planets, in commensurable orbits.
• This premise permits the original Solar System to have only contained "star-like" components and objects. No exotic explanation of initial small rocky planets would be necessary, thereby simplifying the theories of initial formation.
• It barely needs to be mentioned that this premise would necessarily cause all the inner planets to revolve about the Sun in orbits that are in roughly the same plane as Jupiter and in the same direction.
Rather than the Earth being formed as part of the original Solar System, this premise suggests that it developed as an "afterthought"! The initial solar system seems logically to have had a preferred coordinate frame of the Sun's axis, given that it represents virtually all of the mass. Yet, all of the planets tend to lie in a rather different plane, relatively close to the Invariable Plane. Assuming that there was a good deal of stray material left over from the initial formation of the Solar System, the rate of meteoritic influx would have been much higher than today. As to how high, it may never be possible to know. But with the combination of Jupiter "aligning" space debris in and near the orbital path of the Earth, and the Earth's self-accretion, the Earth might have formed in a relatively reasonable period of time.
Note: A careful study of particle densities in the Solar System could constitute a confirmation of this premise. If this is valid, then there should be a number of exceedingly faint rings around the Sun, in orbits that are approximately commensurate with Jupiter, such as at radii that would have slightly different than 3 year orbital periods, or 4 year periods, or 6 year periods, or other simple fractions of Jupiter's orbital period. Many of these are known to exist, as they are within the asteroid belt. The exactly commensurate periods, being unstable, do not contain asteroids, and are known as Kirkwood Gaps in the asteroid belt. But there are many asteroids that have orbital periods that are quite close, the meta-stable situation described above.
Note: Another possible confirmation of this premise would be a thorough analysis of Saturn's ring system, as to orbital elements of the ring particles, particularly the dynamical changes that occur over years. If this premise is valid, statistically those ring particles should be trending toward more organized patterns, possibly in physically smaller volumes. This would be occurring because of the presence of the several massive satellites of Saturn, and the resultant forced-vibration resonant effects. | 5,157 | 26,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-22 | longest | en | 0.956147 |
http://wuyudong.com/2016/11/16/3026.html | 1,719,237,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00022.warc.gz | 29,706,540 | 8,028 | # 工学1号馆
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## 基于C语言的类型安全的动态数组的实现
Wu Yudong November 16, 2016 C 666
## vec.h 代码：
#define vec_t(T)\
struct { T *data; int length, capacity; }
#define vec_init(v)\
memset((v), 0, sizeof(*(v)))
#define vec_deinit(v)\
( free((v)->data),\
vec_init(v) )
#define vec_unpack_(v)\
(char**)&(v)->data, &(v)->length, &(v)->capacity, sizeof(*(v)->data)
#define vec_pop(v)\
(v)->data[--(v)->length]
#define vec_splice(v, start, count)\
( vec_splice_(vec_unpack_(v), start, count),\
(v)->length -= (count) )
#define vec_swapsplice(v, start, count)\
( vec_swapsplice_(vec_unpack_(v), start, count),\
(v)->length -= (count) )
#define vec_push(v, val)\
( vec_expand_(vec_unpack_(v)) ? -1 :\
((v)->data[(v)->length++] = (val), 0), 0 )
#define vec_insert(v, idx, val)\
( vec_insert_(vec_unpack_(v), idx) ? -1 :\
((v)->data[idx] = (val), 0), (v)->length++, 0 )
#define vec_clear(v)\
((v)->length = 0)
#define vec_truncate(v, len)\
((v)->length = (len) < (v)->length ? (len) : (v)->length)
#define vec_first(v)\
(v)->data[0]
#define vec_last(v)\
(v)->data[(v)->length - 1]
#define vec_reserve(v, n)\
vec_reserve_(vec_unpack_(v), n)
#define vec_compact(v)\
vec_compact_(vec_unpack_(v))
#define vec_pusharr(v, arr, count)\
do {\
int i__, n__ = (count);\
if (vec_reserve_po2_(vec_unpack_(v), (v)->length + n__) != 0) break;\
for (i__ = 0; i__ < n__; i__++) {\
(v)->data[(v)->length++] = (arr)[i__];\
}\
} while (0)
#define vec_extend(v, v2)\
vec_pusharr((v), (v2)->data, (v2)->length)
#define vec_find(v, val, idx)\
do {\
for ((idx) = 0; (idx) < (v)->length; (idx)++) {\
if ((v)->data[(idx)] == (val)) break;\
}\
if ((idx) == (v)->length) (idx) = -1;\
} while (0)
#define vec_remove(v, val)\
do {\
int idx__;\
vec_find(v, val, idx__);\
if (idx__ != -1) vec_splice(v, idx__, 1);\
} while (0)
#define vec_sort(v, fn)\
qsort((v)->data, (v)->length, sizeof(*(v)->data), fn)
#define vec_swap(v, idx1, idx2)\
vec_swap_(vec_unpack_(v), idx1, idx2)
#define vec_reverse(v)\
do {\
int i__ = (v)->length / 2;\
while (i__--) {\
vec_swap((v), i__, (v)->length - (i__ + 1));\
}\
} while (0)
#define vec_foreach(v, var, iter)\
if ( (v)->length > 0 )\
for ( (iter) = 0;\
(iter) < (v)->length && (((var) = (v)->data[(iter)]), 1);\
++(iter))
#define vec_foreach_rev(v, var, iter)\
if ( (v)->length > 0 )\
for ( (iter) = (v)->length - 1;\
(iter) >= 0 && (((var) = (v)->data[(iter)]), 1);\
--(iter))
#define vec_foreach_ptr(v, var, iter)\
if ( (v)->length > 0 )\
for ( (iter) = 0;\
(iter) < (v)->length && (((var) = &(v)->data[(iter)]), 1);\
++(iter))
#define vec_foreach_ptr_rev(v, var, iter)\
if ( (v)->length > 0 )\
for ( (iter) = (v)->length - 1;\
(iter) >= 0 && (((var) = &(v)->data[(iter)]), 1);\
--(iter))
## vec.c 代码
1、容量扩充函数vec_expand_()：
int vec_expand_(char **data, int *length, int *capacity, int memsz) {
if (*length + 1 > *capacity) { //当length==capacity,扩充容量
void *ptr;
int n = (*capacity == 0) ? 1 : *capacity << 1;
//当起始容量为0，分配一个单位，否则为原来的一倍
ptr = realloc(*data, n * memsz); //memsz为元素大小
if (ptr == NULL) return -1;
*data = ptr;
*capacity = n;
}
return 0;
}
2、数组容量预留函数vec_reserve_()：
int vec_reserve_(char **data, int *length, int *capacity, int memsz, int n) {
(void) length;
if (n > *capacity) {
void *ptr = realloc(*data, n * memsz);
if (ptr == NULL) return -1;
*data = ptr;
*capacity = n;
}
return 0;
}
3、数组容量预留函数vec_reserve_po2_()：
//数组预留空间不小于n的空间n2，且n2为2的幂
int vec_reserve_po2_(char **data, int *length, int *capacity, int memsz, int n) {
int n2 = 1;
if (n == 0) return 0;
while (n2 < n) n2 <<= 1;
return vec_reserve_(data, length, capacity, memsz, n2);
}
4、在数组的指定位置插入元素函数vec_insert_()：
int vec_insert_(char **data, int *length, int *capacity, int memsz, int idx) {
int err = vec_expand_(data, length, capacity, memsz);
if (err) return err;
memmove(*data + (idx + 1) * memsz,
*data + idx * memsz,
(*length - idx) * memsz);
return 0;
}
void *memmove(void *dst, const void *src, size_t count);
memmove的处理措施：
（1）当源内存的首地址等于目标内存的首地址时，不进行任何拷贝
（2）当源内存的首地址大于目标内存的首地址时，实行正向拷贝
（3）当源内存的首地址小于目标内存的首地址时，实行反向拷贝
vec_insert_函数的原理是将数组的索引idx位置到数组最后元素之间的元素全部后移一个元素位，空出一个位来存放待插入的元素
5、元素覆盖函数vec_splice_()：
void vec_splice_(char **data, int *length, int *capacity, int memsz, int start, int count) {
(void) capacity;
memmove(*data + start * memsz,
*data + (start + count) * memsz,
(*length - start - count) * memsz);
}
6、交换数组指定位置的两个元素vec_swap_()：
void vec_swap_(char **data, int *length, int *capacity, int memsz, int idx1, int idx2) {
unsigned char *a, *b, tmp;
int count;
(void) length;
(void) capacity;
if (idx1 == idx2) return;
a = (unsigned char*) *data + idx1 * memsz;
b = (unsigned char*) *data + idx2 * memsz;
count = memsz;
while (count--) {
tmp = *a;
*a = *b;
*b = tmp;
a++, b++;
}
} | 1,783 | 4,715 | {"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} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.321382 |
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# At a certain picnic, each of the guest was served either a
Author Message
Director
Joined: 27 Dec 2004
Posts: 901
Followers: 1
Kudos [?]: 45 [0], given: 0
At a certain picnic, each of the guest was served either a [#permalink]
### Show Tags
28 Mar 2005, 20:21
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.
At a certain picnic, each of the guest was served either a single scoop or a double scoop of ice cream. How many of the guests were served a double scoop of ice cream?
(1) At the picnic, 60 percent of the guests were served a double scoop of ice cream.
(2) A total of 120 scoops of ice cream were served to all the guests at the picnic.
Explanation is very critical please. Thanks
Director
Joined: 18 Feb 2005
Posts: 670
Followers: 1
Kudos [?]: 6 [0], given: 0
### Show Tags
28 Mar 2005, 20:36
Looks like C
option 1 gives a percent but no total scoops.Not suff
option2 gives 120 scoops in total but how do we get the single and double scoops.Not suff
Taking both we get the double sccops as 60% of 120 is 72
I take C
SVP
Joined: 03 Jan 2005
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### Show Tags
28 Mar 2005, 20:43
Your choice seems to be correct to me, but not the answer. Good thing DS question doesn't require a number as the answer. But do you agree that your number is not right?
Director
Joined: 18 Feb 2005
Posts: 670
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### Show Tags
28 Mar 2005, 20:48
Oh .......I realize now...Its the number of guest v/s no of scoops right? Then should it be E?
SVP
Joined: 03 Jan 2005
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### Show Tags
28 Mar 2005, 20:54
Well I do believe the answer is C. And the number is 45.
If there are 100 people then we need 60*2+40=160 scoop. But we only have 120 scoop so total number of people is 75. 60% of 75 is 45. (calculate as 3/4 of 60 to get faster result.)
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28 Mar 2005, 21:50
gmat2me2 wrote:
Looks like C
option 1 gives a percent but no total scoops.Not suff
option2 gives 120 scoops in total but how do we get the single and double scoops.Not suff
Taking both we get the double sccops as 60% of 120 is 72
I take C
hmm, u said taking both we get the double scoops as 60% of 120, but 60% is the number og guests served double scoops not the percentage of double scoops.
i don't seem to understand why 60%(of people) is being used with 120(number of scoops) to compute the number of guests who were served a double scoop of ice cream.
Does 60% of people mean 60% of 120 was served to them?
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29 Mar 2005, 00:33
(1) means 60% got double scoops, and 40% single scoops. Doesn't give the number of guests. Isnsufficient.
(2) means total scoops. Since we don't know the distribution, it could be anywhere between 60 (all served single scoops) and 120 people (all served double scoops). Insufficient.
Combining both.
if there're 10 people, 6 get double scoops, and 4 get single scoops.
Thus, 10 people require 16 scoops.
Therefore 120 scoops are sufficient for (120/16)x10 = 75 people.
Of course, from 1, the number of people served double scoop = 60% of 75 = 45.
Hope that helps.
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29 Mar 2005, 07:31
kapslock wrote:
(1) means 60% got double scoops, and 40% single scoops. Doesn't give the number of guests. Isnsufficient.
(2) means total scoops. Since we don't know the distribution, it could be anywhere between 60 (all served single scoops) and 120 people (all served double scoops). Insufficient.
Combining both.
if there're 10 people, 6 get double scoops, and 4 get single scoops.
Thus, 10 people require 16 scoops.
Therefore 120 scoops are sufficient for (120/16)x10 = 75 people.
Of course, from 1, the number of people served double scoop = 60% of 75 = 45.
Hope that helps.
Thanks Kapslock. I didn't know it was okay to assmume the number of people.
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04 Apr 2005, 11:34
Quote:
Thanks Kapslock. I didn't know it was okay to assmume the number of people.
I don't think it's ok to assume the number of pple. after all isn't DS about finding the sufficiency of the information given. As such if one has to assume then the information provided is clearly not sufficient.
I go for E.
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04 Apr 2005, 18:41
"C"
X - single scoop ppl....Y - double scoop ppl
state 1: Y = (X+Y)3/5 .....==> 2Y/5 = 3X/5 .....2Y = 3X ....eq 1
state 2: X + 2Y = 120.....insuff to calculate Y.....eq 2
combine:
from eqn 1 and 2 we can find Y....suff
4X = 120....X = 30
Y = 3/2X = 45
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05 Apr 2005, 00:27
Can't say i'm very clear, but i'll take it like that.
Thanks.
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05 Apr 2005, 00:46
(1) At the picnic, 60 percent of the guests were served a double scoop of ice cream.
- Not sufficient. We do not know # of guests.
(2) A total of 120 scoops of ice cream were served to all the guests at the picnic
- Not sufficient. We do not know how many gets double helping and how many single helping
(1) + (2)
Sufficient. we can equate and calculate the total number of guests, and from there the number of guests with double scoop of ice cream.
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05 Apr 2005, 07:39
Folaa3 wrote:
At a certain picnic, each of the guest was served either a single scoop or a double scoop of ice cream. How many of the guests were served a double scoop of ice cream?
(1) At the picnic, 60 percent of the guests were served a double scoop of ice cream.
(2) A total of 120 scoops of ice cream were served to all the guests at the picnic.
Explanation is very critical please. Thanks
Stmt 1 : Not suff as we don't know the total number of guests.
Stmt 2: Obviously not sufficient
1 & 2 together:
Let total number of guests = 100
from Stmt 1: 60 of them get double scoops ---> 60 * 2 = 120 scoops
So 40 of them shld get single scoops
Total scoops = 120 + 40 = 160
Therefore, now using Stmt 2, # of double scoops when total scoops is 120 = 120 /160 * 120 = 90
Thus, number of people getting double scoops = 90 /2 = 45
Re: DS-OG #1 [#permalink] 05 Apr 2005, 07:39
Display posts from previous: Sort by | 2,386 | 7,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-22 | longest | en | 0.927632 |
https://tomcircle.wordpress.com/tag/sheaf/ | 1,501,094,402,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00142.warc.gz | 713,882,793 | 37,952 | # Tutorial on Sheaves in Data Analytics
Background: What is Sheaf (束) ?
https://tomcircle.wordpress.com/2014/05/05/sheaf-sheaves/
Lecture 1
Lecture 2
Lecture 2: What is Topology ?
Lecture 3: What is Sheaf ?
Terms : Stalks (茎) together = Sheaf (束 / 梱)
Lecture 4: Data Structure as Sheaf
Lecture 5: Categorification
Lecture 7 & 8: Cohomology, Noise, Persistence Cohomology
<span More… (Total 8 video lectures)
Tutorial on Sheaves in Data Analytics: http://www.youtube.com/playlist?list=PLSekr_gm4hWLvFtJX0WUueVO65uhvBPrA
# Grothendieck’s Sheaf (束)
Natural Numbers (N) = {1,2,3, 4…}
1-dimension: a Line
2-dimension: a plane
n-dimensional flat space: a Vector Space
Now imagine in a world where we replace every natural number by vector space:
1 by a Line
2 by a Plane
n by a flat space Vector Space
Sum of numbers = Direct sum of vector space.
E.g. Add a 1-D Line to a 2-D Plane = 3-D Space
Product of numbers = Tensor Product (of two vector spaces of respective dimension m & n) with dimension m.n
This new world would be much interesting and richer than the Natural Number world: vector spaces have symmetries, whereas numbers are just numbers with no symmetries.
(Interesting): we can add 1 to 2 in only ONE way (1+2), but there are many ways to embed (add) a Line in a Plane (perpendicular, slanting in any angle, etc).
(Richer): the Lie Group SO(3) is the rotation of a 3-dimensional space. The number 3 is just an attribute of its dimensionality.
Numbers forms a Set
Vector Space forms a Category
Category not only contains ‘objects’ the vector spaces, also ‘morphisms’ from any object to any other object.
Note 1: Morphism of an object to itself = symmetries
Note 2: The Functional Language: Haskel is the next generation of computer languages based on Category, rather than on Set theory.
Categorification: The paradigm shift in modern math, by creating a new world in which the familiar concepts are elevated to a higher level — 欲穷千里路, 更上一层楼。( 6th CE Tang Dynasty Poem: “To see further thousand miles away, climb up one level higher”)
Examples: from sets to categories:
$\boxed { \text {Set} \rightarrow \text {Category} }$ Categorification (1)
Function (sine) to 30 degrees: eg.
$\sin (30 ^\circ) = \frac {1}{2}$
The function ‘sine’ assigns to a point of the circle (manifold) a number 1/2.
A Sheaf assigns to a point in the manifold a vector space.
$\boxed { \text {Number} \rightarrow \text {Vector Space} }$ Categorification (2)
$\boxed { \text {Function} \rightarrow \text {Sheaf} }$ Categorification (3)
Functions were the concepts of archaic math, and sheaves are the concepts of modern math. Sheaf can have symmetries. By elevating a function to a sheaf, we can explore these symmetries to learn more than we could ever learn using functions.
Sheaves applied in the right column of Langrands Program:
Number Theory | Curves over Finite Fields | Riemann Surfaces
$\begin{array}{ |l|l|l|l|} \hline : & Number \: Theory & Curves \: over \: Finite \: Fields & Riemann \: Surfaces \\ \hline L: & Galois \: Group & Galois \: Group & Fundamental \: Group \\ \hline R: & Automorphic \: functions & Automorphic \: functions \: (sheaves) & Automorphic \: sheaves\\ \hline \end{array}$
The 2 rows (L & R) below each column contains the names of the objects on the 2 sides of the Langrands relation specific to that column.
Love and Math by Edward Frenkel http://www.amazon.co.uk/dp/0465050743/ref=cm_sw_r_udp_awd_53swtb16779PY
# The Gap of Today’s Math Education: Rigor 严谨
This professor criticized the lack of rigor in today’s math education, in particular, there exists universally a prevalent ‘ambiguous’ gap between high school and undergraduate math education.
I admire his great insight which is obvious to those postwar baby boomer generation.
I remember I was the last Singapore batch or so (early 1970s) taking the full Euclidean Geometry course at 15 years old, and strangely in that year of Secondary 3 Math (equivalent to 3ème in Baccalaureate) my (Chinese) school had 2 separate math teachers for Geometry and Elementary/Additional (E./A.) Math.
Guess what ? the Geometry teacher was an Art teacher. It turned out it was a blessing in disguise, as my class of average Math students who hated E./A. Maths all scored 90% distinctions in Geometry. We did not treat Geometry like the other boring maths. The lady Art teacher started on the first day from Euclid’s 5 axioms, one by one as the days went by through the year, she derived each theorem with rigourous proof based on axioms and previous proven theorems. When a tough problem seemed too difficult, we were amazed by her astute ‘dotted line’ (补助线) technique, and voilà, we suddenly saw the relationship of the angles or triangles.
(This ‘dotted line’ once was used by my university French professor in Physics, he was very proud of finding a similar triangle relationship in the trajectory of a satellite revolving the earth – he was ‘awed’ by the French classmates who skipped the traditional Euclidean Geometry in lieu of the modern abstract Affine Geometry).
It may be a wise decision that Euclidean Geometry, now excluded from the main stream Math syllabus, be embedded in Art Class. Afterall, the great artists like Leonardo da Vinci applied Perspective Geometry in their art.
The professor challenged the development of these Math coursewares from the “First Principles”, a la Euclid’s Rigor: Group Theory, Combinatorics, etc.
Don’t forget history tells us that Mathematics originated from the Greek Geometry (Plato Math School “Lycée” : “Let no one who ignores Geometry enters this (school) gate. “), Euclid’s rigor influenced us for 3,000 years until 1900. The Modern Math conjectures like The Fermat’s Last Theorem from Number Theory was finally solved after 380 years by Geometry (Elliptic Curve from Shimura-Taniyama-Weil Theorem) in 1994.
The most difficult Math Challenge of today is the Langrands Program in unifying all Mathematics, with the hope of using the André Weil’s “Rosetta Stone” to ‘translate’ 3 distinct Math languages: Algebra, Analysis via the bridge of Geometry (Alexander Grothendieck’s Sheaf).
From the past IMO (International Math Olympiad) champion countries, notice that the Russian and Chinese school of Math do better than the western countries, because out of 6 IMO questions there are often 2 in Euclidean Geometry – a weakness in western school math syllabus (including Singapore which models on UK) for dropping Euclidean Geometry in schools, but the Russian and Chinese wisely keep it. | 1,665 | 6,542 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.851304 |
http://www.metric-conversions.org/length/miles-to-parsecs.htm | 1,516,262,043,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00267.warc.gz | 539,411,467 | 8,463 | Miles to Parsecs
Bookmark Page Parsecs to Miles (Swap Units)
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Show formula
Convert Miles to Parsecs
pc =
mi * 0.000000000000052155
Show working
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Miles
A unit of length equal to 1760 yards
Convert Miles to Parsecs
pc =
mi * 0.000000000000052155
Parsecs
The parsec is a unit of length equivalent to around 20 trillion (20,000,000,000,000) miles, 31 trillion kilometres, or 206,264 times the distance from the earth to the sun.
A parsec is also equivalent to approximately 3.26 light years (the journey distance if you travelled at the speed of light for three years and three months).
Miles to Parsecs table
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< Smaller Values Larger Values >
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59mi 0.00pc | 768 | 1,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-05 | latest | en | 0.702264 |
http://cacasa2.info/advanced-econometrics-takeshi-amemiya/2015/07/maximum-score-estimator-a-binary-case | 1,585,986,700,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370521574.59/warc/CC-MAIN-20200404073139-20200404103139-00531.warc.gz | 29,066,460 | 13,602 | Maximum Score Estimator—A Binary Case
Manski (1975) considered a multinomial QR model, but here we shall define his estimator for a binary QR model and shall prove its consistency. Our proof will be different from Manski’s proof.21 We shall then indicate how to extend the proof to the case of a multinomial QR model in the next subsection. Consider a binary QR model
Р(Уі = 1) = FfriPo), *’“1.2,. . • ,n, (9.6.1)
and define the score function
Sn(fi) = 2 Ш*І0 S 0) + (1 – У,)ХЫ0 < 0)], (9.6.2)
1—1
where
X(E) = 1 if event E occurs (9.6.3)
= 0 otherwise.
Note that the score is the number of correct predictions we would make if we predicted уj to be 1 whenever Э 0. Manski’s maximum score estimator Д,
is defined by
S„(P„) = sup (9.6.4)
ДЄВ
where the parameter space В is taken as
B = (W=1).
Clearly, (9.6.5) implies no loss of generality because S„(c0) = Sn(fi) for any positive scalar c.
Because is not continuous in fi, we cannot use Theorem 4.1.1 without
a modification. However, an appropriate modification is possible by generalizing the concept of convergence in probability as follows:
Definition 9.6.1. Let (ft, A, P) be a probability space. A sequence of not necessarily measurable functions gjico) for rw Є ft is said to converge to 0 in probability in the generalized sense if for any e > 0 there exists Ат Є A such that
AtQ{o)І |£г(й>)|<є} and limr-^„ P(AT) = 1.
Using this definition, we can modify Theorem 4.1.1 as follows:
Theorem 9.6.1. Make the following assumptions:
(A) The parameter space 0 is a compact subset of the Euclidean ЛГ-space
(*’).
(B) Qr(y, в) is a measurable function of у for all 0 Є 0.
(C) T-‘QM converges to a nonstochastic continuous function Q(6) in probability uniformly in в є 0 as Гgoes to <*, and Q(0) attains a unique global maximurn at в0.
Define вт as a value that satisfies
йтівт) = sup QM – (9.6.6)
Then §T converges to в0 in probability in the generalized sense.
We shall now prove the consistency of the maximum score estimator with the convergence understood to be in the sense of Definition 9.6.1.
Theorem 9.6.2. Assume the following:
(A) F is a distribution function such that F(x) = 0.5 if and only if x = 0.
(B) (xf) are i. i.d. vector random variables with a joint density function g(x) such that g(x) > 0 for all x.
Then any sequence (Д,) satisfying (9.6.4) converges to Д, in probability in the generalized sense.
Proof Let us verify the assumptions of Theorem 9.6.1. Assumptions A and В are clearly satisfied. The verification of assumption C will be done in five steps.
First, define for Я > 0
SM = 2 ІУЛГа(*Ї0) + 0 ~ (9-6.7)
f-i
where
^д(х) = 0 if (9.6.8)
= Ax if 0 < x < A’1 = 1 if Г’ёх
Because each term of the summation in (9.6.7) minus its expected value satisfies all the conditions of Theorem 4.2.1 for a fixed positive A, we can conclude that for any є, S > 0 there exists n^A), which may depend on A, such that for all n ё «/A)
14 (9.6.9) where QxiP) = EF(x’А0)у/л(х’Р) + E[ 1 – F{x’ Р0)]ц/х(—х’P). (9.6.10) Second, we have sup Іл-‘ЗД) – n-‘SM 1 (9.6.11)
where
Ых) = 0 if A-‘SM
= 1+Ax if — A~‘ <дг<0 = 1 ~ Ax if 0 ё x < A"1.
Applying Theorem 4.2.1 to A,, we conclude that for any e, S > 0 there exists n2(A), which may depend on A, such that for all пШ л2(А)
(9.6.13)
We have
A2 =§ sup P[(x’P)2 < Г2]. P
But, because the right-hand side of (9.6.14) converges to 0 as A —* <» because of assumption B, we have for all A § Aj
(9.6.15)
Therefore, from (9.6.11), (9.6.13), and (9.6.15), we conclude that for all n ^ n2(A) and for all A S Aj
Therefore, using the same argument that led to (9.6.15), we conclude that for all А ё Ai
P [sup IGO?) – Ся(Д)І > f] = °-
Fourth, because
sup n~lS„(P) – QW £ supn~ls„(fi) – n-lSMI P P
+ sapn~lSM-Q).(P)
P
+sup іа(л – q(P) i.
fi
we conclude from (9.6.9), (9.6.16), and (9.6.19) that for any e, S > 0 we have for all и ё max[/j,(A,), n2(A,)]
Fifth and finally, it remains to show that Q(fi) defined in (9.6.17) attains a unique global maximum at Д,. This is equivalent to showing
f [1-2F(x’fi0)]g(x)d* (9.6.2)
Jx’fia<0
> f [1 – 2F(x%)]g(x) dxi[0¥=fio.
But, because 1 — 2F(x’fi0) >0 in the region {х|х’Д, < 0} and 1 — 2F(x’fi0) < 0 in the region (х|х’Д> > 0} by assumption A, (9.6.22) follows immediately from assumption B. | 1,517 | 4,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-16 | latest | en | 0.785474 |
https://7puzzleblog.com/40/ | 1,652,865,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00574.warc.gz | 131,746,937 | 8,017 | # day/dydd 40 at 7puzzleblog.com
T he Main Challenge
In all four groups below, it is possible to make 24 by using the four numbers once each, with + – × ÷ available. Can you show how to achieve the target number of 24 in each case?
• 3 3 3 3
• 4 4 4 4
• 5 5 5 5
• 6 6 6 6
The 7puzzle Challenge
The playing board of the 7puzzle game is a 7-by-7 grid containing 49 different numbers, ranging from up to 84.
The 1st & 4th rows contain the following fourteen numbers:
2 3 9 10 14 15 22 32 35 40 44 54 60 72
What is the sum of the multiples of 6 listed above?
The Lagrange Challenge
Lagrange’s Four-Square Theorem states that every positive integer can be made by adding up to four square numbers.
For example, 7 can be made by 2²+1²+1²+1² (or 4+1+1+1).
There are just TWO ways of making 40 when using Lagrange’s Theorem. Can you find them both?
The Mathematically Possible Challenge
Using 26 and 11 once each, with + – × ÷ available, which is the ONLY number it is possible to make from the list below?
70 71 72 73 74 75 76 77 78 79
#NumbersIn70s
The Target Challenge
Can you arrive at 40 by inserting 2, 4, 5 and 10 into the gaps on each line?
• (◯+◯–◯)×◯ = 40
• (◯–◯)×◯×◯ = 40
• ◯²×◯²×◯÷ = 40
Answers can be found here.
Click Paul Godding for details of online maths tuition.
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### How much is growth worth?
In my last post, I looked at the price being paid for growth by valuing the assets in place in a business. To make this judgment, I assumed that the business would pay its entire operating income to claim holders (as dividends to stockholders and interest expenses to lenders). The value of assets in place then becomes the value of the earnings in perpetuity, discounted back at the cost of capital.
So, what is the effect of growth on value? To grow in the long term, you have to reinvest some or a big portion of your earnings back into the business, and the amount you have to reinvest will depend upon the return on capital you earn on your new investments:
Reinvestment Rate = Expected Growth rate/ Return on Capital
Thus, a firm with a return on capital of 15% that wants to grow 3% a year will have to reinvest 20% (= 3%/15%) of its operating income back each year. Investors will thus get less in cash flows up front but have higher cash flows in future years.
Consider an example. A firm that generates \$ 10 million in after-tax operating income, and has a cost of capital of 10%, will have a value of assets in place of \$100 million, if it pursues a "no growth" policy:
Value of assets in place = \$10 million/ .10 = \$100 million
If it decides to pursue a 3% growth rate and invest 20% of its after-tax income (based upon the return on capital of 15%), its value can be computed as follows:
Value of firm = 10 million (1-.20) / (.10-.03) = \$114.28 million
The difference between the two values then becomes the value added by growth:
Value of growth = Value of firm – Value of assets in place = \$114.28 - \$100 = \$14.28 million
Determinants of the value of growth
If you accept the proposition that growth creates a trade off of lower cash flows today for higher ones in the future, you have the three ingredients that determine the value of growth. The first is the level of growth, with higher growth rates in the future generating higher earnings over time. The second is how long these high growth rates can be sustained before the company becomes too big to keep growing (at least at rates higher than that of the economy). The third and most critical is the return on capital you generate on new investments.
To see why the last ingredient is so critical, revisit the last example and make the return on capital = cost of capital. If you do so, the reinvestment rate has to be 30% to sustain the expected growth rate of 3%. The value of growth then becomes zero:
Value of firm = 10 million (1-.30) / (.10-.03) = \$100 million
Value of growth = Value of firm – Value of assets in place = \$100 - \$100 = \$ 0
In fact, if the return on capital generated on new investments is less than the cost of capital, growth can destroy value.
The process of valuing growth does get a little more complicated when you set higher growth rates, but the logic and conclusions do not change. If the return on capital > cost of capital, the value of growth will increase as the growth rate increases and the length of the growth period expands. If the return on capital = cost of capital, neither the growth rate nor the length of the growth period affect value and if the return on capital < cost of capital, the value will move inversely with the growth rate and the length of the growth period. If you want to take this concept out for a trial run, this spreadsheet can help you.
Comparing the value of growth to the price paid for growth
If you are paying a price for growth, it is always useful to know the value of this growth. If you accept the reasoning in the last section, it follows that it is not growth that you should be paying a premium for but “quality growth”, with quality defined as the excess return you generate over and above the cost of capital. To illustrate this concept, we compute “intrinsic” PE ratios at varying growth rates for three firms, all of which share a cost of capital of 10% but vary in the returns on capital that they earn on new investments (one has a return on capital of 8%, the second has a return on capital of 10% and the third has a return on capital of 12%).
The PE ratio for just the assets in place is 11.63 and remains unchanged, even if you introduce growth, for a firm that earns its cost of capital. For the firm that generates a return on capital < cost of capital, the PE ratio decreases as growth increases, reflecting value destruction in action. For a firm that generates a return on capital > cost of capital, the PE ratio does increase (the growth premium) as growth increases. It is this premium that you would compare to what you actually pay to make a judgment on whether the added PE you are paying for growth is justified.
Price of Growth versus Value of Growth
Using the spreadsheet on growth as a device for deconstructing growth (and its value), I looked at Microsoft, Kraft, Google and Linkedin. In the table below, I have listed my base assumptions for each company and the value of assets in place and expected growth in each one:
This table can be used to address several issues relating to growth:
a. Price of growth versus value of growth: You can compare the price you are paying for growth with the value of growth, and you come to different conclusions. For Microsoft, where the value of assets in place covers the market price you are paying, the value of growth is a pure bonus. For Kraft, the value of growth is negative, since the firm earns less than its cost of capital, and the price you are paying for growth is therefore too high. For Google, the price of growth is almost exactly equal to the value of growth, making it the only fair priced stock in this grouping. Finally, for Linkedin, the price paid for growth is more than twice the value of that growth, making the stock over valued. For investors who believe in growth at a reasonable price (GARP), this is the statistic worth watching.
b. Implied growth rates: An alternative approach is to solve for that growth rate (Look at the spreadsheet and follow the instructions), holding the return on capital and length of growth period fixed, that would yield the price you paid for that growth. Linkedin, for instance, would have to maintain a compounded growth rate of 73% a year (instead of the estimated growth rate of 60% a year) for the next ten years to justify the price you are paying for the growth. (The spreadsheet provides instructions on how to back out the implied growth rate using the Goal seek function in Excel.)
Growth, in summary, does not yield itself easily to rules of thumb or broad generalizations. In some firms, it can be worth nothing, as is argued by strict value investors, whereas in others, it can be worth a great deal, lending credence to the arguments of growth investors.
### How much are you paying for growth?
The debate about Facebook’s valuation is interesting on many dimensions, but one that is worth focusing on is how much growth is worth, and what you are paying for it. At one extreme are some value investors who argue that growth is “speculative” and that it is worth very little or nothing. At the other are those who argue that growth is priceless and that you should therefore be willing to pay a “fortune” for it. Both groups seem to be in agreement that valuing growth is pointless, because it requires estimates that will be wrong in hindsight. I had a series of posts on growth a few months ago looking at the limits of growth, the scaling up of growth, the value of growth and how management credibility affects that value. In this post, I offer a simple technique for assessing how much you are paying for growth in a company. In the next one, I address how to value that growth.
Growth Assets and Assets in Place
To provide a perspective on how growth and value interact, it is best to start with what I would call a financial balance sheet.
While it is structured like an accounting balance sheet, it is different on two counts. First, rather than break assets down into fixed, current and financial assets, as accounting balance sheets do, assets are broken down into two categories: “assets in place”, representing the value of investments already made and “growth assets”, measuring the value added by expected future growth. Second, accounting balance sheets are rooted in the past, with numbers representing capital originally invested in assets, whereas financial balance sheets are forward looking, with the values of these assets being based on their capacity to generate cash flows or on market values.
The value of assets in place
To understand the price you are paying for growth, consider a simple experiment. A business that has existing assets that are generating earnings has two choices. It can pay the entire income out to claimholders (as dividends to stockholders and interest to lenders ) and forsake future growth. Alternatively, it can reinvest some (or all) of its earnings back into new investments and generate growth for the future. If you adopt the no growth alternative, your earnings from the most recent period will be your cash flow each year in perpetuity. The value of these cash flows can be computed by discounting back at a cost of capital to yield a value for assets in place:
Value of assets in place = After-tax Operating Income from most recent period/ Cost of capital
Note that the depreciation & amortization from the most recent period is reinvested back into the business to keep its earnings power intact.
There are only three estimation inputs that you need to derive this value. The first is the operating income. While it is convenient to use the operating income from the most resent year as the base value, you may choose to use an average over a few years for cyclical and commodity companies. The second is the tax rate. Again, while the effective tax rate is the easiest to access, you may decide to replace it with a marginal tax rate, if you feel that the company will revert to that rate over time. The third is the cost of capital. While you can compute the cost of capital for the firm in question, it may be far simpler to use the average cost of capital for the sector in which the firm operates. There is one variant worth considering. If you feel that the assets of the face obsolescence, you may decide to assume that the earnings from these assets will be available only for a finite period rather than forever. The equation for value of assets in place has to be modified to be an annuity, instead of a perpetuity.
Price paid for growth: DCF
Once you have derived a value for assets in place, you can estimate what you are paying for growth by looking at the traded value of the firm, computed as the enterprise value of the business (market value of equity plus debt minus cash). The difference between the traded enterprise value and the value of the assets in place can be considered the price paid for growth.
In the table below, we look at four firms, Microsoft, Kraft, Google and Linkedin, to illustrate this concept.
For each firm, we report the after-tax operating income and the cost of capital used to derive the value of the assets in place. By comparing this number to the enterprise value of the firm, we then compute, on a percent basis, the proportion of the price that goes towards growth. What are we to make of these numbers? For Microsoft, you can justify the entire market value of the firm with the value of just assets in place. For Kraft and Google, about 40% of the price paid is for expected future growth. For Linkedin, it is almost 99% of the value. Does the fact that Microsoft's entire value is justified by assets in place make it a better investment than Linkedin? Not necessarily, since we have not valued growth explicitly and growth can destroy value. In my next post, I will look at the value of growth at each of these companies and consequences for investors who have paid much higher prices.
Note that this entire analysis can also be done in purely equity terms, with net income divided by cost of equity to derive the growth value in equity in assets in place. If you do so, you can compare the market capitalization (rather than enterprise value) of the firm to the assets in place. The difference will be the price paid for growth.
Price paid for growth: Relative valuation
High growth companies often trade at high multiples of earnings, book value or revenues and the “premium’ is usually justified as the price for growth. This premium can be in enterprise value multiples, such as EV/EBITDA, EV/Sales or EV/Invested capital:
EV growth premium = Actual EV multiple - EV multiple for assets in place
With Google, for instance, the EV/EBIT multiple for just assets in place can be computed to be 7.90, obtained by dividing the intrinsic value of assets in place (\$92,761 million) by the operating income (\$11,742 million). It's actual EV/EBIT multiple is 13.01, estimated by dividing the actual enterprise value of \$152,784 million by the same operating income. The growth premium in the EV/EBIT multiple is therefore 5.11 (13.01- 7.90).
The premium can also be stated in terms of price earnings ratios, as the difference between the PE ratio that you actually pay compared to the PE ratio that you would pay for just the assets in place.
PE premium = Actual PE ratio - PE ratio for assets in place
You can estimate the PE ratio for assets in place, either from the cost of equity directly (PE ratio for assets in place = 1/ Cost of equity) or by backing the equity value from the intrinsic value of assets in place (and subtracting out the debt and adding back cash). Using Google as an example again (with debt of \$4,204 million, cash of \$44,460 million and net income of \$9,737 million):
Intrinsic value of equity in assets in place = \$ 92,761 - \$4,204 + \$ 44,460 = \$132,818
PE for assets in place = \$132,818/ \$9,737 = 13.64
Actual PE = \$192,840/\$9,737 = 19.80
Growth premium in PE = 19.80 - 13.64 = 6.26
## Wednesday, May 23, 2012
### Facebook: Sowing the wind, reaping the whirlwind
Last Thursday, about 24 hours prior to the initial public offering, I posted on what I thought would happen on the opening day. I argued that this was the most pre-priced IPO in history, with transactions in the private share market providing information on what investors would be willing to pay for the stock. That was the basis for my view that those expecting a large jump on the opening day were likely to be disappointed and that this would be the Goldilocks IPO, with a 10-15% bump at open. I also felt that the stock was overvalued by about a third and that what happened on the opening day would be revealing not just for Facebook, but for all social media companies. The stock did open up about 12% and faded very quickly to the offering price by the end of the day. In fact, without active support from the investment banks, it would have dropped below. In the last few days, the stock has cratered, declining to about \$32 at the time of this post. Here are the lessons I am taking away from this process:
Pricing versus Valuation
Pricing is an exercise of gauging demand and supply, reading investor moods and determining what people will pay for an asset, rather than what it is worth. Valuation is about estimating what an asset is worth, given its earning potential, growth and risk. You can tell whether an investor or analyst is a “pricer” or “valuer” by looking at the tools that he or she uses. The tools of choice for most pricers are relative valuation (multiples such as PE or EV multipes), where you assess how much you will pay for an asset by looking at what others are paying for similar assets (usually other companies in the same business), and technical analysis (where you use charts and indicators to gauge shifts in demand). The tools of choice for “valuers” are either discounted cash flow (DCF) or accounting based (building off book value) models.
Momentum is fragile and requires illusions
Momentum is a strong force in markets but it is one that we don’t understand yet and don’t believe that we ever will. It is after all not only the basis for the madness of crowds and behavioral finance, but also of that most feared phenomenon in markets, the dreaded bubble. Not only is momentum driven by market moods and perceptions, but it is fragile and based ultimately upon an illusion. After all, most momentum investors don’t view themselves as such, and choose to rationalize their behavior using “fundamental’ factors. Thus, in the midst of every bubble, investors delude themselves that it is not a bubble by looking for a good reason: that tulip bulbs would become scarcer in the future, that dot com companies would dominate every business that the operated in and that the demand for real estate would always outstrip supply.
In their ideal scenario, I am sure that the investment banks hoped that the momentum that they were detecting in the private share markets and in their conversations with institutional investors would continue into the opening day and the weeks after. So, what happened on the opening day? I believe that the momentum shifted and that the hubris of the company and the bankers in the days leading up to opening day contributed significantly to it happening. Rather than maintain the illusion that the offering price was justified by fundamentals, nebulous though they might have been, the parties involved seemed to completely abandon any illusions about value and made it a starkly momentum game. This was manifested in the hiking of the offering price to \$ 38 on Thursday evening and in insiders in the company publicly bailing out at the offering price. Even the maddest of crowds, when constant confronted with proof that they are being viewed as suckers, will wake up, and to the dismay of the company and the banks, it happened an hour into the offering.
What now?
Much as I would like to believe that what has happened in the last couple of days to Facebook stock is a vindication of valuation, I am a realist. There is no fury that matches that of a disillusioned crowd and I believe that what you are seeing is momentum investors, who were promised quick riches if they bought Facebook stock, bailing out. Will they stop selling at fair value? Since they have no idea what it is, why should they? If momentum shifts in the past are any indicator, you should see the price of Facebook drop not just to its intrinsic value (you have mine, but yours may be different), but to below that value. Since the company is the poster-child for the “social media” sector, I think that you will see this momentum shift play out on other social media companies.
Would I buy Facebook, Linkedin, Groupon or any other social media company? Social media is an umbrella under which you have diverse firms, some with more clearly defined business models than others and some with stronger barriers to entry than others, and when momentum shifts, investors tend not be discriminating. In the words of that eminent philosopher, Justin Bieber, you “never say never” and some of these stocks are likely to be bargains, sooner rather than later. If you are a value investor, you should be ready.
## Thursday, May 17, 2012
### Facebook and "Field of Dreams": Hoodies, Hubris and Hoopla
In mid-February, I posted my valuation of Facebook and my thoughts on what would happen at the IPO. Since the actual offering date is tomorrow and the frenzy mounts, I thought it would make sense to revisit those posts.
1. Valuation Update
In my February 16 post on the company, I attached my valuation of the company, based on the S-1 filing as of that date. Quickly reprising that valuation, I valued the equity in the company at \$29/share (assigning an overall value of about \$72 billion for Facebook's equity), with the following key assumptions:
a. Revenues growing to \$44 billion in ten years, with a compounded revenue growth rate of 40% for the next 5 years, fading down over time
b. A pre-tax operating margin of about 35%, higher than Google's 30% and on par with Apple
c. Reinvestment (in internal projects and acquisitions) that generate a \$1.50 in revenue for every dollar in capital.
d. A cost of capital of 11.42% initially, fading down to 6.50% in steady state
So, what have we learned about Facebook in the last three months that may change this valuation?
a. Facebook wants growth and will pay for it: Facebook has acquired three companies in the last couple of months, Instagram, Tagtile and Glancee. While the Tagtile and Glancee deals were a continuation of a long term strategy of buying small firms with technologies that augment the Facebook experience, Instagram represented a new front: a "more" expensive acquisition of a company that brought with it potential "new users". My guess is that a publicly traded Facebook, with access to far more capital, will continue making acquisitions with the intent of delivering promised revenue growth and that the pace (and the size) of acquisitions will pick up if (or as) internal growth slackens. That is mixed news for investors: the good news is that it increases the odds that the predicted growth in revenues will be delivered but the bad news is that Facebook may pay more for this growth than anticipated.
b. Mark Zuckerberg is lord and master of this company: While there has never been any doubt about the autocratic power structure at Facebook, the last three months have brought home that this is Zuckerberg's company. If news stories are to be believed, the decision to buy Instagram (at least at the final price) was made by Zuckerberg, with little input from the board. If you are going to be a stockholder in Facebook, you should get used to this scene being played out in small and big ways over the next few years.
c. The "Field of Dreams" business model: Finally, Facebook's value still lies in its promise, rather than in actual numbers. Remember the line from the movie, "Field of Dreams", where Kevin Costner wanders through a corn field and hears a voice that tells him that "if you build it, he will come". With Facebook and other social media companies, this line can paraphrased as "if you get the users, they (products, advertising) will come". While I do not want too much of a single story, the news story of GM abandoning its Facebook advertising should provide a cautionary note to the optimistic view that Facebook can easily convert its monstrously large user base into advertising fodder.
Bottom line: Revisiting the valuation, there is not a great deal I would change as a result of news over the last few weeks: a higher revenue growth rate (45% compounded with revenues growing to \$ 56 billion) accompanied by lower margins (30%) and more reinvestment (\$1.25 of revenues for every dollar invested) delivers an estimate of value that stays at the \$70-\$80 billion range.
2. Pricing (IPO) Update
When I labeled this the "IPO of the century" in February, I was speaking tongue in cheek. After all, the century is young and there are other IPOs to come. While there is little that you will learn about the value of the company from the IPO process, there is a great deal that we can learn about human behavior and the ecosystem that feeds off big deals.
a. The bankers will do anything to be part of a "big deal": As you track the news stories, it is quite clear that the bankers need the Facebook deal more than Facebook needs the bankers. In fact, I am quite surprised that Facebook did not follow the Google model and bypass the investment bankers entirely and set up an auction. I think that the only reason that they chose to follow the conventional route is because investment banks are essentially doing this deal at cut rate prices and bending to Facebook's will at every turn..
b. And Mark Zuckerberg know it: As someone who has never been comfortable wearing a tie or a suit, I must confess that I found the brouhaha over Zuckerberg's hoodie to be hilarious. I don't particularly care for Zuckerberg's corporate governance, but I, for one, have never believed that your professionalism is determined by what you wear. I am sure that Bruno Iskil, who lost billions for JP Morgan, wore a very expensive suit, while making his trades. I think Zuckerberg, in addition to mimicking one of his idols, Steve Jobs, was sending a message to Wall Street about who has the upper hand in this game.
c. Investors are replaying an age-old phenomenon: Individual investors are clearly caught up in the mood of the moment, lining up to get allotments of Facebook shares. Is it a bubble? Who knows? If those who forget history are destined to repeat it, it sure looks like a replay of events from the past, and for those who do no remember them, I have a reading suggestion.
d. The insiders: While I don't assume that insiders are infallible, it is telling that they are heading for the exits at the same time as individuals are piling in. Is it possible that they think that the stock is being priced at the top end of the value range? Do they not trust Mark Zuckerberg? Inquiring minds want to know and i guess we will find out as events unfold.
Bottom line: I don't think that there has ever been an IPO where investment bankers have had more information (from private share market prices to institutional investor feedback) to work with, when pricing the stock, than this one. I would be very surprised, if the stock were overpriced; the bankers and the company have too much too lose. I would be equally surprised if the stock were dramatically under priced; a pop of 50% or even 25% would reflect very badly on the bankers' pricing skills. In short, this is shaping up to be a Goldilocks IPO, at least in the initial hours: a pop of about 10-15% (just right for both the bankers and the company). The question is how long the pop will last. This company is too big and too public to stage manage in the weeks after the IPO. If the pop fades quickly, perhaps even by the end of trading tomorrow, I think it is a very bad sign for the momentum game in all social media stocks.
3. Investment strategies
So, what should investors do about Facebook? You can play the IPO game, and I have described some of the ways you could do it, in an earlier post. Generically, here are the four strategies you can adopt:
a. Short term buy: It may be too late for you to get in at the offering price, but if you believe in the short term momentum story, you can buy right as the market for Facebook opens tomorrow morning, hope to ride the crest of the price move up as other investors doing the same and exit before they do.
b. Short term sell: If you think that the hype is overdone and that disappointment will set in very soon, you can sell short right after the market opens tomorrow, especially if it does not open with a significant pop, with the intent of covering in the next week or two.
c. Long term buy: You may be a believer in Facebook's potential and its capacity to dominate the advertising market and to sell products to its users. If so, you should buy sometime in the near future and hold for the long term. How long will you have to wait to see profits? It depends on how quickly Facebook converts its potential to large revenues and profits... could be a year.. could be five..
d. Long term sell: If you do buy into my "Goldilocks IPO" scenario and come up with an estimate of intrinsic value close to mine, though, the investment with the best odds of success on Facebook would be a "long term, short" position on the stock.
Bottom line: I think that the hype is overdone, that disappointment will set in sooner or later and that the stock has far more downside than upside. You can put me in the last group (long term sell) though I am still searching for the most efficient (and least costly) way to execute this.
Does the Facebook IPO have broader implications for the overall equity market? I have heard arguments that a successful Facebook IPO will lead to a rebirth of faith in equities among investors and be a shot in the arm for financial service firms. I think that is nonsense.
• If Facebook does launch successfully tomorrow and the stock price goes up 10%, 20% or even 30%, I don't see how it will cause risk averse investors to come back to stocks. In fact, it will probably feed into their suspicion that the stock market has become a casino that they cannot trust their savings in.
• As for investment banks, a successful Facebook IPO may bring in some fees and commissions but it will not be a reflection of their skills at pricing or deal making. This is a stock that priced and marketed itself, with little or no help from the investment bankers.
In the same vein, a failed IPO (and I will leave you to define what failure means) will have implications for the pricing of social media companies but not much more.
Bottom line: Facebook, in spite of its ubiquitous presence in our lives, is just one company and not a very big one (at least in terms of revenues and earnings) yet. The market will obsess about it tomorrow but it will move on very quickly to the next worry, fear or fad.
## Monday, May 7, 2012
### Lessons learned, unlearned and relearned: A semester of online class
In January, I posted of my intent to put my valuation and corporate finance classes online. As I finished my last sessions in both classes today, I thought it would be a good time to take stock of what the experience taught me about the future of education and how online education can evolve.
A quick review. I teach corporate finance and valuation classes to MBAs at the Stern School of Business at NYU and have done so for 26 years. While I have put my webcasts and material online for many years, I decided to both formalize and organize the online class this year, using a start up firm called Coursekit. It has since been able to attract more funds and has a new name, Lore.
The site allowed me to place the resources for the class (lecture notes, assignments, handouts and exams) and the webcasts of the class in one place, together with a social media add-on where anyone (me or any student registered in the class) could post links, thoughts or questions about the class. Here are the links to the classes:
Valuation: Link to Lore Valuation class
Corporate Finance: Link to Lore Corporate Finance class
I had a lot of fun, teaching these classes, and I am glad I did it, but I did get some valuable lessons in online learning.
1. Discipline is critical on both sides. I had to learn to be disciplined and organized, posting lectures on the site, as soon as I had the links, as well as any other resources I used in my regular class. On the participant side, I recognized how difficult it is for someone (often several time zones away), with a regular job and family commitments, to spend 80 minutes twice a week, watching lectures and then spending more time preparing for the class. While I don't have the statistics, I am sure that of the 2500+ people signed up for the corporate finance class and the 1900 people in the valuation class, relatively few will finish the class on time (today) and that many will not finish the class at all.
What I plan to do about it
(1) I plan to leave the class online for at least the summer and perhaps longer. Hopefully, that will allow those whose constraint is time to catch up on the lectures and even do the projects for the class.
(2) I also plan to create a shorter (20 minute) version of each 80-minute lecture, delivering the high points of the session for those who really cannot spend the time needed for the full lecture.
2. Diverse backgrounds/experiences: I know that some people struggled more than others, partly due to language differences and partly because of diverse backgrounds (some were more well versed in statistics and accounting than others). I feel badly for those who struggled and wish I could have done more to help.
What I plan to do about it
(1) I am looking for online resources that I can direct people who are in search of basic accounting/ statistics classes to. If I cannot find anything, I will attempt to create my own makeshift versions.
(2) Next semester, I hope to rope in a few people who have been gracious enough to offer their help and start a tutorial component to the class. Together, we may be able to fill the gap.
3. Technology: Technology is still evolving and that there were roadblocks, on both sides. On my side, there were at least two sessions in my corporate finance class where the recording system failed, and I had to make recordings to fill in. On the other side, those users who tried to watch the webcasts on Google Chrome were stymied (Don't even ask...) and some had trouble with bandwidth.
What I plan to do about it
(1) Fix the recording system at Stern so that there are no recording failures. Improve the audio and video quality.
(2) Use more conduits for the lectures (iTunes U, YouTube) to allow those who have trouble downloading to be able to go elsewhere to get the material.
In short, there is a great deal that I can do to make the online learning experience a richer, more interactive one, and I plan to keep working on it. For those of you who were and are part of this experiment, thank you for giving it a shot. For those of you who were disappointed in it, I am sorry and I will work on making it better.
As a final point, for those who would rather take your classes in person, I am preparing for two executive valuation seminars that I will be delivering during the next month.
a. The first is a two-day valuation seminar in Mumbai, India, on May 24 and 25. You can get details about the seminar by clicking here.
b. The second is an open-enrollment three-day valuation seminar that I teach every year at the Stern School of Business in New York. This year, it will be on June 4, 5 and 6. While it is intense, I manage to cover almost all of the material that I teach in my regular MBA class (which lasts 14 weeks). You can get details about it by clicking here. | 7,367 | 34,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-26 | latest | en | 0.95327 |
https://discourse.edwardlib.org/t/lda-with-collapsed-gibbs-sampling/975 | 1,656,332,390,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00197.warc.gz | 263,096,242 | 4,756 | # LDA with collapsed Gibbs Sampling
Dear All,
I would like to verify my first Edward code which is in fact a transformation of an example of gaussian mixture inference via gibbs sampling. Technically code works however, I am a bit unsatisfied with the inference quality:
from future import absolute_import
from future import division
from future import print_function
from time import time
import edward as ed
import matplotlib.pyplot as plt
import numpy as np
import tensorflow as tf
from edward.models import (
Dirichlet, Categorical, ParamMixture, Multinomial)
D = 10
N = 200
K = 4
sess = ed.get_session()
# model definition
alpha = tf.ones(K)
theta = Dirichlet(alpha)
gamma = tf.ones(N)
beta = Dirichlet(gamma,sample_shape=K)
mix = ParamMixture(theta,{‘probs’ : beta}, Categorical,sample_shape=N)
z = mix.cat
# analytical conditional part. This is calculated by tensorflow symbolically.
theta_cond = ed.complete_conditional(theta)
beta_cond = ed.complete_conditional(beta)
z_cond = ed.complete_conditional(z)
# test distribution of the same shape parameters but with random priors.
arr = tf.ones(K) * 4.0
np.divide(arr,np.sum(arr))
alpha_test = arr
theta_test = Dirichlet(alpha_test)
arr = np.random.rand(N) * 4.0
np.divide(arr,np.sum(arr))
gamma_test = tf.ones(N)
beta_test = Dirichlet(gamma_test,sample_shape=K)
mix_test = ParamMixture(theta_test,{‘probs’ : beta_test}, Categorical,sample_shape=N)
z_test = mix_test.cat
# Initialize
theta_est,beta_est,z_est = sess.run([theta, beta, z])
mix_data, z_data = sess.run([mix_test, z_test])
mix_data = np.reshape(mix_data,(N,))
theta_est = np.reshape(np.asarray(theta_est),(K,))
print(‘Test parameters:’)
print(‘theta:’, sess.run(theta_test))
print(‘Initial parameters:’)
print(‘theta:’, theta_est)
#print(‘gamma:’, gamma_est)
print()
# inference - gibbs sampler
cond_dict = {theta: theta_est, beta: beta_est,
z: z_est, mix: mix_data}
t0 = time()
T = 50000
for t in range(T):
z_est = sess.run(z_cond, cond_dict)
cond_dict[z] = z_est
theta_est = np.reshape(np.asarray(sess.run([theta_cond], cond_dict)),(K,))
cond_dict[theta] = theta_est
beta_est = sess.run(beta_cond, cond_dict)
cond_dict[beta] = beta_est
print(‘took %.3f seconds to run %d iterations’ % (time() - t0, T))
print()
print(‘Final parameters::’)
print(‘theta:’, theta_est)
plt.figure(figsize=[10, 10])
plt.subplot(2, 1, 1)
plt.hist(mix_data, 50)
plt.title(‘Empirical Distribution of x‘)
plt.xlabel(’x’)
plt.ylabel(‘frequency’)
xl = plt.xlim()
plt.subplot(2, 1, 2)
plt.hist(sess.run(mix,{theta: theta_est, beta: beta_est,z: z_est}), 50)
plt.title(“Predictive distribution”)
plt.xlabel(’x’)
plt.ylabel(‘frequency’)
plt.xlim(xl)
plt.show()
mixture of gaussians:
lda:
Hardly gibbs sampler learnt latent z and distribution parameters.
