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https://cs.stackexchange.com/questions/98378/maximizing-entropy-under-constraint | 1,632,038,954,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00583.warc.gz | 229,482,790 | 37,935 | # Maximizing entropy under constraint
How do I prove that entropy is maximal for $$P(A_2) = \cdots = P(A_n) = (1-a) /(n-1)$$ while $$P(A_1) = a$$ (a fixed number) and $$A_1,…, A_n$$ is a partition of the sample space?
• Try to use convexity/concavity. There are also other ways. Oct 9 '18 at 21:03
Suppose that $$X$$ is a random variable over the set $$\{1,\ldots,n\}$$ satisfying $$\Pr[X=1] = a$$. Let $$Y$$ be the indicator of the event $$X=1$$. Then \begin{align*} H(X) &\stackrel{(1)} = H(Y) + H(X|Y) \\ &\stackrel{(2)}= h(a) + aH(X|Y=1) + (1-a)H(X|Y=0) \\ &\stackrel{(3)}\leq h(a) + (1-a) \log (n-1). \end{align*} Here (1) is the chain rule (using $$H(X) = H(X,Y)$$, since $$Y$$ is determined by $$X$$), (2) follows from $$\Pr[Y=1]=a$$ and the definition of $$H(X|Y)$$, and (3) follows from $$H(X|Y=1) = 0$$ (since when $$Y=1$$ we have $$X=1$$) and $$H(X|Y=0) \leq \log (n-1)$$ (since when $$Y=0$$, the r.v. $$X$$ only attains the values $$2,\ldots,n$$).
Furthermore, assuming $$a \neq 1$$, equality holds only if $$X|Y=0$$ is uniform, that is, if $$\Pr[X=i|Y=0] = 1/(n-1)$$ for $$2 \leq i \leq n$$, or in other words, if $$\Pr[X=i] = (1-a)/(n-1)$$ for $$2 \leq i \leq n$$.
• Thank you for your answer, in (1) I think you mean H(X,Y)? Oct 10 '18 at 10:23
• They’re equal, since Y is determined by X. Oct 10 '18 at 15:28 | 537 | 1,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.757634 |
https://robertlovespi.net/tag/symmetrohedra/page/2/ | 1,610,954,369,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00741.warc.gz | 542,395,380 | 31,819 | A 50-Faced Symmetrohedron with Cuboctahedral Symmetry
In this symmetrohedron, the red, blue, and green faces are regular polygons. Only the yellow isosceles trapezoids are irregular.
A Symmetrohedron with Tetrahedral Symmetry
Symmetrohedra are symmetrical, convex polyhedra which contain many faces (not necessarily all) which are regular polygons. In this symmetrohedron, the hexagons and triangles are regular, while the quadrilaterals are isosceles trapezoids. I made this symmetrohedron using Stella 4d, a program you can try for free at this website.
A Symmetrohedron
Symmetrohedra are symmetrical polyhedra which have many, but not all, faces regular. I just found one which has square faces, regular hexagons, regular octagons, and a bunch of scalene triangles as the irregular faces.
To form this polyhedron, I started with the great rhombcuboctahedron, then formed its base/dual compound. I then took the convex hull of this compound, and then formed the dual of that convex hull, producing the polyhedron you see above. All of this was done using Stella 4d: Polyhedron Navigator, which you can try for free at http://www.software3d.com/Stella.php.
Four Symmetrohedra with Tetrahedral Symmetry
Symmetrohedra are polyhedra with some form of polyhedral symmetry, and many (not necessarily all) regular faces. The first two symmetrohedra here each include four regular enneagons as faces.
The next two symmetrohedra each include four regular dodecagons as faces.
All four of these were made using Stella 4d, which you can try out for free at http://www.software3d.com/Stella.php.
Two Symmetrohedra Which Feature Enneagons
These two symmetrohedra were created using Stella 4d, software you can try for free right here.
A Symmetrohedron Featuring Regular Hexadecagons and Regular Hexagons
Image
I made this using Stella4d, which you can try for free right here.
Four Symmetrohedra
Symmetrohedra are polyhedra with some form of polyhedral symmetry, all faces convex, and many (but not all) faces regular. Here are four I have found using Stella 4d, a polyhedron-manipulation program you can try for yourself at http://www.software3d.com/Stella.php.
The second of these symmetrohedra is also a zonohedron, and is colored the way I usually color zonohedra, coloring faces simply by number of sides per face. That is why some of the red octagons in that solid are regular, while others are elongated. The other three symmetrohedra are colored by face type, with the modification that the fourth one’s scalene triangles are all given the same color.
These symmetrohedra were all generated using Stella 4d, a program you may try for yourself at http://www.software3d.com/Stella.php.
Bowtie Cubes in a Polyhedral Honeycomb
This polyhedron has been described here as a “bowtie cube.” It is possible to augment its six dodecagonal faces with additional bowtie cubes. Also, the bowtie cube’s hexagonal faces may be augmented by truncated octahedra.
These two polyhedra “tessellate” space, together which square pyramidal bifrustrums, meeting in pairs, which fill the blue-and-green “holes” seen above. This last image shows more of the “honeycomb” produced after yet more of these same polyhedra have been added.
This pattern may be expanded into space without limit. I discovered it while playing with Stella 4d, software you may try for free at this website.
A Symmetrohedron with 74 Faces
Symmetrohedra have many regular faces, but irregular faces are allowed in them as well. The octagons, hexagons, and squares in this polyhedron are regular, but the 48 triangles are scalene. Here’s what it looks like with these triangles rendered invisible:
Stella 4d was used to make these images; it may be tried for free at this website: http://www.software3d.com/Stella.php.
The Second of Dave Smith’s “Bowtie” Polyhedral Discoveries, and Related Polyhedra
Dave Smith discovered the polyhedron in the last post here, shown below, with the faces hidden, to reveal how the edges appear on the back side of the figure, as it rotates. (Other views of it may be found here.)
So far, all of Smith’s “bowtie” polyhedral discoveries have been convex, and have had only two types of face: regular polygons, plus isosceles trapezoids with three equal edge lengths — a length which is in the golden ratio with that of the fourth side, which is the shorter base.
He also found another solid: the second of Smith’s polyhedral discoveries in the class of bowtie symmetrohedra. In it, each of the four pentagonal faces of the original discovery is augmented by a pentagonal pyramid which uses equilateral triangles as its lateral faces. Here is Smith’s original model of this figure, in which the trapezoids are invisible. (My guess is that these first models, pictures of which Dave e-mailed to me, were built with Polydrons, or perhaps Jovotoys.)
With Stella 4d (available here), the program I use to make all the rotating geometrical pictures on this blog, I was able to create a version (by modifying the one created by via collaboration between five people, as described in the last post) of this interesting icositetrahedron which shows all four trapezoidal faces, as well as the twenty triangles.
Here is another view: trapezoids rendered invisible again, and triangles in “rainbow color” mode.
It is difficult to find linkages between the tetrapentagonal octahedron Smith found, and other named polyhedra (meaning I haven’t yet figured out how), but this is not the case with this interesting icositetrahedron Smith found. With some direct, Stella-aided polyhedron-manipulation, and a bit of research, I was able to find one of the Johnson solids which is isomorphic to Smith’s icositetrahedral discovery. In this figure (J90, the disphenocingulum), the trapezoids of this icositetrahedron are replaced by squares. In the pyramids, the triangles do retain regularity, but, to do so, the pentagonal base of each pyramid is forced to become noncoplanar. This can be difficult to see, however, for the now-skewed bases of these four pyramids are hidden inside the figure.
Both of these solids Smith found, so far (I am confident that more await discovery, by him or by others) are also golden polyhedra, in the sense that they have two edge lengths, and these edge lengths are in the golden ratio. The first such polyhedron I found was the golden icosahedron, but there are many more — for example, there is more than one way to distort the edge lengths of a tetrahedron to make golden tetrahedra.
To my knowledge, no ones knows how many golden polyhedra exist, for they have not been enumerated, nor has it even been proven, nor disproven, that their number is finite. At this point, we simply do not know . . . and that is a good way to define areas in mathematics in which new work remains to be done. A related definition is one for a mathematician: a creature who cannot resist a good puzzle. | 1,688 | 6,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.939718 |
http://mathhelpforum.com/calculus/162654-calculus-check-my-work-please-print.html | 1,513,179,787,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00161.warc.gz | 182,880,065 | 3,017 | • Nov 9th 2010, 10:23 AM
zhupolongjoe
http://www.math.rutgers.edu/~greenfi...fstuff/w3N.pdf
To that problem
I have
a) increasing for [0,C) and (D,E] since f'(x) is positive there, and so, decreasing for (C,D). Local extrema would just be C as a local max? (Not sure though since it isn't continuous)
b) concave up=f''(x)>0 for [0,A)U[B,C)U(C,E] and concave down only for (A,B) since that is the only place where slope is negative. So point of inflection would be both A and B?
c) Not sure how I could draw the graph....Do I just draw anything that fits those specification or how can I know where things start and stop?
• Nov 9th 2010, 11:41 AM
Opalg
Quote:
Originally Posted by zhupolongjoe
http://www.math.rutgers.edu/~greenfi...fstuff/w3N.pdf
To that problem
I have
a) increasing for [0,C) and (D,E] since f'(x) is positive there, and so, decreasing for (C,D). Local extrema would just be C as a local max? (Not sure though since it isn't continuous) Yes, local max at C. But what happens at D?
b) concave up=f''(x)>0 for [0,A)U[B,C)U(C,E] and concave down only for (A,B) since that is the only place where slope is negative. So point of inflection would be both A and B? Yes.
c) Not sure how I could draw the graph....Do I just draw anything that fits those specifications Yes!
or how can I know where things start and stop? The only information you have cincerns the derivative. So f can only be determined up to a constant (like a constant of integration).
..
• Nov 9th 2010, 12:17 PM
zhupolongjoe
thanks and so D-local min | 462 | 1,541 | {"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} | 3.265625 | 3 | CC-MAIN-2017-51 | longest | en | 0.95113 |
http://www.statisticshowto.com/shifting-data/ | 1,524,186,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937090.0/warc/CC-MAIN-20180420003432-20180420023432-00180.warc.gz | 518,570,199 | 20,508 | # Shifting Data
Shifting data is adding a constant k to each member of a data set, where k is a real number. In visual terms, it is lifting the entire distribution of data points and shifting en masse a distance of k.
## Shifting Data and the Mean & Median
When data is shifted the mean(μ) and median are shifted by the constant k.
That is to say, when you shift a data set by k, so that f(x)= x+k for every x in your data set,
• f(μ)= μ + k
• f(med)= med + k
Note that k can be either positive or negative.
## Measures of Spread and Relative Standing
Your standard deviation, variance, z scores and percentile values all remain unchanged when your data set is shifted. Since every point in your data set moves the exact same distance, there is no change in their relations to each other.
## An Example of Shifting Data
Suppose that you were running a research project on the items people packed into USPS #4 Priority Mail Box. You collect your data and do your analysis, and then realize that although you’ve worked with the net weights only the contents of the boxes are significant to your research. Suppose your measurements are in ounces. Since each box weighs 5 ounces, k= -5.
• If your mean was 41 before the shift, it is now 36.
• If your median was 28, it is now 23.
• If your standard deviation was 16, it is still 16.
• Your variance will stay the same, as will your z score.
## Shifting Data and Rescaling
There are some cases when you may have to do both a data shift and a rescale—for instance, when you need to change temperature readings from Fahrenheit to Celsius degrees, where the temperature in Fahrenheit, F = [(9/5)*C]+32.
In this case, first shift your data by k= -32, and apply this additive constant to your mean and median. The standard deviation and variance will remain unchanged for this step. Then rescale, and multiply your mean and standard deviation by the rescaling constant 5/9 to find the mean and standard deviation for your data set in Celsius.
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Shifting Data was last modified: March 17th, 2018 by | 566 | 2,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-17 | longest | en | 0.908284 |
https://community.failbettergames.com/t/on-the-subject-of-money/17056 | 1,657,069,826,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00280.warc.gz | 222,878,666 | 10,342 | On the subject of money
I wonder… how much could an Echo be worth in modern pounds… could be hard… I myself for example assume, that the payment you get, is actually what is left AFTER living expenses like food, rent and such and actually represents your savings… so, I checked, to see how much a echo could be. A goldfish costs… 40 pence. A REAL goldfish, costs 4 pounds, according to E-bay. So, an echo could be… 10 pounds? Seems like a lot, especially given, that would make a hotel room cost 16000 pounds or 8000, potentially, more than a house, which costed something around 500 Pounds… are the masters trying to mix up my mind? Can somebody help me with counting the value of an echo? Any other ideas?
Well, I assume the hotel you’re referencing is the Royal Bethlehem, which costs so much because it is no ordinary hotel…
True… still, the worth of money gets confusing… makes dead rats cost the same as metal, which COULD probably be used for heating… perfect for a sauna, actually… not to forget WINE costs deffinitely more than four pounds, even if it is really awful one… that would make it be… 10 pence per bottle? Unlikely… So, we need more comparisons… and I agree to the Bethlehem. It is an EXTRAORDINARY one… so, prices like a 7-star hotel (if there is one)
Ok, the bethlehem seems quite cheap… given a 5 star in London costs 359 GPB… per night am I starting at the wrong corner? So, a reservation there costs either… 20 nights or a bit over a month… no… how else could I find the value of the Echo?
I think that it’d be a very hard conversion, in part because many items have more value in London than they do on the surface, regardless of currency exchange. On the surface, for instance, dead rats aren’t worth anything, let alone used as a currency substitute by disreputable sorts. If I were you, I’d find an object that seems to be valued about as much in Fallen London as it was in London on the surface - maybe one of the gemstones? - and work out the currency conversion with that.
The echoes don’t really seem to have a good conversion rate to pounds, this is even more insane if you compare Fallen Londont to Sunless sea, oh, and the developers confirmed Sunless Sea doesn’t take place in an alternate universe(except when you get endings, I would guess), so you can’t just use that as an argument since it was confirmed by Failbetter.
Either way, even if mushrooms were common I don’t think that buying a cane is the same as buying 500 bottles of even crappy wine, and I don’t think even exceptional drums are worth… the same amount as buying 2560 bottles of wine.
Let’s not forget that rats are a currency both sold and accepted by the Bazaar, and apparently it’s worth just as much as things like moon pearls…
Well rats in Fallen London can talk and stuff, so I would hope they’re more expensive.
They’re also one of the few non-mushroom sources of food that can be grown locally, which probably impacts their value a great deal. Keep in mind the entire economy was greatly upheaved, due to the mushroom agricultural economy being totally different and anything from the surface being a good deal harder to import.
I can get wine for \$2 that is surprisingly fit for human consumption, and high-end drum kits definitely run in the thousands. So a drum being worth 2,560 bottles of wine is well within the bounds of reason.
Not when they’re dead they don’t
I thought the dead rats (the cheap ones that are worth a penny each) were just ordinary rats, not the talking kind.
Nope! Not unless rattus faber can’t tell the difference. There’s a card in watchmakers hill where you can mourn a dead rat and all the other rats there act like it was one of them.
[quote=The Master]
Either way, even if mushrooms were common I don’t think that buying a cane is the same as buying 500 bottles of even crappy wine, and I don’t think even exceptional drums are worth… the same amount as buying 2560 bottles of wine.[/quote]
That would explain though why London has so many alcoholics and nonstop drinking parties.
Fallen London and the Neath in general have their own micro-economy with some rahter exotic goods for the normal sundweller. Some coveted trade goods are only of value there.
Which brings me to the question: Has London some important imports and exports? Do they have some common accepted currency for it? I can’t imagine paying some manufacturer or trade company for ship engine parts or caskets of french wine in horror stories (tales of terror) or screams of elder gods (aeolien screams).
I guess diamonds and spider silk as exports still work?
[quote=The Maltese Raven]Fallen London and the Neath in general have their own micro-economy with some rahter exotic goods for the normal sundweller. Some coveted trade goods are only of value there.
Which brings me to the question: Has London some important imports and exports? Do they have some common accepted currency for it? I can’t imagine paying some manufacturer or trade company for ship engine parts or caskets of french wine in horror stories (tales of terror) or screams of elder gods (aeolien screams).
I guess diamonds and spider silk as exports still work?[/quote]
If Sunless Sea is anything to go by, the trade in Darkdrop Coffee is incredibly lucrative.
[quote=The Maltese Raven]Fallen London and the Neath in general have their own micro-economy with some rahter exotic goods for the normal sundweller. Some coveted trade goods are only of value there.
Which brings me to the question: Has London some important imports and exports? Do they have some common accepted currency for it? I can’t imagine paying some manufacturer or trade company for ship engine parts or caskets of french wine in horror stories (tales of terror) or screams of elder gods (aeolien screams).
I guess diamonds and spider silk as exports still work?[/quote]
…echoes are the currency?
There is part of the problem. A great deal of trade down in the Neath is done by barter, or at least, shall we say, with alternative media of exchange. How many economic interactions in this town actually involve Echoes? Most of the time it’s jade or moon pearls or rostygold or what have you. You can’t even get a place with Echoes. I would go so far as to say that the Echo is not the currency of Fallen London; it is the currency of the Bazaar, and the Bazaar alone. London hasn’t had its own currency since the pound went the way of all Surface-dreams. This makes the situation yea more complicated, as the Bazaar treats all of London’s alternative currencies as simple commodities, and there’s a pretty stark difference between what they’ll give you for them and what you’ll pay. This scenario does explain why items bought exclusively at the Bazaar are often so out-of-line with prices in the city outside. If, through whatever labyrinthine restrictions the Bazaar imposes, Nassos is the only place licensed to sell you an Araby fighting-weasel or a cheery goldfish, then their prices will reflect the Bazaar’s own bizarre micro-economy, not that of London as a whole.
With all these complications, the best way to compare Echoes to pounds (ignoring the century’s difference in prices and commodities) would be a cost-of-living analysis, and the only thing I know of in London stable enough to do that with is probably housing. Converting the necessary items into Echoes based on their sale price gets the ranges of 2.0-3.0 E for terribly poor lodgings, 3.0-4.0 E for a just-out-of-town cottage, 4.2-4.5 E for a decent flat, and a spike to 50.0 E for a townhouse, which from the price I hope is somewhere in Mayfair (or whatever Mayfair is called down here). Converting based on purchase price at the Bazaar yields ranges of 6.0 E for terribly poor lodgings, 6.0-8.0 E for a just-out-of-town cottage, 9.0-12.6 E for a decent flat, and 100 E for our townhouse. (I had to figure the townhouse from honey, as the Bazaar sells romantic notions like the Brass Embassy sells souls.) There’s another massive price spike when you get to the top lodgings, but I don’t know that the Brass Embassy, the Bazaar, or the Royal Beth have any real equivalents in modern London with which to compare.
I realize this doesn’t give perfect grounds for comparison. For one thing, you have to choose which scale to use. For another, what does purchase in London mean, in comparative pricing? In effect it’s a perpetual lease, but I don’t know that everyone’s favorite bookseller is all that likely to lease in perpetuity up topside. If it’s outright sale, well, a few years back the Canadian embassy sold 1 Grosvenor Square for £306m. Is an Echo really worth £3m in modern pounds? Even excluding such an upscale town address, it’s probably fair to say that either lodging prices should be compared to a period lease, or else the Neathy real estate market has bottomed out.
edited by Siankan on 1/31/2017
[quote=Siankan]it’s probably fair to say that either lodging prices should be compared to a period lease, or else the Neathy real estate market has bottomed out.
edited by Siankan on 1/31/2017[/quote]
Real estate values have probably seen a steady deterioration over the past few decades as London’s population has apparently been falling since, well, the Fall? I know there are quite a few abandoned buildings that have just been left to the sorrow spiders and the Contrarian mentions this population loss in his debates with the Photographer?
Is it not canon that prices fluctuate all the time according to the whims of the bazaar? So the price of an echo would just be whatever the bazaar decides it is?
Even if that were canon, that’s not canon. The value of an echo is what it will buy, but it’s also what you can get one for. If echoes really have no stable values- which i kind of doubt- then the question is still a valid question, just in the vein of stocks in google, or something similar.
More likely, it just means prices move more frantically than in normal london; like the difference between a stock market and a grocery store.
I may be mistaken, though.
edited by Grenem on 2/1/2017
I always assumed it was kind of a double edged sword. In the neath, a lot of things are in constant flux. Transportation is expensive and I’m not sure goldfish either thrive in or are native to the neath. On the flip side, I suspect that the old saw of "buy land, god’s not making any more of it" doesn’t apply quite as much down here; london may be huge and sprawling, but it doesn’t have the constant pulse of new immigrants that it did on the surface, or not in quantity. In addition, while land is often unkind or even hostile, there’s more of it unclaimed than you’d think.
So, I’ve been assuming land is cheaper and accomodations are what sells a building. Keep in mind that the brass embassy comes with free heating, the bazaar probably has some features I can’t explain besides "central", and the beth has prestige. I suspect that the majority of the price comes from the facilities.
A [unknown time period] labor for a man or clay man goes for around 13.50 echoes, though? If you want to estimate the cost of living, guess what the time period for strong backed labor is, and then you have a reasonable daily income. In addition, a skilled doctor makes 70 echoes a [time period] by doing the job, where that’s your guess as to how long the TTH represents.
I suspect food is overpriced, and many other things are underpriced; the markets have been thrown into flux. I mean, rubies are 0.12 pence down here; that might align with your estimate our world now, but I don’t believe rubies have always been that cheap.
edited by Grenem on 2/1/2017
That’s certainly cheaper than rubies go now, even synthetic, even if they’re very small. Natural rubies of any size are extraordinarily expensive, and in larger sizes they pass up diamonds.
Of course, the rubies up here don’t kill you if you lick them, so that too must be taken into account. | 2,736 | 11,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.952102 |
https://stats.stackexchange.com/questions/588078/what-does-1-unit-change-mean-when-centering-predictors-in-regression | 1,716,460,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00110.warc.gz | 464,606,513 | 42,816 | # What does "1-unit change" mean when centering predictors in regression?
I know that when we standardize a predictor, "one unit change" becomes one standard deviation in the predictor, but what if we only center the data on the mean (i.e. only subtract all values by the mean) ?
• edit below: bounty
This example is from Winter (2019: 139):
mutate(SER_c = SER - mean(SER, na.rm = T)
term estimate
intercept 0.66
SER_c 0.11
POSVerb 0.72
SER_c:POSVerb 0.50
Obs: SER is a continuous predictor and POSVerb is a categorical predictor with two levels (0: Noun, 1: Verb)
The author concludes the following: "POSVerb is the difference between nouns and verbs for words with average SER, rather than some arbitrary 0". All right, that makes sense. The intercept should be, then, the value of the outcome variable for the mean of SER. Ok.
But, still, what does one-unit change in SER mean? If it wasn't centered, I'd say "one unit change in SER implies + 0.11 points to the outcome variable for the noun category", but I now that the intercept means the value at SER mean, how do I interpret this?
this post was helpful, but not enough. Thanks in advance!
• edit : bounty:
Hello, again. Now I have a follow-up question:
• 1: My model has a 2-level categorical predictor and a continuous one centered around its mean (no interaction). Should I say "one-unit change in the continuons predictor" or "one-unit change in the mean of the continous predictor" adds X to the mean of each categorical level?
• Centering alone does not change the interpretation of the non-intercept parameters. Sep 7, 2022 at 19:15
• I am confused by your provided notation: since POSVerb does not appear in the command mutate(SER_c = SER - mean(SER, na.rm = T): how can an estimate of its effect on the dependent variable be in the output? Also why is there no effect of SER on the dependent variable in the output? Sep 7, 2022 at 19:21
• @Alexis , the model is: lm(OutComeVariable ~ SER * POS). SER is continuos and POS is a categorical with 2 predictors which are dummy-coded: 1: Noun, 0: Verb), the mutate function was how the author centered SER into SER_c , is it clearer now? Sep 7, 2022 at 19:44
• In a science experiment you could measure temperature in degrees Kelvin, or recenter those to degrees Centigrade, or even recenter them to a difference relative to any reference temperature you like. Would that change the meaning or magnitude of one degree?
– whuber
Sep 7, 2022 at 20:49
• @LarissaCury I think editing a question to add a second part, awarding a bounty, after accepting an answer is gauche. You should probably ask a different question. Dec 14, 2022 at 19:28
As mentioned in the article that you linked, centering a feature does not impact its scale, so it should not change the interpretation of your model coefficients.
For example, consider the following linear model, where $$x_1$$ is continuous and $$\delta_1$$ is a dummy variable {0, 1}:
$$𝑦=B_0 + B_1x_1 + B_2\delta_1 + B_3 x_1 \delta_1 + u$$ In the equation above, the partial effect of $$x_1$$ on $$y$$ is
$$\frac{\Delta y}{\Delta x_1} = B_1 + B_3\delta_1$$ So, one unit of change in $$x_1$$ leads to an increase of $$B1+B3$$ in $$y$$ if $$d_1 = 1$$ and an increase of $$B_1$$ otherwise.
Suppose that we now center $$x_1$$ around its mean $$\bar{x_1}$$, let's call it $$\hat{x}_1$$, with the following operation:
$$\hat{x_1} = x_1 - \bar{x_1}$$
Notice that we can replace $$x_1$$ in the initial model by $$\hat{x_1} + \bar{x_1}$$, such that $$𝑦=B_0 + B_1(\hat{x_1} + \bar{x_1}) + B_2\delta_1 + B_3 (\hat{x_1} + \bar{x_1}) \delta_1 + u$$ Which you can expand to
$$𝑦=B_0 + B_1\hat{x_1} + B_1\bar{x_1} + B_2\delta_1 + B_3\hat{x_1}\delta_1 + B_3\bar{x_1}\delta_1 + u$$ So now, the partial effect of $$\hat{x_1}$$ on $$y$$ is:
$$\frac{\Delta y}{\Delta \hat{x_1}} = B_1 + B_3\delta_1$$
Which is the same as before. In your particular case:
$$\frac{\Delta y}{\Delta SER_c} = \frac{\Delta y}{\Delta SER} = 0.11 + 0.50 POSVerb$$
In the case of noun ($$POSVerb = 0$$) one unit change of $$SER$$ or $$SER_c$$ yields a 0.11 increase to the outcome. In the case of a verb ($$POSVerb = 1$$), then one unit of change in $$SER$$ or $$SER_c$$ yields a 0.11 + 0.50 = 0.61 increase to the outcome.
Hope this helps!
• Can you expand your answer to address the interaction term in Larissa Cury's question? Sep 7, 2022 at 19:46
• Thanks for your comment @Alexis! I would love to elaborate, but I wasn't able to track down the source. Googling for Winter 2019 didn't bring anything up. From what I can understand of Larissa's problem, I'd say the statement "one unit change in SER implies +0.11 points to the outcome variable for the noun category" is still valid, but having the full source would probably help provide a richer interpretation. Sep 7, 2022 at 19:58
• Ah, yes I get your point now @Alexis, thanks. Let me try to adapt the answer to include interaction effects. Sep 7, 2022 at 20:37
• @Alexis, here's an expanded answer with interaction terms. Hopefully, I didn't make any mistakes, but I would appreciate a quick review. Thanks for the feedback! Sep 7, 2022 at 21:13
• Thanks @Alexis! @LarissaCury I think the best way to intuit the full model is to see as two separate models: 1 model for nouns and 1 model for verbs. The model for nouns is y = 0.66 + 0.11 SER_c = 0.66 + 0.11 (SER - mean(SER)) The model for verbs is y = 1.38 + 0.61 SER_c = 1.38 + 0.61 (SER - mean(SER)) So, when SER = mean(SER), y = 0.66 for nouns, and y = 1.38 for verbs. Then any changes to SER will result in changes of y around mean(SER). Sep 7, 2022 at 23:13
I know that when we standardize a predictor, "one unit change" becomes one standard deviation in the predictor, but what if we only center the data on the mean (i.e. only subtract all values by the mean) ?
If you don't standardize then one unit change will relate to whatever the units are that are used to measure the predictor variable.
For instance if the predictor is temperature measured in Kelvin, then on unit change means an increase by one Kelvin.
In your example you have 'sensory experience ratings' as predictor and one-unit change will relate to whatever the scale is used to obtain those ratings.
Possibly your sensory experience rating is the one from
Juhasz, B.J., Yap, M.J. Sensory experience ratings for over 5,000 mono- and disyllabic words. Behav Res 45, 160–168 (2013). https://doi.org/10.3758/s13428-012-0242-9
Please rate each word on a 1 to 7 scale, with 1 meaning the word evokes no sensory experience for you, 4 meaning the word evokes a moderate sensory experience, and 7 meaning the word evokes a strong sensory experience.
In that case the ratings are based on averages of questionnaires with a 7 point scale.
The meaning of one-unit is the change of the score on that questionnaire by one.
• hi, @Sextux Empiricus, thanks! but how could we expand that to non-Likert-scale results, tho I'm also a linguistic as Winter, I'm from a completly different area, I've never worked with this 'iconicity' measure. I usially work with language exams and proficiency, for example. Right now I have proficiency as a predictor (exam from 0 to 120) and I've centered that. I have SCORE ~ Proficiency_c * type of exam (A or B) Dec 19, 2022 at 21:10 | 2,093 | 7,281 | {"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": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.898683 |
https://stevehacks.com/50c-fahrenheit | 1,708,826,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00340.warc.gz | 551,905,623 | 13,815 | # 50C FAHRENHEIT RECIPES
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Aug 10, 2021 · To get the most accurate temperature, deduct 32 from the Fahrenheit temperature listed in the recipe, and you will get 48. Then, multiply 48 by 5 and divide it by 9. Now, 80 Fahrenheit to Celsius is 26.666. As temperature dials are usually divided into 10s, you can round that up to the nearest tenth and set your oven temperature to 30 degrees C ...
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Using the Celsius to Fahrenheit formula: Fahrenheit (°F) = (Celsius x 1.8) + 32, this example shows how to convert a temperature of 50 degrees Celsius to Fahrenheit (50 C to F). Worked Example How To Convert 50 °C to °F
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35 Party Appetizer Recipes Your Guests Will Love Whether you're looking for classic bite-sized party appetizers like bruschetta, deviled eggs, and cocktail meatballs to gather with family and friends, or need ideas for festive finger food or holiday platters that you can make ahead of time, we've got all the inspiration you'll need in this special collection of our best party appetizer recipes.
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50c to f in Canada, the UK, and several other European countries, temperatures are measured in degrees Celsius (° C).. Luckily, it’s easy to convert between the 50c to f celsius to fahrenheit when needed. Just plug the known temperature into the appropriate equation to convert it like 50 c to f.. 50Celsius to Fahrenheit is the conversion of temperature from one unit Celsius to one ...
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##### 50 FAHRENHEIT TO CELSIUS. CONVERT 50 FAHRENHEIT TO CELSIUS
What is the formula to convert from 50 °F to °C? Fahrenheit to Celsius formula: [°F] = ( [°C]-32) × 5/9. The final formula to convert 50 Fahrenheit to Celsius is: [°F] = ( 50 -32) ×5/9 = 10. The Fahrenheit scale, although very popular in the US has an intriguing history and varied worldwide adoption scenarios.
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Fahrenheit is a scale commonly used to measure temperatures in the United States. Celsius, or centigrade, is used to measure temperatures in most of the world. Water freezes at 0° Celsius and boils at 100° Celsius. Inverse Conversion. Conversion Formula °Celsius = 5 * (°Fahrenheit - 32) 9: Nearby Values
From calculateme.com
##### TEMPERATURE CONVERSION CALCULATOR
For example, to convert 50 degrees Celsius (centigrade) to Fahrenheit, we plug our numbers into the formula as shown below: F = C * 9/5 + 32. F = 50 * 9/5 + 32 F = 90 + 32 F = 122 50 degrees Celsius is equal to 122 degrees Fahrenheit. Formulas for Converting Temperature Scales:
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##### SAMSUNG OVEN DEHYDRATOR RECIPES
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Check it out » | 2,263 | 8,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-10 | latest | en | 0.802147 |
https://syllabus.unict.it/insegnamento.php?id=12598&eng | 1,726,146,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00574.warc.gz | 513,404,549 | 3,359 | # APPLIED PHYSICS
FIS/07 - 6 CFU - 1° Semester
PAOLO MUSUMECI
## Learning Objectives
The purpose of the Applied Physics course is to teach basic physics and mathematics with simple applications to biomedical problems. The homogeneity of the preparation for all students is to be obtained in view of the specific knowledge required in the continuation of the Degree Course. In particular, the student must acquire knowledge of some basic laws and physical techniques for the understanding of physiological, biological and medical processes and will have to learn basic concepts useful for the correct use of the instrumentation used in the professional field.
## Detailed Course Content
PHYSICS
Physical quantities and their measurement - Physical quantities, units and measurement systems, dimensional equations. Measurement tools. Systematic errors and random errors. Average and standard deviation. Functional relationships and graphical representations. Scalar and vector quantities. Operations between vectors. Recalls of mechanics and notions of Biomechanics - Kinematics. Circular motion and harmonic motion. Amount of motion. Principles of dynamics. Work. Power. Power and efficiency. Moment. Static. Elasticity. Physiological stasis. Bone fractures (generalities). Recalls on fluids and applications in biological systems - Density. Viscosity. Hydrostatic pressure. Static of fluids. Stevino's law. Pascal's principle. Principle of Archimedes. Drip. Transfusion. Withdrawal. Drainage. Dynamics of ideal liquids. Bernoulli's theorem. Aneurysm and stenosis. Real liquids. Poiseuille report. Hydraulic resistance and Reynolds number, Sphygmomanometry. Thermometry and thermoregulation - temperature and heat. Temperature measurement. Thermometric scales. Clinical thermometers. Equivalence principle. Specific heat. Thermal balance. State steps. Heat transmission. Energy balance in the human body. Basal metabolic power. The electric and bioelectric phenomena - charges and electric fields. Capacitance and capacitors. Electric current. Laws of Ohm. Elementary circuits. Joule effect. RC circuits. Pacemaker. Defibrillator. Risks related to the use of electricity. Waves and radiations - Wave phenomena. Period and frequency. Amplitude and energy. Mechanical waves. The sound. Intensity of sound. Sound pressure and decibels. Stethoscope. Ultrasound in medicine. Electromagnetic waves. The electromagnetic spectrum. Eye and color vision. Radiation in diagnostics and in therapy. X-ray diagnostics. Radioisotopes and nuclear medicine. Radiotherapy. Biological effects of ionizing radiation. Overview of dosimetry and radioprotection.
## Textbook Information
FISICA
D. Scannicchio - Fisica Biomedica - EdiSES, Napoli 2013
Open in PDF format Versione in italiano | 578 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.867121 |
https://www.techylib.com/fr/view/libyantawdry/evolutionary_computation_model-based_generate_test | 1,553,204,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.7/warc/CC-MAIN-20190321213516-20190321235516-00421.warc.gz | 924,551,985 | 14,746 | # Evolutionary Computation: Model-based Generate & Test
IA et Robotique
23 oct. 2013 (il y a 5 années et 5 mois)
233 vue(s)
23 October, 2013
2
Evolutionary Computation:
Model
-
based Generate & Test
Model
(To Evolve)
Candidate
Solution
Observed
Performance
Test
Feedback
(To update model)
A
Candidate Solution
could be a
vector of
variables
, or a
tree
A
Model
could be a
population
of solutions,
or a
probability model
The
Fitness
of a solution
is application
-
dependent,
e.g. drug testing
Generate:
select
, create/mutate
vectors
/
trees
23 October, 2013
3
Motivation
Idea from natural selection
To contain
combinatorial explosion
E.g. the travelling salesman problem
To evolve quality solutions
E.g. to find hardware configurations
23 October, 2013
4
Terminology in GA
Example of a candidate solution
in binary representation (vs
real coding
)
A
population
is maintained
1
0
0
0
1
1
0
Chromosome
with
Genes
with
alleles
String with
Building blocks
with values
fitness
evaluation
23 October, 2013
5
Example Problem For GA
maximize f(
x
) = 100 + 28
x
x
2
optimal solution: x = 14
(f(x) = 296)
Use 5 bits representation
e.g. binary 01101 = decimal 13
f(
x
) = 295
Note: representation issue can be tricky
Choice of representation is crucial
23 October, 2013
6
Example Initial Population
To maximize f(
x
) = 100 + 28
x
x
2
23 October, 2013
7
Selection in Evolutionary Computation
Roulette wheel
method
The fitter a string,
the more chance it
has to be selected
24%
10101
28%
01010
29%
01101
19%
11000
23 October, 2013
8
Crossover
1
0
1
0
1
0
1
0
1
0
1
0
0
1
0
0
1
1
0
1
18
13
280
295
F(
x
)
0
1
1
0
1
0
1
0
1
0
0
1
1
1
0
0
1
0
0
1
14
9
296
271
1,142
285.5
Sum of fitness:
Average fitness:
23 October, 2013
9
A Control Strategy For GA
Initialize a population
Repeat
Reproduction
Select strings randomly but reflecting fitness
Crossover
Mutation
Replace old population with offspring
Until termination condition
23 October, 2013
10
Discussion
New population has higher average fitness
Is this just luck?
Fundamental theorem
Is an offspring always legal?
Epistasis
A fundamental question: can candidate
solutions be represented by string?
23 October, 2013
11
Simple to program
General
--
wide range of applicability
Efficient
--
finding solutions fast
Effective
--
finding good solutions
Robust
--
finding solutions consistently
Are these claims justified?
Different people have different opinions
Epistasis
(interaction between Genes)
Constraints
Penalties, Repair
Guided Genetic Algorithm
23 October, 2013
23
Constraints
Task: find optimal solution in constrained
satisfaction
Difficulties:
Before one can attempt to optimization, one
needs to find legal solutions
Crossover may generate illegal solutions
Sometimes satisfying constraints alone is hard
When problem is tightly constrained
23 October, 2013
24
Epistasis, Example in TSP
Travelling
Salesman Problem
After crossover,
offspring may not
be legal tours
some cities may be
visited twice,
others missing
1
3
6
8
4
5
2
7
6
8
5
1
4
2
7
3
1
3
6
8
4
6
8
5
1
4
2
7
3
5
2
7
crossover
23 October, 2013
25
Penalty Method
If a string violates certain constraints
Then a
penalty
is deducted from the fitness
E.g. in TSP, if penalties are high, GA may
attempt to satisfy constraints before finding
short tours
Problem with tightly constrained problems:
most strings are illegal
23 October, 2013
26
Repair Method
If a string violates certain constraints
Then attempt to
repair
it
E.g. in TSP, replace duplicated cities with
missing cities
possibly with complete search or heuristics
Make sure that a population only contains
legal candidate solutions
One can then focus on optimization
GA for Machine Learning
Classifiers
23 October, 2013
31
Production System Architecture
Working Memory
Production Rules
Scheduler
Retrieve data
Change data
(
firing
)
Conditions
Actions
Conditions
Actions
Conditions
Actions
…..
Facts
Sensor inputs
Internal states
…..
23 October, 2013
32
Classifier System Components
Classifiers
: Condition
-
Action rules
Special type of
production system
A Credit System
Allocating rewards to fired rules
Genetic Algorithm
for evolving classifiers
23 October, 2013
33
Classifier System Example
Message List
0111
0000
1100
Classifiers
01##:0000
00#0:1100
….
Detectors
0
1
1
1
Effectors
1
1
0
0
Info
Payoff
Action
Classifiers bid to fire
Classifiers have
fixed length
conditions and
actions
23 October, 2013
34
Apportionment of Credit
The set of classifiers work as one system
They react to the environment
The system as a whole gets feedback
How much to credit each classifier?
E.g.
Each classifier
i
has a strength
S
i
which form the basis for credit apportionment
as well as fitness for evolution
23 October, 2013
35
A classifier may make a bid to fire
B
i
=
S
i
*
C
bid
where
C
bid
is the bid coefficient
Effective bid:
EB
i
=
S
i
*
C
bid
+ N(
bid
)
where N(
bid
) is a noise function
with standard deviation
bid
Winner of an auction pays the bid value to
source of the activating message
23 October, 2013
36
Classifiers In Action
C
bid
= 0.1
C
tax
= 0
S
3
220
218
180
162
20
Msg
1000
0001
B
3
16
S
4
220
218
196
146
20
Msg
0001
B
4
R
4
50
Classifiers
01##:0000
00#0:1100
11##:1000
##00:0001
Environment
S
0
200
200
200
200
0
Msg
0111
B
0
20
S
1
180
200
200
200
20
Msg
0000
B
1
20
20
S
2
220
180
200
180
20
Msg
1100
0001
B
2
20
18
S
5
220
218
196
196
20
23 October, 2013
37
Each classifier is “taxed”
S
i
(t+1) = S
i
(t)
-
C
bid
S
i
(t)
-
C
tax
S
i
(t) + R
i
(t)
where S
i
(t) is strength of classifier
i
at time
t
C
bid
S
i
(t) is the bid accepted
C
tax
is
tax
R
i
(t)
is payoff
For
stability
, the
bid value
should be
comparable to
receipts
from environment
23 October, 2013
38
Classifiers: Genetic Algorithms
T
ga
= # of steps between GA calls
GA called once every T
ga
cycles; or
GA called with probability reflecting T
ga
A proportion of the population is replaced
Selection: roulette wheel
weighted by strength S
i
Mutation: 0
{1,#}, 1
{0,#}, #
{0,1}
Genetic Programming
Building decision trees
GP for Machine Learning
23 October, 2013
40
Genetic Programming, Overview
Young field
Koza: Genetic Programming, 1992
Langdon & Poli: Foundations of GP, 2001
Diverse definitions
Must use trees? May use lists?
Must one evolve programs?
Suitable for LISP
Machine learning: evolving trees
dynamic data structure
23 October, 2013
41
Terminals and Functions
Terminal set:
Inputs to the program
Constants
Function set:
Statements, e.g. IF
-
THEN
-
ELSE, WHILE
-
DO
Functions, e.g. AND, OR, +, :=
Arity sensitive
23 October, 2013
42
Functions: Statements (e.g. IF
-
THEN
-
ELSE) or functions (e.g. AND, OR, +,
)
Terminals: Input t the prgram r cnstants
GP: Example Tree (1)
Last race time
Won last time
If
-
then
-
else
Not
-
win
Win
If
-
then
-
else
Win
5 min
<
23 October, 2013
43
GP Application
A tree can be anything, e.g.:
a program
a decision tree
Choice of terminals and functions is crucial
Domain knowledge helps
Larger grammar
larger search space
harder to search
23 October, 2013
44
GP: Example Tree (2)
Win
Not
-
win
Win
Last raced 3 months ago
Same Jockey
Not
-
win
Win
Won last time
Same Stable
Boolean decisions only
(limited functions)
Terminals: input to the program or constants
Functions: Statements (e.g. IF
-
THEN
-
ELSE) or functions (e.g. AND, OR, +,
)
23 October, 2013
45
GP Operators
1
4
3
2
7
6
5
9
8
a
d
c
b
g
f
e
i
h
1
4
3
2
7
6
5
9
8
a
d
c
b
g
f
e
i
h
Crossover
Mutation
:
change a branch
23 October, 2013
46
Fitness in GP
Generating Programs
How well does the program meet the
specification?
Machine Learning
:
How well can the tree predict the outcome?
Function Fitting
:
How great/small the error is
23 October, 2013
47
Generational GP Algorithm
Initialize population
Evaluate individuals
Repeat
Repeat
Select parents, crossover, mutation
Until enough offspring have been generated
Until termination condition fulfilled
23 October, 2013
48
-
state GP Algorithm
Initialize population P
Repeat
Pick random subset of P for tournament
Select winners in tournament
Crossover on winners, mutation
Replace loser(s) with new offspring in P
Until termination condition fulfilled
Estimation of Distribution Algorithms
(EDAs)
Population
-
based Incremental Learning
(PBIL)
Building Bayesian networks
23 October, 2013
51
Population
-
based Incremental Learning
(PBIL)
Statistical approach
Related to ant
-
colonies, GA
Model M:
x
= v1 (0.5)
x
= v2 (0.5)
y
= v3 (0.5)
y
= v4 (0.5)
Sample from M
solution X, eg
<x,v1><y,v4>
Evaluation X
Modify the probabilities
0.6
0.4
0.6
0.4 | 2,716 | 9,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.688144 |
https://scioly.org/forums/viewtopic.php?f=290&t=13840&view=print | 1,590,709,895,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00592.warc.gz | 536,935,780 | 2,181 | Page 1 of 1
### Boomilever height
Posted: February 23rd, 2019, 6:32 pm
So I'm new to science olympiad this year and I was put in Boomilever. For division c I'm getting that the maximum height may only be 15 cm, but I'm not sure if that's the ideal height to build at because if it bends a little below the 15 cm line you will be done testing right? So basically I wanted to ask if I should make my build shorter than 15 cm and by how much?
### Re: Boomilever height
Posted: February 23rd, 2019, 6:43 pm
nyu12345 wrote:So I'm new to science olympiad this year and I was put in Boomilever. For division c I'm getting that the maximum height may only be 15 cm, but I'm not sure if that's the ideal height to build at because if it bends a little below the 15 cm line you will be done testing right? So basically I wanted to ask if I should make my build shorter than 15 cm and by how much?
As long as when you begin loading your boomi is above the 15cm line it should be fine. Once you start to load, there will be lots of friction from the wall onto your boomilever and there is a very small chance that your boomi's contact point with the wall will change.
### Re: Boomilever height
Posted: February 23rd, 2019, 7:05 pm
Carrot wrote:
nyu12345 wrote:So I'm new to science olympiad this year and I was put in Boomilever. For division c I'm getting that the maximum height may only be 15 cm, but I'm not sure if that's the ideal height to build at because if it bends a little below the 15 cm line you will be done testing right? So basically I wanted to ask if I should make my build shorter than 15 cm and by how much?
As long as when you begin loading your boomi is above the 15cm line it should be fine. Once you start to load, there will be lots of friction from the wall onto your boomilever and there is a very small chance that your boomi's contact point with the wall will change.
Oh ok, so its not like when for example the loading part of your boomilever tips below the contact point? Its only the part thats touching the wall
### Re: Boomilever height
Posted: February 23rd, 2019, 7:07 pm
nyu12345 wrote:
Carrot wrote:
nyu12345 wrote:So I'm new to science olympiad this year and I was put in Boomilever. For division c I'm getting that the maximum height may only be 15 cm, but I'm not sure if that's the ideal height to build at because if it bends a little below the 15 cm line you will be done testing right? So basically I wanted to ask if I should make my build shorter than 15 cm and by how much?
As long as when you begin loading your boomi is above the 15cm line it should be fine. Once you start to load, there will be lots of friction from the wall onto your boomilever and there is a very small chance that your boomi's contact point with the wall will change.
Oh ok, so its not like when for example the loading part of your boomilever tips below the contact point? Its only the part thats touching the wall
Yes, that is correct. As long as the point of contact with the wall is above the 15cm line you are good.
### Re: Boomilever height
Posted: February 23rd, 2019, 7:26 pm
Carrot wrote:
nyu12345 wrote:
Carrot wrote:
As long as when you begin loading your boomi is above the 15cm line it should be fine. Once you start to load, there will be lots of friction from the wall onto your boomilever and there is a very small chance that your boomi's contact point with the wall will change.
Oh ok, so its not like when for example the loading part of your boomilever tips below the contact point? Its only the part thats touching the wall
Yes, that is correct. As long as the point of contact with the wall is above the 15cm line you are good.
Thanks
### Re: Boomilever height
Posted: February 23rd, 2019, 7:33 pm
nyu12345 wrote:
Carrot wrote:
nyu12345 wrote:
Oh ok, so its not like when for example the loading part of your boomilever tips below the contact point? Its only the part thats touching the wall
Yes, that is correct. As long as the point of contact with the wall is above the 15cm line you are good.
Thanks
Just remember about the physics in play here so keep your boom as close to 15cm as possible. (Btw that's good )
### Re: Boomilever height
Posted: February 23rd, 2019, 8:11 pm
TheChiScientist wrote:
nyu12345 wrote:
Carrot wrote:
Yes, that is correct. As long as the point of contact with the wall is above the 15cm line you are good.
Thanks
Just remember about the physics in play here so keep your boom as close to 15cm as possible. (Btw that's good )
I'll try my best to. Thanks | 1,185 | 4,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-24 | latest | en | 0.948 |
https://www.mbatious.com/topic/942/cat-question-bank-cubes-and-cuboids/23 | 1,632,669,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00510.warc.gz | 899,453,414 | 24,988 | # CAT Question Bank - Cubes and Cuboids
• Q13) In how many different ways can a cube be painted if each face has to be painted either red or blue?
• Q14) A cube has been cut into 150 smaller cubes of equal volume. What is the minimum number of straight cuts required to do this?
• Q15) Two opposite faces of a cube are painted in blue, another pair of opposite faces are painted green and the remaining faces are painted in red. The cube is now cut into 210 smaller but identical pieces using minimum possible number of cuts. How many smaller pieces have exactly two colours on them?
• Q16) A solid cube of side 1 cm is cut into two identical solid parts by a plane surface. The total surface area of both the solid parts is maximum possible. Two points are then selected on one of these solid parts. What is the maximum possible distance between these two points
• Q17) What is the least possible number of cuts required to divide a cube into 34 identical pieces ?
• Q18) A big cube is formed of 125 smaller cubes of which m are such that all their faces are coloured. If 60% of the area of the big cube is coloured, what is the maximum possible value of m
• Q19) How many distinct cuboids can be formed using 64 cubes of area 1cm^2
• Q20) A cube of dimension 4 cm × 4 cm × 4 cm is painted red on all six faces. Now this cube is cut to form 1 cm × 1 cm × 1 cm identical cubes. What is the ratio of total area of painted surfaces to the total area of unpainted surfaces
• Q21) A cube is painted red and all its faces and is then divided into 343 equal cubes.
1. How many cubes will have three faces painted ?
2. How may cubes will have two faces painted?
3. How many cubes will have one face painted?
4. How many cubes have no face painted ?
5. How many cubes will have at least one face painted?
• Q22) A cuboid of size 5.2 m, 13 m & 39 m is completely cut into n smaller identical cubes. If n has the minimum possible value, what is the total surface area of all the small cubes?
• Q23) The areas of three mutually adjacent faces of a cuboid C1 were found to be 200 sq. cm, 100 sq. cm and 50 sq. cm. Another cuboid C2 has the same volume as C1. Find the minimum value of the sum of the length, the breadth and the height of C2 (in cm).
a) 15
b) 30
c) 40
d) 50
• Q24) There is a rectangular wooden block of length 4 cm, height 3 cm and breadth 3 cm. The two opposite surfaces of 4 cm x 3 cms are painted yellow on the outside. The other two opposite surfaces of 4 cm x 3 cm are painted red on the outside and the remaining two surfaces of 3 cm x 3 cm are painted green on the outside. Now, the block is cut in such a way that cubes of 1cm x 1cm x 1cm are created.
1. How many cubes will have only one colour?
2. How many cubes will have no colour?
3. How many cubes will have any two colours?
• Q25) A square has a perimeter of 108 cm. Four small squares are cut from each of its corners and the side of each square is x cm. The sheet remaining after the squares are cut is folded to form a cuboid. Find the value of x which maximizes the volume of the cuboid formed.
• Q26) A set of 120 small 1 × 1 × 1 cubes are arranged to form a 6 × 5 × 4 cuboid. How many cuboids (including the original) can be formed with vertices chosen from these points such that their faces are parallel to those of the original cuboid?
• Q27) How many right angle triangles can be obtained by joining the vertices of a cuboid?
• Q28) A painted cuboid of dimensions 5 x 6 x 7 is cut into unit cubes. What is the number of unit cubes that have 0 faces painted?
a) 50
b) 60
c) 40
d) 210
• Q29) If you have a 4 * 6 * 8 cuboid from which the largest possible cube is cut out, what would be the minimum number of cubes into which the remaining figure could be divided assuming that there is no piece left over and the all the smaller cubes are of equal dimension
• Q30) A wooden cube of n unit on a side is painted red on all six faces and then cut into n^3 unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is n ?
• @rowdy-rathore can u plz tell d ans to this?
• @Rowdy-Rathore I am not able to find the answer key
61
61
47
145
65
67
106
47 | 1,097 | 4,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.946524 |
https://thevirtualinfo.com/calculating-derivatives-in-pytorch/ | 1,685,792,097,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00629.warc.gz | 595,805,490 | 29,507 | # Calculating Derivatives in PyTorch
0
10
Final Up to date on November 15, 2022
Derivatives are probably the most elementary ideas in calculus. They describe how adjustments within the variable inputs have an effect on the perform outputs. The target of this text is to supply a high-level introduction to calculating derivatives in PyTorch for many who are new to the framework. PyTorch gives a handy technique to calculate derivatives for user-defined capabilities.
Whereas we all the time need to cope with backpropagation (an algorithm identified to be the spine of a neural community) in neural networks, which optimizes the parameters to attenuate the error so as to obtain increased classification accuracy; ideas discovered on this article will likely be utilized in later posts on deep studying for picture processing and different laptop imaginative and prescient issues.
After going via this tutorial, you’ll study:
• Easy methods to calculate derivatives in PyTorch.
• Easy methods to use autograd in PyTorch to carry out auto differentiation on tensors.
• Concerning the computation graph that entails completely different nodes and leaves, permitting you to calculate the gradients in a easy attainable method (utilizing the chain rule).
• Easy methods to calculate partial derivatives in PyTorch.
• Easy methods to implement the by-product of capabilities with respect to a number of values.
Let’s get began.
Calculating Derivatives in PyTorch
Image by Jossuha Théophile. Some rights reserved.
The autograd – an auto differentiation module in PyTorch – is used to calculate the derivatives and optimize the parameters in neural networks. It’s meant primarily for gradient computations.
Earlier than we begin, let’s load up some crucial libraries we’ll use on this tutorial.
Now, let’s use a easy tensor and set the requires_grad parameter to true. This enables us to carry out automated differentiation and lets PyTorch consider the derivatives utilizing the given worth which, on this case, is 3.0.
We’ll use a easy equation \$y=3x^2\$ for instance and take the by-product with respect to variable x. So, let’s create one other tensor based on the given equation. Additionally, we’ll apply a neat methodology .backward on the variable y that varieties acyclic graph storing the computation historical past, and consider the outcome with .grad for the given worth.
As you possibly can see, we’ve obtained a worth of 18, which is appropriate.
## Computational Graph
PyTorch generates derivatives by constructing a backwards graph behind the scenes, whereas tensors and backwards capabilities are the graph’s nodes. In a graph, PyTorch computes the by-product of a tensor relying on whether or not it’s a leaf or not.
PyTorch is not going to consider a tensor’s by-product if its leaf attribute is ready to True. We received’t go into a lot element about how the backwards graph is created and utilized, as a result of the objective right here is to present you a high-level data of how PyTorch makes use of the graph to calculate derivatives.
So, let’s test how the tensors x and y look internally as soon as they’re created. For x:
and for y:
As you possibly can see, every tensor has been assigned with a specific set of attributes.
The information attribute shops the tensor’s information whereas the grad_fn attribute tells concerning the node within the graph. Likewise, the .grad attribute holds the results of the by-product. Now that you’ve got learnt some fundamentals concerning the autograd and computational graph in PyTorch, let’s take just a little extra sophisticated equation \$y=6x^2+2x+4\$ and calculate the by-product. The by-product of the equation is given by:
\$\$frac{dy}{dx} = 12x+2\$\$
Evaluating the by-product at \$x = 3\$,
\$\$left.frac{dy}{dx}rightvert_{x=3} = 12times 3+2 = 38\$\$
Now, let’s see how PyTorch does that,
The by-product of the equation is 38, which is appropriate.
## Implementing Partial Derivatives of Capabilities
PyTorch additionally permits us to calculate partial derivatives of capabilities. For instance, if we’ve to use partial derivation to the next perform,
\$\$f(u,v) = u^3+v^2+4uv\$\$
Its by-product with respect to \$u\$ is,
\$\$frac{partial f}{partial u} = 3u^2 + 4v\$\$
Equally, the by-product with respect to \$v\$ will likely be,
\$\$frac{partial f}{partial v} = 2v + 4u\$\$
Now, let’s do it the PyTorch approach, the place \$u = 3\$ and \$v = 4\$.
We’ll create u, v and f tensors and apply the .backward attribute on f so as to compute the by-product. Lastly, we’ll consider the by-product utilizing the .grad with respect to the values of u and v.
## Spinoff of Capabilities with A number of Values
What if we’ve a perform with a number of values and we have to calculate the by-product with respect to its a number of values? For this, we’ll make use of the sum attribute to (1) produce a scalar-valued perform, after which (2) take the by-product. That is how we are able to see the ‘perform vs. by-product’ plot:
Within the two plot() perform above, we extract the values from PyTorch tensors so we are able to visualize them. The .detach methodology doesn’t enable the graph to additional monitor the operations. This makes it simple for us to transform a tensor to a numpy array.
## Abstract
On this tutorial, you discovered tips on how to implement derivatives on varied capabilities in PyTorch.
Significantly, you discovered:
• Easy methods to calculate derivatives in PyTorch.
• Easy methods to use autograd in PyTorch to carry out auto differentiation on tensors.
• Concerning the computation graph that entails completely different nodes and leaves, permitting you to calculate the gradients in a easy attainable method (utilizing the chain rule).
• Easy methods to calculate partial derivatives in PyTorch.
• Easy methods to implement the by-product of capabilities with respect to a number of values. | 1,334 | 5,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-23 | latest | en | 0.874801 |
http://observatoriodederechoshumanosdelospueblos.org/multiplication-chart-to-1000/ | 1,547,705,555,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658702.9/warc/CC-MAIN-20190117041621-20190117063621-00151.warc.gz | 173,482,807 | 9,214 | # 70 Good Pics Of Multiplication Chart to 1000
70 Good Pics Of Multiplication Chart to 1000
Welcome to our blog, in this moment I will explain to you in relation to multiplication chart to 1000.
printable multiplication worksheets tons of printable worksheets that work on the math skill of multiplication these are very helpful for students and teachers free multiplication worksheets for second sixth grade free multiplication worksheets for second sixth grade great ready made multiplication worksheets for parents teachers and homeschoolers times table – 2 times table free printable worksheets content filed under the times table – 2 times table category multiplication – repeated addition free printable content filed under the multiplication – repeated addition category printable interactive 100 number chart worksheets color free interactive 100 chart worksheets children find number patterns and color code them with this interactive math game great way to reinforce skip multiplication number fact triangles by tim caird a set of smart iwb slides to support a lesson exploring multiplication facts 2x 3x 4x 5x 6x 10x using number fact triangles or number trios& multiplication worksheets this multiplication worksheet may be configured for 2 3 or 4 digit multiplicands being multiplied by multiples of ten you may vary the numbers of printable interactive blank 100 number chart worksheets free interactive blank 100 chart worksheets children find number patterns and color code them with this interactive math game great way to reinforce skip rounding code breaker to nearest 10 100 1000 by this pp starts by talking about napier s bones and how you can use them and links this to the lattice or chinese method for multiplication nthe decimal place value chart math ly math decimal place value chart are discussed here the first place after the decimal is got by dividing the number by 10 it is called the tenths place
1000 images about Math on Pinterest from multiplication chart to 1000 , source:www.pinterest.com
1000 images about Kids on Pinterest from multiplication chart to 1000 , source:www.pinterest.com | 404 | 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} | 2.546875 | 3 | CC-MAIN-2019-04 | latest | en | 0.820399 |
https://stats.stackexchange.com/questions/288034/fitting-tensor-product-p-splines-penalty-parameters | 1,713,243,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00032.warc.gz | 498,682,679 | 40,480 | Fitting Tensor Product P-splines, Penalty Parameters
I am working with the mgcv package in r and I am fitting tensor product P-splines.
test1<-function(x,z,sx=0.3,sz=0.4){ x<-x*20
(pi**sx*sz)*(1.2*exp(-(x-0.2)^2/sx^2-(z-0.3)^2/sz^2)+
0.8*exp(-(x-0.7)^2/sx^2-(z-0.8)^2/sz^2))
}
n<-500
x<-runif(n)/20;z<-runif(n);
xs<-seq(0,1,length=30)/20;zs<-seq(0,1,length=30)
pr<-data.frame(x=rep(xs,30),z=rep(zs,rep(30,30)))
truth<-matrix(test1(pr$x,pr$z),30,30)
f <- test1(x,z)
y <- f + rnorm(n)*0.2
b1<-gam(y~t2(x,z,bs=c("ps","ps")),drop.intercept=T)
I think that is the code for how to fit it with two variables and no intercepts.
My question is how do I find the P-spline smoothing parameters. I am under the assumption there should be two, one for each variable, and possibly one for the interaction between the two P-splines, but this should be the product of the first two. I could be wrong about this.
Anyway reading the documentation this should give me my smoothing parameters:
b1\$sp
# t2(x,z)rr t2(x,z)nr t2(x,z)rn
# 4.280221e-04 3.884871e+06 1.386065e+06
However, these are not products of eachother and I don't know how they relate to the x and z fitted p-spline.
The reason for me thinking this way is due to equation five in this paper.
Any help would be much appreciated.
edit
I have found these definitions for rr, nr, and rn
## label "rr" indicates interaction (range space times range space)
## label "nr" (null space for x0 times range space for x1) is main effect for x1.
## label "rn" is main effect for x0
x0 is x in our case and x1 is z.
Then according to my understanding nr*rn=rr should be the case, I wonder if my interpretation of equation 5 in the paper is wrong?
• Looking at "Straightforward intermediate rank tensor product smoothing in mixed models" page 348 bottom of left col, I read "if the smoothing parameter for the interaction penalty → ∞, the smooth tends to the additive model" which seems to imply that the smoothing parameter of the interaction term is not a product of the other two (otherwise they would have to go to infinity together). However, I am not sure this is referring to the t2 basis. Jun 30, 2017 at 12:51
• @MatteoFasiolo Thanks for the references. They should be helpful. I am including a link to the paper here. Jul 5, 2017 at 13:29 | 699 | 2,322 | {"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} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.846221 |
https://multi-converter.com/square-meters-to-square-hectometers | 1,675,340,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00394.warc.gz | 425,118,783 | 6,514 | # Square Meters to Square Hectometers
Convert m2 to hm2
Change to Square Hectometers to Square Meters
Share:
## How to convert Square Meters to Square Hectometers
1 [Square Meters] = 0.0001 [Square Hectometers]
[Square Hectometers] = [Square Meters] / 10000
To convert Square Meters to Square Hectometers divide Square Meters / 10000.
## Example
50 Square Meters to Square Hectometers
50 [m2] / 10000 = 0.005 [hm2]
## Conversion table
Square Meters Square Hectometers
0.01 m21.0E-6 hm2
0.1 m21.0E-5 hm2
1 m20.0001 hm2
2 m20.0002 hm2
3 m20.0003 hm2
4 m20.0004 hm2
5 m20.0005 hm2
10 m20.001 hm2
15 m20.0015 hm2
50 m20.005 hm2
100 m20.01 hm2
500 m20.05 hm2
1000 m20.1 hm2 | 261 | 675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | longest | en | 0.489104 |
http://alpha.math.uga.edu/~pete/Mazur-Rubin.html | 1,701,325,896,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00523.warc.gz | 2,239,123 | 3,266 | This fall, some UGA number theorists (students, postdocs, junior and senior faculty) are meeting to work through the recent preprint of B. Mazur and K. Rubin, Ranks of twists of elliptic curves and Hilbert's 10th problem. This paper proves some extremely striking results on 2-Selmer ranks of elliptic curves in families of quadratic twists, with (e.g.) the following applications:
Corollary 1.9: For every number field K, there is an elliptic curve E over K with E(K) = 0.
Theorem 1.11: Assuming BSD, for every Galois extension L/K of number fields of prime degree, there exists an elliptic curve E over K such that the rank of E over K and the rank of E over L are both equal to one.
This last result, when combined with work of B. Poonen and A. Shlapentokh, yields:
Theorem 1.12: Assuming BSD, then for every number field K, Hilbert's 10th problem over the ring of integers of K has a negative answer.
Better yet, the technology employed by Mazur and Rubin is "middlebrow" compared to what is needed for most other breakthrough papers in 21st century elliptic curve theory (e.g. other papers by the authors). We wish to capitalize on this opportunity to learn an important new result with a relatively low startup cost.
Mazur and Rubin make use of theorems from the following papers:
J.W.S. Cassels, Arithmetic on curves of genus 1. VIII: On conjectures of Birch and Swinnerton-Dyer
T. Dokchitser and V. Dokchitser, Elliptic curves with all quadratic twists of positive rank
F. Gouvea and B. Mazur, The Square-Free Sieve and the Rank of Elliptic Curves
K. Kramer, Arithmetic of elliptic curves upon quadratic base extension
B. Mazur, Rational Points of Abelian Varieties with Values in Towers of Number Fields
B. Mazur and K. Rubin, Finding large Selmer rank via an arithmetic theory of local constants
L. Merel, Bornes pour la torsion des courbes elliptiques sur les corps de nombres
J. Milne, Arithmetic Duality Theorems (big ups to Jim Milne for making this book available on his homepage!)
A. Wintner, On the prime number theorem
Other especially relevant citations:
S. Chang, On the arithmetic of twists of superelliptic curves
S. Chang, Quadratic twists of elliptic curves with small Selmer rank
D.R. Heath-Brown, The size of Selmer groups for the congruent number problem
K. Ono, Nonvanishing of quadratic twists of modular L-functions and applications to elliptic curve
K. Ono and C. Skinner, Nonvanishing of quadratic twists of modular L-functions
H.P.F. Swinnerton-Dyer, The effect of twisting on the 2-Selmer group
My MathSciNet review of Swinnerton-Dyer's paper: pdf
PROGRESS REPORT:
Week 0 (Friday, August 28th): Main speaker: Pete L. Clark
Discussion of some consequences of the main results -- Corollaries 1.8, 1.9 and 1.10. Definition of quadratic twists; invariance of rational 2-torsion under quadratic twists. Statement of Hilbert's 10th problem over Z and over other rings: positive results (Rumely) and negative results (Davis-Putnam-Robinson-Matiyasevich, Eisentrager, Shlapentokh, Poonen).
Week 1 (Friday, September 4th): Main speaker: Alex Rice
Review of Galois cohomology, the Kummer sequence of an isogeny, weak Mordell-Weil group, twist, principal homogeneous space, Weil-Chatelet group, Selmer group.
Week 2 (Friday, September 11th): Main speaker: Pete L. Clark
Further discussion of Section 1 of the paper. Introduction to the phenomenon of constant 2-Selmer parity (D&D).
Week 3 (Wednesday, September 23rd): Main speaker: Pete L. Clark
Beginning of Section 2: statement and proof of Cassels' Lemma on the image of the local Kummer map.
Week 4 (Friday, September 25th): Main speaker: Seyfi Turkelli (notes)
Review of root numbers of elliptic curves (assuming the conjectured analytic continuation and functional equation), connection with analytic rank. The D&D phenomenon.
Week 5 (Friday, October 2nd): Main Speaker: Nathan Walters
Week 6 (Friday, October 9th): Main Speaker: Jim Stankewicz
Week 7 (Friday, October 16th): Main Speaker: Jim Stankewicz
Week 8 (Friday, October 23rd): Main Speaker: Bob Rumely
Week 9 (Wednesday, October 28th): Main Speaker: Bob Rumely (notes) | 1,072 | 4,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.878505 |
https://ubraintv-jp.com/how-much-does-a-gram-weight/ | 1,660,348,975,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00410.warc.gz | 544,752,830 | 5,985 | The measurement of a gram is just one of the most widely offered measurements, but the variety of items the weigh precisely a gram is fairly short. If you jump back to your elementary college years, you might remember your teacher do the efforts to explain to you just how tiny a gram was, and also there was probably a little of a battle to identify a referral object the weighed a gram.
You are watching: How much does a gram weight
How plenty of things weigh a gram? This inquiry is quite challenging to answer since a gram is a really light load to measure, and also there are few things through the specific measurement. Through some research and also trial and also error, I was able to compile a perform of 10 items that weigh a gram.