Any hints would be welcomed.
M. | 781 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-27 | latest | en | 0.58014 |
http://www.math-drills.com/baseten.shtml | 1,386,763,715,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164035500/warc/CC-MAIN-20131204133355-00097-ip-10-33-133-15.ec2.internal.warc.gz | 435,203,588 | 9,051 | # Base Ten Blocks Worksheets
Welcome to the base ten blocks page at Math-Drills.com where blocking your students' learning is the best approach!
On this page, you will find several worksheets for base ten block manipulatives. Base ten blocks are an excellent tool for teaching children math number concepts because they allow children to touch and manipulate something real while learning important skills that translate well into paper and pencil addition. They are also proportional representations of numbers, so that a thousand block is actually 1000 times greater in size than a one block.
## Counting Base Ten Blocks Worksheets
Although we recommend using actual base ten blocks, not everyone has access to them, and some students like to use them as projectiles or for other purposes not related to learning. Counting various base ten blocks and turning them into numbers is a necessary skill to have in order to trade and represent numbers with base ten blocks.
## Trading Base Ten Blocks Worksheets
Trading blocks both ways will help a great deal when adding and subtracting numbers with base ten blocks, so don't skip this part.
## Representing Numbers With Base Ten Blocks Worksheets
Representing numbers with base ten blocks is a skill that is completely necessary if you want to use base ten blocks for addition, subtraction, multiplication or division. Use the worksheets below as a starting point where students can continue by making their own representations of base ten blocks given only a number. Encourage them to make each number out of the fewest blocks in each case. For example, they would model the number 14 with a rod and four units rather than 14 units.
To add numbers with base ten blocks, students model both numbers and combine the piles. They trade any piles ten or greater for larger blocks, staring with the units pile. To subtract numbers, students model the first number, then remove enough blocks to make the second number. This often involves "borrowing" or "regrouping" where a larger block must be traded for smaller blocks to facilitate the removal. | 404 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2013-48 | longest | en | 0.917539 |
http://conversion-website.com/speed/nautical-mile-per-hour-to-knot.html | 1,534,665,954,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214713.45/warc/CC-MAIN-20180819070943-20180819090943-00055.warc.gz | 89,079,210 | 4,538 | Nautical miles per hour to knots (nmi/h to kn)
Convert nautical miles per hour to knots
Nautical miles per hour to knots converter on this page calculates how many knots are in 'X' nautical miles per hour (where 'X' is the number of nautical miles per hour to convert to knots). In order to convert a value from nautical miles per hour to knots (from nmi/h to kn) just type the number of nmi/h to be converted to kn and then click on the 'convert' button.
Nautical miles per hour to knots conversion factor
1 nautical mile per hour is equal to 0.99999999914471 knots
Nautical miles per hour to knots conversion formula
Speed(kn) = Speed (nmi/h) × 0.99999999914471
Example: How many knots are in 255 nautical miles per hour?
Speed(kn) = 255 ( nmi/h ) × 0.99999999914471 ( kn / nmi/h )
Speed(kn) = 254.9999997819 kn or
255 nmi/h = 254.9999997819 kn
255 nautical miles per hour equals 254.9999997819 knots
Nautical miles per hour to knots conversion table
nautical miles per hour (nmi/h)knots (kn)
109.9999999914471
1514.999999987171
2019.999999982894
2524.999999978618
3029.999999974341
3534.999999970065
4039.999999965788
4544.999999961512
5049.999999957235
5554.999999952959
6059.999999948683
6564.999999944406
7069.99999994013
7574.999999935853
8079.999999931577
8584.9999999273
9089.999999923024
9594.999999918747
10099.999999914471
105104.99999991019
nautical miles per hour (nmi/h)knots (kn)
250249.99999978618
350349.99999970065
450449.99999961512
550549.99999952959
650649.99999944406
750749.99999935853
850849.999999273
950949.99999918747
10501049.9999991019
11501149.9999990164
12501249.9999989309
13501349.9999988454
14501449.9999987598
15501549.9999986743
16501649.9999985888
17501749.9999985032
18501849.9999984177
19501949.9999983322
20502049.9999982467
21502149.9999981611
Versions of the nautical miles per hour to knots conversion table. To create a nautical miles per hour to knots conversion table for different values, click on the "Create a customized speed conversion table" button.
Related speed conversions
Back to nautical miles per hour to knots conversion
TableFormulaFactorConverterTop | 696 | 2,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-34 | latest | en | 0.570306 |
https://www.crazy-numbers.com/en/12880 | 1,716,413,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00622.warc.gz | 638,355,201 | 3,668 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 12880: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 12880: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 12880
Is 12880 a prime number? No
Is 12880 a perfect number? No
Number of divisors 40
List of dividers
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1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 23, 28, 35, 40, 46, 56, 70, 80, 92, 112, 115, 140, 161, 184, 230, 280, 322, 368, 460, 560, 644, 805, 920, 1288, 1610, 1840, 2576, 3220, 6440, 12880
Sum of divisors 35712
Prime factorization 24 x 5 x 7 x 23
Prime factors
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2, 5, 7, 23
## How to write / spell 12880 in letters?
In letters, the number 12880 is written as: Twelve thousand eight hundred and eighty. And in other languages? how does it spell?
12880 in other languages
Write 12880 in english Twelve thousand eight hundred and eighty
Write 12880 in french Douze mille huit cent quatre-vingts
Write 12880 in spanish Doce mil ochocientos ochenta
Write 12880 in portuguese Doze mil oitocentos oitenta
## Decomposition of the number 12880
The number 12880 is composed of:
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
2 iterations of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
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1 iteration of the number 0 : ... Find out more about the number 0 | 820 | 2,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.731767 |
https://www.smartstudy.com/ielts/article/2777366.html | 1,618,757,105,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00025.warc.gz | 1,102,779,201 | 39,970 | 在雅思写作考试小作文部分，主要是要求大家根据所给出的图表来进行描述，其中柱形图是比较容易考到的一类题型。今天小编为大家准备了雅思写作柱状图模板句型，各位考生可以在备考中多多练习，这样可以帮助我们实战考试中灵活的运用。
1.the table shows the changes in the number of...over the period from...to...
该表格描述了在...年之...年间...数量的变化。
2.the bar chart illustrates that...
该柱状图展示了...
3.the graph provides some interesting data regarding...
该图为我们提供了有关...有趣数据。
4.the diagram shows (that)...
该图向我们展示了...
5.the pie graph depicts (that)....
该圆形图揭示了...
6.this is a cure graph which describes the trend of...
这个曲线图描述了...的趋势。
7.the figures/statistics show (that)...
数据(字)表明...
8.the tree diagram reveals how...
该树型图向我们揭示了如何...
9.the data/statistics show (that)...
该数据(字)可以这样理解...
10.the data/statistics/figures lead us to the conclusion that...
这些数据资料令我们得出结论...
11.as is shown/demonstrated/exhibited in the diagram/graph/chart/table...
如图所示...
12.according to the chart/figures...
根据这些表(数字)...
13.as is shown in the table...
如表格所示...
14.as can be seen from the diagram,great changes have taken place in...
从图中可以看出，...发生了巨大变化。
15.from the table/chart/diagram/figure,we can see clearly that...or it is clear/apparent from the chart that...
从图表我们可以很清楚(明显)看到...
16.this is a graph which illustrates...
这个图表向我们展示了...
17.this table shows the changing proportion of a & b from...to...
该表格描述了...年到...年间a与b的比例关系。
18.the graph,presented in a pie chart, shows the general trend in...
该图以圆形图形式描述了...总的趋势。
19.this is a column chart showing...
这是个柱型图，描述了...
20.as can be seen from the graph,the two curves show the flutuation of...
如图所示，两条曲线描述了...的波动情况。
21.over the period from...to...the...remained level.
在...至...期间，...基本不变。
22.in the year between...and...
在...年到...期间...
23.in the 3 years spanning from 1995 through 1998...
1995年至1998三年里...
24.from then on/from this time onwards...
从那时起...
25.the number of...remained steady/stable from (month/year) to (month/year).
...月(年)至...月(年)...的数量基本不变。
26.the number sharply went up to...
数字急剧上升至...
27.the percentage of...stayed the same between...and...
...至...期间...的比率维持不变。
28.the figures peaked at...in(month/year)
...的数目在...月(年)达到顶点，为... | 731 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-17 | latest | en | 0.421946 |
http://bilakniha.cvut.cz/en/predmet5965706.html | 1,611,741,414,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00268.warc.gz | 11,835,798 | 3,673 | CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2020/2021
# Introduction to Curves and Surfaces
Code Completion Credits Range
02UKP1 Z 2 1P+1C
Lecturer:
Tutor:
Supervisor:
Department of Physics
Synopsis:
The goal of the lecture is an introduction to the differential geometry of simple manifolds - curves and two-dimensional surfaces. The basic concepts for the curves are introduced Frenets formulae are explained. In the surface theory we introduce first and second fundamental forms and mean and Gaussian curvature. Essential part of the lecture are the examples calculated by students.
Requirements:
Syllabus of lectures:
Outline of the lecture:
1. Examples and definition of curves
2. Plane curves, natural equation of a curve
3. Space curves, curvature, torsion
4. Frenet formulas
5. Examples and definition of surfaces
6. The first fundamental form, lenght of a curve on the surface
7. Transformation properties of the first fundamental form
Outline of the exercises:
1. Curvature and length of the curve
2. Curvature and area of a surface
3. Metric tensor
Syllabus of tutorials:
Outline of the exercises:
1. Curvature and length of the curve
2. Curvature and area of a surface
3. Metric tensor
Study Objective:
Knowledge:
To provide the simplest examples of manifolds and their properties.
Acquired skills:
Solve mathematical problems defined on manifolds.
Study materials:
Key references:
[1] L. Hlavatý, Úvod do křivek a ploch (in Czech)
www.fjfi.cvut.cz > katedra fyziky > studentský servis > Doprovod přednášek > Úvod do křivek a ploch
Recommended references:
[2] B. Hostinský, Diferenciální geometrie křivek a ploch, Přírodovědecké nakladatelství v Praze, 1949 (in Czech)
[3] W. Kuehnel, Diferential Geometry, AMS2006
[4] T. Banchoff, S Lovett , Diferential Geometry of Curves and Surfaces, CRC Press 2016
Note:
Time-table for winter semester 2020/2021:
Time-table is not available yet
Time-table for summer semester 2020/2021:
Time-table is not available yet
The course is a part of the following study plans:
Data valid to 2021-01-27
For updated information see http://bilakniha.cvut.cz/en/predmet5965706.html | 570 | 2,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-04 | latest | en | 0.720462 |
https://crypto.stackexchange.com/questions/18454/public-key-in-fully-homomorphic-encryption-over-the-integers | 1,717,085,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00300.warc.gz | 147,060,101 | 39,662 | # Public key in fully homomorphic encryption over the integers
I'm reading “Fully Homomorphic Encryption over the Integers” by van Dijk et al.
I wonder why $x_0$, which is a component of the public key, should be an odd number?
• What happens if you set it even? Aug 4, 2014 at 12:23
• Thanks for giving me a hint. (even number mod even number) is always even number and (odd number mod even number) is always odd number. If m=0, the result of modular $x_0$ will be always even number. If m=1, the result will be odd number. So $x_0$ should be a odd number. Aug 4, 2014 at 15:04
• @JongHyunKim, you could post that as an answer, so if someone else arrives here wondering about it, they can easily find it.
– otus
Aug 5, 2014 at 7:46
## 1 Answer
(even number mod even number) is always even number and (odd number mod even number) is always odd number. If $m=0$, the result of modular $x_0$ will be always even number. If $m=1$, the result will be odd number. So $x_0$ should be a odd number. | 292 | 996 | {"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} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.849087 |
https://questioncove.com/updates/524ca8c8e4b0e0b6559942a9 | 1,627,184,284,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00298.warc.gz | 477,719,370 | 5,059 | Mathematics
OpenStudy (osanseviero):
If this cancels out, why is it 1? http://newton.matem.unam.mx:8080/ejercicios/racionales/p0507.gif
zepdrix (zepdrix):
Hmm, example: $$\Large \dfrac{5}{5}\quad=\quad ?$$
OpenStudy (osanseviero):
Oh...it makes complete sense now...I am dividing by the same thing
zepdrix (zepdrix):
Yah, the factors are not cancelling per-say. Maybe we use that word too loosely in math. They're dividing out.
OpenStudy (osanseviero):
Thanks :) | 149 | 471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-31 | latest | en | 0.554456 |
https://slidetodoc.com/multivariate-statistics-pca-principal-component-analysis-correspondence-analysis/ | 1,638,419,389,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00559.warc.gz | 589,780,085 | 14,235 | # Multivariate statistics PCA principal component analysis Correspondence analysis
• Slides: 22
Multivariate statistics • • • PCA: principal component analysis Correspondence analysis Canonical correlation Discriminant function analysis Cluster analysis MANOVA Xuhua Xia Slide 1
PCA • Given a set of variables x 1, x 2, …, xn, – find a set of coefficients a 11, a 12, …, a 1 n, so that PC 1 = a 11 x 1 + a 12 x 2 + …+ a 1 nxn has the maximum variance (v 1) subject to the constraint that a 1 is a unit vector, i. e. , sqrt(a 112+ a 122 …+ a 1 n 2) = 1 – find a 2 nd set of coefficients a 2 so that PC 2 has the maximum variance (v 2) subject to the unit vector constraint and the additional constraint that a 2 is orthogonal to a 1 – find 3 rd, 4 th, … nth set of coefficients so that PC 3, PC 4, … have the maximum variance (v 3, v 4, …) subject to the unit vector constraint and that ai is orthogonal to all ai-1 vectors. – It turns out that v 1, v 2, … are eigenvalues and a 1, a 2, … are eigenvectors of the variance-covariance matrix of x 1, x 2, …, xn • PCA is to find the eigenvalues and eigenvectors. Slide 2
Typical Form of Data A data set in a 8 x 3 matrix. The rows could be species and columns sampling sites. 100 97 99 96 90 90 80 75 60 75 85 95 X= 62 40 28 77 80 78 92 91 80 75 85 100 A matrix is often referred to as a nxp matrix (n for number of rows and p for number of columns). Our matrix has 8 rows and 3 columns, and is an 8 x 3 matrix. A variance-covariance matrix has n = p, and is called n-dimensional square matrix. Xuhua Xia Slide 3
What are Principal Components? PC = a 1 X 1 + a 2 X 2 + … an. Xn • Principal components are linear combinations of the observed variables. The coefficients of these principal components are chosen to meet four criteria • What are the four criteria? Xuhua Xia Slide 4
What are Principal Components? • The four criteria: – There are exactly p principal components (PCs), each being a linear combination of the observed variables; – The PCs are mutually orthogonal (i. e. , perpendicular and uncorrelated); – The components are extracted in order of decreasing variance. – The components are in the form of eigenvalues and eigenvector of unit length. Xuhua Xia Slide 5
A Simple Data Set X 1 X 2 -1. 2649 -1. 7889 -0. 6325 -0. 8944 0 0 0. 6325 0. 8944 1. 2649 1. 7889 X Y Correlation matrix X 1 1 Covariance matrix Y 1 1 Xuhua Xia X Y X 1 1. 414 Y 1. 414 2 Slide 6
General observations • The total variance is 3 (= 1 + 2) • The two variables, X and Y, are perfectly correlated, with all points fall on the regression line. • The spatial relationship among the 5 points can therefore be represented by a single dimension. • For this reason, PCA is often referred to as a dimension-reduction technique. What would happen if we apply PCA to the data? Xuhua Xia Slide 7
Graphic PCA 2 1. 5 1 Y 0. 5 0 -0. 5 -1 -1. 5 -2 -1. 5 Xuhua Xia -1 -0. 5 0 0. 5 1 1. 5 X Slide 8
R functions X 1 X 2 -1. 2649 -1. 7889 -0. 6325 -0. 8944 0 0 0. 6325 0. 8944 1. 2649 1. 7889 options("scipen"=100, "digits"=6) obj. PCA<-prcomp(~X 1+X 2) obj. PCA<-prcomp(md, scale. =T) predict(obj. PCA, md) predict(obj. PCA, data. frame(X 1=0. 3, X 2=0. 5) screeplot(obj. PCA) Don’t use scientific notation. Requesting the PCA to be carried out on the covariance matrix (default) rather than the correlation matrix. Use scale. =TRUE to request PCA on correlation matrix Help decide how many PCs to keep when there are many variables Xuhua Xia Slide 9
A positive definite matrix • When you run the SAS program, the log file will warn that “The Correlation Matrix is not positive definite. ”. What does that mean? • A symmetric matrix M (such as a correlation matrix or a covariance matrix) is positive definite if z’Mz > 0 for all nonzero vectors z with real entries, where z’ is the transpose of z. • Given our correlation matrix with all entries being 1, it is easy to find z that lead to z’Mz = 0. So the matrix is not positive definite: Replace the correlation matrix with the covariance matrix and solve for z. Xuhua Xia Slide 10
SAS Output Standard deviations: [1] 1. 7320714226 0. 0000440773 Rotation: PC 1 PC 2 X 1 0. 577347 0. 816499 X 2 0. 816499 -0. 577347 [1, ] [2, ] [3, ] [4, ] [5, ] better to output in variance (eigenvalue) accounted for by each PC eigenvectors: PC 1 = 0. 57735 X 1+0. 81650 X 2 PC 1 PC 2 -2. 19092 0. 0000278767 -1. 09545 -0. 0000557540 0. 0000000000 1. 09545 0. 0000557540 2. 19092 -0. 0000278767 Principal component scores What’s the variance in PC 1? Xuhua Xia Slide 11
PCA on correlation matrix (scale. =T) Standard deviations: [1] 1. 4142135619 0. 0000381717 Rotation: PC 1 PC 2 X 1 0. 707107 X 2 0. 707107 -0. 707107 PC 1 PC 2 [1, ] -1. 788850 0. 0000241421 [2, ] -0. 894435 -0. 0000482837 [3, ] 0. 0000000000 [4, ] 0. 894435 0. 0000482837 [5, ] 1. 788850 -0. 0000241421 Xuhua Xia Slide 12
Crime Data in 50 States STATE ALABAMA ALASKA ARIZONA ARKANSAS CALIFORNIA COLORADO CONNECTICUT DELAWARE FLORIDA GEORGIA HAWAII IDAHO ILLINOIS. . MURDER 14. 2 10. 8 9. 5 8. 8 11. 5 6. 3 4. 2 6. 0 10. 2 11. 7 7. 2 5. 5 9. 9. . RAPE 25. 2 51. 6 34. 2 27. 6 49. 4 42. 0 16. 8 24. 9 39. 6 31. 1 25. 5 19. 4 21. 8. . ROBBE 96. 8 138. 2 83. 2 287. 0 170. 7 129. 5 157. 0 187. 9 140. 5 128. 0 39. 6 211. 3. . ASSAU 278. 3 284. 0 312. 3 203. 4 358. 0 292. 9 131. 8 194. 2 449. 1 256. 5 64. 1 172. 5 209. 0. . BURGLA 1135. 5 1331. 7 2346. 1 972. 6 2139. 4 1935. 2 1346. 0 1682. 6 1859. 9 1351. 1 1911. 5 1050. 8 1085. 0. . LARCEN 1881. 9 3369. 8 4467. 4 1862. 1 3499. 8 3903. 2 2620. 7 3678. 4 3840. 5 2170. 2 3920. 4 2599. 6 2828. 5. . AUTO 280. 7 753. 3 439. 5 183. 4 663. 5 477. 1 593. 2 467. 0 351. 4 297. 9 489. 4 237. 6 528. 6. . PROC PRINCOMP OUT=CRIMCOMP; Xuhua Xia Slide 17
STATE MURDER RAPE ROBBE ASSAU BURGLA LARCEN AUTO Alabama 14. 2 25. 2 96. 8 278. 3 1135. 5 1881. 9 280. 7 Alaska 10. 8 51. 6 96. 8 284. 0 1331. 7 3369. 8 753. 3 Arizona 9. 5 34. 2 138. 2 312. 3 2346. 1 4467. 4 439. 5 Arkansas 8. 8 27. 6 83. 2 203. 4 972. 6 1862. 1 183. 4 California 11. 5 49. 4 287. 0 358. 0 2139. 4 3499. 8 663. 5 Colorado 6. 3 42. 0 170. 7 292. 9 1935. 2 3903. 2 477. 1 Connecticut 4. 2 16. 8 129. 5 131. 8 1346. 0 2620. 7 593. 2 Delaware 6. 0 24. 9 157. 0 194. 2 1682. 6 3678. 4 467. 0 Florida 10. 2 39. 6 187. 9 449. 1 1859. 9 3840. 5 351. 4 Georgia 11. 7 31. 1 140. 5 256. 5 1351. 1 2170. 2 297. 9 Hawaii 7. 2 25. 5 128. 0 64. 1 1911. 5 3920. 4 489. 4 Idaho 5. 5 19. 4 39. 6 172. 5 1050. 8 2599. 6 237. 6 Illinois 9. 9 21. 8 211. 3 209. 0 1085. 0 2828. 5 528. 6 Indiana 7. 4 26. 5 123. 2 153. 5 1086. 2 2498. 7 377. 4 Iowa 2. 3 10. 6 41. 2 89. 8 812. 5 2685. 1 219. 9 Kansas 6. 6 22. 0 100. 7 180. 5 1270. 4 2739. 3 244. 3 Kentucky 10. 1 19. 1 81. 1 123. 3 872. 2 1662. 1 245. 4 Louisiana 15. 5 30. 9 142. 9 335. 5 1165. 5 2469. 9 337. 7 Maine 2. 4 13. 5 38. 7 170. 0 1253. 1 2350. 7 246. 9 Maryland 8. 0 34. 8 292. 1 358. 9 1400. 0 3177. 7 428. 5 Massachusetts 3. 1 20. 8 169. 1 231. 6 1532. 2 2311. 31140. 1 Michigan 9. 3 38. 9 261. 9 274. 6 1522. 7 3159. 0 545. 5 Minnesota 2. 7 19. 5 85. 9 85. 8 1134. 7 2559. 3 343. 1 Mississippi 14. 3 19. 6 65. 7 189. 1 915. 6 1239. 9 144. 4 Missouri 9. 6 28. 3 189. 0 233. 5 1318. 3 2424. 2 378. 4 Montana 5. 4 16. 7 39. 2 156. 8 804. 9 2773. 2 309. 2 Nebraska 3. 9 18. 1 64. 7 112. 7 760. 0 2316. 1 249. 1 Nevada 15. 8 49. 1 323. 1 355. 0 2453. 1 4212. 6 559. 2 New Hampshire 3. 2 10. 7 23. 2 76. 0 1041. 7 2343. 9 293. 4 New Jersey 5. 6 21. 0 180. 4 185. 1 1435. 8 2774. 5 511. 5
Crime data (cont. ) New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virginia Washington West Virginia Wisconsin Wyoming 8. 8 10. 7 10. 6 0. 9 7. 8 8. 6 4. 9 5. 6 3. 6 11. 9 2. 0 10. 1 13. 3 3. 5 1. 4 9. 0 4. 3 6. 0 2. 8 5. 4 39. 1 29. 4 17. 0 9. 0 27. 3 29. 2 39. 9 19. 0 10. 5 33. 0 13. 5 29. 7 33. 8 20. 3 15. 9 23. 3 39. 6 13. 2 12. 9 21. 9 109. 6 472. 6 61. 3 13. 3 190. 5 73. 8 124. 1 130. 3 86. 5 105. 9 17. 9 145. 8 152. 4 68. 8 30. 8 92. 1 106. 2 42. 2 52. 2 39. 7 343. 4 319. 1 318. 3 43. 8 181. 1 205. 0 286. 9 128. 0 201. 0 485. 3 155. 7 203. 9 208. 2 147. 3 101. 2 165. 7 224. 8 90. 9 63. 7 173. 9 1418. 7 1728. 0 1154. 1 446. 1 1216. 0 1288. 2 1636. 4 877. 5 1489. 5 1613. 6 570. 5 1259. 7 1603. 1 1171. 6 1348. 2 986. 2 1605. 6 597. 4 846. 9 811. 6 3008. 6 2782. 0 2037. 8 1843. 0 2696. 8 2228. 1 3506. 1 1624. 1 2844. 1 2342. 4 1704. 4 1776. 5 2988. 7 3004. 6 2201. 0 2521. 2 3386. 9 1341. 7 2614. 2 2772. 2 259. 5 745. 8 192. 1 144. 7 400. 4 326. 8 388. 9 333. 2 791. 4 245. 1 147. 5 314. 0 397. 6 334. 5 265. 2 226. 7 360. 3 163. 3 220. 7 282. 0 md<-read. fwf("crime. txt", c(14, 6, 5, 6, 6, 7, 7, 6), header=T) attach(md) cor(md[, 2: 8] If you copy the data to a text file, add a top line obj. PCA<-prcomp(md[, 2: 8], scale. =T) with a comment sign #, otherwise you need to obj. PCA summary(obj. PCA) specify the 'sep=' with read. fwf PCScore<-predict(obj. PCA, md)
Correlation Matrix MURDER RAPE ROBBERY ASSAULT BURGLARY LARCENY AUTO 1. 0000 0. 6012 0. 4837 0. 6486 0. 3858 0. 1019 0. 0688 RAPE ROBBERY ASSAULT BURGLARY LARCENY 0. 6012 1. 0000 0. 5919 0. 7403 0. 7121 0. 6140 0. 3489 0. 4837 0. 5919 1. 0000 0. 5571 0. 6372 0. 4467 0. 5907 0. 6486 0. 7403 0. 5571 1. 0000 0. 6229 0. 4044 0. 2758 0. 3858 0. 7121 0. 6372 0. 6229 1. 0000 0. 7921 0. 5580 0. 1019 0. 6140 0. 4467 0. 4044 0. 7921 1. 0000 0. 4442 AUTO 0. 0688 0. 3489 0. 5907 0. 2758 0. 5580 0. 4442 1. 0000 If variables are not correlated, there would be no point in doing PCA. The correlation matrix is symmetric, so we only need to inspect either the upper or lower triangular matrix. Xuhua Xia Slide 20
Eigenvalues > summary(obj. PCA) Importance of components: PC 1 PC 2 PC 3 PC 4 PC 5 PC 6 PC 7 Standard deviation 2. 029 1. 113 0. 852 0. 5625 0. 5079 0. 4712 0. 3522 Proportion of Variance 0. 588 0. 177 0. 104 0. 0452 0. 0369 0. 0317 0. 0177 Cumulative Proportion 0. 588 0. 765 0. 869 0. 9137 0. 9506 0. 9823 1. 0000 screeplot(obj. PCA, type = "lines") Xuhua Xia Slide 21
Eigenvectors PC 1 PC 2 PC 3 PC 4 PC 5 PC 6 PC 7 MURDER -0. 3003 -0. 62918 0. 17824 -0. 23216 0. 53810 0. 25912 0. 267589 RAPE -0. 4318 -0. 16944 -0. 24421 0. 06219 0. 18848 -0. 77327 -0. 296490 ROBBE -0. 3969 0. 04224 0. 49588 -0. 55793 -0. 52002 -0. 11439 -0. 003902 ASSAU -0. 3966 -0. 34353 -0. 06953 0. 62984 -0. 50660 0. 17235 0. 191751 BURGLA -0. 4402 0. 20334 -0. 20990 -0. 05757 0. 10101 0. 53599 -0. 648117 LARCEN -0. 3574 0. 40233 -0. 53922 -0. 23491 0. 03008 0. 03941 0. 601688 AUTO -0. 2952 0. 50241 0. 56837 0. 41922 0. 36980 -0. 05729 0. 147044 • Do these eigenvectors mean anything? – All crimes are negatively correlated with the first eigenvector, which is therefore interpreted as a measure of overall safety. – The 2 nd eigenvector has positive loadings on AUTO, LARCENY and ROBBERY and negative loadings on MURDER, ASSAULT and RAPE. It is interpreted to measure the preponderance of property crime over violent crime…. . . Xuhua Xia Slide 22
biplot(obj. PCA) Xuhua Xia Slide 23
Plot PC 1 and PC 2 Massachusetts 2 Rhode Island Hawaii Connecticut 1 Delaware Arizona New Jersey Colorado Minnesota Vermont New Iowa. Hampshire Wisconsin Utah Washington Oregon Maine PC 2 0 New York Alaska Michigan California Illinois Ohio Wyoming Indiana Kansas Idaho Maryland Nevada -1 Florida Pennsylvania Texas North Dakota Montana Nebraska Missouri Oklahoma Virginia South Dakota West Virginia New Mexico Tennessee Kentucky Arkansas Georgia -2 North Carolina Louisiana South Carolina Alabama Mississippi -4 -2 0 PC 1 2 4
PC Plot: Crime Data Maryland North and South Dakota Nevada, New York, California Mississippi, Alabama, Louisiana, South Carolina Xuhua Xia Slide 25
Steps in a PCA • • • Generate a correlation or variance-covariance matrix Obtain eigenvalues and eigenvectors Generate principal component (PC) scores Choose the number of PCs Plot the PC scores in the space with reduced dimensions Xuhua Xia Slide 26 | 5,429 | 11,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-49 | latest | en | 0.814035 |
http://mathhelpforum.com/differential-geometry/94689-tests-convergences-print.html | 1,529,949,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00442.warc.gz | 194,323,789 | 2,887 | # Tests for convergences
• Jul 8th 2009, 07:57 PM
Juancd08
Tests for convergences
SUM(arctan(a_n) converges if and only if SUM(a_n) converges.
a_n in this case is a sequence of positive numbers.
• Jul 12th 2009, 10:09 AM
mathemanyak
[quote=Juancd08;337083]SUM(arctan(a_n) converges if and only if SUM(a_n) converges.
preposition
1.Let arctan(a_n)=b_n => tan(b_n)=a_n
2.ıf any sequence is converges than its subsequences also converge.
Solution
tan(b_n) is a subsequence of b_n. By the comparison princeple if b_n is converge then tan b_n also converge. Conversly now show that if tan(b_n) is converge than b_n converges. Similiarly arctan(a_n) is subsequence of a_n then right handed also satisfies. then proof is done
• Jul 12th 2009, 11:12 AM
HallsofIvy
[QUOTE=mathemanyak;338106]
Quote:
Originally Posted by Juancd08
SUM(arctan(a_n) converges if and only if SUM(a_n) converges.
preposition
1.Let arctan(a_n)=b_n => tan(b_n)=a_n
2.ıf any sequence is converges than its subsequences also converge.
Solution
tan(b_n) is a subsequence of b_n. By the comparison princeple if b_n is converge then tan b_n also converge. Conversly now show that if tan(b_n) is converge than b_n converges. Similiarly arctan(a_n) is subsequence of a_n then right handed also satisfies. then proof is done
How is tan(b_n) a subsequence of b_n? You are saying that, for any positive integer m, there exist a positive integer n such that b_n= tan(b_m)? I don't see that. Or are you simply asserting that tan(b_n) is dominated by bn: that tan(b_n)< b_n for all n? | 472 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-26 | latest | en | 0.856183 |
http://mathhelpforum.com/calculus/213306-help-plotting-maximum-minimum-values-graph.html | 1,526,945,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00224.warc.gz | 184,134,095 | 10,377 | # Thread: Help plotting maximum and minimum values on graph
1. ## Help plotting maximum and minimum values on graph
Hello, I am going back to an unfinished homework assignment and I realized that I skipped this problem because my professor never covered it in class. I could use your help with the following problem:
Sketch the graph of the function f that is continuous on [1,5] and has:
• absolute minimum at 2
• absolute maximum at 3
• local minimum at 4
I understand that the [1,5] means that I will be plotting points between 1 and 5 on the x-axis, and that the maxima/minima are basically x values, but how do I figure out what their y values are?
Thank you!
Danni
2. ## Re: Help plotting maximum and minimum values on graph
There is more than one such function, so your question should probably read "... the graph of a function ...".
You are correct that [1,5] means all x, $\displaystyle 1\le{x}\le5$. I would draw the graph by first drawing the minimum and maximum at x=2 and 3, then the minimum (with y value between the two others) at x=4, then connecting them and extending to the whole interval (staying between the minimum and maximum at x=2 and 3).
I think you have to just pick convenient y-values - there's not a single correct answer (unless there's some additional information you didn't put in your post).
- Hollywood
3. ## Re: Help plotting maximum and minimum values on graph
Hi dannibambi!
You can pick any y values you like, as long as they fit your criteria.
Your graph would for instance look something like this graph. | 366 | 1,559 | {"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} | 3.53125 | 4 | CC-MAIN-2018-22 | latest | en | 0.93315 |
https://warreninstitute.org/label-the-the-tissues-and-structures-on-the-histology-slide/ | 1,713,492,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00675.warc.gz | 568,618,279 | 22,659 | # Identify histology slide tissues and structures with labels.
Welcome to Warren Institute's latest blog post! Today, we'll dive into the intricate world of histology as we explore the art of labeling the tissues and structures on a histology slide. Understanding the microscopic intricacies of biological specimens is crucial for students and professionals in the field of Mathematics education. We'll unravel the subtle details that define each tissue type and learn how to precisely identify and label them. Join us as we embark on this fascinating journey through the hidden world of histology!
## Understanding the Histology Slide
Identifying Different Tissues and Structures on a histology slide is essential for understanding the microscopic anatomy of organisms. This section will guide you through the process of labeling and interpreting the various components present in the slide.
## Key Tissue and Structure Labels
Each tissue and structure on the histology slide plays a crucial role in the overall function of the organism. Understanding their labels will provide insight into their significance and contribution to the organism's physiology.
## Connecting Histology to Mathematics
While histology may seem unrelated to mathematics education, there are interdisciplinary connections between the two fields. Understanding the structures and patterns within tissues can serve as a practical application of mathematical concepts such as geometry and spatial reasoning.
## Enhancing Learning with Interactive Labeling Activities
To engage students in the histology learning process, interactive labeling activities can be employed. These activities not only reinforce the identification of tissues and structures but also promote critical thinking and analytical skills.
### How can labeling the tissues and structures on a histology slide be incorporated into a mathematics education lesson?
Labeling the tissues and structures on a histology slide can be incorporated into a mathematics education lesson by using the activity to teach students about coordinate systems and measuring distances between different points on the slide. This can help students develop their spatial reasoning skills and understanding of graphing and mapping in a real-world context.
### What are some effective strategies for students to accurately label the tissues and structures on a histology slide using mathematical concepts?
Using grid systems and coordinate planes can help students accurately label tissues and structures on a histology slide by applying mathematical concepts such as Cartesian coordinates and measurements.
### What mathematical skills can be developed through the process of labeling the tissues and structures on a histology slide in a mathematics education setting?
The process of labeling the tissues and structures on a histology slide can develop spatial reasoning, pattern recognition, and measurement skills in a mathematics education setting.
### Are there any research studies demonstrating the benefits of integrating histology slide labeling into mathematics education curriculum?
There are no research studies demonstrating the benefits of integrating histology slide labeling into mathematics education curriculum.
### How can technology be utilized to enhance the learning experience when labeling the tissues and structures on a histology slide within a mathematics education context?
Technology can be utilized to enhance the learning experience in Mathematics education by using interactive software and apps that allow students to label tissues and structures on a histology slide.
In conclusion, labeling the tissues and structures on the histology slide is a crucial aspect of Mathematics education, as it provides students with a hands-on understanding of the mathematical concepts underlying biological processes. Incorporating histology slides into the curriculum can enhance students' comprehension and appreciation of the quantitative aspects of biology, fostering a more holistic approach to STEM education. By engaging with real-world examples and applying mathematical principles to the analysis of biological specimens, students can develop a deeper understanding of both mathematics and the natural world. | 692 | 4,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.9084 |
http://www.learningresources.com/category/intervention/intervention/math/counting+-+cardinality.do | 1,397,699,222,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00622-ip-10-147-4-33.ec2.internal.warc.gz | 529,415,357 | 14,207 | Counting & Cardinality - Learning Resources®
Untitled Document #container2 { position: relative; float:left; width:960px; font-family:Verdana, Geneva, sans-serif; color: black; height: 500px; } #mainimg { position: relative; float: left; margin-left: 5px; margin-top: 5px; } #row1 { position: relative; float:left; width:950px; margin-top: 10px; font-size: 12px; margin-left: 10px; } #topimage1 { position: relative; float: left; margin-left: 00px; margin-top: 20px; line-height: 25px; } #topimage2 { position: relative; float: left; margin-top: 20px; margin-left: 20px; line-height: 25px; } #column1 { position: relative; float: left; margin-left: 35px; marign-top: 40px; width: 400px; height: 200px; margin-top: 20px; } #column2 { position: relative; float: right; marign-top: 40px; width: 400px; height: 200px; margin-top: 20px; margin-right: 45px; } #link1 { position: relative; float:left; height: 35px; color: #003; clear: left; width: 300px; } #link2 { position: relative; float:left; height: 30px; color: #003; margin-left: 20px; clear: left; } In order to succeed in math, students must thoroughly understand the meaning of numbers. Our multi-sensory manipulatives, games and floor mats are Common Core (CCSS) aligned and engage struggling kindergarteners as they learn to identify numbers, count in sequence, count sets and compare numbers. Tier 1 Tier 2 Tier 3 | 417 | 1,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | latest | en | 0.510164 |
https://www.extendoffice.com/excel/formulas/excel-lookup-the-first-partial-match-number.html | 1,718,381,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00433.warc.gz | 704,290,984 | 40,733 | ## Lookup the first partial match number
There are cases that you need to get the position of the first partial match that contains specific number in a range of numeric values in Excel. In this case, a MATCH and TEXT formula that incorporates asterisk (*), the wildcard that matches any number of characters, will do you a favor. And if you also need to know the exact value at that position, you can add the INDEX function to the formula.
#### Get the position of the first partial match number
To get the position of the first partial match number containing “345” from the number range as shown above, a MATCH and TEXT formula together with wildcards will help you in this way: In order to match any numbers that contain 345, you will wrap the number 345 between two asterisks (*). However, the operation will convert the numeric value to a text value. So, you will have to use the TEXT function to convert numbers in the number range to text. Only in this case, the MATCH function will be able to find the position of the partial match properly.
#### Generic syntax
=MATCH("*"&number&"*",TEXT(lookup_array,"0"),0)
√ Note: This is an array formula that requires you to enter with Ctrl + Shift + Enter.
• number: The number you specified to lookup its first partial match.
• lookup_array: The range of numeric values to retrieve the position of the first partial match from.
To get the position of the first match number that contains “345”, please copy or enter the formulas below in the cell E6, and press Ctrl + Shift + Enter to get the result:
=MATCH("*"&345&"*",TEXT(B5:B16,"0"),0)
Or, use a cell reference to make the formula dynamic:
=MATCH("*"&E5&"*",TEXT(B5:B16,"0"),0)
√ Note: To stick a cell reference/number and text together, you must add an ampersand (&) in between them. And the text should be enclosed in double quotes.
#### Explanation of the formula
=INDEX("*"&345&"*",TEXT(B5:B16,"0"),0)
• TEXT(B5:B16,"0"): The TEXT function converts all the numeric values in B5:B16 to text with the format code “0”. So, we will get a text array like this: {"56445";"21354";"84265";"54342";"34545";"45632";"87954";"68546";"34567";"75681";"33587";"16467"}.
• INDEX("*"&345&"*",TEXT(B5:B16,"0"),0) = INDEX("*"&345&"*",{"56445";"21354";"84265";"54342";"34545";"45632";"87954";"68546";"34567";"75681";"33587";"16467"},0): The lookup value "*"&345&"*" can match any text strings that contain the string “345”, no matter what position 345 is in the text strings. The match_type 0 asks the MATCH function to find the position of the first exact lookup value in the array. So, the MATCH will return 5.
#### Retrieve the first partial match number
To retrieve the first partial match number based on the position provided by MATCH as shown below, we can put a spotlight on the INDEX function.
#### Generic syntax
=INDEX(MATCH(return_range,"*"&number&"*",TEXT(lookup_array,"0"),0))
√ Note: This is an array formula that requires you to enter with Ctrl + Shift + Enter.
• return_range: The range where you want the combination formula to return the first partial match from.
• number: The number you specified to lookup its first partial match.
• lookup_array: The range of numeric values to retrieve the first partial match from.
To get the first match number that contains “345”, please copy or enter the formulas below in the cell E7, and press Ctrl + Shift + Enter to get the result:
=INDEX(MATCH(B5:B16,"*"&345&"*",TEXT(B5:B16,"0"),0))
Or, use a cell reference to make the formula dynamic:
=INDEX(MATCH(B5:B16,"*"&E5&"*",TEXT(B5:B16,"0"),0))
#### Explanation of the formula
=INDEX(B5:B16,MATCH("*"&E5&"*",TEXT(B5:B16,"0"),0))
• MATCH("*"&E5&"*",TEXT(B5:B16,"0"),0) = 5:
• INDEX(B5:B16,MATCH("*"&E5&"*",TEXT(B5:B16,"0"),0)) = INDEX(B5:B16,5): The INDEX function returns the 5th value in the return range B5:B16, which is 34545.
#### Related functions
Excel MATCH function
The Excel MATCH function searches for a specific value in a range of cells, and returns the relative position of the value.
Excel TEXT function
The TEXT function converts a value to text with a specified format in Excel.
Excel INDEX function
The Excel INDEX function returns the displayed value based on a given position from a range or an array.
#### Related Formulas
Locate first partial match with wildcards
To get the position of the first partial match that contains specific text string in a range in Excel, you can use a MATCH formula with wildcard characters - the asterisk (*) and question mark (?).
Lookup closest match
To look for the closest match of a lookup value in a numeric dataset in Excel, you can use the INDEX, MATCH, ABS and MIN functions together.
Lookup closest match value with multiple criteria
In some cases, you may need to lookup the closest or approximate match value based on more than one criteria. With the combination of INDEX, MATCH and IF functions, you can quickly get it done in Excel.
Two-way approximate match with multiple criteria
In this tutorial, we will talk about how to look for an approximate match based on multiple criteria listed both in columns and rows in an Excel spreadsheet, with the help of INDEX, MATCH, and IF functions.
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• Brings Efficient Tabs to Office (include Excel), Just Like Chrome, Edge and Firefox. | 1,790 | 7,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.783121 |
https://www.pcreview.co.uk/threads/three-condition-if-statements.3929841/ | 1,656,175,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00411.warc.gz | 998,079,255 | 14,331 | # Three condition "if" statements
T
#### Tel
I'm trying to write an IF statement that needs to look at the input of three
cells (D4, D5 and D6).
If any of them are "YES" then I need cell E4 to = cell G1. However,
If they are all "no" then I need cell E4 to = cell G2
Finally, if they are all blank then I need cell e4 to be blank
Also, if I do this can I conditionally format cell E4 so if it = G1 then it
is White text on a red background, and if it = G2 then it is Black text on a
green background (I'm okay with CF just need to know if it's possible)
TVM
Terry
M
#### Ms-Exl-Learner
Try this in E4 Cell
=IF(OR(D4="YES",D5="YES",D6="YES"),G1,IF(AND(D4="NO",D5="NO",D6="NO"),G2,IF(AND(D4="",D5="",D6=""),"","Value is not Matching the Criteria")))
If this post helps, Click Yes!
T
#### Tel
Many thanks for your reply, I've astounded myself by being able to work it
out, as follows:
=IF(AND(D4=G5,D5=G5,D6=G5),G11,IF(OR(D4=G4,D5=G4,D6=G4),G2,""))
The CF was simple too, essentially Condition 1 Cell Value is - equal to -
=G1 (Set format)
Condition 2 Cell Value is - equal to - -G2 (Set format)
Probably basic to others but someone might find it useful.
Terry | 381 | 1,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.909238 |
https://documen.tv/a-swing-set-is-going-to-be-placed-over-a-region-of-mulch-that-is-shaped-like-a-trapezoid-the-bas-27704819-20/ | 1,679,775,264,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00563.warc.gz | 246,655,752 | 15,893 | Question
A swing set is going to be placed over a region of mulch that is shaped like a trapezoid. The bases of the trapezoid have a length of 12 and 15 feet, and the perpendicular distance between the bases is 8.5 feet. What is the area of the region under the swing set? If necessary, round your answer to the nearest tenth.
The area is
square feet.
114.8 ft²
Step-by-step explanation:
Understanding the given values :
• bases (a and b) = 12 feet and 15 feet
• perpendicular distance (height, h) = 8.5 feet
Inputting into the formula for the area of a trapezoid :
• A = 1/2 x (a + b) x h
• A = 1/2 x (12 + 15) x 8.5
• A = 4.25 x 27
• A = 114.8 ft² (nearest tenth) | 212 | 668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.884329 |
https://academickids.com/encyclopedia/index.php/Spectrum_of_an_operator | 1,716,446,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00030.warc.gz | 58,769,480 | 7,327 | # Spectrum of an operator
In functional analysis, the concept of the spectrum of an operator is a generalisation of the concept of eigenvalues, which is much more useful in the case of operators on infinite-dimensional spaces. For example, the bilateral shift operator on the Hilbert space [itex]\ell^2(\mathbf Z)[itex] has no eigenvalues at all; but we shall see below that any bounded linear operator on a complex Banach space must have non-empty spectrum.
The study of the properties of spectra is known as spectral theory.
Contents
## Definition
Let X be a complex Banach space, and B(X) the Banach algebra of bounded linear operators on X. Then if I denotes the identity operator, and TB(X) then the spectrum of T (normally written as σ(T) ) consists of λ such that λ I - T is not invertible in the algebra of bounded linear operators on X. Note that by the closed graph theorem, this condition is equivalent to asserting λ I - T fails to be bijective.
## Basic properties
Theorem: The spectrum is non-empty, bounded, and closed.
Proof: Suppose the spectrum is empty; then the function R(λ) = (λI - T)-1 is defined everywhere on the complex plane. So if Φ is any linear functional on B(X), F(λ) = Φ(R(λ)) is a continuous function C[itex]\to[itex]C. It is not hard to see that
[itex]\lim_{\mu \to \lambda} \frac{F(\lambda) - F(\mu)}{\lambda - \mu} = -\Phi( R(\lambda)^2 )[itex]
so F is an analytic function. However, F(λ) is O-1) for large λ so F is a bounded analytic function, and hence constant by Liouville's theorem, and thus everywhere zero as it is zero at infinity. However, by the Hahn-Banach theorem this implies that R(λ) is zero for all λ, which is obviously a contradiction.
The boundedness of the spectrum is immediate from the Neumann series expansion (named after the German mathematician Carl Neumann),
[itex](I - A)^{-1} = \sum_{n = 0}^\infty A^n[itex],
which is valid for any AB(X) with ||A|| < 1. This implies that if |λ| > ||T||, (λ I - T) is invertible (taking A = T/λ). So σ(T) is bounded, and the spectral radius
[itex]r(T) = \sup \{|\lambda| : \lambda \in \sigma(T)\}[itex]
is bounded above by ||T||.
Furthermore, the Neumann series implies that for any two operators A, B with A invertible and ||A - B|| < ||A-1||-1, B must also be invertible. It follows that the set of invertible operators is open, and hence, since the function CB(X) defined by λ → λ I - T is continuous, the set of λ for which λ I - T is invertible is open, so its complement is closed; but this complement is exactly σ(T).
## Classification of points in the spectrum
Loosely speaking, there are a variety of ways in which an operator S can fail to be invertible, and this allows us to classify the points of the spectrum into various types.
### Point spectrum
If an operator is not injective (so there is some nonzero x with S(x) = 0), then it is clearly not invertible. So if λ is an eigenvalue of T, we necessarily have λ ∈ σ(T). The set of eigenvalues of T is sometimes called the point spectrum of T.
### Approximate point spectrum
More generally, S is not invertible if it is not bounded below; that is, if there is no 'c' > 0 such that ||Sx|| > c||x|| for all xX. So the spectrum includes the set of approximate eigenvalues, which are those λ such that T - λ I is not bounded below; equivalently, it is the set of λ for which there is a sequence of unit vectors x1, x2, ... for which
[itex]\lim_{n \to \infty} \|Tx_n - \lambda x_n\| = 0[itex].
The set of approximate eigenvalues is known as the approximate point spectrum.
For example, in the example in the first paragraph of the bilateral shift on [itex]\ell^2(\mathbf{Z})[itex], there are no eigenvectors, but every λ with |λ| = 1 is an approximate eigenvector; letting xn be the vector
[itex]\frac{1}\sqrt{n}(\dots, 0, 1, \lambda, \lambda^2, \dots, \lambda^{n-1}, 0, \dots)[itex]
then ||xn|| = 1 for all n, but
[itex]Tx_n - \lambda x_n = \frac{2}\sqrt{n} \to 0[itex].
### Compression spectrum
The unilateral shift on [itex]\ell^2(\mathbf{N})[itex] gives an example of yet another way in which an operator can fail to be invertible; this shift operator is bounded below (by 1; it is obviously norm-preserving) but it is not invertible as it is not surjective. The set of λ for which λ I - T is not surjective is known as the compression spectrum of T.
This exhausts the possibilities, since if T is surjective and bounded below, T is invertible.
## Further results
The spectral radius formula states that
[itex]r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}[itex].
This can be proved using similar methods to the above theorem, considering the power series expansion of F(1/λ); this must converge for all λ > r(T), and applying the uniform boundedness principle to the series coefficients gives the result.
If T is a compact operator, then it can be shown that any nonzero approximate eigenvalue is in fact an eigenvalue.
If X is a Hilbert space and T is a normal operator, then a remarkable result known as the spectral theorem gives an analogue of the diagonalisation theorem for normal finite-dimensional operators (Hermitian matrices, for example).
An account of the spectral theorem (http://www.srcf.ucam.org/~dl267/writeups/spectral_measures.pdf)de:Spektrum (Operatortheorie)
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• Solar System (http://www.academickids.com/encyclopedia/index.php/Solar_System) | 1,670 | 6,286 | {"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} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.914914 |
http://www.nugorise.top/how-to-make-a-triangular-pyramid-out-of-paper/ | 1,656,417,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00037.warc.gz | 102,365,752 | 11,326 | # how to make a triangular pyramid out of paper
Trim rough or loose pieces of paper from the pyramid with scissors. Try to mount the pyramid with a triangular base before putting on glue so that you are clear of where each tab will go. Apply glue to the bottom edges of your pyramid and attach the triangle piece to form the shapes base.
• ### How do you make a pyramid pattern with 4 triangles?
• A proper pyramid template pattern will have a square base, and of each side of that base, there will be an attached triangle. Two or all four of these triangles will have tabs on them. Once cut out, the four triangles will come together and join at the top to form the pyramid faces. Cut out your pyramid pattern.
• ### How to make an origami pyramid?
• Making an Origami Pyramid 1. Find a square piece of paper. To make a pyramid, you need to start with a paper that has equal dimensions in length… 2. Fold and unfold the paper. First, fold and unfold it diagonally through the center from top right to bottom left, and… 3. Lay the paper flat on a …
• ### What is the shape of a triangular pyramid?
• A triangular pyramid is a 3D shape in which all the faces are triangles. It is a pyramid with a triangular base connected by four triangular faces where 3 faces meet at one vertex. If it is a right triangular pyramid, the base is a right-angled triangle while the other faces are isosceles triangles.
• ### How do you make a simple triangle with paper?
• You want the paper in front of you so that the triangle labeled side D and A has its bottom edge facing you. Fold the paper into a smaller triangle. Start by folding the left side of the triangle in half in on itself, so the outer edges of sides C and D meet. Repeat on the other side, so that the outer edges of sides A and B meet. | 402 | 1,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | latest | en | 0.906625 |
https://programmer.group/1001-a-b-format-1002-a-b-for-polynomials.html | 1,582,879,020,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147116.85/warc/CC-MAIN-20200228073640-20200228103640-00407.warc.gz | 515,195,562 | 6,344 | Keywords: less
# 1001 A+B Format
Calculate a+b and output the sum in standard format – that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where −10
6
≤a,b≤10
6
. The numbers are separated by a space.
Output Specification:
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input:
-1000000 9
Sample Output:
-999,991
The first time I wrote 18 points, one of them was my negligence. This question is mainly to record my understanding of to [string, = = I didn't have any understanding. I found it later when I searched for the error. It felt very useful. When I write, I think it's better to turn it into a string, and then I can simulate it again (there's an oversight during the simulation). My simulation is different from most of the online ones. This simulation idea is accumulated by doing problems these days, but it's nothing
Here's a 20 point code
1. Sscanf and ssprintf are all contacted before character conversion, so the first thing I think about when I do this is this. It's very comfortable to master this. For example, the only trouble I think is that the parameter must be char array,
```#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
#include<map>
const int N = 205;
const int INF = 1000000000;
using namespace std;
int main(){
int a,b,c;
cin>>a>>b;
c = a + b;
if(c<0){
c = -c;
cout<<'-';
}
if(c < 1000) cout<<c;
else{
char s[9];
sprintf(s,"%d",c);
int flag = 0;
string ans;
for(int i=strlen(s)-1;i>=0;i-=3){
if(i !=strlen(s) - 1) ans+=',';
for(int j=0;j<3;j++){
if(i-j>=0) ans+=s[i-j];
else{
flag = 1;
break;
}
}
if(flag == 1) break;
}
reverse(ans.begin(),ans.end());
cout<<ans;
}
return 0;
}
```
The second type is to_string, the converted String type. In fact, when comparing later, I want to use stringstream. First, I don't know. Second, I haven't reached that point. At that time, I have to use this method to master it. I believe there will be some later, and I don't know whether to_string can be used in the examination room. It's very convenient. The idea is the same as my first one
```#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
const int maxn = 205;
const int INF = 0x3f3f3f3f;
using namespace std;
int main(){
int a,b,c;
cin>>a>>b;
c = a+b;
if(c<0) {
c = -c;
cout<<'-';
}
if(c<1000) cout<<c;
else{
string ans = to_string(c);
string s;
int flag = 0;
for(int i=ans.length() -1;i>=0;i-=3){
if(i!=ans.length()-1) s+=',';
for(int j=0;j<3;j++){
if(i-j>=0) s+=ans[i-j];
else {
flag = 1;
break;
}
}
if(flag ==1) break;
}
reverse(s.begin(),s.end());
cout<<s;
}
return 0;
}
```
This question simply simulates and grasps to string. Later, I also looked at the ideas of other gods, and I was too lazy to record them. This kind of question is based on my first feeling
# 1002 A+B for Polynomials
This time, you are supposed to find A+B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K N
1
a
N
1
N
2
a
N
2
... N
K
a
N
K
where K is the number of nonzero terms in the polynomial, N
i
and a
N
i
(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N
K
<⋯<N
2
<N
1
≤1000.
Output Specification:
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
Sample Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output:
3 2 1.5 1 2.9 0 3.2
For the first time, 17 points. How to say, the problem is not comprehensive. If the coefficient is zero when adding polynomials, it should be removed at this time,
I use map,
In fact, this problem is not made independently, because I forgot how to write the reverse iterator of map,
I have learned how to write on the Internet
First kind
```tree
1020
1053
1043
1086
1102
1079
1090
1094
1106
1004
1064
1099
//Union checking set
1107 Not familiar (check collection)
1118 It's hard to do1107 Same, but I don't know why not
1034
1114
//heap
1147
1155
//chart
1076
1034
1013
1021
1018
1030
1072
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
const int maxn = 205;
const int INF = 0x3f3f3f3f;
using namespace std;
bool vis[maxn]={false};
int main(){
map<int,double,greater<int> > m1;
map<int,double>::iterator it;
int n;
cin>>n;
for(int i=1;i<=n;i++){
int c;
double e;
cin>>c>>e;
m1[c] = e;
}
cin>>n;
for(int i=1;i<=n;i++){
int c;
double e;
cin>>c>>e;
if(m1[c] == 0) m1[c] = e;
else {
m1[c]+=e;
if(m1[c] == 0 ){
it = m1.find(c);
m1.erase(it);
}
}
}
cout<<m1.size();
for(it = m1.begin();it!=m1.end();it++){
printf(" %d %.1f",it->first,it->second);
}
return 0;
}
```
Second kinds
```#include<iostream>
#include<cstring>
#include<vector>
#include<map>
const int maxn = 205;
const int INF = 0x3f3f3f3f;
using namespace std;
bool vis[maxn]={false};
int main(){
map<int,double> m1;
map<int,double>::reverse_iterator rit;
map<int,double>::iterator it;
int n;
cin>>n;
for(int i=1;i<=n;i++){
int c;
double e;
cin>>c>>e;
m1[c] = e;
}
cin>>n;
for(int i=1;i<=n;i++){
int c;
double e;
cin>>c>>e;
if(m1[c] == 0) m1[c] = e;
else {
m1[c]+=e;
if(m1[c] == 0 ){
it = m1.find(c);
m1.erase(it);
}
}
}
cout<<m1.size();
for(rit = m1.rbegin();rit!=m1.rend();rit++){
printf(" %d %.1f",rit->first,rit->second);
}
return 0;
}
```
I feel that I haven't learned a lot about containers, let alone the usage of related containers. I can only slowly find out the shortcomings when I do the questions again. 1003 1004 was done when I learned trees and graphs in order, so I won't write it out alone,
143 original articles published, praised 3, 9228 visitors
Posted by dacio on Mon, 27 Jan 2020 06:43:40 -0800 | 1,922 | 6,120 | {"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} | 2.546875 | 3 | CC-MAIN-2020-10 | latest | en | 0.85402 |
https://community.qlik.com/t5/QlikView-App-Dev/Set-Analysis-To-exclude-an-item/td-p/444439 | 1,716,791,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00598.warc.gz | 147,935,769 | 47,069 | Announcements
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Anonymous
Not applicable
## Set Analysis To exclude an item
Hello! What's wrong with my formula ...I need to sum all the Total Value of the field Item NO but to exlcude item '5126'...but seems the <> doesn't work...