When looking in ~ grams, we frequently think of measurement for medicine because of the tiny dosages. However, there space some concrete objects that sweet a mere gram.
## Ten Objects that sweet a Gram
Finding objects that weigh a gram can be challenging. Once I was a class teacher, we often referred come the old standby the a paperclip weighing a gram. Not only was it readily available in the classroom, yet it was additionally a known visual reference for the students. as an adult or teacher, possibly you are curious to know what other objects sweet a gram.
### Dollar bill
We all recognize sugar is a nice lightweight material in and of itself. To acquire a one-gram measurement, girlfriend will require ¼ that a teaspoon. The teaspoon need to be level and not rounded up. This is referring to continual fine serial sugar, no confectioners, brown or life sugar. As soon as you rise or diminish the thickness of the sugar granules, the weight will certainly change.
### A Chinese Yen
There room no coins in the United states that weigh specifically a gram, yet the Chinese Yen weighs precisely one gram.
## How do I measure a gram?
When you begin looking at the tiny measurement of grams, you might be wondering how you can measure something for this reason small. In order come accurately measure up a gram, it is finest to usage a digital scale. Digital scales administer a much an ext accurate measure than manual scales.
You will notification pharmacists use scales as soon as measuring the end powder for medicine. If a pharmacist tried come eyeball a gram, it could be really dangerous and also even deadly to the patient.
## Using a Balance Scale
If friend aren’t fortunate to have an electronic scale, you can use a balance range to measure grams. Most likely you won’t have actually a balance scale simply laying around, so you might need to do one. Over there are number of tutorials online that can assist you with making a balance scale. Friend will require to complying with materials to do a working scale:
Ruler2 cardboard squaresA pencilTapeStringItems that weigh 1 gramAttach the strings and cardboard piece to each finish of the ruler.Balance the center of the ruler on the pencil point.Place the thing you want to measure up on one cardboard square, which will certainly throw the range off balance.Place 1-gram objects on the cardboard attached to the various other side of the leader until the scale is well balanced out.Remember, this is no going to be very accurate measurement of grams because there room a lot of of factors playing into the weight, such as the cardboard, string, and also tape. However, this will give you one idea of how numerous grams an item is.
While this might not be the most scientific technique of measurement, yet it might be a fun technique to usage in a class to check out the load of a gram.
## What room Grams typically Used For?
You room not walk to uncover the measurement of grams in usual places. Because that example, you likely won’t uncover a recipe the calls because that grams of one item. However, if you room working in any kind of science or chemistry field, friend will most likely see the measurement of grams that are included in the metric measure system.
Who provides Grams in daily Measurement?
So currently that you recognize a couple of items that weigh a gram, below are some amazing facts that might leave girlfriend astounded. Follow to the people Atlas, there are three countries in the civilization that don’t usage the metric system.
United StatesMyanmarLiberia
All the other countries in the world currently use the metric device as their type of measurement.
Gram Equivalencies
Perhaps you room wondering how numerous grams would make up some the the standard dimensions we room accustomed to in the unified States.
See more: What Are The 3 Parts Of Interphase ? — Overview & Diagrams Why Is The Cell Cycle Important To Organisms
1 cup = 201.6 grams1 pint = 473.2 grams1 quart = 946.4 grams1 gallon = 3,785 grams
For someone in the united States, these measurements may seem much an ext confusing than simply using the words we space accustomed to, but approximately the world, this is the accepted kind of measurement.
Now the you know several objects the weight a gram, take part time, look around and see if friend can discover anything else that weighs a gram. Happy Measuring! | 1,055 | 5,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-33 | latest | en | 0.965399 |
https://nickadamsinamerica.com/tag/fill-in-calendar-template-2016/ | 1,611,366,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00017.warc.gz | 501,460,155 | 18,326 | # Fill In Calendar Template 2016
## Kindergarten Readiness WorksheetsKindergarten Readiness Worksheets
Published at Tuesday, March 24th 2020, 16:10:34 PM by Natuche Chauvin. Worksheet. These multiplication worksheets may be configured for either single or multiple digit horizontal problems. The factors may be selected to be positive, negative or mixed numbers for these multiplication worksheets. You may vary the numbers of multiplication problems on the multiplication worksheets from 12 to 30. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
## Educational Worksheets For KindergartenEducational Worksheets For Kindergarten
Published at Sunday, February 23rd 2020, 23:41:37 PM. Worksheet By Maddy Faure. These Radical Worksheets will produce problems for adding and subtracting radical expressions. You may select what type of radicals you want to use. These Radical Worksheets are a good resource for students in the 5th Grade through the 8th Grade.
## Senior Kindergarten Reading WorksheetsSenior Kindergarten Reading Worksheets
Published at Saturday, February 22nd 2020, 16:25:54 PM. Worksheet By Leona Renault. These kindergarten worksheets will produce adding shape problems where the children will add the two different groups of shapes for the total number of shapes. These kindergarten worksheets will generate 8 problems per worksheet.
### Printable Kindergarten Math WorksheetsPrintable Kindergarten Math Worksheets
Published at Saturday, February 22nd 2020, 07:33:19 AM. Worksheet By Solaine Delorme. These fractions worksheets are perfect for practicing subtracting fractions from whole numbers. You can select from five different degrees of difficulty. The easiest will keep the denominators and the numerators between 1 and 9. The hardest will keep the numerators between 1 and 20. The answer worksheet will show the progression on how to solve the problems. These fraction worksheets will generate 10 or 15 problems per worksheet.
#### Easy Math Worksheets For KindergartenEasy Math Worksheets For Kindergarten
Published at Friday, February 21st 2020, 08:56:06 AM. Worksheet By Ivonne Barthelemy. These time word problems worksheets will produce questions with elapsed days, weeks, months, and years, with ten problems per worksheet. These word problems worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
##### Free Printable Games For KindergartenFree Printable Games For Kindergarten
Published at Sunday, February 16th 2020, 13:24:31 PM. Worksheet By Tallis Boulanger. This section contains all of the graphic previews for the Constructions Worksheets. We have constructing line segments, perpendicular segments, angles, triangles, median of triangles, altitudes of triangles, angle bisectors, and circles. These geometry worksheets are a good resource for children in the 5th Grade through the 10th Grade.
###### Is And Are Worksheets For KindergartenIs And Are Worksheets For Kindergarten
Published at Saturday, February 15th 2020, 01:44:04 AM. Worksheet By Celeste Lamy. Have the students calculate their score then determine who stacked the most candies and who had the highest score—due to the scoring, this may not be the same person. Discuss any unusual approaches students used, if any.
## How To Make Worksheets For KindergartenHow To Make Worksheets For Kindergarten
Published at Thursday, February 13th 2020, 23:21:44 PM. Worksheet By Sydnee Auger. These Radical Worksheets will produce problems for multiplying radical expressions. You may select the difficulty for each expression. These Radical Worksheets are a good resource for students in the 5th Grade through the 8th Grade.
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Any content, trademark/s, or other material that might be found on this site that is not this site property remains the copyright of its respective owner/s. | 879 | 4,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.842196 |
https://byjus.com/question-answer/in-the-algebraic-expression-4x-2-2-the-value-of-unknown-is/ | 1,723,019,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00668.warc.gz | 113,893,548 | 21,794 | 1
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Question
# In the algebraic expression, 4x−2=2, the value of unknown is
A
1
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2
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Solution
## The correct option is A 1Given algebraic expression : 4x−2=2 Add 2 on both sides of the equation 4x−2+2=2+2 ⇒4x=4 Divide by 4 on both sides of the equation ⇒4x4=44⇒x=1
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Join BYJU'S Learning Program | 209 | 629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.806837 |
http://forums.wolfram.com/mathgroup/archive/2001/Jul/msg00159.html | 1,524,261,919,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944742.25/warc/CC-MAIN-20180420213743-20180420233743-00391.warc.gz | 133,295,361 | 7,718 | Can this be made cleaner and more efficient?
• To: mathgroup at smc.vnet.net
• Subject: [mg29807] Can this be made cleaner and more efficient?
• From: Flip at safebunch.com
• Date: Wed, 11 Jul 2001 01:26:56 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Hi All,
finally I got this algorithm working, but it is "ugly" and "slow" code.
Can an experienced Mathematica user help clean this up and make it more efficient.
By the way, this is an implementation of the Lucas Primality Test.
Notes:
1. Is it possible to move the sequence inside of the module and calculate it one
step at time? (it usually converges on the 2nd or 3rd try in finding D).
2. A check should be done on "n" and verify it is not a perfect odd sqaure
(i.e., n = 9 or n = 25, for example, will cause no solution for finding D and
this should really be checked). If n is a perfect square, terminate the test.
3. If one encounters a JacobiSymbol = 0 while searching for a d with
JacobiSymbol = -1, terminate the test as this means that n has factors (i.e., it
is composite).
Can someone here help? Thank you for any inputs ... Wilson
*** Here is the code ***
In[206]:=
Clear[a]
In[207]:=
a[1] =5;
In[208]:=
a[n_]:= a[n] = (-1)^(n+1) (Abs[a[n-1]]+2)
In[209]:=
(*find the first D in a[n] for which JS[a,n] = -1 *)
In[210]:=
findD[a_,b_]:=Module[{n=a, lst=b},
i=1;
While[JacobiSymbol[lst[[i]],n]
!= -1, i++];lst[[i]]]
In[211]:=
lpt1[a_,b_,c_]:=Module[{d=a, p=b, n=c},
Clear[u,v,k,w,r];
k = n+1;w = Reverse[IntegerDigits[k,2]];
r = Length[w]-2;q = (1-d)/4;
Print["n, k, r, d, q = ",{n,k,r,d,q}];
Print["w = ", w];t= 2n;
u = 1;v = p;
Print["u, v = ", {u,v}];
While[r > -1 ,
x = Mod[u*v, t];
y = Mod[(v^2+ d*u^2)/2, t];
{u,v}={x,y};
Print["u, v = ", {u,v}]; Print["r = ",r];
Print["Kr = ",w[[r+1]]];
If[w[[r+1]]==1,
u = Mod[(p*x+y)/2, t];
v = Mod[(p*y+ d*x)/2, t]];
(*Print["u, v = ", {u,v}]*);{u,v}={u,v};
r=r-1];
Print["u, v = ", {u,v}];
Print["For n = ",n];
If[Mod[u,n]==0, Print["LPT says Prime"],Print["LPT says Not Prime"]]]
In[212]:=
c=Table[a[n],{n,1,100}];
In[213]:=
p=1;
In[214]:=
n=5;
In[215]:=
d = findD[n,c]
Out[215]=
-7
In[216]:=
lpt1[d,p,n]
```
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• Next by thread: Re: Exponential Equations | 854 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-17 | latest | en | 0.757046 |
http://tdittrich.blogspot.com/2015/11/decimal-operations-addsubtract-money.html | 1,519,509,082,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815951.96/warc/CC-MAIN-20180224211727-20180224231727-00710.warc.gz | 331,763,948 | 15,821 | Welcome to my math blog! The purpose of this blog is to help you stay informed about our learning and experiences that have taken place during our math class. I have also included links your child (and you) may want to use in order to supplement math learning in 5th grade.
Tuesday, November 3, 2015
As usual, we began our class with a few math warm-ups.
• First, we completed a "Which One Doesn't Belong?" Students look at four items (numbers, shapes) and choose the one that does not belong. There are many answers to these problems, students just have to be able to justify their thinking mathematically.
• Next, we completed our estimation180 for the day. Today we were estimating the value of a roll of nickels.
After warm-ups, we move into framing our lesson by looking at the class and student objectives for the day. I also explained the TEK we are working on for the day.
Now it is time for a mini-review (formative assessment). I have assigned students a problem on goformative.com to solve. Today's problem combined the addition and subtraction of decimals with the Order of Operations.
Below is a student example of "my favorite no" (on the grid paper). In blue, you will see the steps the students needed to take to have the correct answer (please excuse the writing... it is not easy writing with a mouse!).
From there, we reviewed our vocabulary from yesterday and then moved into practicing solving decimal addition and subtraction problems. We focused on word problems that had to do with visiting a barber shop, finding total costs, paying bills, and determining change. We used a page from The Mailbox Teacher's Helper Feb/Mar 2008 entitled "Checking Out."
As part of the activity, I required the students to write the number sentence they were to solve and show their work on grid paper. To challenge the students' higher thinking skills, I required them to create their own word problem, write its number sentence, and solve it.
To view my instructions, please view the video: Checking Out. | 452 | 2,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-09 | longest | en | 0.968496 |
https://www.mathtransformations.com/7th-grade | 1,720,783,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00724.warc.gz | 556,977,062 | 175,557 | top of page
### Online Resources
The Number System
The Area Model:
​
Area Model: Multiplication with Whole Numbers
Interactive Simulation - PhET
Area Model: Decimals
Interactive Simulation - PhET
Area Model - Algebra
Interactive Simulation - PhET
​
Expressions and Equations:
Equality Explorer Basics
Interactive Simulation - PhET
Equality Explorer
Interactive Simulation - PhET
Equality Explorer: Two Variables
Interactive Simulation - PhET
Expression Exchange
Interactive Simulation - PhET
​
Number Sentence - Equations (Desmos)
The title doesn’t capture what this task is about. It focuses on matching algebraic expressions to contexts and word descriptions. It is a worthwhile task.
Equivalent Expressions (Desmos)
Students explore different ways to find the number of tiles in the border of a square swimming pool. They discover that the expressions they generate are all equivalent. (You would probably need to introduce this during a synchronous lesson and then have students work on it.)
Enter algebraic or numeric expressions and compare. **Cool modeling!**
NCTM Algebra Tiles
Poses problems, has students model with algebra tiles and solve, tells them if they are correct or incorrect.
(PBS Media Learning)
Interactive opportunity to balance a scale.
Inequalities Polygraph (Desmos)
Video Game: Dragon Box (algebra 12+)
Teaches all the rules for solving algebraic equations but does it visually in a virtual video game environment..
The App is \$7.99
​
Obstacle Course
Video to Give you Inspiration
bottom of page | 337 | 1,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.78927 |
https://bestincentivetours.com/best-time-to-travel/how-did-mary-and-joseph-travel-from-nazareth-to-bethlehem.html | 1,656,661,343,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00002.warc.gz | 172,648,782 | 18,253 | # How Did Mary And Joseph Travel From Nazareth To Bethlehem?
There are several legends that explain why Joseph and Mary, who was quite pregnant at the time, rode donkeys the five or more days it took them to go from Nazareth to Bethlehem, a route that is now commonly known as the Nativity Trail.
## How long would it have taken Mary and Joseph to walk from Nazareth to Bethlehem?
It is estimated that it took seven days to get from Nazareth to Bethlehem. If the hypothesis that is currently held by biblical scholars regarding the length of the journey from Nazareth to Bethlehem is correct, then Mary and Joseph would have needed to travel approximately 90 miles in four days, traveling at an average speed of 2.5 miles per hour for approximately eight hours a day.
## What mode of transportation did Mary use to get to Bethlehem?
There is no evidence that Mary traveled to Bethlehem on the back of a donkey.There is not a single place in any of the Gospels where it is said that Mary did anything other than walk.The entire trip is summarized in three lines as follows: Joseph and Mary traveled to Bethlehem, and while they were there, Mary went into labor.While they were there, they were in Bethlehem.There is no indication of the mode of conveyance.
## How far apart are Bethlehem and Nazareth?
The distance from Nazareth to Bethlehem, measured in a direct line, is exactly 9135 kilometers (kilometers) long and 914.85 meters in height. It is 5676.8 miles from Nazareth to Bethlehem if we are to measure the distance in kilometers.
We recommend reading: Quick Answer: How To Play The Journey On Fifa 17?
## How long did it take for the Three Kings to get to Bethlehem?
Following the North Star across the desert is how the three wise men made their way to Bethlehem in order to find the divine infant, as described in the Gospel of Matthew. According to texts that were composed much later, the three kings, Melchior from Europe, Caspar from Arabia, and Balthazar from Africa, appeared in Jerusalem twelve days after the birth of Jesus.
## How many miles did Joseph and Mary travel to Bethlehem?
To get to the city where Joseph’s ancestors lived, they had to travel around 90 miles: first to the south along the lowlands of the Jordan River, then to the west across the hills that encircle Jerusalem, and finally into Bethlehem.″It was a really arduous trip,″ said Strange, who regularly heads an excavation team in the ancient city of Sepphoris, which is located close to Nazareth in Israel.″It was a fairly arduous trip.″
## How long did it take Mary and Joseph to travel from Bethlehem to Egypt?
EXPLAINING THE GOSPEL IN ACCORDANCE WITH ST. MATTHEW Joseph did what he was told. It is possible that it took the fugitives three days to make it to Egypt, where they were no longer subject to Herod’s authority.
## What was the donkey’s name that carried Mary?
Here is another Rankin/Bass Christmas special that I really like seeing every year. It tells the story of a young donkey by the name of Nestor who had extremely large ears and who would eventually become the donkey that would assist Joseph and Mary on their journey to Bethlehem.
We recommend reading: How To Travel With Frozen Breast Milk By Car?
## How old was Joseph when he married Mary?
It wasn’t the only paper that had such language either. Christ himself narrates the story of his step-father in an additional early text called The History of Joseph the Carpenter, which was written in Egypt between the 6th and 7th centuries and claims Joseph was 90 years old when he married Mary and died at 111. This text was composed between the 6th and 7th centuries.
## Did Mary ride a donkey to Egypt?
Joseph goes beside Mary and Jesus as they travel to Egypt while Mary breastfeeds Jesus while riding on a donkey.
## How far did Jesus walk from Nazareth to Jerusalem?
Pilgrims go from Nazareth to the Sea of Galilee while following the Jesus Trail. JERUSALEM, May 1, 2009 — The Jesus Trail is sixty-five kilometers long and begins in Nazareth, which was Jesus’ hometown. Its construction was just finished not too long ago, and its path takes travelers through the cities and villages of the Galilee area in northern Israel.
## How long did Joseph and Mary stay in Bethlehem?
Luke also adds that eight days after Jesus’ birth, Joseph, Mary, and Jesus traveled to Jerusalem and then on to Nazareth. This journey took place after Jesus was born at Bethlehem. Although they have certain things in common, Matthew and Luke are quite different books, and it is extremely difficult to reconcile the disparities between them.
## Why did Joseph and Mary travel to Bethlehem right before Jesus was born?
Joseph and Mary’s journey to Bethlehem is described in the Gospel of Luke as being motivated by the need to comply with an imperial mandate that demanded all persons return to the places of their ancestors ″that the whole world should be taxed.″ This explains why Jesus was born in the town of since Mary was pregnant with him at the time when the instruction had to be carried out.
We recommend reading: How Fast Can A Tsunami Travel On Land?
## How old was Joseph when Jesus was born?
At one point in time, people believed that Joseph was an old man when he married Mary. However, current research suggests that Mary and Joseph were both in their teens when Jesus was born, with Mary being around sixteen and Joseph being approximately eighteen years old. During that time period, newlywed Jewish couples typically did things in this manner.
## What is the distance between Bethlehem and Jerusalem?
Distance between Jerusalem and Bethlehem is 6713 KM / 4171.7 miles.
## How many days did the three kings travel?
It takes place on January 6th. The Gospel of Matthew in the Bible has a passage regarding the visit of the magi. According to this passage, Melchior, Caspar, and Balthazar journeyed for a total of twelve days in order to approach Jesus. They came with one one-of-a-kind present apiece. | 1,293 | 5,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.979423 |
onlineclonidine.us.com | 1,713,617,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00285.warc.gz | 21,692,984 | 16,405 | # On the Web Casinos: Mathematics of Bonuses
On-line casino players know that the latter kinds offer you various bonuses. “Free-load” appears attractive, however, are that they really useful those bonuses? Are you currently profitable for gamblers? The answer to the question depends on lots of requirements. Mathematics will help people answer this particular question.
Let us start out with an ordinary reward on deposit: you transfer £ 100 and obtain £ 100 longer, and that it ssiplay will be possible to get with staked \$3000. It’s a common illustration of reward on the very first deposit. The size of the bonus and deposit can be different, as well as the required stake prices, however a very important factor stays unchangeable – the sum of the bonus is accessible for withdrawal soon after the required bet. Till this moment it’s not possible to get money, usually.
If you’re เว็บพนันบอล going to play in the online casino to get quite a very long time and rather insistently, this bonus will help you, it can really be thought spare money. If you play slots using 95 percent pay-outs, a plus will allow one to make typically extra 2000 \$ of stakes (\$100/(1-0,95)=\$2000), after that the sum of reward will probably soon be finished. But there might be complications, for instance, in the event that you only wish to have a peek in a casino, even minus needing quite a while, in the event that you would like roulette or other games, banned by casinos’ rules for successful rear bonuses. In many casinos that you wont be authorized to draw dollars or may only return a deposit, even if a wager isn’t made on the games enabled in the match . If you are enthusiastic on blackjack or roulette, along with a bonus can be earned straight back just by playing with slots, earn the mandatory £ 3000 of stakes, at the class of 95 percent of payouts you could lose generally \$3000*(1-0,95)=£ 150. Since you see, you not only miss the incentive but also eliminate your own pocket \$50, in this situation it’s advisable to refuse your bonus. Anyway, if blackjack and poker are allowed for successful back the bonus with a casino benefit simply about 0,5 percent, so it could be anticipated after winning the bonus back you have \$100-3000*0,005=85 of this casino dollars.
“sticky” or”phantom” bonuses: | 515 | 2,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | longest | en | 0.944503 |
https://www.coursehero.com/file/1652137/Phasor20analysis20of20mechanical20systems/ | 1,498,251,497,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320174.58/warc/CC-MAIN-20170623202724-20170623222724-00385.warc.gz | 848,884,873 | 39,297 | Phasor%20analysis%20of%20mechanical%20systems
# Phasor%20analysis%20of%20mechanical%20systems - Phasor...
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Phasor Analysis of Linear Mechanical Systems and Linear Differential Equations ME104, Prof. B. Paden In this set of notes, we aim to imitate for linear mechanical systems and linear differential equations, the phasor analysis we learned for electric circuits. Recall how we derived the complex impedance for an inductor. Starting with the differential equation for the V-I characteristic for an inductor I dt d L V = ( 1 ) we substitute complex sinusoids t j e V V ω ˆ ( 2 ) t j e I I ˆ ( 3 ) So that equation (1) becomes ( t j t j e I dt d L e V ˆ ˆ = ) ( 4 ) Differentiating and solving yields I L j V ˆ ˆ = ( 5 ) and the impedance of the inductor is defined by ˆ ˆ V Z jL I = ± ( 6 ) where “ ” denotes “defined equal to”. Having done this calculation once, we see that we can jump directly from (1) to (5) by making the substitution ± j dt d ( 7 ) Phasor Analysis of Linear Mechanical Systems Consider a mechanical damper (a.k.a. shock absorber) which produces a velocity- dependent force according to the linear differential equation x dt d b f = ( 8 ) Making the substitution j dt d , we get
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x b j f ˆ ˆ ω = ( 9 ) And defining the mechanical impedance to be the ratio of force to displacement, we have ˆ ˆ f Z jb x = ± (Newtons/meter) (10) Note that the damper has a low stiffness at low frequencies, and a high stiffness at high frequencies. The units of impedance are Newtons/meter in mechanical systems and volts/amp = Ohms in electrical systems. For a mass, m , we have 2 2 dd d f mx m dt dt dt ⎛⎞ == ⎜⎟ ⎝⎠ x ( 1 1 ) Substituting j dt d yields () 2 2 ˆ ˆ f j m x ωω ( 1 2 ) The impedance of a mass increases very rapidly with frequency. This explains why anvils
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## This note was uploaded on 03/02/2009 for the course ME 104 taught by Professor Staff during the Fall '08 term at UCSB.
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# week2da - Spring 2010 DecisionAnalysis MS405 Modeling...
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Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 Week 2 1 Decision Analysis MS 405 Modeling Decisions Kemal Kılıç Faculty of Engineering and Natural Sciences
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Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 Week 1 - Review What is Decision Analysis? Decision Analysis Process Elements of Decision Problems Values and Objectives Decisions to Make Uncertain Events Consequences Week 2 2
Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 What you were supposed to do! Before today: Read the first two chapters, solve the suggested problems, or even better as many problems as possible, from the back of the book. Suggested Problems: Page 36 2.1 Page 37 2.10 Page 38 2.12 This week : Chapter 3 – Structuring Decisions Decisions Trees and Influence Diagrams. Read the chapter. Week 2 3
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Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 Week 2 Objectives: Having identified the elements of a Decision Problem, how should one begin the modeling process? Creating a decision model requires three fundamental steps 1. Filter & operationalize the objectives ( Revisit ) 2. Structure the elements in a logical framework Influence Diagrams and Decision Trees 1. Refinement and precise definition of all of the elements How to Evaluate the Decisions Trees and Influence Diagrams? (Chapter 4) Week 2 4
Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 Step 1:Filter & Operationalize the Objectives Process of Identifying Objectives 1. Develop a wish list. What do you want? What should you want? 2. Identify alternatives. What is the perfect alternative, a terrible alternative, some reasonable alternatives, what is good or bad about each? 3. Consider problems and shortcomings. What is wrong or right with your organization? What needs fixing? 4. Predict consequences. What has occurred that was good or bad? What might occur that you care about? Week 2 5
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Kemal Kılıç, Sabancı Üniversitesi Kemal Kılıç, Sabancı Üniversitesi Spring, 20 Spring, 2010 Process of Identifying Objectives (Cont.) 5. Identify goals, constraints, and guidelines. What are your aspirations? What limitations are placed on you? 6. Consider different perspectives. What would your competitor or constituency be concerned about? At some time in the future, what would concern you? 7. Determine strategic objectives. What are your ultimate objectives? What are your values that are absolutely fundamental? 8. Determine generic objectives. What objectives do you have for your customers, your employees, your shareholders, yourself? What environmental, social, economic, or health and safety objectives are important?
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Ask a homework question - tutors are online | 829 | 3,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.763531 |
http://stackoverflow.com/questions/2939933/rock-paper-scissors-for-arbitrary-odd-number-of-elements | 1,455,354,703,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166261.11/warc/CC-MAIN-20160205193926-00147-ip-10-236-182-209.ec2.internal.warc.gz | 207,346,876 | 20,005 | # Rock Paper Scissors for arbitrary odd number of elements
How do I efficiently create a rock-scissors-paper game for n elements, where n is any odd number >=3.
In other words, I want a non-transitive complete ordering of n elements such that each element is greater than (n-1)/2 other elements and each element is lesser than (n-1)/2 other elements.
-
Efficient in what way? Code density? Executable size or speed? Memory use? How big can N be? – Emile Cormier May 30 '10 at 19:31
Assume your items are numbered 0,1,2,...,n-1.
Item i beats item j iff `i - j (mod n) > (n-1)/2`.
In other words you can rotate the list such that your chosen item is in the middle of the list:
``````i - (n-1) / 2, ..., i-2, i-1, i, i+1, i+2, ..., i + (n-1) / 2
``````
Then item i beats all the items below it in the list.
A matrix of i vs j would look like this:
`````` 0 1 2 3 4
0 - L L W W
1 W - L L W
2 W W - L L
3 L W W - L
4 L L W W -
``````
This is not the only possibility, but it is probably the simplest. You can construct any matrix that obeys these rules:
• All values on the diagonal are zero.
• The other values are 1 or -1 (win, lose).
• It is a skew symmetric matrix.
• It has exactly (n-1)/2 wins and losses in every row and column.
Here is another more complex example:
`````` 0 1 2 3 4
0 - L W W L
1 W - W L L
2 L L - W W
3 L W L - W
4 W W L L -
``````
Or phrased another way:
```0 beats 2 and 3.
1 beats 0 and 2.
2 beats 3 and 4.
3 beats 1 and 4.
4 beats 0 and 1.
```
In this example it may be possible to relabel the items to give the same logic as in the previous game. I doubt that holds in general though.
-
+1. Nice construction! – Aryabhatta May 30 '10 at 19:26
btw, if you are interested, this is a tournament graph: en.wikipedia.org/wiki/Tournament_(graph_theory) with score sequence s_i = (n-1)/2. – Aryabhatta May 30 '10 at 19:33
Fantastic, thanks!
As another approach (inspired by yours), k beats k+1 (mod n-1), k+2 (mod n-1), etc... for the next (n-1)/2 elements.
- | 654 | 2,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-07 | latest | en | 0.865608 |
http://shitohichiumaya.blogspot.com/2010/11/eigenvalue-and-transfer-function-4.html | 1,532,284,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00163.warc.gz | 313,378,759 | 10,951 | ## 2010-11-03
### Eigenvalue and transfer function (4)
Eigenvalue and Eigenvector
Scalar case
3 x 4 = 12
If I could use variables a,x,b:
ax = b.
The first example shows x=4 times a=3 equals b =12. Here, I multiply x a times. I didn't multiply a times x. In the scalar case, these have no difference since the following commutative law works on scalars.
ax = xa
Please note, this is not always correct. Even we can not exchange the meaning in the scalar numbers. For example, assume there is a chocolate box that costs five Euro. We can buy two packages. This is 2 times 5 = 10 Euro. This is not two's 5 Euro time. We can double the 5 Euro chocolate, but we can not see five Euro time doubles. Usually it doesn't make sense: five Euro times (five times works, five Euro times has a problem). So I remind the order of operator is also important since vector is more strict about the operations. | 230 | 901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-30 | latest | en | 0.92168 |
findy.vn | 1,713,495,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00893.warc.gz | 224,789,688 | 32,764 | Net Worth
# Understanding What is Net Worth: An Easy Guide for Beginners
If you have an interest in finance and economics then you might have come across a term called “Net Worth”. The concept of Net Worth is very important in understanding the financial status of a person.
Unfortunately, the fundamentals of net worth are often misunderstood by many of us. It is not just a number on a balance sheet. We can simply put it up as the sum of all the assets minus liabilities a person owns. In today’s post, we will try to solve the mystery of Net Worth. We will explore its definition, significance, and how to calculate it.
Nội dung bài viết
## Definition of a Person’s “Net Worth”
In simple language, we can say that Net Worth is a term that describes the total value of a person’s financial assets minus his liabilities. It is the most popular measure used to determine the overall wealth of a person.
A positive net worth indicates that the value of assets exceeds liabilities which means the person has surplus wealth. On the other hand, a negative net worth suggests that liabilities outweigh assets, indicating a potential debt burden on a person. Monitoring net worth is essential for assessing financial progress and making informed financial decisions.
READ Justin Jones: Biography, Age, Parents, Protests, Arrests, and Restoration to House of Representatives
## How to Calculate Net Worth?
Calculating Net Worth is easy. First, you need to find the total value of everything you have. Then, from that total value, you must minus your liabilities or everything you owe. Let’s break down this process with an easy example.
## 1st Step – Identify your assets
The first of calculating your net worth is identifying your assets. It may include the following things.
• Cash – The money you have in your bank accounts, wallet, or saved at home. Example: Let’s say you have \$500 in your checking account and \$200 in a savings account.
• Investments: Any stocks, bonds, mutual funds or retirement accounts you have. Example: 100 shares of XYZ stock valued at \$10 per share and a Roth IRA worth \$5,000.
• Real Estate: The value of any properties you own such as your primary residence or rental properties. Example: Your house at a value of \$250,000.
• Vehicles: The current market value of your cars, motorcycles, or any other vehicles you own. Example: Your car having a value of \$15,000.
• Personal Property: The value of valuable items like jewelry, artwork, electronics, or furniture. Example: Jewelry worth \$2,000 and a laptop worth \$1,500.
• Other Assets: Any other assets you may have like business ownership or collectibles. Example: A small business ownership worth \$10,000.
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## 2nd Step – Calculate the Total Value of Your Assets
Add up the values of all your assets from the previous step to find your total asset value.
Example: \$500 (cash) + \$200 (savings) + \$1,000 (stocks) + \$5,000 (IRA) + \$250,000 (house) + \$15,000 (car) + \$2,000 (jewelry) + \$1,500 (laptop) + \$10,000 (business) = \$285,200.
READ Nikki Haley Net Worth- Explore her path from an entrepreneur to a successful politician
## 3rd Step – Identify your liabilities
Make a list of all of your liabilities or debts you owe.
• Mortgage: The amount you owe on your mortgage. Example: \$200,000.
• Credit Card Debt: The outstanding balance on all your credit cards. Example: \$5,000.
• Student Loans: The amount you owe in student loans. Example: \$10,000.
• Auto Loans: The balance on any car loans. Example: \$8,000.
• Other Debts: Any other debts or loans you have. Example: Personal loan of \$2,000.
## 4th Step – Calculate the total value of your liabilities
Add the value of all your liabilities from the previous step to find your total liability value. Example: \$200,000 (mortgage) + \$5,000 (credit card debt) + \$10,000 (student loans) + \$8,000 (auto loans) + \$2,000 (personal loan) = \$225,000.
## 5th Step – Calculate your net worth
Now, subtract your total liabilities from your total assets. In this way, you can get your net worth. Example: \$285,200 (total assets) – \$225,000 (total liabilities) = \$60,200.
So, as per this example, your net worth is \$60,200.
## Significance of Net Worth
Here are some of the major significance of knowing the net worth of a person or company.
How can I check my Net Worth?
READ Janet Protasiewicz: Biography, Husband, Early Life, Age, Wisconsin Supreme Court Election, & more
Can Net Worth be negative?
Yes, if liabilities are more than assets then Net Worth can be negative.
How can I increase my net worth?
You can do it by increasing your savings, investing wisely, and budgeting effectively.
What are assets?
They are the valuable things that you own like money in bank accounts, investments, real estate, vehicles, jewelry, etc.
How often should I track my net worth?
It is recommended to track your net worth on a regular basis such as monthly, quarterly, and yearly.
Post by: Findy
category by: Net Worth
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Close | 1,173 | 5,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.929074 |
imdcdemo.sgssys.info | 1,585,576,954,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00402.warc.gz | 92,852,140 | 7,734 | # An Ordinary Mean in Z
I frequently wondered exactly what exactly did normal mean, Earlier I began analyzing algebra. I received a book to look in plus it seemed the same as plenty of work merely to figure out exactly what that word meant.
You see, my very first idea was”normal” simply means normal. Exactly what does mean in math? Wellthere are two meanings to this word.
When thesis writing help speaking about the standard the word is employed in math. Conventional as in”regular distribution” is some thing which everybody else knows about. When there is a variety of our standard deviation at an example the distribution is, also it’ll be spread normally by chance.
The normal distribution is used to help us choose how we want to divide all of our time between how much time we will put into chores, how much time we will spend on things like sports or hobbies, and how much time we will put into the leisure activities. In the middle is where our priorities lie.
But, normal means something different if we are dealing https://paramountessays.com/ with utmost price. Today subsequently distributions are used in finding how many kids are suitable to be foster kids for. In picking from a variety of alternative replies to a standardized test distributions are all applied.
In a normal distribution, normal means so that it is distributed, meaning that the amounts will fall in line as a result of 33, which each sample is going to probably be dispersed. This means that they will not change from each other, and also will have exactly the identical mean the exact same standard deviation.
So, what does normal mean in math when we are dealing with extreme values in the distribution of numbers? Well, the normal distribution is used when we want to determine how many people are suitable to be foster parents. When we are talking about the results of a normal distribution, normal means that the result will follow a normal distribution, meaning that the numbers will follow a normal distribution by chance.
Normal distributions are used in choosing from https://ufdc.ufl.edu/AA00023024/1 several different alternative answers to a standardized test. Normal distributions are used in choosing from several different alternative answers to a standardized test. So, what does normal mean in math when we are dealing with extreme values in the distribution of numbers?
Well, when you have a normal distribution and you are trying to decide whether or not to pass a child on to adoption for reasons that are far-fetched, normal means that you are going to pass them because they are suitable. The percentage of children who are suitable to be foster parents will be very close to 100%. We can also use this data when we are dealing with numbers that we may need to use in our homework, such as how many people are going to lose their jobs, how many people are going to have their hours cut, how many people are going to be laid off due to something that is going to happen with another company, and so on.
The normal distribution is used when we are trying to choose among various possible answers to a standardized test. Normal means that the answer to every question will be distributed according to what we would expect the average person to do. The difference between a normal distribution and the normal distribution is the distribution that will vary over the number of degrees of freedom.
We can think of the normal distribution as something like a bell curve. The normal distribution has two tails, and when we use it for an equation, we can see that the normal distribution will form two tails. What does normal mean in math?
OK, so now you know that the normal distribution is a bit different than normal means. Why would you want to know this? | 742 | 3,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-16 | latest | en | 0.958765 |
https://math.dartmouth.edu/~m54x15/day-to-day.html | 1,516,736,137,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892238.78/warc/CC-MAIN-20180123191341-20180123211341-00712.warc.gz | 753,923,464 | 2,562 | ## Math 54: Topology
### Summer 2015
General Info
Diary
The homework problems are to be found in the textbook; 'a.b' means 'Problem b in Paragraph a'. It is strongly suggested that you treat all the problems listed here.
Policy: homework should be submitted no later than 5pm on the due date. Let your instructor know in advance if you feel that you need extra time.
wk date reading topic homework / comments 1 06/26 Introduction 1.1, 1.3, 1.9, 2.2, 2.4, 2.5 2 06/29 §1 to §6 Basic set theory: finite and infinite sets 3.3, 3.9, 3.11, 4.1, 4.2, 4.9, 4.11 06/30 (x) §1 to §6 Basic set theory: countable sets 5.4, 6.5, 6.7 07/01 No class PC away 07/03 No class College Holiday Homework #1, due Tuesday 07/07. 3 07/06 Examples of metric spaces 07/07 (x) Metric spaces and continuous maps Homework #2, due Monday 07/13 and solution. 07/08 §12, §13 Topological spaces: topologies and bases 07/10 §13 Topological spaces: bases 13.1, 13.3, 13.4, 13.5, 13.6, 13.8 4 07/13 §14 The order topology 07/14 (x) §15 The product topology on X × Y 16.2, 16.3, 16.6, 16.9 07/15 §16 The subspace topology Homework #3, due Monday 07/20 and solution. 07/17 §17 Closed sets 5 07/20 §17 Closure, accumulation points 17.3, 17.5, 17.6, 17.7, 17.8 07/21 (x) §17 Hausdorff and T1 spaces 17.12, 17.14, 17.15, 17.16 07/22 Midterm 1 (solution) 07/24 Guest lecture The problem with Topology, by James Binkoski (after T. Maudlin) Homework #4, due Wednesday 07/29 and solution. 6 07/27 §18 Continuous maps 18.2, 18.3, 18.5, 18.8 07/28 (x) §18 Continuous maps Guest lecture by D. Freund 07/29 Not in text Homeomorphisms and the category Top An introduction to categories can be found in course notes by P. Schapira. 07/31 §19 Products of topological spaces 19.1, 19.2, 19.3, 19.4, 19.7, 19.8, 19.10 Homework #5, due Friday 08/07 and solution. 7 08/03 §20 The metric topology 20.1, 20.3, 20.4, 20.5, 20.6 08/04 (x) §20 The metric topology 21.3, 21.6, 21.7, 21.8 08/05 §21 Sequences in metric spaces - Functors Homework #6, due Monday 08/10 and solution. 08/07 §23 Connected spaces Guest lecture by D. Freund 8 08/10 §24, §25 Path connectedness, connected components 23.4, 23.5, 23.7, 24.1, 24.2, 24.3 08/11 (x) §26 Compact spaces 08/12 Midterm 2: in-class (solution), take-home (solution) 08/14 §26 Compact spaces 26.1, 26.2, 26.3, 26.5, 26.7, 26.8 9 08/17 §27 Compact subspaces of the real line 08/18 (x) §28 Limit point and sequential compactness 28.2, 28.6, 28.7 08/19 §29 Local compactness Homework #7, due Tuesday 08/25 and solution. 08/21 §31 - §34 Separation properties and metrizability 10 08/24 Not in text Topological properties of matrix groups 08/25 (x) No class 08/26 Not in text Topological properties of matrix groups 08/29 Final examination and solution. | 1,020 | 2,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-05 | latest | en | 0.774933 |
http://www.contrib.andrew.cmu.edu/~hanlai/?m=201205 | 1,518,936,864,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811794.67/warc/CC-MAIN-20180218062032-20180218082032-00385.warc.gz | 429,259,714 | 5,657 | Pre-phase What is statistics
In my opinion, statistic is ALL about estimation. Estimating the probability of some events that happen over the universe of all events.
Test of significance
When people first learn about statistics, they are probably learned from stats 101, where the professor told them how to do a t-test, or chi-square-test where they can decide a certain judgement is significant or not. Well, this is an estimation too, in fact, these tests are estimating the probability of you making a mistake by saying the judgement is significant. For example, if you are doing a t-test of two samples, and the p-value is 0.01, that is saying if you say these two samples are significant, the probability that you are incorrect is 0.001. It's pretty much means that you are almost correct.
Then why is there a whole area of statistics if the only goal of statistics is to estimate?
It's because there are so many models, that each have its own strength when estimating a probability. There's parametric, non-parametric to estimate a probability distribution over continuous, or discrete interval. There's also graphical models, multivariate models if the dataset you have got more than one variables, and you want to estimate conditional probability.
When you are estimating something, there are also many measurements of how good the estimator is. There are always trade-off between properties of an estimator, if your estimator is unbiased, it's probably going to have high variance.
There are many questions to ask when you want to estimate something.
Would you like an estimator that is generally good, but can make is very bad mistake or you'd like an estimator that is not as good, but is guaranteed to not make a very bad mistake?
Would you like an estimator that is unbiased when sample size is infinity with high variance or you'd like an estimator that is little biased but with very small variance?
etc.
So before you get into the field of statistics, these questions are definitely important to keep in mind, and when you use statistics to solve problems in research, you'll always have to state how/why you chose such estimator.
Matlab: Double precision problem
In matlab, Sometimes when you try to compare two numbers, they don't usually gives you the answer you excepted. When you compare two integers,
```a=1; b=1; a==b ```
will gives you 1.
but when you compare double, sometimes it doesn't work. Simple cases that if you do
``` a=0.001; b=0.001; a==b```
will give you 1 still. But if you save a into a file, and use ` textread(filename)` to get the value, the value may still look like 0.001, but if you do
` a==0.001`
It might give you 0 because the a was read from a file and it was in some weird format. This might be a bug. Some people fix it by doing
` abs(a-0.001)<0.000001`
it basically means if a and 0.001 is very close, then they are equal.
I personally have a quicker fix.
` a+1==0.001+1`.
For some reason, after any operation on the variable, the value no long have anything weird going on inside. | 693 | 3,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-09 | latest | en | 0.964516 |
https://web2.0calc.com/members/quinn/ | 1,550,651,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494485.54/warc/CC-MAIN-20190220065052-20190220091052-00471.warc.gz | 737,422,634 | 8,214 | # quinn
+5
674
3
+238
### can someone prove sin2x=2*sinx * cosx for me?Thank you
quinn Oct 10, 2014
+5
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6
+238
### In a trail,supposea lawyer wants 2 men and 5 womento make up a special panel. so the formlar is nCr=n!/[r!*(n-r)]!,right?So,5C2=5!/[2!(5-2)]
quinn Oct 9, 2014
+5
331
3
+238
### in the figure ,cntral ganle BAC is 30 degrees. If circle A has a diamter of 6 units , what is the area of the shaded segment
quinn Oct 8, 2014
+8
1770
5
+238
### if a,b,care the 3 sides of the triangle ABC,and a^2+b^2+c^2=ab+bc+ac.what is the shape of the triangle?
quinn Sep 30, 2014
+5
577
10
+238
quinn Sep 28, 2014
+5
256
1
+238
### does the equation 4^x=x have a soloution?
quinn Sep 26, 2014
+8
1197
11
+238
### In the triangle ABC, a,b,c are the 3 sides of the triangle.if a(1-x^2)+2bx+c(1+x^2)=2 is a qudratic equation. The equation have two same sol
quinn Sep 26, 2014
+8
759
14
+238
### if x^2-2011x+1=0 is true,then waht is the value of x-2010x+2011/(x^2+1)?
quinn Sep 25, 2014 | 439 | 1,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-09 | latest | en | 0.774698 |
https://www.doubtnut.com/qna/102371051 | 1,718,516,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00811.warc.gz | 657,774,664 | 41,747 | # If A=[2−335] then find A−1 by adjoint method.
Video Solution
Text Solution
Verified by Experts
## Given, A=[2−335] |A|=2×5−(−3)(3)=10+9=19≠0 ∴A−1 exists. To find adjoint of A : M11=5,A11=(−1)2⋅5=5 M12=3,A12=(−1)3⋅3=−3 M21=−3,A21=(−1)3⋅(−3)=3 M22=2,A22=(−1)4⋅(2)=2 ∴ Matrix of cofactors is [5−332] ∴ Adj. A=[53−32] We have, A−1=1|A|⋅Adj. A=119[5,3−3,2]
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If A=[2−341], then adjoint of matrix A is
A[1342]
B[1342]
C[1342]
D[1342]
• Question 2 - Select One
## If A=⎡⎢⎣1−212λ−213−3⎤⎥⎦ be the adjoint matrix of matrix B such that |B|=9, then the value of λ is equal to
A1
B774
C232
D392
• Question 3 - Select One
## If B = ⎡⎢⎣320240110⎤⎥⎦, then what is adjoint of B equal to ?
A000000218
B002001008
C002001000
DIt does not exist
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 644 | 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} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.753001 |
https://panicjanet.com/how-do-you-find-the-independent-and-dependent-variables-in-a-research-article/ | 1,653,554,097,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00126.warc.gz | 514,753,788 | 14,485 | # How do you find the independent and dependent variables in a research article?
## How do you find the independent and dependent variables in a research article?
You can use this typical form to determine the independent and dependent variables from the title of the study. If the study title is in the form “The effects of X on Y in Z”. X is the independent variable and Y is the dependent variable – the outcome, and Z is the type of subjects represented.
## Is dependent and independent variable always present in a study?
Of the two, it is always the dependent variable whose variation is being studied, by altering inputs, also known as regressors in a statistical context. In an experiment, any variable that the experimenter manipulates can be called an independent variable.
## Can location be an independent variable?
For example, a student might change the position of a plane’s wing to see how it affects the average speed of a model plane. The wing position would be the independent variable- because the student purposely changes its location.
## What is a constant independent and dependent variable?
A controlled variable is one that is constant and is unchanged in an experiment. It is held constant in order to observe the result of the independent variable. An independent variable is the variable that is being changed in the experiment in each trial, while a dependent variable is the one that is being measured.
## What do independent variables mean?
Answer: An independent variable is exactly what it sounds like. It is a variable that stands alone and isn’t changed by the other variables you are trying to measure. For example, someone’s age might be an independent variable.
## Which is the dependent variable?
The dependent variable is the one that depends on the value of some other number. If, say, y = x+3, then the value y can have depends on what the value of x is. Another way to put it is the dependent variable is the output value and the independent variable is the input value.
## What is a dependent variable in research?
Definitions. Dependent Variable. The variable that depends on other factors that are measured. These variables are expected to change as a result of an experimental manipulation of the independent variable or variables. It is the presumed effect.
## Is pressure a dependent or independent variable?
If you manipulated (changed) the pressure and measured the volume as a result, the pressure would be the independent variable and the volume would be the dependent variable because the volume was dependent on the pressure.
## Is temperature a dependent or independent?
So, the temperature depends on the time. So, the temperature is dependent, and the time is independent. If you’re looking at a graph, the independent variable is generally on the horizontal axis.
## How is time an independent variable?
The independent variable is always displayed on the x-axis of a graph, while the dependent variable appears on the y-axis. Time is a common independent variable, as it will not be affeced by any dependent environemental inputs.
## Is weather a dependent variable?
In a scientific experiment, a researcher wants to have controlled set of conditions, and then change one parameter (independent variable) only to see if there is a difference in results (dependent variable). You could compare weather results at the same location using different methods of measuring the weather. | 664 | 3,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.936594 |
http://mathhelpforum.com/statistics/34320-permutations-bottles-print.html | 1,506,361,447,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818692236.58/warc/CC-MAIN-20170925164022-20170925184022-00584.warc.gz | 221,602,451 | 3,039 | # Permutations of Bottles
• Apr 13th 2008, 11:05 AM
mixtapevanity
Permutations of Bottles
In how many different ways can I arrange 7 green and 8 brown bottles so that exactly one pair of green bottles are side by side?
I've been trying to do this all day, am I right for separating the answers into 14 cases of where the 2 green bottles can be across 15 places?
• Apr 13th 2008, 11:38 AM
Soroban
Hello, mixtapevanity!
Quote:
In how many different ways can I arrange 7 green and 8 brown bottles
so that exactly one pair of green bottles are side by side?
Duct-tape two green bottles together.
Now we have 14 units to arrange:
. . $\boxed{GG}\,, G, G, G, G, G, B, B, B, B, B, B,B,B$
Now place the eight brown bottles in a row.
. . Note that there are spaces before, after and between them.
. . . $\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\ _\,B\,\_$
We will take the six green units and place them in six of the nine spaces.
And there are: . ${8\choose6} \:=\:\boxed{84}$ ways.
• Apr 13th 2008, 11:48 AM
mixtapevanity
Thanks for your reply! That is along the lines of what I was thinking. But can I ask what did you mean by (8 6) = 84?
• Apr 13th 2008, 12:29 PM
Plato
That is a mere typo. It should be $\binom{9}{6}=84$
Nine blanks choosing six can be done in 84 ways. | 419 | 1,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-39 | longest | en | 0.910622 |
https://www.hitpages.com/doc/4515562721902592/9/ | 1,485,133,531,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281659.81/warc/CC-MAIN-20170116095121-00027-ip-10-171-10-70.ec2.internal.warc.gz | 930,545,361 | 5,454 | X hits on this document
# Chapter 3 Mathematics of Finance - page 9 / 20
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Barnett/Ziegler/Byleen Finite Mathematics 12e
9
# Interest Earned
The present value of all payments is \$22,457.78. The total amount of money withdrawn over two years is
3000(4)(2)=24,000.
Thus, the accrued interest is the difference between the two amounts:
\$24,000 – \$22,457.78 =\$1,542.22.
Document views 78 Page views 78 Page last viewed Sun Jan 22 20:43:44 UTC 2017 Pages 20 Paragraphs 127 Words 1100 | 159 | 518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | latest | en | 0.818634 |
https://www.knowpia.com/knowpedia/Standard_wire_gauge | 1,643,030,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304570.90/warc/CC-MAIN-20220124124654-20220124154654-00648.warc.gz | 889,631,662 | 15,056 | BREAKING NEWS
Standard wire gauge
Summary
A standard wire gauge.
British Standard Wire Gauge (often abbreviated to Standard Wire Gauge or SWG) is a unit for denoting wire size given by BS 3737:1964 (now withdrawn). It is also known as the Imperial Wire Gauge or British Standard Gauge. Use of SWG sizes has fallen greatly in popularity, but they are still used as a measure of thickness in guitar strings and some electrical wire. Cross sectional area in square millimetres is now the more usual size measurement for wires used in electrical installation cables. The current British Standard for metallic materials such as wire and sheet is BS 6722:1986, which is a solely metric standard.
SWG was fixed by Order of Council August 23, 1883. It was constructed by improving the Birmingham Wire Gauge. It was made a legal standard on March 1, 1884 by the British Board of Trade. SWG is not to be confused with American Wire Gauge, which has a similar but not interchangeable numbering scheme.
A table of the gauge numbers and wire diameters is shown below.[1][2] The basis of the system is the thou (or mil in US English), or 0.001 in. Sizes are specified as wire diameters, stated in thou and tenths of a thou (mils and tenths). The wire diameter diminishes with increasing size number. No. 7/0, the largest size, is 0.50 in. (500 thou or 12.7 mm) dia., No. 1 is 0.30 in. (300 thou), and the smallest, No. 50, is 0.001 in. (1 thou or about 25 µm).
The system as a whole approximates a (constant-ratio)[further explanation needed] exponential curve. The weight per unit length diminishes by an average of approximately 20% at each step. Because the weight per unit length is related to the cross sectional area, and therefore to the square of the diameter, the diameter diminishes by approximately 10.6%:
${\displaystyle {\mbox{Diameter Ratio}}=1-{\sqrt {1-0.2}}\approx 10.6\%}$
But the system is piecewise linear, only approximating the exponential curve loosely. Thus it runs in constant steps of 0.4 thou (0.4 mil) through the range No. 49 - No. 39 and 0.8 thou (0.8 mil) through No. 39 - No. 30.
British Standard Wire Gauge (SWG) diameters
SWG (in) (mm) Step
7/0 0.500 12.700 0.036"/gauge
6/0 0.464 11.786 0.032"/gauge
5/0 0.432 10.973
4/0 0.400 10.160 0.028"/gauge
3/0 0.372 9.449 0.024"/gauge
2/0 0.348 8.839
0 0.324 8.230
1 0.300 7.620
2 0.276 7.010
3 0.252 6.401 0.020"/gauge
4 0.232 5.893
5 0.212 5.385
6 0.192 4.877 0.016"/gauge
7 0.176 4.470
8 0.160 4.064
9 0.144 3.658
10 0.128 3.251 0.012"/gauge
11 0.116 2.946
12 0.104 2.642
13 0.092 2.337
14 0.080 2.032 0.008"/gauge
15 0.072 1.829
16 0.064 1.626
17 0.056 1.422
18 0.048 1.219
19 0.040 1.016 0.004"/gauge
20 0.036 0.914
21 0.032 0.813
22 0.028 0.711
23 0.024 0.610 0.002"/gauge
24 0.022 0.559
25 0.020 0.5080
26 0.018 0.4572 0.0016"/gauge
27 0.0164 0.4166
28 0.0148 0.3759 0.0012"/gauge
29 0.0136 0.3454
30 0.0124 0.3150 0.0008"/gauge
31 0.0116 0.2946
32 0.0108 0.2743
33 0.0100 0.2540
34 0.0092 0.2337
35 0.0084 0.2134
36 0.0076 0.1930
37 0.0068 0.1727
38 0.0060 0.1524
39 0.0052 0.1321 0.0004"/gauge
40 0.0048 0.1219
41 0.0044 0.1118
42 0.004 0.1016
43 0.0036 0.0914
44 0.0032 0.0813
45 0.0028 0.0711
46 0.0024 0.0610
47 0.0020 0.0508
48 0.0016 0.0406
49 0.0012 0.0305 0.0002"/gauge
50 0.0010 0.0254 | 1,296 | 3,273 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.933046 |
https://www.physicsforums.com/threads/how-to-integrate-this.794024/ | 1,657,170,489,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00672.warc.gz | 1,013,518,623 | 15,729 | # How to integrate this?
## Homework Statement
How to integrate the double integral cos(x+y) dy dx from 0 to pi and from 0 to pi again.
## The Attempt at a Solution
Here's the work:
u=x+y
du=dy
cos(u)du=sin(x+y)
integral of [sin(x+y)] evaluate from 0 to pi dx from 0 to pi
integral of (sin(x+pi)-sin(x))dx from 0 to pi
And what's next?
LCKurtz
Homework Helper
Gold Member
## Homework Statement
How to integrate the double integral cos(x+y) dy dx from 0 to pi and from 0 to pi again.
## The Attempt at a Solution
Here's the work:
u=x+y
du=dy
cos(u)du=sin(x+y)
integral of [sin(x+y)] evaluate from 0 to pi dx from 0 to pi
integral of (sin(x+pi)-sin(x))dx from 0 to pi
And what's next?
Integrate again, similar to what you did for the first integral.
Er...doesn't the final step that you list give you the answer already?
But I can't. Because sin(x+pi)-sin(x)=sin(x)+sin(pi)-sin(x)=sin(pi)=0. The integral of 0 is?
LCKurtz | 290 | 932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-27 | latest | en | 0.903346 |
http://math.stackexchange.com/questions/180052/question-on-the-relation-between-sheaves-over-an-object-and-sheaves-on-a-categor | 1,467,263,985,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398075.47/warc/CC-MAIN-20160624154958-00112-ip-10-164-35-72.ec2.internal.warc.gz | 194,564,260 | 18,142 | # Question on the relation between sheaves over an object and sheaves on a category over that object
The Proposition appearing in the wonderful answer of Zhen Lin to this other question states that for a small category $\mathbb{C}$ and an object $X\in \mathbb{C}$
(*)$$\widehat{\mathbb{C}}\downarrow Y(X) \cong \widehat{\mathbb{C}\downarrow X}$$
i.e. the category of presheaves on $\mathbb{C}$ over the Yoneda embedded $X$ is equivalent (even isomorphic?) to the category of presheaves on $\mathbb{C}\downarrow X$. This appears also in Sheaves in Geometry and Logic by MacLane and Moerdijk, Exercise III.8.
There is an obvious functor $(-)|_X:\widehat{\mathbb{C}}\to \widehat{\mathbb{C}\downarrow X}$ by precomposing a presheaf $\mathbb{C}^{op}\to Set$ with the opposite of the forgetful functor $\mathbb{C}\downarrow X\to \mathbb{C}$.
• Can this functor $(-)|_X$ be identified with the isomorphism $F$ from the left to the right in (*)? If not, how do $F$ and $(-)|_X$ relate?
If you start with an object $p:Z\to Y(X)$ of $\widehat{\mathbb{C}}\downarrow Y(X)$ and apply $(-)|_X$, you get $Z|_X\to Y(X)_X$ but $Y(X)_X=Hom_{\mathbb{C}\downarrow X}(-,X)$ is the terminal presheaf on $\mathbb{C}\downarrow X$. Hence it seems to me that the structure morphism $p$ is not respected by $(-)|_X$, we only get $Z|_X$.
On the other hand, the last page of chapter VII of the above mentioned book (or Exercise III.8.b) states the equivalence (*) for sheaves where $\mathbb{C}\downarrow X$ carries the obvious Grothendieck topology induced from $\mathbb{C}$. Doesn't this somehow suggest that the functor $(-)_X$ plays a role in the equivalence?
-
Good observation. The answer is that $(-)|_X$ does induce the equivalence of categories, but in a subtle way.
Let $\mathbb{D} = (\mathbb{C} \downarrow X)$ and let $u = (-)|_X$. First of all, there is an induced functor $u^* : \hat{\mathbb{C}} \to \hat{\mathbb{D}}$ defined by precomposing a presheaf on $\mathbb{C}$ with the functor $u$. Let $u_! : \hat{\mathbb{D}} \to \hat{\mathbb{C}}$ be the left Kan extension. Given a presheaf $F$ on $\mathbb{D}$, $u_! F$ is defined to be the colimit in $\hat{\mathbb{C}}$ of a canonical diagram of shape $\int^{\mathbb{D}} P$, and one may verify that in fact we have the explicit formula $$u_! F (C) = \{ (f, z) : f \in \mathbb{C}(C, X), z \in F (f) \}$$ In particular, if $F$ is the terminal object in $\hat{\mathbb{D}}$, $u_! F$ is isomorphic to $H_X = \mathbb{C}(-, X)$. Thus $u_!$ lifts to a functor $\hat{\mathbb{D}} \cong (\hat{\mathbb{D}} \downarrow 1) \to (\hat{\mathbb{C}} \downarrow H_X)$, and this is the required equivalence of categories.
Wow, that's tricky. So there is a composition of two left adjoints $u_{!}=forget\circ u'_{!}:\widehat{\mathbb{D}}\to (\widehat{\mathbb{C}}\downarrow X)\to \widehat{\mathbb{C}}$ and I guess that either $forget$ and $u'_{!}$ do have right adjoints. For $forget$, I would suppose $(-\times X)$ (I ingonre the Yoneda embedding in the notation) and for $u'_{!}$ it's the one you describe in Proposition to the cited answer, let's denote it by $v$. But does this mean that $u^{*}$ is the composition $v\circ (-\times X)$? All very confusing... – geometrystudent Aug 8 '12 at 15:55 | 1,045 | 3,211 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-26 | latest | en | 0.755136 |
https://economics.stackexchange.com/questions/9560/why-are-imports-subtracted-from-gdp/12974 | 1,723,322,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00177.warc.gz | 168,132,449 | 44,195 | Why are imports subtracted from GDP?
GDP is a measure of a country's production.
$$GDP = C + I + G + X_n$$
$C$ = Consumer Consumption
$I$ = Gross Investment
$G$ = Government Expenditures
$X_n$ = Exports - Imports
Exports are what we produce and make a profit from by selling to buyers outside our country. Imports are not produced by our country, so it shouldn't be included in the GDP, so it makes sense to exclude it from the calculation; ie. there should be no "- imports" in the calculation.
However, the calculation subtracts imports from the GDP. Imports somehow take away from what we produced? That seems to say, calculate how much I have produced, X, and then don't count some of it because I imported Y. Which doesn't make sense given that importing doesn't remove goods and services that have already been produced!
For example, let's say I can take apples and make pies... I produced value in the form of the "pie" quality. Importing the "pie" quality and tacking it on an apple creates an apple pie. However, I didn't make that "pie" so I can understand how it doesn't get included in the GDP: the value of the "pie" quality was not produced in this country, so it isn't included in the GDP.
Let's also say my GDP = "made a cat meow" + "turned a tree into a fountain". Somehow, by importing the "pie" quality, I am to ignore some of the value of making a cat meow?
When we import something, we consume it. So when calculating consumption we are bound to count import as a positive component of GDP. Since it is not (we did not produce imports domestically), we subtract it to make it neutral.
• You're spot on. The accounting identity (exports-imports) is done because C+I+G includes domestic consumption on foreign goods and GDP is the value of domestic production. Commented Dec 3, 2015 at 10:29
One way to navigate rather safely in National Accounting Identities, is to put on the one side "what you have available" and on the other side "what you do with it". In the specific case ($M$ being imports) we would have
$$GDP + M = C + I + G + X$$
"What we have available" is what we produce and what we import.
"What we do with it" is a) consume it b) invest it c) let the government incur expenses and d) export it (note that we can export not only what we produce but also what we have imported).
I believe it is all reasonable. Then, rearranging,
$$GDP = C + I + G + (X-M)$$
the term in the parenthesis being "net exports".
GDP means gross domestic product. It the sum of all services and goods produced within the boundary of nation. Since imports are produced in other countries so we donor include imports in our GDP I mean to say that we pay for it... So we take a negative sign with it
Net exports means total exports-total imports. Export represents domestic production selling to another country. That's why it is included in GDP (as GDP means the total market value of all final goods and services produced in a country within a given period). Import is subtracted because it's the production of a foreign country purchased by domestic country.
• This is the same answer than the accepted one. Commented Aug 27, 2017 at 8:48
This is the expenditure method. It includes all the expenditures in the economy. Indeed, when we import goods, we spend to consume. Even if you exclude imports from calculations, you will not exclude its part from consumption. Thus, you need to substract it. | 797 | 3,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.966335 |
https://socratic.org/questions/the-sum-of-the-measures-of-the-interior-angles-of-a-polygon-is-720-what-type-of- | 1,638,597,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00603.warc.gz | 560,950,420 | 6,807 | # The sum of the measures of the interior angles of a polygon is 720°. What type of polygon is it?
Jul 18, 2018
Hexagon
#### Explanation:
Let's do this two ways: Using a pattern we know and using exterior angles.
Pattern
For a triangle, the sum of interior angles is 180 degrees.
For a quadrilateral, the sum of interior angles is 360 degrees.
We know that there is some pattern, so we can deduce quite logically that a pentagon has a sum of 540 degrees and a hexagon has a sum of 720 degrees, hence the answer is a hexagon.
Exterior Angles
The real question is how do we find that there is a pattern? We can recall that the sum of all exterior angles in any polygon is 360 degrees. If you don't remember learning this, you can imagine driving along the perimeter of any polygon and seeing that, in the end, you've taken a single turn.
Anyway, in a normal polygon with $n$ sides, each of these exterior angles is, therefore, ${360}^{\setminus} \frac{\circ}{n}$. Since an external angle plus an interior angle is a line, we know that an interior angle will have measure ${180}^{\setminus} \circ - {360}^{\circ} / n$.
With $n$ of those angles, we get the total internal angle of
${\left(180 n - 360\right)}^{\circ}$. Setting this equal to ${720}^{\circ}$, we easily find $n = 6$.
Hexagon
#### Explanation:
A polygon of $n$ number of sides can be divided into $n - 2$ number of triangles.
Let the polygon have $n$ number of sides then the sum of its interior angles will be equal to the sum of interior angles of $n - 2$ triangles
$= \left(n - 2\right) {180}^{\setminus} \circ$
${720}^{\setminus} \circ = \left(n - 2\right) {180}^{\setminus} \circ$
$n - 2 = \frac{720}{180}$
$n - 2 = 4$
$n = 6$
Hence the polygon has $6$ sides hence it's a hexagon | 497 | 1,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2021-49 | latest | en | 0.860588 |
https://www.lmfdb.org/L/ModularForm/GL2/Q/holomorphic/13/2/e/a/4/1/ | 1,576,130,381,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00403.warc.gz | 775,695,324 | 7,183 | # Properties
Degree 2 Conductor 13 Sign $0.964 + 0.265i$ Motivic weight 1 Primitive yes Self-dual no Analytic rank 0
# Related objects
## Dirichlet series
L(s) = 1 + (−1.5 − 0.866i)2-s + (−1 + 1.73i)3-s + (0.5 + 0.866i)4-s − 1.73i·5-s + (3 − 1.73i)6-s + 1.73i·8-s + (−0.499 − 0.866i)9-s + (−1.49 + 2.59i)10-s − 2·12-s + (−2.5 − 2.59i)13-s + (2.99 + 1.73i)15-s + (2.49 − 4.33i)16-s + (1.5 + 2.59i)17-s + 1.73i·18-s + (−3 + 1.73i)19-s + (1.50 − 0.866i)20-s + ⋯
L(s) = 1 + (−1.06 − 0.612i)2-s + (−0.577 + 0.999i)3-s + (0.250 + 0.433i)4-s − 0.774i·5-s + (1.22 − 0.707i)6-s + 0.612i·8-s + (−0.166 − 0.288i)9-s + (−0.474 + 0.821i)10-s − 0.577·12-s + (−0.693 − 0.720i)13-s + (0.774 + 0.447i)15-s + (0.624 − 1.08i)16-s + (0.363 + 0.630i)17-s + 0.408i·18-s + (−0.688 + 0.397i)19-s + (0.335 − 0.193i)20-s + ⋯
## Functional equation
\begin{aligned} \Lambda(s)=\mathstrut & 13 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & (0.964 + 0.265i)\, \overline{\Lambda}(2-s) \end{aligned}
\begin{aligned} \Lambda(s)=\mathstrut & 13 ^{s/2} \, \Gamma_{\C}(s+1/2) \, L(s)\cr =\mathstrut & (0.964 + 0.265i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
$$d$$ = $$2$$ $$N$$ = $$13$$ $$\varepsilon$$ = $0.964 + 0.265i$ motivic weight = $$1$$ character : $\chi_{13} (4, \cdot )$ primitive : yes self-dual : no analytic rank = 0 Selberg data = $(2,\ 13,\ (\ :1/2),\ 0.964 + 0.265i)$ $L(1)$ $\approx$ $0.298115 - 0.0402203i$ $L(\frac12)$ $\approx$ $0.298115 - 0.0402203i$ $L(\frac{3}{2})$ not available $L(1)$ not available
## Euler product
$L(s) = \prod_{p \text{ prime}} F_p(p^{-s})^{-1}$ where, for $p \neq 13$, $$F_p$$ is a polynomial of degree 2. If $p = 13$, then $F_p$ is a polynomial of degree at most 1.