= sum({\$<ItemNO<>{'5126'}>}[Total Value])
I really appreciate if you could help me with this problem...I'm still a newbie in Qlikview.
Tks.
1 Solution
Accepted Solutions
Partner - Specialist III
Hi
Try this
= sum({\$<ItemNO ={"*"}-{5126}>}[Total Value])
Let me know if it works.
Best Regards,
Gabriel
29 Replies
Not applicable
Author
Using =- instead of <> should do the trick
Not applicable
Author
Instead of <> try replacing it with -=
= sum({\$<ItemNO-={'5126'}>}[Total Value])
Not applicable
Author
Hi,
you can't use <> inside set analysis^^
try this
sum({\$<ItemNO -= {5126}>}[Total Value])
OR
sum({\$<ItemNO = {*}-{5126}>}[Total Value])
those two will trigger red underline but it will still work
Regards,
Alex
Anonymous
Not applicable
Author
= sum({\$<ItemNO-={'5126'}>}[Total Value])
Creator III
Try this one..
= sum({\$<ItemNO-={'5126'}>}[Total Value])
or
= sum({\$<ItemNO-={5126}>}[Total Value])
in expression window, It will show underlined with red color but you will get the right result.
Let me know if you a have any concerns/issues
Thanks,
Sibin Jacob
Not applicable
Author
Hi Cacostao,
Try This expression
= sum({1-\$<ItemNO={'5126'}>}[Total Value])
if you have any doubt please find the below attachment
Regards,
Nirmal raj.
Message was edited by: nirmal.j
Partner - Creator
hi
im japanese!! thats why i can't explain well in english...sorry
but you can try this expression!!
--------------------------------------------------------------------
= sum({\$<ItemNO={'*'}-{'5127'}>}[Total Value])
--------------------------------------------------------------------
'*' is all of ItemNO!!
the expression means all of ItemNO minus '5127'!!
Not applicable
Author
Hi Cacostao,
Try This expression
= sum({1-\$<ItemNO={'5126'}>}[Total Value])
if you have any doubt please find the below attachment
Regards,
Nirmal raj.
Creator II
Hello,
Тry use -= instead of <>
= sum({\$<ItemNO-={'5126'}>}[Total Value])
Community Browser | 603 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-22 | latest | en | 0.796164 |
https://educegh.com/definition-how-to-find-equivalent-fractions-with-exam/ | 1,721,876,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00498.warc.gz | 190,932,912 | 14,062 | # Definition, how to find equivalent fractions? Steps, examples
By | September 18, 2022
Equivalent fractions refer to fractions that look different but have the same value. When two or more fractions have the same value but look different are called equivalent fractions. Let’s look at this example below
\text{Example}\space\frac{2}{3} = \frac{4}{6}\\
\text{In this example\space} \frac{4}{6} \text {\space is an equivalent fraction of \space } \frac{2}{3}
But how can you do this? This is simply done by following the Steps below. Let’s look at how to find equivalent fractions of a given fraction.
## How to find equivalent fractions
To find the equivalent fractions of a given fraction, there are two main ways.
• Multiplying
• Dividing (Cancellation)
Steps in finding Equivalent fractions using multiplication
In this method, Equivalent fractions are formed by Multiplying the numerator (top) number and the denominator (bottom) number by the same number. Let’s look at an example
\text{Example 1: find the three Equivalent fraction of \space} \frac{3}{4}\\
\space Ans: \text{To do this Multiply the top number and the down number by 2}\\\space
\space \frac{2×3}{2×4}= \frac{6}{8}\\\text{\space} \\
\text{Multiply}\space \frac{3}{4} \space\text{again by } 3=\space\frac{3×3}{3×4}=\frac{9}{12}\\
\text{Then multiply} \space \frac{3}{4} \space\text{by}\space4=\frac{4×3}{4×4}=\frac{12}{16}
\text{Hence, the equivalent fraction of } \space \frac{3}{4}=\frac{6}{8}=\frac{9}{12}.
Therefore, to find the equivalent fraction of a given fraction, multiply the numerator (top) number and the denominators (bottom) by 2,3,4,5… In turns.
This method is pretty cool, right?
Let’s look at the second method
## Finding equivalent fractions Using the cancellation method
To find an equivalent fraction using the cancellation method, divide the top and the bottom number by the same number and this will also give you the equivalent fraction of a given number.
\text{Example1; find the equivalent fraction of }\frac{18}{30}\\
\text{ Answer: divide the top and bottom of the Fraction by}\space2\space=\frac{18\div2}{30\div2}=\frac{9}{15}
\text{Divide}\space\frac{9}{15}\space\text{by}\space3 = \frac{9÷3}{15÷3}=\frac{3}{5}\\
\text{Hence, the Equivalent of fractions of }\space\frac{18}{30}=\frac{9}{15}=\frac{3}{5}
READ: How to find the area of a rectangle with examples | 673 | 2,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.783054 |
http://mathhomeworkanswers.org/7127/how-many-2-5-are-in-1 | 1,398,411,239,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00027-ip-10-147-4-33.ec2.internal.warc.gz | 211,912,539 | 19,634 | # how many 2/5 are in 1
Studying reciprocals, What is the correct answer when asked how many 2/5 are in 1? Is the answer 5/2 or 1 1/2?
Also, when asked how many 2/5 are in 2, is the answer 10/2 or 5?
asked Nov 4, 2011
1/(2/5) = 5/2
answered Nov 4, 2011 by Level 6 User (23,120 points)
if a,b,c,d,e are 5 natural prime number and a
answered Jun 18, 2013 by anonymous
i dont know the answer to it because im asking yall whats the answer to it so how should i know
answered Oct 24, 2013 by nunya
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130 views | 652 | 1,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2014-15 | longest | en | 0.897936 |
http://edbodmer.com/corporate-finance/edbodmer-wikispaces-com-advancedvaluationandcorporatefinancemodeling/understanding-multiples/ | 1,585,885,649,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00359.warc.gz | 54,173,363 | 21,048 | # Understanding P/E and EV/EBITDA Multiples
P/E ratios and EV/EBITDA ratios are used in valuation in all sorts of contexts and often discussed by experts on television. The question of why a multiple for an industry or a company in an industry should be high or low and whether the overall level of the multiple is high or low is not addressed in a rigorous manner in either finance texts or by those horrible people who shout so loudly on the TV. The exercises in this section demonstrate that both the P/E ratio and EV/EBITDA ratio are affected by future returns, growth and the cost of capital which are the standard drivers of value. The initial videos describe how P/E ratios can be computed from the value drivers with changes in growth and cost of capital. The EV/EBITDA analysis shows how the EV/EBITDA ratio is different from the P/E ratio because the EV/EBITDA ratio is affected by tax, capital expenditure and depreciation relative to EBITDA and working capital adjustments. A problem I have with all of this is that the true economic drivers of changes in return are not discussed, meaning that you cannot just predict returns but you should carefully think about how price can move to short-run marginal cost; how margins are driven by supply and demand; how technological obsolescence can ruin a business; how some industries are more exposed to surplus capacity than others; how monopoly or oligopoly profits can disappear; how valuable are brands and the ability of companies like Apple, Starbucks, Coke, Disney and McDonalds can continue to earn returns through making the unsuspecting public become strongly addicted to their products.
## Files Used in Valuation Concepts Lesson Set 2: Computation of P/E, EV/EBITDA and P/B from Economic Drivers
PE and EV to EBITDA Multiples – Linear.xlsm
Proof of Timing in DCF Model.xlsx
PE and EV to EBITDA Multiples -Growth.xlsm
Corporate and Project Model.xlsm
Variable WACC in DCF Model.xlsm
## EV/EBITDA and Required Income to Pay for Your House versus Your Masaritti
EV/EBITDA will be lower for companies that need to replace assets more quickly. If the net investment is the same and the EBITDA is the same, the company with shorter lived assets will have to replace assets more quickly and have more capital expenditures for the same EBITDA. This means EV/EBITDA will be lower for companies with higher depreciation rates.This means you cannot theoretically compare the EV/EBTIDA across different industries with different types of assets (and also within the industry) | 544 | 2,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.929 |
https://hollows.info/what-is-the-difference-between-fourier-transform-and-fourier-series/ | 1,709,473,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00823.warc.gz | 299,351,963 | 10,118 | ## What is the difference between Fourier transform and Fourier series?
The Fourier series is used to represent a periodic function by a discrete sum of complex exponentials, while the Fourier transform is then used to represent a general, nonperiodic function by a continuous superposition or integral of complex exponentials.
What is the difference between Fourier transform and Fast Fourier transform?
Main Differences Between FFT and DFT FFT is a much efficient and fast version of Fourier transform whereas DFT is a discrete version of Fourier transform. FFT is an implementation of DFT whereas DFT establishes a relationship between the time domain and the frequency domain representation.
What is the Fourier transform used for?
The Fourier Transform is an important image processing tool which is used to decompose an image into its sine and cosine components. The output of the transformation represents the image in the Fourier or frequency domain, while the input image is the spatial domain equivalent.
### What is Fourier series and Fourier transform in signal and system?
The main drawback of Fourier series is, it is only applicable to periodic signals. To overcome this shortcoming, Fourier developed a mathematical model to transform signals between time (or spatial) domain to frequency domain & vice versa, which is called ‘Fourier transform’.
What exactly is Fourier transform?
What is the Fourier transform? At a high level the Fourier transform is a mathematical function which transforms a signal from the time domain to the frequency domain. This is a very powerful transformation which gives us the ability to understand the frequencies inside a signal.
What are the applications of Fast Fourier Transform?
It covers FFTs, frequency domain filtering, and applications to video and audio signal processing. As fields like communications, speech and image processing, and related areas are rapidly developing, the FFT as one of the essential parts in digital signal processing has been widely used.
## What do you mean by Fourier transformation?
A Fourier transform is a mathematical technique for converting a time function into one expressed in terms of frequency. A Fourier transform is a circuit analysis technique that decomposes or separates a waveform or function into sinusoids of different frequency which sum to the original waveform.
What does Fourier series represent?
A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. It is analogous to a Taylor series, which represents functions as possibly infinite sums of monomial terms. A sawtooth wave represented by a successively larger sum of trigonometric terms.
Is the Fourier transform unique?
It is unique. If the function f(t) is piecewise continious and square integrable the fourier coffiecients are unique.
### What are the disadvantages of Fourier tranform?
The major disadvantage of the Fourier transformation is the inherent compromise that exists between frequency and time resolution. The length of Fourier transformation used can be critical in ensuring that subtle changes in frequency over time, which are very important in bat echolocation calls, are seen.
What are the different types of the Fourier transform?
aperiodic spectrum This is the most general form of continuous time Fourier transform.
• discrete aperiodic spectrum This is the Fourier series expansion of a periodic signal with time period .
• III.
• IV.
• Why there is a need of Fourier transform?
Fourier Transform is used in spectroscopy, to analyze peaks, and troughs. Also it can mimic diffraction patterns in images of periodic structures, to analyze structural parameters. Similar principles apply to other ‘transforms’ such as Laplace transforms, Hartley transforms.
## What is the philosophical meaning of Fourier series?
A Fourier series is a way to represent complex waves, such as sound, as a series of simple sine waves. The series breaks down a wave into a sum of sines and cosines. This means that elements of a wave can be isolated from each other. | 762 | 4,120 | {"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} | 3.640625 | 4 | CC-MAIN-2024-10 | latest | en | 0.913186 |
https://www.gnu.org/software/fisicalab/manual/en/fisicalab/Example-6-_0028cal_0029.html | 1,713,733,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00063.warc.gz | 714,967,936 | 2,842 | Next: , Previous: , Up: Examples calorimetry [Contents][Index]
### 15.6 Example 6
How much heat in calories must be added to transform a block of 3 kg of ice at -5 C, in water at + 22 C? (specific heat of ice = 0.50 cal/g*K, specific heat of water = 1 cal/g*K, and heat of fusion of water = 79.7 cal/g).
Solution with FisicaLab
Select the Thermodynamics group and, inside this, the Calorimetry and gases module. Erase the content of the chalkboard. And add one element Process, one element Applied heat (applied to the element Process), one element Block, one element Change of state solid-liquid and one element Liquid. As show the image below (the yellow arrows are only to indicate the direction of the process):
To the element Applied heat, with the conversion factor to kilocalories, we have:
Q
Q @ cal
The element Process, whose name in this issue is irrelevant, represents the change of the water from its solid state at -5 C until liquid state at 22 C. This element must contain the element Block, which represents the water in solid state, the element Change of state solid-liquid, which represents the change of state of the water, and the element Liquid, which represents the final state of the water. These elements gonna be called Solid, Fusion and Liquid, respectively. Then to the element Process we have:
Name
0
Object 1
Solid
Object 2
Fusion
Object 3
Liquid
Object 4
0
Object 5
0
Now to the element Block, which represents the water in solid state and that we will call Solid, because this name have in the element Process, we have:
Name
Solid
m
3
c
0.5 @ cal/g*K
Ti
-5 @ C
Tf
0 @ C
Now the element Change of state, called Fusion, is:
Name
Fusion
m
3
cf
79.7 @ cal/g
Sense
>
And to the element Liquid, which represents the final state of the water and called Liquid, we have:
Name
Liquid
m
3
c
1 @ cal/g*K
Ti
0 @ C
Tf
22 @ C
Now click in the icon Solve to get the answer:
```
Q = 312.600 kcal ;
Status = success.
```
Next: Example 7, Previous: Example 5, Up: Examples calorimetry [Contents][Index] | 548 | 2,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.858806 |
https://upjoke.com/calorie-jokes | 1,670,660,731,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00514.warc.gz | 631,925,111 | 17,691 | UPJOKE
This joke may contain profanity. 🤔
### How many calories are in eating pussy?
It depends which way she wipes.
This joke may contain profanity. 🤔
### How many calories does poop have?
This joke may contain profanity. 🤔
### They say having sex burns as many calories as running 8 miles…
I think that’s ridiculous, who can run 8 miles in 30 seconds
### Just burned 2,000 calories.
I promise that's the last time I leave brownies in oven while I nap.
### I just burned 1200 calories...
I left the pizza in the oven too long.
### I've been counting calories.
I'm trying to beat my high score.
This joke may contain profanity. 🤔
### What's the difference between a calorie and a dick?
Your mom can tell you how many calories she eats per day.
### I just burned 81,500 calories
Now to hide the remains and the flamethrower
### My New Years resolution was to eat 1200 calories a day. I’ve been doing so great!
I’ve surpassed my goal every day so far!
### Just burnt 2,000 calories...
That'll be the last time I bake a pizza while I'm asleep!
empty.
### How many calories are in a communion bread?
Nun!
This joke may contain profanity. 🤔
### They say that, during sex, you burn off as many calories as running right miles.
Who the hell runs eight miles in 30 seconds?
Edit: I appreciate all the real jokes in the comments.
As for the guy who invented autocorrect, well, there's a special place in she'll for him.
### 1400 calories a day?
Pfft, piece of cake.
### You burn 26 calories a minute kissing.
That's probably why I'm overweight.
### What do you call a vampire that checks the calorie content of these between meals?
Count Snackula.
This joke may contain profanity. 🤔
### My wife and I just discovered an easy way to burn 2000 calories an hour during sex
Note to self: Leave bedroom door open so we can hear the oven timer next time.
This joke may contain profanity. 🤔
### Run.
Scientists say that making love use's the same amount of calories as running a mile, I think that is bullshit, who the fuck can run a mile in fifteen seconds.?
This joke may contain profanity. 🤔
### It's said that you can burn up to a 150 calories during sex.
That's a very impressive amount to burn in 2 seconds.
### How do you burn a lot of calories quickly?
Set a fat kid on fire
This joke may contain profanity. 🤔
### A recent study says sex burns 3.6 calories a minute...
So that’s why I’m fat.
### What’s the best way to burn 1,000 calories?
Leave the pizza in the oven.
### I burnt 800 calories this morning
Forgot the pizza in the oven.
### I just burnt 2500 calories
This is the last time I take a nap while baking cookies
This joke may contain profanity. 🤔
### As a sex education teacher, I know that the semen in the average male ejaculation has about 20 calories.
But I tell my daughter that there are 350 calories in it.
This joke may contain profanity. 🤔
### You can burn up to 150 calories through one vigorous session of masturbation...
Still got me kicked out of my weight watchers meeting though.
### U.S. vending machines to begin displaying calorie information to encourage smarter snack choices.
Machines’ reflective glass surface not doing the trick.
### Does the body burn fewer calories in a day in amputees?
Yes. By an arm and a leg.
### Managed to lose 1000 calories in five seconds...
...by dropping my cheeseburger :(
### How to burn 3500 calories ( a pound of fat) in your sleep!
Sleep for 55 hours.
This joke may contain profanity. 🤔
### Lady: How many calories are there is Semen?
Doctor: Trust me.
If you swallow he won't give a fuck how fat you are.
### Calories are just like the Viet Cong
Even if you burn them, they come back
### I wonder how many calories women burn by...
... jumping to conclusions.
### My doctor told me to reduce my calories.
So I went home, raided the cupboards, and ate half of eight muffins.
The next time I saw him, he looked me up and down, and said, "Have you been reducing your calories?"
I said, "Yes. Just the other day I ate half of eight muffins."
"What! You haven't lis...
### I just burnt 400 calories.
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Weird flax but 0k
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After doing some extensive calculations, this year I burned roughly 5 calories.
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It also gets you removed from your local gym.
In kill-o-bites.
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..is that they have the figures to prove it
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### Doctor walks into a restaurant to try some new dishes
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### The Russian and American generals are talking about their troops..(Old Joke)
The Russian general says, "we feed our troops 1,500 calories a day." The American general says "that's nothing. We feed our troops 5,000 calories a day, at least." "Impossible!" says the Russian general. "No man can eat an entire sack of potatoes in 24 hours." | 2,778 | 11,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.949742 |
https://web2.0calc.com/questions/if-where-represents-a-base-8-digit-and-represents | 1,624,323,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00001.warc.gz | 542,979,637 | 5,544 | +0
# If , where represents a base-8 digit (and represents a four-digit number whose second digit is ), then what is the
0
117
1
If \(n=1d41_8\), where \(d\) represents a base-8 digit (and \(1d41_8\) represents a four-digit number whose second digit is \(d\)), then what is the sum of all possible values of \(n\) in base 10?
Feb 9, 2021
#1
+113739
+1
\(n= 1+4*8 + d*8^2 + 1*8^3 \) (base 10) where d is and integer and \(0\le d \le7\)
n= 1 + 32 +68d + 512
you can take it from there.
Feb 9, 2021 | 204 | 512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.527954 |
https://www.physicsforums.com/threads/heat-sinking-to-aluminum-block-and-dissipation.666110/ | 1,726,215,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00416.warc.gz | 886,408,925 | 19,054 | # Heat sinking to Aluminum block and dissipation
• fastline
In summary, the individual is seeking assistance in estimating heat sinking and dissipation in an Aluminum structure for the purpose of cooling a hot liquid. The structure has a power input of 1000w, a volume of 30 ci, and a surface area of 352in2. The ambient temperature should not exceed 110F and the structure temperature should not cause burns. The specific heat of aluminum will need to be referenced to determine the amount of heat energy contained in the liquid. The dissipation of heat is also a concern, with a determined U value of 36 (btu/ft2 F). The individual is unsure if the dT above ambient should be used in this calculation and is also seeking to roughly estimate the U value
fastline
I am curious if someone can help be estimate heat sinking and dissipation in an Aluminum structure? What I am trying to do is is cool a hot liquid by just allowing it to heat the structure it is in since there is a reasonable amount of mass. Though the Aluminum structure is not designed specifically to dissipate, this becomes just a function of dT and surface area mostly.
power in is about 1000w max, structure is estimated at about 30 ci, and 352in2 of area. There is not active fan but that can be added really needed. Ambient temp not to exceed 110F and structure temp just should not "burn you". Vague, I know...
I am just trying to figure out the energy required to heat the structure, and it's ability to dissipate the heat.
As a start, you'll need to look up tables of specific heat, to find the value for aluminum.
You can figure out how much heat energy is contained in the hot liquid by multiplying its temperature by its specific heat by its mass.
Thanks for that. From the calcs I have run, this structure of Al will not hold much energy as heat.
However, more importantly, the dissipation. I determined the U value as 36 (btu/ft2 F)but I cannot remember if I would use the dT above ambient? IE Ambient is 100F structure is 110F, surface area is 2.5ft2 so heat loss is 900btu?
Determining forced air U values seems much more complex so I am just trying to roughly estimate that.
Power = UValue * Area * ΔT
where ΔT is the temperature gradient between the same two points used to calculate the UValue.
I can definitely help you estimate the heat sinking and dissipation in your Aluminum structure. Heat sinking refers to the transfer of heat from one object to another, while dissipation is the process of releasing or dispersing heat into the surrounding environment. In your case, the hot liquid will transfer heat to the Aluminum structure, and the structure will then dissipate the heat into its surroundings.
To estimate the heat sinking and dissipation in your Aluminum structure, we need to consider several factors. First, the amount of power (1000w) being transferred to the structure will determine the rate at which it heats up. This heat transfer will also depend on the specific heat capacity of the liquid and the thermal conductivity of Aluminum.
Next, we need to consider the size and surface area of the structure. The 30 ci volume and 352in2 surface area will play a significant role in determining the rate of heat dissipation. A larger surface area will allow for more heat to be released into the environment, while a smaller surface area will trap more heat in the structure.
Additionally, the ambient temperature (not to exceed 110F) and the desired temperature of the structure (should not "burn you") will also affect the heat dissipation. The larger the temperature difference between the structure and its surroundings, the faster heat will dissipate.
It is also important to note that the addition of an active fan can greatly enhance the heat dissipation process. The fan will increase air flow over the surface of the structure, allowing for more efficient heat transfer and dissipation.
In conclusion, the energy required to heat the structure and its ability to dissipate heat will depend on the factors mentioned above. It would be helpful to have more specific information, such as the specific heat capacity of the liquid and the thermal conductivity of the Aluminum, in order to provide a more accurate estimate. But overall, the factors of power, volume, surface area, and temperature difference will all play a role in determining the heat sinking and dissipation in your Aluminum structure.
## 1. How does heat sinking to an aluminum block work?
Heat sinking to an aluminum block works by transferring heat from a heat-generating component to the aluminum block, which has a large surface area to dissipate the heat. The aluminum block then conducts the heat away from the component and into the surrounding environment.
## 2. What are the benefits of using an aluminum block for heat sinking?
Aluminum is a highly conductive material, meaning it can quickly transfer heat away from a component. It is also lightweight, inexpensive, and readily available, making it a popular choice for heat sinking.
## 3. How do I choose the right size and shape of an aluminum block for heat sinking?
The size and shape of the aluminum block will depend on the heat-generating component and the amount of heat it produces. Generally, a larger block with a greater surface area will be more effective at dissipating heat. It is also important to consider the airflow and ventilation around the block to ensure efficient heat dissipation.
## 4. Can I use other materials besides aluminum for heat sinking?
Yes, there are other materials that can be used for heat sinking, such as copper and graphite. However, aluminum is often the preferred choice due to its high thermal conductivity and cost-effectiveness.
## 5. How can I improve the heat sinking performance of an aluminum block?
To improve the heat sinking performance of an aluminum block, you can use thermal interface materials, such as thermal paste or pads, between the component and the block to improve heat transfer. Additionally, ensuring proper ventilation and airflow around the block can also help improve its performance.
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Perspective projection matrix incorrect (Read 431 times) 0 Members and 1 Guest are viewing this topic.
Troubleshoots
JGO Knight
Medals: 36
Exp: 7-9 months
Damn maths.
« Posted 2014-02-04 11:53:00 »
I spent most of yesterday debugging because I couldn't implement perspective projection. Anyway, I thought my issue was the fact that my matrix multiplication was incorrect, which I fixed (I then spent an hour this morning fixing orthographic projection). Now I'm having trouble getting perspective projection to work, which I've tried to debug but I'm not sure how to fix it.
Orthographic projection:
1 2 3 4 `projection.overwrite(new float[] { 2f / width, 0f, 0f, 0f, 0f, 2f / height, 0f, 0f, 0f, 0f, -2f, 0f, -((x * 2f + width) / width), -((y * 2f + height) / height), -1f, 1f });`
t = top, b = bottom, l = left, r = right
Debugging results (it works):
Quote
Projection: [0.0025, 0.0, 0.0, 0.0, 0.0, 0.0033333334, 0.0, 0.0, 0.0, 0.0, -2.0, 0.0, -1.0, -1.0, -1.0, 1.0]
Scale: [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]
Rotation: [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]
Translation: [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]
trsp: [0.0025, 0.0, 0.0, 0.0, 0.0, 0.0033333334, 0.0, 0.0, 0.0, 0.0, -2.0, 0.0, -1.0, -1.0, -1.0, 1.0]
[0.25, 0.33333334, 0.0, -199.0] <--- viewProjection(trsp) * (100, 100, 0, 1)
Perspective projection:
1 2 3 4 ` projection.overwrite(new float[] { (2f * zNear) / width, 0f, 0f, 0f, 0f, (2f * zNear) / height, 0f, 0f, (2f * x + width) / width, (2f * y + height) / height, -(-zFar + zNear) / (zFar - zNear), -1f, 0f, 0f, -(2f * zFar * zNear) / (zFar - zNear), 0f });`
t = top, b = bottom, l = left, r = right, n = zNear, f = zFar
Debugging results - n = 1, f = 100 (nothing is displayed):
Quote
Projection: [0.0025, 0.0, 0.0, 0.0, 0.0, 0.0033333334, 0.0, 0.0, 1.0, 1.0, 1.0, -1.0, 0.0, 0.0, -2.020202, 0.0]
Scale: [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]
Rotation: [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]
Translation: [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]
trsp: [0.0025, 0.0, 0.0, 0.0, 0.0, 0.0033333334, 0.0, 0.0, 1.0, 1.0, 1.0, -1.0, 0.0, 0.0, -2.020202, 0.0]
[0.25, 0.33333334, 199.0, 0.0] <-- I'm guessing it's something to do with the Z or with the W coordinate?
What could the problem be? I'm not really sure what I'm doing wrong. I've tried translating the Z coordinate and extending the far pane, but I see nothing. I'm clearing the depth buffer and I tried enabling depth testing. None of those things worked.
EDIT: I think it must be something to do with the Z coordinate, since when I translate it, the W coordinate changes but nothing is displayed.
Why are all OpenGL tutorials written in Brainf**k?
Gef
« Reply #1 - Posted 2014-02-04 12:02:50 »
Because we don't see the code around this one, perhaps try disabling other stuffs like lighting, texture...
EDIT : here is a site that gives a well explanation on perspective projection, if that could help.
Troubleshoots
JGO Knight
Medals: 36
Exp: 7-9 months
Damn maths.
« Reply #2 - Posted 2014-02-04 12:21:59 »
Because we don't see the code around this one, perhaps try disabling other stuffs like lighting, texture...
I've not got anything else enabled. The code:
The viewport is being set as the width and the height of the screen.
This is the render method (it uses my LWJGL wrapper/library). This is the same for orthographic projection (except for clearing depth buffer):
1 2 3 4 5 6 7 8 `GL.glClearColor(colour);GL.glClear(Buffer.COLOR_DEPTH); // combined colour and depth bufferprogram.attach(); // simply calls glUseProgramprogram.setUniformMatrix4x4("mvp", camera.update()); // calls setUniformMatrix4, the matrix isn't transposedvao.attach();vao.render(Primitive.TRIANGLE_STRIP, 4); // calls glDrawArraysVertexArrayObject.detach();ProgramObject.detach();`
I'm updating my view*Projection uniform variable with this method (you can see the values of the matrices (column major) in my OP), this is the same for orthographic projection:
1 2 3 4 ` public Matrix4 update() { return scale.copy().multiply(rotation).multiply(translation) .multiply(projection); }`
This is how I'm setting the data for the attribute (it shouldn't really matter since orthographic projection works and uses the same data):
1 2 3 4 5 6 7 8 9 10 `data = FloatBufferIO.create(0f, 0f, 100f, 0f, 0f, 100f, 100f, 100f);vertices = new BufferObject(Target.ARRAY_BUFFER, data, Usage.STATIC_DRAW);vao = new VertexArrayObject();vao.attach();vertices.attach();vao.formatAttributeData(program.getAttributeIndex("position"), 2, // enables the attribute & calls glVertexAttribPointer DataType.SIGNED_FLOAT, false, 8, 0);VertexArrayObject.detach();`
When resized (and when the camera is created), this is called in the PerspectiveCamera class:
1 2 3 4 `projection.overwrite(new float[] { (2f * zNear) / width, 0f, 0f, 0f, 0f, (2f * zNear) / height, 0f, 0f, (2f * x + width) / width, (2f * y + height) / height, -(-zFar + zNear) / (zFar - zNear), -1f, 0f, 0f, -(2f * zFar * zNear) / (zFar - zNear), 0f });`
Sorry if it's unclear. If you want to see anything else just ask.
Edit: This is my matrix multiplication method (it seems to work correctly, judging by the debugging results):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 ` float m00 = (components[0] * matrix.components[0]) + (components[1] * matrix.components[4]) + (components[2] * matrix.components[8]) + (components[3] * matrix.components[12]); float m01 = (components[0] * matrix.components[1]) + (components[1] * matrix.components[5]) + (components[2] * matrix.components[9]) + (components[3] * matrix.components[13]); float m02 = (components[0] * matrix.components[2]) + (components[1] * matrix.components[6]) + (components[2] * matrix.components[10]) + (components[3] * matrix.components[14]); float m03 = (components[0] * matrix.components[3]) + (components[1] * matrix.components[7]) + (components[2] * matrix.components[11]) + (components[3] * matrix.components[15]); float m10 = (components[4] * matrix.components[0]) + (components[5] * matrix.components[4]) + (components[6] * matrix.components[8]) + (components[7] * matrix.components[12]); float m11 = (components[4] * matrix.components[1]) + (components[5] * matrix.components[5]) + (components[6] * matrix.components[9]) + (components[7] * matrix.components[13]); float m12 = (components[4] * matrix.components[2]) + (components[5] * matrix.components[6]) + (components[6] * matrix.components[10]) + (components[7] * matrix.components[14]); float m13 = (components[4] * matrix.components[3]) + (components[5] * matrix.components[7]) + (components[6] * matrix.components[11]) + (components[7] * matrix.components[15]); float m20 = (components[8] * matrix.components[0]) + (components[9] * matrix.components[4]) + (components[10] * matrix.components[8]) + (components[11] * matrix.components[12]); float m21 = (components[8] * matrix.components[1]) + (components[9] * matrix.components[5]) + (components[10] * matrix.components[9]) + (components[11] * matrix.components[13]); float m22 = (components[8] * matrix.components[2]) + (components[9] * matrix.components[6]) + (components[10] * matrix.components[10]) + (components[11] * matrix.components[14]); float m23 = (components[8] * matrix.components[3]) + (components[9] * matrix.components[7]) + (components[10] * matrix.components[11]) + (components[11] * matrix.components[15]); float m30 = (components[12] * matrix.components[0]) + (components[13] * matrix.components[4]) + (components[14] * matrix.components[8]) + (components[15] * matrix.components[12]); float m31 = (components[12] * matrix.components[1]) + (components[13] * matrix.components[5]) + (components[14] * matrix.components[9]) + (components[15] * matrix.components[13]); float m32 = (components[12] * matrix.components[2]) + (components[13] * matrix.components[6]) + (components[14] * matrix.components[10]) + (components[15] * matrix.components[14]); float m33 = (components[12] * matrix.components[3]) + (components[13] * matrix.components[7]) + (components[14] * matrix.components[11]) + (components[15] * matrix.components[15]); components[0] = m00; components[1] = m01; components[2] = m02; components[3] = m03; components[4] = m10; components[5] = m11; components[6] = m12; components[7] = m13; components[8] = m20; components[9] = m21; components[10] = m22; components[11] = m23; components[12] = m30; components[13] = m31; components[14] = m32; components[15] = m33;`
Edit2: I'll try transposing the projection matrix, since the page you linked me to has the exact same matrix transposed. It doesn't display, however this is the result:
Quote
Projection: [0.0025, 0.0, 1.0, 0.0, 0.0, 0.0033333334, 1.0, 0.0, 0.0, 0.0, 1.0, -2.002002, 0.0, 0.0, -1.0, 0.0]
Scale: [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]
Rotation: [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]
Translation: [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]
trsp: [0.0025, 0.0, 1.0, 0.0, 0.0, 0.0033333334, 1.0, 0.0, 0.0, 0.0, 1.0, -2.002002, 0.0, 0.0, -1.0, 0.0]
[0.25, 0.33333334, -2.002002, 0.0]
Edit3: Never-mind, that was based on row major order. It gives the exact same matrix. I think that the issue must be related to something else.
Why are all OpenGL tutorials written in Brainf**k?
Troubleshoots
JGO Knight
Medals: 36
Exp: 7-9 months
Damn maths.
« Reply #3 - Posted 2014-02-04 13:01:06 »
--> --> --> --> --> --> -->
That's the last 60 seconds of my life. ^^
The problem: -(-zFar + zNear) / (zFar - zNear).
The solution: -(zFar + zNear) / (zFar - zNear).
FFS@@~#'#;'#;'@~:@{:[';[p';[p#@:@:@:@:{l:[] #;{@~:~{@:?:@:@:@:';[p];[;]-p]=;[]=+{}= ='[];[#';];'[#;
Ten hours of debugging just for that...
Edit: It requires me to translate the camera 1 on the Z axis. Should I add 1 to -(2f * zFar * zNear) / (zFar - zNear)?
Why are all OpenGL tutorials written in Brainf**k?
Gef
« Reply #4 - Posted 2014-02-04 17:54:24 »
Well done !
(I would not have imagined being so useless )
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How to: JGO Wikiby Mac702015-02-17 20:56:162D Dynamic Lighting2015-01-01 20:25:42How do I start Java Game Development?by gouessej2014-12-27 19:41:21Resources for WIP gamesby kpars2014-12-18 10:26:14Understanding relations between setOrigin, setScale and setPosition in libGdx2014-10-09 22:35:00Definite guide to supporting multiple device resolutions on Android (2014)2014-10-02 22:36:02List of Learning Resources2014-08-16 10:40:00List of Learning Resources2014-08-05 19:33:27
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https://oeis.org/A100986 | 1,624,290,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00286.warc.gz | 387,227,887 | 3,859 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A100986 Smallest k such that concatenation of r*k and 1 is a prime for all r = 1 to n but not prime for r = n+1, or smallest k such that 10*r*k+1 is a prime for all r = 1 to n but not prime for r = n+1. 9
1, 3, 21, 33, 1083, 2541, 822486, 51282, 1296060612 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS a(10) > 10^10. - Robert Price, Apr 02 2019 LINKS EXAMPLE a(4)=33 because 331, 661, 991 and 1321 (1321=10*4*33+1) are all prime, but 1651 (1651=10*5*33+1) is not prime. - Robert Price, Apr 02 2019 MATHEMATICA Table[k = 1; While[! AllTrue[Table[10*r*k + 1, {r, 1, n}], PrimeQ] || PrimeQ[10*(n + 1)*k + 1], k++]; k, {n, 1, 9}] (* Robert Price, Apr 02 2019 *) PROG (PARI) isok(k, n) = {for (r=1, n, if (! isprime(10*r*k+1), return (0)); ); !isprime(10*(n+1)*k+1); } a(n) = {my(k=1); while(! isok(k, n), k++); k; } \\ Michel Marcus, Apr 03 2019 CROSSREFS Cf. A089323. Sequence in context: A287930 A287799 A089323 * A213141 A075732 A087690 Adjacent sequences: A100983 A100984 A100985 * A100987 A100988 A100989 KEYWORD more,nonn AUTHOR Ray G. Opao, Jan 13 2005 EXTENSIONS Corrected a(7) and added a(9) by Robert Price, Apr 02 2019 STATUS approved
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Last modified June 21 11:31 EDT 2021. Contains 345363 sequences. (Running on oeis4.) | 608 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-25 | latest | en | 0.69203 |
https://www.solarempower.com/blog/how-many-solar-panels-to-run-air-conditioner/ | 1,713,300,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00097.warc.gz | 893,815,932 | 38,344 | # How Many Solar Panels To Run Air Conditioner?
Updated on March 9, 2023
For most home appliances it’s quite easy to work out exactly how many solar panels you would need to run them.
This is because they pull a constant current, so once you know an appliance power rating in watts, you can match that with solar panel output.
Some appliances, like fridges, freezers and air conditioners, are not so straight-forward, as they contain compressors with motors. This means that the power they draw varies and needs to be averaged out.
## How Many Solar Panels To Run Air Conditioner?
An air conditioner would need 1200 watts of solar panels for each Ton of cooling capacity, assuming irradiance of 4 Peak-sun-hours/day. A 100Ah battery (such as this one by Renogy) is recommended per Ton for each hour anticipated running time. This will allow running when irradiance is low and also provide a reservoir to supply motor surge current.
These are the things we need to know before building a solar system to run an air conditioner:
• Why air conditioners are rated in tons
• AC power rating (kW)
• What is surge current and power factor?
• How long will the air conditioner be running?
• Are batteries needed to run your air conditioner?
• What type of batteries are best suited for solar energy storage?
• How much energy do solar panels generate?
• Solar panel sizes you’re going to use
• Inverter sizing (and why you need one)
### Video – How many solar panels are needed to run AC on an RV?
Solar powered air conditioning is a really good thing to have, specially in countries with hot humid conditions and plenty of sun!
Solar AC makes a great DIY project, but we need to explore AC electrical and cooling characteristics before answering the question ‘how many solar panels do I need to run an air conditioner unit?’
## Can you run air conditioner off solar panels?
When talking about d.c. loads, power calculations are quite easy. Power rating is found by multiplying volts x amps.
For example, when a d.c. resistive load has a 12 volt supply and pulls 12 amps, we use this formula to find the power:
12V x 12A = 144watts
In a.c. circuits, volts x amps doesn’t always equal watts (although sometimes it does!) Read on …
A.C. power calculations take into account the voltage and current waveforms
Home appliances use a.c. power and the calculation is handled in a different way. Voltage and current waveforms constantly move from positive to negative in a sinusoidal shape.
If V and I are maximum at the same time, they are said to be ‘in phase’ and volts x amps will really equal watts.
If volts and amps are not in phase, then the equation doesn’t work. We’ll take a look at this later.
## How much electricity does a solar powered air conditioner use?
### Why is AC rated in tons and not watts?
All air conditioning units have different efficiencies, so if we used their electrical kilowatt ratings for comparison, it wouldn’t reflect their cooling power.
This means that 2 AC units rated at 1.5 kW might not have the same cooling power.
So what is a ton in AC ratings? It represents the amount of cooling required to melt 2000 pounds of ice in one day. It’s commonly known in the trade as a short ton.
The bigger the tonnage, the more air can be cooled each hour. One ton can also be represented as 12,000 BTU/h (British Thermal Units per hour) or 3.5 kW (kilowatt).
This means that a 1 Ton air conditioner has a cooling capacity in Watts of:
(12000BTU x 1055) / 3600 = 3516 Joules per sec = 3516W (cooling)
### What is EER value in AC?
EER means ‘energy efficiency ratio’ and it tells you how well an air conditioning unit cools. If you divide an ac unit’s BTU number by it’s electrical rating in Watts you come up with a ratio which indicates it’s efficiency.
An efficient air conditioning unit could have an EER of 11, for example. AC units with an EER rating of more than 10 are recommended for countries or locations with hotter climates.
The EER rating system is based on an outdoor temperature of 95°F, a temperature inside of 80°F, and 50% humidity.
The SEER rating is more common nowadays, as it takes into account an air conditioner’s cooling capacity over a range of outside temperatures – the S means the ratings are ‘seasonal’.
The table below shows the SEER/EER for popular home AC units.
Lennox are making particular gains in supplementing their designs with solar panel support and this will probably become the industry-standard in the not-too distant future.
### Table – SEER and EER ratings compared for 5 popular AC brands
Air Conditioner Brand and Model SEER Rating EER Rating Notes: Lennox Signature Series XC25 26 16.5 Comes with solar option Amana AVXC20 & Daikin DX20VC 24.5 14 Lennox Signature Series XC21 21.2 15 Maytag M1200 PSA1BG 20 13.5 Carrier Infinity 20 24VNA0 20.5 15.5
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### How much does it cost per hour to run an air conditioning unit?
This can vary quite a lot and depends on several factors:
• what is the efficiency of the AC unit?
• how often during the day you run it
• how long the AC unit runs
• the outside temperature
• the cost per kWh of your electricity
As you can see, it can get complicated but is it worth going into all this detail? Let’s examine the factors you need to be aware of.
Air conditioning and solar installers have rule-of-thumb guides for sizing systems. One of these rules states that an air conditioning unit pulls around 7 amps for every 1 Ton of it’s cooling capacity rating.
However, this changes when the Tonnage changes. It’s widely accepted that the relationship of amps vs tons changes as the cooling power increases:
• 2 ton equates to 15 amps
• 3 ton equates to 18 amps
• 4 ton equates to 21 amps
For sizing and installation purposes, this is good. In my case, when I’m designing a solar panel installation, I really want to know exactly what’s going on electrically!
You can get confused about RLA (Rated Load Amps), which can be found on the air conditioning label tag – see image below:
What does it cost to run AC per hour?
The RLA value is not the AC normal running current – it is the maximum current the compressor motor will pull while the AC unit is running.
In addition, this is not the same as the LRA value, which is Locked Rotor Amps. This occurs is when the compressor motor is stalled, can’t rotate and pulls maximum current.
This value is the same as inrush or surge current when the compressor motor starts from standstill.
The real running current depends on other things.
### Air conditioner surge current
When an a.c. motor starts it pulls several times more current than when it runs continuously.
This running current will be dependent on the load, but the surge or inrush current is the highest amps the motor pulls, unless the motor is locked solid and can’t move at all (this is very bad.)
The size of the wiring is based on maximum current when running plus 10%. Surge currents only last for a few seconds normally and the heating effect in the wiring isn’t significant.
However, breakers and fuses need to be sized accordingly, so that they can accommodate this surge without tripping out.
This also has important implications for any solar panel system designed to run air conditioning.
If running during the day without a battery bank used as an energy reservoir, the solar panels wattage rating needs to be big enough to supply this motor surge current.
Solar panel systems designed to run home appliances make use of a device called an inverter to change d.c. (direct current) voltage produced by solar panels into the a.c. (alternating current) power for domestic use.
Solar panels can only run air conditioning using an inverter
RENOGY are fast becoming the preferred source for solar panels, kits, batteries and solar control accessories. Based in the US, where the products are manufactured, they are widely known and respected for innovation and quality.Check the latest RENOGY Prices
## How many solar panels to power ac unit?
Now we’ve explored what happens in an air conditioning unit on the the load side, we can take a look at the supply side in some depth.
I’m going to base the rest of the post on using 300 watt fixed solar panels, because the higher panel ratings are a better buy in terms of installed dollars/watt.
*This is the 300 watt solar panel I’m talking about.
### How much power can a 300 watt solar panel generate?
A solar panel’s output power in watts depends on these factors:
• solar panel efficiency (monocrystalline panel efficiency is between 20% and 22%)
• solar panel area in square meters
• orientation (what is the angle to the horizontal and vertical)
• will you be using fixed or auto-tracking for the solar installation
Solar panel ratings given by the manufacturer follow the guidelines laid out in the Standard Test Conditions or STC.
Basically, the panel rating is stipulated when the solar irradiance is 1000 watts/m2 and the temperature is 25 degrees C.
This assumes that the panel is directly under the sun at 90 degrees, but this is hardly ever the case!
Solar panel output reduces when the angle to the sun becomes more acute. Unfortunately, it also reduces at other angles, if the temperature is above 25 degrees C.
The power output goes down at a rate of 0.5 watts for every degree increase in temperature.
If the sun’s energy is low because of the irradiance level in your geographical location, then solar panel output will go down proportionately.
### How does a solar inverter work? Some basics
Solar panels produce direct current (d.c.) but home appliances need alternating current (a.c.) – that’s why you need an inverter to convert d.c. to a.c.
An inverter chops up the direct current into packets or pulses, and re-shapes them electronically into the smooth sinusoidal waveform, just like the power fed into your home by utility companies.
## How many solar panels to run 5 ton ac unit?
Air conditioner is needed when it’s hottest, which is the four hours around noon-time in most locations.
This is a lucky break, because this is the same window of time when the sun’s energy is high and solar panel power output is at a maximum.
1 Ton of air conditioning cooling capacity is about 1.2kW, so if you run it for 4 hours it would consume 4.8kWh over that time.
Logically, a 5 Ton air conditioning unit will consume:
4.8kWh x 5 = 24kWh
A 300 watt solar panel can produce 300 watts x 4 hours for the same time period (if the sunshine conditions are ideal) which equal 1.2kWh.
The basic formula for finding the number of solar panels to run a 5 ton ac unit is:
AC energy consumption (24kWh)/energy produced (1.2kWh) = 20 solar panels (300 watt rating)
## How many solar panels to run a 2 ton ac unit?
Let’s say we’re going to run the AC for 4 hours again in the hottest part of the day.
The basic calculation appears below, and assumes continuous running, surge current ignored:
9.6kWh/1.2kWh = 8 solar panels (300 watts)
## Can I run a 1.5 ton AC on solar without batteries?
Small air conditioning units are easier to power using solar panels. The basic calculation for a 1.5 ton air conditioning unit is:
• 7.2kWh/1.2kWh = 6 (300 watts rating)
So it’s possible run a 1.5 Ton AC unit on 6 x 300 watt solar panels, provided the sun is shining!
What about the AC inrush current?
Inrush current is pulled when the compressor motor starts from being stationary, for example when it starts a compression cycle when the temperature drops.
Starting when the motor is cold is the worst possible situation and in this case the motor could pull 3 or 4 times it’s normal current in. In fact,it’s best to assume 4 times for safety.
This means a 5 ton air conditioner could need 4 times more energy to get the cycle started, which would mean installing 80 solar panels!
Clearly, this isn’t practical, so what is the answer? And what about running the AC unit on a cloudy day or at night?
The solution is to install a bank of batteries for energy storage. This ensures a reservoir of extra power when it is needed and means you can run air conditioning when there’s no sun.
## Key takeaways about solar air conditioners
Unlike an RV air conditioner a central air conditioner has a much higher power consumption and a solar array needs to have enough power to accommodate the current surges.
Generally, it’s better for a solar power system to incorporate solar batteries, even for a relatively low power appliance such as a window air conditioner. (A 5000 BTU air conditioner is considered small for window use.)
Battery systems also have the advantage of delivering energy during a power outage, so the home isn’t deprived of cool air. In general, 1000 watts of solar panels is needed for every 1000 Btu of ac. The size of the battery bank will determine the number of hours air conditioning can run. A solar charge controller is required for any solar system using batteries and lithium batteries are strongly recommended as solar battery storage.
A single solar panel will not have enough power to run AC, even in direct sunlight. The heavier the AC then the higher number of panels needed.
An off-grid system may be used and adequate roof space is needed, which should be sized up to make sure the amount of power generated is adequate, which depends on the energy consumption of your air conditioner.
The exact number of solar panels is determined by the average annual energy consumption taken from the manufacturer’s name plate. As a general rule, 2.5kw of power with batteries is adequate for a small AC unit. | 3,060 | 13,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | longest | en | 0.934283 |
https://www.ato.gov.au/forms-and-instructions/research-and-development-tax-concession-2010-schedule-and-instructions/list-of-tables/table-14-calculation-of-the-adjusted-increase-in-expenditure-on-rd-by-the-group | 1,713,988,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00151.warc.gz | 588,421,644 | 73,648 | Search Suggestion:
# Table 14: Calculation of the adjusted increase in expenditure on R&D by the group
Last updated 1 June 2010
Row f of column G in table 1 A \$ Row f of column G in table 4 B \$ A + B (Write 0 if negative.) C \$ Adjustment balance (from column G in table 7) D \$ Adjusted increase in expenditure by the group (C - D) (Write 0 if negative.) E \$ Transfer the result at E to: column H in table 7 E in table 15 E in table 17.
## Table 15: Calculation of your company's share of the Australian owned part of the adjusted increase in expenditure on R&D by the group
Increase in expenditure on Australian owned R&D by the eligible company (row a of column H in table 1) A \$ Total increase in expenditure on Australian owned R&D by the eligible companies in the group (row f of column H in table 1) B \$ Net increase in expenditure on Australian owned R&D by the group (row f of column G in table 1) Note: If the figure at row f of column G in table 1 is negative, write 0 at this label. C \$ Net increase in expenditure on foreign owned R&D by the group (row f of column G in table 4) Note: If the figure at row f of column G in table 4 is negative, write 0 at this label. D \$ Adjusted increase in expenditure on R&D by the group (from E in table 14) E \$ A B (Do not round this number.) F \$ C + D G \$ C G (Do not round this number.) H \$ The company's share of the Australian owned part of the adjusted increase (E F H) I \$ Transfer the amount at I to: A in table 16.
Note: If the figures at A, B, C or E above are zero, leave M item 3 in part D of the Research and development tax concession schedule 2010 blank as you are ineligible for the Australian owned R&D incremental tax concession. Also print X in the No box at the top of part D of the Research and development tax concession schedule 2010. If you are ineligible, skip step 5.2 and go to step 6.
QC22870 | 493 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.875175 |
https://www.devtreks.org/commontreks/preview/commons/resourcepack/Monitoring%20and%20Evaluation%20Calculation%201/476/none | 1,500,856,860,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424645.77/warc/CC-MAIN-20170724002355-20170724022355-00659.warc.gz | 759,697,511 | 4,573 | #### Go to CommonTreks Home
Note: Going home clears all of these panels.
#### Monitoring and Evaluation Calculation 1
Pdf 01, Monitoring and Evaluation, Introduction
This reference (3.4 MB) iuses a malnutrition intervention to introduce DevTreks monitoring and evaluation framework. The resource was saved successfully on: 11/23/2016 12:29:14 AM
Pdf 02, Monitoring and Evaluation Calculation
Clubs using DevTreks can use ME calculators (1.2 MB) to track generic indicators that support the basic monitoring and evaluation of projects, programs, and technologies. Better monitoring and evaluation of work progress and performance may help people to improve their lives and livelihoods. The resource was saved successfully on: 11/23/2016 12:29:32 AM
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United Nations Monitoring and Evaluation
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Video: Video 03, M and E Calculations
3 of 3, (6.0 MB, 1:20). This video demonstrates how Indicators for Investment, Budget, and Time Period base elements measure ME Impacts. That is, what evidence exists that money has been, or is being, spent well? Have lives and livelihoods actually improved? A useful experiment is to try to find this evidence for just about any government expenditure, anywhere. As another example of “doing it right”, interpret the following question: “Other than institutional factors, is there any technical reason that this data can’t be completed online, stored uniformly online, and easily accessed online by people and machines, for every major government expenditure in the world?” The resource was saved successfully on: 11/9/2016 1:05:18 AM | 599 | 2,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.844996 |
https://math.answers.com/algebra/What_is_the_equation_for_a_line_that_passes_through_the_points_%28_8_2_%29_and_%28_-7_2%29 | 1,726,374,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00172.warc.gz | 350,063,035 | 48,985 | 0
# What is the equation for a line that passes through the points ( 8 2 ) and ( -7 2)?
Updated: 1/12/2024
Wiki User
8y ago
Points: (8, 2) and (-7, 2)
Slope: 0
Equation: y = 2
Wiki User
8y ago
Chinwendu
Lvl 1
8mo ago
pls can you show how you solved it
Chinwendu
Lvl 2
8mo ago
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Jul 22, 2019
Part one: Toward a theory of everything
Regression to market
Uncertainty, market efficiency and the Kelly Criterion are three of the most common discussion points amongst bettors. In his latest guest contribution for Pinnacle, @PlusEVAnalytics tries to answer the questions that often arise out of these discussions. Read on to find out more.
In physics, a “theory of everything” is an attempt to conceptualise all aspects of the known universe in a single theoretical framework. My aim here is far less grandiose but along the same lines. Over the past few weeks I’ve been involved in several discussions (and at times heated debates) on Twitter on topics that have more in common than people may realize:
- What sample size of results is sufficient to convince a modeler that their edge is or is not real?
- What is “regression to market” and how should it be done?
- How important is closing line value (CLV)?
- What is better, full Kelly or fractional Kelly? If fractional, what’s the best way to determine the fraction?
In this article I will tie all of these ideas together into the betting version of a theory of everything.
The generator
Let’s start by defining a “generator” as a process that creates a sequence of random outcomes. We can divide generators into two categories:
The quantification of uncertainty is the center of all gambling mathematics. What people often neglect is that uncertainty comes from two distinct sources, and the nature of each is quite different.
Artificial generators like dice, cards or roulette wheels are carefully controlled so that everything there is to know about their probability distribution is already known. (For the sake of this discussion we’ll ignore things like loaded dice, marked cards and biased wheels!)
Natural generators like sporting events, elections and weather result from complex interactions of many factors. They too have a probability distribution, but it is unknown and unknowable – the best we can do is build models to estimate it. Sports betting exists entirely in the realm of natural generators.
For a model used to estimate the behaviour of a natural generator, let’s define the “true probability” as the thing that’s unknown and unknowable, the “model probability” as the modeler’s best estimate, and the “model error” as the difference between them.
Model probability =
true probability + model error
As the difference between a known quantity and an unknown quantity, the model error is also unknown and unknowable.
Process uncertainty and parameter uncertainty
Next, let’s use this idea of generators to discuss uncertainty. The quantification of uncertainty is the center of all gambling mathematics. What people often neglect is that uncertainty comes from two distinct sources, and the nature of each is quite different.
“Process uncertainty” comes from the randomness that’s built into the generator itself. It’s why repeated spins of the same wheel don’t necessarily give the same result, and it’s also why the same two tennis players can play each other multiple times under identical conditions and have one player win some and the other win some.
“Parameter uncertainty” comes from our incomplete understanding of how the generator works. In some more high-brow publications it may be referred to as “epistemic uncertainty”, rooted in epistemology – the philosophy of knowledge.
For example, suppose you give a soccer team a 60% probability of winning, you bet on them at even money, and they lose. Why did you lose your bet? Perhaps you were correct in your assessment, but you were unlucky – the 40% event happened, and you lost your bet. This is process uncertainty – good bet, unlucky result.
On the other hand, perhaps you were incorrect in your assessment – the true probability may have been 50%, or 30%, or even 1%. You made a bet that you thought was a good bet but in reality was a bad bet. This is parameter uncertainty. Because the true probability is unknown, it’s very difficult to figure out how much of your results – both good and bad – are driven by process uncertainty as opposed to parameter uncertainty.
Let’s pause here and make some statements that will prove useful as we continue this exploration into generators and uncertainty:
1. Artificial generators include process uncertainty only. Natural generators include both process uncertainty and parameter uncertainty.
2. Recall that we defined model error as the difference between the true probability and the model probability. Therefore, model error is a reflection of parameter uncertainty.
3. Process uncertainty impacts each result in a series independently of the others. This is why roulette scoreboards provide useless information.
4. Parameter uncertainty impacts each result in a series in a correlated manner. If you use the same model to produce probabilities for 100 different games, any errors in your model will likely propagate in a similar way across some or all of the 100.
5. Corollary of #3: Over a large sample size, the impact of process uncertainty will shrink towards zero. The proportion of snake eyes in repeated rolls of two dice will approach the theoretical probability of 1/36. This is known as the “law of large numbers”.
6. Corollary of #4 and #5: Repeatedly observing the results of a natural generator allows one to continually learn from the observations to improve one’s understanding of how the generator works. This is the basis of Bayesian modelling.
Regression to market
Model error is, as we’ve shown, quite a slippery creature. We know it exists, but it’s impossible to quantify. There is, however, one important property of parameter uncertainty that we can infer from the theory of market efficiency:
In a market that is at least somewhat efficient, the parameter uncertainty will tend to “push” in the direction of the probabilities that are implied by the market price. The more efficient the market, the stronger the push.