$p$$F_p$
bad13 $$1 + (2.5 + 2.59i)T$$
good2 $$1 + (1.5 + 0.866i)T + (1 + 1.73i)T^{2}$$
3 $$1 + (1 - 1.73i)T + (-1.5 - 2.59i)T^{2}$$
5 $$1 + 1.73iT - 5T^{2}$$
7 $$1 + (3.5 - 6.06i)T^{2}$$
11 $$1 + (5.5 + 9.52i)T^{2}$$
17 $$1 + (-1.5 - 2.59i)T + (-8.5 + 14.7i)T^{2}$$
19 $$1 + (3 - 1.73i)T + (9.5 - 16.4i)T^{2}$$
23 $$1 + (-3 + 5.19i)T + (-11.5 - 19.9i)T^{2}$$
29 $$1 + (1.5 - 2.59i)T + (-14.5 - 25.1i)T^{2}$$
31 $$1 - 3.46iT - 31T^{2}$$
37 $$1 + (-7.5 - 4.33i)T + (18.5 + 32.0i)T^{2}$$
41 $$1 + (4.5 + 2.59i)T + (20.5 + 35.5i)T^{2}$$
43 $$1 + (4 + 6.92i)T + (-21.5 + 37.2i)T^{2}$$
47 $$1 + 3.46iT - 47T^{2}$$
53 $$1 + 3T + 53T^{2}$$
59 $$1 + (-6 + 3.46i)T + (29.5 - 51.0i)T^{2}$$
61 $$1 + (0.5 + 0.866i)T + (-30.5 + 52.8i)T^{2}$$
67 $$1 + (-3 - 1.73i)T + (33.5 + 58.0i)T^{2}$$
71 $$1 + (-3 + 1.73i)T + (35.5 - 61.4i)T^{2}$$
73 $$1 - 1.73iT - 73T^{2}$$
79 $$1 - 4T + 79T^{2}$$
83 $$1 - 13.8iT - 83T^{2}$$
89 $$1 + (6 + 3.46i)T + (44.5 + 77.0i)T^{2}$$
97 $$1 + (-6 + 3.46i)T + (48.5 - 84.0i)T^{2}$$
\begin{aligned} L(s) = \prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1} \end{aligned} | 1,590 | 2,802 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-51 | latest | en | 0.439452 |
https://www.coursehero.com/file/9349209/Regression-Analysis-Practice-Questions/ | 1,542,223,141,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114212809-00057.warc.gz | 837,404,984 | 90,557 | Regression Analysis Practice Questions
# Regression Analysis Practice Questions - ECO 221 Spring...
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ECO 221, Spring 2014 Rollins College Prof. M. Vidovic Practice Questions – Least Squares Regression 1. A consumer organization has reported test data for 50 car models. We will examine the associations between the x = weight of the car (in thousands of pounds) and the y = fuel efficiency (in miles per gallon). a. Does a scatterplot of the data suggest a linear relationship between x and y ? b. Calculate the slope and intercept of the least squares line and write out the equation of the regression line. c. What is the predicted fuel efficiency when vehicle weight is 3 thousand pounds? d. Compute the correlation coefficient and interpret. e. Compute the coefficient of determination and interpret. f. Compute the first 3 residuals. g. What is the value of s e ? Interpret the s e . h. Below is the residual plot. Are there any features of the plot that indicate that a line is not an appropriate description of the relationship? -5 0 5 Residuals 2 2.5 3 3.5 4 Weight 2.
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https://oeis.org/A215892 | 1,652,932,215,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00659.warc.gz | 498,824,198 | 4,164 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A215892 a(n) = 2^n - n^k, where k is the largest integer such that 2^n >= n^k. 2
0, 5, 0, 7, 28, 79, 192, 431, 24, 717, 2368, 5995, 13640, 29393, 0, 47551, 157168, 393967, 888576, 1902671, 3960048, 1952265, 8814592, 23788807, 55227488, 119868821, 251225088, 516359763, 344741824, 1259979967, 3221225472, 7298466623, 15635064768 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 2..1000 FORMULA a(n) = 2^n - n^floor(n*log_n(2)), where log_n is the base-n logarithm. EXAMPLE a(2) = 2^2 - 2^2 = 0, a(3) = 2^3 - 3 = 5, a(4) = 2^4 - 4^2 = 0, a(5) = 2^5 - 5^2 = 7, a(6)..a(9) are 2^n - n^2, a(10)..a(15) are 2^n - n^3, a(16)..a(22) are 2^n - n^4, and so on. MATHEMATICA Table[2^n - n^Floor[n*Log[n, 2]], {n, 2, 35}] (* T. D. Noe, Aug 27 2012 *) PROG (Python) for n in range(2, 100): a = 2**n k = 0 while n**(k+1) <= a: k += 1 print(a - n**k, end=', ') (MAGMA) [2^n - n^Floor(n*Log(n, 2)): n in [2..40]]; // Vincenzo Librandi, Jan 14 2019 CROSSREFS Cf. A000325, A024012, A024013, A024014, A024015, A024016. Cf. A024017, A024018, A024019, A024020, A024021, A024022. Cf. A060508. Sequence in context: A201417 A147666 A343071 * A200643 A200231 A124914 Adjacent sequences: A215889 A215890 A215891 * A215893 A215894 A215895 KEYWORD nonn AUTHOR Alex Ratushnyak, Aug 25 2012 STATUS approved
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Last modified May 18 23:31 EDT 2022. Contains 353826 sequences. (Running on oeis4.) | 766 | 1,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.517889 |
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x | 2,634 | 10,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.923279 |
https://tsfa.co/how-to-find-slope-intercept-equation-26 | 1,680,379,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00442.warc.gz | 656,945,958 | 5,382 | # How to find slope intercept equation
We can write the formula for the slope-intercept form of the equation of line L whose slope is m and x-intercept d as: y = m(x – d) Here, m = Slope of the line. d = x-intercept of the line.
Decide mathematic problems
Explain mathematic equations
Solve word questions too
## Identify the Slope and Intercept of a Line in Slope-Intercept Form
Find the slope-intercept equation of a line that passes through the point ( − 6, 2) and has a slope of 5. First, write out the general form of a slope-intercept equation. y = m x + b
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Explain mathematic problems
Math can be a difficult subject for many people, but it doesn't have to be! By taking the time to explain the problem and break it down into smaller pieces, anyone can learn to solve math problems.
## Slope Intercept Form Y=mx+b
Writing equations from -intercept and another point. Let's write the equation of the line that passes through the points and in slope-intercept form. Recall that
## Using Two Points to Write an Equation
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## Slope-Intercept and Point-Slope Forms of a Linear Equation
{{@N-H2TEXT@}}
No matter what math task you're trying to solve, there are always a few basic steps you can follow to help you figure it out. | 349 | 1,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-14 | latest | en | 0.928332 |
https://www.physicsforums.com/threads/convolution-with-unit-step.773387/ | 1,511,465,272,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806856.86/warc/CC-MAIN-20171123180631-20171123200631-00533.warc.gz | 843,480,419 | 16,482 | # Convolution with unit step
1. Sep 28, 2014
### redundant6939
1. The problem statement, all variables and given/known data
Find convolution of x[n] (graph in attachment) and h[n] where h[n] = u[n]
2. Relevant equations
3. The attempt at a solution
- flipped the h[n] to have h[-n]
- moved to the left once (h[-1-n]) to align
- multiplied h and x and it gives me all zeros
Is this correct or I'm missing something?
2. Sep 29, 2014
### rude man
I don't know about your graphical algorithm but if you use the definition
y[n] = x[n]*h[n] = ∑ (k = -∞ to +∞) x[k] h[n-k]
then I get a non-zero sequence of numbers:
a+0.5, b+0.5, c+0.5, c+0.5, d+0.5, d+0.5, d+ 0.5, ...
wher a, b , c, and d are positive integers for you to find.
3. Sep 29, 2014
4. Oct 1, 2014 | 272 | 768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-47 | longest | en | 0.871465 |
https://examsumo.com/jee-main-vs-jee-advanced-vs-bitsat-exam-comparision/ | 1,604,142,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00107.warc.gz | 310,694,018 | 19,592 | ## BITSAT: Fierce Competition vs Easy Paper
JEE Mains vs JEE Advanced vs BITSAT
Often students ask the difference between JEE Mains and BITSAT Some of them ask the difference between IIT JEE and BITSAT. Another similar question is Which is harder JEE Mains or BITSAT.
You will find answers to all such question here. I am going to discuss the statistics and also the view of the students who have taken the exam.
## Number of Seats and Colleges in BITSAT, NIT, IIIT, CFTI and IIT
BISAT 3 2,200 IIT 23 12,079 NIT 31 17,390 IIIT 23 2,228 GFTI 23 3,741
As you can see clearly number of seats in BITS are very less than other institutes. Looking at this data will give you an idea about the competition between BITSAT and other Colleges.
## Number of your competitor per seat for BITSAT and other Colleges
Total Number Of Seats Total Number of Applicatns Number of your Competitor Per Seat BITSAT 2000 200000 100 NIT+IIIT+GFTI 18620 + 4023 + 4703 = 1000000 36 IIT 12079 231000 19
In the above table, you can see that a greater number of students compete against you for a single seat in BITSAT. The competition for a seat in NIT’s, IIIT’s, GFTI’s is one-third of BITSAT. And most surprising competition for IIT’s is one-fifth.
## Number Of Question in BITSAT, JEE MAINS, JEE ADVANCED and the time for every question
Total Number Of Question Duration Time for every Question BTSAT 150 3 hour = 180 1.2 minute = 72 sec JEE Mains 90 3 hour = 180 2 minute = 120 sec JEE Advanced 54*2=108 (Paper 1+2 ) 6 hour = 360 3.33 minute = 200 sec
You get a shorter time to solve a BITSAT question than a problem in JEE Mains Paper or JEE Advanced. And it is such a big difference to ignore. You will get almost double time in Mains and triple time in JEE Advanced than in BITSAT to solve a problem. From this perspective, BITSAT seems more competitive than any other Engineering Entrance Exam.
## Reservation in BITSAT, JEE Mains and JEE Advanced
There is no reservation available for any category in BITSAT. Students from all category are considered to be the same in this exam. Each student will have to do better from other students of all the category. This means there will be more competition for every seat of BITS.
## Comparison of Cut-Off
This part is essential to understand. In BITSAT 2018 almost 29000 students scored above 225. But in these marks, you will get no branch in any of the three Colleges of BITS. While if you get such marks in JEE Mains, you will quickly get a rank under 2000. This is also true for JEE Advanced.
Therefore you can conclude that BITSAT has the highest cut off among all the three exams.
## Level Of Questions
Now here is the catch. A lot of students who have qualified JEE Mains & Advanced and also BITSAT feel that BITSAT is a lot easier.
Questions in BITSAT are so comfortable that they are mostly about speed.
Most of the questions in BITSAT are usually straightforward and need the direct application of formulae or concepts.
Here is a direct comparison between the questions of JEE Mains and BITSAT.
Indefinite Integration Question Of BITSAT
Indefinite Integration Question Of JEE Mains Differential Equation Question Of BITSAT
Differential Equation Question Of Mains
As you can see questions in BITSAT are pretty easy than JEE Mains.
In JEE Mains you will have time to think about all the possible solutions. Then you will apply the shortest possible method.
But questions will have only one straightforward solution. And if you can think about it apply it at once. You can not think about the different method of the question for more than 30 seconds.
## English Proficiency and Mental Ability Question
There will be 15 question from English Proficiency and ten questions from Mental ability. Questions from English can be difficult for some students who are weak in English, which is not right for most of you.
Start reading English newspaper from today, and you will be able to do at least 8-10 questions.
On the other hand you can buy this book to cover English Proficiency and Mental Ability.
Questions from Mental ability are also easy to medium level. In the right mindset, you can attempt these question under 10 minutes.
## Verdict
Question Difficulty: Easy
Competition: Very High
Questions in BITSAT are easier than JEE Mains and IIT JEE. You need to remember every formula in your fingertips. Your speed is vital in BITSAT. You can not think twice for a question.
But competition in BITSAT is remarkably higher than JEE Mains even higher from IIT JEE Advanced.
### One comment
• Varalaksmi
Very informative and good analysis. Thank you sooooo much sir.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,136 | 4,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-45 | latest | en | 0.890211 |
https://xcas.univ-grenoble-alpes.fr/forum/viewtopic.php?p=9750 | 1,603,449,103,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00480.warc.gz | 955,752,954 | 7,524 | ## Getting the monomials of a polynomial
jocaps
Messages : 118
Inscription : lun. avr. 17, 2017 4:32 pm
### Getting the monomials of a polynomial
Hi,
What is the easiest (and not very slow) way to get a list of monomials of a polynomial. For instance if I have 3x*y+3x^2+y
I want the list [3x*y,3x^2,y]. I can write a code that takes the total degree and walks through all possible monomial not exceeding this degree, but this will not be efficient. I can also convert the polynomial into string and then use python string manipulators that delimit all "+" and "-" and get the monomials, but I think giac has something more efficient.
Edit: I think I found a way that i think would be relatively fast.
Code : Tout sélectionner
``````from giacpy import giac,part
f=giac("3x*y+3*x**2+y")
monomials=[]
for i in xrange(len(f)):
monomials.append(part(f,i+1))
``````
This will yield [3*x*y,3x**2,y]
Jose
frederic han
Messages : 1113
Inscription : dim. mai 20, 2007 7:09 am
Localisation : Paris
Contact :
### Re: Getting the monomials of a polynomial
In giac your formula is a tree, you want the operands when sommet is +
but with symbolic coefficients it won't work.
related tools but different:
coeff(P,[x,y,z],[1,2,3]) to have the coeff of x^1*y^2*z^3 in P
or symb2poly(P,x,y,z)
Code : Tout sélectionner
``````from giacpy import *
x,y,z=giac('x,y,z')
P=(x+2*y-z+'1/2')**55
from time import time
Q=(P.ratnormal()) # here sommet is * (numerator/denominator)
t=time()
L=(Q.op(0)).op() # a giac list
L0=list(L) # a python list of Pygen elements
print(time()-t)
#A=((sum(L)/Q).simplify()) # sum is long
monomials=[]
t=time()
for i in xrange(len(Q)):
monomials.append(part(part(Q,1),i+1)) # here I need the numerator
print(time()-t)
``````
jocaps
Messages : 118
Inscription : lun. avr. 17, 2017 4:32 pm
### Re: Getting the monomials of a polynomial
I know this is an old thread, but I want to point out on something in the last answer of frederic. The proposed method does not necessarily give me the correct list. Consider the following for instance
Code : Tout sélectionner
``````from giacpy import giac
x,y,z=giac('x,y,z')
P=x**2 + x +y
Q=(P.ratnormal()) # here sommet is * (numerator/denominator)
L=(Q.op(0)).op() # a giac list
L0=list(L) # a python list of Pygen elements
print L0 #this gives me [x,2] , it should give me [x**2,x,y]
``````
frederic han
Messages : 1113
Inscription : dim. mai 20, 2007 7:09 am
Localisation : Paris
Contact :
### Re: Getting the monomials of a polynomial
The problem is that ratnormal can output a symbol with sommet * or +
So if the sommet is + you just have to do Q.op()
Code : Tout sélectionner
``````>>> P=x**2+3*x+y
>>> Q=P.ratnormal()
>>> Q.sommet()
'+'
>>> Q.op()
x**2,3*x,y
>>> P2=x**2+3*x/2+y # warning 3/2*x is x in python2 but '3/2'*x or 3*x/2 are OK
>>> Q2=P2.ratnormal()
>>> Q2
(2*x**2+3*x+2*y)/2
>>> Q2.op(-1)
'*'
>>> Q2.op(0).op()
2*x**2,3*x,2*y
>>> Q3=P2.expand() # expand is slower but output a sommet +
>>> Q3
x**2+3/2*x+y
>>> Q3.op(-1)
'+'
>>> Q3.op()
x**2,3/2*x,y
``````
jocaps
Messages : 118
Inscription : lun. avr. 17, 2017 4:32 pm
### Re: Getting the monomials of a polynomial
Thanks fréderic. Though it seems to be a lot of work just for a list of monomials. I guess I will use it if the number of such monomials are rather long otherwise it will not be too expensive to just use "part". | 1,144 | 3,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-45 | latest | en | 0.633108 |
https://www.r-bloggers.com/world-cup-2010-statistics-plotted-with-r/ | 1,590,924,513,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00004.warc.gz | 875,371,540 | 51,503 | # World Cup 2010 Statistics Plotted with R
July 11, 2010
By
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Opta agreed to let the UK Guardian Data Blog publish 2010 World Cup Team and Player statistics. The data is available in a Google Docs spreadsheet. There are two tabs on this spreadsheet – one is PLAYERS the other is TEAM statistics. I chose File -> Download As -> CSV and downloaded the files through a web browser, then moved them to my working R directory. I named the first World Cup 2010 data.csv (Player Data) and the second World Cup 2010 TEAM data.csv (Team Data).
By the way, if anyone knows how individual Google Docs spreadsheets can be downloaded as CSVs via URLs, please let me know by commenting on this post. I could not figure out how to do this straight from R by reading a URL (which is my preference).
The following are a few charts that can be created with the data. You might also want to do more sophisticated predictive analysis, by I will leave that to Paul.
The sheet with player data can be read in as a CSV
The following attributes are available for each player.
names(DF)
Player.Surname
Team
Position
Time.Played
Total.Shots.Attempted
Total.Passes
Tackles.Attempted
The base graphics package can be used to produce the following chart of the USA team’s shots attempted by player.
# Create a smaller data frame that
# contains only USA player names
# and shots attempted.
PS=DF[DF\$Team==’USA’,c(‘Player.Surname’,’Total.Shots.Attempted’)]
# Make the player Names the rownames
rownames(PS)=PS[,1]
PS=PS[-1]
# Flip the X axis labels and provide enough room in the margins to print the names
par(las=2,mar=c(8, 4, 1, 2) + 0.1)
# Pivot the table, print the barplot and add a title
barplot(t(PS))
title(‘2010 World Cup USA Total Shots Attempted’)
Now an example with the Team data. In this case, the column names are actually the names of the countries.
names(DF2)
The attributes about each team are available in the first column.
DF2[1]
Games Played
Goals
Ave Goals per game
Shots (excl blocked shots)
% Shots on Target
% Goals to Shots
Overall Pass Completion %
Cross Completion %
Goals Conceded
Ave goals conceded per game
Tackles Won %
Fouls
Yellow Cards
Red Cards
I prefer these attributes as row names – so moved them there using the following:
rownames(DF2)=DF2[,1]
DF2=DF2[,-1]
This time, we will use qqplot and create a horizontal barchart that includes a gradient that increases to highlight the countries with the most fouls. I think you will agree – qqplot produces much better results. The author of the (Hadley Wickham) just released a new version of this package. He also has written a book on it which goes into greater depth about its use and design (based upon Leland Wilkison’s Grammar of Graphics). The example that follows uses the simpler qplot call, the team names as the x axis, and the number of fouls as the y axis. The “Geometry” specified indicates that we are using a bar chart, and we specify coord_flip to switch the x and y axis.
library(qqplot2)
FOULS=t(DF2)[,c(‘Fouls’)]
qplot(names(FOULS), as.numeric(FOULS), geom=”bar”, stat=’identity’, fill=Fouls) + xlab(‘Country’) + ylab(‘Fouls’) + coord_flip() + scale_fill_continuous(low=”black”, high=”red”) + labs(fill=’Fouls’)
When t was used to pivot the data.frame, it changed it to a matrix and the type of the numeric values became character. The as.numeric function was used to cast it back.
The following is the same type of plot for Goals. This chart also includes a title. It appears at the top of this post.
GOALS=t(DF2)[,c(‘Goals’)]
qplot(names(GOALS), as.numeric(GOALS), geom=”bar”, stat=’identity’, fill=as.numeric(GOALS)) + xlab(‘Country’) + ylab(‘Goals’) + coord_flip() + scale_fill_continuous(low=”yellow”, high=”blue”) + labs(fill=’Goals’) + opts(title = “2010 World Cup Goals (as of 07/10/2010)”)
Hope you enjoyed this little excursion.
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 1,132 | 4,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.862694 |
https://www.physicsforums.com/threads/vector-addition-calculate-magnitude-of-sum.433503/ | 1,725,811,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00309.warc.gz | 889,886,270 | 17,962 | Vector Addition: Calculate Magnitude of Sum
• joej24
In summary, when a vector of magnitude 20 is added to a vector of magnitude 25, the magnitude of the resulting vector can range from 5 to 45. The correct answer is 12.
joej24
Homework Statement
A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude of this sum might be:
a. zero
b. 3
c. 12
d. 47
e. 50
c ^2 = a^2 + b^2
The Attempt at a Solution
I was not entirely sure on what to do. I tried using the Pythagorean theorem to find the resultant of the 2 vectors and got 32. ( c ^2 = 20^2 + 25^2).
The answer by the way is C.
Thanks!
Joe
Try do add the vectors in different ways. Then you will get a possible maximal and a minimal value for the magnitude of the resulting vector. All but one possible answers are outside of this region.
okay. I got it. The max magnitude of their addition would be when they are parallel in the same direction. mag would = 45. The min would be when they are parallel in the opposite direction. mag would = 5. Therefore the answer is 12. Thanks!
Good.
I would like to clarify and explain the concept of vector addition and how to calculate the magnitude of the sum. Vector addition is the process of combining two or more vectors to obtain a single vector, known as the resultant. The magnitude of the resultant is calculated using the Pythagorean theorem, as you correctly attempted. However, it is important to note that the magnitude of the resultant is not simply the sum of the magnitudes of the individual vectors.
In this case, the magnitude of the sum would be the square root of (20^2 + 25^2) = 32.0156. This means that the answer is not exactly 12, but rather a value close to it. Therefore, the correct answer is option C.
It is also important to mention that the direction of the resultant vector is determined by the direction of the individual vectors being added. This can be calculated using trigonometric functions such as sine and cosine.
In conclusion, when adding vectors, it is crucial to understand the concept of vector addition and how to calculate the magnitude and direction of the resultant. I hope this explanation helps in understanding the solution to this problem.
What is vector addition and why is it important?
Vector addition is the process of combining two or more vectors to determine the magnitude and direction of their sum. It is important because it allows us to accurately calculate the resulting motion or force when multiple vectors are acting on an object.
What is the formula for calculating the magnitude of vector addition?
The formula for calculating the magnitude of vector addition is the Pythagorean theorem, where the magnitude is equal to the square root of the sum of the squares of the individual vector components.
How do you represent vectors in a coordinate system?
Vectors can be represented in a coordinate system by using arrows. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector.
Vector addition involves combining both the magnitude and direction of vectors, while scalar addition only involves adding the numerical values of scalars. Vectors also follow the laws of vector addition, such as the commutative and associative properties, while scalars do not.
What are some real-world applications of vector addition?
Vector addition is used in various fields, such as physics, engineering, and navigation. It can be used to calculate the net force on an object, determine the resultant velocity of a moving object, and plot the path of a moving object. It is also used in navigation systems to determine the direction and speed of an object's movement.
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2K | 898 | 3,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.950467 |
https://www.physicsforums.com/threads/how-can-you-determine-the-normal-force.262477/ | 1,542,574,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744649.77/warc/CC-MAIN-20181118201101-20181118223101-00446.warc.gz | 979,454,987 | 13,783 | # Homework Help: How can you determine the normal force?
1. Oct 7, 2008
### davidelete
1. The problem statement, all variables and given/known data
The diagram below shows the forces acting on a block moving at a constant velocity down an inclined plane. The weight of the block, Fw, can be resoled into two components, one parallel to the surface of the plane, Fll, and one perpendicular to the surface of the plane, F_l_. If you know the weight of the block and the value of angle θ when the block moves at constant velocity, you can calculate the coefficient of sliding friction. You can also use a computer to do the calculations for you.
http://img352.imageshack.us/img352/5617/physicstrianglethingac9.png [Broken]
1. What are the two forces the values of which must be known before the coefficient of sliding friction can be calculated?
2. What data must you have in order to determine these two forces?
3. How can you determine the normal force?
4. How can you determine the force of friction?
5. What equation is used to calculate the coefficient of sliding friction?
6. In the space below, draw a flowchart to calculate the coefficient of sliding friction.
7. If you have access to a computer, follow your flowchart and write a computer program to calculate the coefficient of sliding friction. The measure of angle θ must be expressed in radians. Let Ff represent the force of friction, and FN represent the normal force. Attach a printout of your program to this page.
8. Choose a weight between 25 N and 74 N and an angle θ between 15° and 30°. Enter these data and run your program. Attach a printout of your results to this page.
9. After you have run several different values for θ, try to determine a relationship between the coefficient of friction and trig functions of angle θ.
2. Relevant equations
N/A as of now. I first have to figure out the first few before I get into any sort of math.
3. The attempt at a solution
1. Frictional Force (Ff)and Normal Force (FN)
2. The weight of the box, gravity, and the degree of θ.
3. Multiply the mass by negative gravity as in FN.
My teacher said something about the cosine of θ, but I don't really know how to implement this, being that this is the first day that we are going over vectors and she bombarded us with this.
4. The force of friction can be determined by normal force as in Ff=μ* FN
5. Ff=μ* FN or μ=Ff/FN
6. I have no idea. My teacher just said to put our problem solving steps down for this one. First I need some problem solving steps...
7. She said to use Microsoft Excel for this purpose. As soon as I figure out the first five, this shouldn't be too hard. Any guidance would help, of course.
8. ???
9. ???
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited by a moderator: May 3, 2017
2. Oct 7, 2008
### barato
Hmmmmm. Ill try to work it out, I am also just beginning physics. Ill tell you what I got.
3. Oct 7, 2008
Help!!!
4. Oct 7, 2008 | 792 | 3,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-47 | latest | en | 0.905141 |
http://www.idsia.ch/~juergen/toesv2/node23.html | 1,410,852,276,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657114105.77/warc/CC-MAIN-20140914011154-00211-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 570,561,973 | 5,563 | Theorems for EOMs and GTMs
Next: Tighter Bounds? Up: Probability vs Descriptive Complexity Previous: Probability vs Descriptive Complexity
## Theorems for EOMs and GTMs
Theorem 5.3 For with PE(x) > 0,
(38)
Using for y interpreted as an integer -- compare Def. 2.8 -- this yields
(39)
That is, objects that are hard to describe (in the sense that they have only long enumerating descriptions) have low probability.
Proof. The left-hand inequality follows by definition. To show the right-hand side, one can build an EOM T that computes using not more than KPE(x) + KT(KPE(x)) + O(1) input bits in a way inspired by Huffman-Coding [#!Huffman:52!#]. The claim then follows from the Invariance Theorem. The trick is to arrange T's computation such that T's output converges yet never needs to decrease lexicographically. T works as follows:
(A) Emulate UE to construct a real enumerable number 0.s encoded as a self-delimiting input program r, simultaneously run all (possibly forever running) programs on UE dovetail style; whenever the output of a prefix q of any running program starts with some for the first time, set variable V(x) := V(x) + 2-l(q) (if no program has ever created output starting with x then first create V(x) initialized by 0); whenever the output of some extension q' of q (obtained by possibly reading additional input bits: q'=q if none are read) lexicographically increases such that it does not equal x any more, set V(x) := V(x) - 2-l(q').
(B) Simultaneously, starting at the right end of the unit interval [0,1), as the V(x) are being updated, keep updating a chain of disjoint, consecutive, adjacent, half-open (at the right end) intervals IV(x) = [LV(x), RV(x)) of size V(x) = RV(x) - LV(x) in alphabetic order on x, such that the right end of the IV(x) of the largest x coincides with the right end of [0,1), and IV(y) is to the right of IV(x) if . After every variable update and each change of s, replace the output of T by the x of the IV(x) with .
This will never violate the EOM constraints: the enumerable s cannot shrink, and since EOM outputs cannot decrease lexicographically, the interval boundaries RV(x) and LV(x) cannot grow (their negations are enumerable, compare Lemma 4.1), hence T's output cannot decrease.
For the IV(x) converge towards an interval I(x) of size PE(x). For with PE(x) > 0, we have: for any there is a time t0 such that for all time steps t>t0 in T's computation, an interval of size will be completely covered by certain IV(y) satisfying and . So for the also converge towards an interval I(x) of size PE(x). Hence T will output larger and larger y approximating x from below, provided .
Since any interval of size c within [0,1) contains a number 0.z with l(z) = -lg c, in both cases there is a number 0.s (encodable by some r satisfying ) with l(s) = -lg PE(x) + O(1), such that , and therefore .
Less symmetric statements can also be derived in very similar fashion:
Theorem 5.4 Let TM T induce approximable CPT(x) for all (compare Defs. 4.10 and 4.12; an EOM would be a special case). Then for , PT(x) > 0:
(40)
Proof. Modify the proof of Theorem 5.3 for approximable as opposed to enumerable interval boundaries and approximable 0.s.
A similar proof, but without the complication for the case , yields:
Theorem 5.5 Let denote an approximable semimeasure on ; that is, is describable. Then for :
(41)
(42)
As a consequence,
(43)
Proof. Initialize variables and . Dovetailing over all , approximate the GTM-computable in variables Vx initialized by zero, and create a chain of adjacent intervals IVx analogously to the proof of Theorem 5.3.
The IVx converge against intervals Ix of size . Hence x is GTM-encodable by any program r producing an output s with : after every update, replace the GTM's output by the x of the IVx with . Similarly, if 0.s is in the union of adjacent intervals Iy of strings y starting with x, then the GTM's output will converge towards some string starting with x. The rest follows in a way similar to the one described in the final paragraph of the proof of Theorem 5.3.
Using the basic ideas in the proofs of Theorem 5.3 and 5.5 in conjunction with Corollary 4.3 and Lemma 4.2, one can also obtain statements such as:
Theorem 5.6 Let denote the universal CEM from Theorem 4.1. For ,
(44)
While PE dominates PM and PG dominates PE, the reverse statements are not true. In fact, given the results from Sections 3.2 and 5, one can now make claims such as the following ones:
Corollary 5.1 The following functions are unbounded:
Proof. For the cases and PE, apply Theorems 5.2, 5.6 and the unboundedness of (12). For the case PG, apply Theorems 3.3 and 5.3.
Next: Tighter Bounds? Up: Probability vs Descriptive Complexity Previous: Probability vs Descriptive Complexity
Juergen Schmidhuber
2001-01-09
Related links: In the beginning was the code! - Zuse's thesis - Life, the universe, and everything - Generalized Algorithmic Information - Speed Prior - The New AI | 1,304 | 5,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.882028 |
http://de.metamath.org/mpeuni/sge0pnfmpt.html | 1,721,643,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00228.warc.gz | 7,946,237 | 5,925 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > sge0pnfmpt Structured version Visualization version GIF version
Theorem sge0pnfmpt 39338
Description: If a term in the sum of nonnegative extended reals is +∞, then the value of the sum is +∞ (Contributed by Glauco Siliprandi, 3-Mar-2021.)
Hypotheses
Ref Expression
sge0pnfmpt.k 𝑘𝜑
sge0pnfmpt.a (𝜑𝐴𝑉)
sge0pnfmpt.b ((𝜑𝑘𝐴) → 𝐵 ∈ (0[,]+∞))
sge0pnfmpt.p (𝜑 → ∃𝑘𝐴 𝐵 = +∞)
Assertion
Ref Expression
sge0pnfmpt (𝜑 → (Σ^‘(𝑘𝐴𝐵)) = +∞)
Distinct variable group: 𝐴,𝑘
Allowed substitution hints: 𝜑(𝑘) 𝐵(𝑘) 𝑉(𝑘)
Proof of Theorem sge0pnfmpt
StepHypRef Expression
1 sge0pnfmpt.a . 2 (𝜑𝐴𝑉)
2 sge0pnfmpt.k . . 3 𝑘𝜑
3 sge0pnfmpt.b . . 3 ((𝜑𝑘𝐴) → 𝐵 ∈ (0[,]+∞))
4 eqid 2610 . . 3 (𝑘𝐴𝐵) = (𝑘𝐴𝐵)
52, 3, 4fmptdf 6294 . 2 (𝜑 → (𝑘𝐴𝐵):𝐴⟶(0[,]+∞))
6 sge0pnfmpt.p . . . 4 (𝜑 → ∃𝑘𝐴 𝐵 = +∞)
7 eqcom 2617 . . . . 5 (𝐵 = +∞ ↔ +∞ = 𝐵)
87rexbii 3023 . . . 4 (∃𝑘𝐴 𝐵 = +∞ ↔ ∃𝑘𝐴 +∞ = 𝐵)
96, 8sylib 207 . . 3 (𝜑 → ∃𝑘𝐴 +∞ = 𝐵)
10 pnfex 9972 . . . 4 +∞ ∈ V
1110a1i 11 . . 3 (𝜑 → +∞ ∈ V)
124, 9, 11elrnmptd 38361 . 2 (𝜑 → +∞ ∈ ran (𝑘𝐴𝐵))
131, 5, 12sge0pnfval 39266 1 (𝜑 → (Σ^‘(𝑘𝐴𝐵)) = +∞)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 Ⅎwnf 1699 ∈ wcel 1977 ∃wrex 2897 Vcvv 3173 ↦ cmpt 4643 ‘cfv 5804 (class class class)co 6549 0cc0 9815 +∞cpnf 9950 [,]cicc 12049 Σ^csumge0 39255 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-fal 1481 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-sup 8231 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-seq 12664 df-sum 14265 df-sumge0 39256 This theorem is referenced by: voliunsge0lem 39365
Copyright terms: Public domain W3C validator | 2,061 | 3,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.121051 |
https://www.spreadsheetweb.com/excel-weekday-function/ | 1,701,627,426,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00779.warc.gz | 1,121,174,587 | 13,148 | Select Page
The Excel WEEKDAY function is a Date & Time formula that finds the day of the week of a given date. The WEEKDAY function returns an integer indicating the day's index according to the specified day of the week mapping scheme. In this guide, we’re going to show you how to use the Excel WEEKDAY function and also go over some tips and error handling methods.
Supported versions
• All Excel versions (return_type numbers between 11 and 17 were added in 2010 version)
Excel WEEKDAY Function Syntax
WEEKDAY(serial_number, [return_type])
Arguments
serial_number The date of the day you want. Microsoft recommends using the DATE or other similar functions that return a date as a serial number. [return_type] Optional. A number that determines the day of week mapping scheme. The dDefault value is 1. See the table below for available types.
Return Types
Return_type Number returned by Excel WEEKDAY 1 or omitted Numbers 1 (Sunday) through 7 (Saturday). Behaves like in previous versions of Microsoft Excel. 2 Numbers 1 (Monday) through 7 (Sunday). 3 Numbers 0 (Monday) through 6 (Sunday). 11 Numbers 1 (Monday) through 7 (Sunday). 12 Numbers 1 (Tuesday) through 7 (Monday). 13 Numbers 1 (Wednesday) through 7 (Tuesday). 14 Numbers 1 (Thursday) through 7 (Wednesday). 15 Numbers 1 (Friday) through 7 (Thursday). 16 Numbers 1 (Saturday) through 7 (Friday). 17 Numbers 1 (Sunday) through 7 (Saturday).
Example
To find the day of the week of a date, provide a date as the serial_number argument of the Excel WEEKDAY function. Microsoft recommends using serial numbers, instead of date strings like "2/5/2019". A serial number date can be generated via functions like DATE.
You can also set a number for the [return_type] argument. This number specifies the numbering scheme. You can omit the argument to return a number based on a week starts on a Sunday and ends on a Saturday.
=WEEKDAY(DATE(2018,1,13),11) returns 6. The function assumes a week starts on Monday, and Monday equals 1.
Tips
• Use the DATE function to generate a date serial number.
• [return_type] numbers between 11 and 17 were added in Excel 2010. You can still use the Excel WEEKDAY function with numbers between 1 and 3.
• [return_type] 2 and 11 work identically.
• Excel keeps date and time values as numbers. Excel assumes that Jan 1st, 1900 is 1, and every subsequent date value is based on this. While whole numbers represent days, decimals represent time values. For example; 1/1/2018 is equal to 43101, and 12:00 is equal to 5.
Issues
If the start_date is not a valid date, Excel WEEKDAY function returns the #VALUE! error value. Microsoft recommends using a formula to generate a valid date serial value. | 660 | 2,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.83182 |
https://www.statisticshowto.com/winsorize/ | 1,726,687,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00484.warc.gz | 924,952,088 | 15,632 | # Winsorize: Definition, Examples in Easy Steps
Outliers > Winsorize
## What is Winsorization?
Winsorization is a way to minimize the influence of outliers in your data by either:
• Assigning the outlier a lower weight,
• Changing the value so that it is close to other values in the set.
The data points are modified, not trimmed/removed (as in the trimmed mean).
The Winsorize technique was first introduced by Dixon (1960), who attributed it to Charles P. Winsor.
Statistics such as the mean and variance are very susceptible to outliers; Winsorization can be an effective way to deal with this problem, improve statistical efficiency and increase the robustness of statistical inferences.
The downside is that bias is introduced into your results, although the bias is a lot less than if you had simply deleted the data point. The alternative is to keep the data point as-is, but that may not be the best choice as it could dramatically skew your results. Either way, you should always have a good justification for Winsorizing your data; Never run the procedure arbitrarily in the hopes of getting more significant results.
## A Basic Method to Winsorize by Hand
1. Analyze your data to make sure the outlier isn’t a result of measurement error or some other fixable error.
2. Decide how much Winsorization you want. This is specified as a total percentage of untouched data. For example, if you want to Winsorize the top 5% and bottom 5% of data points, this is equal to 100% – 5% – 5% = 90% Winsorization. A 80% Winsorization means that 10% is modified from each tail area (see Tips on Cut-Off Point Selection below).
3. Replace the extreme values by the maximum and/or minimum values at the threshold. For example:
• The following data set has several (bolded) extremes:
{0.1, 1, 12, 14, 16, 18, 19, 21, 24, 26, 29, 32, 33, 35, 39, 40, 41, 44, 99, 125}
Mean = 33.405.
• After modifying the top and bottom 10% (I matched those values to the nearest extreme):
{12, 12,12, 14, 16, 18, 19, 21, 24, 26, 29, 32, 33, 35, 39, 40, 41, 44, 44, 44}
80% Winsorized mean = 24.95.
You could choose to add a little more to the larger/smaller values to account for their weights. for example, the values 99 and 125 were modified, but 125 is approximately 125% larger than 99. Therefore, instead of replacing those values with 44 and 44, I could replace them with 44 and 55 (because 125% * 44 = 55).
Need help? Check out our tutoring section!
## Tips on Cut-Off Point Selection
A poor choice in Step 2 above can result in estimators with inflated mean squared errors(MSE). A few suggestions for cut-off point choice and avoiding this problem:
• Compare the MSE from modified and unmodified results. If a classical estimator (like the arithmetic mean) has a much smaller MSE, this may indicate a poor cut-off point choice. Note: it stands to reason that you should probably choose the cut-off point that minimizes the MSE compared to the classical estimator, but in practice this is very difficult to do.
• If in doubt, refer to published literature to see if your data type (i.e. cholesterol levels, intelligence, rock minerals or something else) is commonly Winsorized and what percentage is usually used in your particular field.
• Do not set your cut-off point before collecting your data. Wait until you actually have the data in front of you before making your choice.
Reference:
W. J. Dixon (1960). Simplified Estimation from Censored Normal Samples, The Annals of Mathematical Statistics, 31, 385–391.
Kotz, S.; et al., eds. (2006), Encyclopedia of Statistical Sciences, Wiley. | 904 | 3,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.914952 |
https://www.mathworks.com/matlabcentral/cody/problems/44948-calculate-a-damped-sinusoid/solutions/2912227 | 1,606,181,951,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00477.warc.gz | 759,645,153 | 18,887 | Cody
# Problem 44948. Calculate a Damped Sinusoid
Solution 2912227
Submitted on 4 Sep 2020 by Anastasios Katsaounis
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
y_correct = [1.0000 0.7246 0.1554 -0.4232 -0.7524 -0.7118 -0.3583 0.1177 0.4912 0.6065]; y_test = damped_cos(0.5, 1, 10); assert( all ( abs(y_correct(:) - y_test(:)) < 1e-4 ) )
t = 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 y = 1.0000 0.7246 0.1554 -0.4232 -0.7524 -0.7118 -0.3583 0.1177 0.4912 0.6065
2 Pass
y_correct = [1.0000 -3.4903 12.1825]; y_test = damped_cos(-0.5, 5, 3) assert( all ( abs(y_correct(:) - y_test(:)) < 1e-4 ) )
t = 0 2.5000 5.0000 y = 1.0000 -3.4903 12.1825 y_test = 1.0000 -3.4903 12.1825
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Start Hunting! | 410 | 954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.519176 |
https://bartoszmilewski.com/tag/foundations-of-mathematics/ | 1,685,598,134,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00613.warc.gz | 151,970,337 | 28,151 | This is my 100th WordPress post, so I decided to pull all the stops and go into some crazy stuff where hard math and hard physics mix freely with wild speculation. I hope you will enjoy reading it as much as I enjoyed writing it.
## It’s a HoTT Summer of 2013
One of my current activities is reading the new book, Homotopy Type Theory (HoTT) that promises to revolutionize the foundations of mathematics in a way that’s close to the heart of a programmer. It talks about types in the familiar sense: Booleans, natural numbers, (polymorphic) function types, tuples, discriminated unions, etc.
As do previous type theories, HoTT assumes the Curry-Howard isomorphism that establishes the correspondence between logic and type theory. The gist of it is that any theorem can be translated into a definition of a type; and its proof is equivalent to producing a value of that type (false theorems correspond to uninhabited types that have no elements). Such proofs are by necessity constructive: you actually have to construct a value to prove a theorem. None if this “if it didn’t exist then it would lead to contradictions” negativity that is shunned by intuitionistic logicians. (HoTT doesn’t constrain itself to intuitionistic logic — too many important theorems of mathematics rely on non-constructive proofs of existence — but it clearly delineates its non-intuitionistic parts.)
Type theory has been around for some time, and several languages and theorem provers have been implemented on the base of the Curry-Howard isomorphism: Agda and Coq being common examples. So what’s new?
## Set Theory Rant
Here’s the problem: Traditional type theory is based on set theory. A type is defined as a set of values. Bool is a two-element set {True, False}. Char is a set of all (Unicode) characters. String is an infinite set of all lists of characters. And so on. In fact all of traditional mathematics is based on set theory. It’s the “assembly language” of mathematics. And it’s not a very good assembly language.
First of all, the naive formulation of set theory suffers from paradoxes. One such paradox, called Russell’s paradox, is about sets that are members of themselves. A “normal” set is not a member of itself: a set of dogs is not a dog. But a set of all non-dogs is a member of itself — it’s “abnormal”. The question is: Is the set of all “normal” sets normal or abnormal? If it’s normal that it’s a member of normal sets, right? Oops! That would make it abnormal. So maybe it’s abnormal, that is not a member of normal sets. Oops! That would make it normal. That just shows you that our natural intuitions about sets can lead us astray.
Fortunately there is an axiomatic set theory called the Zermelo–Fraenkel (or ZF) theory, which avoids such paradoxes. There are actually two versions of this theory: with or without the Axiom of Choice. The version without it seems to be too weak (not every vector space has a basis, the product of compact sets isn’t necessarily compact, etc.); the one with it (called ZFC) leads to weird non-intuitive consequences.
What bothers many mathematicians is that proofs that are based on set theory are rarely formal enough. It’s not that they can’t be made formal, it’s just that formalizing them would be so tedious that nobody wants to do it. Also, when you base any theory on set theory, you can formulate lots of idiotic theorems that have nothing to do with the theory in question but are only relevant to its clunky set-theoretical plumbing. It’s like the assembly language leaking out from higher abstractions. Sort of like programming in C.
## Donuts are Tastier than Sets
Tired of all this nonsense with set theory, a group of Princeton guys and their guests decided to forget about sets and start from scratch. Their choice for the foundation of mathematics was the theory of homotopy. Homotopy is about paths — continuous maps from real numbers between 0 and 1 to topological spaces; and continuous deformations of such paths. The properties of paths capture the essential topological properties of spaces. For instance, if there is no path between a and b, it means that the space is not connected — it has at least two disjoint components — a sits in one and b in another.
Two paths from a to b that cannot be continuously deformed into each other
If two paths between a and b cannot be deformed into each other, it means that there is a hole in space between them.
Obviously, this “traditional” formulation of homotopy relies heavily on set theory. A topological space, for instance, is defined in terms of open sets. So the first step is to distill the essence of homotopy theory by getting rid of sets. Enter Homotopy Type Theory. Paths and their deformations become primitives in the theory. We still get to use our intuitions about paths as curves inscribed on surfaces, but otherwise the math is totally abstract. There is a small set of axioms, the basic one asserting that the statement that a and b are equivalent is equivalent to the statement that they are equal. Of course the notions of equivalence and equality have special meanings and are very well defined in terms of primitives.
## Cultural Digression
Why homotopy? I have my own theory about it. Our mathematics has roots in Ancient Greece, and the Greeks were not interested in developing technology because they had very cheap labor — slaves.
Euclid explaining geometry to his pupils in Raphael’s School of Athens
Instead, like all agricultural societies before them (Mesopotamia, Egypt), they were into owning land. Land owners are interested in geometry — Greek word γεωμετρία literally means measuring Earth. The “computers” of geometry were the slate, ruler and compass. Unlike technology, the science of measuring plots of land was generously subsidized by feudal societies. This is why the first rigorous mathematical theory was Euclid’s geometry, which happened to be based on axioms and logic. Euclid’s methodology culminated in the 20th century in Hilbert’s program of axiomatization of the whole of mathematics. This program crashed and burned when Gödel proved that any non-trivial theory (one containing arithmetic) is chock full of non-decidable theorems.
I was always wondering what mathematics would be like if it were invented by an industrial, rather than agricultural, society. The “computer” of an industrial society is the slide rule, which uses (the approximation of) real numbers and logarithms. What if Newton and Leibniz never studied Euclid? Would mathematics be based on calculus rather than geometry? Calculus is not easy to axiomatize, so we’d have to wait for the Euclid of calculus for a long time. The basic notions of calculus are Banach spaces, topology, and continuity. Topology and continuity happen to form the basis of homotopy theory as well. So if Greeks were an industrial society they could have treated homotopy as more basic than geometry. Geometry would then be discovered not by dividing plots of land but by studying solutions to analytic equations. Instead of defining a circle as a set of points equidistant from the center, as Euclid did, we would first define it as a solution to the equation x2+y2=r2.
Now imagine that this hypothetical industrial society also skipped the hunter-gather phase of development. That’s the period that gave birth to counting and natural numbers. I know it’s a stretch of imagination worthy a nerdy science fiction novel, but think of a society that would evolve from industrial robots if they were abandoned by humanity in a distant star system. Such a society could discover natural numbers by studying the topology of manifolds that are solutions to n-dimensional equations. The number of holes in a manifold is always a natural number. You can’t have half a hole!
Instead of counting apples (or metal bolts) they would consider the homotopy of the two-apple space: Not all points in that space can be connected by continuous paths.
There is no continuous path between a and b
Maybe in the world where homotopy were the basis of all mathematics, Andrew Wiles’s proof of the Fermat’s Last Theorem could fit in a margin of a book — as long as it were a book on cohomology and elliptic curves (some of the areas of mathematics Wiles used in his proof). Prime numbers would probably be discovered by studying the zeros of the Riemann zeta function.
Industrial robot explaining to its pupils the homotopy of a two-apple space.
## Quantum Digression
If our industrial robots were very tiny and explored the world at the quantum level (nanorobots?), they might try counting particles instead of apples. But in quantum mechanics, a two-particle state is not a direct product of two one-particle states. Two particles share the same wave function. In some cases this function can be factorized when particles are far apart, in others it can’t, giving rise to quantum entanglement. In quantum world, 2 is not always equal to 1+1.
In Quantum Field Theory (QFT — the relativistic counterpart of Quantum Mechanics), physicist calculate the so called S matrix that describes idealized experiments in which particles are far away from each other in the initial and final states. Since they don’t interact, they can be approximated by single-particle states. For instance, you can start with a proton and an antiproton coming at each other from opposite directions. They can be approximated as two separate particles. Then they smash into each other, produce a large multi-particle mess that escapes from the interaction region and is eventually seen as (approximately) separate particles by a big underground detector. (That’s, for instance, how the Higgs boson was discovered.) The number of particles in the final state may very well be different from the number of particles in the initial state. In general, QFT does not preserve the number of particles. There is no such conservation law.
Counting particles is very different from counting apples.
## Relaxing Equality
In traditional mathematics, the notions of isomorphism and equality are very different. Isomorphism means (in Greek, literally) that things have the same shape, but aren’t necessarily equal. And yet mathematicians often treat isomorphic things as if they were equal. They prove a property of one thing and then assume that this property is also true for all things isomorphic. And it usually is, but nobody has the patience to prove it on the case-by-case basis. This phenomenon even has its own name: abuse of notation. It’s like writing programs in a language in which equality ‘==’ does not translate into the assembly-language CMP instruction followed be a conditional jump. We would like to work with structural identity, but all we do is compare pointers. You can overload operator ‘==’ in C++ but many a bug was the result of comparing pointers instead of values.
How can we make isomorphism more like equality? HoTT took quite an unusual approach by relaxing equality enough to make it plausibly equivalent to isomorphism.
HoTT’s homotopic version of equality is this: Two things are equal if there is a path between them. This equality is reflexive, symmetric, and transitive, just like equality is supposed to be. Reflexivity, for instance, tells us that x=x, and indeed there is always a trivial (constant) path from a point to itself. But there could also be other non-trivial paths looping from the point to itself. Some of them might not even be contractible. They all contribute to equality x=x.
There could be several paths between different points, a and b, making them “equal”: a=b. We are tempted to say that in this picture equality is a set of paths between points. Well, not exactly a set but the next best thing to a set — a type. So equality is a type, often called “identity type”, and two things are equal if the “identity type” for them is inhabited. That’s a very peculiar way to define equality. It’s an equality that carries with it a witness — a construction of an element of the equality type.
## Relaxing Isomorphism
The first thing we could justifiably expect from any definition of equality is that if two things are equal they should be isomorphic. In other words, there should be an invertible function that maps one equal thing to another equal thing. This sound pretty obvious until you realize that, since equality is relaxed, it’s not! In fact we can’t prove strict isomorphism between things that are homotopically equal. But we do get a slightly relaxed version of isomorphism called equivalence. In HoTT, if things are equal they are equivalent. Phew, that’s a relief!
The trick is going the other way: Are equivalent things equal? In traditional mathematics that would be blatantly wrong — there are many isomorphic objects that are not equal. But with the HoTT’s notion of equality, there is nothing that would contradict it. In fact, the statement that equivalence is equivalent to equality can be added to HoTT as an axiom. It’s called Voevodski’s axiom of univalence.
It’s hard (or at least tedious), in traditional math, to prove that properties (propositions) can be carried over isomorphisms. With univalence, equivalence (generalized isomorphism) is the same as equality, and one can prove once and for all that propositions can be transported between equal objects. With univalence, the tedium of proving that if one object has a given property then all equivalent (“isomorphic”) object have the same property is eliminated.
Incidentally, where do types live? Is there (ahem!) a set of all types? There’s something better! A type of types called a Universe. Since a Universe is a type, is it a member of itself? You can almost see the Russel’s paradox looming in the background. But don’t despair, a Universe is not a member of itself, it’s a member of the higher Universe. In fact there are infinitely many Universes, each being a member of the next one.
## Taking Roots
How does relaxed equality differ from the set-theoretical one? The simplest such example is the equality of Boolean types. There are two ways you can equal the Bool type to itself: One maps True to True and False to False, the other maps True to False and False to True. The first one is an identity mapping, but the second one is not — its square though is! (apply this mapping twice and you get back to original). Within HoTT you can take the square root of identity!
So here’s an interesting intuition for you: HoTT is to set theory as complex numbers are to real numbers (in complex numbers you can take a square root of -1). Paradoxically, complex numbers make a lot of things simpler. For instance, all quadratic equations are suddenly solvable. Sine and cosine become two projections of the same complex exponential. Riemann’s zeta function gains very interesting zeros on the imaginary line. The hope is that switching from sets to homotopy will lead to similar simplifications.
I like the example with flipping Booleans because it reminds me of an interesting quantum phenomenon. Imagine a quantum state with two identical particles. What happens when you switch the particles? If you get exactly the same state, the particles are called bosons (think photons). If you don’t, they are called fermions (think electrons). But when you flip fermions twice, you get back to the original state. In many ways fermions behave like square roots of bosons. For instance their equation of motion (Dirac equation) when squared produces the bosonic equation of motion (Klein-Gordon equation).
## Computers Hate Sets
There is another way HoTT is better than set theory. (And, in my cultural analogy, that becomes more pertinent when an industrial society transitions to a computer society.) There is no good way to represent sets on a computer. Data structures that model sets are all fake. They always put some restrictions on the type of elements they can store. For instance the elements must be comparable, or hashable, or something. Even the simplest set of just two elements is implemented as an ordered pair — in sets you can’t have the first or the second element of a set (and in fact the definition of a pair as a set is quite tricky). You can easily write a program in Haskell that would take a (potenitally infinite) list of pairs and pick one element from each pair to form a (potentially infinite) list of picks. You can, for instance, tell the computer to pick the left element from each pair. Replace lists of pairs with sets of sets and you can’t do it! There is no constructive way of creating such a set and it’s very existence hinges on the axiom of choice.
This fact alone convinces me that set theory is not the best foundation for the theory of computing. But is homotopy a better assembly language for computing? We can’t represent sets using digital computers, but can we represent homotopy? Or should we start building computers from donuts and rubber strings? Maybe if we keep miniaturizing our computers down to the Planck scale, we could find a way to do calculations using loop quantum gravity, if it ever pans out.
## Aharonov-Bohm Experiment
Even without invoking quantum gravity, quantum mechanics exhibits a lot of interesting non-local behaviors that often probe the topological properties of the surroundings. For instance, in the classic double-slit experiment, the fact that there are paths between the source of electrons and the screen that are not homotopically equivalent makes the electrons produce an interference pattern. But my favorite example is the Bohm-Aharonov experiment.
First, let me explain what a magnetic potential is. One of the Maxwell’s equations states that the divergence of the magnetic field is always zero (see a Tidbit at the end of this post that explains this notation):
This is the reason why magnetic field lines are always continuous. Interestingly, this equation has a solution that follows from the observation that the divergence of a curl is zero. So we can represent the magnetic field as a curl of some other vector field, which is called the magnetic potential A:
It’s just a clever mathematical trick. There is no way to measure magnetic potential, and the solution isn’t even unique: you can add a gradient of any scalar field to it and you’ll get the same magnetic field (the curl of a gradient is zero). So A is totally fake, it exists only as a device to simplify some calculations. Or is it…?
It turns out that electrons can sniff the magnetic potential, but only if there’s a hole in space. It turns out that, if you have a thin (almost) infinite linear coil with a current running through its windings, (almost) all magnetic field will be confined to its interior. Outside the coil there’s no magnetic field. However, there is a nonzero curl-free magnetic potential circling it. Now imagine using this coil as a separator between the two slits of the double-slit experiment. As before, there are two paths for the electron to follow: to the left of the coil and to the right of the coil. But now, along one path, the electron will be traveling with the lines of magnetic potential; along the other, against.
Aharononv-Bohm experiment. There are two paths available to the electron.
Magnetic potential doesn’t contribute to the electron’s energy or momentum but it does change its phase. So in the presence of the coil, the interference pattern in the two slit experiment shifts. That shift has been experimentally confirmed. The Aharonov-Bohm effect takes place because the electron is excluded from the part of space that is taken up by the coil — think of it as an infinite vertical line in space. The space available to the electron contains paths that cannot be continuously deformed into each other (they would have to cross the coil). In HoTT that would mean that although the point a, which is the source of the electron, and point b, where the electron hit the screen, are “equal,” there are two different members of the equality type.
## The Incredible Quantum Homotopy Computer
The Aharonov-Bohm effect can be turned on and off by switching the current in the coil (actually, nobody uses coils in this experiment, but there is some promising research that uses nano-rings instead). If you can imagine a transistor built on the Aharonov-Bohm principle, you can easily imagine a computer. But can we go beyond digital computers and really explore varying homotopies?
I’ll be the first to admit that it might be too early to go to Kickstarter and solicit funds for a computer based on the Aharonov-Bohm effect that would be able to prove theorems formulated using Homotopy Type Theory; but the idea of breaking away from digital computing is worth a thought or two.
Or we can leave it to the post apocalyptic industrial-robot civilization that doesn’t even know what a digit is.
## Acknowledgments
I’m grateful to the friendly (and patient) folks on the HoTT IRC channel for answering my questions and providing valuable insights. | 4,461 | 20,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.930863 |
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# The shaded region in the figure above represents a
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The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the lenght and width of the frame, what is the length of the picture, in inches?
A. $$9\sqrt2$$
B. $$\frac {3}{2}$$
C. $$\frac {9}{\sqrt2}$$
D. $$15 ( 1 - \frac {1}{\sqrt2})$$
E. $$\frac {9}{2}$$
[Reveal] Spoiler: OA
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Last edited by Bunuel on 06 Jan 2014, 03:24, edited 3 times in total.
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Stiv wrote:
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The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the lenght and width of the frame, what is the length of the picture, in inches?
A. $$9\sqrt2$$
B. $$\frac {3}{2}$$
C. $$\frac {9}{\sqrt2}$$
D. $$15 ( 1 - \frac {1}{\sqrt2}$$
E. $$\frac {9}{2}$$
Say the length and the width of the picture are $$x$$ and $$y$$ respectively. Since they have the same ratio as the lenght and width of the frame, then $$\frac{x}{y}=\frac{18}{15}$$ --> $$y=\frac{5}{6}x$$.
Next, since the frame encloses a rectangular picture that has the same area as the frame itself and the whole area is $$18*15$$, then the areas of the frame (shaded region) and the picture (inner region) are $$\frac{18*15}{2}=9*15$$ each.
The area of the picture is $$xy=9*15$$ --> $$x*(\frac{5}{6}x)=9*15$$ --> $$x^2=2*81$$ --> $$x=9\sqrt{2}$$.
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The total area is 15*18=270, the area of the picture is half of the whole area = 135. the ration of the width and length of the picture is the same as the frames 15/18 or 5/6. We need to find the length of the picture 5x*6x=135, 30x^2=135, x^2=135/30, x=3/sqrt2, so the length = 6*3/sqrt2=9sqrt2
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Yes, I understand now. Thank you for the explanation!
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28 Jul 2013, 01:04
l-length of the pic and w-width
18*15-LW - Sof the frame
LW - S of the pic
18/15=L/W --> W= 15L/18
according to the statement 18*15-LW=LW --> 18*15-L*15L/18=L*15L/18 --> L=9\sqrt{2}
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11 Jan 2014, 22:36
Why do I get a different answer if I just left at l/w=18/15 than simplifying to l/w=6/5 ?? Aren't the ratios the same?
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12 Jan 2014, 05:01
b00gigi wrote:
Why do I get a different answer if I just left at l/w=18/15 than simplifying to l/w=6/5 ?? Aren't the ratios the same?
To point out why you are getting a different answer you have to show your work. Anyway, yes, the ratio is the same but we need to find the length of the picture, which can be done as explained here: the-shaded-region-in-the-figure-above-represents-a-135095.html#p1100419
Hope this helps.
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Total area of the given figure= 18*15 = 270
Area of frame = Area of the picture => We need to divide the total area into two parts, 270/2 = 135. The frame and picture have 135 inch^2 area each.
l(pic) l(frame)
----- = ---------- = 6/5 ==> Area of picture = 135= 6k * 5k ==> 30k^2=135 ==> k =3/sqrt(2). So, l(pic)= 6* 3/sqrt(2) = 9*sqrt(2)
w(pic) w(frame)
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Display posts from previous: Sort by | 2,351 | 7,560 | {"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} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.841246 |
http://highalphabet.com/systemdynamicsprb36/ | 1,653,004,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00325.warc.gz | 25,461,370 | 1,529 | Consider the mechanical system shown in the figure. The system is initially at rest. The displacements x1 and x2 are measured from their respective equilibrium positions before the input u is applied. Assume that b1=Ns/m, b2=10Ns/m, k1=4N/m, and k2=20N/m. Obtain the displacement x2(t) when u is a step force input of 2 N. Derive the transfer function and then plot the response for 5 s using step command. The step input force u has a magnitude of 2 N.
Highalphabet Home Page dynamics dynamics System Dynamics Page 1
Consider the mechanical system shown in the figure. The system is initially at rest. The displacements x1 and x2 are measured from their respective equilibrium positions before the input u is applied. Assume that b1=Ns/m, b2=10Ns/m, k1=4N/m, and k2=20N/m. Obtain the displacement x2(t) when u is a step force input of 2 N. Derive the transfer function and then plot the response for 5 s using step command. The step input force u has a magnitude of 2 N.
System Dynamics Page 2 dynamics dynamics dynamics dynamics dynamics dynamics dynamics System dynamics Page 3 | 266 | 1,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-21 | latest | en | 0.8399 |
http://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=110791&nr_roman=(C)(X)DCCXCI&lang=ro | 1,539,745,523,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510969.31/warc/CC-MAIN-20181017023458-20181017044958-00080.warc.gz | 547,332,534 | 8,450 | # 110.791 = (C)(X)DCCXCI110.790 = ? ... 110.792 = ?Explicații mai jos.
## Ultimele conversii de numere în numerale romane
110.791 = (C)(X)DCCXCI 17 Oct, 06:05 EET (UTC +2) 169.723 = (C)(L)(X)M(X)DCCXXIII 17 Oct, 06:05 EET (UTC +2) 15.843 = (X)(V)DCCCXLIII 17 Oct, 06:05 EET (UTC +2) 169.723 = (C)(L)(X)M(X)DCCXXIII 17 Oct, 06:05 EET (UTC +2) 100.063 = (C)LXIII 17 Oct, 06:04 EET (UTC +2) 169.723 = (C)(L)(X)M(X)DCCXXIII 17 Oct, 06:04 EET (UTC +2) 169.722 = (C)(L)(X)M(X)DCCXXII 17 Oct, 06:04 EET (UTC +2) 169.722 = (C)(L)(X)M(X)DCCXXII 17 Oct, 06:04 EET (UTC +2) 169.722 = (C)(L)(X)M(X)DCCXXII 17 Oct, 06:04 EET (UTC +2) 169.722 = (C)(L)(X)M(X)DCCXXII 17 Oct, 06:03 EET (UTC +2) 97 = XCVII 17 Oct, 06:03 EET (UTC +2) 4.578 = M(V)DLXXVIII 17 Oct, 06:03 EET (UTC +2) 169.722 = (C)(L)(X)M(X)DCCXXII 17 Oct, 06:03 EET (UTC +2) numere convertite, vezi mai mult... | 463 | 861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.698714 |
https://learnool.com/tef6-lewis-structure/ | 1,719,120,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00005.warc.gz | 303,771,737 | 24,504 | # TeF6 Lewis structure
The information on this page is ✔ fact-checked.
TeF6 (tellurium hexafluoride) has one tellurium atom and six fluorine atoms.
In TeF6 Lewis structure, there are six single bonds around the tellurium atom, with six fluorine atoms attached to it, and on each fluorine atom, there are three lone pairs.
Contents
## Steps
Here’s how you can easily draw the TeF6 Lewis structure step by step:
#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms
Now, let’s take a closer look at each step mentioned above.
### #1 Draw a rough skeleton structure
• First, determine the total number of valence electrons
In the periodic table, tellurium lies in group 16, and fluorine lies in group 17.
Hence, tellurium has six valence electrons and fluorine has seven valence electrons.
Since TeF6 has one tellurium atom and six fluorine atoms, so…
Valence electrons of one tellurium atom = 6 × 1 = 6
Valence electrons of six fluorine atoms = 7 × 6 = 42
And the total valence electrons = 6 + 42 = 48
• Second, find the total electron pairs
We have a total of 48 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
Total electron pairs = total valence electrons ÷ 2
So the total electron pairs = 48 ÷ 2 = 24
• Third, determine the central atom
We have to place the least electronegative atom at the center.
Since tellurium is less electronegative than fluorine, assume that the central atom is tellurium.
Therefore, place tellurium in the center and fluorines on either side.
• And finally, draw the rough sketch
### #2 Mention lone pairs on the atoms
Here, we have a total of 24 electron pairs. And six Te — F bonds are already marked. So we have to only mark the remaining eighteen electron pairs as lone pairs on the sketch.
Also remember that tellurium is a period 5 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.
So for each fluorine, there are three lone pairs, and for tellurium, there is zero lone pair because all eighteen electron pairs are over.
Mark the lone pairs on the sketch as follows:
### #3 If needed, mention formal charges on the atoms
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For tellurium atom, formal charge = 6 – 0 – ½ (12) = 0
For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0
Here, both tellurium and fluorine atoms do not have charges, so no need to mark the charges.
In the above structure, you can see that the central atom (tellurium) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.
Therefore, this structure is the stable Lewis structure of TeF6. | 760 | 3,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.862597 |
https://discuss.leetcode.com/topic/57923/my-c-13ms-solution | 1,513,312,862,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948563629.48/warc/CC-MAIN-20171215040629-20171215060629-00563.warc.gz | 566,928,850 | 8,340 | # My C 13ms solution.
• although my solution is 13ms,i think it’s not reach the problem request.
Could anybody tell me the algorithm which time complexity is O(log (m+n))??
``````double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
int i = 0,j = 0;
int *NewArray = (int *)malloc((nums1Size + nums2Size) * sizeof(int));
int ArraySize = 0;
int inx;
double Result = 0;
while (i < nums1Size && j < nums2Size) {
if (nums1[i] < nums2[j]) {
NewArray[ArraySize] = nums1[i];
i ++;
ArraySize ++;
}else{
NewArray[ArraySize] = nums2[j];
j ++;
ArraySize ++;
}
}
while (i < nums1Size) {
NewArray[ArraySize] = nums1[i];
i ++;
ArraySize ++;
}
while (j < nums2Size) {
NewArray[ArraySize] = nums2[j];
j ++;
ArraySize ++;
}
if(ArraySize % 2 == 0){
inx = ArraySize/2;
Result = (NewArray[inx] + NewArray[inx - 1])*1.0 / 2;
}else{
inx = ArraySize/2;
Result = NewArray[inx];
}
return Result;
}
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 323 | 1,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-51 | latest | en | 0.467757 |
https://wizedu.com/questions/914227/explain-the-difference-between-the-principle-of | 1,701,694,377,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00199.warc.gz | 733,880,269 | 8,623 | ##### Question
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##### (a) Explain, using examples, the difference between a ‘top-down’ approach and a ‘bottom-up’ approach to equity...
(a) Explain, using examples, the difference between a ‘top-down’ approach and a ‘bottom-up’ approach to equity valuation. (b) There are four principles that underlie the concept of efficient markets. Outline, using examples, each principle. (c) Write out the formula for the constant growth dividend valuation model. What key assumptions are required? (d) You are interested in buying a share that paid its last annual dividend 9 months ago. You can assume that the next dividend payment (3 months from today)... | 835 | 4,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.906167 |
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Simple Harmonic Motion
Aston University
Engineering and Applied Science – Physics Lab Report
01/11/2014
Determination by simple harmonic motion of the acceleration due to gravity
Introduction:
A system undertaking simple harmonic motion (SHM) can be restrained very accurately. The period of the SHM depends both on the mass of the system and the strength of the force tending to restore the system to its equilibrium state). Oscillations are a common part of life, for instance the vibrations of a musical instrument which helps make sounds or the foundations of a car suspension which are assisted by oscillations. The main aims of this experiment was to determine if the oscillation of a mass which hung vertically from a spring; this oscillating system was used to measure the acceleration of earth due to gravity and to determine the accuracy of experimental results precisely. (http://www.pgccphy.net/1020/phy1020.pdf
Theory: | | |
Acceleration due to gravity
The value of 9.8m/s/s acceleration is given to a free falling object, directing downwards towards Earth. Any object moving solely under the influence of gravity is known as acceleration of gravity and this vital quantity is denoted by Physicians as the symbol g. (http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity)
Simple harmonic motion
This is everyplace where the acceleration is proportional and opposite to displacement to the continuous amplitude from the position of equilibrium.