For all the numerical examples that follow, we will use a point spread bet with market odds of -105 / -105 for the sake of simplicity. The math works equally well for any odds, but the calculations become more difficult.
If you find a betting angle that nobody else in the world has, you may have a great deal of positive expected value, but your CLV will hover around zero.
In this case, the market price implies a win probability of 50% for each side. Suppose your model projects a win probability of 55% for one side. Market efficiency dictates that the model error is much more likely to be +2% (meaning a true probability of 53%) than -2% (meaning a true probability of 57%). To restate this same example in terms of expected values – if your model projects a theoretical expected value of +7.4%, your true expected value is much more likely to be +3.5% than +11.3%.
This asymmetry means that using the Kelly Criterion with an assumed edge of +7.4% is likely to cause you to over-bet your true edge, which is an extremely dangerous thing to do from a bankroll management perspective.
Bettors generally have two strategies to guard against this – either they use a fractional Kelly approach, or they take their model’s projected probabilities and “regress them to market” by using a weighted average of the model probabilities and the market implied probabilities. The weights can be chosen subjectively, or they can be estimated using a method such as maximum likelihood.
Note that because bet size under Kelly is directly proportional to the size of the edge, the two methods above are mathematically equivalent. The formula to convert between them is
Regression weight =
(Kelly multiple * model prob + (1 – Kelly multiple) * market prob * (1 + bookmaker’s margin) – market prob / (model prob – market prob)
In our example, the bookmaker’s margin is 1 – 0.5 * (205/105) = 2.4%. So, a bet size of 1/4 Kelly would be equivalent to a regression weight of (0.25 * 0.55 + (0.75 * 0.50 * 1.024) – 0.50) / (0.55 – 0.50) = 0.429, meaning the equivalent regression would be 0.429 x model probability + 0.571 x market probability.
Suppose you give a soccer team a 60% probability of winning, you bet on them at even money, and they lose. Why did you lose your bet? Perhaps you were correct in your assessment, but you were unlucky – the 40% event happened, and you lost your bet.
The “regression to market” approach has clear advantages, but it also has some disadvantages. The proper weights may change over time as the market evolves – this is true both on a micro level (from opening line to the time you evaluate your bet to the closing line) and on a macro level (as individuals enter and exit the market).
Also, the logic is a bit of a tightrope walk – it’s based on the idea that the market is efficient, but in a fully efficient market it’s impossible to even have an edge to begin with. This exact criticism can be applied to the idea of measuring one’s edge using “closing line value” (CLV) – that is, measuring a model’s success by how correlated it is with the movement from the line at the time the bet is placed to the closing line.
This is a good measure of success if, and only if, the market is reacting to the same “signal” as your model, just later in time. If you find a betting angle that nobody else in the world has, there is no reason why the market would catch up to you – you may have a great deal of positive expected value, but your CLV will hover around zero.
This guest contribution was made by one of our Twitter followers - @PlusEVAnalytics. You can read part two of his contribution here.
Betting Resources - Empowering your betting
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• ankitrathi
# Inferential Statistics for Data Science
Inferential Statistics for Data Science
This is the 4th post of blog post series ‘Probability & Statistics for Data Science’, this post covers these topics related to inferential statistics and their significance in data science.
1. Inferential Statistics
2. Sampling Distributions & Estimation
3. Hypothesis Testing (One and Two Group Means)
4. Hypothesis Testing (Categorical Data)
5. Hypothesis Testing (More Than Two Group Means)
6. Quantitative Data (Correlation & Regression)
7. Significance in Data Science
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Inferential Statistics
Inferential statistics allows you to make inferences about the population from the sample data.
Population & Sample
A sample is a representative subset of a population. Conducting a census on population is an ideal but impractical approach in most of the cases. Sampling is much more practical, however it is prone to sampling error. A sample non-representative of population is called bias, method chosen for such sampling is called sampling bias. Convenience bias, judgement bias, size bias, response bias are main types of sampling bias. The best technique for reducing bias in sampling is randomization. Simple random sampling is the simplest of randomization techniques, cluster sampling & stratified sampling are other systematic sampling techniques.
Sampling Distributions
Sample means become more and more normally distributed around the true mean (the population parameter) as we increase our sample size. The variability of the sample means decreases as sample size increases.
Central Limit Theorem
The Central Limit Theorem is used to help us understand the following facts regardless of whether the population distribution is normal or not: 1. the mean of the sample means is the same as the population mean 2. the standard deviation of the sample means is always equal to the standard error. 3. the distribution of sample means will become increasingly more normal as the sample size increases.
Confidence Intervals
A sample mean can be referred to as a point estimate of a population mean. A confidence interval is always centered around the mean of your sample. To construct the interval, you add a margin of error. The margin of error is found by multiplying the standard error of the mean by the z-score of the percent confidence level:
The confidence level indicates the number of times out of 100 that the mean of the population will be within the given interval of the sample mean.
Hypothesis Testing
Hypothesis testing is a kind of statistical inference that involves asking a question, collecting data, and then examining what the data tells us about how to proceed. The hypothesis to be tested is called the null hypothesis and given the symbol Ho. We test the null hypothesis against an alternative hypothesis, which is given the symbol Ha.
When a hypothesis is tested, we must decide on how much of a difference between means is necessary in order to reject the null hypothesis. Statisticians first choose a level of significance or alpha(α) level for their hypothesis test.
Critical values are the values that indicate the edge of the critical region. Critical regions describe the entire area of values that indicate you reject the null hypothesis.
left, right & two-tailed tests
These are the four basic steps we follow for (one & two group means) hypothesis testing:
1. State the null and alternative hypotheses.
2. Select the appropriate significance level and check the test assumptions.
3. Analyze the data and compute the test statistic.
4. Interpret the result.
Hypothesis Testing (One and Two Group Means)
Hypothesis Test on One Sample Mean When the Population Parameters are Known
We find the z-statistic of our sample mean in the sampling distribution and determine if that z-score falls within the critical(rejection) region or not. This test is only appropriate when you know the true mean and standard deviation of the population.
Hypothesis Tests When You Don’t Know Your Population Parameters
The Student’s t-distribution is similar to the normal distribution, except it is more spread out and wider in appearance, and has thicker tails. The differences between the t-distribution and the normal distribution are more exaggerated when there are fewer data points, and therefore fewer degrees of freedom.
Estimation as a follow-up to a Hypothesis Test
When a hypothesis is rejected, it is often useful to turn to estimation to try to capture the true value of the population mean.
Two-Sample T Tests
Independent Vs Dependent Samples
When we have independent samples we assume that the scores of one sample do not affect the other.
unpaired t-test
In two dependent samples of data, each score in one sample is paired with a specific score in the other sample.
paired t-test
Hypothesis Testing (Categorical Data)
Chi-square test is used for categorical data and it can be used to estimate how closely the distribution of a categorical variable matches an expected distribution (the goodness-of-fit test), or to estimate whether two categorical variables are independent of one another (the test of independence).
goodness-of-fit
degree of freedom (d f) = no. of categories(c)−1
test of independence
degree of freedom (df) = (rows−1)(columns−1)
Hypothesis Testing (More Than Two Group Means)
Analysis of Variance (ANOVA) allows us to test the hypothesis that multiple population means and variances of scores are equal. We can conduct a series of t-tests instead of ANOVA but that would be tedious due to various factors.
We follow a series of steps to perform ANOVA:
1. Calculate the total sum of squares (SST )
2. Calculate the sum of squares between (SSB)
3. Find the sum of squares within groups (SSW ) by subtracting
4. Next solve for degrees of freedom for the test
5. Using the values, you can now calculate the Mean Squares Between (MSB) and Mean Squares Within (MSW ) using the relationships below
6. Finally, calculate the F statistic using the following ratio
7. It is easy to fill in the Table from here — and also to see that once the SS and df are filled in, the remaining values in the table for MS and F are simple calculations
8. Find F critical
ANOVA formulaes
If F-value from the ANOVA test is greater than the F-critical value, so we would reject our Null Hypothesis.
One-Way ANOVA
One-way ANOVA method is the procedure for testing the null hypothesis that the population means and variances of a single independent variable are equal.
Two-Way ANOVA
Two-way ANOVA method is the procedure for testing the null hypothesis that the population means and variances of two independent variables are equal. With this method, we are not only able to study the effect of two independent variables, but also the interaction between these variables.
We can also do two separate one-way ANOVA but two-way ANOVA gives us Efficiency, Control & Interaction.
Quantitative Data (Correlation & Regression)
Correlation
Correlation refers to a mutual relationship or association between quantitative variables. It can help in predicting one quantity from another. It often indicates the presence of a causal relationship. It used as a basic quantity and foundation for many other modeling techniques.
Pearson Correlation
Regression
Regression analysis is a set of statistical processes for estimating the relationships among variables.
Regression
Simple Regression
This method uses a single independent variable to predict a dependent variable by fitting the best relationship.
Multiple Regression
This method uses more than one independent variable to predict a dependent variable by fitting the best relationship.
It works best when multicollinearity is absent. It’s a phenomenon in which two or more predictor variables are highly correlated.
Nonlinear Regression
In this method, observational data are modeled by a function which is a nonlinear combination of the model parameters and depends on one or more independent variables.
Significance in Data Science
In data science, inferential statistics is used is many ways:
1. Making inferences about the population from the sample.
2. Concluding whether a sample is significantly different from the population.
3. If adding or removing a feature from a model will really help to improve the model.
4. If one model is significantly better than the other?
5. Hypothesis testing in general.
References:
Ankit Rathi is an AI architect, published author & well-known speaker. His interest lies primarily in building end-to-end AI applications/products following best practices of Data Engineering and Architecture. | 1,825 | 8,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-16 | latest | en | 0.889641 |
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Burning ropes (Posted on 2002-03-28)
You have two ropes. You know that each rope takes one hour to burn from one end to the other, but that the burning does not take place at a constant rate (i.e. half the rope may burn in one minute, and the other half will take 59 more minutes to burn).
How would you go about measuring a 45 minute time interval with just these two ropes?
See The Solution Submitted by levik Rating: 3.7308 (26 votes)
Subject Author Date First puzzle Math Man 2012-05-11 21:03:09 Solution K Sengupta 2007-05-16 06:17:05 Wow jbcf05 2007-03-07 16:26:16 a new curve on things. gabbo 2004-09-15 21:46:54 re: a try Dacre 2003-10-24 09:28:49 cool Talon5000 2003-07-31 11:04:58 Saw John Tate 2003-06-08 11:16:11 Give up. John Tate 2003-06-08 11:14:28 try this mark hartman 2003-05-28 18:03:33 re: a try Emon Hunte 2003-02-02 16:49:24 a try ubergeek 2002-12-26 00:23:16 re: Cool dex 2002-12-18 13:29:51 Cool Eric 2002-05-07 06:38:30
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https://cosmolearning.org/courses/introduction-riemann-surfaces-algebraic-curves/video-lectures/ | 1,586,510,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00202.warc.gz | 402,372,019 | 20,683 | # Introduction to Riemann Surfaces and Algebraic Curves
## Video Lectures
Displaying all 48 video lectures.
I. Definitions and Examples of Riemann Surfaces
Lecture 1
Play Video
The Idea of a Riemann Surface
Goals of the Lecture: To develop a suitable definition of a structure of a Riemann Surface on a 2-dimensional surface that will allow us to carry out Complex Analysis (i.e., study of holomorphic (or) analytic functions) on the given surface. Topics: Complex plane, open set, analytic (or) holomorphic function, Cauchy-Riemann equations, complex differentiable, convergent power series, Taylor expansion, Taylor coefficients, open map, biholomorphic map (or) holomorphic isomorphism, homeomorphism (or) topological isomorphism, complex coordinate chart, compatibility of charts, transition functions, Riemann surface structure
Lecture 2
Play Video
Simple Examples of Riemann Surfaces
Goals of the Lecture: - To see how the real plane can be equipped with two different Riemann Surface structures - To see how the real 2-dimensional sphere can be equipped with a Riemann surface structure. Topics: Complex coordinate chart, compatible charts, transition function, complex atlas, Riemann surface, holomorphic function, unit disc, complex plane, Uniformisation theorem, simply connected, Riemann sphere, Riemann mapping theorem
Lecture 3
Play Video
Maximal Atlases and Holomorphic Maps of Riemann Surfaces
Goals of the Lecture: - To give a better definition of a Riemann surface using equivalence of atlases - To define the notion of a holomorphic (or) analytic map from one Riemann surface into another Riemann surface and in particular an isomorphism of Riemann surfaces Topics: Complex atlas, equivalent atlases, union of equivalent atlases, maximal atlas, holomorphic (or) analytic mapping between Riemann surfaces, isomorphism of Riemann surfaces
Lecture 4
Play Video
Riemann Surface Structure on a Cylinder
Goals of the Lecture: - To interpret a cylinder as a suitable quotient of the complex plane; - To use the above interpretation to give a Riemann surface structure on a cylinder and to raise the question as to how many such non-isomorphic structures exist. Topics: Translation by a complex number, equivalence relation, equivalence class, set modulo an equivalence relation, glueing edges of a strip, inverse image of an equivalence class, quotient topology, quotient map, open map, homeomorphism, Möbius transformation, group action on a set, orbits of an action, set modulo (action of) a group
Lecture 5
Play Video
Riemann Surface Structure on a Torus
Goals of the Lecture: - To interpret the torus as a suitable quotient of the complex plane; - To use the above interpretation to give a Riemann surface structure on a torus and to raise the question as to how many such non-isomorphic structures exist. Topics: Translation by a complex number, equivalence relation, equivalence class, set modulo an equivalence relation, glueing edges of a parallelogram, inverse image of an equivalence class, quotient topology, quotient map, open map, homeomorphism, Möbius transformation, group action on a set, orbits of an action, set modulo (action of) a group
II. Classification of Riemann Surfaces
Lecture 6
Play Video
Riemann Surface Structures on Cylinders and Tori via Covering Spaces
Goals of the Lecture:- To look at the set of all possible Riemann surface structures on a cylinder and the need for a method to distinguish between them- To explain the motivation for the use of the Theory of Covering Spaces to distinguish Riemann surface structures- To motivate the notion of a covering map by examples- To get introduced to the fact (called General Uniformisation) that any Riemann surface is the quotient (via a covering map) of a suitable simply connected Riemann surface- To understand the idea of the Fundamental group and where it fits into our discussionKeywords for Lecture 6:Cylinder, punctured plane, punctured unit disc, annulus, Riemann's Theorem on removable singularities, covering map, covering space, pathwise connected, locally pathwise connected, admissible neighbourhood or admissible open set, universal covering space or simply connected covering space, fundamental group, uniformisation of a general Riemann surface
Lecture 7
Play Video
Möbius Transformations Make up Fundamental Groups of Riemann Surfaces
Goals of the Lecture: - Every good topological space possesses a unique simply connected covering space called the Universal covering space - The fundamental group of the topological space shows up as a subgroup of automorphisms of its universal covering space - The universal covering map expresses the target space as the quotient of the universal covering space of the target, by the fundamental group of the target - A covering map can be used to transport Riemann surface structures from source to target and vice-versa, thus making it into a holomorphic covering map - Any Riemann surface is the quotient of the complex plane, or the upper half-plane, or the Riemann sphere by a suitable group of Möbius transformations isomorphic to the fundamental group of the Riemann surface - The study of any Riemann surface boils down to studying suitable subgroups of Möbius transformations Topics: Covering map, covering space, admissible open set or admissible neighborhood, simply connected covering or universal covering, local homeomorphism, Riemann surface structure inherited by a topological covering of a Riemann surface, uniformisation, fundamental group, Möbius transformation
Lecture 8
Play Video
Homotopy and the First Fundamental Group
Goals of the Lecture: - To understand the notion of homotopy of paths in a topological space - To understand concatenation of paths in a topological space - To sketch how the set of fixed-end-point (FEP) homotopy classes of loops at a point becomes a group under concatenation, called the First Fundamental Group - To look at examples of fundamental groups of some common topological spaces - To realise that the fundamental group is an algebraic invariant of topological spaces which helps in distinguishing non-isomorphic topological spaces - To realise that a first classification of Riemann surfaces can be done based on their fundamental groups by appealing to the theory of covering spaces. Topics: Path or arc in a topological space, initial or starting point and terminal or ending point of a path, path as a map, geometric path, parametrisation of a geometric path, homotopy, continuous deformation of maps, product topology, equivalence of paths under homotopy, fixed-end-point (FEP) homotopy, concatenation of paths, constant path, binary operation, associative binary operation, identity element for a binary operation, inverse of an element under a binary operation, first fundamental group, topological invariant.
Lecture 9
Play Video
A First Classification of Riemann Surfaces
Goals of the Lecture: - To get an idea of the classification of Riemann surfaces that can be arrived at based on the fundamental group, using the theory of covering spaces - To get introduced to the notions of: moduli problem, moduli space, number of moduli, fine and coarse classification, and to write these down for simple Riemann surfaces Topics: Biholomorphic map or isomorphism of Riemann surfaces, classification of Riemann surfaces, universal covering of a Riemann surface, abelian fundamental group, complex plane, unit disc, upper half-plane, punctured plane, punctured unit disc, cylinder, complex torus, annulus, Riemann sphere, g-torus, coarse classification, fine classification, moduli problem, moduli theory, moduli space, number of moduli
III. Universal Covering Space Theory
Lecture 10
Play Video
The Importance of the Path-lifting Property
Goals of Lecture 10: - To explore the reasons for the fundamental group occurring both as the inverse image of any point under the universal covering map as well as a subgroup of automorphisms of the universal covering space - To understand the notions of lifting property, unique-lifting property and uniqueness-of-lifting property - To understand the Covering Homotopy Theorem - To note that surjective local homeomorphisms have the uniqueness-of-lifting property - To note that a surjective local homeomorphism is a covering iff it has the path-lifting property - To deduce that covering maps have the unique path-lifting property Topics: Lifting of a map, lifting of a path, lifting property, unique-lifting property, uniqueness-of-lifting property, Covering Homotopy Theorem, local homeomorphism, unique path-lifting property, existence of lifting, fundamental group, universal covering
Lecture 11
Play Video
Fundamental groups as Fibres of the Universal covering Space
Goals of Lecture 11: - To see that the formation of the fundamental group is a covariant functorial operation, from the category whose objects are pointed topological spaces and whose morphisms are base-point-preserving continuous maps, to the category whose objects are groups and whose morphisms are group homomorphisms - To deduce the path-lifting property for a covering map as a consequence of the Covering Homotopy Theorem - To deduce from the Covering Homotopy Theorem that the fundamental group of a covering space can be identified naturally with a subgroup of the fundamental group of the space being covered - To note that the inverse image of a point (fibre over a point) under a covering map may be identified with the space of cosets of the fundamental group (based at a point fixed above) inside the fundamental group at the point below - To note that the universal covering of a space may be pictured as a fibration consisting of fundamental groups over that space Topics: Covering Homotopy Theorem, stationary homotopy, lifting of a homotopy, path-lifting property, category, objects of a category, morphisms of a category, covariant functor, functorial operation, fundamental group as a covariant functor, pointed topological space, group action, transitive action, fundamental group, universal covering, subgroup, cosets of a subgroup in a group
Lecture 12
Play Video
The Monodromy Action
Goal of Lecture 12: To understand how the fundamental group based at a point of the target of a covering map acts naturally on the fiber of the covering map over that point, the fiber being thought of as embedded inside the source of the covering map. Topics: Path, lifting of a path, unique-path-lifting property, Covering Homotopy Theorem, surjective local homeomorphism, universal covering space, injective group homomorphism, fundamental group, simply connected space, trivial group, fiber of a covering map, coset space, group action, orbit, orbit map, stabilizer subgroup, fibration.
Lecture 13
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The Universal covering as a Hausdorff Topological Space
Goals of Lecture 13: - To ask the question as to why the fundamental group of a space occurs as a subgroup of automorphisms of its universal covering - To define the universal covering space intuitively as a space of paths - To give a natural topology on the space defined above and to show that this topology is Hausdorff Topics: Path, Fixed-end-point (FEP) homotopy equivalence class, fundamental group, pathwise or arcwise connected, Hausdorff, locally simply connected, universal covering, basic open set, base for a topology, sub-base for a topology
Lecture 14
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The Construction of the Universal Covering Map
Goals of Lecture 14: - In the previous lecture, the universal covering space was constructed for a given space as a Hausdorff topological space along with a natural map into the given space. In this lecture, we show that this natural map is a covering map - It would follow that if the given space is locally arcwise connected and locally simply connected, then the same properties hold for the universal covering space as well Topics: Path, Fixed-end-point (FEP) homotopy equivalence class, fundamental group, pathwise or arcwise connected, Hausdorff, locally simply connected, universal covering, basic open set, base for a topology, sub-base for a topology, admissible neighborhood
Lecture 15
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Universality of the Universal Covering
Goals of the Lecture: - In the previous couple of lectures, the universal covering space was constructed for a given space as a Hausdorff topological space along with a natural map into the given space. That natural map was shown to be a covering map. In this lecture, we show that the universal covering space we constructed is indeed simply connected and has a universal property - We show that the universal covering space we have constructed is also a covering space for any other covering space. We further show that any covering space which is simply connected is homeomorphic to the universal covering space we have constructed. It follows that any two simply connected covering spaces thereby are not only just homeomorphic, but homeomorphic by a map that respects the covering projections, i.e., are isomorphic as covering spaces; in fact, even the isomorphism becomes unique if a point of the source and one of the target are fixed. These results show the universality of a simply connected covering space, which is why such a space is called "the" universal covering space Topics: Path, Fixed-end-point (FEP) homotopy equivalence class, fundamental group, pathwise or arcwise connected, Hausdorff, locally simply connected, universal covering, basic open set, base for a topology, sub-base for a topology, admissible neighborhood, isomorphism of covering spaces, universal property
Lecture 16
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The Fundamental Group of the base as the Deck Transformation Group
Goals of the Lecture: - In the previous couple of lectures, the universal covering space was constructed for a given space as a Hausdorff topological space along with a natural map into the given space. That natural map was shown to be a covering map. In the first part (part A) of this lecture, we showed that the universal covering space we constructed is indeed simply connected and has a universal property. In this lecture (part B), we show that we can naturally identify the fundamental group of the base space with a subgroup of self-isomorphisms of the universal covering space called the Deck Transformation Group - For any covering space, we may define the so-called Deck Transformation Group. This is the subgroup of self-homeomorphisms of the covering space that respect the covering projection map. If the covering space is the universal covering space, then the fundamental group of the base space (the space whose coverings we are concerned with) gets naturally identified with the deck transformation group. Thus the fundamental group of the base acts on the universal covering via the so-called deck transformations. These act along the fibers of the covering projection map. This action is called the Monodromy Action Topics: Path, Fixed-end-point (FEP) homotopy equivalence class, fundamental group, pathwise or arcwise connected, Hausdorff, locally simply connected, universal covering, basic open set, base for a topology, sub-base for a topology, admissible neighborhood, isomorphism of covering spaces, universal property, deck transformation, deck transformation group, monodromy action
IV. Classifying Moebius Transformations and Deck Transformations
Lecture 17
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The Riemann Surface Structure on the Topological Covering of a Riemann Surface
Goals: - To extend the theory of topological coverings to that of holomorphic (complex analytic) coverings - To show that any Riemann surface structure on the base space of a topological covering induces a Riemann surface structure on the covering space in such a way that the covering projection map is holomorpic. To achieve this using the technique of "pulling back charts from below" - To see why the Riemann surface structure induced above is essentially unique - In particular, we get a unique Riemann surface structure on the topological covering of a Riemann surface. The deck transformations therefore become holomorphic automorphisms of this Riemann surface structure Topics: Topological covering, holomorphic covering, admissible neighborhood, chart, pulling back charts by local homeomorphisms, locally biholomorphic, pulling back Riemann surface structures, holomorphicity or complex analyticity of continuous liftings, deck transformation
Lecture 18
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Riemann Surfaces with Universal Covering the Plane or the Sphere
Goals: - To see how the topological quotient of the universal covering of a space by the deck transformation group (which is isomorphic to the fundamental group of the space) gives back the space - In particular, the topological universal covering of a Riemann surface (which inherits a unique Riemann surface structure as shown in the previous lecture) modulo (or quotiented by or divided by) the fundamental group gives back the Riemann surface - To see that nontrivial deck transformations are fixed-point free - To see why any Riemann surface with universal covering the Riemann sphere is isomorphic to the Riemann sphere itself - To get a characterization of discrete subgroups of the additive group of complex numbers - To use the above characterisation to deduce that a Riemann surface with universal covering the plane has to be isomorphic to either the plane itself, or to a complex cylinder, or to a complex torus Topics: Holomorphic covering, holomorphic universal covering, group action on a topological space, orbit of a group action, equivalence relation defined by a group action, quotient by a group, topological quotient, quotient topology, quotient map, transitive action, deck transformation, open map, Riemann sphere, one-point compactification, stereographic projection, Möbius transformation, unique lifting property, group of translations, admissible neighborhood, module, submodule, subgroup, discrete submodule, discrete subgroup
Lecture 19
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Classifying Complex Cylinders Riemann Surfaces
Goals: - To characterize discrete subgroups of the additive group of complex numbers and to use this characterization to classify Riemann surfaces whose universal covering is the complex plane - To see how the twisting of the universal covering space by an automorphism (of the universal covering space) leads to the identification of the fundamental group (of the base of the covering) with a conjugate of the deck transformation group of the original covering - To show that the natural Riemann surface structures, on the quotient of the complex plane by the group of translations by integer multiples of a fixed nonzero complex number does not depend on that complex number; in other words that all such Riemann surfaces are isomorphic Topics: Upper-triangular matrix, complex plane, universal covering, deck transformation, abelian fundamental group, additive group of translations, module, submodule, discrete submodule, discrete subgroup
Lecture 20
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Möbius Transformations with a Single Fixed Point
Goals: - To realize that in order to study Riemann surfaces with abelian fundamental group and having universal covering the upper half-plane, one needs to first classify Möbius transformations in general and in particular study among those that are automorphisms of the upper half-plane - To motivate how the classification of Möbius transformations can be done using two seemingly unrelated aspects: one of them being the set of fixed points in the extended complex plane and the other being the value of the square of the trace of the transformation. - To show that these two aspects, though one of them is geometric while the other numeric, are in fact precisely related to each other - To characterize Möbius transformations with exactly one fixed point in the extended complex plane as precisely those that are conjugate to a translation; to show that such transformations are also precisely the so-called parabolic transformations, where parabolicity is defined as the square of the trace being equal to four Topics: Upper half-plane, unit disc, abelian fundamental group, deck transformation group, Möbius transformation, universal covering, holomorphic automorphism, group isomorphism, linear fractional transformation, bilinear transformation, fixed point of a map, square of the trace of a Möbius transformation, parabolic Möbius transformations, translations, conjugation by a Möbius transformation, special linear group, projective special linear group, upper-triangular matrix
Lecture 21
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Möbius Transformations with Two Fixed Points
Goals: - To analyze Möbius transformations with more than one fixed point in the extended complex plane - To continue with the classification of Möbius transformations begun in the previous lecture by defining the notions of loxodromic, elliptic and hyperbolic Möbius transformations using the values of the square of the trace of the transformation - To characterize geometrically the loxodromic, elliptic and hyperbolic Möbius transformations by showing that they can be conjugated by suitable Möbius transformations to multiplication by a complex number - To show that the elliptic Möbius transformations are precisely those that are conjugate to a rotation about the origin - To show that the hyperbolic Möbius transformations are precisely those that are conjugate to a real scaling Topics: Parabolic, elliptic, hyperbolic and loxodromic Möbius transformations, fixed point of a Möbius transformation, square of the trace of a Möbius transformation, translation, conjugation by a Möbius transformation, special linear group, projective special linear group
Lecture 22
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Torsion-freeness of the Fundamental Group of a Riemann Surface
Goals: - To analyze what the conditions of loxodromicity, ellipticity or hyperbolicity imply for an automorphism of the upper half-plane, i.e., to characterize the automorphisms of the upper half-plane. This is required for the classification of Riemann surfaces with universal covering the upper half-plane - To show that the fundamental group of a Riemann surface is torsion free i.e., that it has no non-identity elements of finite order - To show that the Deck transformations of the universal covering of a Riemann surface have to be either hyperbolic or parabolic in nature - To deduce that the fundamental group of a Riemann surface is torsion free Topics: Möbius transformation, special linear group, projective special linear group, parabolic, elliptic, hyperbolic, loxodromic, fixed point, conjugation, translation, Riemann sphere, extended complex plane, upper half-plane, square of the trace (or trace square) of a Möbius transformation, torsion-free group, element of finite order of a group, torsion element of a group, universal covering, fundamental group, Deck transformations
Lecture 23
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Characterizing Riemann Surface Structures on Quotients of the Upper Half
Goals: - To show that any Riemann Surface with nonzero abelian fundamental group and universal covering the upper half-plane has fundamental group isomorphic to the additive group of integers i.e., that it is cyclic of infinite order - To classify the Riemann surface structures naturally inherited by annuli in the complex plane, and to show that there is a family of such distinct (i.e., non-isomorphic) structures parametrized by a real parameter - To deduce that if a Riemann surface has fundamental group isomorphic to the product of the additive group of integers with itself, then it has to be isomorphic to a complex torus, and hence in particular that it has to necessarily be compact Topics: Upper half-plane, unit disc, annulus, torus, simply connected, abelian fundamental group, additive group, translation, deck transformation, Möbius transformation, universal covering, holomorphic automorphism, parabolic, elliptic, hyperbolic, loxodromic, fixed point, commuting Möbius transformations, conjugation, translation, universal covering, discrete subgroup, discrete submodule, generator of a group
Lecture 24
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Classifying Annuli up to Holomorphic Isomorphism
Goals: To show that the various annuli with inner radii in the real open unit interval and with outer radius unity are all non-isomorphic as Riemann surfaces Topics: Upper half-plane, universal covering, fundamental group, deck transformation group, Möbius transformations, real special linear group, real projective (special) linear group, simply connected, biholomorphic map, holomorphic isomorphism, infinite cyclic group, parabolic Möbius transformation, hyperbolic Möbius transformation, fixed point, extended plane, abelian fundamental group, commuting Möbius transformations, commuting deck transformations, punctured unit disc, annulus, unique lifting property
V. The Riemann Surface Structure on the Quotient of the Upper Half-Plane by the Unimodular Group
Lecture 25
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Orbits of the Integral Unimodular Group in the Upper Half-Plane
Goals: - To ask for a description of the set of holomorphic isomorphism classes of complex tori - To state the Theorem on the Moduli of Elliptic Curves that not only answers the question above but also shows that the set above has a beautiful God-given geometry - To see how the upper half-plane and the unimodular group (integral projective special linear group) enter into the discussion - To use the theory of covering spaces to prove a part of the Theorem on the Moduli of Elliptic Curves, namely that the set of holomorphic isomorphism classes of complex 1-dimensional tori is in a natural bijective correspondence with the set of orbits of the unimodular group in the upper half-plane Topics: Real torus, complex torus, Möbius transformation, translation, abelian group, holomorphic universal covering, admissible neighborhood, fundamental group, deck transformation group, biholomorphism class (or) holomorphic isomorphism class, locally biholomorphic map, upper half-plane, projective special linear group, unimodular group, orbits of a group action, action of a subgroup, underlying fixed geometric structure, superimposed (or) overlying (or) extra geometric structure, variation of extra structure for a fixed underlying structure (or) moduli problem, quotient by a group, equivalence relation induced by a group action, universal property of the universal covering, unique lifting property, moduli of elliptic curves, forming the fundamental group is functorial
Lecture 26
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Galois Coverings are precisely Quotients by Properly Discontinuous Free Actions
Goals: - To ask the question as to when the quotient of a space, by a subgroup of automorphisms (self-isomorphisms) of that space, becomes again a space with good properties. For example: when does the quotient of a Riemann surface, by a subgroup of holomorphic automorphisms, again become a Riemann surface? - To define properly discontinuous (free) actions and note that they are fixed-point-free - To see that the action of the Deck transformation group on the covering space is properly discontinuous - To define Galois (or) Regular (or) Normal coverings and characterize them precisely as quotients by properly discontinuous actions Topics: upper half-plane, biholomorphism class (or) holomorphic isomorphism class, complex torus, projective special linear group, unimodular group, quotient by a subgroup of automorphisms, quotient by the Deck transformation group, orbits of a group action, quotient topology, properly discontinuous action, action without fixed points, transitive action, admissible neighborhood, Galois covering (or) Normal covering (or) Regular covering, covariant functor, normal subgroup, equivalence relation induced by a group action, open map, unique lifting property, covering homotopy theorem, Riemann sphere, stabilizer (or) isotropy subgroup, ramified (or) branched covering
Lecture 27
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Local Actions at the Region of Discontinuity of a Kleinian Subgroup
Lecture 28
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Quotients by Kleinian Subgroups give rise to Riemann Surfaces
Goals: - To see how the quotient of the region of discontinuity by a Kleinian subgroup of Möbius transformations is a union of Riemann surfaces - To see how the quotients above are ramified (or) branched coverings of Riemann surfaces, with ramifications at the points with nontrivial isotropies (stabilizers) - To see in detail how to get a complex coordinate chart at the image point of a point of ramification Topics: Upper half-plane, unimodular group, fixed point, projective special linear group, quotient by a subgroup of Möbius transformations, holomorphic automorphisms, extended plane, properly discontinuous action, stabilizer (or) isotropy subgroup, region of discontinuity of a subgroup of Möbius transformations, limit set of a subgroup of Möbius transformations, elliptic Möbius transformations, isolated point, discrete subset, Kleinian subgroup of Möbius transformations, quotient topology, ramification (or) branch points, ramified (or) branched covering, unramified (or) unbranched covering, branch cut, slit disc
Lecture 29
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The Unimodular Group is Kleinian
Goals: - To see that a Kleinian subgroup of Möbius transformations is a discrete subspace of the space of all Möbius transformations and also that such a subgroup is either finite or countable as a set - To define a subgroup of Möbius transformations to be Fuchsian if it maps a half-plane or a disc onto itself - To see that a discrete Fuchsian subgroup is Kleinian. For example, the unimodular group is thus Kleinian - To conclude using the results of the previous lecture that the quotient of the upper half-plane by the unimodular group is a Riemann surface Topics: Schwarz's Lemma, Riemann Mapping Theorem, properly discontinuous action, Kleinian subgroup of Möbius transformations, region of discontinuity of a subgroup of Möbius transformations, upper half-plane, unimodular group, projective special linear group, discrete subgroup of Möbius transformations, Fuchsian subgroup of Möbius transformations, holomorphic automorphisms, extended plane, stabilizer (or) isotropy subgroup, orbit map, second countable metric space, space of matrices, space of invertible matrices, space of determinant one matrices
VI. Doubly-Periodic Meromorphic (or) Elliptic Functions
Lecture 30
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The Necessity of Elliptic Functions for the Classification of Complex Tori
Goals: - In the last few lectures, we have shown that the quotient of the upper half-plane by the unimodular group has a natural Riemann surface structure. In order to show that this Riemann surface is isomorphic to the complex plane, we have to realize that we need to look for invariants for complex tori - To motivate how the search for invariants for complex tori leads us to the study of doubly-periodic meromorphic functions (or) elliptic functions, the stereotype of which is given by the famous Weierstrass phe-function Topics: Upper half-plane, unimodular group, projective special linear group, set of orbits, quotient Riemann surface, lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, complex torus associated to a lattice, translation, j-invariant for a complex torus, Felix Klein, group-invariant function, bounded entire function, Liouville's theorem, singularity of an analytic function, poles, meromorphic function, doubly-periodic meromorphic function (or) elliptic function, Karl Weierstrass, algebraic curve, elliptic curve, Weierstrass phe-function, Residue theorem, double pole
Lecture 31
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The Uniqueness Property of the Weierstrass Phe-function
Goals: - To show that the Weierstrass Phe-function is the unique doubly-periodic meromorphic function (i.e., the unique elliptic function) with residue-zero double poles precisely at each point of the lattice and with constant term zero in the Laurent development at the origin Topics: Upper half-plane, lattice (or) grid in the plane, fundamental parallelogram (or) period parallelogram associated to a lattice, complex torus associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, isolated double pole, singular part of the Laurent expansion, deleted neighborhood, even function, entire function, algebraic elliptic cubic curve
Lecture 32
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The First Order Degree Two Cubic Ordinary Differential Equation satisfied by the Weierstrass Phe-function
Goals: - To show that the Weierstrass phe-function associated to a lattice satisfies a first order degree two cubic ODE - The ODE mentioned above is the key to studying the geometry of the complex torus associated to the lattice and eventually leads to the classification (moduli) theory of complex tori Topics: Upper half-plane, invariants for complex tori, lattice (or) grid in the plane, fundamental parallelogram (or) period parallelogram associated to a lattice, complex torus associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, isolated double pole, singular part of the Laurent expansion, analytic part of the Laurent expansion, antiderivative for the Weierstrass phe-function, Identity theorem for power series or Laurent series, differentiating term-by-term and integrating term-by-term under uniform convergence, even function, odd function, entire elliptic functions are constants, algebraic elliptic cubic curve
Lecture 33
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The Values of the Weierstrass Phe function at the Zeros of its Derivative
Goals: - To find the zeros of the derivative of the Weierstrass phe-function associated to a lattice - To use the ODE established in the previous lecture to analyze the values of the Weierstrass phe-function at the zeros of its derivative and to show that these values are nonvanishing analytic (holomorphic) functions on the upper half-plane - To introduce the notion of order for an elliptic function, namely the finite positive integer which is the number of times the function assumes any value in the extended complex plane (Riemann sphere) Topics: Upper half-plane, invariants for complex tori, lattice (or) grid in the plane, fundamental parallelogram (or) period parallelogram associated to a lattice, complex torus associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, simple zero, pole of order three, isolated double pole, Argument Principle, Residue theorem, order of an elliptic function, automorphic function (or) automorphic form, modular function (or) modular form, congruence mod two subgroup of the unimodular group, even function, odd function
VII. A Form Modular for the Congruence / Mod 2: Subgroup of the Unimodular Group on the Upper Half-Plane
Lecture 34
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The Construction of a Modular Form of Weight Two on the Upper Half-Plane
Goals: - To construct an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group Topics: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, universal covering, fundamental group, deck transformation group, holomorphic lifting, conjugation by a Möbius transformation, conjugate subgroup, additive group of translations, isomorphism of lattices, generator of a group, locally invertible map, locally biholomorphic map, conformal map, zeros of the derivative of the Weierstrass phe-function
Lecture 35
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The Fundamental Functional Equations satisfied by the Modular Form of Weight
Goals: - In the previous lecture, we constructed an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group. We ask what the effect of a general element of the unimodular group is on this weight two modular form - To see that in order to answer the question above, it is enough to compute the effect under each of five unimodular elements representing pre-images of the five non-trivial elements in the quotient by the congruence-mod-2 subgroup - To see that the computations above result in five simple and beautiful functional equations satisfied by the weight two modular formKeywords: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, zeros of the derivative of the Weierstrass phe-function
Lecture 36
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The Weight Two Modular Form assumes Real Values on the Imaginary Axis
Goals: - In the last few lectures, we constructed an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group. We saw that the effect of a general element of the unimodular group on this weight two modular form can be understood by just computing the effect under each of five unimodular elements representing pre-images of the five non-trivial elements in the quotient by the congruence-mod-2 subgroup. We saw that these computations resulted in five simple and beautiful functional equations satisfied by the weight two modular form. Using these functional equations diligently, in the present and the forthcoming lectures, we obtain a suitable region in the upper half-plane on which the mapping properties of the weight two modular form can be studied. In this lecture we show that the weight two modular form assumes only real values on the imaginary axis, which will turn out to be a boundary for such a region Topics: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, zeros of the derivative of the Weierstrass phe-function
Lecture 37
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The Weight Two Modular Form Vanishes at Infinity
Goals: - In the last few lectures, we constructed an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group. We saw that the effect of a general element of the unimodular group on this weight two modular form can be understood by just computing the effect under each of five unimodular elements representing pre-images of the five non-trivial elements in the quotient by the congruence-mod-2 subgroup. We saw that these computations resulted in five simple and beautiful functional equations satisfied by the weight two modular form. Using these functional equations, we want to find a suitable region in the upper half-plane on which the mapping properties of this weight two modular form may be easily studied. In the previous lecture we saw that the weight two modular form assumes only real values on the imaginary axis, which will turn out to be a boundary for such a region. In this lecture, we show that the weight two modular form vanishes at infinity Topics: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, zeros of the derivative of the Weierstrass phe-function, singular part of the Laurent expansion, pole of order two, uniform convergence, Weierstrass M-test, removable singularity, entire function, periodic function, period of a function, singly periodic function, Liouville's theorem
Lecture 38
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The Weight Two Modular Form Decays Exponentially in a Neighbourhood of Infinity
Goals: - In the last few lectures, we constructed an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group. We saw that the effect of a general element of the unimodular group on this weight two modular form can be understood by just computing the effect under each of five unimodular elements representing pre-images of the five non-trivial elements in the quotient by the congruence-mod-2 subgroup. We saw that these computations resulted in five simple and beautiful functional equations satisfied by the weight two modular form. Using these functional equations, we want to find a suitable region in the upper half-plane on which the mapping properties of this weight two modular form may be easily studied. - In the last couple of lectures we saw that the weight two modular form assumes only real values on the imaginary axis, which will turn out to be a boundary for such a region, and that the weight two modular form vanishes at infinity. In part A of this lecture (Lecture 38) we estimate that this vanishing at infinity is in fact an exponential decay. This estimation is actually a computation of the Fourier coefficient that matters most in the Fourier development of the weight two modular form which has period two. This estimation is critical for the study of the mapping properties which will be completed in part B (Lecture 39) of this lecture Topics: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, zeros of the derivative of the Weierstrass phe-function, singular part of the Laurent expansion, pole of order two, uniform convergence, Weierstrass M-test, removable singularity, entire function, periodic function, period of a function, singly periodic function, Liouville's theorem, Fourier coefficient, Fourier development
Lecture 39
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Suitable Restriction of the Weight Two Modular Form is a Holomorphic Conformal Isomorphism onto the Upper Half-Plane
Goals: - In the last few lectures, we constructed an analytic function on the upper half-plane which is a modular form of weight two, i.e., an analytic function that is invariant under the action of the congruence-mod-2 subgroup of the unimodular group. We saw how the effect of a general element of the unimodular group on this weight two modular form can be understood by just computing the effect under each of five unimodular elements representing pre-images of the five non-trivial elements in the quotient by the congruence-mod-2 subgroup. We saw that these computations resulted in five simple and beautiful functional equations satisfied by the weight two modular form. Using these functional equations, we want to find a suitable region in the upper half-plane on which the mapping properties of this weight two modular form may be easily studied - In the last couple of lectures we saw that the weight two modular form assumes only real values on the imaginary axis, which will turn out to be a boundary for such a region, and further that the weight two modular form vanishes at infinity. In part A of this lecture (Lecture 38) we estimated, by calculating the Fourier coefficient that mattered most, that this vanishing at infinity is in fact an exponential decay. This estimation is critical for the study of the mapping properties which we complete in part B (Lecture 39) of this lecture. We show that the weight two modular form assumes every value on the upper half-plane, and that when restricted to a suitable region it actually gives a holomorphic conformal isomorphism onto the upper half-plane with a continuous monotonic conformal extension to the boundary on the Riemann Sphere so that every real value and the point at infinity is also assumed precisely once Topics: Upper half-plane, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) modular form, congruence-mod-2 subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, normal subgroup, zeros of the derivative of the Weierstrass phe-function, monotonic function, contour, winding number, Fourier coefficient, Fourier development
VIII. The Elliptic Modular J-invariant and the Moduli of Complex 1-dimensional Tori (or) Elliptic Curves
Lecture 40
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The J-Invariant of a Complex Torus (or) of an Algebraic Elliptic Curve
Goals: - To associate to each complex 1-dimensional torus a complex number, called the j-invariant of the complex torus, which depends only on the holomorphic isomorphism class of the torus. This j-invariant will be shown in the forthcoming lectures to completely classify all complex tori - In the previous unit of lectures, we constructed a weight two modular form on the upper half-plane and studied its mapping properties. In this lecture we use this weight two modular form to define a full modular form, i.e., a holomorphic function on the upper half-plane that is invariant under the action of the full unimodular group. It is this modular form that goes down to give the j-invariant function on the Riemann surface of holomorphic isomorphism classes of complex tori with underlying set consisting of the orbits of the unimodular group in the upper half-plane Topics: Upper half-plane, quotient by the unimodular group, orbits of the unimodular group, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) weight two modular form, full modular function (or) full modular form, congruence-mod-2 normal subgroup of the unimodular group, special linear group, finite group, kernel of a group homomorphism, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, universal cover, neighborhood of infinity, lower half-plane, rational function, kernel of a group homomorphism, functional equations satisfied by the weight two modular form, meromorphic functions are holomorphic functions to the Riemann Sphere, j-invariant of a complex torus (or) j-invariant of an algebraic elliptic curve
Lecture 41
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Fundamental Region in the Upper Half-Plane for the Elliptic Modular J-Invariant
Goals: - To show that there exists a complex torus with j-invariant any prescribed complex number; in other words, to show that the j-invariant is surjective as a map onto the complex numbers - To use the functional equations satisfied by the weight two modular form as well as the mapping properties of that form, as studied in the previous unit of lectures, to find a suitable region in the upper half-plane where the mapping properties of the full modular form given by the j-invariant can be clearly studied Topics: Upper half-plane, quotient by the unimodular group, orbits of the unimodular group, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) weight two modular form, full modular function (or) full modular form, period two modular form, congruence-mod-2 normal subgroup of the unimodular group, projective special linear group with mod-2 coefficients, finite group, kernel of a group homomorphism, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, universal cover, neighborhood of infinity, lower half-plane, rational function, kernel of a group homomorphism, functional equations satisfied by the weight two modular form, j-invariant of a complex torus (or) j-invariant of an algebraic elliptic curve, Fundamental theorem of Algebra, complex field is algebraically closed, fundamental region for the full modular form, fundamental region for the unimodular group.
Lecture 42
Play Video
The Fundamental Region in the Upper Half-Plane for the Unimodular Group
Goals: - To introduce the notion of a fundamental region for a group-invariant surjective holomorphic map, for example for a holomorphic map that is invariant under the action of a subgroup of holomorphic automorphisms - To describe a suitable region in the upper half-plane and to show that it is a fundamental region for the unimodular group. Topics: Upper half-plane, quotient by the unimodular group, orbits of the unimodular group, representative of an orbit, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) weight two modular form, full modular function (or) full modular form, period two modular form, congruence-mod-2 normal subgroup of the unimodular group, projective special linear group with mod-2 coefficients, finite group, kernel of a group homomorphism, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, universal cover, neighborhood of infinity, lower half-plane, rational function, kernel of a group homomorphism, functional equations satisfied by the weight two modular form, j-invariant of a complex torus (or) j-invariant of an algebraic elliptic curve, Fundamental theorem of Algebra, complex field is algebraically closed, fundamental region for the full modular form, fundamental region for the unimodular group, ramified (or) branched covering, group-invariant holomorphic maps, fundamental region for a group-invariant holomorphic map, fundamental parallelogram associated to a lattice in the plane, fundamental region associated to the quotient map defining a complex torus.
Lecture 43
Play Video
A Region in the Upper Half-Plane Meeting Each Unimodular Orbit Exactly Once
Goals of the Lecture: - We prove the fact that a suitable region in the upper half-plane, which was described in the previous lecture and which was shown there to intersect each orbit of the unimodular group, meets each unimodular orbit at precisely one point. All this amounts to showing that the region is indeed a fundamental region for the unimodular group as claimed in the previous lecture Topics: Upper half-plane, quotient by the unimodular group, orbits of the unimodular group, representative of an orbit, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) weight two modular form, full modular function (or) full modular form, period two modular form, congruence-mod-2 normal subgroup of the unimodular group, projective special linear group with mod-2 coefficients, finite group, kernel of a group homomorphism, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, universal cover, neighborhood of infinity, lower half-plane, rational function, kernel of a group homomorphism, functional equations satisfied by the weight two modular form, j-invariant of a complex torus (or) j-invariant of an algebraic elliptic curve, Fundamental theorem of Algebra, complex field is algebraically closed, fundamental region for the full modular form, fundamental region for the unimodular group, ramified (or) branched covering, group-invariant holomorphic maps, fundamental region for a group-invariant holomorphic map, fundamental parallelogram associated to a lattice in the plane, fundamental region associated to the quotient map defining a complex torus.
Lecture 44
Play Video
Moduli of Elliptic Curves
Goals of Lecture 42: - To complete the proof of the fact that a suitable region in the upper half-plane, described in the previous lecture and shown there to be a fundamental region for the unimodular group, is also a fundamental region for the elliptic modular j-invariant function - In view of the above, we complete the proof of the theorem on the Moduli of Elliptic Curves: the natural Riemann surface structure, on the set of holomorphic isomorphism classes of complex 1-dimensional tori (complex algebraic elliptic curves) identified with the set of orbits of the unimodular group in the upper half-plane, is holomorphically isomorphic via the j-invariant to the complex plane Topics: Upper half-plane, quotient by the unimodular group, orbits of the unimodular group, representative of an orbit, invariants for complex tori, complex torus associated to a lattice (or) grid in the plane, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, automorphic function (or) automorphic form, weight two modular function (or) weight two modular form, full modular function (or) full modular form, period two modular form, congruence-mod-2 normal subgroup of the unimodular group, projective special linear group with mod-2 coefficients, finite group, kernel of a group homomorphism, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, universal cover, neighborhood of infinity, lower half-plane, rational function, kernel of a group homomorphism, functional equations satisfied by the weight two modular form, j-invariant of a complex torus (or) j-invariant of an algebraic elliptic curve, Fundamental theorem of Algebra, complex field is algebraically closed, fundamental region for the full modular form, fundamental region for the unimodular group, ramified (or) branched covering, group-invariant holomorphic maps, fundamental region for a group-invariant holomorphic map, fundamental parallelogram associated to a lattice in the plane, Galois theory, Galois group, Galois extension of function fields of meromorphic functions on Riemann surfaces, symmetric group, Galois covering.
IX. Complex 1-dimensional Tori are Projective Algebraic Elliptic Curves
Lecture 45
Play Video
Punctured Complex Tori are Elliptic Algebraic Affine Plane
Goals of the Lecture: - In this and the forthcoming lectures, our aim is to show that complex tori are algebraic, i.e., that they are actually elliptic algebraic projective curves. This is the reason that complex tori exhibit a rich geometry which involves a beautiful interplay between their complex analytic properties and the algebraic geometric and number theoretic properties of the elliptic curves they are associated to. It is a deep and nontrivial theorem that any compact Riemann surface is algebraic, so such Riemann surfaces exhibit a rich geometry as in the case of complex tori - Towards the above end, in this lecture we begin by identifying any punctured complex torus with a plane curve in complex 2-space. This plane curve is called the associated elliptic algebraic affine plane cubic curve. For this identification we make use of the Weierstrass phe-function associated to the complex torus, its derivative, their properties and the first order degree two cubic ordinary differential equation that they satisfy Topics: Upper half-plane, complex torus associated to a lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, triple pole (or) pole of order three, cubic equation, elliptic algebraic cubic curve, zeros of a polynomial equation, bicontinuous map (or) homeomorphism, open map, order of an elliptic function, elliptic integral, Argument principle, even function, odd function, analytic branch of the square root, simply connected, punctured torus, elliptic algebraic affine cubic plane curve, projective plane cubic curve, complex affine two-space, complex projective two-space, one-point compactification by adding a point at infinity
Lecture 46
Play Video
The Natural Riemann Surface Structure on an Algebraic Affine Nonsingular Plane Curve
Goals: - To show that the graph of a holomorphic function is naturally a Riemann surface embedded in complex affine 2-space - To use the Implicit Function Theorem to show that the zero locus of a nonsingular polynomial in two complex variables is naturally a Riemann surface embedded in complex affine 2-space - To show that the elliptic algebraic affine cubic plane curve associated to a punctured complex torus, as described in the previous lecture, has a natural Riemann surface structure which is holomorphically isomorphic to the natural Riemann surface structure on the punctured complex torus (inherited from the complex torus) Topics: Upper half-plane, complex torus associated to a lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, triple pole (or) pole of order three, cubic equation, elliptic algebraic cubic curve, zeros of a polynomial equation, bicontinuous map (or) homeomorphism, open map, order of an elliptic function, elliptic integral, Argument principle, even function, odd function, analytic branch of the square root, simply connected, punctured torus, elliptic algebraic affine cubic plane curve, projective plane cubic curve, complex affine two-space, complex projective two-space, one-point compactification by adding a point at infinity, Implicit function theorem, graph of a holomorphic function, nonsingular (or) smooth polynomial in two variables, zero locus of a polynomial, solving an implicit equation locally for an explicit function, nonsingular (or) smooth point of the zero locus of a polynomial in two variables, Hausdorff, second countable, connected component, nonsingular cubic polynomial, discriminant of a polynomial, cubic discriminant.
Lecture 47
Play Video
Complex Projective 2-Space as a Compact Complex Manifold of Dimension Two
Goals: - In Part A (Lec 47) of this lecture, we define complex projective 2-space and show how it can be turned into a two-dimensional complex manifold. In Part B (Lec 48), we show that any complex torus is holomorphically isomorphic to the natural Riemann surface structure on the associated elliptic algebraic cubic plane projective curve embedded in complex projective 2-space Topics: Upper half-plane, complex torus associated to a lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, triple pole (or) pole of order three, cubic equation, elliptic algebraic cubic curve, zeros of a polynomial equation, bicontinuous map (or) homeomorphism, open map, order of an elliptic function, elliptic integral, Argument principle, even function, odd function, analytic branch of the square root, simply connected, punctured torus, elliptic algebraic affine cubic plane curve, projective plane cubic curve, complex affine two-space, complex projective two-space, one-point compactification by adding a point at infinity, Implicit function theorem, graph of a holomorphic function, nonsingular (or) smooth polynomial in two variables, zero locus of a polynomial, solving an implicit equation locally for an explicit function, nonsingular (or) smooth point of the zero locus of a polynomial in two variables, Hausdorff, second countable, connected component, nonsingular cubic polynomial, discriminant of a polynomial, cubic discriminant, homogeneous coordinates on projective 2-space, punctured complex 3-space, quotient topology, open map, complex two-dimensional manifold (or) complex surface, complex one-dimensional manifold (or) Riemann surface, complex coordinate chart in two complex variables, holomorphic (or) complex analytic function of two complex variables, glueing of Riemann surfaces, glueing of complex planes, zero set of a homogeneous polynomial in projective space, degree of homogeneity, Euler's formula for homogeneous functions, homogenisation, dehomogenisation, a complex curve is a real surface, a complex surface is a real 4-manifold
Lecture 48
Play Video
Complex Tori are the same as Elliptic Algebraic Projective Curves
Goals: - In Part A of this lecture (Lec 47), we defined complex projective 2-space and showed how it can be turned into a two-dimensional complex manifold. In Part B (Lec 48), we show that any complex torus is holomorphically isomorphic to the natural Riemann surface structure on the associated elliptic algebraic cubic plane projective curve embedded in complex projective 2-space Topics: Upper half-plane, complex torus associated to a lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, triple pole (or) pole of order three, cubic equation, elliptic algebraic cubic curve, zeros of a polynomial equation, bicontinuous map (or) homeomorphism, open map, order of an elliptic function, elliptic integral, Argument principle, even function, odd function, analytic branch of the square root, simply connected, punctured torus, elliptic algebraic affine cubic plane curve, projective plane cubic curve, complex affine two-space, complex projective two-space, one-point compactification by adding a point at infinity, Implicit function theorem, graph of a holomorphic function, nonsingular (or) smooth polynomial in two variables, zero locus of a polynomial, solving an implicit equation locally for an explicit function, nonsingular (or) smooth point of the zero locus of a polynomial in two variables, Hausdorff, second countable, connected component, nonsingular cubic polynomial, discriminant of a polynomial, cubic discriminant, homogeneous coordinates on projective 2-space, punctured complex 3-space, quotient topology, open map, complex two-dimensional manifold (or) complex surface, complex one-dimensional manifold (or) Riemann surface, complex coordinate chart in two complex variables, holomorphic (or) complex analytic function of two complex variables, glueing of Riemann surfaces, glueing of complex planes, zero set of a homogeneous polynomial in projective space, degree of homogeneity, Euler's formula for homogeneous functions, homogenisation, dehomogenisation, a complex curve is a real surface, a complex surface is a real 4-manifold | 14,203 | 66,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.751131 |
https://www.physicsforums.com/threads/total-internal-reflection-and-refraction-index-in-optical-fiber-and-speed-of-light.300979/ | 1,532,321,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00353.warc.gz | 976,379,103 | 14,751 | # Total internal reflection and refraction index in optical fiber and speed of light
1. Mar 19, 2009
### ATH500
I did an experiment on the speed of light with different length of optical fibers.