Period
The time that it takes to complete one full cycle of oscillation which is measured quantitatively
Hooke’s law
‘Hooke’s law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load. Under these conditions the object returns to its original shape and size upon removal of the load. Elastic behaviour of solids according to Hooke’s law can be explained by the fact that small displacements of their constituent molecules, atoms, or ions from normal positions is also proportional to the force that causes the displacement.’ (http://www.britannica.com/EBchecked/topic/271336/Hookes-law)
Springs constant
Newton’s third law states that for every force, there is an equal and opposite force. This refers to the spring which change shape when there is any kind of exertion of pressure, either stretching or compressing divided by the distance the spring stretches or compresses. The unit of spring constant can be worked out by using Hooke’s law. (http://www.education.com/science-fair/article/springs-pulling-harder/)
Equipment Required * One spring
* One mass hanger (approx. 10g)
* Nine 10g masses
* An electric timer
* A clamp stand
* One measuring ruler
Procedure
The clamp was set up on the table safely as this was needed for part 1 and part 2 of the lab, we then attached the spring to the clamp. The timer was set to 0 for part 2 of the lab.
We proceeded with the lab
PART 1 - Measurement of spring length
Once the experiment was set and safe. Part 1 of the lab consisted of a range of masses from 10g to 100g which was attached to the spring and the amount of stretch of the spring was measured using a measuring stick each time. It consisted if measuring the original spring length with no mass and we also did three trials for each mass added.
.Part 2 - Measuring the period of oscillation
Part 2 of the experiment consisted of the same 0g-100g of masses which was also attached 10g at time to the spring, and then, with a stopwatch we timed the time it took for the spring to make 10 complete periods of oscillations. The timer always started on 0 seconds. The experiment was also repeated thrice to ensure accuracy and reliability.
Theoritical Analysis:
Figure 1. section of relationship with mass and the pull of the string which leads to displacement of the spring form equilibrium level
Figure 1. shows the immediate position of mass hung vertically from a spring. The position where force of gravity is exactly balanced and an equilibrium position is reached is represented by X. The stretch on the spring was measured using a measuring ruler.
i.e. Fs = k. L
where Fs is the restoring force from the spring k is the spring constant L is the distance the spring has stretched
Fg = m . g
Where m is the mass g is the acceleration due to gravity
so, k . L = m . g or L = (g/k) . m
equation L = (g/k) . m tells us a straight line should pass through the origin with gradient g/k on a graph of a spring stretch versus mass.
Part 2 we measured the spring oscillations with an electric stopwatch. We started from the zero mass which hung vertically from a spring, then added a known mass starting from 10g, increasing the value of masses each time by 10g, until a final mass of 100g (1kg) was reached and counting the number of oscillations each time through observation, which depended on the spring constant and the mass.
i.e T = 2 . π / w
Where T is the time taken for one complete oscillation
So, T = 2 , π √ w
T = 2 . π . √ (m/k) or T² = (4 . π² / k) . m
T² = (4 . π² / k) . m tells us that a graph of period squared versus mass should be a straight line passing through the origin with the gradient of (4 . π² / k)
Fs; the restoring force is the cause that makes the mass oscillate. The mass will remain to travel up and down in a number of oscillations due to the conservation of momentum.
Table 1 shows the results for part one of the experiment, where different lengths of mass was added to the spring and the distance the spring stretched was measured with the measuring stick.
Mass(kg) | Spring length (m) | Spring stretch (m) | | | Trial 1 | Trial 2 | Trial 3 | average | | | 0 | 0.36 | 0.36 | 0.360 | 0.360 | 0 | | 0.01 | 0.36 | 0.36 | 0.360 | 0.360 | 0 | | 0.02 | 0.357 | 0.357 | 0.356 | 0.357 | 0.003 | | 0.03 | 0.354 | 0.355 | 0.354 | 0.354 | 0.006 | | 0.04 | 0.348 | 0.348 | 0.349 | 0.348 | 0.O12 | | 0.05 | 0.341 | 0.342 | 0.342 | 0.342 | 0.018 | | 0.06 | 0.337 | 0.338 | 0.337 | 0.337 | 0.023 | | 0.07 | 0.332 | 0.332 | 0.333 | 0.332 | 0.028 | | 0.08 | 0.330 | 0.330 | 0.330 | 0.330 | 0.030 | | 0.09 | 0.329 | 0.328 | 0.328 | 0.328 | 0.032 | 0.1 | 0.326 | 0.326 | 0.326 | 0.326 | 0.034 |
The relationship between spring stretch and mass
Graph 1 is plotted using Table 1
Table 1 shows the spring stretch for each individual mass when it is hung to the spring. In this experiment we have observed that as the mass increases the length of the spring increases. The original length of the spring is shown is 0.36 at 0kg. When the mass of 10g is added the spring the length remains constant. We can see an increase in spring length from 0.02 kg till 0.1 kg. The total average of all the spring lengths after their stretch is 0.343. The graph shows a positive correlation.
Table 2 shows the period of oscillation against mass. From the data we have collected the results shows that as mass increases, the period of oscillation increases. The lowest stretch was from 0 to 0.02kg of mass, which was a stretch of 0.003m. In table 2 the oscillation period was at its maximum rate when the weight of 0.02kg was added and the oscillation period was at its lowest when a weight of 0.1kg was added, oscillation at 0.454 seconds (Table 2). The average oscillation period of the total of the oscillations are 0.307 seconds. The graph again shows a positive correlation.
Table 1 and 2 include three trials to ensure accuracy and reliability so that the data can be reproduced, for example referring back to table 1 when we added on a mass of 0.05kg to the spring, at first, we measured the stretch at 0.0341 and then the second and third trials made it obvious that the actual length the spring stretched was 0.0342, meaning the first trial was likely to be an error. The average values vary between 0.362 and 0.360 in Table 1 and the average period of oscillation varies between 0.277 seconds and 0.454 seconds in table 2. The highest value that the mass was stretched was when the mass of 0.1kg when the spring was stretched from 0 to 0.034m (Table 1) The lowest stretch was from 0 to 0.02kg of mass, which was a stretch of 0.003m.
The graph was plotted to make it easier to observe and analyse results and even with the anomalies it showed the general pattern that the results indicated. The quantity the spring stretches intrigued against the weight put on to the spring in Table 1 gives relatively a straight line that goes through the origin, meaning that the extension of the spring directly depends on the mass force applied to it
Table 2. shows the results of the relationship between mass an the period of oscillation | Measurement Time (s) | | Mass (kg) | Trial 1 | Trial 2 | Trial 3 | average | Period (s) | Preriod2 (S2) | 0.00 | 0 | 0 | 0 | 0 | 0 | 0 | 0.01 | 0 | 0 | 0 | 0 | 0 | 0 | 0.02 | 2.73 | 2.79 | 2.76 | 2.77 | 0.277 | 0.077 | 0.03 | 3.10 | 3.16 | 3.24 | 3.17 | 0.317 | 0.100 | 0.04 | 3.45 | 3.40 | 3.30 | 3.38 | 0.338 | 0.114 | 0.05 | 3.50 | 3.42 | 3.39 | 3.44 | 0.344 | 0.118 | 0.06 | 3.90 | 3.88 | 3.78 | 3.85 | 0.385 | 0.148 | 0.07 | 4.05 | 3.91 | 3.88 | 3.95 | 0.395 | 0.118 | 0.08 | 4.30 | 4.10 | 4.20 | 4.20 | 0.420 | 0.148 | 0.09 | 4.43 | 4.56 | 4.47 | 4.49 | 0.449 | 0.202 | 0.1 | 4.52 | 4.48 | 4.62 | 4.54 | 0.454 | 0.206 |
K = 4.π2 2.2767
K = 17.34
The relationship between mass and oscillation period Graph 2. Is the plotted data from Table 2 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Calculating the value of gravity
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
Table 1 shows the spring stretch for each individual mass when it is hung to the spring. In this experiment we have observed that as the mass increases the length of the spring increases. The original length of the spring is shown is 0.36 at 0kg. When the mass of 10g is added the spring the length remains constant. We can see an increase in spring length from 0.02 kg till 0.1 kg. The total average of all the spring lengths after their stretch is 0.343. The graph shows a positive correlation.
Table 2 shows the period of oscillation against mass. From the data we have collected the results shows that as mass increases, the period of oscillation increases. The lowest stretch was from 0 to 0.02kg of mass, which was a stretch of 0.003m. In table 2 the oscillation period was at its maximum rate when the weight of 0.02kg was added and the oscillation period was at its lowest when a weight of 0.1kg was added, oscillation at 0.454 seconds (Table 2). The average oscillation period of the total of the oscillations are 0.307 seconds. The graph again shows a positive correlation.
Table 1 and 2 include three trials to ensure accuracy and reliability so that the data can be reproduced, for example referring back to table 1 when we added on a mass of 0.05kg to the spring, at first, we measured the stretch at 0.0341 and then the second and third trials made it obvious that the actual length the spring stretched was 0.0342, meaning the first trial was likely to be an error. The average values vary between 0.362 and 0.360 in Table 1 and the average period of oscillation varies between 0.277 seconds and 0.454 seconds in table 2. The highest value that the mass was stretched was when the mass of 0.1kg when the spring was stretched from 0 to 0.034m (Table 1) The lowest stretch was from 0 to 0.02kg of mass, which was a stretch of 0.003m.
The graph was plotted to make it easier to observe and analyse results and even with the anomalies it showed the general pattern that the results indicated. The quantity the spring stretches intrigued against the weight put on to the spring in Table 1 gives relatively a straight line that goes through the origin, meaning that the extension of the spring directly depends on the mass force applied to it
Analysis Under Hooke’s law the spring has returned to its original shape and proves that the gravitational force exerted on the mass is the same as the force exerted by the spring, the displacement of the size of the spring was directly proportional to the force in this case as the stretch increased with the increase of masses. A true relationship between the period, mass, and the spring constant was shown by our results which match the theory that the period of oscillation and the amount a spring stretches is dependent on the mass. The dots have been connected correctly as I have a positive correlation which is dependent to the value of acceleration. These results can be put into application in real if engineers who are trying to build a trampoline try to apply the theory into building a safe trampoline so when a person with a normal weight pushes up and down on it they are able to oscillate pulling from a position of equilibrium, allowing the springs to acquire potential which assists the upward bounce without anything going wrong as all anomalous results can be identified through such lab tests and then avoided in real life. The little differences between experimental and theoretical theory and anomalous results maybe due to certain things that have not been taken into account, such as stopping the stop watch couple of seconds later, which is a human error or because the spring length was not quantified beforehand and the spring could have possibly be compressed which had an effect on the size, and also the surface the experiment was carried out on could have influenced measurements. Taking different settings into account would give precision with our results as not all SHM systems in real life are applied to the settings that were used in the Lab, it is important to be able to generalise the information gained. Low cost materials were used throughout the experiment, hence why it was easier to analyse the connections between the variables that shape the SMH of a spring mass system.
Percentage error
Gradient = 6.2866 % Error = Original value – Result g value X 100 Original value
= 9.8 - 6.28 X 100 9.8
= 35.92%
Precaution
* Ensure equipment works properly e.g. stopwatch.
* Avoid human error by being very careful repeating experiments many times and working out averages to deducting anomalous results and ensuring accuracy
* The spring should return to the original shape after the mass is taken off, hence why the limit of mass attached should not exceed the limit.
* Using the identical spring throughout with the same mass, length, coils etc. to make the experiment fair. CONCLUSION
The hypothesis which states that ‘the length of spring is directly proportional to the applied force’ is correct as it caused a larger change in length of spring as a greater force was applied as the masses increased. The oscillations are characterised by the principle variables. The safeguarding of the identical spring throughout the experiment meant that the spring constant was always maintained for any value of force and spring extension. To improve this experiment similar studies in different setting could be carried out.
The lab has aided me to understand that the elasticity force means that if the spring is stretched within its limit, it will always return to its original shape. This can be applied in real life which has been reviewed earlier in lab; the trampoline apparatus which is sufficient for any person with the weight that is the maximum limit of the SMH system of a trampoline will be able to jump on without the trampoline losing shape. Other real life examples are the clock pendulum and a swing. The restoring forces in the swing are sufficient enough to make the swing move forward and travel upwards when it is being pushed; the opposite direction gives out a restoring force which pushes the swing back down. Again, this force is not great enough to overpower equilibrium, and travel in the other side. The swing will swing in a constant period and amplitude as long as the process is repeated, oscillations will carry on.
As discussed earlier experimental errors and my anomalies could be due to not measuring correctly or pulling the spring at different strengths each time, this errors need to be corrected in real life application.
REFERENCES Peters Bird (2011); Oscillations and harmonic oscillations – exam tutor; (http://www.examstutor.com/physics/resources/studyroom/waves_and_oscillations/oscillations_and_harmonic_oscillations/); Accessed 28 Nov 2014
D.G. Simpson, Ph.D (2013); Waves, Acoustics, Electromagnetism, Optics, and Modern Physics; Introductory Physics; 11; (2); P.8
No author (2014); Acceleration of gravity – Physics Gravity; http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity (Accessed 26 Nov 2014)
The Editors of Encyclopædia Britannica (2014) – Britain Enclopaedia; Accessed 28 Nov 2014
Erin Bjornsson (2010); Hooke's Law; Calculating Spring Constants; 6; (4); P.4
Jane Malboo. (2006). forces are central to physics. Available: http://drbonesshow.com/links/funphysics.html. Last accessed 29th nov 2014.
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Words: 363 - Pages: 2 | 5,441 | 19,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.909797 |
https://chestofbooks.com/crafts/metal/Metal-Pattern/Drawing-Tools-And-Material-Part-4.html | 1,669,747,851,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00047.warc.gz | 208,940,802 | 7,098 | Triangles, or Set Squares. - In the selection of triangles, the draftsman has the choice in material between pear wood; mahogany, ebony lined; hard rubber; German silver, and steel, silver or nickel plated. In style he has the choice between open work, of the form shown in Fig. 107, and the solid, as in Fig. L08. In shape, the two kinds which are adapted to the pattern cutter's use are shown in Figs. 107 and 108, the latter being described as 30, 60 and 90 degrees, or 30 by 60 degrees, and the former as 45, 45 and 90 degrees, or simply 45 degrees. The special uses of each of these two tools are shown in the chapter on Geometrical Problems (Chap. IV). In size, the pattern cutter requires large rather than small ones. If he can have two sizes of each, the smaller should measure from 4 to 6 inches on the side, and the larger from 10 to 12 inches; but if only a single size is to be had, one having dimensions intermediate to those named will be found the most serviceable.
Fig. 107 - Open Hard Rubber Triangle or Set Square, 40 x 45 x 90 Degrees.
Fig. 108.. - Hard Wood Triangle or Set Square, SO x GO x 00 Degrees.
The value of a triangle, for whatever purpose used, depends on its accuracy. Particularly is this to be said of the right angle, which is used more than either of the others. A method of testing the accuracy of the right angle is shown in Fig. 109. Draw the line A B with an accurate ruler or straight-edge. Place the right angle of a triangle near the center of this line, as shown by D C B, and make one of the edges coincide with the line, and then draw the line D C against the other edge. Turn the triangle into the position indicated by D C A. If it is found that the sides agree with A C and C D, it is proof that the angle is a right angle and that the sides are straight.
Fig. 109. - Testing the Right Angle of a Triangle.
Besides the kinds of triangles described above, a fair article can be made by the mechanic from sheet zinc or of heavy tin. Care must, however, be taken in cutting to obtain the greatest possible accuracy. For many of the purposes for which a large size 45 degree triangle would be used the steel square is available, but as the line of the hypothenuse is lacking, it cannot be considered a substitute.
Fig. 110. - Compasses with Interchangeable Parts.
Fig. 111 - Plain Dividers.
Fig. 112. - Hair-Spring Dividers.
Fig. 113. - Steel Spring Spacers.
Compasses and Dividers. - The term compasses is applied to those tools, of various sizes and descriptions, which hold a pencil and pen in one leg, and are used for drawing circles, while dividers are those tools which, while of the same general form as compasses, have both legs ending in fixed points, and are used for measuring spaces. A special form of dividers - used exclusively for setting off spaces, as in the divisions of a profile line - is called spacers, as illustrated and described below.
A pair of compasses consists of the parts shown in Fig. 110, being the instrument proper with detachable points, and extras comprising a needle point, a pencil point, a pen and a lengthening bar, all as shown to the left. In selection, care should be given to the workmanship; notice whether the parts fit together neatly and without lost motion, and whether the joint works tightly and yet without too great friction. A good German silver instrument, although quite expensive at the outset, will be found the cheapest in the end. A pencil point of the kind shown in our engraving is to be preferred over the old style which clamps a common pencil to the leg. The latter is not nearly so convenient and is far less accurate.
Of dividers there are two general kinds, the plain dividers, as shown in Fig. 111, and the hair-spring dividers, as shown in Fig. 112. The latter differ from the former simply in the fact of having a fine spring and a joint in one leg, the movement being controlled by the screw shown at the right. In this way, after the instrument has been set approximately to the distance desired, the adjustable leg is moved, by means of the screw, either in or out, as may be required, thus making the greatest accuracy of spacing possible. Both instruments are found desirable in an ordinary set of tools. The plain dividers will naturally be used for larger and less particular work, while the hairspring dividers will be used in the finer parts. It frequently happens that two pairs of dividers, set to different spaces, are convenient to have at the same time. A pair of spacers, shown in Fig. 113, is almost indispensable in a pattern cutter's outfit. He will find advantageous use for this tool, even though possessing both pairs of dividers described above. In size they are made less than that, of the dividers. The points should be needle-like in their Oneness, and should be capable of adjustment to within a very small distance of each other. It is sometimes desirable to divide a given profile into spaces of an eighth of an inch. The spacers should be capable of this, as well as adapted to spaces of three-quarters of an inch, without being too loose. As will be seen from the engraving, this instrument is arranged for minute variations in adjustment. | 1,181 | 5,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-49 | latest | en | 0.94474 |
https://inches-to-cm.appspot.com/5930-inches-to-cm.html | 1,721,436,113,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00786.warc.gz | 258,964,115 | 6,637 | Inches To Centimeters
# 5930 in to cm5930 Inches to Centimeters
in
=
cm
## How to convert 5930 inches to centimeters?
5930 in * 2.54 cm = 15062.2 cm 1 in
A common question is How many inch in 5930 centimeter? And the answer is 2334.64566929 in in 5930 cm. Likewise the question how many centimeter in 5930 inch has the answer of 15062.2 cm in 5930 in.
## How much are 5930 inches in centimeters?
5930 inches equal 15062.2 centimeters (5930in = 15062.2cm). Converting 5930 in to cm is easy. Simply use our calculator above, or apply the formula to change the length 5930 in to cm.
## Convert 5930 in to common lengths
UnitLength
Nanometer1.50622e+11 nm
Micrometer150622000.0 µm
Millimeter150622.0 mm
Centimeter15062.2 cm
Inch5930.0 in
Foot494.166666667 ft
Yard164.722222222 yd
Meter150.622 m
Kilometer0.150622 km
Mile0.0935921717 mi
Nautical mile0.0813293737 nmi
## What is 5930 inches in cm?
To convert 5930 in to cm multiply the length in inches by 2.54. The 5930 in in cm formula is [cm] = 5930 * 2.54. Thus, for 5930 inches in centimeter we get 15062.2 cm.
## Alternative spelling
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https://library.fiveable.me/symplectic-geometry/unit-6/linear-symplectic-transformations/study-guide/SSwKaSbyKAQynzke | 1,726,623,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00558.warc.gz | 323,651,488 | 54,199 | # 6.2 Linear symplectic transformations
Linear symplectic transformations are the backbone of symplectic geometry. They preserve the , maintaining the structure of symplectic vector spaces. These transformations are crucial in physics, especially in .
The encompasses these transformations. It's a with fascinating properties, like being non-compact and connected. Understanding this group is key to grasping the mathematical framework of classical mechanics and beyond.
## Linear symplectic transformations
### Defining properties and characteristics
###### Top images from around the web for Defining properties and characteristics
• Linear symplectic transformations preserve the symplectic form on a
• Defining property expressed mathematically $ω(Tv, Tw) = ω(v, w)$ for all vectors v and w in the symplectic vector space, where ω represents the symplectic form
• Always invertible and volume-preserving ()
• Composition of two linear symplectic transformations results in another
• Preserve structure fundamental in Hamiltonian mechanics
• Form a group called the linear symplectic group denoted as $Sp(2n, R)$ for a 2n-dimensional real symplectic vector space
• Maintain canonical structure of equations in physics (Hamilton's equations)
• Linearizations of symplectomorphisms which act as canonical transformations in Hamiltonian mechanics
### Examples and applications
• in 2D phase space preserve area and symplectic structure
• Scaling transformations that maintain ()
• in phase space that preserve symplectic form
• Time evolution of generates linear symplectic transformations
• Optical systems modeled using (ABCD matrices) in paraxial optics
• Transformations between different in classical mechanics (position-momentum to action-angle variables)
## Group structure of transformations
### Properties of the linear symplectic group
• Linear symplectic group $Sp(2n, R)$ classified as a Lie group with both group and
• Dimension of $Sp(2n, R)$ equals $n(2n + 1)$, where n represents half the dimension of the symplectic vector space
• Closed subgroup of the general linear group $GL(2n, R)$
• Non-compact and connected group but not simply connected
• of $Sp(2n, R)$, denoted $sp(2n, R)$, consists of 2n × 2n matrices X satisfying $XJ + JX^T = 0$, where J represents the
• Exponential map from $sp(2n, R)$ to $Sp(2n, R)$ proves surjective allowing every element of $Sp(2n, R)$ to be expressed as the exponential of an element in $sp(2n, R)$
• Contains important subgroups like the unitary symplectic group $USp(2n) = Sp(2n, R) ∩ U(2n)$ relevant in quantum mechanics
### Group operations and structure
• Composition of linear symplectic transformations serves as the group operation
• Identity element represented by the identity matrix
• Inverse of a symplectic transformation always exists and remains symplectic
• Group multiplication non-commutative for dimensions greater than 2
• Topology of $Sp(2n, R)$ diffeomorphic to $R^{n(2n+1)} × S^1$ for $n > 1$
• Fundamental group of $Sp(2n, R)$ isomorphic to the integers $Z$ for $n > 1$
## Matrix representations of transformations
### Symplectic bases and matrix forms
• for 2n-dimensional space consists of vectors $\{e1, ..., en, f1, ..., fn\}$ satisfying $ω(ei, fj) = δij$ with all other pairings zero
• Linear symplectic transformation represented by 2n × 2n matrix S satisfying $S^T J S = J$, where J denotes the standard symplectic matrix
• Standard symplectic matrix J expressed as 2n × 2n block matrix $[0 \quad I; -I \quad 0]$, with I representing the n × n identity matrix
• Matrix representation preserves block structure $[A \quad B; C \quad D]$, where A, B, C, and D represent n × n matrices satisfying specific relations
• always equals 1
• S given by $J^{-1} S^T J$, simplifying inverse computations
• Set of all 2n × 2n symplectic matrices forms matrix Lie group isomorphic to $Sp(2n, R)$
### Properties and computations
• Symplectic matrices preserve symplectic inner product between vectors
• occur in reciprocal pairs (λ, 1/λ)
• Symplectic matrices have even-dimensional eigenspaces
• Trace of symplectic matrix remains invariant under similarity transformations
• Symplectic Gram-Schmidt process used to construct symplectic bases
• states any symmetric positive definite matrix can be diagonalized by a symplectic congruence transformation
## Transformations in Hamiltonian mechanics
### Canonical transformations and symplectic flows
• Linear symplectic transformations preserve canonical form of Hamilton's equations of motion
• Flow of linear Hamiltonian system generates one-parameter subgroup of linear symplectic transformations
• Preserve symplectic structure of phase space maintaining physical properties of Hamiltonian systems
• and eigenvectors provide information about stability of equilibrium points
• allows unique factorization of linear symplectic transformation into product of symplectic rotation and symplectic dilation
• Key role in study of normal forms for Hamiltonian systems simplifying analysis of nonlinear dynamics near equilibrium points
### Applications in physics and mechanics
• Used in perturbation theory for nearly integrable Hamiltonian systems ()
• Describe evolution of
• Model linear optical systems in laser physics and beam propagation
• Analyze stability of periodic orbits in
• Study invariant tori in of dynamical systems
• Characterize in symplectic topology and geometry
## Key Terms to Review (35)
Canonical Coordinate Systems: Canonical coordinate systems are specific sets of coordinates used in symplectic geometry that simplify the representation of Hamiltonian systems. These coordinates, usually expressed in pairs of position and momentum, facilitate the analysis of dynamical systems by transforming the equations of motion into a more manageable form. This transformation helps to reveal the underlying geometric structures and symmetries present in the system, making them crucial in the study of linear symplectic transformations.
Canonical Transformation: A canonical transformation is a change of coordinates in phase space that preserves the symplectic structure of Hamiltonian mechanics. This means that if you transform the coordinates and momenta of a dynamical system, the new coordinates still satisfy Hamilton's equations, reflecting the underlying physics. These transformations are crucial because they allow for the simplification of problems, reveal conserved quantities, and maintain the relationships defined by symplectic geometry.
Celestial mechanics: Celestial mechanics is the branch of astronomy that deals with the motions and gravitational interactions of celestial bodies, such as planets, moons, and stars. This field uses mathematical models and physical principles to predict the behavior of these bodies over time, which is crucial for understanding orbital dynamics and stability in space.
Coherent States in Quantum Optics: Coherent states in quantum optics are specific quantum states of a harmonic oscillator that closely resemble classical states, exhibiting minimum uncertainty in phase space. These states are vital for understanding the behavior of laser light and have applications in various fields including quantum information and quantum computing, where they help bridge the gap between classical and quantum descriptions of light.
Determinant of symplectic matrix: The determinant of a symplectic matrix is a specific value that reflects the volume-preserving properties of linear transformations in symplectic geometry. In particular, a symplectic matrix, which is a square matrix that preserves a symplectic form, has a determinant equal to 1 or -1, indicating that it either preserves or reverses orientation while conserving the area in the phase space.
Dilation: Dilation is a geometric transformation that alters the size of an object while maintaining its shape and proportions. In the context of linear symplectic transformations, dilation refers to a specific type of transformation that scales vectors in a symplectic vector space, affecting the area while preserving the symplectic structure.
Eigenvalues of symplectic matrices: Eigenvalues of symplectic matrices are complex numbers that arise from the study of linear transformations preserving a symplectic structure. These eigenvalues are crucial in understanding the behavior of dynamical systems and solutions to Hamiltonian equations, reflecting the geometric properties of phase space transformations that maintain the area and symplectic form.
Ergodic theory: Ergodic theory is a branch of mathematics that studies the long-term average behavior of dynamical systems. It connects statistical mechanics and deterministic systems, showing how a system's time evolution relates to its space configuration over time. This concept is crucial for understanding the stability and predictability of symplectic transformations, conservation laws, and geometric structures in various contexts.
Hamiltonian Mechanics: Hamiltonian mechanics is a reformulation of classical mechanics that emphasizes the use of Hamiltonian functions, which describe the total energy of a system, to analyze the evolution of dynamical systems. This framework connects deeply with symplectic geometry and offers insights into the conservation laws and symmetries that govern physical systems.
Inverse of symplectic matrix: The inverse of a symplectic matrix is another symplectic matrix that, when multiplied with the original matrix, yields the identity matrix. This property is essential in the study of linear symplectic transformations, as it ensures that the transformation can be reversed, preserving the underlying symplectic structure. The existence of the inverse highlights the robustness of symplectic matrices in preserving geometric properties in Hamiltonian mechanics and other areas.
KAM Theory: KAM Theory, or Kolmogorov-Arnold-Moser theory, is a mathematical framework that addresses the stability of integrable systems under small perturbations, demonstrating that many Hamiltonian systems exhibit quasi-periodic behavior. This concept is crucial for understanding how certain Hamiltonian vector fields maintain their structure despite small changes, thus connecting it to the behavior of dynamical systems and their conservation laws.
Lie Algebra: A Lie algebra is a mathematical structure that consists of a vector space equipped with a binary operation called the Lie bracket, which satisfies certain properties such as bilinearity, alternativity, and the Jacobi identity. Lie algebras are fundamental in the study of symmetries and are closely connected to Lie groups, providing a framework to analyze linear symplectic transformations.
Lie Group: A Lie group is a mathematical structure that combines algebraic and geometric properties, specifically a group that is also a differentiable manifold. This dual nature allows for the study of continuous transformations, making Lie groups essential in understanding symmetries and conservation laws in various fields, including physics and geometry.
Linear hamiltonian systems: Linear Hamiltonian systems are a special class of dynamical systems governed by linear differential equations that respect Hamiltonian mechanics. These systems are defined by a Hamiltonian function, which encapsulates the total energy of the system, and exhibit a symplectic structure, ensuring that the geometry of phase space is preserved over time. They play a crucial role in understanding the behavior of physical systems, especially in mechanics and optics.
Linear symplectic group: The linear symplectic group, denoted as $Sp(2n, \mathbb{R})$, is the group of $2n \times 2n$ matrices that preserve a symplectic form on a real vector space. This group consists of all linear transformations that maintain the structure of the symplectic vector space, making it crucial in areas like classical mechanics and Hamiltonian systems.
Linear symplectic transformation: A linear symplectic transformation is a linear map between symplectic vector spaces that preserves the symplectic structure, meaning it maintains the area in phase space under transformation. These transformations are represented by matrices that satisfy specific conditions, particularly that the matrix and its transpose must yield a symplectic matrix, ensuring that the inner product remains unchanged. They play a crucial role in the study of Hamiltonian mechanics and the geometric structure of phase spaces.
Liouville Theorem: The Liouville Theorem states that in a Hamiltonian system, the volume of phase space is preserved under the flow generated by Hamilton's equations. This concept highlights the symplectic structure of phase space, which is fundamental in understanding how systems evolve over time and connects deeply with transformations that preserve geometric properties and behaviors in dynamics.
Non-compact group: A non-compact group is a mathematical group that does not satisfy the compactness property, meaning it cannot be covered by a finite number of open sets. This implies that the group is either infinite or lacks the necessary boundedness and closure conditions to be classified as compact. Non-compact groups play an important role in various mathematical contexts, especially in understanding linear symplectic transformations where these groups may arise when analyzing symmetries and transformations in infinite-dimensional spaces.
Phase Space Volume: Phase space volume refers to the multi-dimensional space in which all possible states of a system are represented, with each state corresponding to a unique point in this space. This concept is crucial in understanding how physical systems evolve over time, especially under linear symplectic transformations, as it relates to the conservation of volume during these transformations.
Poisson bracket: The Poisson bracket is a binary operation defined on the algebra of smooth functions over a symplectic manifold, capturing the structure of Hamiltonian mechanics. It quantifies the rate of change of one observable with respect to another, linking dynamics with the underlying symplectic geometry and establishing essential relationships among various physical quantities.
Ray transfer matrices: Ray transfer matrices are mathematical tools used to describe the propagation of rays through optical systems in a symplectic geometry framework. They provide a systematic way to analyze how light beams transform as they pass through different optical elements, allowing for the study of their behavior and interactions in complex configurations.
Rotation Matrices: Rotation matrices are special orthogonal matrices used to perform rotation transformations in Euclidean space. They preserve the inner product, meaning they maintain distances and angles between vectors, which is essential in understanding linear symplectic transformations. These matrices have applications in various fields, including physics and computer graphics, and they play a critical role in preserving the symplectic structure of a vector space.
Shear Transformations: Shear transformations are linear mappings that displace points in a fixed direction, effectively slanting the shape of an object without altering its area. These transformations can be understood as changes where one direction is scaled by a factor while others remain unchanged, resulting in a deformation of the geometric figure. They are essential in the study of symplectic geometry, particularly in how symplectic forms interact with such transformations.
Smooth manifold structures: Smooth manifold structures refer to the mathematical framework that defines a smooth manifold, which is a topological space that locally resembles Euclidean space and is equipped with a smooth structure allowing for calculus. These structures are essential for analyzing geometric properties and behaviors in symplectic geometry, particularly in the context of transformations and mappings that preserve certain symplectic properties.
Sp(2n, r): The term sp(2n, r) refers to the symplectic group of degree 2n over the field of real numbers r, which consists of all 2n x 2n matrices that preserve a symplectic form. These matrices are significant as they represent linear symplectic transformations that maintain the geometry defined by the symplectic structure, which is crucial in understanding the behavior of dynamical systems in physics and mathematics.
Standard symplectic matrix: A standard symplectic matrix is a square matrix that preserves the symplectic structure, which means it maintains the properties of a symplectic form under linear transformations. Specifically, for a matrix to be considered standard symplectic, it must satisfy the condition that its transpose multiplied by a given symplectic form results in the same symplectic form. This type of matrix plays a key role in linear symplectic transformations, enabling the study of systems such as Hamiltonian dynamics and phase spaces.
Symplectic basis: A symplectic basis is a specific kind of basis for a symplectic vector space that consists of pairs of vectors which are related through the symplectic form. This unique structure highlights the interplay between geometry and linear algebra, where each pair represents a canonical symplectic pairing. Understanding symplectic bases is crucial for analyzing the properties of symplectic vector spaces and the behavior of linear transformations that preserve this structure.
Symplectic Capacities: Symplectic capacities are numerical invariants that measure the 'size' of a symplectic manifold in a way that is compatible with the symplectic structure. They help to classify symplectic manifolds and can be used to compare different manifolds based on their geometric and topological properties. This concept connects deeply with the applications of foundational theorems, linear transformations in symplectic spaces, implications of fundamental results like Gromov's theorem, and the interplay between geometric optics and symplectic structures.
Symplectic eigenvalues: Symplectic eigenvalues are specific values associated with a linear symplectic transformation that characterize how the transformation affects the symplectic structure of a vector space. They arise from the study of symplectic matrices, where the eigenvalues provide insights into the geometric properties of the transformations and their behaviors under symplectic bases and normal forms. Understanding these eigenvalues helps in classifying symplectic forms and in analyzing the stability of dynamical systems.
Symplectic flows: Symplectic flows refer to the continuous evolution of a system in a symplectic manifold, characterized by preserving the symplectic structure over time. This concept is essential in understanding how Hamiltonian systems operate, as it emphasizes the conservation of geometric properties during the flow induced by Hamiltonian dynamics. Symplectic flows ensure that phase space volume is preserved, which is a fundamental aspect of classical mechanics and dynamical systems.