I know that the fiber has a refraction index of 1.49.
I would like to know if I need to take in account the total internal reflection of the optical fibers too.
Because when I use long fibers like for exemple 50 meters or 20 meter, it shows value
inferior to the speed of light even after having multiplied the refraction index of 1.49.
I get 218x10^6 m/s or 198 x 10^6....
But when I use optical fiber of short length, for example when I compare the difference between a 10 meters optical fiber and a 0.60 meter one, I got a result that is almost the same as the speed of light: 291326041 meters/second.
I will post the results I got when doing the experiment with the optical fibers and the oscilloscope.
Here are four results for a length of 10 meters:
Here is the result for 20 meters:
For 50 meters:
For 0.60 meter:
For 0.615 meter:
I would like to know why do I get value of the speed of light that are inferior the more the length of the fiber increase.
Thank you,
Pascal
2. Mar 19, 2009
### Oerg
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
I think you do need to take into account of the total internal reflection since this means that the ray travels for a longer distance than the length of the fiber.
3. Apr 1, 2009
### ATH500
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
4. Apr 1, 2009
### Danger
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
Keep in mind that I'm no expert. The speed of individual photons, however, is always c. It's the interactions with atoms that slows down the propagation speed in a medium. It would make sense, then, that a longer path involves more interactions.
5. Apr 1, 2009
### ATH500
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
Are you sure that's how it works ? I thought that the refraction index took care of all this and that it already take in account the interactions inside the same material. Because my optical fiber is made of the same material all along it must the refraction index should not change because every particle of that material already slows down 1.49 time the speed of light.
I wonder if the longer it gets the more impurities it has and if those impurities can slow it down...
6. Apr 2, 2009
### Danger
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
No, I'm not; that's why I told you to keep in mind that I'm no expert. I was just giving you my best high-school drop-out's guess about the situation.
7. Apr 2, 2009
### ATH500
Re: Total internal reflection and refraction index in optical fiber and speed of ligh
It's a red LED input flashing at 100Hz. And when the receptor perceives light from the LED, it drops in voltage and we can see the effect on the oscilloscope. The LED correspond to the yellow graph and the receptor to the blue graph. I wonder if the impurities or the total internal reflection in the fiber has an effect. | 779 | 3,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-30 | latest | en | 0.89265 |
http://www.antonine-education.co.uk/Pages/Physics_2/Waves/WAV_01/Waves_Page_1.htm | 1,512,977,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00471.warc.gz | 321,650,139 | 12,866 | # Waves Tutorial 1 - Progressive Waves
Waves are caused by oscillations. Oscillations are complete to-and-fro movements, of which vibrations are one example. Another example is the oscillation of electrons, which cause radio waves. We will study oscillations in more detail with simple harmonic motion.
Write down three different examples of oscillations.
Waves occur when a disturbance at the source of the wave causes particles to oscillate about a fixed central point. There is a maximum displacement from the central point, which is called the equilibrium position, or average level. When particles reach that maximum displacement, they start to move towards the central point. They pass through the central point as they move to the maximum displacement on the other side.
Question 2 What are the features of wave motion?
We can show this on a water wave. The particles of water oscillate up and down from the equilibrium position. The wave is travelling from left to right. P is going down, Q is at the maximum displacement, and R is going up.
The wave is called a progressive wave because it is moving in a particular direction. It is transferring energy from the point of disturbance, but the particles are not travelling with the wave, merely going up and down.
Question 3 How is energy transferred if particles do not travel with the wave?
Ripples in solid materials like sand are NOT waves. They are caused by the piling up of sand in the wind. They do not oscillate.
Waves can be considered to travel either as plane wave-fronts, from a plane source or as circular wave-fronts from a point source:
In 3 dimensions, the waves would propagate spherically from a point source.
##### Terms Used with Waves
Displacement of a particle is the distance at any given moment from the central or equilibrium position, i.e. the undisturbed position. It is given the Physics Code s or x, and the SI unit is metre (m). The displacement decreases the further the wave progresses from its source.
• Intensity of waves at a point is the power per unit area at that point. The energy of a wave increases as the square of its amplitude. However the energy decreases as the square of the distance from the source, which is known as the inverse square law. The physics code for intensity is I and the units are watts per square metre (W m-2).
• Amplitude of a wave, code A or r, units metres (m), is the maximum displacement of a particle from its equilibrium position. In other words it is the height of the wave from the average level. It is NOT the height from crest to trough. (NB: Be careful of the code. Here amplitude is given the code A, but in many texts you will see a. This could be confused with acceleration.)
• Wavelength is defined as the distance between any two points on adjacent cycles that are in phase, in other words the distance between adjacent peaks or troughs. The code for wavelength is l (lambda, a Greek letter ‘l’). The units for wavelength are metre (m).
• Frequency, code f, has the unit hertz (Hz), and is the number of waves passing a given point every second.
• Period is the time taken for one complete oscillation. The code is T and the units seconds (s). Frequency is the reciprocal of period and is related to period by the simple equation:
• Wave velocity, code v, units metres per second (m s-1), tells us the speed of propagation of the wave, i.e. how fast it travels. For water waves this is a few cm s-1. In air, sound waves propagate at 340 m s-1. For light the speed is 3 x 108 m s-1. The speed of light is given the code c.
• Mechanical waves are produced by a disturbance in a material, or a medium, and can be longitudinal or transverse. Mechanical waves need a medium or material to travel in. In electromagnetic waves the disturbances are in the form of oscillating electrical and magnetic fields. They are always transverse. Electromagnetic waves can travel in a vacuum.
• The phase of a particle is the fraction of the cycle a particle has passed through relative to a given starting point. We describe the difference in the motion of particles in terms of the phase difference. This is the fraction of a wavelength by which their motions are different.
• The path difference between two waves is the number of cycles difference there is in the distance they have to travel.
Graphical Representation of Waves
We can show the features of waves in two ways:
• A displacement - displacement graph which shows how the wave varies in space.
• A displacement - time graph which shows how the wave varies with time.
The displacement - displacement graph shows the physical features of a wave:
The displacement - time graph shows how the displacement varies with time and it is like this:
You need to note these points:
• The displacement time graph does not show the physical features of the graph.
• The wavelength should not be confused with the period.
• It is not correct to call the positive amplitude a crest, not the negative amplitude a trough.
Alternating currents vary sinusoidally like the wave above.
Question 4
What do you understand by these terms? Displacement Period Frequency
Question 5
On a copy of the picture below, add the following features:
• Wavelength
• Amplitude
• Crest
• Trough
• Direction of disturbance.
The Wave Equation
The frequency, speed, and wavelength of any wave can be linked by the simple equation:
v = fl
where v is the speed of the wave in m s-1, f is the frequency in Hz, l is the wavelength in m.
When we use this equation, we do not always have to use SI units, but it is important to be consistent.
Worked Example What is the frequency of water waves of wavelength 4 cm travelling at a speed of 1.6 m s-1? Answer Formula first: v = fl Rearrange: f = v/l f = 1.60 m s-1 ÷ 0.04 m = 40 Hz (Note that the 4 cm was converted to 0.04 m to keep the units consistent. The speed (1.6 m s-1) could have been changed to 160 cm s-1)
The wave equation is used for longitudinal and transverse waves.
Note that when we use the equation for light or radio waves, we use the code c for the speed of light - c = 3.0 ´ 108 m s-1. So the equation is written c = fl.
Question 6 A ripple tank dipper makes 8 water waves in a time of 2 s. When it is just about to make the 9th wave, the first wave has travelled 48 cm from the dipper. (a) What is the frequency of the waves? (b) What is the wavelength of the waves? (c) What is the wave speed?
Energy in a Wave
If there is a bad storm, the news reports show spectacular waves smashing against the coastline. Often there are shots of the damage that has been done to sea walls and other buildings. This is because the wind disturbs the water to make high amplitude waves.
The energy is related to the amplitude of a mechanical wave:
This can be converted into an equation:
Where:
• E - energy (J);
• k - some constant (J m-2);
• A - amplitude (m).
Therefore, when the amplitude doubles, the energy carried by the wave goes up by four times.
A more detailed treatment is studied at university level, but the energy in waves along a string can be found HERE.
The physics code A can stand for amplitude or area. Be careful that you know the context in which it's being used. The photon energy in an electromagnetic wave is given by E = hf. Amplitude is not involved.
Intensity
Intensity is defined as:
Energy per unit area
The physics code is I and the units are joule per square metre (J m-2). The equation is:
The energy has a maximum value, often written I0, at the source. As waves propagate, they spread out. For each doubling of radius, the intensity goes down by 4 times:
This is called the Inverse Square Law. The equation for the inverse square law is:
This is true for all waves.
Phase
When a wave is travelling, all the particles are in continuous motion. The different particles have different displacements, velocities and directions. Indeed this is true even of adjacent particles. The phase of a particle is the fraction of the cycle a particle has passed through relative to a given starting point. We describe the difference in the motion of particles in terms of the phase difference. This is the fraction of a wavelength by which their motions are different.
Consider the two particles X and Y X is at the trough of a wave, whereas Y is at the crest. Their directions are upwards and downwards respectively. They are half a wavelength (l/2) out of phase. By linking oscillation to rotary movement, we can also describe X and Y as being 180o or p radians out of phase. We say that these particles are in antiphase.
W and Z are one wavelength, 360o or 2p radians apart. They are both at the starting point of a cycle. Their motion, including displacement, velocity and direction, is identical. We can therefore say that they are in phase. Particles can be any amount out of phase.
If we have two waves, we can measure their path difference. If the waves have a path difference of 1 wavelength (or any other whole number of wavelengths), they are in step or in phase. Waves with a path difference of 1/4 wavelength have a phase difference of 1/4 of a cycle (90o or p/2 rad). If they have a path difference of 1/2 a wavelength, the waves are in antiphase. Path difference is important when we analyse how waves superpose.
On the picture below, draw a second wave that is lagging the first wave by 90o, i.e. it’s behind the first wave. Draw another wave that is p radians out of phase. Is it leading or lagging?
The phase can be calculated for any points on the wave that are d metres apart, using the relationship:
The term d is the distance between the two points. The curious symbol f is ‘phi’, a Greek letter ‘f’, used as Physics code for phase angle.
The angle in this relationship is in radians, where 2p rad = 360 o. Hence 1 rad ≈ 57 o
When you use radians, you must ensure that your calculator is set to radians. The easiest way to lose marks is put your angle in radians while you are set to degrees… It is up to you to make sure how to work your calculator.
Radians is a dimensionless unit, and some people leave it out. In these notes I will always use the shorthand rad
The graph is a summary of phase relationships:
Modelling Waves (Extension)
Questions that use this concept are not likely to be asked in AS exams. It is entirely possible that they will come up in A-level.
The simplest type of wave is called a sine wave. This is because the displacement varies with the sine of the time. The equation is shown below:
The terms are:
• x - the displacement (m);
• A - the amplitude (m) which is the maximum displacement;
• w - the angular velocity (rad s-1);
• t - the time (s).
Note that x is often used for the displacement in waves. The code s can be used as well. The symbol w is omega, a Greek lower case letter long 'o' (ō). It represents the frequency of the wave, and is linked to the frequency by the equation:
So we can also write:
We can use an Excel spreadsheet (other spreadsheets are available) to model the sine waves. You can try it for yourself if you are a dab hand at spreadsheets.
Here is a screen shot with the formulae:
This spreadsheet model has a time interval of 0.01 s for a time period of 1.0 s. The sine function in Excel works with angles in radians. It does not work with degrees. Here is the sine wave produced with the model:
Velocity of the particles that form the wave can be worked out using the gradient of the sine wave. In calculus notation, we can write:
Maths Window The derivatives of trigonometrical functions are as follows: And We won't look at the tangent functions here. If there is a constant being processed by the function, then the derivative is multiplied by the constant. In this case, we'll make the constant B. Therefore: And:
For acceleration, we can write:
The minus sign tells us that the acceleration is towards the average level (zero displacement).
A wave has an amplitude of 0.25 m and a frequency of 6.0 Hz. Calculate: a. The angular velocity; b. The displacement at 1.1 s; c. The velocity at 1.1 s; d. The acceleration at 1.1 s.
Sine waves are very closely related to:
• Circular motion;
• Simple harmonic motion.
Both of these are in the second year A level and will be discussed in Further Mechanics. | 2,859 | 12,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2017-51 | latest | en | 0.910735 |
https://math.stackexchange.com/questions/1792139/how-many-words-can-be-formed-given-4-letters-and-in-each-word-there-must-be | 1,581,955,867,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875142603.80/warc/CC-MAIN-20200217145609-20200217175609-00048.warc.gz | 481,317,747 | 30,906 | # How many words can be formed, given $4$ letters, and in each word there must be at least two letters are the same?
How many words can be formed, given $4(a,b,c,d)$ letters, and in each word from $4$ letters there must be at least two letters are the same? The position of the letter doesn't matter. The answer is $232,$ I don't know how to "attack" it.
• How long are the words? – almagest May 19 '16 at 20:16
• 4 letters, forgot the add this. – Planet_Earth May 19 '16 at 20:16
• What do you mean "position doesn't matter"? Is "abbc" and "abcb" equal? – Fred Yang May 19 '16 at 20:24
• There are $4^4$ possible words in total. There are $4!=24$ words with no repeats. Hence $256-24=232$ with at least one repeat. – almagest May 19 '16 at 20:25
• @almagest, give this as answer so I can mark it as best. – Planet_Earth May 19 '16 at 20:26
There are $4^4=256$ possible words in total. There are $4!=24$ words with no repeats. Hence $256−24=232$ with at least one repeat. | 309 | 974 | {"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} | 3.640625 | 4 | CC-MAIN-2020-10 | latest | en | 0.923172 |
https://byjus.com/question-answer/which-of-the-following-is-a-true-statement-about-quadrilaterals-2/ | 1,721,357,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00169.warc.gz | 129,241,785 | 22,682 | 1
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In a quadrialateral, all angles are right angles.
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Solution
## The correct option is D A quadrilateral has 4 sides.The first option is, "A quadrilateral has 4 sides". For a closed figure to be a quadrilateral, it should have 4 sides'. Hence, this is true. The second option is, "All sides are equal in a quadrilateral". Only squares and rhombus have all sides equal. Hence, this is not true. The third option is, "In a quadrilateral, all angles are right angles". In parallelograms other than squares and rectangles, adjacent angles will vary. Hence, this is not true. The last option is, "A quadrilateral has 5 sides". This is not true, since, all quadrilaterals have 4 sides.
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# Solving 2 Step Equations Game Activity for 6th, 7th, 8th, 9th Grade
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Two Step Equations: These two step equations cootie catchers are a great way for students to have fun while they practice solving two step equations. How to Play and Assembly Instructions are included.
Two Step Equations Cootie Catchers Contents:
There are 2 cootie catchers in this product, each one having 8 problems for a total of 16 problems. Each cootie catcher contains problems where the student must either add, subtract, multiply or divide. Step by step answers are included.
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
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Vlanemg_cows
Registered: 05.07.2003
From:
Posted: Friday 29th of Dec 11:44 Hello there I have almost taken the decision to look fora algebra private teacher, because I've been having a lot of stress due to math homework this year. Every day when I come home from school I waste all the afternoon with my algebra homework, and in the end I still seem to be getting the wrong answers. However I'm also not completely sure whether a algebra tutor is worth it, since it's very costly , and who knows, maybe it's not even so good . Does anyone know anything about "college algebra worksheets"+"binomials" that can help me? Or maybe some explanations about binomials,percentages or function range? Any ideas will be much appreciated .
AllejHat
Registered: 16.07.2003
From: Odense, Denmark
Posted: Sunday 31st of Dec 08:28 Once upon a time those were the only ways out. But thanks to technology today , we have something known as the Algebrator! It’s an easy to use software , something which even a newcomer would enjoy working on and the best part is that it would solve all your questions and also explain the steps it took to reach that answer! Isn’t that just great ? Well I think it is, and I’m sure after , you’ll be on the same page as me.
3Di
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Posted: Monday 01st of Jan 13:02 Algebrator is very useful, but please never use it for copy pasting solutions. Use it as a guide to understand and clear your concepts only.
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Posted: Wednesday 03rd of Jan 09:43 equation properties, geometry and system of equations were a nightmare for me until I found Algebrator, which is really the best algebra program that I have ever come across. I have used it through several math classes – Algebra 1, Basic Math and Remedial Algebra. Just typing in the algebra problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I truly recommend the program.
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Posted: Thursday 04th of Jan 12:05 Sounds like something I need to purchase right away. Any links for ordering this program over the internet?
Jot
Registered: 07.09.2001
From: Ubik
Posted: Saturday 06th of Jan 08:42 You can get all the details about the program here http://www.mathfraction.com/equivalent-fractions-1.html.
Forum | 795 | 3,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-13 | latest | en | 0.953205 |
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### Introduction
The Correlation Matrix Calculation window calculates and visualizes the rank order correlation matrix of a data set.
The correlation matrix contains Spearman's rank order coefficient (also known as rho) for each pair of datasets.
It is symmetric because correlation between A and B is the same as correlation between B and A. It's elements lie in the [-1,1] interval.
As the correlation of a variable with itself is 1, the diagonal elements of the matrix are all equal to 1.
The output of the function is an nxn array where n is the number of variables in the data set.
To see the output functions of this window, click here.
### Window Elements
In the location field you can indicate the range of spreadsheet cells that contain the data. You can specify whether these are orientated in columns (selected by default, as this is usually the case) or rows. The number of columns (respectively rows) is the n mentioned above.
The correlation matrix of the data is shown. Its elements are the pairwise correlation of each of the variables within the dataset.
Optionally, in the Labels field you can specify where the labels of the dataset are in the spreadsheet. If no labels are selected, the datasets will be named Var1, Var2, etc..
In the Output location field you can specify where the correlation matrix should be placed in the spreadsheet. It will be inserted there upon pressing the OK button.
The number of data values for each variable (source points) is displayed below the graphs.
Selecting any white square in the correlation matrix will generate a scatter plot of the data for the corresponding two variables. You can choose whether to display the actual data values or the percentiles. In the first case, the horizontal and vertical axes are adjusted to the range of the data. When showing percentiles, the range of the axes is [0,1].
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##### Submissions: 1037 Accuracy: 59.73% Difficulty: Basic Marks: 1
Given a number N, the task is to find XOR of count of 0s and count of 1s in binary representation of a given number.
Input:
The first line of the testcase contains an integer T, denoting number of testcases. The next T lines contains one integer N each.
Output:
Print XOR of count of 0s and 1s in binary representation of the given number in separate lines
Constraints:
1<=T<=10^5
1<=N<=10^9
Example:
```Input : 5
Output : 3
Binary representation : 101
Count of 0s = 1,
Count of 1s = 2
1 XOR 2 = 3.
Input : 7
Output : 3
Binary representation : 111
Count of 0s = 0
Count of 1s = 3
0 XOR 3 = 3.```
#### ** For More Input/Output Examples Use 'Expected Output' option **
Author: ShivamKD
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Need help with your code? Please use ide.geeksforgeeks.org, generate link and share the link here.
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I'm not sure how to do part a) and b). For a) I don't get how the hint helps.
Eg, I did $\displaystyle P_1P_2 = \frac{1}{||a||^2}aa^T\frac{1}{||b||^2}bb^T$ and we know that a.b = 0 but there are no dot products in the former equation...
Cheers!
2. Originally Posted by usagi_killer
I'm not sure how to do part a) and b). For a) I don't get how the hint helps.
Eg, I did $\displaystyle P_1P_2 = \frac{1}{||a||^2}aa^T\frac{1}{||b||^2}bb^T$ and we know that a.b = 0 but there are no dot products in the former equation...
In fact, there is a (disguised) dot product in that equation, because $\displaystyle a^{\textsc t}b$ is just another way of writing $\displaystyle a.b$. | 240 | 702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-22 | latest | en | 0.929054 |
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## 1678 Reputation
19 years, 150 days
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I am a research associate at Simon Fraser University and a member of the Computer Algebra Group at the CECM.
## Absolutely...
Use the Gcdex command mod n to compute inverses in your ring. For example, the Maple commands below assign the inverse of x^2+x+1 in Z[x]/(x^4-1) mod 11 to the variable s.
```n := 11;
r := x^4-1;
f := x^2+x+1;
g := Gcdex(f, r, x, 's', 't') mod n;
Rem(s*f, r, x) mod n;
```
The Gcdex command runs the extended Euclidean algorithm to compute the gcd along with cofactors s and t with f*s + r*t = g mod n. Note that if the gcd is not 1 then your element is a zero-divisor and can't be inverted. The Rem command divides by r, producing a canonical representative for elements of your ring. To do these computations in characteristic zero, use the lower-case gcdex and rem commands.
## Vector graphics software...
You need a good vector graphics program, like Adobe Illustrator. Other programs are listed here. Note that there are some good free programs, like Inkscape, and Xara Xtreme has been opened up recently as well. One problem that you will find is that eps sucks, and Maple really needs to export SVG. For 3D plots your options are somewhat better. Maple can export POVRay files and Autocad DXF, however both will require you to re-render the plot in one of those programs.
## multiplying brackets...
One problem is that you need a multiplication sign between pairs of brackets. Maple interprets `(f)(g)` as the function `f` with argument `g`. Assuming your system is a set of rational functions, the numerators have the following sets of solutions: ```{{d121 = 1/b, d111 = 0, c = 0, d112 = 0, d122 = 0, d212 = 0, d221 = 0, s212 = 0, d222 = 1/b, d211 = 1/b, b = b}, {d221 = 0, a = 0}, {c = 0, d221 = 0, a = 0}, {d112 = -1/4*d111^2/d121, a = a, d111 = d111, d121 = d121, b = 1/d121, d122 = 0, d212 = 0, d221 = 0, s212 = 0, d222 = d121, c = -1/2*a*d111/d121, d211 = d121}, {d221 = 0, a = 0, b = 0}, {c = 0, a = 0, b = 0}} ```
## Groebner bases and the solve command...
The easiest way to solve a system of equations is to use the solve command. For example:
```sys := { 3*x1 + 2*x2 + x3^2, x2 - x3 + x4^2 };
solve(sys);
```
Try this first. Because many of your equations are linear, it should work. If it doesn't work (ie: it takes a long time) then you might try computing a Groebner basis. Groebner bases are a generalization of Gaussian elimination to nonlinear systems in multiple variables. In Maple 10 (only) you can do the following:
```sys := {2*x1 + 3*x2 + ....}; # your system
G := PolynomialIdeals:-GroebnerBasis(sys, 'tord', 'plex'): # compute lex basis
solve(G);
```
This will compute a Groebner basis using lexicographic order, automatically ordering the variables and so on. This basis G is much easier to solve, and the solve command should have no trouble with it. You can use this technique to solve moderately hard polynomial systems in Maple 10. In earlier versions of Maple Groebner basis computations are not as robust, and you have to choose the variable order. A good rule of thumb is to put the variables that appear in the highest degrees last in the order. So for example, if x1, x2, x3, ..., x14 are linear variables and x15 and x16 appear with degree 2, then put x15 and x16 last. The syntax to compute a Groebner basis is:
```G := Groebner[gbasis](sys, plex(x1,x2,x3,...,x15,x16));
solve(G);
```
Hopefully something in here works for you. If not, you can post your system and I'll try to solve it.
## Maple lists and vectors...
Here are some examples to get you started. L := [0, 0.02, 0.5]; This creates a Maple list and assigns it to the variable L. Lists in Maple are actually stored internally as arrays (they are not linked lists), so accessing elements is constant time. The problem with lists is that you can't efficiently reassign elements one by one - it rebuilds the list each time. If you want to change all of the elements at once, you can use a seq or the map command and it will be efficient. In the example below, we will square all the elements of L, first by mapping a function onto L, then by using a sequence. L2 := map( x->x^2, L); L2 := [seq(i^2, i=L)]; L2[2]; # access second element Because you can't modify lists in place they are not appropriate for general linear algebra. Maple uses Vector and Matrix, which construct 1D or 2D arrays. Unlike lists, you can modify these structures in place. V := Vector(L); # make a Vector out of the list V := Vector([0, 0.02, 0.5]); # input a Vector directly A := Matrix([[1, 3, 0], [2, 0, 4]]); # make a Matrix A[1,3] := 2; # reassign element in row 1 column 3 A; # print You can also use the map command to apply functions to every element of a Vector or Matrix. If you are coming from Matlab probably the most important thing for you is using fast hardware floats. The Vector and Matrix above can contain any type of Maple object, for example, exact rational numbers or polynomials. Maple will detect when floating point numbers are present and use dedicated floating point routines, but that involves the overhead of checking every element in the matrix or vector. You can avoid this by declaring a datatype, as shown below. V := Vector(L, datatype=float[8]); When Maple encounters matrices and vectors with float[8] datatype, it doesn't waste any time doing anything. It just calls a compiled routine from the NAG (Numerical Algorithms Group) Library. You should get performance similar to Matlab for dense matrices. There is also some support for sparse matrices (in NAG format, with float[8]). I think there is an iterative algorithm for solving systems of full rank.
## Emacs...
Joe Riel has this for Emacs: http://www.k-online.com/~joer/maplev/maplev.html
## you can't make a progress bar...
There is no way to make a progress bar like how you want, because Maple's output method is designed around printing lines of text or entire expressions at once. What I have found useful is to have an inner loop print different characters (- and + work well) for different types of things that can happen, and have an outer loop break the lines. You can do this using printf("-"); to print the characters and printf("\n"); to break a line. This works especially well for double loops containing an if statement. Another possibility is to print some sort of count and add indents for recursions.
## we lost some of your procedure...
We lost some of your code, due to the fact that html interprets < and > as tags and nukes anything in between. One way to get around that is to select "plain text" as the input format when you post your comment. Can you please post your procedure again ?
## Eigenvalues and Eigenvectors....
It's worth pointing out that your problem may have no solution, unless 1 is an eigenvalue of M. In that case the solution v is the corresponding eigenvector. You can compute eigenvalues and eigenvectors using commands in the LinearAlgebra package. Type with(LinearAlgebra); to load the package, and ?Eigenvalues to see help on the Eigenvalues command. For background information, consult any textbook on linear algebra. A number of free textbooks are available online, such as this one.
## This is not an easy problem,...
This is not an easy problem, to be sure. Your best bet is to try to express the function as a piecewise function. Maple might be able to do this for you, ie: convert(foo, piecewise), I'm not sure. In any case, how you proceed will depend strongly on what the "patches" are. You should try the Optimization package (added in Maple 9.5, improved in Maple 10).
## solution for polynomials...
If the equations are all polynomials, do the following:
```kernelopts(opaquemodules=false); # access internal commands
eqns := {x1 = f1(theta1, theta2), x2 = f2(theta1, theta2), ...}: # set of equations
X := indets(map(lhs, eqns)); # should be {x1,x2,...}
U := indets(map(rhs, eqns)); # should be {theta1, theta2}
with(PolynomialIdeals):
J := < map(lhs-rhs, eqns) > ; # construct an ideal
V := [PolynomialIdeals:-SuggestVariableOrder(J, X union U)];
X := select(member, V, X); # order the variables optimally
U := select(member, V, U);
infolevel[':-GroebnerBasis'] := 4: # monitor computation
PolynomialIdeals:-GroebnerBasis(J, lexdeg(X,U), method='bb'): # precompute an easy GB
K := EliminationIdeal(J, {op(X)}); # use Groebner Walk to eliminate U and keep X
result := Generators(K);
```
I apologize - it's a little overcomplicated - but given the structure of your equations it should be the fastest possible way. If the result is {0} then theta1 and theta2 can't be eliminated from the polynomials. If the result is {1} then your original equations had no solution.
## Non-commutative Groebner...
Maple can do it, provided your non-commutative relations have the following form: D[i]*x[j] = sigma[i](x[j])*D[i] + delta[i](x[j]) where delta[i](p*q) = sigma[i](p)*delta[i](q) + delta[i](p)*q. This covers many common cases, such as differential algebras and various shift algebras. See the help page for Ore_algebra for details. Maple can not handle arbitrary non-commutative relations between the variables, but such computations usually aren't feasible anyway. If you need that, try the program Bergman (http://servus.math.su.se/bergman/). Otherwise, let me know if you need a hand getting Ore_algebra and Groebner working for you. There is a non-commutative example on the help page ?Groebner,Basis_details.
## 10.3.9 is broken...
You aren't doing anything wrong. Apple broke most Java applications in OS X 10.3.9 with a Quicktime update six months ago. It caused a big uproar so they released an uninstaller. The issue still exists in Quicktime 7.1, but this time there is no uninstaller because it was a big security update. So as unbelievable as it seems, OS X 10.3.9 is broken (apps don't run!). The Apple discussion boards are full of people who had to hack their systems, reinstall 10.3.9, or upgrade to 10.4 to get their programs to run again. Apple shows no intention of fixing it. Maplesoft however wrote a new installer for Maple 10.04 which works on an updated 10.3.9 system. You will have to contact Maplesoft support and download Maple 10.04. As a 10.3.9 user myself, I really appreciated that they went to this extra effort to support their customers on a broken platform.
## system properties...
Mostly out of curiousity, how large are your linear systems and how many vectors are in the right hand side ?
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https://math.answers.com/basic-math/What_would_6.87_be_in_mixed_numbers | 1,726,815,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00518.warc.gz | 342,983,051 | 47,305 | 0
# What would 6.87 be in mixed numbers?
Updated: 4/28/2022
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11y ago
6.87 = 687/100
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Q: What would 6.87 be in mixed numbers?
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Related questions
### What numbers go into 687?
1, 3, 229 and 687.
### In what situation would not be a good idea with estimating mixed numbers?
it would not be a good time to estimate mixed numbers is when both of the fractions are not a mixed number
### Desccride a situation in which estimating mixed numbers would not be good idea?
When both of the fractions are not even mixed numbers!
### Can a mixed numbers only have a whole number?
No, because then it would not be a mixed number.
### What is 5129 as a mixed number?
Whole numbers such as 5129 are not normally expressed as mixed numbers
### Can you have fraction in integers?
No. They would then be mixed numbers, not integers.
### How would you write 0.755 as a mixed number?
You can't. Mixed numbers are greater than one.
you dont
### When in life would you use mixed numbers?
all the time but not when baby
### Are decimals in mixed numbers?
No. Mixed numbers are a number and a fraction.
### What is 22 percent of 687?
22% of 687= 22% * 687= 0.22 * 687= 151.14
### What the definition for dividing mixed numbers?
what is the definition for dividing mixed numbers | 346 | 1,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.931821 |
http://math.stackexchange.com/questions/151936/finding-points-on-ellipse | 1,469,378,801,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00219-ip-10-185-27-174.ec2.internal.warc.gz | 159,111,128 | 18,124 | Finding points on ellipse
I have ellipse in 2D. I want to compute fixed number of points on this ellipse with constant angular separation between those points.
My first idea was to generate line equations from center of the ellipse and then solve equations of these lines with ellipse equation. But it's not efficient computationally. Second idea is to use equation in polar coordinates and vary $t$ from $0$ to $2\pi$ with overlap ${2\pi}/{NumberOfPoints}$.
What is the simplest (less computationally absorbing) method to get fixed ellipse points number?
-
How are you measuring the distance? Straight line, or along the ellipse? (Or angular separation?) – TonyK May 31 '12 at 9:01
Distance is measured by angular seperation. – krzych May 31 '12 at 9:10
Are you stuck on using the center as the point of reference as opposed to one of the focal points? In the latter case the equation takes a relatively simple form in polar coordinates. If the axes of the ellipse are not aligned with $x/y$-axes, then just subtract a constant from $\theta$. – Jyrki Lahtonen May 31 '12 at 11:05
I can't figure out x and y from these equation. Also some good graphic representation of this equation would be helpful. I found some but still can't obtain what I need. – krzych Jun 1 '12 at 6:56
If you insist on having equal angular increments, the equations are:
$$k = \frac{1}{\sqrt{ {b^2}\cos^2{\theta} + {a^2}\sin^2{\theta} }}$$
$$x = kab\cos{\theta}$$
$$y = kab\sin{\theta}$$
So, you have to compute a sine, a cosine, and a square root for each given value of $\theta$. Not too much for a typical modern computer, I would think.
But, since you are going to use equal increments of $\theta$, there's an old-time computer graphics trick that reduces the computation. Suppose we let $\delta\theta$ denote the angular increment. Then we know that
$$\cos(\theta + \delta\theta) = \cos\theta\cos\delta\theta - \sin\theta\sin\delta\theta$$
$$\sin(\theta + \delta\theta) = \sin\theta\cos\delta\theta + \cos\theta\sin\delta\theta$$
You pre-compute $\cos\delta\theta$ and $\sin\delta\theta$. Then you can compute each incremented value of $\cos(\theta + \delta\theta)$ and $\sin(\theta + \delta\theta)$ with just four multiplies and two adds.
- | 591 | 2,236 | {"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} | 3.828125 | 4 | CC-MAIN-2016-30 | latest | en | 0.904312 |
https://www.warsquid.com/blog/kmeans/ | 1,713,278,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00326.warc.gz | 971,754,297 | 3,484 | Colour reduction with k-means
Reducing an image down to its most dominant colours gives us a good sense of its unique colour palette, and can also produce some aesthetically-pleasing results.
To find the dominant colours in an image, we can use a technique called K-means clustering. We'll use the RGB colour space to represent colours, so we can think of each colour as 3 numbers representing the amount of red, green, and blue respectively1. This allows us to do things like calculate the distance from one colour to the next, or average the colours.
To find the K dominant colours in an image, the process is basically:
• Start with K randomly generated colours - we'll call these the centroids
• Group every colour in the image by matching it with the centroid it's closest to - we'll call these groups clusters.
• Calculate new centroids by finding the average colour in each cluster.
• Repeat the grouping and averaging process until the centroids stabilize.
(The RGB colour space doesn't necessarily match up well to how we perceive colours, the LAB space might do a better job.)
I've implemented this process below. Each time, we start with a random set of centroids. Click the "Update centroids" button to carry out one step of the K-means algorithm, which will automatically update the colour maps and images below. Keep clicking to see how To start over, click the "New random centroids" button (just mapping the images to these random centroids can give great-looking results on its own).)
We'll calculate colours from an original source image:
• Number of centroids (K):
The current centroids:
Their hues: (These are only approximate, we're trying to visualize 3 dimensional colours in 1 dimension, and when colours get close to black their hues are a bit arbitrary)
The image reduced down to its dominant colours:
Applying the same palette to another image:
(photo credit: Petersham Reservoir by OzRocky, CC-BY-SA) | 402 | 1,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.905121 |
https://jp.mathworks.com/matlabcentral/answers/1805130-fitting-a-hyperbola-through-a-set-of-points?s_tid=prof_contriblnk | 1,679,730,315,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00408.warc.gz | 376,607,731 | 32,024 | # Fitting a hyperbola through a set of points
26 ビュー (過去 30 日間)
Ravi Mudragada 2022 年 9 月 14 日
コメント済み: Ravi Mudragada 2022 年 9 月 15 日
I'm trying to fit a hyperbola through given data points. The code works fine for the first set of points but does not work for the other sets of points. The code based on this answer (https://in.mathworks.com/matlabcentral/answers/355943-how-can-we-fit-hyperbola-to-data#answer_280986) is below and the accompanying data is attached. Any suggestions would be highly encouraged!
clear all;
x1=data.P1;
y1=data.I1;
%x1=data.P2; % Uncomment to replicate issue
%y1=data.I2; % Uncomment to replicate issue
hyprb = @(b,x1) b(1) + b(2)./(x1+b(3)); % Generalised Hyperbola
NRCF = @(b) norm(y1 - hyprb(b,x1)); % Residual Norm Cost Function
B0 = [0, 0, 0];
%options = optimset('MaxFunEvals',800000);
B1 = fminsearch(NRCF, B0);%options); % Estimate Parameters
figure(1)
%plot(y1, x1, '+r')
%x2 = [0:1:30];
hold on
plot(y1, x1, 'ko', hyprb(B1,x1), x1, '-r')
xlabel('Impulse (MPa-msec)')
ylabel('Pressure (MPa)')
hold on
%plot(y1,y1-hyprb(B1,x1))
grid on
text(15, 10, sprintf('y = %.4f %+.4f/(x %+.4f)', B1))
hold off
%x0 = [0,100];
%[B,resnorm,residual,exitflag,output] = lsqcurvefit(hyprb,B0,x1,y1);
%plot(x1,hyprb(B, x1),'b-')
%grid on
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### 採用された回答
Torsten 2022 年 9 月 14 日
Remove the NaN value in the last position of data.P2 and data.I2:
clear all;
%x1=data.P1;
%y1=data.I1;
x1=data.P2; % Uncomment to replicate issue
x1 = x1(1:end-1);
y1=data.I2; % Uncomment to replicate issue
y1 = y1(1:end-1);
hyprb = @(b,x1) b(1) + b(2)./(x1+b(3)); % Generalised Hyperbola
NRCF = @(b) norm(y1 - hyprb(b,x1)); % Residual Norm Cost Function
B0 = [0, 0, 0];
%options = optimset('MaxFunEvals',800000);
B1 = fminsearch(NRCF, B0);%options); % Estimate Parameters
figure(1)
%plot(y1, x1, '+r')
%x2 = [0:1:30];
hold on
plot(y1, x1, 'ko', hyprb(B1,x1), x1, '-r')
xlabel('Impulse (MPa-msec)')
ylabel('Pressure (MPa)')
hold on
%plot(y1,y1-hyprb(B1,x1))
grid on
text(15, 10, sprintf('y = %.4f %+.4f/(x %+.4f)', B1))
hold off
%x0 = [0,100];
%[B,resnorm,residual,exitflag,output] = lsqcurvefit(hyprb,B0,x1,y1);
%plot(x1,hyprb(B, x1),'b-')
%grid on
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### その他の回答 (1 件)
William Rose 2022 年 9 月 14 日
The columns in the workbook are 12 long for P1, I1. The columns are 11 long for P2,I2.
When readtable() reads the workbook, it assigns NaN values to the final row of I2,P2. These Nan values are throwing off the fit. Do not include the last row of values in I2, P2, and it works fine.
%x1=data.P1;
%y1=data.I1;
x1=data.P2(1:end-1); % Uncomment to replicate issue
y1=data.I2(1:end-1); % Uncomment to replicate issue
hyprb = @(b,x1) b(1) + b(2)./(x1+b(3)); % Generalised Hyperbola
NRCF = @(b) norm(y1 - hyprb(b,x1)); % Residual Norm Cost Function
B0 = [0, 0, 0];
%options = optimset('MaxFunEvals',800000);
B1 = fminsearch(NRCF, B0);%options); % Estimate Parameters
figure(1)
%plot(y1, x1, '+r')
%x2 = [0:1:30];
hold on
plot(y1, x1, 'ko', hyprb(B1,x1), x1, '-r')
xlabel('Impulse (MPa-msec)')
ylabel('Pressure (MPa)')
hold on
%plot(y1,y1-hyprb(B1,x1))
grid on
text(15, 10, sprintf('y = %.4f %+.4f/(x %+.4f)', B1))
hold off
Are you aware that you are plotting x on the vertical axis and y on the horizontal axis? That is up to you, of course.
I would define the fitting funciton as
hyprb = @(b,x1) b(1) + b(2)./(x1-b(3)); % note that I changed the sign on b(3)
because if you do, then b(1) is the coordinate of the vertical asymptote and b(3) is the coordinate of the horizontal asymptote, and with this change, the equation for the hyperbola can be rewritten
(y-b1)(x-b3)=b2
which has a nice symmetry. That version of the formula is not useful in your code, but it is nice for explaining the equation.
Good luck.
##### 3 件のコメント表示非表示 2 件の古いコメント
Ravi Mudragada 2022 年 9 月 15 日
Did that! However your explanation of the issue helped a lot too. Much appreciated!
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Translated by | 1,505 | 4,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | longest | en | 0.353187 |
https://oeis.org/A005042 | 1,603,329,361,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878662.15/warc/CC-MAIN-20201021235030-20201022025030-00113.warc.gz | 451,101,463 | 4,694 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A005042 Primes formed by the initial digits of the decimal expansion of Pi. (Formerly M3129) 28
3, 31, 314159, 31415926535897932384626433832795028841 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The next term consists of the first 16208 digits of Pi and is too large to show here (see A060421). Ed T. Prothro found this probable prime in 2001. A naive probabilistic argument suggests that the sequence is infinite. [Michael Kleber, Jun 23 2004] REFERENCES M. Gardner, personal communication. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS M. Gardner, Letter to N. J. A. Sloane, Nov 16 1979. Ed T. Prothro, How I Found the Next Pi Prime Eric Weisstein's World of Mathematics, Pi-Prime FORMULA a(n) = floor(10^(A060421(n)-1)*A000796), where A000796 is the constant Pi = 3.14159... . - M. F. Hasler, Sep 02 2013 MAPLE Digits := 130; n0 := evalf(Pi); for i from 1 to 120 do t1 := trunc(10^i*n0); if isprime(t1) then print(t1); fi; od: MATHEMATICA a = {}; Do[k = Floor[Pi 10^n]; If[PrimeQ[k], AppendTo[a, k]], {n, 0, 160}]; a (* Artur Jasinski, Mar 26 2008 *) nn=1000; With[{pidigs=RealDigits[Pi, 10, nn][[1]]}, Select[Table[FromDigits[ Take[pidigs, n]], {n, nn}], PrimeQ]] (* Harvey P. Dale, Sep 26 2012 *) PROG (PARI) c=Pi; for(k=0, precision(c), isprime(c\.1^k) & print1(c\.1^k, ", ")) \\ - M. F. Hasler, Sep 01 2013 CROSSREFS See A060421 for further terms. Cf. A198018, A198019, A195834, A047777, A053013, A064467. Sequence in context: A118913 A297480 A282973 * A317482 A136582 A173649 Adjacent sequences: A005039 A005040 A005041 * A005043 A005044 A005045 KEYWORD nonn,base AUTHOR STATUS approved
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Last modified October 21 20:32 EDT 2020. Contains 337925 sequences. (Running on oeis4.) | 714 | 2,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.585219 |
https://nanonaren.wordpress.com/2014/08/25/hungarian-algorithm-part-ii/ | 1,529,761,990,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00103.warc.gz | 666,960,966 | 18,552 | ## Hungarian Algorithm (Part II)
View source file on Github
[part 1 is not necessary to follow this] When I first looked up the Hungarian algorithm on Wikipedia, I was immediately drawn away from what seemed like a tortured graphical explanation and went straight to the rather simpler matrix represented solution. I was tempted to lookup the explanation for the algorithm in the original paper but I resisted; I wanted to work out the reason for the steps in the algorithm myself; and – joy upon joy – the explanation is beautifully symmetrical.
## Symmetry
Let $S$ be the set of $n \times n$ matrices with non-negative entries. $S$ comprises of all the problem sets definable for $n$ tasks and $n$ people. Let $\mathcal{A}(s)$ be the cost of the minimal-cost assignment for matrix $s$. Let’s see if there is a transformation of $s \in S$ that doesn’t affect the solution.
Proposition Let $s \in S$ and $(\cdot)^T$ be the transpose. Then $\mathcal{A}(s) = \mathcal{A}(s^T)$.
Proof: If $(i_k,j_k)$ make up the indices of a minimal assignment, then $(j_k,i_k)$ make up those in $s^T$. Thus, $\sum_k s_{i_k j_k} = \sum_k s^T_{i_k j_k}$.
Proposition Let $s \in S$ and let $R(k,x) : S \rightarrow S$ for $1 \le k \le n$ and $x \in \mathbb{R}$ be the operation where $R(k,x)$ adds $x$ to row $k$. Then, $\mathcal{A}(s) = \mathcal{A}(R(k,x)(s)) - x$.
Proof In a minimal assignment, exactly one cell is selected from every row. Therefore, adding $x$ to row $k$ means every possible assignment (minimal or not) increases in total cost by $x$.
Being symmetrical to the problem, these operations $(\cdot)^T$ and $R(k,x)$ can be composed. Not only that but $(\cdot)^T$ is its own inverse and $R(k,-x)$ is the inverse for $R(k,x)$. A group is formed. For our purposes, composition is going to allow us to perform steps in sequence while inverses ensure that our steps are reversible – taking us back to the original problem.
## Steps 1 and 2
Step 1 is the composition of the transformations $R(k_i,-x_i)$ for $1 \le i \le n$, $k_i = i$, and $x_i = \min \{ s_{i(\cdot)} \}$.
Step 2 is the composition of the above transformations applied to the transpose of the matrix at the end of step 1.
## Step 3
Now that we have at least one zero in each row, the task now is to assign as many tasks as possible such that the total cost is zero. How can we do this without resorting to brute-force? The explanations I found were confusing to say the least and I decided to see if I can work it out myself. Here’s what I came up with.
1. Consider that the first $k$ people have been given the best assignment such that $a_i = j$ meaning the $i$th person is assigned task $j$ (obviously $M_{ij}=0$) and $a_i = -1$ meaning no task was assigned to the $i$th person.
2. We encounter two possible cases when trying to assign a task for the $(k+1)$th person.
3. CASE 1: There exists $j$ with $M_{(k+1)j}=0$ such that $\nexists i \le k$ where $a_i = j$. In which case, set $a_{k+1} = j$.
4. CASE 2: No such $j$ exists. The question is, can we re-adjust the assignments for the first $k$ people such that $k+1$ can be given an assignment? The solution can be written in terms of the simple linear algorithm that determines if a vertex in a graph is reachable from another. See if you can spot how it works in the following example.
Explaining the example: Column 4 is not used which can be used by row 1 (i.e. acting as the starting vertex), 2) the next row can take the column currently taken by row 1 (i.e. an edge connects rows $x \rightarrow y$ if $x$ takes column $c$ and $y$ has a $0$ in the same column), which recursively repeated leads to 3) row 4 that can finally take the column of row 3.
In general, the assignment for the $(k+1)$th row is determined by finding a path from a row with a zero in a free column to the $(k+1)$th row by traversing edges connecting two rows if the second can take the column currently taken by the first. If no such path exists, then the \$(k+1)th row has no assignment.
## Step 4
If every person was given a task the algorithm stops. Otherwise, we create a new zero in the matrix by first minimally covering the existing zeros (see the Wikipedia link), computing the minimum $m$ from the uncovered cells, and then applying the row operations to uncovered rows $r$, $R(r,-m)$, and the column operation, $C(c,m)$, to covered columns $c$.
There you have it. A problem solved purely by reducing it with symmetrical transforms (the cleverness going into determining which symmetries to apply!). With the Hungarian algorithm, I decided to make it my contribution to the newly developing Statistics.Matrix module – right now I use hmatrix for my matrix needs which plugs-in to the C LAPACK/BLAS libraries – that’s part of the excellent statistics package. The pull request is here. | 1,278 | 4,803 | {"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": 63, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-26 | latest | en | 0.867369 |
http://www.romannumerals.co/date/february-10-1800-in-roman-numerals/ | 1,718,284,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00467.warc.gz | 56,349,884 | 14,019 | ## What is February 10, 1800 in Roman Numerals?
### A: II・X・MDCCC
Your question is, "What is February 10, 1800 in Roman Numerals?", the answer is 'II・X・MDCCC'. Here we will explain how to convert and write the date 2/10/1800 with the correct Roman numeral figures.
February 10, 1800 =
II・X・MDCCC
## How is February 10, 1800 converted to Roman numerals?
To convert February 10, 1800 to Roman Numerals the conversion involves you to split the date into place values (ones, tens, hundreds, thousands), like this:
MonthDayYear
DateFebruary101800
Number Place Values2101000 + 800
Numeral Place ValuesIIXM + DCCC
=IIXMDCCC
## How do you write February 10, 1800 in Roman numerals?
To write February 10, 1800 in Roman numerals correctly, combine the converted values together. The highest numerals must always precede the lowest numerals for each date element individually, and in order of precedence to give you the correct written date combination of Month, Day and Year, like this:
II・X・MDCCC
## More from Roman Numerals.co
February 11, 1800
Learn how February 11, 1800 is translated to Roman numerals.
Dates in Roman Numbers
Select another date to convert in to Roman Numbers | 316 | 1,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.858002 |
https://mathoverflow.net/questions/173541/some-types-of-diophantine-equations-and-their-decidability | 1,618,148,987,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00473.warc.gz | 481,093,426 | 32,731 | # Some types of diophantine equations and their decidability
The MDRP theorem – which answers Hilbert's tenth problem in the negative – says:
There is no algorithm for determining whether an arbitrary diophantine equation has a solution.
In other words: there is no algorithm for determining whether an arbitrary diophantine equation $p(n_1,\dots,n_k,x)=0$ has a non-empty solution set
$$N = \lbrace x \in \mathbb{N}\ |\ (\exists n_1,\dots,n_k)\ p(n_1,\dots,n_k,x)=0 \rbrace$$
There are essentially four ways for a diophantine equation to have a non-empty solution set:
1. a finite non-empty solution set $N$
which necessarily has an infinite complement $\overline{N} = \mathbb{N} \setminus N$
(type $\tau_{1/\omega}$)
2. an infinite solution set with an infinite complement
(type $\tau_{\omega/\omega}$)
3. an infinite solution set with a finite complement
(type $\tau_{\omega/1}$)
4. an infinite solution set with an empty complement
(type $\tau_{\omega/0}$)
Accordingly the MRDP theorem says:
There is no algorithm for determining whether an arbitrary diophantine equation $p(n_1,\dots,n_k,x)=0$ is not of type $\tau_{0/\omega}$ (i.e. does not have an empty solution set).
But this is equivalent with:
There is no algorithm for determining whether an arbitrary diophantine equation $p(n_1,\dots,n_k,x)=0$ is of type $\tau_{1/\omega}$ or of type $\tau_{\omega/\omega}$ or of type $\tau_{\omega/1}$ or of type $\tau_{\omega/0}$.
My first question is:
($*$) Is it decidable (or semi-decidable) whether the solution set of an diophantine equation is of type $\tau_{\omega/0}$, i.e. for all $x \in \mathbb{N}$ it holds that $$(\exists n_1,\dots,n_k)\ p(n_1,\dots,n_k,x)=0$$
Naively, one might believe that the answer is "yes" because
Conjecture: The solution set of an diophantine equation $p(n_1,\dots,n_k,x)=0$ is of type $\tau_{\omega/0}$ iff there are $n_i, k$ and another polynomial $p'(n_1,\dots,n_k,x)$ such that $$p(n_1,\dots,n_k,x) = p'(n_1,\dots,n_k,x)(x-n_i)^k$$
Because in this case $x$ – via $n_i$ – can take every value. But even when this conjecture is too naive and false, the question ($*$) might be answered in the positive in another way.
But if the question ($*$) is to be answered in the negative, this post stops here.
Otherwise it continues. In this case we can omit the type $\tau_{\omega/0}$ from the disjunction above and obtain:
There is no algorithm for determining whether an arbitrary diophantine equation $p(n_1,\dots,n_k,x)=0$ is of type $\tau_{1/\omega}$ or of type $\tau_{\omega/\omega}$ or of type $\tau_{\omega/1}$.
And another question arises naturally:
($*\!*$) Is it decidable (or semi-decidable) whether the solution set of an diophantine equation is of type $\tau_{\omega/1}$, i.e. for all but finitely many $x \in \mathbb{N}$ it holds that $$(\exists n_1,\dots,n_k)\ p(n_1,\dots,n_k,x)=0$$
And so on. Which results in the question:
Can the undecidability of a diophantine equation to be not of type $\tau_{0/\omega}$ be reduced to the undecidability of a diophantine equation to (positively) be of another type?
My first remark is that your question really has little to do with diophantine solutions sets, since by the MRDP theorem these are precisely the computably enumerable sets. So all your questions amount to exactly equivalent questions about the c.e. sets. In other words, every Turing machine program $e$ computes a c.e. set $W_e$, and you are asking about deciding whether $W_e=\mathbb{N}$? whether $W_e$ is infinite? whether $W_e$ is finite? whether $W_e$ is co-finite? whether $W_e$ is empty?
All these questions are undecidable, and furthermore, none of them are even c.e. So there are no such decision procedures to decide the problems you mention in your question. The MRDP theorem explains exactly that every c.e. set $W_e$ is diophantine and conversely.
Let's begin by showing that the set $\text{Inf}$ of programs $e$ such that $W_e$ is infinite, is not computable, and indeed, it is not even semi-decidable; it is a $\Pi^0_2$-complete set, and thus equivalent to the double jump. One cannot answer this question even with an oracle for the halting problem.
To see that there is no computable procedure to decide if a given c.e. set $W$ contains every number, suppose toward contradiction that you could decide this. Then the halting problem would be decidable by the following procedure: given a Turing machine program $p$, define $W$ to consist of all numbers, if $p$ halts on $0$, and otherwise nothing. This is a c.e. set, and $p$ halts on $0$ just in case $W$ is everything, and so if we could decide if $W$ is everything, then we could decide the halting problem, a contradiction.
Essentially similar reasoning shows that you cannot decide whether a given c.e. set is infinite, whether it is co-finite, whether it is finite, and so on.
Let's now prove the further property that $\text{Inf}$ is not even c.e. One can get a hint of this by looking at the quantifier complexity. Namely, to say that a c.e. set is everything is to assert $\forall x W(x)$, where $W(x)$ is c.e. and hence of the form $\exists n\phi(n,x)$, so we have $\forall x\exists n\phi(n,x)$, which is complexity $\Pi^0_2$. One can show that the set $\text{Inf}$ of all programs $e$ that accept every input (which is equivalent to your decision problem) is $\Pi^0_2$-complete, and hence not $\Sigma^)_1$ and therefore not c.e. To see this, suppose that we have any given $\Pi^0_2$ question $\forall x\exists n\phi(x,n,y)$. Define a c.e. set $W$ that starts checking this: it starts with $x=0$ and searches for a witness $n$, enumerating $x$ into $W$ when the witness is found; and then it moves to $x+1$. This set $W$ will be everything just in case the $\Pi^0_2$ statement is true, and it will be finite otherwise, including all the $x$ up to the first failing instance $x$. So this reduces the $\Pi^0_2$ question to the question of whether a given c.e. set is everything or not. So that property is $\Pi^0_2$-complete and hence cannot be c.e.
Similar argument shows that the set of programs $e$ which compute finite sets is $\Sigma^0_2$-complete, and therefore not c.e.; the set of programs $e$ that compute the empty set is $\Pi^0_1$-complete, and therefore not c.e., the set of programs $e$ that compute infinite sets is $\Pi^0_2$-complete, and therefore not c.e. These are all explained in detailed in Soare's book on the Computably Enumerable sets and degrees, or any other decent book on computability theory.
And each of these facts about the c.e. sets corresponds to an analogous fact about the diophantince sets, in light of the MRDP theorem.
• It might be useful to mention that, in converting the question from Diophantine sets to c.e. sets, you've tacitly used that the MRDP theorem holds uniformly, in the sense that, from an index $e$, one can effectively compute a Diophantine definition for $W_e$. – Andreas Blass Jul 8 '14 at 2:03
• Yes, that is completely right. We can go back and forth computably from any of the usual presentations of the c.e. sets, in terms of diophantine equations, Turing machines, definability complexity, and so on, in a completely uniform manner. – Joel David Hamkins Jul 8 '14 at 2:32
• Thanks for the elaborate answer! But why do you say, that the question has little to do with diophantine sets? Since c.e. and diophantine sets are the same, it has to do with diophantine sets - like your answer does. You prefer Turing machines, I prefer diophantine equations. So why did you insist on it in your first remark? – Hans-Peter Stricker Jul 8 '14 at 6:37
• Oh, I had just meant that when it comes to decidability questions about general diophantine sets, most of what we know comes from computability theory and the characterization of the diophantine sets as the c.e. sets, rather than from the general theory of diophantine sets. I think of the diophantine characterization as just another characterization on the list, like (many versions of) Turing machines, register machines, numerous other machine characterizations, group presentations, (failures of) tiling problems, game of life, and so on. – Joel David Hamkins Jul 8 '14 at 10:45
• @Andreas: Once you know that the $W_e$’s are well-defined in the first place (i.e., there exists a universal r.e. predicate), the uniform version of the MRDP theorem is a trivial corollary of the plain MRDP theorem, so this does not really make a difference. – Emil Jeřábek Jul 8 '14 at 11:02 | 2,266 | 8,446 | {"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} | 3.5 | 4 | CC-MAIN-2021-17 | longest | en | 0.774338 |
https://farfromhomemovie.com/what-is-a-research-hypothesis/ | 1,723,780,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00631.warc.gz | 196,748,785 | 12,251 | 2021-04-19
## What is a research hypothesis?
A hypothesis is a statement that can be tested by scientific research. If you want to test a relationship between two or more things, you need to write hypotheses before you start your experiment or data collection.
## What is working hypothesis in research?
A working hypothesis is a hypothesis that is provisionally accepted as a basis for further research in the hope that a tenable theory will be produced, even if the hypothesis ultimately fails.
What is research hypothesis example?
The precursor to a hypothesis is a research problem, usually framed as a question. It might ask what, or why, something is happening. For example, we might wonder why the stocks of cod in the North Atlantic are declining. The research hypothesis is a paring down of the problem into something testable and falsifiable.
What can be used to test the hypothesis?
Hypothesis testing is used to assess the plausibility of a hypothesis by using sample data. The test provides evidence concerning the plausibility of the hypothesis, given the data. Statistical analysts test a hypothesis by measuring and examining a random sample of the population being analyzed.
### How do you test alternative hypothesis?
If you are performing a two-tailed hypothesis test, the alternative hypothesis states that the population parameter does not equal the null hypothesis value. For example, when the alternative hypothesis is HA: μ ≠ 0, the test can detect differences both greater than and less than the null value.
### What are the 6 parts of a hypothesis?
SIX STEPS FOR HYPOTHESIS TESTING.HYPOTHESES.ASSUMPTIONS.TEST STATISTIC (or Confidence Interval Structure)REJECTION REGION (or Probability Statement)CALCULATIONS (Annotated Spreadsheet)CONCLUSIONS.
What is hypothesis and its steps?
Hypothesis testing is a scientific process of testing whether or not the hypothesis is plausible. The first step is to state the null and alternative hypothesis clearly. The null and alternative hypothesis in hypothesis testing can be a one tailed or two tailed test. The second step is to determine the test size.
How do you set up a hypothesis?
6:36Suggested clip · 120 secondsAn Easy Rule to Setting Up the Null & Alternate Hypotheses …YouTubeStart of suggested clipEnd of suggested clip
## How do you write an alternative hypothesis?
Always write the alternative hypothesis, typically denoted with Ha or H1, using less than, greater than, or not equals symbols, i.e., (≠, >, or hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false.
## What do you mean by alternative hypothesis?
In statistical hypothesis testing, the alternative hypothesis is a position that states something is happening, a new theory is preferred instead of an old one (null hypothesis).
What is the purpose of the alternative hypothesis?
Alternative hypothesis purpose An alternative hypothesis provides a chance of discovering new theories that can disprove an existing one that might not be supported by evidence.
How do you write an alternative hypothesis in research?
Converting research questions to hypothesis is a simple task. Take the questions and make it a positive statement that says a relationship exists (correlation studies) or a difference exists between the groups (experiment study) and you have the alternative hypothesis.
### What is null and alternative hypothesis example?
They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints….Learning Outcomes.H 0H aequal (=)not equal (≠) or greater than (>) or less than (greater than or equal to (≥)less than (less than or equal to (≤)more than (>)
### What is another name for research hypothesis?
Hypothesis Synonyms – WordHippo Thesaurus….What is another word for hypothesis?premisepropositionthesisassumptionpostulatepostulationpresuppositionsurmiseconceptconjecture235 | 802 | 3,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.923107 |
https://www.physicsforums.com/threads/copper-wire-resistance-problem.388182/#post-2637236 | 1,632,602,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00139.warc.gz | 901,954,134 | 16,367 | # Copper wire resistance problem
A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
Dale
Mentor
2020 Award
This should probably go into the homework section, and you will need to show your work so far.
A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
its resistance will be increased.because R is always inversely proportional to length
its resistance will be increased.because R is always inversely proportional to length
Thanks, could you please mention by how much the resistance will rise ?
This should probably go into the homework section, and you will need to show your work so far.
I am really sorry to confuse you but this question appeared for my board exams recently . Could you please help me out .
What equations do you know that relate resistance to length, and amount of substance?
Thanks, could you please mention by how much the resistance will rise ?
it resistance will be increased by two times.
Is diameter the only thing changing?
$$R = \rho \frac{h}{A}$$
We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
$$V = Ah = \pi r^2 h$$ and $$V$$ is a constant.
Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by half). So the new area is:
$$A' = \pi (\frac{r}{2})^2 = \pi \frac{r}{4}$$
Therefore, to destroy no matter and to keep volume constant, we must multiply $$h'$$ by a constant to keep $$V'$$ equal to $$V$$:
$$V = V' = \pi r^2 h = (\pi \frac{r^2}{4})(Kh)$$
Seen above, K must equal four to cancel out that fractional four, thus bringing both left and right equations equal and thus keeping volume constant.
To sum:
$$h' = 4h$$
$$A' = \pi \frac{r^2}{4} = (\pi r^2) \frac{1}{4} = \frac{A}{4}$$
We now plug in our new height and our new area into the equation for resistance:
$$R' = \rho \frac{4^2h}{A}= 4^2(\rho \frac{h}{A}) = 4^2 R = 16R$$
Last edited:
$$R = \rho \frac{h}{A}$$
We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
$$V = Ah = \pi r^2 h$$ and $$V$$ is a constant.
Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by a fourth). So the new area is:
$$A' = \pi (\frac{r}{4})^2 = \pi \frac{r}{16}$$
Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?
Therefore, to destroy no matter and to keep volume constant, we must multiply $$h'$$ by a constant to keep $$V'$$ equal to $$V$$:
$$V = V' = \pi r^2 h = (\pi \frac{r^2}{16})(Kh)$$
Seen above, K must equal sixteen to cancel out that fractional sixteen, thus bringing both left and right equations equal and thus keeping volume constant.
To sum:
$$h' = 16h$$
$$A' = \pi \frac{r^2}{16} = (\pi r^2) \frac{1}{16} = \frac{A}{16}$$
We now plug in our new height and our new area into the equation for resistance:
$$R' = \rho \frac{16^2h}{A}= 16^2(\rho \frac{h}{A}) = 16^2 R$$
Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?
you're right | 981 | 3,649 | {"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} | 4.4375 | 4 | CC-MAIN-2021-39 | latest | en | 0.959818 |
http://zsgabin.pl/journal/archive.php?id=most-popular-nc-pick-4-numbers-423d9b | 1,718,215,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00495.warc.gz | 59,185,368 | 16,422 | # most popular nc pick 4 numbers
Entire website contents are Copyright© 2006-2010 A Lottery Number Picker Co. I might have also missed or deleted some drawings.
And to make things much easier, the Lotterycodex calculator can also generate lotto combinations for you so that everything will be delivered to you on a silver platter. Let’s take a look at hot and cold numbers of other lotteries. However, the combination will turn out to be the worst because a line composed of all even numbers rarely occur in lottery draws. What do you do when there's a certain 'hot' number, or a favorite
Double (1123) Straight: There are 4,320 straight doubles.
Most cards (playslips) offer 5 or more ways to play each Pick 4 number permutation. If you think that picking the most common winning lottery numbers will provide a better advantage, then you are wasting your time and money. I might have deleted quadruple combinations, such as 0000, 1111, etc. Fortunately, you don’t need a degree in math to make it work. If you do get lucky, and these tips for how to win the lottery work, you’ll definitely want to know these 13 things lotto winners won’t tell you. The title of this program or list is Boxed Key Chart (Pick 4). These are the lottery numbers that have come up the most across the pick 3 and pick 4 lottery drawings in all the states during 2019. The results will keep accumulating until you either leave the Pick 4 screen or click The image below shows the frequency of each number from April 2004 to October 2017. Fourteen and 62 were drawn 24 times; 10 was drawn 23 times; and 42 and 48 were drawn 22 times. My best advice to you, please read it. Reports created by FrequencyRank, lottery, lotto, Pick-4 statistical software..
In the event of a discrepancy between the numbers posted on this website and the official winning numbers, the official winning numbers as certified by the Multi-State Lottery Association and/or the NCEL shall control.
So even a straight 1-2-3-4-5 combination is equally likely as any balanced combinations in the universe of EuroMillions combinations.
The most popular midsize truck in America, the Toyota Tacoma comes in extended- and crew-cab styles and in six trim levels, including the off-road-ready TRD Pro. Therefore, the hot and cold numbers don’t count. In this regard, I would like to share with you a little known secret that only people in-the-know do in the lottery. Let’s find out: Using combinatorial computation, we can see that this pattern (two-green-and-three-magenta) belongs to one of the worst probability groups in EuroMillions. Admittedly, mistakes are likely. information about our privacy practices please click here. Conclusively, this combination does not belong to the best probability group in EuroMillions. This command will list for you just those doubles that have the digit 7 in them. There are ONLY 210 single box numbers in Pick 4 (wow!). row of numbers on your 'Quick Picks'?. Let’s prove that.
The 36 Most Dominant Singles . Even though those cold numbers will tend to even out, it’s not a good idea to use cold numbers. 0000 through 9999. All materials on this Website are owned by or licensed to the NCEL. So in a finite structure like the lottery, logical analysis is what works best. Digit-Wise Hit Frequency and Rank Analysis Digit-Wise Elapse Time Analysis Digit-Wise Low-High Analysis how to, If you want to change Preselections and/or Rejections as you Instead of picking a meaningful date, like a birthday or anniversary, it’s a no-brainer to select the most common lottery numbers to boost your odds.
I posted the list for each key digit so it can be here as a reference. It means that those less frequently appearing numbers in the initial draws eventually catch up as the draw continues to take place. No Match/Singles (1234) Boxed: There are ONLY 210 single box numbers in Pick 4 (wow!). 0127 0137 0147 0157 0167 0178 0179 0237 0247 0257 0267 0278 0279 0347 0357 0367 0378 0379 0457 0467 0478 0479 0567 0578 0579 0678 0679 0789 1237 1247 1257 1267 1278 1279 1347 1357 1367 1378 1379 1457 1467 1478 1479 1567 1578 1579 1678 1679 1789 2347 2357 2367 2378 2379 2457 2467 2478 2479 2567 2578 2579 2678 2679 2789 3457 3467 3478 3479 3567 3578 3579 3678 3679 3789 4567 4578 4579 4678 4679 4789 5678 5679 5789 6789 Items: 84. Right?
The 36 numbers in group 1 are the 36 most dominant SINGLES in the Pick 3 game. According to USA Mega, the most common Mega Millions numbers drawn since October 31, 2017, are 14, 62, 10, 42, and 48. Don’t they?”. That is a very big statement to make but I know what I am talking about. Most people would be shocked to learn that there are only 210 no match singles in Pick 4. We have crunched all off the lottery numbers and compiled the 2020 list of the most popular pick 3 and pick 4 lottery numbers. A statistical report as such is not of much use. Using combinatorial mathematics, we know that the probability of this pattern is 0.0009557477. I agree to receive
Those are much better odds of winning at Pick 4.
Don’t forget. a first-time user's uncanny ability to find the 'bugs'. The 36 Most Dominant Singles . First off, let me show you this message that I received from one lotto player: Indeed, Mr. H.R’s frustration of playing the lottery for 20 years and getting a measly prize of \$20 at the most reflects the widespread dissatisfaction of the majority of lotto players. Let’s start with Irish Lottery 6/47 from September 2015 to October 2017. Below is a breakdown of each type.
Please check the checkbox in order to get your free download. One caveat: The Pennsylvania pick 4 game has a very long history.
What I have showed you here in only a few minutes is how to cut that number down dramatically to a 1 in 108 or a 1 in 84 chance of getting a hit. There is an excellent chance that one of these 84 numbers will hit soon after.
But, Chux Ball Buster - A Lottery Number Picker will always be
That is a pretty short list considering that there are 10,000 possibilities in Pick 4. Side Note: The good ol' fashioned way for the Lottery to pick 4 digits, is from four 10-ball (I will discuss more advanced strategies in future articles.). But before I discuss that, let me talk about some basic facts about the game. For example, a balanced low-high combination such as 2-14-28-30-44-48 may be considered best, according to Gianella’s method.
Rarely a week goes by without one of these numbers hitting. Well, it’s about time, you change the way you play lotto. The most common Mega Ball number is 10, which has come up 13 times in the past two years. You can download the full list of numbers with 10 numbers in each section by entering your email address below.
The Most Common Winning Lottery Numbers Don’t Work According To Math, The Law Of Large Numbers In Layman’s Term, The Most Common Winning Lottery Numbers Of The Australian TattsLotto 6/45, The Hot and Cold Numbers Of U.S. Powerball 5/69, Make an intelligent decision when you play the lottery, The law of large numbers is a principle of probability, In probability theory, the law of large numbers (LLN), The Lottery and the Winning Formula According to Math, increase your chances of hitting the jackpot prize, The Geometry of Chance: Lotto Numbers Follow a Predicted Pattern, Top 10 Lottery Strategy Myths Debunked (Perhaps You’re Doing #4 or #10), Lottery Calculator: Knowing the Best Lotto Combinations Without a Math Degree, Using Birth Dates in PLaying the Lottery? Both the Preselections and Rejections address the matter of picking your lucky lottery numbers from 'virtual' ball machines. and Ones', 'Reset
As the number of experiments increases, the actual ratio of outcomes will converge on the theoretical, or expected, ratio of outcomes. Let me show you why I do not recommend people using either hot or cold numbers. Data recorded in Pennsylvania Lottery, evening drawings only. Meanwhile, stay clear of 08, 15, 25, 24, and 18 when choosing your first five numbers; these appeared only between four and six times. The short parpalucks create staggering biases in pick-4 digit frequency.
1. Welcome to Chux Ball Buster - A Lottery Number Picker! Here’s how to win the lottery (or at least boost your chances) by picking the most common lottery numbers. The underlying principle of probability still provides accuracy and precision, no matter how different the structure is. We will treat your information with respect. 2020-01-05 . Naturally, you demand a better selection of numbers with a better ratio of success to failure. If you are following my study of applied mathematics in the lottery, you will understand that the lottery follows certain immutable laws. But I like the way Wikipedia puts it out which says: In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. (See how Lotterycodex Calculator works). link in the footer of any email you receive from us, or by contacting us at So to win playing Pick 4 you really have to sort the list and only focus on what you need. Just because some numbers appear more frequently in lottery draws doesn’t mean they are the best numbers to play. Nothing can be further from the truth. It’s a priceless piece of information, and the reason you should read it is that you’ll keep losing money if you don’t. Neither do the least appearing ones. Thank you for signing up. The position-by-position frequency will always show the exact amount of digits that appeared in a number of drawings analyzed. I agree. In fact, you may accidentally play numbers that follow the pattern with the worst probability.
You will read more about it in this little special report called “Inverse Lotto Strategy.”. We are no longer supporting IE (Internet Explorer) as we strive to provide site experiences for browsers that support new web standards and security practices. Most recent drawing in this file: May 15, 2017 (8 3 3 4) In the lottery, this means that given enough number of draws, each ball in the number field exhibits the same probability. 3. With a probability of 0.0025486605, any number combination that follows this color pattern will appear once in every 392 draws. You can change your mind at any time by clicking the unsubscribe The most basic strategy I can give you right now is this; find the most due digit or a digit that you think will hit (use Mike's Key Number Finder (Pick 4) on the p34sug website for help on this) and focus on the list of 84 numbers for this key digit. | 2,551 | 10,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.957263 |
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How to factorise several kinds of algebra expressions.
This topic is part of the TCS FREE high school mathematics 'How-to Library', and will help you to factorise several kinds of algebra expressions.   (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Algematics software. Click here for instructions.
### Theory:
Algematics uses six different methods to factorise an algebraic expression. They are:
#### 1. common factors
8p² + 12pq - 4pr factorises to: 4p(2p + 3q - r)
#### 2. difference of perfect squares
a² - b²Â factorises to: (a + b)(a - b)
#### 3. sum of perfect cubes
a³ + b³Â factorises to: (a + b)(a² - ab + b²)
#### 4. difference of perfect cubes
a³ - b³Â factorises to: (a - b)(a² + ab + b²)
#### 5. split middle term of quadratics
2x² + 3x - 5Â becomes:Â 2x² - 2x + 5x - 5
(While this has not factorised the expressions, this is just the first of several steps needed to factorise the quadratic. In this case you will need to use the factorise command two more times. )
[ To find out how to split the middle term of a quadratic yourself, click here. ]
#### 6. group and factorise four terms
2x² - 2x + 5x - 5Â becomes:Â 2x(x - 1) + 5(x - 1)
This does two things. First, the four terms are swapped around and regrouped if necessary, then the pairs of terms are factorised in such a way that a common factor results. In this example, (x-1) is now a common factor, so that if the factorise command is used one more time the expression will be fully factorised.
The software examines the expression and chooses the method to use. All you have to do is keep on clicking the  (factorise) button until the expression no longer changes.
If you click  and nothing changes, then the software cannot factorise the expression. Press Ctrl+Delete to delete the duplicate step.
### Method:
IMPORTANT: This topic assumes that you know how to enter mathematical formulas into Algematics. Find out how by completing the three simple tutorials in the 'Getting Started' section of the Algematics program 'Help'.
We will use the expression: to demonstrate how to use the factorise command.
#### Step 1Â Enter the data
Click  and type the expression to be factorised into the maths box in the
data entry dialog box.
If the ‘EMPTY’ message is not displayed between the blue buttons, click the  button until the message: ‘INSERT’ appears. If required, use the ‘ * ’ symbol for multiply, and the ‘ / ’ symbol for divide.
      Maths...
  x[6] - 1
Â
Â
Â
and then click
#### Step 2Â Factorise
Click the  (factorise) button. This will generate a new step.
If this new step is the same as the last, then this means that Algematics cannot factorise the expression any further. In that case, delete the duplicate step by pressing Ctrl+Delete on the keyboard.
Keep clicking  until the expression cannot be factorised any further.
The example expression factorises in the following sequence:
Click  : (Difference of perfect squares)
(x³ + 1)(x³ - 1)
Click  : (Sum of perfect cubes)
(x + 1)(x² - x´1 + 1)(x³ - 1)
Click  : (Difference of perfect cubes)
(x + 1)(x² - x´1 + 1)(x - 1)(x² + x´1 + 1)
This expression is now fully factorised, and can be tidied up by clicking  (simplify all), giving:
(x + 1)(x² - x + 1)(x - 1)(x² + x + 1)
Go back to step 1 to factorise another expression.
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AO3
Time: 4560 minutes
Instructions
Use black ink or ball-point pen.
centre number and candidate number.
Answer the questions in the spaces provided
there may be more space than you need.
Calculators must not be used in questions marked with as asterisk (*).
Diagrams are NOT accurately drawn, unless otherwise indicated.
Information
This gold test is aimed at students targeting grades 7-9.
This test has 10 questions. The total mark for this paper is 40.
The marks for each question are shown in brackets
use this as a guide as to how much time to spend on each question.
Keep an eye on the time.
1. The diagram shows a cuboid ABCDEFGH.
AB = 7 cm, AF = 5 cm and FC = 15 cm.
Let x = BC.
(a) The square of the diagonal of a cuboid is equal to the sum of the squares of its three
sides.
Write an equation connecting AB, AF, BC and FC.
.......................................................
(b) Rearrange your equation to make x the subject.
.......................................................
(2)
(c) Write an equation for the volume of the cuboid.
.......................................................
2
(d) Calculate the volume of the cuboid.
....................................................... cm3
(2)
(Total for Question 1 is 4 marks)
___________________________________________________________________________
3
2. The diagram shows a solid cone.
The diameter of the base of the cone is 24x cm.
The height of the cone is 16x cm.
The curved surface area of the cone is 2160 cm2.
The volume of the cone is V cm3, where V is an integer.
(a) Draw a suitable right-angled triangle and find l, the slant height of the cone, in
terms of x.
.......................................................
(1)
(b) Write an equation for the curved surface area of the cone in terms of x.
.......................................................
(1)
4
(c) Solve your equation to find the value of x.
.......................................................
(1)
h = .......................................................
r = .......................................................
(e) Use the formula for the volume of a cone to find the value of V.
.......................................................
(2)
(Total for Question 2 is 5 marks)
___________________________________________________________________________
5
*3. The diagram shows a solid hemisphere.
4 3
Volume of sphere = r
3
Surface area of sphere = 4r2
250
The volume of the hemisphere is
3
(a) Write a formula for the volume of a hemisphere.
....................................................... cm
(b) Use your answer to part (a) and the given volume of the hemisphere to find the radius of
the hemisphere.
....................................................... cm
(1)
(c) Write a formula for the curved surface area of a hemisphere.
....................................................... cm
6
(d) Use your answers to parts (b) and (c) to work out the curved surface area of the
hemisphere.
....................................................... cm2
(1)
(e) Work out the surface area of the flat circular face of the hemisphere.
....................................................... cm2
(1)
(f) Find the total surface area of the hemisphere.
...................................................... cm2
(1)
(Total for Question 3 is 4 marks)
___________________________________________________________________________
7
4. Thelma spins a biased coin twice.
The probability that it will come down heads both times is 0.09
(a) Calculate the probability that the coin will come down heads when it is spun once.
....................................................... cm
(1)
(b) Calculate the probability that the coin will come down tails when it is spun once.
....................................................... cm
(c) Calculate the probability that the coin will come down tails both times when it is spun
twice.
....................................................... cm
(2)
(Total for Question 4 is 3 marks)
___________________________________________________________________________
8
5. A pendulum of length L cm has time period T seconds.
T is directly proportional to the square root of L.
(a) Write an equation to show the relationship between L and T.
.......................................................
(1)
Let L = 1 represent the original length of the pendulum.
(b) What value can you use to represent L after the increase?
.......................................................
(c) Write a pair of simultaneous equations to show the relationship between L and T before
and after the increase.
.......................................................
(1)
(d) Solve your simultaneous equations to find the percentage increase in the time period.
.......................................................%
(1)
(Total for Question 5 is 3 marks)
___________________________________________________________________________
9
6. Here is a right-angled triangle.
All measurements are in centimetres.
The area of the triangle is 2.5 cm2.
(a) Write an equation for the area of the triangle in terms of x.
.......................................................
(1)
(b) Show that x2 2x 5 = 0.
.......................................................
(1)
.......................................................
(1)
10
(d) Choose the appropriate value of x and use it to find the base and height of the triangle.
....................................................... cm
....................................................... cm
(e) Use Pythagoras theorem to find the length of the hypotenuse.
....................................................... cm
(1)
(f) Find the perimeter of the triangle.
....................................................... cm
(2)
(Total for Question 6 is 6 marks)
___________________________________________________________________________
11
*7. There are 10 pens in a box.
There are x red pens in the box.
All the other pens are blue.
Jack takes at random two pens from the box.
(a) Write an expression, in terms of x, for the probability that the first pen Jack takes is red.
.......................................................
(1)
(b) Write an expression, in terms of x, for the probability that the first pen Jack takes is red
and the second pen is blue.
.......................................................
(1)
(c) Write an expression, in terms of x, for the probability that the first pen Jack takes is blue.
.......................................................
(d) Write an expression, in terms of x, for the probability that the first pen Jack takes is blue
and the second pen is red.
.......................................................
(1)
12
(e) Use your answers to parts (b) and (d) to from an expression, in terms of x, for the
probability that Jack takes one pen of each colour.
.......................................................
.......................................................
(2)
(Total for Question 7 is 5 marks)
___________________________________________________________________________
13
8. Mark has made a clay model.
He will now make a clay statue that is mathematically similar to the clay model.
The model has a base area of 6 cm2.
The statue will have a base area of 253.5 cm2.
(a) Find the linear scale factor between the model and the statue.
.......................................................
(1)
(b) Find the volume scale factor between the model and the statue.
.......................................................
Mark used 2 kg of clay to make the model.
(c) Work out how much clay Mark needs to make the statue.
.......................................................
Clay is sold in 10 kg bags.
Mark has to buy all the clay he needs to make the statue.
(d) How many bags of clay will Mark need to buy?
.......................................................
(2)
(Total for Question 8 is 3 marks)
___________________________________________________________________________
14
9. The number of bees in a beehive at the start of year n is Pn.
The number of bees in the beehive at the start of the following year is given by
Pn + 1 = 1.05(Pn 250)
At the start of 2015 there were 9500 bees in the beehive.
(a) How many bees will there be in the beehive at the start of 2016?
.......................................................
(1)
(b) How many bees will there be in the beehive at the start of 2017?
.......................................................
(c) How many bees will there be in the beehive at the start of 2018?
.......................................................
(2)
(Total for Question 9 is 3 marks)
___________________________________________________________________________
15
*10.
(a) Find the gradient of line AB.
.......................................................
(1)
(b) Find the gradient of a line that is perpendicular to AB.
.......................................................
(1)
(c) Substitute your answer to part (b) and the coordinates of C into the equation y = mx + c
and solve it to find c.
.......................................................
(1)
(d) Write the equation of the line that passes through C and is perpendicular to AB.
.......................................................
(1)
(Total for Question 10 is 4 marks)
TOTAL FOR PAPER IS 40 MARKS
16
Question Origin Question Origin
1 3H qu.12 6 2H qu.19
2 3H qu.17 7* 1H qu.21
3* 1H qu.18 8 3H qu.20
4 3H qu.18 9 2H qu.21
5 2H qu.15 10* 1H qu.23
Specimen papers set 1 problem solving: Gold Test Grades 7-9
1 (a)-(b) 431 B1 for use of Pythagoras involving the unknown length
P1 for setting up an equation equivalent to
(c)-(d)
P1 for finding the volume using their
A1awrt 430.5
2 (a) 20736
P1 for a method to find the slant height of the cone eg or by similar
(b) triangles and Pythagorean triples
P1 for setting up an equation for the curved surface area in terms of x eg
(c)
(d)-(e) P1 for complete method to find the value of x
P1 for a method to find the volume
A1 cao
3 (a)-(b) 75 P1 250 1 4
starts process by using p and p r 3 to find radius as 5
3 2 3
(c)-(d) P1 starts process using curved surface area eg (4 52 ) 2
(e) P1 complete process shown eg (4 52 ) 2 + ( 52 )
(f) A1 for 75
4 (a) 0.49 P1 for
(b)-(c)
P1 for (1- )
Specimen papers set 1 problem solving: Gold Test Grades 7-9
A1 cao
5 (a) 18.3 P1 for a start to the process interpreting the information correctly,
eg. T = k oe
(b)-(c) P1
for next stage in process to find percentage change in T,
eg. 1.4
(d) A1
for 18.3 to 18.4
6 (a) 8.63 to 8.65 P1 for a start of process, eg.
(b) P1 for rearranging to give a quadratic equation,
eg x2 2x 5 = 0 oe.
(c) P1 for a process to solve the quadratic equation, condoning one sign error in
use of formula (x = 3.449... and x = 1.449...)
(d)-(e) P1 for selecting the positive value of x and applying Pythagoras to find the
hypotenuse,
eg. (3.4492 + 1.4492) (= 3.74...)
(f) P1 for complete process to find perimeter
A1 for answer in the range 8.63 to 8.65
7 (a) 10 x - x 2 P1 x 10 - x x -1 10 - x x 9- x
for or or or or or seen on diagram
45 10 10 9 9 9 9
or in a calculation
(b) P1 x 10 - x x x - 1 10 - x 9 - x
for or for +
10 9 10 9 10 9
10 - x x
10 9
(c)-(d) P1 x 10 - x 10 - x x x - 1 10 - x 9 - x
for + for 1 ( + )
10 9 10 10 9 10 9
x
9
18
Specimen papers set 1 problem solving: Gold Test Grades 7-9
(e)-(f) P1 for beginning to process the algebra
A1 10 x - x 2
oe
45
8 (a) 55 253.5
P1 for (=6.5)
6
(b)-(d) P1 for 2 6.53 10 (=54.925)
A1 cao
9 (a) 10169 or 10170 P1 for correct use of formula to find number in 2016,
eg. 1.05(9500 250) (= 9712.5)
(b)-(c) P1 for complete iterative process,
eg. 2017: 1.05(9712.5 250) (= 9935.625)
2018: 1.05(9935.625 250)
C1 for answer of 10169.90... correctly rounded or truncated to nearest whole
number
10 (a) 1 3 P1 for a process to find the gradient of the line AB
y =- x+
2 2
(b) P1 (dep) for a process to find the gradient of a perpendicular line
eg use of 1/m
(c) P1 (dep on P2) for substitution of x=5, y=1
(d) A1 equation stated oe
19 | 2,933 | 12,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-39 | latest | en | 0.798782 |
http://slideplayer.com/slide/764996/ | 1,547,663,594,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657557.2/warc/CC-MAIN-20190116175238-20190116201238-00053.warc.gz | 215,749,702 | 22,290 | 5.5 Properties and Laws of Logarithms
Presentation on theme: "5.5 Properties and Laws of Logarithms"— Presentation transcript:
5.5 Properties and Laws of Logarithms
Do Now: Solve for x. x = 3 x = 1/3 x = 6 x = 12
Consider some more examples…
Without evaluating log (678), we know the expression “means” the exponent to which 10 must be raised in order to produce 678. log (678) = x 10x = 678 If 10x = 678, what should x be in order to produce 678? x = log(678) because 10log(678) = 678
And with natural logarithms…
Without evaluating ln (54), we know the expression “means” the exponent to which e must be raised in order to produce 54. ln (54) = x ex = 54 If ex = 54, what should x be in order to produce 54? x = ln(54) because eln(54) = 54
Basic Properties of Logarithms
Common Logarithms Natural Logarithms 1. log v is defined only when v > 0. 1. ln v is defined only when v > 0. 2. log 1 = 0 and log 10 = 1 2. ln 1 = 0 and ln e = 1 3. log 10k = k for every real number k. 3. ln ek = k for every real number k. 4. 10logv=v for every v > 0. 4. elnv=v for every v > 0. ** NOTE: These properties hold for all bases – not just 10 and e! **
Example 1: Solving Equations Using Properties
Use the basic properties of logarithms to solve each equation.
Laws of Logarithms aman=am+n
Because logarithms represent exponents, it is helpful to review laws of exponents before exploring laws of logarithms. When multiplying like bases, add the exponents. aman=am+n When dividing like bases, subtract the exponents.
Product and Quotient Laws of Logarithms
For all v,w>0, log(vw) = log v + log w ln(vw) = ln v + ln w
Using Product and Quotient Laws
Given that log 3 = and log 4 = , find log 12. Given that log 40 = and log 8 = , find log 5. log 12 = log (3•4) = log 3 + log 4 = log 5 = log (40 / 8) = log 40 – log 8 =
Power Law of Logarithms
For all k and v > 0, log vk = k log v ln vk = k ln v For example… log 9 = log 32 = 2 log 3
Using the Power Law Given that log 25 = 1.3979, find log .
Given that ln 22 = , find ln 22. log (25¼) = ¼ log 25 = ln (22½) = ½ ln 22 =
Simplifying Expressions
Logarithmic expressions can be simplified using logarithmic properties and laws. Example 1: Write ln(3x) + 4ln(x) – ln(3xy) as a single logarithm. ln(3x) + 4ln(x) – ln(3xy) = ln(3x) + ln(x4) – ln(3xy) = ln(3x•x4) – ln(3xy) = ln(3x5) – ln(3xy) =
Simplifying Expressions
Simplify each expression. log 8x + 3 log x – log 2x2 log 4x2
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MGEA06H3 (157)
Iris Au (146)
Lecture 5
# Lecture notes for week 5
7 Pages
175 Views
Department
Economics for Management Studies
Course
MGEA06H3
Professor
Iris Au
Semester
Winter
Description
FISCAL POLICY AS AN AUTOMATIC STABILIZER AND THE AS-AD MODEL Outline N Discuss the concepts like inflationary gap, deflationary gap, and full-employment level of output. N The use of fiscal policy to smooth out business cycles. N Develop the AS-AD modelbringing price into the model. Our Extended Modelthe Budget Balance Again N Our extended model (let IM = 0 and hold r & E constant): 0 C = C 0+ c1DI T = T 0+ t1Y, 1 > 1 > 0 TR = TR tr Y, 1 > tr > 0 0 1 1 I = 0 G X = X 0 IM = im 1 AE = AE 0 + cYY, where AE 0 C +0I +0G + X + c0TR 1c T0 1 0 cY= [c 11 t1 tr1) im1] BB = (T 0 TR 0 G) + (t 1 tr 1 Y N Question: What happens to the budget balance when Y changes? N Answer: Holding all else constant (T , TR , and G are all held constant), when Y increases, BB increases. Similarly, when Y 0 0 decreases, BB decreases. The budget balance is ENDOGENOUS (it responds to a change in Y)! o A change in G, T 0, or TR0would have TWO effects on the BB. Why? BB = (T 0 TR 0G) + (t + 1r )1Y o When T 0 decreases, TR 0ncreases, or G increases, this will cause (0 TR0 G) to decrease BB decreases. (FIRST ROUND EFFECT) o When T decreases, TR increases, or G increases (change in AE ), this leads to a higher level of AE Y increases (t + 0 0 0 1 tr1) Y increases BB increases. (SECOND ROUND EFFECT) o First round effect dominates the second round effect. Full-Employment Level of Output and Output Gaps Full-Employment Level of Output, Y FE N It is the level of output consistent with full employment. N The term full employment means o Good (productive) workers find it easy to find a job. o Young (inexperienced) workers find it easy to find their first job. o Old (experienced) workers are not forced out of the labour force if they lose their jobs (i.e., no discouraged workers). N In other words, when the economy is operating at its normal pace (not too fast and not too slow), then we can say that the economy is at its full employment. N The unemployment rate at Y FE is called theon-accelerating inflation rate of unemployment (NAIRU) or natural rate of unemployment (NRU). N Full employment DOES NOT mean that everyone is working because o Frictional unemploymentunemployment that results from the turnover in the labour market as workers move between jobs. o Structural unemploymentlong-term and chronic unemployment arising from imbalances between the skills and other characteristics of workers in the market and the needs of employers. o The labour market is dynamic (people move in and out of the labour market). o There are frictions in the economy (some industriesregions expand while some industriesregions contract). Recessionary or Deflationary Gap N It occurs when the equilibrium level of output (Y*) is less than the full-employment level, i.e., Y* < Y . FE N This means that the economy is producing less output than is normally associated with full employment. N How? Firms lay off their workers (i.e. many unemployed workers). www.notesolution.com
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Submit | 987 | 3,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.826806 |
https://www.physicsforums.com/threads/pion-decay-lorentz-transforms.377204/ | 1,544,517,602,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823614.22/warc/CC-MAIN-20181211083052-20181211104552-00332.warc.gz | 948,528,420 | 14,186 | Homework Help: Pion Decay & Lorentz Transforms
1. Feb 10, 2010
ian2012
1. The problem statement, all variables and given/known data
A pi+, produced by a cosmic ray, has an energy of 280MeV in the laboratory frame S. It undergoes a decay: pi+ ---> muon+ & muon-neutrino. The muon produced continues to travel along the same direction as the pi+ and has a mean lifetime of 2.2mirco seconds in the muon's rest frame (S''). Calculate the expected distance it would have travelled in the laboratory frame S before decaying.
2. Relevant equations
In the rest frame of the pion (S'):
$$E^{'}_{\mu}=\frac{(m^{2}_{\pi}+m^{2}_{\mu})c^{2}}{2m_{\pi}}$$
$$P^{'}_{\mu}=\frac{(m^{2}_{\pi}-m^{2}_{\mu})c}{2m_{\pi}}$$
and other relativistic relations and Lorentz Transforms
$$m_{\pi}=139.6\frac{MeV}{c^{2}}, m_{\mu}=105.7\frac{MeV}{c^{2}}$$
3. The attempt at a solution
I have attempted the problem, I don't know if i am correct.
The obvious thing to do is find the mean lifetime in the frame S' and the distance in S' in order to solve the equation:
$$x_{\mu}=\gamma(x^{'}_{\mu}+vt^{'}_{\mu})$$
To find t' I have used the fact that time dilates in moving frames therefore t' = gamma x t'' (2.2 mircoseconds)
I then used the relativistic momentum equation to find the velocity of muon in the pion rest frame S': v' = 0.28177/gamma
Then i used LT to find the distance travelled in the pion frame = gamma x v x t'' = 0.28177 x t''.
Then I used the energy of the pion in the lab frame to find the velocity of the pion in the lab frame using the relativistic energy equation and the momentum equation to give = 1.739/gamma.
Subbing these into the above equation I got the expected distance = gamma x 4.45x10^-6 metres.
Is this correct? Is the gamma supposed to be there?
2. Feb 10, 2010
vela
Staff Emeritus
Looks okay so far.
So just to be clear: this gamma is for the pion moving relative to S, not the gamma above.
I don't follow what you did here. Could you provide more detail, like exactly what quantities you plugged into what equations?
The answer isn't right. You left out a factor of c, for one thing. Even after multiplying your answer by c, it's still off. You should get 1565 meters.
3. Feb 13, 2010
ian2012
Sorry, but I have been away for a while. Looking back at this question, I've realised what mistakes I've made:
1. I didn't realise gamma can be calculated using Energy and Mass (very trivial).
2. I didn't realise there are two different gamma's, since there are three different frames.
3. I also missed out the speed of light in the final answer, as you pointed out.
The final answer I get is: 1551 metres with minimal rounding of answers. (x' = 179m, v[lab/pion] = 0.867c, t' = 2.29 microseconds). This is a little bit off your 1565 metres.
4. Feb 13, 2010
vela
Staff Emeritus
You miscalculated x'. It should be 186 m.
5. Feb 14, 2010
ian2012
Oh right yeah of course. I divided by an extra gamma term. Yeah i get x' = 186m and x = 1.56km with rounding. Thanks vela | 884 | 2,990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-51 | latest | en | 0.856739 |
https://www.fearpk.com/cs604-assignment-2-solution-fall-2019/ | 1,660,473,009,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00221.warc.gz | 679,479,007 | 14,747 | # cs604 assignment 2 solution fall 2019
CS604: Operating Systems
Assignment 2 solution file
Question# 1:
Assume you have to apply Shortest Job First (SJF) scheduling algorithms on the set of different processes given in the table below. The CPU burst time is also given for each process. Consider that all the processes arrive in the ready queue within time 0 seconds except P2 that arrive in ready queue within time 8 seconds. You are required to show the Gantt Chart to illustrate the execution sequence of these processes and calculate the Total Waiting Time and Average Waiting Time for the given processes by using SJF algorithm.
Process CPU Burst Time (seconds) P0 2 P1 6 P2 1 P3 4 P4 3 P5 8
Solution:
Process CPU Burst time (seconds) Arrival Time P0 2 0 P1 6 0 P2 1 8 P3 4 0 P4 3 0 P5 8 0
GANTT CHART:
Consider that all the processes arrive in the ready queue within time 0 seconds except P2 that arrive in ready queue within time 8 seconds.
Waiting time = Starting time – Arrival time
Waiting time for P0= 0 – 0 = 0
Waiting time for P1= 10 – 0 = 10
Waiting time for P2= 9 – 8 = 1
Waiting time for P3= 5 – 0 = 5
Waiting time for P4= 2 – 0 = 2
Waiting time for P5= 16 – 0 = 16
Total Waiting Time = (P0 + P1 + P2 + P3 + P4 + P5 )= (0+10+1+5+2+16)= 34
Average Waiting time = Total Waiting Time / Total number of values
Average Waiting time = 34 / 6 = 5.66 seconds
Question# 2:
Consider a scenario where you have to apply Round Robin scheduling algorithm on the below given set of processes with each having a quantum size=8 milliseconds. The CPU burst time and arrival time for each process is also provided in the given table. You are required to show the Gantt Chart to illustrate the execution sequence of these processes. Moreover, calculate the Average Turnaround Time and Average Waiting Time for given processes by using round robin algorithm.
Process CPU Burst Time (Milliseconds) Arrival Time (Milliseconds) P0 15 0 P1 8 4 P2 25 18 P3 18 5
Solution:
A set of processes with each having a quantum size=8 milliseconds
Process CPU Burst Time (Milliseconds) Arrival Time (Milliseconds) Remaining Time Remaining Time Remaining Time Remaining Time P0 15 0 15 – 8 = 7 0 0 0 P1 8 4 8 – 8 = 0 0 0 0 P2 25 18 25 25 – 8 = 17 17 – 8 = 9 9 – 8 = 1 P3 18 5 18 – 8 =10 10 – 8 = 2 0 0
GANTT CHART:
Turnaround Time = Time A Process Terminate – Arrival Time
Turnaround Time for P0= 31 – 0 = 31
Turnaround Time for P1= 16 – 4 = 12
Turnaround Time for P2= 66 – 18 = 48
Turnaround Time for P0= 31 – 0 = 31
Turnaround Time forP3 = 57 – 5 = 52
Total Turnaround Time = P0 + P1 +P2 + P3 = 31 + 12 + 48 + 52 = 143
Average Turnaround Time = Total Turnaround Time /Total number of values
Average Turnaround Time = 143 / 4 = 35.75 Milliseconds
Waiting time = Turnaround Time – CPU Burst Time
Waiting time for P0= 31 – 15 = 16
Waiting time for P1= 12 – 8 = 4
Waiting time for P2= 48 – 25 = 23
Waiting time for P3= 52 – 18 = 34
Total Waiting Time = P0 + P1 + P2 + P3= 16 + 4 + 23 + 34 = 77
Average Waiting time = Total Waiting Time / Total number of values
Average Waiting time = 77 / 4 = 19.25 Milliseconds | 978 | 3,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.842728 |
http://web.archive.org/web/20100815101042/http:/ryba4.com/python/ramer-douglas-peucker | 1,411,405,810,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657137108.99/warc/CC-MAIN-20140914011217-00118-ip-10-234-18-248.ec2.internal.warc.gz | 298,483,447 | 9,262 | http://ryba4.com
# Ramer-Douglas-Peucker Algorithm in Python
© Dmitri Lebedev (http://ryba4.com) [http://ryba4.com], April 2010, version 1.1.
This is a Python implementation of Ramer-Douglas-Peucker algorithm that simplifies polylines.
Image by Jitse Niesen [http://en.wikipedia.org/wiki/User:Jitse_Niesen].
The algorithm works the following way:
• Make a straight line between the beginning and the end of the polyline.
• Calculate the distance between each point and the straight.
• If the farthest point is further than the threshold value, it's included in the end result, and the operation is repeated on the parts to the left and to the right of it.
To simplify the code, my implementation uses object-oriented approach, though it is not hard to re-write it in procedural style. The positive side-effect of OO style is that the function can work with points of any number of dimensions (as long as all the points have it the same).
In this code I use a special formula to calculate the distance between a point and a line. Here's how it looks:
We know X, Y and Z (begin, end and curr in the code). We can calculate automatically a and b vectors and need vector c's length (module). To find the longest vector, we may use the squares of the lengths. Now see the way of thought:
Note that |a|2 actually takes less calculations than |a|, so this formula gets rid of few x**.5 operators.
The code:
```def ramerdouglas(line, dist):
"""Does Ramer-Douglas-Peucker simplification of
a line with `dist` threshold.
`line` must be a list of Vec objects,
all of the same type (either 2d or 3d)."""
if len(line) < 3:
return line
begin, end = line[0], line[-1]
distSq = [begin.distSq(curr) -
((end - begin) * (curr - begin)) ** 2 /
begin.distSq(end) for curr in line[1:-1]]
maxdist = max(distSq)
if maxdist < dist ** 2:
return [begin, end]
pos = distSq.index(maxdist)
return (ramerdouglas(line[:pos + 2], dist) +
ramerdouglas(line[pos + 1:], dist)[1:])
```
Basically, that's all the algorithm's code. The rest is service code with line and vector classes.
```class Line:
"""Polyline. Contains a list of points and outputs
a simplified version of itself."""
def __init__(self, points):
pointclass = points[0].__class__
for i in points[1:]:
if i.__class__ != pointclass:
raise TypeError("""All points in a Line
must have the same type""")
self.points = points
def simplify(self, dist):
if self.points[0] != self.points[-1]:
points = ramerdouglas(self.points, dist)
else:
points = ramerdouglas(
self.points[:-1], dist) + self.points[-1:]
return self.__class__(points)
def __repr__(self):
return '{0}{1}'.format(self.__class__.__name__,
tuple(self.points))
class Vec:
"""Generic vector class for n-dimensional vectors
for any natural n."""
def __eq__(self, obj):
"""Equality check."""
if self.__class__ == obj.__class__:
return self.coords == obj.coords
return False
def __repr__(self):
"""String representation. The string is executable as Python
code and makes the same vector."""
return '{0}{1}'.format(self.__class__.__name__, self.coords)
if not isinstance(obj, self.__class__):
raise TypeError
return self.__class__(*map(sum, zip(self.coords, obj.coords)))
def __neg__(self):
"""Reverse the vector."""
return self.__class__(*[-i for i in self.coords])
def __sub__(self, obj):
"""Substract object from self."""
if not isinstance(obj, self.__class__):
raise TypeError
return self + (- obj)
def __mul__(self, obj):
"""If obj is scalar, scales the vector.
If obj is vector returns the scalar product."""
if isinstance(obj, self.__class__):
return sum([a * b for (a, b) in zip(self.coords, obj.coords)])
return self.__class__(*[i * obj for i in self.coords])
def dist(self, obj = None):
"""Distance to another object. Leave obj empty to get
the length of vector from point 0."""
return self.distSq(obj) ** 0.5
def distSq(self, obj = None):
""" Square of distance. Use this method to save
calculations if you don't need to calculte an extra square root."""
if obj is None:
obj = self.__class__(*[0]*len(self.coords))
elif not isinstance(obj, self.__class__):
raise TypeError('Parameter must be of the same class')
# simple memoization to save extra calculations
if obj.coords not in self.distSqMem:
self.distSqMem[obj.coords] = sum([(s - o) ** 2 for (s, o) in
zip(self.coords, obj.coords)])
return self.distSqMem[obj.coords]
class Vec3D(Vec):
"""3D vector"""
def __init__(self, x, y, z):
self.coords = x, y, z
self.distSqMem = {}
class Vec2D(Vec):
"""2D vector"""
def __init__(self, x, y):
self.coords = x, y
self.distSqMem = {}
```
Highlighted with Pygments [http://pygments.org].
This code at work:
Figure 1. Original line in 10*10 km square. 492 points.
Figure 2. Simplified line, threshold = 50 m. ~150 points.
Figure 3. Simplified line, threshold = 100 m. ~90 points.
Figure 3. Simplified line, threshold = 250 m. ~45 points.
The data is a coastline from this place [http://osm.org/go/2ttZPN0-] in OpenStreetMap [http://osm.org]. Coordinates are in kilometres.
I know this code isn't the fastest, and procedural approach (which I actually like more) can work faster on larger amount of data, but I finally chose OO approach, as it allows readable and maintainable code. Here is a procedural-style implementation [http://mappinghacks.com/2008/05/05/douglas-peucker-line-simplification-in-python/]. | 1,400 | 5,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2014-41 | longest | en | 0.859603 |
https://landsurveyorsunited.com/group/excel/forum/topics/vertical-curve-and-stting-out?commentId=544331%3AComment%3A449361&xg_source=activity&groupId=544331%3AGroup%3A19425 | 1,550,889,978,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249434065.81/warc/CC-MAIN-20190223021219-20190223043219-00072.warc.gz | 638,868,263 | 28,029 | # Land Surveyors United
Earth's Largest Land Surveyor Community
# Vertical curve and setting out spreadsheet utility
Fellow Surveyors,
enjoy
This Content Originally Published by a land surveyor to Land Surveyors United Network
Views: 2385
### Replies to This Discussion
Nice work..beautiful tool AAMIR!
thanks
Good Job thank u
superb job. thanks
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41 members | 812 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.838625 |
https://lostinamericafilm.com/a-brief-guide-to-the-rules-of-poker/ | 1,720,928,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00027.warc.gz | 324,658,104 | 10,065 | # A Brief Guide to the Rules of Poker
Before you can play poker, you must know its rules. These include betting rounds, Hand rankings, and Game theory. If you have no idea about the basics, read the following articles. They will help you understand the game of poker and help you make the right move. Here is a brief guide to the rules of poker:
## Rules
Depending on the type of game you play, there are many rules to consider. The rules of poker determine what happens when a player makes the best possible poker hand. A Royal Flush is the strongest hand, while a high card is the weakest. Learn what the rules are for different poker hands and how they’re calculated to determine who wins. You’ll also find more information on hand rankings and specific types of poker, such as draw poker.
## Betting rounds
Poker betting rounds are a necessary part of the game. If you’re just starting out, you’ll want to know the basics of the game. You should know the difference between high hands and low hands and how many betting rounds are required. There are also several rules to follow when it comes to the betting rounds. You can find more information about them below. Here are a few common ones. Betting rounds in poker depend on whether someone has already bet or not.
## Hand rankings
To understand the hand rankings when playing poker, you must know how to raise or fold your hand. When you raise, you add chips to the pot, while when you fold, you match your opponents’ bet. A player can win the pot only with a high-quality hand. Hand rankings are based on the rank of the top card, the second and third cards, and the kicker. The top two hands have the highest ranking.
## Game theory
The game theory of poker describes the relationships between three different orders of players in a card game. It offers an analytical framework for studying poker and other games. Essentially, poker is an interplay between three orders that affects the outcome of the game. In other words, the game can be described as a mathematical model. There are many similarities between poker and other games, and this theory can help you better understand the nuances of the game.
## Poker lingo
Learning about the lingo of poker is essential if you want to win. Poker players speak a variety of different terms, and it is helpful to know the most common ones. For example, the sandbag strategy refers to a player who checks and raises with three tens and an ace. Another common poker term is scare card, which refers to the next card dealt that may produce a better hand. | 532 | 2,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.960911 |
https://www.exceldemy.com/value-error-in-excel/ | 1,709,305,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00556.warc.gz | 743,258,082 | 79,448 | # [Fixed!] VALUE Error in Excel (8 Reasons with Solutions)
Get FREE Advanced Excel Exercises with Solutions!
Excel is a powerful tool for data analysis and manipulation, but it can be frustrating when errors occur in your formulas or functions. One of the most common errors is the #VALUE! error in Excel, which occurs when a formula or function references a cell with an invalid data type.
This error can prevent your spreadsheet from producing accurate results and may require troubleshooting to fix.
Understanding the causes of the #VALUE! error and how to resolve it can help you avoid data errors and improve the accuracy of your Excel workbooks.
In this article, we’ll show 8 reasons and solutions for the #VALUE! error in Excel with clear images.
## [Fixed!] VALUE Error in Excel: 8 Reasons with Solutions
To explore the reasons and solutions, we’ll use the following dataset that represents some fruit products’ order dates, quantities, and prices.
### Reason 1: If Text Is Used in Arithmetic Operations During Calculation
Here, we used the basic arithmetic operation to calculate the total price for each order. But for some unit prices, it is returning a #VALUE! error.
Have a look at the highlighted cells, the cells contain text values instead of numbers, arithmetic operations can’t deal with text values so it’s returning a #VALUE! error.
#### Solution: Use PRODUCT Function to Solve VALUE Error
To avoid this situation, use functions instead of direct arithmetic operations.
Here, we used the PRODUCT function to calculate the total price and it ignored the text values. The formula is-
`=PRODUCT(D5,E5)`
Also, you can use SUM, SUMIF, or any other functions that replace the particular kind of arithmetic Operations.
### Reason 2: Blank Cell Contains Hidden Spaces Work as Text
While performing an arithmetic operation, if we multiply a value by an empty cell then it returns zero because a blank cell means zero. But in the following calculation, we got the #VALUE! error instead. This happens when spaces remain in the cells, and without selecting the cell it’s not possible to identify it.
Look, there is a space character in cell E5.
#### Solution: Use Functions Instead of Arithmetic Formula
Using functions can also rescue us from this kind of situation. Space character works like text values and we already described before functions can ignore text values.
So to multiply the quantity and unit price we used the PRODUCT function again-
`=PRODUCT(D5,E5)`
### Reason 3: Cell Contains Leading Space Before Number or Dates
Here, we subtracted the first order date from the last order date to calculate the total working days of taking orders. Unfortunately, the formula returned a #VALUE! error in Excel although the format of the dates is correct.
The problem is for the space character again. See, the first date contains a leading space that’s why the formula considers it as a text value.
You will face the same issue while using the other date-related functions like EDATE, and NETWORKDAYS functions.
#### Solution: Remove Spaces Manually or Use TRIM Function
When you doubt facing this kind of situation, use the TRIM function before every cell reference. It will remove all leading and trailing spaces from the reference cells.
We used the following formula and then it returned the desired result-
`=TRIM(B18)-TRIM(B5)`
### Reason 4: Not Applying Proper Keys While Performing an Array Formula
Before Microsoft Excel 365, we needed to use the CTRL + SHIFT + ENTER keys instead of pressing the ENTER key to perform any array operations in the earlier versions of Excel.
In an earlier version of Excel, we typed the following array formula with the SUM function and pressed only the ENTER key-
`=SUM(D5:D18*E5:E18)`
And look, the formula returned a #VALUE! error in Excel.
#### Solution: Press CTRL+SHIFT+ENTER Altogether to Perform Array Formula
Never forget to press the CTRL + SHIFT + ENTER keys together while using an array formula. After applying it, we got the desired result.
### Reason 5: Incorrect Date Format Creating VALUE Error in Excel
If we use any date format that Excel doesn’t support then it will return a #VALUE! error.
We used the following arithmetic formula to get the total working days, but as the first date is not valid for Excel it returned the #VALUE! error.
#### Solution: Choose the Correct Date Format
The solution is simple, just insert the date using the correct date format.
Look, after using the right format, we got the output without facing any errors.
### Reason 6: Inappropriate Range in Arguments of a Function
Some logical functions in Excel need to match the ranges by maintaining the same size, otherwise, they will return a #VALUE! error.
Here, we applied the SUMIFS formula to return the total price for the carrot but got a #VALUE! error.
`=SUMIFS(F5:F15,C5:C18,D20)`
Because the criteria range and the sum range are not the same sizes. This issue may happen and return a #VALUE! error in Excel while using some other functions like COUNTIFS, SUMPRODUCT, and XLOOKUP functions.
#### Solution: Match Ranges According to Arguments
Always maintain the ranges in the arguments appropriately. After changing the wrong range we got the correct output.
### Reason 7: Incorrect Column Index Number in Lookup Functions
When we use some lookup functions in Excel like the VLOOKUP, and INDEX-MATCH functions to return output for a lookup value, we need to specify the column index number. According to that index number, the functions return corresponding output from the array.
If you mistakenly insert the wrong or invalid index number then you will get the #VALUE! error.
Look at the below image, we used the two following formulas to get the total price for the fruit and they returned a #VALUE! error.
`=VLOOKUP(D20,C5:F18,0)`
`=INDEX(C5:F18,MATCH(D20,C5:C18,0),-1)`
The reason is- that the column index number must have been greater than 0, as we inserted 0 and -1 so the #VALUE! error occurred.
#### Solution: Insert the Proper Index Number According to Array
Insert the correct index number relative to the array.
As we wanted to get the total price our index number is 4 according to the selected array, after modifying we got the correct result.
### Reason 8: Formula Referring a Cell with VALUE Error Returns VALUE Error
Suppose you have a large dataset or referring to a cell from another sheet, if that cell contains a #VALUE! error then you will get the #VALUE! error again if you use that cell in a formula.
We had some VALUE errors in column F and we applied an arithmetic formula to get 5% VAT and got #VALUE! errors again.
#### Solution: Eliminate VALUE Errors from the Reference Cells
Remove the #VALUE! errors by following the described other solutions that match your scenario.
## How to Find All Cells with VALUE Error in Excel
To avoid unwanted situations while using a formula or to fix errors, we need to find them first where they are remaining, especially for a large sheet. The Find and Replace tool is very useful for it because it can find out the errors too. Let’s find the errors from the following dataset using this tool.
• Go to the Home ribbon, then click as follows: Editing > Find & Select > Find…
• After appearing in the Find and Replace tool dialog box, click on Options >> It will open up some other find options.
• Type VALUE! in the Find what box.
• Select the Sheet option from the Within box.
• Choose Values from the Look in the dropdown.
• Finally, click on the Find All tab.
• Soon after, the dialog box will show you the result like the following image. It will provide you with the Excel file name, sheet name, cell name, and formula.
## What Is the Ultimate Solution for All Errors? (Using IFERROR Function)
Suppose you already know that there are some errors in your sheet or may make errors after applying the formula but you want to hide them or detect them by a specific text then you can easily handle it by applying the IFERROR function.
Here, we used the following formula to return the text ‘Çheck the Value’ instead of the #VALUE! errors-
`=IFERROR(D5*E5,"Check the Value!")`
Note: there is a drawback of using the IFERROR function, it can’t detect the other error- #N/A or #DIV/0, #VALUE, #REF, etc independently. It considers all the errors in the same way.
• Why does Excel say #VALUE! error for no reason?
Actually, if the #VALUE! error occurs then there must be a reason. The #VALUE! error in Excel typically occurs when a formula or function in a cell references an invalid data type, such as trying to perform a mathematical operation on text or a non-numeric value. This error can also occur if a function expects a certain type of input, such as a range of cells, but is instead provided with a single cell or a value that cannot be interpreted.
• Can we face #VALUE! error while summing, multiplication, or dividing in Excel?
Yes, if you use a direct arithmetic formula for summing, multiplication, or dividing in Excel instead of using functions then the #VALUE! error can occur.
• Can #VALUE! error happens in Excel IF function?
Yes, it is possible to get a #VALUE! error when using the IF function in Excel. This can happen if the formula within the IF statement returns a non-numeric or invalid value, or if one or more of the arguments provided to the IF function are not of the expected data type.
• Can number stored as text return VALUE error in Excel?
Yes, some earlier versions of Excel can return a #VALUE! error if you insert a number in a formula that is stored as text.
## Conclusion
In conclusion, the #VALUE! error in Excel is a common issue that can occur when a formula or function references an invalid data type or value. This error can be frustrating and can cause inaccurate calculations and analysis.
However, by carefully checking formulas and ensuring that all inputs are correct and in the expected format, users can avoid this error and improve the accuracy of their Excel spreadsheets.
Additionally, using error handling functions and other Excel tools can help to catch and correct any errors that may occur, making it easier to work with large datasets and complex formulas. Leave a comment if you have further queries or have something to add.
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Advanced Excel Exercises with Solutions PDF | 2,427 | 11,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.799895 |
https://nl.mathworks.com/matlabcentral/answers/304242-how-can-i-delete-some-specific-rows-from-a-cell | 1,713,505,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00086.warc.gz | 398,322,374 | 26,682 | Info
# How can I delete some specific rows from a cell?
1 view (last 30 days)
Thimiod Athan on 23 Sep 2016
Closed: MATLAB Answer Bot on 20 Aug 2021
How can I delete som specific rows from a cell according to the matrix(see the pic). When the elementw of the Matrix are1 I must delete the same rows from the cell. Any ideas?
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##### 3 CommentsShow 1 older commentHide 1 older comment
Thimiod Athan on 24 Sep 2016
Edited: Thimiod Athan on 24 Sep 2016
Hello Adam. I want to delete the 2 raws (with the red color) from the following cell(6X3)
Thimiod Athan on 24 Sep 2016
I think I find it Table(find(A==1),:)=[]. It was sooo easy...
Image Analyst on 23 Sep 2016
Edited: Image Analyst on 23 Sep 2016
What is "elementw" and "Matrix"? All we see are "Table" and "A".
If you have a cell array called ca, and you have a logical column vector "A", you can remove cells from ca where A is 1 or true like this:
caNew = ca(A==0); % Extract cells where A=0 (i.e. remove cells where A=1).
Or, equivalently
caNew = ca; % Initialize
caNew(A) = []; % Remove cells where A=1
##### 3 CommentsShow 1 older commentHide 1 older comment
Thimiod Athan on 24 Sep 2016
I think I find it Table(find(A==1),:)=[]. It was sooo easy...
Image Analyst on 24 Sep 2016
Even easier is to use logical indexing. No need for find():
Table(A==1,:)=[];
table is a built in function in MATLAB. Even though MATLAB is case sensitive, so table is different than table, it's a good idea not to use variable names that are the "same" as function names.
I could have told you this . . . if you'd been a little more careful about your question. For example you never mentioned the "Table" variable at all in your question. And you used the word "cell" in a non-MATLAB-ish way. Go here http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F for a very good discussion of cells in MATLAB.
Vignesh Murugavel on 3 Aug 2021
Using Logical Indexing
Table(A==1,:)=[]; | 562 | 1,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.885307 |
http://dllahr.blogspot.com/2012/04/moving-faster-than-speed-of-light.html | 1,532,360,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596542.97/warc/CC-MAIN-20180723145409-20180723165409-00361.warc.gz | 95,815,871 | 18,918 | ## Saturday, April 7, 2012
### Moving faster than the speed of light
I've been thinking about physics a lot lately, and I'm starting to jot things down, so I don't keep going over the same ground again but also to help me iron out the logical inconsistencies that can creep in when you do a problem in Caput. A lot of this is really me just thinking about physics that I've read, and doing thought experiments so I can understand it.
This post is thoughts about what would happen if you moved faster than the speed of light. Start with these premises:
• The only way we know about a particle (anything really) is the effect / force that particle exerts on other particles.
• You push a block with your hand: the electric / magnetic forces from the electrons in the atoms in the proteins / molecules in the cells in your hand interact with the electrons in the atoms in the molecules (cellulose) in the block of wood
• The above example is for electricity and magnetism, but (I've read) applies equally well to other, more exotic forces - e.g. the strong nuclear force between quarks in an atom's nucleus
• The forces between particles can be represented as fields. Fields are vectors that exist through space that indicate the force (magnitude, direction) that a "test" particle would experience if it were at that location
• Imagine 2 charged particles. From Coulomb's law, we can calculate the force between them. Or, for each particle we could determine the field it generates throughout space. Then, the force on each particle is determined by the field generated by the other particle.
• Movement of particles causes changes in the fields
• As the location of 2 particles gets closer together, the force they exert on each other increases. Similarly, the field strength increases.
• The changes in the fields propagates at the speed of light
The above might sound crazy, but they are well established physics, with tons of experimental evidence. Given the above it is almost nonsensical to talk about a particle moving faster than the speed of light. Which is somewhat expected - the above description of reality is based on the tenet that nothing travels faster than light. But the exercise of investigating what would happen if something moved faster than light helps me understand the relationships. So, 2 scenarios to imagine:
#### Particle approaches at faster than light
The particle will arrive at a location before the effect of the particle being at the location does. This is just logically inconsistent.
#### Particle moves away faster than light
This situation is harder to rule out. As the particle recedes, it is not arriving before its effect. The problem with this one occurs for two situations I can think of:
1. Imagine another particle, chasing this one. The "effective" location, based on the fields, is only moving at the speed of light. In this case, the particle has effectively "disappeared". The chasing particle sees only the location represented by the field
2. Imagine instead of a single particle, an atom moving faster than the speed of light. Background: for a stationary atom emitting radiation, the frequency is intrinsic to the motion of the oscillation of the electron(s) within the atom. The radiation, regardless of the relative velocity between the emitting atom and the observer, propagates at the speed of light. The wavelength is determined by the frequency and the speed of light.
Now, for the atom moving faster than the speed of light: Take the period of oscillation, imagine the first cycle has occurred. Now, in this period of time, the atom has traveled a distance greater than the wavelength of the radiation, and a new cycle occurs. So the separation in peaks / troughs between the first and second cycle is greater than the wavelength (as it would be defined for regular sub-luminal speeds). Furthermore, for the third cycle, the discrepancy is even greater. So, even though the atom is travelling at constant velocity, the radiation is continuously increasingly red-shifted (chirped down!). Effectively, as time goes on, the emission of radiation is red-shifted until it would disappear completely. Now, this description is discrete, but it could be made continuous.
Why would the above be impossible or inconsistent? Well, the particle, in this case, has effectively disappeared from the universe, since internally it is emitting radiation, but this vanishes / does not appear anywhere else.
The reverse of this is also possible to imagine, in which an atom emitting radiation approaches at faster than the speed of light, and the radiation is continuously increasingly blue shifted. In this case, leaving aside the issue from above of the particle arriving before its effect, the radiation observed would be increasingly blue shifted over time (chirped up!). Where is the increased power / energy coming from? Again, the internal state of the atom is disconnected from the rest of the universe.
1. If you want to use "caput" analogous to the use of vitrum, as in "in vitro" - then you should use the ablative - "in capite."
2. (This is Joel. Megan pointed me to the site.)
For the first part of the post, that's an interesting point but I might have a simple counterexample. Just imagine that instead of a vacuum, we're doing the problem in water. The speed of light in water is about 0.75*c. EM waves should propagate at that speed as well. But of course we know that an electron can travel through water at a speed which exceeds 0.75*c (the reason for Cherenkov radiation). So there is clearly no logical contradiction there; I'm not sure if I see where one would come in once we move back into a vacuum.
(Not that superluminal particles are possible; they of course aren't. Also, I'll mention that my inspiration for thinking about the effect in water comes from a paper that Shel Glashow put out before the OPERA superluminal neutrino results were conclusively put down that pointed out that a superluminal particle would emit Cherenkov-like radiation leading to an energy spectrum inconsistent with the OPERA results. I'll confess that I didn't read the paper, only the APS synopsis:
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.181803 )
So, when you write "The particle will arrive at a location before the effect of the particle being at the location does" is that right? Won't the EM field arising from when the particle is at point 'A' arrive at point 'A' at the exact same time as the particle (sure the particle is moving faster than the field, but the distance form point 'A' to point 'A' is 0). It is only true to say that the particle will arrive before the effects of where it previously was, but that might not be a problem.
Anyway, those were just the thoughts that popped into my head and I certainly might have missed something. Haven't thought too much yet about the red/blueshifting ideas you mention at the end.
3. Hi Joel! Thanks for the thoughts. Good point about Cherenkov radiation. Reading the wikipedia page (http://en.wikipedia.org/wiki/Cherenkov_radiation), I'm reminded that when light propagates through a material it is by causing the electrons / nuclei in that material to respond to the electric field of the light. So, for example, when light strikes a surface, you get an initial response from the electrons at the surface. That initial response then triggers further electric fields that propagate into the material, and the process repeats. The slower than c speed is due to the "responsiveness" of the material's electrons to the applied electric fields.
In that respect, if we zoom in and view the subatomic area around the charged Cherenkov particle, the electric fields in its immediate vicinity are still travelling at c. It seems that they are "damped" out at further distances - at any further distance, the electric field is dominated by the intervening matter. It is only when that matter (electrons / nuclei) move in response to the cherenkov particle that the field from the cherenkov particle propagates.
That's my initial thoughts, remembering what I've read about dielectric responsiveness of materials, I'll need to go back and see what I can pull out of my texts about Cherenkov. | 1,800 | 8,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.943727 |
https://r-lang.com/how-to-convert-r-matrix-to-vector/ | 1,653,183,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00031.warc.gz | 546,909,984 | 24,323 | # How to Convert R Matrix to Vector
To convert Vector to Matrix in the R language, use the matrix() function. The vector can hold multiple elements, but all the elements are of the same data type. R Matrix is a vector with attributes of a dimension and optionally, dimension names attached to a Vector.
## Convert R Matrix to Vector
To convert Matrix to Vector in R, use the c() function. The c() function combine values into a Vector or List. The c() is an inbuilt generic function that combines its arguments to form a vector. All arguments are coerced to a standard type which is the type of the returned value and all attributes except names are removed.
### Example
To create a matrix in R, use the matrix() function.
``````mat <- matrix(1:12, nrow = 4)
mat``````
#### Output
`````` [,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12``````
You can see that we created a 4 X 3 matrix.
To convert this matrix to a vector, use the c() function and pass this matrix in the argument.
``````mat <- matrix(1:12, nrow = 4)
mat
rv <- c(mat)
rv``````
#### Output
`````` [,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
[1] 1 2 3 4 5 6 7 8 9 10 11 12``````
That is it. We get the vector from the matrix. The c() function converts a matrix into a one-dimensional vector.
## Converting Matrix to Vector Using as.vector() Function
The as.vector() function converts a distributed matrix into a non-distributed vector. The as.vector() takes a matrix as an argument and returns the vector.
### Syntax
``as.vector(x, mode = "any", proc.dest = "all")``
### Parameters
x: numeric distributed matrix.
mode: A character string giving an atomic mode or “list“, or (except for ‘vector’) “any”.
proc.dest: destination process for storing the matrix.
### Example
``````mat <- matrix(1:12, nrow = 4)
mat
cat("After converting from matrix to vector", "\n")
rv <- as.vector(mat)
rv``````
#### Output
`````` [,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
After converting from matrix to vector
[1] 1 2 3 4 5 6 7 8 9 10 11 12``````
You can see that we get the vector using as.vector() function. The as.vector() method in R Language is used to convert an object into a vector. That is it for this conversion tutorial.
Categories R | 751 | 2,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.708929 |
http://booksreadr.net/ppt/simple-machines-powerpoint | 1,408,807,342,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500826259.53/warc/CC-MAIN-20140820021346-00053-ip-10-180-136-8.ec2.internal.warc.gz | 28,627,912 | 13,757 | # Simple Machines Powerpoint ppts
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simple machines powerpoint - Science Education at Jefferson Lab
The 6 Simple Machines Wedge Screw Inclined Plane Pulley Wheel and Axle Lever Energy: Ability to do work Work= Force x Distance Force: A Push or a Pull Definitions: Inclined Plane Inclined Plane The Egyptians used simple machines to build the pyramids.
http://education.jlab.org/jsat/powerpoint/0708_simple_machines_8.ppt
Date added: October 14, 2011 - Views: 312
Simple Machines - Patton Museum Online: Official General ...
Simple Machines What is a Simple Machine? A simple machine has few or no moving parts. Simple machines make work easier Wheels and Axles The wheel and axle are a simple machine The axle is a rod that goes through the wheel which allows the wheel to turn Gears are a form of wheels and axles ...
http://www.generalpatton.org/education/lesson_plans/Simple_machines.ppt
Date added: October 14, 2011 - Views: 177
Simple Machines - Salem City Schools in Salem, Virginia
Simple Machines SOL 3.2 Machine – a tool that helps us do work Machines help us by: Changing the amount of force on an object. Changing the direction of the force.
http://salem.k12.va.us/staff/hroutt/powerpoints/Simple%20Machines.ppt
Date added: October 14, 2011 - Views: 102
Simple machines and work power point - Science Education at ...
Work and Simple Machines What is work? In science, the word work has a different meaning than you may be familiar with. The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)
http://education.jlab.org/jsat/powerpoint/work_and_simple_machines.ppt
Date added: October 12, 2011 - Views: 434
Simple Machines - Curriculum
a lesson for: Lonna Hanson’s Third Grade Science Classes Lincoln Elementary Introduction Simple machines are machines with few or no moving parts.
http://curriculum.k12.sd.us/LH055/Simple%20Machines-PowerPoint.net.ppt
Date added: November 11, 2011 - Views: 71
Simple Machine PowerPoint - Quia
Simple Machines SOL 3.2 By Ms. Weinberg Simple machines are tools used to make work easier. You have probably used some simple machines, but did not realize that they were actually called simple machines!
http://www.quia.com/files/quia/users/amiew/Simple-Machines
Date added: October 3, 2011 - Views: 92
PowerPoint Presentation - Work, Power, & Simple Machines
Title: PowerPoint Presentation - Work, Power, & Simple Machines Author: Freeman High School Last modified by: wwprsd Created Date: 2/7/2002 3:20:07 PM
http://wwpms.sharpschool.com/UserFiles/Servers/Server_10640393/File/Milman/Work%20Power%20Simple%20Machines.ppt
Date added: December 27, 2012 - Views: 69
Simple Machines Review - TeachEngineering
Title: Simple Machines Review Author: Franklin D. Roosevelt Last modified by: denise Created Date: 4/19/2005 1:11:32 AM Document presentation format
http://www.teachengineering.org/collection/cub_/lessons/cub_simple/cub_simple_lesson01_presentation.ppt
Date added: October 14, 2011 - Views: 68
Simple Machines Review - RCPS: Always Connected, 100% Charged
January 30, 2001 Ms. Ashby’s Third Grade Class A machine is anything that helps us do work. A simple machine has few or no moving parts. There are 6 types of simple machines.
http://www.rockingham.k12.va.us/resources/elementary/files/3SimpleMachines.ppt
Date added: October 14, 2011 - Views: 69
Simple Machines - The Global Classroom
Simple machines are useful because ... all of the simple machines are shown. Create some clouds around each picture and add some examples for each simple machine ... Times New Roman SchoolBoy Batang Tahoma Default Design PowerPoint Presentation PowerPoint Presentation ...
http://www.globalclassroom.org/together/simple_machines.ppt
Date added: November 7, 2011 - Views: 94
PowerPoint Presentation
Click on this link: http://www.solpass.org/3s/newscience3.asp Choose a simple machine game or ... friction work force gravity A machine made up of multiple simple machines is ... Times New Roman simple machines Bitmap Image PowerPoint Presentation PowerPoint Presentation PowerPoint ...
http://www.schools.ccps.k12.va.us/sites/K5/PowerPoint%20%20Its%20Elementary/simple%20machines.ppt
Date added: December 4, 2011 - Views: 24
Simple Machines - LaVergne Middle School
Simple Machines What do machines do? Why do we need machines? Common Simple Machines… First, rank the following words from 0-5 (0= you have never heard of this word & 5= you know the word VERY well).
http://www.lms.rcs.k12.tn.us/TEACHERS/FloydA/web/documents/JoyceSimpleMachines.ppt
Date added: May 6, 2013 - Views: 18
Simple Machines (2) - Suffolk Teaching Activities & Resources
Simple Machines SOL 3.2 By Ms. Weinberg Adapted by K. Green ITRT – Suffolk City Schools 6 Simple Machines Simple Machine Video & Quiz Simple Machine Websites Information More Links & Information EdHead Click the Green, Start Box DirtMeister – Simple Machines 6 Simple Machines Simple Machine ...
http://star.spsk12.net/science/k3/SimpleMachines.pps
Date added: May 14, 2012 - Views: 148
Simple and Compound Machines - Marshall Technology Education
Simple and Compound Machines Simple and Compound Machines Gateway To Technology Unit 5 – Lesson 5.3 – Applied Physics Simple and Compound Machines Gateway To Technology Unit 5 – Lesson 5.3 – Applied Physics Simple and Compound Machines Gateway To Technology Unit 5 – Lesson 5.3 ...
http://marshallteched.com/Simple_and_Compound__Machines.ppt
Date added: December 12, 2011 - Views: 23
Simple Machines and Mechanical Advantage - South Kingstown ...
Simple Machines and Mechanical Advantage Simple Machines Ancient people invented simple machines that would help them overcome resistive forces and allow them to do the desired work against those forces.
Date added: May 6, 2013 - Views: 14
Introduction to Simple Machines - Lamar R-1
Simple Machines Helping us lift, lower, fasten, split, cut, divide, and move! What is a Simple Machine? A simple machine has few or no moving parts.
http://www.lamar.k12.mo.us/eastwestlmc/powerpoint/5-Simple-machinesfinal.ppt
Date added: October 14, 2011 - Views: 35
Work, Power, & Simple Machines
Work, Power, & Machines Chapter 14 What is work ? The product of the force applied to an object and the distance through which that force is applied.
http://cc040.k12.sd.us/PPTS/work%2Cpower%2Cand%20machines.ppt
Date added: October 22, 2011 - Views: 99
Simple Machines - Florida Science Olympiad
Mike McKee University of Central Florida Orlando, Florida October 7, 2006 Simple Machines What is the purpose of a simple machine? What should you be measuring as evidence of this purpose?
Date added: December 12, 2011 - Views: 49
Mr. King’s Science Classes Introduction Simple machines are machines with few or no moving parts. There are 7 main Simple Machines. Kinds of Simple Machines We will be learning about the seven simple machines.
Date added: November 18, 2011 - Views: 38
Math of Simple Machines - Colorado School of Mines
Math of Simple Machines Tuesday July 24, 2007 Ratio and Proportion Work Anything that pushes or pulls to move an object across a distance. Mechanical Advantage Efficiency of a Machine An Ideal Machine has 100% efficiency Simple Machines The Lever The Pulley The Wheel and Axle The Inclined Plan ...
http://mcs.mines.edu/Research/k12-partnership/workshops/2007-2008/Math%20of%20Simple%20Machines.ppt
Date added: November 5, 2011 - Views: 74
Chapter 12 – Simple Machines - School District of Spring ...
Title: Chapter 12 – Simple Machines Author: HS LMC Last modified by: Ned Hilleren Created Date: 9/12/2006 6:48:12 PM Document presentation format
http://www.springvalley.k12.wi.us/html/hillerenn/documents/C-12notes.ppt
Date added: March 1, 2012 - Views: 56
Simple Machine Test - Welcome to Shawano School District
Simple Machine Test Put your name at the top of your paper. Number your paper from 1 to 10. Good Luck #1 What Simple Machine Is This? #2 What Simple Machine Is This?
http://www.sgsd.k12.wi.us/homework/ballwaj/Science/Simple%20Machine%20Test.ppt
Date added: October 2, 2012 - Views: 37
PowerPoint Presentation
Simple Machines Inclined Plane Lever Pulley Wedge Screw Wheel and Axle . Title: PowerPoint Presentation Author: IMT Last modified by: IMT Created Date: 9/19/2011 7:50:24 PM Document presentation format: On-screen Show (4:3) Company: PUSD Other titles:
Date added: June 19, 2012 - Views: 4
Sunken Millions Simple Machines - Jefferson County Public Schools
You’ve recovered the sunken millions. Sunken Millions Simple Machines Mrs. Price Grade 3 Congratulations! You ... Inclined plane B. Wedge C. Screw D. Pulley \$600 A door is an example of what simple ... Sunken Millions Simple Machines Subject: PowerPak for PowerPoint 1.0 Author: Jefferson ...
http://classroom.jc-schools.net/sci-units/games/Millions-machines.ppt
Date added: September 28, 2011 - Views: 35
Slide 1
Simple Machines Big Question: What great ideas make our lives easier? Author: Allan Fowler Genre: Expository Nonfiction Word Reading choice toy enjoy boy inventor helper avoid Tools How They Are Used Unit 5 Simple Machines - Day 3 One of the tools we learned ...
Date added: September 2, 2011 - Views: 60
PowerPoint Presentation
ACADs (08-006) Covered. Keywords. Levers, gears, cams, pulleys, physics terms, units, mechanical principles, efficiency, machine. Description. This PowerPoint presentation can be used to train people about the basics of simple machines.
Date added: November 2, 2013 - Views: 5
Simple Machines Powerpoint - Duxbury Public Schools / Homepage
Simple Machines What is a Simple Machine? A simple machine has few or no moving parts. Simple machines make work easier Wheels and Axles The wheel and axle are a simple machine The axle is a rod that goes through the wheel which allows the wheel to turn Gears are a form of wheels and axles ...
http://www.duxbury.k12.ma.us/cms/lib2/MA01001583/Centricity/Domain/488/Simple_machines.ppt
Date added: October 9, 2013 - Views: 8
Simple Machines - Sevier
Simple Machines Your Task: Watch this slide show Draw a picture of each kind of simple machine on your answer sheet Write a short description of why we use simple machines Simple Machines Machines do not reduce the amount of work for us, but they can make it easier.
http://www.pbp.sevier.org/Simple%20Machines3.ppt
Date added: October 14, 2011 - Views: 59
Simple Machines - Wikispaces - ksmithscience - home
Simple Machines Group MM PULLEY! Definition Pulley is a simple machine that involves a wheel, that holds a rope or line. It helps to spread out the weight to lessen the force needed and used so that more weight can be lifted FIXED PULLEY MOVABLE PULLEY COMBINED PULLEY Mechanical Advantage The ...
http://ksmithscience.wikispaces.com/file/view/Simple+Machines-Group+M+M.ppt
Date added: May 2, 2013 - Views: 17
PowerPoint Presentation
The 6 Simple Machines Wedge Screw Inclined Plane Pulley Wheel and Axle Lever Energy: Ability to do work Work= Force x Distance Force: A Push or a Pull Definitions: Inclined Plane Inclined Plane The Egyptians used simple machines to build the pyramids.
http://www.asd5.org/cms/lib4/WA01001311/Centricity/Domain/638/0708_simple_machines_abbreviated%20ppt.ppt
Date added: March 17, 2014 - Views: 5
Simple Machines - Tenafly Public Schools
Simple Machines Physics Mrs. Coyle What are some simple machines? Lever Incline Plane Pulley Simple Machines Multiply the effort force (allows you to use a smaller effort force to raise a heavier object).
http://sites.tenafly.k12.nj.us/~hcoyle/Physics%20Powerpoints/Ch%2010%2011%20Work%20Power%20Energy/3%20Simple%20Machines%20and%20MA.ppt
Date added: June 1, 2012 - Views: 33
PowerPoint Presentation
Simple and Complex Machines BRMS 6th Science Standard 6-5.8 Illustrate the ways that simple machines exist in common tools and in complex machines.
Date added: February 21, 2014 - Views: 1
PowerPoint Presentation
10.1 Types of simple machines The lever, wheel and axle, rope and pulleys, screw, ramp, and gears are the most common simple machines. Complex machines, combine many simple machines into mechanical systems.
http://www.cposcience.com/home/Portals/2/Media/post_sale_content/PHY2/PresentationSlides/PHY2Chp10.ppt
Date added: May 4, 2013 - Views: 13
PowerPoint Presentation
Compound Machines A Wedge in a Compound Machine: Stapler Staples are wedges: they cut through paper because their ends are pointed in a wedge shape. Simple machines in a stapler: ... PowerPoint Presentation Last modified by: Tony Saine Created Date:
http://powpak.allencountyesc.org/powpak/data/Saine_t/files/Compound_Machines.ppt
Date added: March 24, 2013 - Views: 6
History of Simple Machines - Carl Wozniak
Title: History of Simple Machines Author: Registered User Last modified by: Carl Wozniak Created Date: 10/29/2006 4:25:01 PM Document presentation format
http://www.carlwozniak.com/MSED250files/LeversInHistory.ppt
Date added: September 3, 2011 - Views: 134
PowerPoint Presentation
Block and Tackle Wheel and axle Machine with two wheels of different sizes rotating together Simple Machines Gears are a type of simple ... Effort and Resistance Energy and Simple Machines Energy and Simple Machines Mechanical Advantage Efficiency Lever PowerPoint Presentation ...
http://www.midlandisd.net/cms/lib01/TX01000898/Centricity/Domain/1897/Simple%20Machines%20Notes.ppt
Date added: April 10, 2014 - Views: 5
Simple Machines - LaVergne Middle School
Simple Machines Common Simple Machines… First, rank the following words from 0-5 (0= you have never heard of this word & 5= you know the word VERY well).
http://www.lms.rcs.k12.tn.us/teachers/campbellp/documents/SimpleMachines.ppt
Date added: December 27, 2013 - Views: 8
Simple Machines Math - Randolph School, a K-12, independent ...
Simple Machines Math (Mechanical Advantage, Efficiency, and Energy) Efficiency Problem 2 John uses 39 J of energy to move four boxes with the handcart.
Date added: November 7, 2011 - Views: 49
PowerPoint Presentation
Simple Machines Make Our Work Easier! What is a Simple Machine? A simple machine has few or no moving parts. Simple machines make work easier. Scientists have identified six different simple machines: lever, screw, pulley, wedge, inclined plane, and wheel and axle.
http://sciencewilmeth5.wikispaces.com/file/view/Simple_machines.ppt/421060736/Simple_machines.ppt
Date added: November 12, 2013 - Views: 4
PowerPoint Presentation
simple machines simple machines lever pulley lever lever screw wheel and axle inclined plane inclined plane hedge clippers lever wedge pliers lever scissors lever wedge screwdriver lever wheel and axle lever wheelbarrow lever wheel and axle * lever pulley wheel and axle screw levers screw screw ...
http://www.sim.rcs.k12.tn.us/teachers/nicholsn/Simple_Machinespictures[1].ppt
Date added: December 23, 2013 - Views: 2
Motion, Forces &Machines PowerPoint presentation
Motion, Forces & Machines PowerPoint Presentation Paige Davis & Andrea Edney Core #3 – Science Introduction You have just watched an short clip of the rocket.
http://staff.fcps.net/kfwhite/MOtion%20powerpoint.ppt
Date added: August 27, 2011 - Views: 97
Gears - Welcome to the Georgia Agriculture Curriculum ...
Understanding Simple Machines Levers Inclined Plane Wedge Wheel and Axle Screws Gears Pulleys Cams Machine Elements Gadget Anatomy Assembled by Craig Tillmann
http://www.gaaged.org/Browseable_Folders/Power_Points/Mechanics/Understanding_Simple_Machines.ppt
Date added: May 5, 2013 - Views: 19
Simple Machines - teacher web
Simple machines. The basic machines that make up other machines are called simple machines. There are six of them. They are lever, pulley, wheel and axle, inclined plane, screw, and wedge.
http://teacherweb.com/GA/LKMossPrimarySchool/MrsMichelleHarris/Simple-Machines-Powerpoint-Harris.pptx
Date added: January 23, 2014 - Views: 6
Work and Simple Machines - Mrs. Bennett's Science Zone
Title: Work and Simple Machines Author: lewisv Last modified by: Windows User Created Date: 10/30/2004 2:17:16 AM Document presentation format: On-screen Show (4:3)
Date added: January 25, 2014 - Views: 6
PowerPoint Presentation
Simple Machines allow users to more easily accomplish work. This PowerPoint presentation is a tool that you can use to review these concepts and prepare for the upcoming FOSS Assessment for the Levers and Pulleys module.
http://silverdale.cksd.wednet.edu/Staff/Drock/Levers%20and%20PulleysPP.pps
Date added: October 13, 2011 - Views: 157
PowerPoint Presentation
Simple machines in a pair of ... Gothic Batang Monaco IMPORT Verdana Times Office Theme Microsoft ClipArt Gallery To do in class Kinematics PowerPoint Presentation Machines and Tools Simple Machines Mechanisms Types of Motion PowerPoint Presentation Linkages Rotating Arms Role of ...
http://web.cecs.pdx.edu/~mperkows/CLASS_479/2014LECTURES/Jan7-2014/2014_10_Lever-screw%20-wedge-MA.ppt
Date added: January 24, 2014 - Views: 12
Simple Machines (1) - Suffolk Teaching Activities & Resources
Simple Machines A simple machine is a machine with few or no moving parts. Simple machines make work easier. There are 6 types of Simple Machines pulley lever wedge wheel and axel inclined plane screw Inclined Plane An inclined plane is a slanting surface, such as a ramp, that connects a lower ...
http://star.spsk12.net/science/k3/SimpleMachines.ppt
Date added: February 12, 2012 - Views: 33
PowerPoint Presentation
Title: PowerPoint Presentation Author: pg Last modified by: pg Created Date: 2/20/2005 3:07:52 AM Document presentation format: On-screen Show Company
http://www.oocities.org/mrstaralynnhernandez/WorkPowerandSimpleMachines.pps
Date added: May 14, 2013 - Views: 11
PowerPoint Presentation
PowerPoint Presentation Author: Media Center Last modified by: Media Center Created Date: ... Energy Calculating Potential Energy Conservative Forces Gravitational PE Mechanical Energy Conservation of Energy Power Power (Simple) Machines ...
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Energy/Energy_in_a_nutshell.ppt
Date added: February 12, 2012 - Views: 52
Simple Machines By Design - eLeMeNTS
Simple Machines By Design ... Here is a simple rubric ... Bold Jazz LET Calibri Bold Calibri Bold Italic Arial Arial Arial Arial Arial Bold Blank Presentation Simple Machines By Design PowerPoint Presentation PowerPoint Presentation PowerPoint Presentation PowerPoint ...
http://www.lmnts.org/SimpleMachinesByDesign.ppt
Date added: December 4, 2011 - Views: 29 | 4,676 | 18,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | longest | en | 0.806563 |
https://gogeometry.com/circle/tangent_circles_theorems_problems_index_6.html | 1,716,443,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00634.warc.gz | 241,426,587 | 3,517 | # Online Geometry: Tangent Circles, Theorems and Problems Page 6
Tangent Circles, Theorems and Problems: Table of Content 6/8 Mascheroni Construction with compass alone. Midpoint of a segment. Intersecting circles. Tangent circles. Proposed Problem 213. Triangle, Incircle, Inradius, Semicircles, Common Tangents. Proposed Problem 190. Tangent circles, Tangent chord, Perpendicular, Distance. Proposed Problem 180. Externally Tangent circles, Common External Tangents, Areas. Proposed Problem 145. Four Triangles, Incircle, Tangent and Parallel to Side, Incenters, Circumcenters. Tangent circles. Proposed Problem 143. Four Triangles, Incircle, Tangent and Parallel to Side, Circumradii. Tangent circles. Proposed Problem 100. Circle Area, Archimedes' Book of Lemmas. Tangent circles. Proposed Problem 99: Circle Area, General Extension to Pythagoras' Theorem. Circumcircles or circumscribed circle Proposed Problem 94. Similar Triangles, Circumcircles, Circumradii. Tangent circles. Go to page: Previous | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Next | 278 | 1,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.736349 |
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# Solve equation
1. I've been told to solve this equation and put it into surd form.
x^2-6x+2=0
I have never done this type of question before. What is the best way to approach and solve this question
2. Do you know the quadratic formula?
3. Nope
4. (Original post by zed963)
Nope
Surely you do.
This one?
5. Oh I've seen that on an exam paper formula sheet.
6. (Original post by zed963)
I've been told to solve this equation and put it into surd form.
x^2-6x+2=0
I have never done this type of question before. What is the best way to approach and solve this question
Asked by whom
For what purpose
There are 2 methods (essentially the same)
Use the formula as given above
Complete the square
If you do not know what either of these are then you cannot do this question
If your teacher has asked you to do this I suggest they teach you material first
7. (Original post by TenOfThem)
Asked by whom
For what purpose
There are 2 methods (essentially the same)
Use the formula as given above
Complete the square
If you do not know what either of these are then you cannot do this question
If your teacher has asked you to do this I suggest they teach you material first
My teacher hasn't taught me this its just in the book that i'm working from.
And why does everything have to depend on a teacher
8. (Original post by zed963)
My teacher hasn't taught me this its just in the book that i'm working from.
And why does everything have to depend on a teacher
Indeed.
If you're working through a book then it's almost definitely the case that the question you're being asked can be solved using material introduced in the preceding chapters. If the quadratic formula hasn't been introduced in the book before this question is asked, then it's likely that, at the very least, the method of completing the square has been introduced. So on that assumption, you could complete the square to get the answer.
If you're still unsure, flick through the book to find out how to solve quadratic equations.
9. (Original post by zed963)
My teacher hasn't taught me this its just in the book that i'm working from.
And why does everything have to depend on a teacher
If you don't want to depend on a teacher then surely you need to actually read the bit in the book that teaches this bit. If it's asking questions without previously explaining how to tackle these questions I suggest you get a new textbook.
10. (Original post by zed963)
And why does everything have to depend on a teacher
What alternative do you suggest
Do you think it is better asking on here rather than asking a teacher
11. (Original post by zed963)
My teacher hasn't taught me this its just in the book that i'm working from.
And why does everything have to depend on a teacher
In your first post you said "I've been told to solve this equation and put it into surd form"
So who TOLD you?
12. (Original post by steve2005)
In your first post you said "I've been told to solve this equation and put it into surd form"
So who TOLD you?
I think by told he means the question in the book says 'solve this equation and write your answer in surd form'.
13. (Original post by zed963)
I've been told to solve this equation and put it into surd form.
x^2-6x+2=0
I have never done this type of question before. What is the best way to approach and solve this question
I suggest you buy a textbook or take out a library book.
14. (Original post by hassi94)
I think by told he means the question in the book says 'solve this equation and write your answer in surd form'.
He is using various phrases to get us to teach him. He seems keen but unfortunately not keen enough to use a textbook or google.
15. (Original post by zed963)
My teacher hasn't taught me this its just in the book that i'm working from.
And why does everything have to depend on a teacher
For the quadratic formula you have A, B and C, and they are in the form:
Maybe that will help. For if there is no number, just for example, take the number as 1.
16. (Original post by steve2005)
He is using various phrases to get us to teach him. He seems keen but unfortunately not keen enough to use a textbook or google.
I know, I agree that he needs to be a bit more independent or use his teacher.
After all OP, this forum is for helping people when they struggle; not for outright teaching things. If you had read up about the quadratic equation but got stuck on this question then that's fine. But it's like you're not trying.
17. x²-6x+2=0
x²-6x=-2
x²-6x = (x-3)²-9 <--- since (x-3)² = x²-6x+9, you have to take 9.
------------------------------ Notice: (x-3)²... half of the 6 from x²-6x.
(x-3)² -9=-2 <---- Equate this back to x²-6x=-2
(x-3)²=7
(x-3)=±√7
x=3±√7
So; x=3+√7 or x=3-√7
18. (Original post by Transcendence)
x²-6x+2=0
x²-6x = (x-3)²-9
(x-3)² -9=-2
(x-3)²=7
(x-3)=±7
x=3±7
x=10 or x=-4
Please don't give full solutions; and definitely don't give incorrect ones.
19. I have this textbook.
http://www.amazon.co.uk/GCSE-Mathema...6023712&sr=8-1
20. You are trying to learn concepts from scratch on TSR, it will be better if you watch video tutorials, to understand this stuff.
If you get a difficult question that you can't solve, then you should come here for help.
By the way, which chapter is this question from?
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Unformatted text preview: ECE 562: Advanced Digital Communications Lecture 5: Rate Efficient Reliable Communication Introduction We now move to rate efficient reliable communication (energy efficiency tends to come for free in this scenario). In this lecture we see that there are block communication schemes smarter than the naive repetition coding seen earlier that promise arbitrarily reliable communication while still having a non-zero data rate. We begin by setting the stage for studying rate efficient reliable communication by carefully dividing the transmitter strategy of mapping the information bits to transmit voltages into two distinct parts: 1. maps information bits into coded bits by adding more redundancy: the number of coded bits is larger than the number of information bits and the ratio is called the coding rate . This process is generally called coding . 2. map coded bits directly into transmit voltages. This is done sequentially : for in- stance, if only two transmit voltages are allowed ( E ) then every coded bit is se- quentially mapped into one transmit voltage. If four possible transmit voltages are allowed ( E, E 3 ), then every two consecutive coded bits are mapped into a single transmit voltage sequentially. This mapping is typically called modulation and can be viewed as a labeling of the discrete transmit voltages with a binary sequence. The receiver could also be potentially broken down into two similar steps, but in this lecture we will continue to focus on the ML receiver which maps the received voltages directly into information bits. Focusing on a simple binary modulation scheme and the ML receiver, we see in this lecture that there are plenty of good coding schemes: in fact, we will see that most coding schemes promise arbitrarily reliable communication provided they are decoded using the corresponding ML receiver! Transmitter Design: Coding and Modulation We are working with an energy constraint of E , so the transmit voltage is restricted to be within E at each time instant. For simplicity let us restrict that only two transmit voltages are possible: + E and- E . 1 If we are using T time instants to communicate, this means that the number of coded bits is T , one per each time instant. With a coding rate of R , the number of information bits (the size of the data packet) is B = RT . Surely, R 1 in this case. The scenario of R = 1 exactly corresponds to the sequential communication scheme studied in Lecture 3. As we saw there, the reliability level approaches zero for large packet sizes. The point is that even though we have spaced the transmit voltages far enough apart (the spacing is 2 E in this case), the chance that at least one of the bits is decoded incorrectly approaches unity when there are a lot of bits. The idea of introducing redundancy between the number of information bits and coded bits (by choosing...
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## This note was uploaded on 01/30/2009 for the course ECE 562 taught by Professor Pramodviswanath during the Fall '08 term at University of Illinois at Urbana–Champaign.
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5 - ECE 562: Advanced Digital Communications Lecture 5:...
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Ask a homework question - tutors are online | 734 | 3,612 | {"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} | 2.71875 | 3 | CC-MAIN-2017-13 | longest | en | 0.921655 |
https://www.physicsforums.com/threads/grade-11-kinematic-question.350932/ | 1,529,508,252,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00428.warc.gz | 890,119,519 | 12,890 | # Homework Help: Grade 11 Kinematic Question.
1. Nov 1, 2009
1. The problem statement, all variables and given/known data
A car is traveling along a highway with a speed of 25m/s when the driver sees an obstruction 180m directly ahead. It takes the driver 0.80s to react and begin breaking.
A) How long will it take the car to stop once brakes are applies, provided the car stops just before the obstruction?
B) What is the value of acceleration of the car just before hitting the obstruction? Assume acceleration is uniform.
2. Relevant equations
d= v (t)
a = v2-v1 / t2 - t1
d = v1(t) + 1/2 a (t2)
3. The attempt at a solution
I don't understand how or know how to use two equations with missing variables to find a variable.
2. Nov 1, 2009
### Seannation
Well the car is going at 25m/s whilst the driver is taking time to react, so you can work out the distance the car goes before the brakes are applied and take this distance away from 180m. This new number is the distance you need to work out the braking time for.
Now you know the distance the car takes to brake, and the change in velocity (25m/s to 0m/s). Do you know a way to work out the acceleration required to stop the car?
$$v^2 = u^2 + 2as$$
Where v = final velocity, u = initial velocity, a = acceleration and s = distance. Rearrange this equation to get acceleration.
Once you have the acceleration (it should be a negative number, since velocity is decreasing with time), you can use your acceleration equation to work out the time taken. Remember to add on the 0.8s taken for the driver to react. | 401 | 1,584 | {"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} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.923423 |
http://mathhelpforum.com/discrete-math/112750-questions-sets.html | 1,527,109,507,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00227.warc.gz | 189,275,692 | 10,467 | 1. ## questions on sets
thanks
1234567
2. For the first consider: $\displaystyle A = \left\{ {1,2,4,6} \right\}\;\& \,B = \left\{ {1,2,3,5} \right\}$.
For the second one, what $\displaystyle A^2$ mean?
I have never seen it defined in a 'set' context.
3. thanks got the help,
A^2 and B^2 means the following
let A={1,2} and B={3,4}
then A^2={{1,1},{1,2},{2,1},{2,2}}
B^2={{3,3},{3,4},{4,3},{4,4}}
4. Originally Posted by 1234567
A^2 and B^2 means the following
let A={1,2} and B={3,4} then A^2={{1,1},{1,2},{2,1},{2,2}}
B^2={{3,3},{3,4},{4,3},{4,4}}
It appears that you mean the cross product. But the notation is incorrect.
$\displaystyle A\times A=\{(1,1),(1,2),(2,1),(2,2)\}$.
5. You might also want to think about the situation in which $\displaystyle A\cap B= \phi$.
6. thanks for the help. | 318 | 806 | {"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} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.818107 |
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# ass 2 - ECONOMICS 5300 ASSIGNMENT 2 SOME PRELIMINARIES ON...
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ECONOMICS 5300 ASSIGNMENT 2 SOME PRELIMINARIES ON MICROECONOMICS AND REVIEW OF CHAPTERS 3 AND 5 1. THE PRODUCTION FUNCTION OF A FIRM IS GIVEN BY: Q = L 0.4 K 0.5 FOR Q = OUTPUT, L = LABOR, AND K = CAPITAL. a) DERIVE THE MARGINAL PRODUCTS OF LABOR AND CAPITAL. b) DOES THIS PRODUCTION FUNCTION HAVE THE PROPERTY OF CONSTANT RETURNS TO SCALE, INCREASING RETURNS TO SCALE, OR DECREASING RETURNS TO SCALE? DERIVE AND SHOW YOUR ANSWER. c) GIVEN YOUR ANSWER TO b) ABOVE, WHAT DOES THE RETURNS TO SCALE PROPERTY OF THE PRODUCTION FUNCTION IMPLY ABOUT COST CONDITIONS? ARE COSTS INCREASING, DECREASING OR CONSTANT AS PRODUCTION IS INCREASED? DERIVE YOUR ANSWER.
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2. SUPPOSE A PRODUCTION FUNCTION WHERE OUTPUT Y IS RELATED TO A SINGLE INPUT, X IN THE FORM, Y = X 0.9 . a) DERIVE THE TOTAL COST FUNCTION THAT IS DUAL TO THIS PRODUCTION FUNCTION IF THE PRICE OF THE INPUT, X, IS r. b) DERIVE THE AVERAGE AND MARGINAL COST FUNCTIONS FROM THE TOTAL COST FUNCTION THAT YOU DERIVED IN a) ABOVE 3. THE DEMAND FUNCTION FOR PIPE WRENCHES IS GIVEN BY: Y = 0.4P -1.05 , FOR Y = QUANTITY OF PIPE WRENCHES DEMANDED, AND P = PRICE OF PIPE WRENCHES.
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{[ snackBarMessage ]} | 432 | 1,511 | {"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} | 2.90625 | 3 | CC-MAIN-2018-17 | latest | en | 0.571717 |
http://mathhelpforum.com/calculus/224925-triple-integral-volume-problem-using-cylidrical-coordinates.html | 1,519,312,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00770.warc.gz | 233,850,807 | 12,357 | # Thread: Triple Integral: Volume problem, Using cylidrical coordinates
1. ## Triple Integral: Volume problem, Using cylidrical coordinates
Problem statement:
A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.
My solution: 0=Theta | = integral (Sorry I don't know know Latex. If you can please direct me to a page with syntax and I will happily reformat my solution). Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.
||| r dzdrd0
Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)
I would like to know why the limits of integration of z is -sqrt(9-r^2) to +sqrt(9-r^2). The radius is at most 3 and at least 1. However if r=3, z=0 which is true when r=3 on the xy-plane but a picture can be used to show that we can find a z=/=0 if r=3.
It seems sqrt(9-r^2)=z doesn't give a correct relation between r and z at all for this problem. I expect to find the max of z (z value when the cylinder intersects the sphere).
2. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
quickly before I even look at the actual problem
LaTeX Help
3. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.
4. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
Originally Posted by Elusive1324
Problem statement:
A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.
My solution: 0=Theta | = integral. Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.
||| r dzdrd0
Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)
My solution:
$\int^{2\pi}_{0}\int^{3}_{1}\int^{\sqrt{9-r^2}}_{-\sqrt{9-r^2}} r\,dz\,dr\,d\theta$
Is the expression correct?
Edited: I used wolfram to do the integration which yields $\frac{64\pi\sqrt{2}}{3}$. I crossed check this value with the online homework system and it says that this is correct.
O.K
Next, I'm going to try to verify this using spherical coordinates.
5. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
Originally Posted by romsek
also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.
I second this. The material removed consists of a cylinder of radius 1 and length $4\sqrt{2}$ and the two "caps" at each end of the cylinder. Calculating their volume is the only "hard" part.
6. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
Finding the volume of the cylinder: $V=\pi(r^2)h=\pi(h) = 4\pi\sqrt{2}$. 'r', the radius of the cylinder is 1. 'h' = 4√2 is given in the diagram in the attachment.
Volume of 1 Cap (Cavalieri's Principle):
$dV=\pi(r^2)*dx$ and $r=f(x)=\sqrt{9-x^2}$
$Vcap = \pi\int^{3}_{2\sqrt{2}} 9-x^2\,dx = 18-\frac{38\sqrt{2}}{3}$
$2*Vcap + 4\pi\sqrt{2} \neq \frac{64\pi\sqrt{2}}{3}$
Did I miss something?
Edited: Oh wait... subtract this from the volume of the sphere.
Edited: I evaluated (4/3)(pi)(27)-4pi*sqrt(2)-2*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3 in wolfram alpha to check if the two values are equal but they aren't. So something in my calculation went wrong.
7. ## Re: Triple Integral: Volume problem, Using cylidrical coordinates
Edited: Forgot the pi in front of a term. This expression is true: (4/3)(pi)(27)-4pi*sqrt(2)-2pi*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3
On the left hand side it's the volume minus the cylinder and the two caps. The right side is obtained by triple integrate in cylindrical coords. | 1,130 | 3,949 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-09 | longest | en | 0.857563 |
http://www.mountainpathmedia.com/ven2ihd/6pd5cnl.php?tag=8354de-incomplete-metric-space | 1,618,681,035,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00055.warc.gz | 158,268,128 | 6,154 | For example, the map. Theorem 19. Hence, we will have to make some adjustments to this initial construction, which we shall undertake in the following sections. By Theorem 13, C b(X;Y) is a closed subspace of the complete metric space B(X;Y), so it is a complete metric space. Shouldn't some stars behave as black holes? If X is one of the spaces l 2, s, l 2 × s and Y is a locally convex linear metric space which is uniformly homeomorphic to X, then X is isomorphic to X. @TöreDenizBoybeyi the definition of trivial is, in this case, rather personal. How does the Dissonant Whispers spell interact with advantage from the halfling's Brave trait? Let ε > 0 be given. The procedure is as follows. 3 Hence the metric space is, in a sense, "complete.". $$\|t^{(i)} - t^{(j)} \|_\infty = \| (0, \ldots, 0, \frac{1}{j+1}, \ldots, \frac{1}{i}, 0, 0, \ldots ) \|_\infty = \frac{1}{j+1},$$ In mathematics, a complete metric space is a metric space in which every Cauchy sequence in that space is convergent. A sequence (x n) in X is called a Cauchy sequence if for any ε > 0, there is an n ε ∈ N such that d(x m,x n) < ε for any m ≥ n ε, n ≥ n ε. Theorem 2. Some Noncontractive Maps on Incomplete Metric Spaces Have Also Fixed Points. Non-examples. In other words, every Cauchy sequence in the metric space tends in the limit to a point which is again an element of that space. To see this, consider the sequences $$t^{(1)} = (1, 0, 0, \ldots),$$ $$t^{(2)} = (1, \frac{1}{2}, 0, 0, \ldots ),$$ $$t^{(3)} = (1, \frac{1}{2}, \frac{1}{3}, 0, 0, \ldots )$$ Want to improve this question? (Recall, from Lecture 3, that this is known as the L. 1. metric on C. 2. This page was last modified 17:23, 4 January 2009. Is there any metric on R with which it is incomplete. Actually, any space with the discrete metric is complete: every Cauchy sequence is constant. The completion has a universal property. Let $$m_0 := \{t \in \mathbb{R}^\mathbb{N} : \{t_1, t_2, \ldots \} \text{ is finite} \}$$ In a space with the discrete metric, the only Cauchy sequences are those which are constant from some point on. Examples of complete and incomplete spaces. “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…. Completeness is not a topological property: it is possible for a complete metric space to be homeomorphic to a metric space which is not complete. Meaning of the Term "Heavy Metals" in CofA? We can define a topological space to be metrically topologically complete if it is homeomorphic to a complete metric space. metric space, but, as will be seen in part (v) of Exercise 1.2, Dfails to even be a metric. Disconnectedness, completeness and compactness. Proof. Examples of metric spaces in which every non-empty open set is uncountable. Then (C b(X;Y);d 1) is a complete metric space. A topological condition for this property is that the space be metrizable and an absolute Gδ, that is, a Gδ in every topological space in which it can be embedded. 2 Department of Mathematics and General Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia. The new space is referred to as the completion of the space. Which of the following metric spaces are complete? If $p,q>N$, we have: Append content without editing the whole page source. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Theorem. Tawseef Rashid, 1 Qamrul Haq Khan, 1 Nabil Mlaiki, 2 and Hassen Aydi 3,4. Exercises 1.1For any a;b2C[E], show that the sequence fd(a n;b n)g n2N is a Cauchy sequence of real numbers and hence converges. Uniform homeomorphisms of locally convex complete metric spaces have been studied by Mankiewicz [1], [2], cf. How do we get to know the total mass of an atmosphere? Let X be a metric space with metric d. Then X is complete if for every Cauchy sequence there is an associated element such that . Given an incomplete metric space M, we must somehow define a larger complete space in which M sits. Let us look at some further examples of complete and incomplete spaces, starting with an incomplete one. this converges to $0$ for $i,j \rightarrow \infty.$ Therefore $(t^{(i)})_{i\in \mathbb{N}}$ is a cauchy sequence in $m_0$. Why is "threepenny" pronounced as THREP.NI? is a homeomorphism between the complete metric space R and the incomplete space which is the unit circle in the Euclidean plane with the point (0,-1) deleted. Any unbounded subset of any metric space. What does “blaring YMCA — the song” mean? This sequence does not converge. I know complete means that every cauchy sequence is convergent. My question is: Can someone give examples of incomplete spaces such that either they have unusual metric or unusual ambient space other than rational numbers etc ? [0;1);having the properties that (A.1) ... compactness implies completeness, but (C) may hold for incomplete X, e.g., X= (0;1) ˆ R. Proposition A.10. The space Q of rational numbers, with the standard metric given by the absolute value of the difference, is not complete.Consider for instance the sequence defined by = and + = +.This is a Cauchy sequence of rational numbers, but it does not converge towards any rational limit: If the sequence did have a limit x, then necessarily x 2 = 2, yet no rational number has this property. Take any complete metric space and remove one (or two) points? Equip it with the sup-norm, i.e. View/set parent page (used for creating breadcrumbs and structured layout). Any convergent sequence in a metric space is a Cauchy sequence. Turns out, these three definitions are essentially equivalent. What is its completion, ((0;1) ;d))? d(p,q)=\bigl\lvert\mathrm e^{-p}-\mathrm e^{-q}\bigr\rvert\le\mathrm e^{-p}+\mathrm e^{-q}<2\cdot\mathrm e^{-N}<\varepsilon. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. also Bessaga [1], §11. The answer is yes. 1) is the space of bounded, continuous functions f: X!Y equipped with the uniform metric d 1. be the sequences of real numbers that only take on finitely many values. But it's limit (in the bigger space $\mathbb{R}^\mathbb{N}$ of sequences of real numbers) is 1. Assume that (x n) is a sequence which converges to x. Is There (or Can There Be) a General Algorithm to Solve Rubik's Cubes of Any Dimension? In a metric space $(M,d)$, any Cauchy sequence $\{a_n\}_{n \in \mathbb{N}}$ in $M$ is convergent? To make space incomplete either i can change the metric or the ambient space. First I’ll describe the process of creating the Cauchy completion of a metric space; and then I’ll … Examples include the real numbers with the usual metric, the complex numbers, finite-dimensional real and complex vector spaces, the space of square-integrable functions on the unit interval L^2([0,1]), and the p-adic numbers. Why should I expect that black moves Rxd2 after I move Bxe3 in this puzzle? rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Theorem: A subset of a complete metric space is itself a complete metric space if and only if it is closed. | 1,990 | 7,540 | {"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} | 3.671875 | 4 | CC-MAIN-2021-17 | latest | en | 0.886786 |
https://stat.ethz.ch/pipermail/r-help/2008-August/171287.html | 1,430,770,083,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455113263.14/warc/CC-MAIN-20150501043833-00030-ip-10-235-10-82.ec2.internal.warc.gz | 909,219,380 | 3,775 | # [R] Design-consistent variance estimate
Doran, Harold HDoran at air.org
Mon Aug 18 16:53:14 CEST 2008
Whoops, the final var estimator var(f(Y)) should have N^4 in the
denominator not N^2
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Doran, Harold
> Sent: Monday, August 18, 2008 10:47 AM
> To: Stas Kolenikov
> Cc: r-help at r-project.org
> Subject: Re: [R] Design-consistent variance estimate
>
> It also turns out that in educational testing, it is rare to
> consider the sampling design and to estimate
> design-consistent standard errors. I appreciate your thoughts
> on this, Stas. As a result, I was able to bring to my mind
> more transparency into what R's survey package as well as SAS
> proc surveymeans are doing. I've copied some minimal latex code below.
> My R code reflecting this latex replicates svymean() and the
> SAS procedures exactly under all conditions that I have
> tested so far for a
> 1 stage cluster sample.
>
> It clearly reduces to a more simple expression when cluster
> sizes are equal.
>
> My hat is off to sampling statisticians, this has got to be a
> lot of fun for you :)
>
> ### LaTeX
>
> \documentclass[12pt]{article}
> \usepackage{bm,geometry}
> \begin{document}
>
> In this scenario, the appropriate procedure is to estimate
> design-consistent standard errors. This is accomplished by
> first defining the ratio estimator of the mean as:
>
>
> f(Y) = \frac{Y}{N}
>
>
> \noindent where $Y$ is the total of the variable and $N$ is
> the population size. Treating both $Y$ and $N$ as random
> variables, a first-order taylor series expansion of the ratio
> estimator $f(Y)$ can be used to derive the design-consistent
> variance estimator as:
>
>
> var(f(Y)) = \left[\frac{\partial f(Y)}{\partial Y},
> \frac{\partial f(Y)}{\partial N}\right] \left [ \begin{array}{cc}
> var(Y) & cov(Y,N)\\
> cov(Y,N) & var(N)\\
> \end{array}
> \right]
> \left[\frac{\partial f(Y)}{\partial Y}, \frac{\partial
> f(Y)}{\partial N}\right]^T
>
> \noindent where
>
>
> \left[\frac{\partial f(Y)}{\partial Y}\right] = \frac{1}{N}
>
>
>
> \left[\frac{\partial f(Y)}{\partial N}\right] = -
> \frac{Y}{N^2}
>
>
> var(Y) = \frac{k}{k-1} \sum_{j=1}^k(\hat{Y}_j-\hat{Y}_{..})^2
>
>
>
> \hat{Y}_j = \sum_{i=1}^{n_j}\hat{Y}_{j(i)}
>
>
> \hat{Y}_{..} = k^{-1} \sum_{j=1}^k \hat{Y}_j
>
>
> var(N) = \frac{k}{k-1} \sum_{j=1}^k(\hat{N}_j-\hat{N}_{..})^2
>
>
>
> \hat{N}_j = \sum_{i=1}^{n_j}\hat{N}_{j(i)}
>
>
> \hat{N}_{..} = k^{-1} \sum_{j=1}^k \hat{N}_j
>
>
> cov(Y,N) = \sum_{j=1}^k(\hat{Y}_j- \hat{Y}_{..}) (\hat{N}_j-
> \hat{N}_{..}) \times \frac{k}{k-1}
>
>
> \noindent where $j$ indexes cluster $(1, 2, \ldots, k)$,
> $j(i)$ indexes the $i$th member of cluster $j$, and $n_j$ is
> the total number of members in cluster $j$.
>
> The estimate of the variance of $f(Y)$ is then taken as:
>
>
> var(f(Y)) = \frac{N^2var(Y) - 2cov(Y,N)NY + var(N)Y^2 }{N^2}
>
>
> The standard error is then taken as:
>
>
> se = \sqrt{var(f(Y))}
>
>
> \end{document}
>
> > -----Original Message-----
> > From: Stas Kolenikov [mailto:skolenik at gmail.com]
> > Sent: Monday, August 18, 2008 10:40 AM
> > To: Doran, Harold
> > Cc: r-help at r-project.org
> > Subject: Re: [R] Design-consistent variance estimate
> >
> > On 8/16/08, Doran, Harold <HDoran at air.org> wrote:
> > > In terms of the "design" (which is a term used loosely)
> the schools
> > > were not randomly selected. They volunteered to participate
> > in a pilot study.
> >
> > Oh, that's a next level of disaster, then! You may have to
> work with
> > treatment effect models, of which there are many:
> > propensity score matching, nearest neighbor matching, instrumental
> > variables, etc.
> > Those methods require asymptotics in terms of number of treatment
> > units, which would be schools -- and I would imagine those are
> > numbered in dozens rather than thousands in your study, so
> > straightforward application of those methods might be problematic...
> > At least I would augment my analysis with propensity score weights:
> > somehow estimate the (school level) probability of participating in
> > the study (I imagine you have the school characteristics at
> hand for
> > your complete universe of schools
> > -- principal's education level, # of computers per student,
> fraction
> > free/reduced price lunch, whatever...
> > you probably know those better than I do :) ), and use
> inverse of that
> > probability as the probability weight. If the selection was
> > informative, you might see quite different results in weighted and
> > unweighted analysis.
> >
> > > In Wolter (1985) he shows the variance of a cluster sample with a
> > > single strata and then extends that to the more general
> example. It
> > > turns out though in many educational assessment studies,
> the single
> > > stage cluster sample is a norm and not so rare.
> >
> > I can see why. Thanks, I'll keep educational statistics examples in
> > mind for those kinds of designs!
> >
> > --
> > Stas Kolenikov, also found at http://stas.kolenikov.name
> Small print:
> > I use this email account for mailing lists only.
> >
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help | 1,627 | 5,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-18 | longest | en | 0.796571 |
https://cs.stackexchange.com/questions/150554/prove-that-l-an-bl-n-leq-l-is-not-regular-by-pumping-lemma | 1,656,773,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00677.warc.gz | 236,771,340 | 64,891 | # Prove that $L=\{a^n b^l : n \leq l\}$ is not regular by pumping lemma
I'm currently trying to prove that $$L=\{a^n b^l : n \leq l\}$$ is not regular by pumping lemma
My proof:
If we choose $$w$$ such that $$w=a^P b^P$$, then since $$|xy| \leq p$$, $$y$$ must be $$a^P$$, meaning it can be pumped any number of times $$i$$, such that $$a^{iP} b^P \in L$$. Since $$iP \leq P$$ is not true for any $$i > 1$$, it cannot exist in $$L$$, therefore $$L$$ cannot be regular.
Is this a valid proof? I don't fully comprehend the pumping lemma, so please let me know if/where I may be going wrong.
Thank you
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher.
– D.W.
Apr 11 at 2:32
• Crosspost with this question on math.stackexchange. Apr 12 at 5:59
• what is $$P$$? If it is the pumping length, then say it so;
• what is $$p$$? Is it a typo and should be a $$P$$?
• what are $$x$$ and $$y$$? Don't hesitate to remind the hypotheses of the pumping lemma;
• why would $$y$$ be $$a^P$$? The only conclusion you can make is that $$y = a^k$$ for $$1\leqslant k \leqslant P$$. | 450 | 1,515 | {"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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-27 | latest | en | 0.906845 |
https://www.gamingzion.com/gambling/gambling-news/how-to-bet-on-ice-hockey-hockey-betting-odds/ | 1,702,230,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00552.warc.gz | 827,513,483 | 44,562 | # How to Bet on Ice Hockey: Hockey Betting Odds
By:
Posted: December 9, 2015
Updated: October 6, 2017
Are you an ice hockey fan interested in placing some bets on your favourite winter sport, but are not sure where to start? Look no further than GamingZion’s guide on how to bet on ice hockey!
This guide is made for people who’d like to make some money off of hockey but are overwhelmed by all of the information and statistics in sportsbooks. In our first installment, we’ll be differentiating the different kinds of hockey betting odds available to sports fans from all over the globe.
## Hockey betting odds in the US: The moneyline
If you’ve ever taken a glance at online sportsbooks in US or ones aimed at an American audience, you’ll see that they use something called the moneyline, which are numerical odds with plus and minus signs. A plus sign will indicate the underdog, while the minus sign indicates which team is most favoured to win. For example:
Washington Capitals: -130
In this example, bettors would be risking \$130 on the Capitals for a chance to win \$100 in profits. This is a safe bet in comparison to the Flyers, where you’d risk more at the chance to win more, with bettors standing to win \$150 in profits for every \$100 that they wager. The same formula applies to smaller bets.
## Hockey betting odds in the UK: Fractional odds
Fractional odds are those that hockey fans will find on British internet betting websites. In fractional odds, the number in the front of the fraction represents how much money you’re trying to win, whereas the number behind the fraction represents how much you’re risking. For example:
New York Rangers: 4/5
San Jose Sharks: 12/5
In this example, the betting lines favour the Rangers, as you’d be risking \$4 to win \$5. In opposition, if you bet on the Sharks, you’d stand to make \$12 for every \$5 that you wagered.
## Hockey betting odds in Europe, Australia and Canada: Decimal odds
Decimal odds are the kind of hockey betting odds that you’ll find in gambling news from Europe, Australia, and Canada. With decimal odds, the bet always start at 2.0. This number moves down for the favourite, and moves up for the underdog. For example:
Vancouver Canucks: 1.80
Boston Bruins: 2.30
In this line, if you bet on the Canucks you will be risking \$1 in order to win \$0.80 in profit, \$1.80 total. Whereas with the Bruins, you’d be wagering \$1 to win \$1.30 in profit, total \$2.30. So in this case the Bruins would be the underdogs because you’d be taking more of a risk, but stand to make more profit than if you chose the Canucks, who are a safer bet.
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Notify of | 618 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-50 | latest | en | 0.912645 |
https://www.unitconverters.net/power/kilojoule-minute-to-mbh.htm | 1,726,202,904,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00888.warc.gz | 978,436,819 | 3,624 | Home / Power Conversion / Convert Kilojoule/minute to MBH
# Convert Kilojoule/minute to MBH
Please provide values below to convert kilojoule/minute [kJ/min] to MBH, or vice versa.
From: kilojoule/minute To: MBH
### Kilojoule/minute to MBH Conversion Table
Kilojoule/minute [kJ/min]MBH
0.01 kJ/min0.0005686903 MBH
0.1 kJ/min0.0056869027 MBH
1 kJ/min0.0568690272 MBH
2 kJ/min0.1137380544 MBH
3 kJ/min0.1706070817 MBH
5 kJ/min0.2843451361 MBH
10 kJ/min0.5686902722 MBH
20 kJ/min1.1373805444 MBH
50 kJ/min2.8434513609 MBH
100 kJ/min5.6869027219 MBH
1000 kJ/min56.8690272188 MBH
### How to Convert Kilojoule/minute to MBH
1 kJ/min = 0.0568690272 MBH
1 MBH = 17.5842642103 kJ/min
Example: convert 15 kJ/min to MBH:
15 kJ/min = 15 × 0.0568690272 MBH = 0.8530354083 MBH | 328 | 770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.318256 |
https://www.coursehero.com/file/5892553/lec32/ | 1,495,873,295,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608877.60/warc/CC-MAIN-20170527075209-20170527095209-00497.warc.gz | 1,081,029,383 | 62,439 | # lec32 - Todays Agenda Arraysbasedproblems...
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1 Computer Programming I TA C162 . 7 Apr 2007 Today’s Agenda Arrays based problems 1. Searching-> Linear Search, Binary Search 2. Sorting -> Bubble Sort (Done), Insertion Sort (Done) Merge Sort 3. Strings-> Word reversing
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2 Computer Programming I TA C162 . 7 Apr 2007 Linear search Each item in the list is examined in the order in which it occurs in the list until the desired item is found or the end of the list is reached Algorithm // precondition : N>=1 1. Read a array of N elements named List[0…N-1] and search element key 2. Repeat for i=0 to i=N-1 1. If key equals to List[i] 1. Display element found and stop 3. Display element not found 4. Stop Post condition : If i<N element found otherwise element not found
3 Computer Programming I TA C162 . 7 Apr 2007 main() // Linear search implementation { int List[10],index,key; printf("Enter Array elements:"); for(index = 0; index<10; index++) scanf("%d",&List[index]); printf("Enter search element"); scanf("%d", &key); for(index=0;index<10; index++) if(key==List[index]) { printf("Element found at %d location",index); return; } printf("Element not found"); }
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## This note was uploaded on 05/14/2010 for the course CS SS ZG653 taught by Professor Shanta during the Spring '09 term at Birla Institute of Technology & Science.
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Ask a homework question - tutors are online | 576 | 1,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-22 | longest | en | 0.295563 |
https://forums.theregister.com/forum/all/2019/04/22/tesla_self_driving_cars/#c_3766331 | 1,719,255,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00232.warc.gz | 225,835,851 | 45,760 | Tesla touts totally safe, not at all worrying self-driving cars – this time using custom chips
Tesla claims to have built a "fully self driving" (FSD) system using custom-designed math processors, allowing its vehicles to potentially drive themselves completely autonomously. At an all-day event at the company's Los Angeles headquarters on Monday, executives were on stage to outline the computer system, which has been …
1. 144 trillion operations per second
"...its chip could do 144 trillion operations per second" - AI, as an autopilot, does not need that much.
1. Re: 144 trillion operations per second
I'd love to see something that would back your statement up.
To be honest, I'm just moving past basic theoretical understanding of neural networks and moving into application. I've been very interested in reducing transform complexity and therefore reducing the number of operations per second for a given "AI" operation. Think of me as the guy who would spend 2 months hand coding and optimizing assembler back in the 90's to draw a few pixels faster. (I did that too)
I don't entirely agree from my current understanding with the blanket statement that it wouldn't need that much. I believe at the moment that there are other bottlenecks to solve first, but at least in my experience processing convolutional networks in real time from multiple high resolution sources at multiple frequency ranges could probably use all 144 trillion operations and then some.
Do you have something that would back up your statement... I'd love to see for better understanding of the topic.
1. Re: 144 trillion operations per second
Very difficult to calculate because I don't know everything, but I'll try.
Let's assume that Tesla AI paragraph is an average of 100 words, 50 patterns before creating synonymous clusters, and 200 patterns after that. Suppose that there are 1,000 subtexts paragraphs that surround it (dictionary definitions + examples, and which define its meaning/ context-subtexts = 100 patterns each = 100.000 patterns in total): 200 + 100,000 = 100,200 patterns.
Tesla computer does not need to create them each time, they are hard coded.
Next, Tesla = eight cameras, 12 ultrasonic sensors, and front-facing radar = 21 sensor;
each the sensor provide 100,200 patterns per second = approximately 2.3 million patterns per second, matched with 100,200 AI patterns.
Thus we get less than 300 billion operations per second, plus 1 trillion (? - have no idea how it works) on commands, but Tesla's processor = 144 trillion.
1. Re: 144 trillion operations per second
Why the definitions? Without them you cannot create synonymous clusters. There is a sentence:
-- After this publication was made, the correspondence was lost and our mail failed to deliver.
There are three synonyms here: "publication", "correspondence" and "mail"; and so "was done", "was lost", "failed" and "deliver" refer to all three: these twelve patterns are the sentence's synonymous cluster.
If you don't get these three synonyms - because you profile/ train on terabytes and don't see the words - you can lose 75% of possible matches and therefore 75% of information cannot be found. Add one more synonym and you lose more than 93%! The same with a paragraphs, if you have more than one sentence (with synonyms) in it.
Are you ready to lose up to 99% of your information? Train by terabytes!
1. Re: 144 trillion operations per second
Dictionary definitions allow unique indexing of data, so that the desired AI paragraph with the desired patterns can be found, by comparison of several hundred bytes, kilobytes the max. Indeed, the computer literally remember and understands what is wanted from it.
The training by terabytes doesn't work! Look how Google Waymo lost years on terabytes.
2. Re: 144 trillion operations per second
A theoretical understanding of ANNs is just the start. The problem is we understand how they work at a high level - data in , matches out - and we understand at a very low level - back propagation changes weightings which changes neuron synapse combinations and/or values required to fire. What no one really understands is what is going on inbetween, eg how exactly do the dozens or even hundreds of hidden linked layers actually achieve their results. The whole seems to be greater than the sum of the parts with ANNs and anyone who tells you they completely grok the processes going on deep inside are deluded or liars. And until we really understand how they do it then we can never be 100% certain how they'll behave in a given situation.
1. Re: 144 trillion operations per second
Pattern weights are almost always irrational numbers that computer always rounds and almost always differently, finding the end result. For example, 0.(3) sometimes 0.4, sometimes 0.3. The number of patterns is huge, and the computational error accumulates: often in the same situations AI gives different answers. That's my observation, what I got from talking to lexical clones.
How to deal with it? Helps statistics, ie training. AI gets feedback and if the result is positive, etc... Basics of Cybernetics. People learn the same way, do you remember your school and College?
2. Re: 144 trillion operations per second
You need an awful lot of that performance just to recognize the outline of a cat and decide that it is a cat licking it's bum and not a carrier bag lying in the road.
1. Re: 144 trillion operations per second
Yes. I meant the textual part of AI, the matching of what they call "labels" and I "annotations".
3. Re: 144 trillion operations per second
"AI, as an autopilot, does not need that much"
In the air? perhaps not. We give aircraft miles (literally) of space to themselves.
On the ground is another matter and reaction times of millliseconds are crucial.
1. Re: 144 trillion operations per second
Ok, I lost, my mistake, forgot about "millliseconds". I thought in the term of Information Retrieval where you can wait a few seconds.
2. Bandwidth & storage
So we have all these Teslas sending data/video to where? That's a lot of cat pics to store somewhere - wonder what they're using?
1. Re: Bandwidth & storage
"So we have all these Teslas sending data/video to where?"
Tesla's backend servers. Funnily enough, bandwidth wasn't discussed, AFAICR.
C.
1. Re: Bandwidth & storage
We had a data network sufficient to transmit real-time position data 30 years ago.
Now we have ... erm, nearly have ... 5g. Do you suppose Trump's hostility to Huawei is really aimed at Musk?
2. Re: Bandwidth & storage
"So we have all these Teslas sending data/video to where?"
The same place your wetware sends most of its petabytes. The bit bucket.
The ability to _FORGET_ and filter is one of the more crucial requirements for intelligence and development. The important part is knowing WHAT to filter and forget.
1. Re: Bandwidth & storage
This is where the Habsburgs went wrong - they famously "learned nothing and forgot nothing".
2. Re: Bandwidth & storage
FWIW - all the data travels over a OpenVPN endpoint hosted in Claifornia.
In terms of pics & video - those get sent once it acquires wifi and is parked.
Mobile data is used (over the same VPN) whilst on the road, however no video uploads, etc.
3. Compton would love it
So if someone wants to steal a Teslla he can use an app and have one unlocked and ready to be stolen and taken apart (Tesla parts are scarce and expensive) promptly delivered to their hood ? Niiice.
1. Re: Compton would love it
It'll be easily spotted - if a Tesla is stolen it automatically rebrands as "Edison".
4. PT Barnum...
Didn't PT Barnum say something about suckers being born every minute? Without any hardcore testing of these chips and the firmware the claim is pure BS. Give me a minimum of 5 years of widespread testing in all conditions with the system performing flawlessly (per the hype) and then I might grudgingly grant you are close. A minimum of 10 years before I believe you have likely seen about every reasonable driving situation someone, somewhere will face.
1. Re: PT Barnum...
Exactly. I'd want to see about a decade of driving results that are substantially better than the typical Human before I accepted any claim out of his gob.
Fully autonomous? Until I, a totally blind person without a drivers license, can climb into the passenger seat, tell it where to go, & not have to take control for *any* reason then you don't get to make that claim. If I have to be in the driver seat, have a license, & be able to take control in an emergency, then it's not a *fully* autonomous ride at all.
You want to offer a robo taxi? Fine. I'll believe it when I hear that insurance companies will insure your cars. If they won't insure them then you won't be able to offer them & nobody will get to ride in them.
Mr. Barnum would marvel at the audacity the MuskRat shows every time he opens his muzzle, amazed at the suckers that believe the BS that comes out of it, & turn green with envy over the amounts of cash the bastard rakes in for having done it.
1. Re: PT Barnum...
From an actuarial perspective each car model can be considered as one driver doing a very very large number of miles. Once they are fully autonomous I expect the manufacturers to insure them for the first few years just like they provide a few years of warranty.
2. Re: PT Barnum...
"I'd want to see about a decade of driving results that are substantially better than the typical Human before I accepted any claim out of his gob."
It isn't that hard to drive better than a human most of the time, and it isn't that hard to drive better than most humans.
The bigger problem is programmatically adhering to rigid rules such as "PEDESTRIANS WILL NOT CROSS THE ROAD EXCEPT AT DESIGNATED POINTS", which sound fine to wetware but when interpreted literally result in automated killing machines roaming the streets mowing down pedestrians who dare to try.
Thankfully only the USA and a couple of other authoritarian countries have such "car is KING" type rules, and even then, "common sense" gets applied by human drivers. A machine programmed with such rules that runs into a parade and marching band won't stop for them, which will make for a whole new interpretation of American (road) Pies.
Google have got it pretty much right. Uber utterly fucked up.
5. This time for sure?
Maybe their new chip will have enough CPU power to recognize giant trucks stopped in the road. Oh, and barriers.
1. Re: This time for sure?
So, then, Level 1.9993 autonomy? Good enough for Elon and (too) many others. I am inclined to think that Silicon Valley success criteria -- "Works much of the time" -- are NOT going to prove in the long run to be adequate for autonomous vehicles.
Recommended reading -- Richard Feynman, Appendix F to the Rodgers Commission Report on the loss of the Space Shuttle Challenger. https://science.ksc.nasa.gov/shuttle/missions/51-l/docs/rogers-commission/Appendix-F.txt
1. Re: This time for sure?
Why wont they be? They don't have to be better than Lewis Hamilton - just better than the average human driver - which is a pretty low bar tbh.
If Tesla can show the insurance companies that a Tesla will have less accidents than a human over a relevant statistical sample then they'll be queuing up - especially since they would still charge an extra "AI Premium" and cream off profits for a few years.
The Feynman report whilst an excellent example of no-BS scientific communication isnt relevant...
1. Re: This time for sure?
They need to be much better than the 'average' driver, machines killing people at the same rate as clumsy people is not going to be acceptable.
One way to make sure they're good enough is to make the CEO (and other Board members) personally liable for their systems mistakes as if they were at the controls.
2. Re: This time for sure?
"Why wont they be? They don't have to be better than Lewis Hamilton - just better than the average human driver - which is a pretty low bar tbh."
The average human drive ris actually very good and orders of magnitude better than any automatic system we can produce. In the UK on average a car occupant is killed once very 4 million miles travelled.
This drags the average down because it includes the drink drivers and boy racers. Middle aged sensible driver do far better.
1. Re: This time for sure?
Also, a fair number of accidents are also caused by distractions whether that be fiddling to find a particular station on the radio, texting, or what have you. The big issue with this assistive technology is that some people are using it as an excuse to be distracted further. As a result the idiot who is flashing back and forth between distraction and road is now allowing the distraction to take up much more of their time which likely results in a much worse accident when the automation misses the mark.
3. Re: less accidents
Fewer.
Can we have a Stannis icon for this sort of thing?
4. Re: This time for sure?
"They don't have to be better than Lewis Hamilton - just better than the average human driver - which is a pretty low bar tbh."
Amusingly I would guess that it's far easier to be faster than Lewis Hamilton on a racing circuit (ignoring overtaking) than to drive around a crowded city without crashing into anything.
1. Re: This time for sure?
Depends on whether auto boxes are allowed or not. An AI with a DSG gearbox perhaps, an AI with a manual gearbox is going nowhere.
5. Re: This time for sure?
"If Tesla can show the insurance companies that a Tesla will have less accidents than a human over a relevant statistical sample then they'll be queuing up - especially since they would still charge an extra "AI Premium" and cream off profits for a few years."
They'll charge the extra premium on repair costs, however that won't last long - because:
As soon as it's shown that Robocars in control have statistically fewer big crashes AND small dings than humans (and the video evidence will be compelling evidence that humans are driving into robocars, therefore insurers will probably start foregoing "knock-for-knock" handling of claims, resulting in humans facing increasingly \$\$LARGE repair/medical liabilities) then you'll see premiums for human-controlled vehicles (or humans taking control of robocars) start to climb steadily whilst robocar premiums either stay steady or decline (There will still be vandalism, hit'n'run whilst parked and other claims to deal with, but that constant surveillance is going to make anonymous damage to an unattended vehicle pretty much a thing of the past and result in a lot of people finding themselves with criminal records for behaviour they've been getting away with for years)
That in turn is going to result in a knee point of adoption.
Think it won't happen? Look at photos of cities in the early 20th century and look at how fast the transition from mostly horse+cart to mostly motor-vehicle was (about 15 years)
The flipside of this is that personal vehicle ownership is likely to start declining. The single most expensive part of a taxi is the driver. Eliminate the wetware and it's far cheaper to use a hire vehicle than to pay the standing costs of a personal one - personal vehicles will become the preserve of the rich again and pedestrians are likely to reclaim the streets as car-park jammed curbs become a thing of the past.
This will have knock-on effects too. Westminster and other councils haven't planned for this change and they stand to lose 30-50% of their income very quickly - even in the short term, a robocar can be instructed to go park somewhere cheaper. They're going to be scrambling to make up the difference, as are podunk shitholes that make money by setting up speedtraps.
2. Re: This time for sure?
"Maybe their new chip will..."
Comparing oranges and applesauce there sunshine. The older Autopilot very explicitly came with warnings that it COULD NOT detect and stop for stationary objects in its path when travelling in excess of fifty miles per hour.
Letting your cruise control drive into such things is on par with putting your hand into a blender and then complaining that your fingers hurt.
1. Re: This time for sure?
"Comparing oranges and applesauce there sunshine. The older Autopilot very explicitly came with warnings that it COULD NOT detect and stop for stationary objects in its path when travelling in excess of fifty miles per hour."
That might wash if some fucking hyping idiot had not called it "Autopilot" and called it "Fancy cruise control" instead.. wonder which self premoting twat that was? Elon?
People have died because of the twats ego..
6. If, if, and more if's....
If all cars/trucks were autonomous, if all roads were well maintained (even the side roads in neighborhoods, etc.) and if pedestrians, motorcycles, bicycles, etc. were restricted for using roads, then maybe autonomous vehicles will work. Until the human factor is removed, there's still too much variability in human actions for the vehicles to contend with.
1. Re: If, if, and more if's....
Autonomous vehicles can probably be made to work in a few situations in the next few years. Long haul trucking is one. Get on the expressway. Stay in your lane. Don't run into things. Don't pass anything that isn't stationary or moving VERY slowly. And a couple of thousand additional rules will probably prove to be good enough. But what's the point? You'll need a human driver on board to handle unexpected situations. Eventually, perhaps the driver can be dispensed with and only dispatched when required. But that'll require vehicles smart enough to recognize situations beyond their ken, pull over safely, and call for help.
The other likely situation is campus shuttles, airport parking lot shuttles and the like. They move very slowly on fixed routes. As long as they don't run into/over stuff/people/pets, maybe they don't really need a full time driver on board.
But I think fully autonomous vehicles are likely a lot further off than most folks think. I suspect insurance underwriters are probably going to share my skepticism.
And something that troubles me. While some common vehicle safety features like seat belts work really well, others like automated braking systems and electronic traction control, notoriously not only don't work very well in some situations -- e.g. some unpaved roads, ice and snow. In the worst cases, they actively endanger those in vehicles as well as bystanders. Yet, the problems are ignored. In some cases, these things are hard to turn off, and in a few it is actually illegal to do so. Are we going to have similar problems with autonomous vehicles?
1. Re: And a couple of thousand additional rules will probably prove to be good enough.
1. "Serve the public trust"
2. "Protect the innocent"
3. "Uphold the law"
4. "Any attempt to arrest a senior officer of OCP results in shutdown"
...
238."Avoid destructive behavior"
239. "Be accessible"
240. "Participate in group activities"
241. "Avoid interpersonal conflicts"
242. "Avoid premature value judgements"
243. "Pool opinions before expressing yourself"
244. "Discourage feelings of negativity and hostility"
245. "If you haven't got anything nice to say don't talk"
246. "Don't rush traffic lights"
247. "Don't run through puddles and splash pedestrians or other cars"
248. "Don't say that you are always prompt when you are not"
249. "Don't be over-sensitive to the hostility and negativity of others"
250. "Don't walk across a ball room floor swinging your arms"
1. Re: And a couple of thousand additional rules will probably prove to be good enough.
I guess those Teslas going up in flames couldn't take their programming any more...
2. Re: If, if, and more if's....
Long haul trucking is one. Get on the expressway. Stay in your lane. Don't run into things. Don't pass anything that isn't stationary or moving VERY slowly.
Sounds like a train, which would be a much better way to move containerized goods long haul, with the trucks reserved just for the endpoint collection/distribution.
Unfortunately the problem there isn't technical, it's truck drivers & their unions.
1. Re: If, if, and more if's....
They do that in the US. But as European trains are for people, it takes longer to get a container from Hamburg to Bari by train than by ship
1. Re: If, if, and more if's....
it takes longer to get a container from Hamburg to Bari by train than by ship
In what way exactly is five days (the transit time for rail container transport Bari-Hamburg) longer than 17 days (same, by sea)?
2. Re: If, if, and more if's....
It's also legislative because we could take it a step further and use ships where possible. The problem in the US is the Jones Act which keeps foreign merchant1 vessels, container ships in particular, from stopping in more than one US port. That means all goods destined for the US must be disgorged in one place rather than hitting several ports before the ship heads back across the seas.
Ideally, ships would handle all coastal movement, trains carry it inward and trucks only make up the short haul local moving. There wouldn't be anywhere near as many trucks on the road which would help both traffic and smog as well as probably save a whole bunch of money. Pity it will never happen because we have to protect the jobs of truck drivers, ship builders, and merchant marines.
1 - It's the Passenger Vessel Services Act for cruise ships.
2. Re: If, if, and more if's....
I get the impression that in your world there are no crashes caused by human error..
In my world virtually all crashes are caused by human error - generally inattention or impatience.
That is what these vehicles are being designed to reduce - and yet somehow they are required to be infinitely better than humans before you'd consider them. That's a very odd train of thought.
1. Re: If, if, and more if's....
What he's saying is that unless the roads are made ideal for them they aren't going to work.
Currently when they get stuck a human driver has to take over.
Human drivers don't get stuck.
1. Re: If, if, and more if's....
Au contraire they get stuck all the time. For example - there are regular examples of Human drivers blindly following a sat nav instruction off the road. Thats a very AI-like failure mode.
1. Re: If, if, and more if's....
By stuck I don't mean careering off the road or colliding with something - which a computer is very good at avoiding, but the car being literally stuck behind something and unable to progress due to being unable to take the initiative needed to proceed.
For example when a driver encounters parked cars on their side of the road or a slow cyclist they have to decide whether and when it is safe to move into the oncoming lane and pass them. That includes being able to move into oncoming traffic based on judging distances. it isn't enough to simply brake behind the parked car ahead and wait for the road to be entirely clear - on some roads at busy times you'd end up stuck.
1. Re: If, if, and more if's....
"when a driver encounters parked cars on their side of the road or a slow cyclist they have to decide whether and when it is safe to move into the oncoming lane and pass them."
I've watched SUV drivers put cyclists into hedges on lanes - breaking bones on more than one occasion.
I've watched those same SUV drivers _SCREAM_ at me to get my car out of the way because they can't get their tanks past me, despite having 8 feet of clearance between my car and the other hedge.
I've watched an elderly boy racer take off every single wing mirror on parked cars for 100 yards, and not stop (he must've been about 70)
I trust a machine to measure such things far more accurately than any human AND to give adequate clearance to other users.
You may THINK you are a better driver than average - but apparently so do 80% of other road users - and the better you think you are, the worse the driving usually is.
2. Re: If, if, and more if's....
Currently when they get stuck a human driver has to take over.
I don't know how true that is (isn't the human there for emergencies - like a driving instructor with dual-controls?), nor what value of "Currently" you're using. But taking that at face value, a human driver could presumably be in a control centre, enabling a pool of drivers to deal with "stuck" situations for a much larger pool of vehicles.
Human drivers don't get stuck.
Oh my. You've led a sheltered life!
1. Re: If, if, and more if's....
"a human driver could presumably be in a control centre, enabling a pool of drivers to deal with "stuck" situations for a much larger pool of vehicles"
yes that could solve it
3. Human drivers don't get stuck.
Tell that to a mini-roundabout with 4 cars sat waiting indefinitely.....
1. Re: Human drivers don't get stuck.
There's always an Audi to break the deadlock.
1. Re: Human drivers don't get stuck.
"There's always an Audi to break the deadlock."
And the cyclist - one of my cow-orkers ended up underneath such a vehicle with a shattered skull and months off work.
The driver's excuse? She didn't see him as she came off the roundabout at 40mph and straight over the top of him - and tried to drive off before being stopped by witnesses.
One of the benefits of having robocars is that their driving standards will rapidly result in MINIMUM human standards being sharply increased - after all, if you're an incompetent driver you can get a machine to do it. This will be driven by insurers and they WILL require periodic retesting just like they did in aviation before it became law. No play, no premium - and you won't be discriminated against if they refuse to insure you as you can always get a machine to drive you.
4. Re: If, if, and more if's....
"Currently when they get stuck a human driver has to take over."
The human driver doesn't have to be onboard.
"Human drivers don't get stuck."
Uh yeah. Right. You haven't been watching the same "bad driving UK/Oz/USA/NZ/wherever" youtube videos I have.
Even Lewis Hamilton has off days. Mere mortals make mistakes every couple of minutes some drivers are just plain asleep at the wheel whilst others have a death wish. Road rules and safety regulations are generally setup so that it takes _at least_ 3 (if not 5) serious errors to cause a crash (sometimes the errors are the road designer's), but they still happen regularly.
Having _consistent_ driving on the roads will bring its own benefits:
Slow rural drivers in particular are a dire statistical menace all of their own as they put everyone trying to pass them on the wrong side of the road (speeders only put themselves on the wrong side) - this is why inconsistent speed limits for different vehicle types is dangerous.
City gridlock is invariably caused by arseholes trying to barge their way through, blocking intersections and choking narrow points.
Then there are the rat runs, etc etc.
1. Re: If, if, and more if's....
"Uh yeah. Right. You haven't been watching the same "bad driving UK/Oz/USA/NZ/wherever" youtube videos I have."
I've seen loads of those videos, both online and in TV programmes dedicated to them. On the other hand, I've averaged in the region of 50,000 miles per year over the last 25 years or so and I've seen far, far fewer incidents in real life. I've seen a fair few idiots doing stupid things and nearly coming a cropper. I've seen the results of accidents (and been stuck in the queues) a number of times too. But in my personal experience, the numbers of idiots on the road is relatively small. I've been the one taking evasive action a number times too, but I can't stress enough that those times and experiences are the exception rather than the rule. All just my personal experience though. YMMV :-)
2. Re: If, if, and more if's....
"That is what these vehicles are being designed to reduce - and yet somehow they are required to be infinitely better than humans before you'd consider them. That's a very odd train of thought."
If someone you know dies in an RTA, you have one or more other drivers to blame and the insurance pays out. Who you gonna blame when it's a robo-car? Especially if \$BigCorp is going to be sending in it's high priced lawyers to fight every case because they don't ever want to admit to faulty hardware or software in their cars.
3. Re: If, if, and more if's....
In my world virtually all crashes are caused by human error... That is what these vehicles are being designed to reduce
The main types of human error that cause crashes are inattention, distraction, overconfidence, intoxication and impaired visibility. What doesn't cause many crashes is complete misinterpretation of the context. That's because we're using a perceptual system that's had millions of years of optimisation for our physical world, and driving in an environment that's been modified to accommodate the limitations of our perceptual system when controlling motor vehicles at speed.
Everything I read about autonomous vehicles makes it clear that they are a long way from this degree of general perception. For example, experienced drivers can infer the direction of an invisible road ahead by checking the line of telegraph posts or trees, and on country lanes at night they become aware of an oncoming car by the loom of its headlights long before they see it.
It's the sort of idea that could only have been dreamed up in a country full of grid-pattern towns and wide, straight highways. In 1911 somebody drove a Ford Model T up Ben Nevis to prove its capability. I can think of plenty of drives in Europe I'd like to see an autonomous car complete before I trust it.
1. Re: If, if, and more if's....
> For example, experienced drivers can infer the direction of an invisible road ahead by checking the line of telegraph posts or trees, and on country lanes at night they become aware of an oncoming car by the loom of its headlights long before they see it.
The presentation shows a car predicting (without maps) the direction a road takes over a crest it cannot see - solved problem.
3. Re: If, if, and more if's....
Until the human factor is removed, there's still too much variability in human actions for the vehicles to contend with.
Never mind the human factor, there's also a huge amount of natural variability. So a cat in the road. Or a deer. Many versions of those available. Or for Australian vehicles, cane toads, kangaroos and cassowarys. All challenges that human drivers can potentially learn to deal with better then 'AI' can.
But more hype. Musk promised this years ago, and failed to deliver. Much like the \$35k 'people's EV'.
7. What Rot
"Earlier this month, researchers outlined how a Tesla could be made to veer off course into oncoming traffic with a few simple stickers on the ground, demonstrating the danger of relying too heavily on video camera inputs."
Nah. That's the danger of relying too heavily on LANE MARKINGS as primary guidance data. Humans rely on a couple video camera inputs and do just fine. I'm still flabbergasted that they were able to trick a system that should have been stress-tested on Highway 101 and its Shittastic Markings by now.
1. Re: What Rot
These companies working on self driving cars are not required to fully disclose their research and have no incentive to admit how impossible the task of making self driving cars actually is. They possibly even delude themselves, or perhaps hope legislation will allow them to sell the technology with no liability for accidents.
Getting a car to follow lane markings is pretty achievable, so is detecting an obstacle ahead and braking before hitting it (couldn't both be done 40 years ago?). But as soon as the situation becomes a little different, a little more complicated, then you find you have to start throwing more complicated rules into the AI, and in turn those complicated rules open up more vulnerabilities, which require even more complex rules to patch up.
IMO you need artificial general intelligence to drive cars autonomously, by which point self driving cars would be the least of our concerns.
1. Re: What Rot
"These companies working on self driving cars are not required to fully disclose their research"
In the USA.
"and have no incentive to admit how impossible the task of making self driving cars actually is."
In the USA.
Thats' the same USA where the aviation regulator allowed Boeing to rat rod the 737 into the NG (which is mildly unstable(*)), shopped whistleblowers to Boeing when they flagged forged documentation on crucial airframe structural members (supposed to be precision CNC machined, actually made sloppily by hand and beaten to fit, broke badly in several minor crashes killing at least 20 people and corroded badly in service, will result in airframes falling out of the sky) and then waved extreme rat rodding of the 737MAX and sloppy software through.
It was a world technology leader once. Now it's leading the world in other areas - such as corruption.
(*) You can't power a NG out of a stall. It's one of the few aircraft you MUST put the nose down _FIRST_ in, then power up. The MAX is even worse - and a stall can happen easily in a low speed turn, so don't think it's just "pulling up"
2. Re: What Rot
humans will also blindly follow road markings and traffic cones even if the consequence is dire.
Luckily most of the time the line markings haven't been tampered with or wrongly placed and the traffic cones aren't rearranged by nefarious individuals.
1. Re: What Rot
Or the lane markings are a bit worn and it's dark and wet and the road has had contraflows or some other reason for different lane markings on it in the past. Which are burnt off but equally as prominent as the current lane markings in the dark and wet.
8. Use of investor's capital?
I've worked in a few environments where we did our own HDL development. We worked almost entirely in the FPGA world because we did too much "special purpose" algorithms which would often require field updates... an area not well suited for ASICs.
But, I believe what Tesla is doing here is a mistake.
Large scale ASIC development is generally reserved for a special category of companies for a reason. Yes, their new tensor processor almost certainly is a bunch of very small tensor cores, which each are relatively easy to get right, and the interconnect is probably either a really simple high speed serial ring bus... so, it's probably not much harder than just "daisy chaining" a bunch of cores. But even with a superstar chip designer on staff, there are a tremendous amount of costs in getting a chip like this right.
Simulation is a problem.
In FPGA, we often just simulate using squiggly lines in a simulator. Then we can synthesize and upload it to a chip. The trial and error cycle is measured in hours and hundreds of dollars.
In ASIC, all the work is often done in FPGA first, but then to route, mask and fab a new chip... especially of this scale, there is a HUGE amount involved. It requires multiple iterations and there are always going to be issues with power distribution, grounding, routing... and most importantly, heat. Heat is a nightmare in this circumstance. Intel, NVidia, Apple, ARM, etc... probably each spend 25-50% of their R&D budgets on simply putting transistors in just the right places to distribute heat appropriately. It's not really possible to properly simulate the process either... and a super-star chip designer probably know most of the tricks of the trade to make it happen, but there's more to just intuition with regards to this.
Automotive processors must operate under extreme environmental conditions... especially those used in trucks traversing mountains and deserts.
If Tesla managed to actually make this happen and they managed to build their own processors instead of paying NVidia, AMD or someone similar to do it for them, I see this as being a pretty bad idea overall.
Of course, I'd imagine that NVidia is raking Tesla over the coals and making it very difficult for Tesla to reach self-driving in a model 3 class car, but there has the be a better solution than running an ASIC design company within their own organization. Investing in another company in exchange for favorable prices would have made more sense I think. Then the development costs could have been spread across multiple organizations.
1. Re: Use of investor's capital?
I'm betting there is very little bespoke about the Tesla chips. Probably 95% off the shelf with some custom config and connectivity. I suspect its closer to Apples "custom SKU's" of Intel chips than their A* series of arm processors.
9. I Call Shotgun!
I can sometimes top two calculations per second in burst mode but probably average less than one (two digit addition), so I expect the Tesla to be 144 trillion times better than me at driving.
1. Re: I Call Shotgun!
You have a lot of stuff going on about that you're unaware off. I'm sure you can recognise both a tree and a road sign at the same time and decide which one needs your attention.
1. Re: I Call Shotgun!
Squirrel!
10. allowing its vehicles to potentially drive themselves completely autonomously
The important word being potentially. With a suitable weight on the acceerator my 40 year old sportscar will driveitself completely autonomously - for a while.
Once people might have listened to Musk, but between the previous failures to deliver and the new blatant overreach (robot taxis making \$\$\$\$ next year?!) no one puts any value on anything he has to say on autonomous vehicles.
And some unkind people might even say he was making a lot up as he went expecting to use the resulting hype to overshadow the imminent release of the Q1 numbers. Pity the reaction seems to have been a combination of 'meh' plus various others (like nVidia) immediately ripping his words apart.
Icon showing live footage from a Shanghai car park. ->
Quite, what have any of Musk's companies ever achieved of any significance?
You're not talking about NASA's skunkworks project established to funnel government money into private hands also known as SpaceX, are you ?
12. Benchmarks and other deceptions
"Tesla claimed its chip could do up to 144 trillion math calculations per second..."
Well, that may be true, but the bottleneck is almost always the interconnect between cores (of which there are undoubtedly many) and the data paths to get the data from the sensors to the appropriate core / buffer.
Paralleling data crunching is also difficult for many algorithms (I have worked in the HPC space where such things come up regularly) and floating point (the IEEE version) is not fully reciprocal (close but some rounding errors may have unfortunate consequences) although it is perfectly possible that some version of fixed point is being used.
That said, there are some places self-driving cars will really struggle to ever become a reality, such as Cornish lanes.
1. Re: Benchmarks and other deceptions
Those lanes may be one place where AI drivers most easily outperform humans.
At least, the substantial minority of human drivers who struggle to reverse in a straight line (let alone round a bend) when you need to get to the nearest passing place.
2. Re: Benchmarks and other deceptions
You don't even need to steer down Cornish/Devon lanes. You just bounce off the sides. Preferably in a beat-up defender just after last orders while wearing wellies.
1. Re: Benchmarks and other deceptions
The local Cornish approach is far too often "it's OK to drink dive if you're a local as you know the roads".
A particular hazard on many of these roads is curiously abandoned cars. In the middle of fecking nowhere, about a metre from the edge of the road, usually just after a blind corner at the top or bottom of a hill. No sign of the driver of course, or even why they would stop and park where they do. Other hazards are:
• mud covered vehicles with no lights doing 10mph on the A30. In the dark.
• caravans. Is any other country in the world blighted by these things as much as the UK?
• tosser trolleys (chelsea tractors, wanker wagons, various other terms for non-offroad "offroad" vehicles) - driven by locals or emmets and always too big to fit on many roads or in parking spaces.
• "Interesting" place names that one feels a need to stop and photograph :)
/(still) honary Cornish I guess...
3. Re: Benchmarks and other deceptions
Remember the Monty Python footnote: "'up to' includes zero."
4. Cornish lanes.
Cornish lanes is not a problem for Tesla AI!
Tesla AI is texts, ordinary everyday human texts.
Why texts? Any text pattern has a context and a huge number of subtexts (e.g. dictionary definitions, explanatory instructions, etc.). These contexts and subtexts, superimposed on the description of a particular situation (which is also text), allow you to find a single pattern and tell the machine what it should do.
That is, is enough to describe once a certain situation - Cornish one - and no problems with Tesla AI will never be.
13. It's one thing to have the hardware
It's quite another to make the software. As a mental exercise, think of all these things that happen quite frequently on the roads - a set of traffic lights that are broken or out, road works, potholes, broken glass or debris on the road, large puddles, a police man giving directions, diversions, some lorry trying to back into a driveway, narrow lanes with oncoming traffic and spots to yield, an icy hill, a pedestrian crossing, a lollipop lady, cyclist on a narrow windy lane, road bumps, box junctions, etc. etc. etc.
These are situations that are not covered by a simple set of rules but a complex understanding of the prevailing conditions, recognition / cooperation with other humans in order to make a decision. These are basically intractible problems, each and every one.
Software in an autonomous car has to cope with all of that sanely, safely while making good progress in every scenario and permutation. It has to do it at least as well as a human can. If it can't do all that then Tesla can forget about self driving cars tooling around without drivers for the forseeable future.
What I expect to happen instead is Tesla will dump some features out there which will improved autonomy in limited situations but still won't be anywhere close to a level 5 vehicles. The car might be level 4 in some scenarios but you'd better believe you will require an alert and attentive driver in every scenario. I actually wonder if they're going to turn some of that the processing power in their new module inwards to monitor the driver and ensure they are being attentive.
1. Re: It's one thing to have the hardware
"It's quite another to make the software."
Exactly what I was going to say. Autonomous cars are not a hardware problem. Sure, there will be benefits, especially in power consumption, from using dedicated chips designed specifically to do the job, but it would already be trivial to get more than enough computing power by just sticking a half-decent workstation laptop under the bonnet. Same for the sensors - arguing about exactly what kind are best and how many you need isn't the big problem, it's figuring how to actually use all that information in a sensible way that is the tricky part.
1. Tesla will not be controlled by algorithms but AI
Tesla will not be controlled by algorithms but AI, where AI is texts, ordinary everyday human texts. This is the highlight that makes possible for a car without a driver to exist!
Thanks to my discovered and patented method of text structuring, is possible to select a pattern from a text, in its explicit context and implicit subtext. And this pattern can be used as a computer command, driving a car.
Why texts? Any textual pattern has a context and a huge number of subtexts (for example, definitions from a dictionary for its words, explanatory instructions, etc). These contexts and subtexts, superimposed on the description of a particular situation (which is a text too), allow you to find a single pattern and tell the car what it should do.
1. Re: Tesla will not be controlled by algorithms but AI
What's text got to do with a runaway, unmarked (read: no text to read) car barrelling towards you on a single-lane road? AND there's another car a short distance behind you? Can an AI think outside the box and realize the only hope is to go OFF the script (and the road)?
1. Re: Tesla will not be controlled by algorithms but AI
Already got one! :- ) Now your text should be tied to a specific situation :-)
This is called machine learning: AI adds structured texts that contain all the necessary patterns for a particular circumstance (sets of commands).
What is a programming language and program?
A programming language is a set of special words (commands) that structure a text/ description of what you want, and make it understandable to computer. In other words, a program is a structured text and programmers are translators.
AI structures texts and makes programs out of them automatically, without the participation of programmers/ highly paid individuals. Can you imagine how many trillion AI saves only on cars without a driver?
Amazon has already hired thousands of people to annotate texts. I think Tesla, like all other companies, has also hired thousands of people to annotate sensor data, which is the answer to your question.
1. Re: Tesla will not be controlled by algorithms but AI
Well then, riddle me this, Pythagoras. How does a computer distinguish between a little and YOUR little kid while the sun is glaring in your lens because it's sunrise or sunset.
1. Re: Tesla will not be controlled by algorithms but AI
It all depends on how well used Tesla or Waymo sensors are, and how many different situations Tesla or Waymo have described by texts.
Indeed, today it is possible to structure texts, that is, to decompose them into patterns and make them understandable to computer, thanks to AI-parsing. Previously, it was done by specially trained people, the so-called programmers. They laid out the texts into patterns using the standards of so-called programming languages, combining their commands and using manually structured data from SQL databases (on the basis of n-gram parsing). Today, the same and much cheaper, faster and better makes the computer.
That is, if, for example, Tesla has the right sensors and textual descriptions of "a little and MY little kid while the sun is glaring in your lens because it's sunrise or sunset" - some proper actions may occur. For example, lollipops and soda will be issued to both the children.
This is AI - it finds answers according to the NIST TREC demand, and acts accordingly.
14. What could possibly go wrong?
"The aim is to push firmware updates over the air to vehicles to bring their software up to level four, which is what you'd call baseline human-free driving, all powered by the supposedly superior FSD hardware."
So this morning my car behaves differently from the way it did when I drove home yesterday. Nevertheless, until it reaches level four I'm still supposed to take over when the car loses it's marbles, but what's going to trigger that today?
15. TLDR
Can I legally be driven by the car whilst asleep drunk in the back seat?
If not, it's not self driving.
If I am in any way, other than maintenance, legally resposible for the actions of the vehicle whilst it is driving, then it's not self-driving.
1. Re: TLDR
Nope, because in many jurisdictions the law is about being drunk whilst in charge of the vehicle. You can be done for just having the keys on you. So your test will always fail.
1. Re: TLDR
I dunno why you've been downvoted, the car clearly fails my personal definition of 'self driving'.
I don't want an augmented cruise control that will probably put me to sleep. I want to go to sleep, comfortably, safely and legally.
16. But is it safe?
"two fully independent math processors on board which both receive full video from the car and make their own independent evaluations before another part of the system compares them to make sure they match. "
Are they running the same software?
How do they avoid the Boeing effect: Self-certification of safety and business pressures work against each other.
Maybe directors need to be in the chair from Marathon Man: https://www.youtube.com/watch?v=kzw1_2b-I7A
17. 1 million self driving taxis means that everyone who spent north of 60-100K+ on a Tesla is dying to turn their luxury car into a taxi.
Hahahahahaha
1. I wonder whilst doing its moonlighting job it it will also clean up the puke which possibly may be liberally spread round the inside of the car?
2. "1 million self driving taxis means that everyone who spent north of 60-100K+ on a Tesla is dying to turn their luxury car into a taxi."
No, it means they spent north of 60-100K to have their own personal, clean space, guaranteed free of other peoples mess, dirt, left over takeaway, shit, piss and vomit while they travel.
They already bought the car. They own it. Tesla can't hijack it for use as a taxi while they are not using it without permission/payments.
18. Musk is losing his mind
Either he's going to be sued into oblivion for his claims about a "robot taxi that will let you keep 75% of the earnings" since that's financial information people would use to base purchases on, or he'll be sued into oblivion when a driverless car that's years from being ready to do that kills its first pedestrian while on "robot taxi" duty.
19. Logic chopping
I reckon I could build a self-driving car and hype it as such.
But I would not pretend that it could avoid accidents.
20. "...totally safe..." self-driving cars
So, no speed limits on them ? If they're "totally safe", then they can drive as fast as they decide. There would be no logical reason whatsoever for externally-mandated speed limits on self-driving vehicles that can provide their own "totally safe" assurance.
Well, perhaps we'd want to avoid sonic booms, so maximum ground speed should be Mach 0.95.
I suspect that we're rapidly heading into a new version of the 'omniscient omnipotent paradox', except with self-driving cars instead of gods.
I'm happy that Elon is happy.
I presume he will therefore have no objections to being required to spend "a while" chained to the front of one of his vehicles, Mad Max Style, as it is tested in a multitude of different driving environments around the world?
Seems reasonable enough to me. . . .
22. Delusional
As others have noted, the speed of the processor seems largely irrelevant to the real issue. Fundamentally the current algorithms that control these “AI” cars cannot, in any sensible way, think for themselves. This is almost certainly deliberate as it would be all but impossible to predict the behaviour of a self driving car that could perform such a feat (and therefore, even if we could build such a thing, hugely reckless to unleash them on public roads). Unfortunately, lacking this “think on your feet” ability, these cars are entirely constrained to handling situations they have been explicitly trained on. Given I cannot see how it is possible to train a car for every possible situation, I find it hard to believe any claims of impending success. My big concern is that many governments will fall for the hype / propaganda and green light them well before they are even vaguely ready. The current safely claims / comparisons are so ridiculously biased it beggars belief. Comparing the safety record of brand new, high end sports cars, driven entirely within posted speed limits, almost entirely on highways in largely dry climates with average human behaviour seems borderline fraudulent. If a comparison were made with the safety record of similar cars, on similar roads, in similar conditions, at similar speeds, excluding intoxicated drivers (as this is illegal, so seems unfair to include in such a comparison) then I might be willing to reconsider (at least for its use on such highways).
23. Tesla's vision is fine but re autonomous driving
How does one deal with the human element, such as 1. Wanting to pull over suddenly to look at a nice view or take a leak? "Car! Quick, pull into that lay-by!" How will car know which lay-by? A human passenger would gesture to the driver, pointing. Can internal camera see what the occupants are doing?
2. And there hasn't been any discussion as to what Tesla's AI will do when faced with two unavoidable collisions:
a) Elderly couple crossing the road or b) Younger mother with child?. If course, this situation is rare even for humans, but owner of vehicle should be able to specify. This will also move the burden of responsibility away from Tesla (or other manufacturer) to the owner for insurance etc purposes.
1. Re: Tesla's vision is fine but re autonomous driving
IOW, there's no real solution to a Trolley Problem, so it's extremely hard to pick between two such scenarios because each case can be different. Say you say run over the old last but what if the next situation involves your gran? Pick the mother, then it's your wife and baby next time.
Two other situations bear considering. First, snap object recognition, which even humans can fail at (You can tell the difference between a brick and a bag? Fine, how about a brick in a bag?).
Second, intuitive situations that humans solve without conscious recognition. Since we don't know we're even doing it, we don't know HOW we're doing it, and without the knowledge of how we do it, it's impossible to teach it to another human, let alone a machine.
I'm waiting for someone to prove a few of these serious automation problems to be physically intractable (except for the Trolley Problem, as that's a dilemma and has no real solution by design).
24. 144 Trillion?
Bet Tesla can do a lot of coin mining on those systems that are sitting in the garage overnight or otherwise not in active driving use!
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Not a member of The Register? Create a new account here. | 11,693 | 54,164 | {"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} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.966528 |
https://educationexpert.net/mathematics/1707857.html | 1,716,366,881,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00059.warc.gz | 181,336,309 | 7,001 | 11 May, 17:28
# Find the unit rate 50 min?20 calls?
+2
1. 11 May, 18:17
0
Step-by-step explanation:
This is a fraction equal to
50 minutes : 20 calls
We want a unit rate where
1 is in the denominator,
so we divide top and bottom by 20
50 minutes : 20
20 calls : 20 = 2.5 minutes / 1 call | 100 | 297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.71975 |
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Chapter Twelve. Some Lessons From Capital Market History. Key Concepts and Skills. Know how to calculate the return on an investment Understand the historical returns on various types of investments Understand the historical risks on various types of investments. Chapter Outline. Returns
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Some Lessons From Capital Market History
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Chapter
Twelve
## Some Lessons From Capital Market History
### Key Concepts and Skills
• Know how to calculate the return on an investment
• Understand the historical returns on various types of investments
• Understand the historical risks on various types of investments
### Chapter Outline
• Returns
• The Historical Record
• Average Returns: The First Lesson
• The Variability of Returns: The Second Lesson
• Capital Market Efficiency
### Risk, Return and Financial Markets
• We can examine returns in the financial markets to help us determine the appropriate returns on non-financial assets
• Lesson from capital market history
• There is a reward for bearing risk
• The greater the potential reward, the greater the risk
• This is called the risk-return trade-off
### Dollar Returns
• Total dollar return = income from investment + capital gain (loss) due to change in price
• Example:
• You bought a bond for \$950 1 year ago. You have received two coupons of \$30 each. You can sell the bond for \$975 today. What is your total dollar return?
• Income = 30 + 30 = 60
• Capital gain = 975 – 950 = 25
• Total dollar return = 60 + 25 = \$85
### Percentage Returns
• It is generally more intuitive to think in terms of percentages than dollar returns
• Dividend yield = income / beginning price
• Capital gains yield = (ending price – beginning price) / beginning price
• Total percentage return = dividend yield + capital gains yield
### Example – Calculating Returns
• You bought a stock for \$35 and you received dividends of \$1.25. The stock is now selling for \$40.
• What is your dollar return?
• Dollar return = 1.25 + (40 – 35) = \$6.25
• What is your percentage return?
• Dividend yield = 1.25 / 35 = 3.57%
• Capital gains yield = (40 – 35) / 35 = 14.29%
• Total percentage return = 3.57 + 14.29 = 17.86%
### The Importance of Financial Markets
• Financial markets allow companies, governments and individuals to increase their utility
• Savers have the ability to invest in financial assets so that they can defer consumption and earn a return to compensate them for doing so
• Borrowers have better access to the capital that is available so that they can invest in productive assets
• Financial markets also provide us with information about the returns that are required for various levels of risk
### Year-to-Year Total Returns
Large-Company Stock Returns
Long-Term Government
Bond Returns
U.S. Treasury Bill Returns
### Average Returns
• The “extra” return earned for taking on risk
• Treasury bills are considered to be risk-free
• The risk premium is the return over and above the risk-free rate
• Large stocks: 13.0 – 3.9 = 9.1%
• Small stocks: 17.3 – 3.9 = 13.4%
• Long-term corporate bonds: 6.0 – 3.9 =2.1%
• Long-term government bonds: 5.7 – 3.9 = 1.8%
### Variance and Standard Deviation
• Variance and standard deviation measure the volatility of asset returns
• The greater the volatility the greater the uncertainty
• Historical variance = sum of squared deviations from the mean / (number of observations – 1)
• Standard deviation = square root of the variance
### Example – Variance and Standard Deviation
Variance = .0045 / (4-1) = .0015 Standard Deviation = .03873
### Work the Web Example
• How volatile are mutual funds?
• Morningstar provides information on mutual funds, including volatility
• Click on the web surfer to go to the Morningstar site
• Pick a fund, such as the Aim European Development fund (AEDCX)
• Enter the ticker, press go and then scroll down to volatility
### Efficient Capital Markets
• Stock prices are in equilibrium or are “fairly” priced
• If this is true, then you should not be able to earn “abnormal” or “excess” returns
• Efficient markets DO NOT imply that investors cannot earn a positive return in the stock market
### What Makes Markets Efficient?
• There are many investors out there doing research
• As new information comes to market, this information is analyzed and trades are made based on this information
• Therefore, prices should reflect all available public information
• If investors stop researching stocks, then the market will not be efficient
• Efficient markets do not mean that you can’t make money
• They do mean that, on average, you will earn a return that is appropriate for the risk undertaken and there is not a bias in prices that can be exploited to earn excess returns
• Market efficiency will not protect you from wrong choices if you do not diversify – you still don’t want to put all your eggs in one basket
### Strong Form Efficiency
• Prices reflect all information, including public and private
• If the market is strong form efficient, then investors could not earn abnormal returns regardless of the information they possessed
• Empirical evidence indicates that markets are NOT strong form efficient and that insiders could earn abnormal returns
### Semistrong Form Efficiency
• Prices reflect all publicly available information including trading information, annual reports, press releases, etc.
• If the market is semistrong form efficient, then investors cannot earn abnormal returns by trading on public information
• Implies that fundamental analysis will not lead to abnormal returns
### Weak Form Efficiency
• Prices reflect all past market information such as price and volume
• If the market is weak form efficient, then investors cannot earn abnormal returns by trading on market information
• Implies that technical analysis will not lead to abnormal returns
• Empirical evidence indicates that markets are generally weak form efficient
### Quick Quiz
• Which of the investments discussed have had the highest average return and risk premium?
• Which of the investments discussed have had the highest standard deviation?
• What is capital market efficiency?
• What are the three forms of market efficiency? | 1,510 | 6,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.859449 |
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