Symplectic Form: A symplectic form is a closed, non-degenerate 2-form defined on a differentiable manifold, which provides a geometric framework for the study of Hamiltonian mechanics and symplectic geometry. It plays a crucial role in defining the structure of symplectic manifolds, facilitating the formulation of Hamiltonian dynamics, and providing insights into the conservation laws in integrable systems.
Symplectic polar decomposition theorem: The symplectic polar decomposition theorem states that any linear symplectic transformation can be uniquely decomposed into a product of a symplectic matrix and a positive symmetric matrix. This decomposition plays a crucial role in understanding the structure of symplectic transformations and their applications in Hamiltonian mechanics. The theorem shows how to separate the 'symplectic' part of a transformation from its 'positive symmetric' part, which is essential for analyzing the geometry of symplectic spaces.
Symplectic vector space: A symplectic vector space is a finite-dimensional vector space equipped with a non-degenerate, skew-symmetric bilinear form called the symplectic form. This structure allows for a geometric framework where concepts like area and volume can be naturally interpreted, making it essential in the study of Hamiltonian mechanics and other areas of mathematics. The symplectic form must satisfy certain properties, like being closed and non-degenerate, which leads to a rich interplay with linear algebra and transformations.
Symplectomorphism: A symplectomorphism is a smooth, invertible mapping between two symplectic manifolds that preserves their symplectic structure. This means that if you have a symplectic form on one manifold, the image of that form under the mapping will still be a symplectic form on the other manifold, ensuring the preservation of geometric and physical properties between these spaces.
Williamson's Theorem: Williamson's Theorem is a fundamental result in symplectic geometry that characterizes the linear symplectic transformations of a symplectic vector space. It establishes that every symplectic transformation can be represented in terms of a certain kind of matrix called a symplectic matrix, which preserves the symplectic structure of the space. This theorem provides a powerful tool for understanding the classification of symplectic forms and the behavior of linear symplectic transformations. | 4,188 | 21,502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 27, "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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.80862 |
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Instructor: Lydia Neptune
If you are shot out of a cannon, how do you know when you've reached your maximum height? When walking through a valley, how do you know when you are at the bottom? In this lesson, use the properties of the derivative to find the maxima and minima of a function.
## Extrema
We know that an extrema is a maximum or minimum value on a graph. If I'm given a graph, I can point out where the extrema are. Here, I've got a global maximum value. It's larger than any other point on this entire region. Here, I've got a local maximum value. It's the largest point in that area. Here, I've got a global minimum, and I've got a local minimum. I also can't forget to include the end points on my graph. But how do I actually calculate these?
Let's consider Super C, Human Cannonball, for a second. Super C is shot up into the air, and we can graph his height as a function of time. If I want to know his maximum height, I can see it's right between when he stops going up, and he starts going down. So it's right at that point. If I put this on a graph, I find that a maximum value would be between where the derivative is positive and the derivative is negative. That's when the slope changes from going up to going down.
A local minimum is where the slope changes from going down to going up. So for a continuous function, when the derivative changes from positive to negative, the derivative is going to go through zero. At this global maximum value, the derivative will be zero at that point exactly. Similarly, here, for this local maximum value, the derivative will be zero at the very top. Super C, at the very top of his trajectory, was not going up, and he was not going down. His height as a function of time, that derivative, was zero right there.
## Finding Extrema
We can use this to our advantage to find extreme values. So for some function y=f(x), the first thing we want to do is find the critical points of this function. (That is, where the derivative is equal to zero.) So we are going to find some x values where the derivative is equal to zero. The second step is that we are going to find what y is at those critical points. We are also going to find y at the end points. So we might realize that Super C reached the pinnacle of his height 1 second into his flight, but now we are going to find exactly how high that was - that's the y value in this case. The third step is that we are going to compare all of those y values that we just calculated, and we are going to see which one is the maximum value that corresponds to our global maximum. We will also see which one is our minimum value that is going to correspond to our global minimum value. All of the other critical points might be local maxima or minima, but not always.
So let's put some numbers on this. Let's look at Super C, the human cannonball. Let's say that his height as a function of time (I'm going to write y=f(x), so x here is time and y is height) - let's say that that function is -x^2 + 2x and his entire flight goes from 0 to 2. OK, so our first step in finding all of the extrema is to find the critical points, that is, where f`(x)=0. So I'm going to differentiate our f(x). The derivative of -x^2 + 2x is -2x + 2. Now I'm going to set that equal to zero, and solve for x. Well, f`(x)=0 when x=1. So I know at what point in time Super C reached the pinnacle, at x=1, but how high was he at x=1? I'm going to calculate f(x=1). I'm going to plug 1 into our original equation here. It's important that it's the original equation and not the derivative. So I have -(1)^2 + 2(1). That's -1 + 2, or just 1. His height at x=1 is 1. So this point here is the point (1, 1).
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Question: How can I shuffle a deck of cards or similar set of sequential numbers? How can I reorder a list of things in random order?
Answer: Think about one way you could do this with a physical deck of cards.
You could put the cards in order (not necessary, but in theory it shouldn't matter). Then you could randomly pick any card from the ones in the deck and either turn it over (to display it immediately) or move it to a second pile (to create a shuffled deck).
If you were really good about picking one card from the remaining ones in a truly random manner, the resulting output (display or deck) would be in truly random order.
Essentially, that's what we will do in the code that is given below.
The only hard part is replicating the physical removal of the card from the deck, so that the deck gets smaller and smaller. We could simply do this using an array and "sliding" all the cards "above" the chosen one "down" by one notch. This works. But it can be *very* slow!
So let's think of a slight modification of that physical scheme: Instead of just picking a random card from the deck, let's pull out the chosen card and--at the same time--take the card on the top of the deck and shove it there in place of the picked card. Does that change anything, really? Of course not! With good picking of random numbers, you get good random results.
And so, on to the code:
`<%' This function will DEAL the "needed"' number of values from the given "inArray"'' If the value for "needed" matches the' upper bound of the "inArray", then the' entire "inArray" is dealt out.'' NOTE: As written, this code never uses' or touches element zero of the inArray' and puts no value in element zero of the' outArray. (Obviously, easy to change.)'Function Shuffle( inArray, needed ) ' find out how many input elements there are... incnt = UBound( inArray ) ' then create the output array to be the size ' requested via the "needed" argument dim outArray redim outArray( needed ) ' now we will select the number of values ' specified as "needed"... For i = 1 To needed ' choose a random number from 1 to our ' current input array usage size... choose = Int( incnt * Rnd(1) ) + 1 ' put that chosen element into the next ' slot in the output array... outArray( i ) = inArray( choose ) ' here's the tricky part: Since we just ' used the "choose" element, we don't need ' it any more...we replace it with the last ' element of the in-use part of the array! inArray( choose ) = inArray( incnt ) ' and then we (effectively) shrink the array! ' Next time through the loop, there will be ' one fewer elements in the array to choose ' from...because we have (effectively) deleted ' the one just chosen! incnt = incnt - 1 Next ' return the shuffled output Shuffle = outArrayEnd Function' This is just a convenience function'' If you need *all* the "cards" in a deck of a given' size shuffled, and the "name" of a card can just be ' its numeric position in the unshuffled deck, then ' just call ShuffleDeck, passing the size of the deck' to be shuffled.'Function ShuffleDeck( deckSize ) Dim i, deck() ReDim deck( deckSize ) For i = 1 To deckSize deck(i) = i Next ShuffleDeck = Shuffle( deck, deckSize )End Function %><% Randomizear = Array(0,"you","can","put","anything","in","the","array","of","course")str = Mid( Join(ar," "), 2 )Response.Write "Picking 4 words from this list: " _ & str & "
" & vbNewLinesh = Shuffle( ar, 4 )For i = 1 to 4 Response.Write "
1. " & sh(i) & vbNewLineNext Response.Write "
" & vbNewLineResponse.Write "Shuffling a 'deck' of 20 numbered cards.
" _ & "The cards were originally numbered from 1 to 20.
" & vbNewLinesh = ShuffleDeck( 20 )str = Mid( Join( sh, "," ), 2 )Response.Write "The shuffled deck: " & str & "" & vbNewLine%>`
I have included a pair of demonstrations of the code. First, by picking four words from a list (array) of words. Second, I use the "auxiliary" function that I show there to generate and shuffle a "deck" of 20 cards. | 1,109 | 4,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-39 | latest | en | 0.893333 |
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Joined: 24 Apr 2006
Posts: 350
Posted: Thu Sep 01, 2011 10:32 am Post subject: CLOP for Noisy Black-Box Parameter Optimization Hi, This is my paper for the Tilburg conference: Title: CLOP: Confident Local Optimization for Noisy Black-Box Parameter Tuning Abstract: Artificial intelligence in games often leads to the problem of parameter tuning. Some heuristics may have coefficients, and they should be tuned to maximize the win rate of the program. A possible approach consists in building local quadratic models of the win rate as a function of program parameters. Many local regression algorithms have already been proposed for this task, but they are usually not robust enough to deal automatically and efficiently with very noisy outputs and non-negative Hessians. The CLOP principle, which stands for Confident Local OPtimization, is a new approach to local regression that overcomes all these problems in a simple and efficient way. It consists in discarding samples whose estimated value is confidently inferior to the mean of all samples. Experiments demonstrate that, when the function to be optimized is smooth, this method outperforms all other tested algorithms. pdf and source code: http://remi.coulom.free.fr/CLOP/ It makes no miracle: you'll have to play a lot of games to get really good parameters. But it is certainly much more efficient than any manual method you could use with bayeselo. It is also more efficient than any other algorithm I am aware of. Compared to the old version of QLR, I solved all the unstability problems. I do not have a mathematical proof of convergence, but I am convinced it always work well, unless the maximum is at a discontinuity, which never happens in practice. Comments and questions are welcome. Rémi
Joona Kiiski
Joined: 18 Jan 2009
Posts: 546
Posted: Fri Sep 02, 2011 8:52 am Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization Very impressive achievement Remi. I will try your new tool at some point in the near future!_________________Joona Kiiski
Rémi Coulom
Joined: 24 Apr 2006
Posts: 350
Posted: Fri Sep 02, 2011 9:45 am Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization Thanks Joona. I added some screenshots of the program to the web page, and a description: These are some screenshots of an old version in action. You can also run the program from the command line, which is more convenient for use on a remote cluster. The program can deal with chess outcomes (win/draw/loss), and integer parameters. The program is written in C++ with Qt, so it can be compiled and run on Windows, Linux, and MacOS. http://remi.coulom.free.fr/CLOP/ Rémi
Zach Wegner
Joined: 08 Mar 2006
Posts: 1922
Location: Earth
Posted: Fri Sep 02, 2011 4:44 pm Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization
zamar wrote: Very impressive achievement Remi. I will try your new tool at some point in the near future!
+1
I am very glad that you decided to re-release your source code, even though you are a "commercial" now. Thanks!
I will try and understand your paper. Do you know how well your ideas could be applied to non-game-playing applications (i.e., a floating point objective function)?
_________________
http://zct.sourceforge.net
http://chessprogramming.wikispaces.com
Marco Costalba
Joined: 14 Jun 2008
Posts: 2090
Posted: Fri Sep 02, 2011 5:34 pm Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization
Rémi Coulom wrote: Thanks Joona. I added some screenshots of the program to the web page, and a description: These are some screenshots of an old version in action. You can also run the program from the command line, which is more convenient for use on a remote cluster. The program can deal with chess outcomes (win/draw/loss), and integer parameters. The program is written in C++ with Qt, so it can be compiled and run on Windows, Linux, and MacOS. http://remi.coulom.free.fr/CLOP/ Rémi
As you perhaps know Stockfish has been tuned in an automatic way by means of an (not disclosed) algorithm written by Joona, so I am very happy that he finds your work useful because he is the expert of tuning in the SF team...and so I hope this will give him good hints to further tune the tuner
Rémi Coulom
Joined: 24 Apr 2006
Posts: 350
Posted: Fri Sep 02, 2011 7:58 pm Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization
Zach Wegner wrote: Do you know how well your ideas could be applied to non-game-playing applications (i.e., a floating point objective function)?
I did not try, but I expect the basic idea of CLOP would work well in many situations, even the completely noiseless case.
I am really enthusiastic about this algorithm, because it is extremely simple, and very universal. The problem of optimizing a function from noisy (or noiseless) observations has really been researched a lot for more than 50 years. It is really difficult to contribute anything significant to this field. I am looking forward to the feedback of optimization specialists. Time will tell if CLOP makes an impact.
Rémi
Rein Halbersma
Joined: 22 May 2007
Posts: 241
Posted: Fri Sep 02, 2011 10:07 pm Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization Hi Rémi, Can CLOP also be applied to LOS as an objective function? Suppose that a tournament has a very skewed prize money distribution (e.g. in poker). In such cases, I can imagine that programs slightly below the absolute top might want to maximize a different mean-variance combination than the best programs. E.g. optimize their chance of winning a tournament, rather than their ELO. Rein
Rémi Coulom
Joined: 24 Apr 2006
Posts: 350
Posted: Sat Sep 03, 2011 9:24 am Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization
Rein Halbersma wrote: Hi Rémi, Can CLOP also be applied to LOS as an objective function? Suppose that a tournament has a very skewed prize money distribution (e.g. in poker). In such cases, I can imagine that programs slightly below the absolute top might want to maximize a different mean-variance combination than the best programs. E.g. optimize their chance of winning a tournament, rather than their ELO. Rein
I don't really understand your question.
The objective function can be the expected value of any random variable that depends on parameters. So, if, instead of playing one game and getting the result, you play a tournament and observe LOS over a specific opponent, you can use CLOP to optimize it. But that would be a strange way to use CLOP. If you wish to optimize a program against a set of opponents instead of just one opponent, you can use the "Replications" option of CLOP to play a game against each opponent, and then CLOP will maximize the average winning rate against all these opponents.
Regarding your example, if I understand correctly, you mean letting the program be more agressive when it needs to win, and safer when a draw is OK. For that, you could tune your evaluation with CLOP, considering that a draw is a loss (resp. a win) to make it play agressively (resp. defensively). Make sure your evaluation function is asymmetric, then. It may be more efficient than using just contempt.
Rémi
Andres Valverde
Joined: 18 Feb 2007
Posts: 466
Location: Almeria. SPAIN
Posted: Sat Sep 03, 2011 10:28 am Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization Can anybody explain in a nutshell how can one use it in engine tunning? I find the doc very interesting but pretty abstract._________________Saludos, Andres
Rémi Coulom
Joined: 24 Apr 2006
Posts: 350
Posted: Sat Sep 03, 2011 10:40 am Post subject: Re: CLOP for Noisy Black-Box Parameter Optimization
Andres Valverde wrote: Can anybody explain in a nutshell how can one use it in engine tunning? I find the doc very interesting but pretty abstract.
Did you manage to open DummyExperiment.clop with CLOP?
Once you manage to run this experiment, you should be all set up. All you have to do is write your own scripts to replace DummyScript.py. Run DummyScript.py without arguments (or look at the source) for an explanation.
Rémi
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You cannot vote in polls in this forum | 2,206 | 8,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.913023 |
https://www.ucl.ac.uk/~ucahmto/0005_2023/Ch4.S16.html | 1,723,449,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00156.warc.gz | 790,118,763 | 7,342 | # 4.16 Matrix nullspace basis
We are ready to prove that the fundamental solutions of $A\mathbf{x}=\mathbf{0}$ are a basis for $N(A)$. To set up the notation, suppose $A$ is a $m\times n$ matrix, $R$ is a RREF matrix obtained by doing row operations to $A$, the number of columns of $R$ with a leading entry is $r$ and the number of columns with no leading entry is $k$, so $r+k=n$. Let the numbers of the columns with no leading entry be
$j_{1}
and the numbers of the columns that do have a leading entry be
$i_{1}
so that the numbers $j_{1},\ldots,j_{k},i_{1},\ldots,i_{r}$ contain all of the numbers $1,\ldots,n$ exactly once. The variable corresponding to a column with no leading entry is called a free parameter because its value can be chosen freely. There are $k$ fundamental solutions to $A\mathbf{x}=\mathbf{0}$, with the $j$th fundamental solution defined to be the one where the $j$th free parameter is 1 and all the other free parameters are 0.
###### Theorem 4.16.1.
The fundamental solutions to $A\mathbf{x}=\mathbf{0}$ are a basis of the nullspace $N(A)$.
Before we do the proof, let’s work an example to illustrate how it will go. Free parameters and their coefficients will be coloured red. Take
$R=\begin{pmatrix}{\color[rgb]{1,0,0}0}&1&{\color[rgb]{1,0,0}2}&0&{\color[rgb]{% 1,0,0}3}\\ {\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}&1&{\color[rgb]{1,0,0}4}\\ {\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}\end{pmatrix}$
so that we have $m=3$ and $n=5$. The columns with no leading entry are 1, 3, and 5 so $k=3$, ${\color[rgb]{1,0,0}j_{1}}=1,{\color[rgb]{1,0,0}j_{2}}=3,{\color[rgb]{1,0,0}j_{% 3}}=5$, and variables ${\color[rgb]{1,0,0}x_{1}}$, ${\color[rgb]{1,0,0}x_{3}}$, and ${\color[rgb]{1,0,0}x_{5}}$ are the free parameters. The columns with leading entries are 2 and 4, so $r=2$ and $i_{1}=2,i_{2}=4$. The equations corresponding to $R\mathbf{x}=\mathbf{0}$ are
$\displaystyle x_{2}+{\color[rgb]{1,0,0}2x_{3}}+{\color[rgb]{1,0,0}3x_{5}}$ $\displaystyle=0$ $\displaystyle x_{4}+{\color[rgb]{1,0,0}4x_{5}}$ $\displaystyle=0$
(missing off the final one, since it just says $0=0$). These equations show that once values for the free parameters are known, the other variables $x_{2}$ and $x_{4}$ are completely determined by those values — we have $x_{2}={\color[rgb]{1,0,0}-2x_{3}-3x_{5}}$ and $x_{4}={\color[rgb]{1,0,0}-4x_{5}}$.
The three fundamental solutions are
$\mathbf{s}_{1}=\begin{pmatrix}{\color[rgb]{1,0,0}1}\\ 0\\ {\color[rgb]{1,0,0}0}\\ 0\\ {\color[rgb]{1,0,0}0}\end{pmatrix},\mathbf{s}_{2}=\begin{pmatrix}{\color[rgb]{% 1,0,0}0}\\ -2\\ {\color[rgb]{1,0,0}1}\\ 0\\ {\color[rgb]{1,0,0}0}\end{pmatrix},\mathbf{s}_{3}=\begin{pmatrix}{\color[rgb]{% 1,0,0}0}\\ -3\\ {\color[rgb]{1,0,0}0}\\ -4\\ {\color[rgb]{1,0,0}1}\end{pmatrix}.$
These are linearly independent, because if
$a_{1}\mathbf{s}_{1}+a_{2}\mathbf{s}_{2}+a_{3}\mathbf{s}_{3}=\mathbf{0}$
then row 1 of this equation shows that $a_{1}=0$, row 3 shows that $a_{2}=0$, and row 5 shows that $a_{3}=0$. To show that the fundamental solutions span, suppose $\mathbf{s}=\begin{pmatrix}s_{1}\\ \vdots\\ s_{5}\end{pmatrix}$ is any vector such that $R\mathbf{s}=\mathbf{0}$. We claim that $\mathbf{s}$ is equal to $s_{1}\mathbf{s}_{1}+s_{3}\mathbf{s}_{2}+s_{5}\mathbf{s}_{3}$. Certainly these two vectors have the same entries in rows 1, 3, and 5, since the entries in these rows are $s_{1},s_{3}$, and $s_{5}$. What about the entries in rows 2 and 4, that is, the values of $x_{2}$ and $x_{4}$? As above, these entries are completely determined by the values of the free parameters. The free parameters are the same for both vectors, so the values of $x_{2}$ and $x_{4}$ are the same as well. We have shown that any solution $\mathbf{s}$ is in the span of the fundamental solutions, so we are done.
###### Proof.
Theorem 3.9.1 shows that $N(A)=N(R)$, so we will show that the fundamental solutions are a basis of $N(R)$.
First we show that the fundamental solutions $\mathbf{s}_{1},\ldots,\mathbf{s}_{k}$ are linearly independent. Suppose that
$\sum_{j}a_{j}\mathbf{s}_{j}=\mathbf{0}.$ (4.6)
The first fundamental solution corresponds to the variable for column $j_{1}$, so $\mathbf{s}_{1}$ has a 1 in row $j_{1}$ and all the other fundamental solutions have a 0 there. Thus the entry in row $j_{1}$ on the left hand side of (4.6) is $a_{1}$, so $a_{1}=0$. Similarly all the other coefficients are zero.
Now let $\mathbf{s}$ be any solution of $A\mathbf{x}=\mathbf{0}$ and let the entry of $\mathbf{s}$ in row $i$ be $s_{i}$. We are going to show that
$\mathbf{s}=\sum^{k}_{i=1}s_{j_{i}}\mathbf{s}_{i}.$
The left hand side and right hand side of this claimed equation are solutions to $R\mathbf{x}=\mathbf{0}$ with the same free parameter values $s_{j_{1}},\ldots,s_{j_{k}}$. We just have to show that the values of the other variables are the same. But the values of the variables that aren’t free parameters are uniquely determined by the values of the free parameters, because by the RREF property, for $1\leqslant a\leqslant r$ the only equation containing $x_{i_{a}}$ is the one coming from row $a$ of $R$ which has the form
$x_{i_{1}}+\cdots=0$
where the only other variables occurring with nonzero coefficient are free parameters. Since $\mathbf{s}$ and $\sum^{k}_{i=1}s_{j_{i}}\mathbf{s}_{i}$ are solutions to $R\mathbf{x}=\mathbf{0}$ with the same free parameter values $s_{j_{1}},\ldots,s_{j_{k}}$, they are equal. | 1,919 | 5,464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 83, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.604861 |
https://www.fxsolver.com/browse/?like=2164&p=19 | 1,670,193,914,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00738.warc.gz | 820,135,659 | 57,555 | '
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Category | 482 | 2,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.896072 |
https://www.physicsforums.com/threads/mathematica-trouble-solving-for-x-in-sinc-x-1-1-01.664532/ | 1,725,963,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00766.warc.gz | 882,338,251 | 17,858 | # Mathematica Trouble: Solving for x in Sinc[x] > (1/1.01)
• Airsteve0
In summary, the conversation is about solving for x in the equation Sinc[x] > (1/1.01) and how to approach it using Mathematica. The user is facing difficulties with the Solve and NSolve commands and is only looking for positive values within the range of 0 and Pi/2. The Wolfram Alpha website provides helpful visualizations and solutions for the equation. The conversation also mentions the Math & Science Software section of the forum for better and quicker answers. A possible solution using FindRoot is also mentioned.
Airsteve0
If anyone on here uses Mathematica maybe you could help me with an issue I am having with computing the following:
Solving for x in: Sinc[x] > (1/1.01)
I am looking only at the positive values and not making any headway with the Solve or NSolve commands. I should also mention that this only makes sense for the angle being between 0 and Pi/2.
Last edited:
This
http://www.wolframalpha.com/input/?i=plot+sinc(x)+for+0+<+x+<+pi/2
shows there will be a single open interval where Sinc[x]>1/1.01 and 0<x<pi/2.
This
http://www.wolframalpha.com/input/?i=solve+sinc(x)==1/1.01+and+x>0
gives you the upper endpoint.
When you know there is a single simple interval for a solution then I would suspect that Mathematica might correctly accept
NSolve[Sinc[x]==1/1.01,x]
to give you that upper endpoint, but unfortunately I cannot verify that for you at the moment.
Note: There is the Math & Science Software section of the forum where many of the small questions about Mathematica syntax might be more quickly and better answered. Poke around several sections below the Math category and you will find it.
In[1]:= FindRoot[Sinc[x]-1/1.01,{x,.25}]
Out[1]= {x->0.244097}
## 1. What does the function Sinc[x] represent?
The Sinc[x] function is a mathematical function that stands for the "sine cardinal" function. It is defined as sin(x)/x and is commonly used in signal processing and Fourier analysis.
## 2. How is the inequality Sinc[x] > (1/1.01) solved for x?
The first step in solving this inequality is to isolate the x term on one side of the equation. This can be done by subtracting (1/1.01) from both sides, giving Sinc[x] - (1/1.01) > 0. Then, using algebraic manipulation and knowledge of the properties of the Sinc function, you can find the range of values for x that satisfy the inequality.
## 3. Is there a specific method or function in Mathematica that can be used to solve this type of inequality?
Yes, Mathematica has a built-in function called Reduce that can be used to solve inequalities. This function takes in an expression and a variable and returns a list of conditions that must be satisfied for the variable to satisfy the expression. In this case, it can be used to solve the inequality Sinc[x] > (1/1.01) for x.
## 4. Can the solution to this inequality be graphically represented?
Yes, the solution to this inequality can be graphically represented by plotting the Sinc[x] and (1/1.01) functions on the same graph and finding the values of x where the Sinc[x] function is greater than the (1/1.01) function. Alternatively, the solution can also be represented by using the RegionPlot function in Mathematica.
## 5. Are there any restrictions on the values of x that can satisfy this inequality?
Yes, since the Sinc[x] function is undefined at x=0, the solution to this inequality will exclude that value. Additionally, the solution may also have restrictions based on the domain of the Sinc[x] function, which is typically the set of all real numbers except for x=0.
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1K | 987 | 3,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.897927 |
https://quizzes.studymoose.com/ch-1-2-quiz/ | 1,713,574,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00054.warc.gz | 438,158,140 | 13,928 | # Ch.1
## Unlock all answers in this set
question
Data
Consists of information coming from observations, counts, measurements, or responses.
question
Statistics
the science of collecting, organizing, analyzing, and interpreting data in order to make decisions.
question
Two types of data sets?
Population, Sample
question
Population
collections of all out comes, responses, measurements, or counts that are of interest.
question
Sample
a subset, or part, of a population
question
Parameter
a numerical description of a population characteristic
question
statistic
a numerical description of a sample characteristic
question
A sample statistic will not change from sample to sample True or False
The statement is false. A sample statistic can change from sample to sample. A parameter will not change*
question
Descriptive statistics
the branch of statistics that involves the organization, summarization, and display of data.
question
Inferential statistics
the branch of statistics that involves using a sample to draw conclusions about a population.
question
Select all the levels of measurement for which data can be qualitative. A. Nominal .B. Interval C. Ratio D. Ordinal
Nominal Ordinal
question
Data at the ordinal level are quantitative only. True or False
False. Data at the ordinal level can be qualitative or quantitative.
question
For data at the interval​ level, you cannot calculate meaningful differences between data entries True or False
The statement is false. A true statement is​ "For data at the interval​ level, you can calculate meaningful differences between data​ entries."
question
More types of calculations can be performed with data at the nominal level than with data at the interval level True or False
False. More types of calculations can be performed with data at the interval level than with data at the nominal level.
question
Data at the ratio level cannot be put in order.
The statement is false. A true statement is​ "Data at the ratio level can be placed in a meaningful​ order."
question
What is the difference between an observational study and an​ experiment?
In an​ experiment, a treatment is applied to part of a population and responses are observed. In an observational​ study, a researcher measures characteristics of interest of a part of a population but does not change existing conditions.
question
What is the difference between a census and a​ sampling?
A census includes the entire population. A sampling includes only part of the population.
question
What is the difference between a random sample and a simple random​ sample?
With a random​ sample, each individual has the same chance of being selected. With a simple random​ sample, all samples of the same size have the same chance of being selected.
question
What is replication in an​ experiment? Why is replication​ important?
Replication is repetition of an experiment under the same or similar conditions. Replication is important because it enhances the validity of the results.
question
A placebo is an actual treatment. True or False
The statement is false. A placebo is a fake treatment.
question
A​ double-blind experiment is used to increase the placebo effect. True or False
The statement is false. Double blinding is used to decrease the placebo effect.
question
Using a systematic sample guarantees that members of each group within a population will be sampled. True or False
False. Using a stratified sample guarantees that members of each group within a population will be sampled.
question
The method for selecting a stratified sample is to order a population in some way and then select members of the population at regular intervals. True or False
False. The method for selecting a systematic sample is to order a population in some way and then select members of the population at regular intervals.
question
Observational studies are sometimes referred to as natural experiments. Explain what this means.
In an observational​ study, a researcher measures characteristics of interest of a part of a population but does not change existing conditions.
question
nominal level of measurment
qualitative only. Data at this level are categorized using names, labels, or qualities. No mathematical computations can be made at this level.
question
Ordinal Level of measurment
Qualitative and quantitative. Data at this level can be arranged in order, or ranked, but differences between data entries are not meaningful.
question
interval level of measurement
can be ordered, and meaning full differences between data entries can be calculated. No inherent zeros
question
ratio level of measurment
can be ordered, and meaning full differences between data entries can be calculated. Has inherent zeros
question
Blinding
a technique where the subjects do not know whether they are receiving a treatment or a placebo
question
Double-blind experiment | 1,012 | 4,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.89707 |
https://lw2.issarice.com/posts/aKcy8428zspgSKjYA/sleeping-beauty-resolved-pt-2-identity-and-betting | 1,628,079,802,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00543.warc.gz | 361,332,248 | 24,372 | # Sleeping Beauty Resolved (?) Pt. 2: Identity and Betting
post by ksvanhorn · 2018-06-02T02:43:24.220Z · LW · GW · 50 comments
## Contents
Introduction
Indexicals and Identity
Betting Arguments
None
## Introduction
This is a followup to my previous article, Sleeping Beauty Resolved? [LW · GW] Some objected to the solution I presented (building on Radford Neal's analysis) on the grounds that it runs afoul of a Thirder betting argument:
If Beauty uses anything other than a probability of exactly for Heads, she will accept certain bets she should not, and reject others she should accept.
Alas, there is an alternative Halfer betting argument that makes the same claim, but replacing with .
I'll show that both arguments are wrong, as they get the effective payoffs wrong; but if Beauty uses the correct effective payoffs, together with a probability of for Heads, she makes the right betting decisions.
To get there requires addressing some questions of unique identity, so that's where I'll start.
## Indexicals and Identity
Thirder arguments often make use of statements such as
today is Monday
and
today is Tuesday,
treating them as mutually exclusive propositions. Probability theory is based on the classical propositional logic, and deals exclusively with classical propositions; are the above legitimate classical propositions?
I argue that they are not. The word "today" is problematic; it is an indexical, which the article "Demonstratives and Indicatives" in the Internet Encyclopedia of Philosophy defines as
...any expression whose content varies from one context of use to another. The standard list of indexicals includes adverbs such as “now”, “then”, “today”, “yesterday”, “here”, and “actually”.
The article furthermore remarks,
Indexicals and demonstratives raise interesting technical challenges for logicians seeking to provide formal models of correct reasoning in natural language...
and goes on to discuss various efforts to construct logics appropriate for reasoning with indexicals.
Clearly indexicals pose a problem for classical logic, else there would be no interest in constructing alternative logics to deal with them. As Richard Epstein writes in Classical Mathematical Logic: The Semantic Foundations of Logic,
...we cannot take sentence types as propositions if we allow the use of indexicals in our reasoning
(p. 4). This is because the meaning of a classical proposition must be definite and stable:
When we reason together, we assume that words will continue to be used in the same way... We will assume that throughout any particular discussion equiform words will have the same properties of interest to logic. We therefore identify them and treat them as the same word...
...if we accept this agreement, we must avoid words such as 'I', 'my', 'now', or 'this', whose meaning or reference depends on the circumstances of their use. Such words, called indexicals, play an important role in reasoning, yet our demand that words be types requires that they be replaced by words that we can treat as uniform in meaning or reference throughout a discussion.
(p. 3). In short:
Every usage of the same proposition in an argument / analysis must mean the same thing and have the same true/false value in all contexts within the scope of the analysis.
The important point is to ensure that any temporal (or spatial, etc.) reference is uniquely defined. If we are having a face-to-face conversation and I use the word "now," it's clear that means the specific, well-defined point in time at which I utter that word. If I refer to "the day on which Julius Caesar took his first sip of wine," then even though nobody knows what day that was, I have uniquely identified a particular day--there cannot be two such days.
But when Beauty asks, "Is today Monday?", how does she identify which "today" she means? Can she find some uniquely identifying descriptor that unambiguously distinguishes "today" from "the other day"?
Maybe. If the experimenters randomly choose to put a black marble on her night stand one day, and a white marble the other day, and Beauty knows this, then as soon as she glances at the night stand and sees (say) a black marble, she can then say that "today" means "the day on which there is a black marble on the nightstand." In this case the usual Thirder arguments hold, and she gets a probability for Heads of .
But if Beauty is an AI and her entire state of mind and stream of experiences from Monday are exactly reproduced on Tuesday, then the term "today" is inescapably ambiguous--Beauty has no way of uniquely identifying "today". Thirder arguments based on using "today is Monday" and "today is Tuesday" as mutually exclusive propositions are invalid in this case. As shown previously, the Halfer argument of "no new relevant information" applies in this case, and Beauty gets a probability of for Heads.
In the intermediate case, where is the stream of perceptions Beauty has experienced since awakening and , , is the probability of experiencing the identical stream of perceptions at some time on the other day, then "today" is partially identified as "the day in which Beauty experiences stream of perceptions ." (The closer is to zero, the more probable it is that uniquely identifies "today.") We then get a probability for Heads that is intermediate between the Halfer and Thirder positions: .
How can receiving a random bit cause Beauty to update her probability, as in the case where Beauty is an AI? If Beauty already knows that she will update her probability no matter what bit she receives, then shouldn't she already update her probability before receiving the bit?
This question references the special case where Beauty is an AI whose only sensory input after awakening on Monday/Tuesday is a sequence of random bits. I showed that her probability of Heads before receiving any bits is , and after receiving the first bit this falls to , no matter which bit is received. The above argument is also one a Halfer could use against the Thirder position: if Beauty already knows on Sunday that her probability for Heads is going to be on Monday, why isn't that already her probability for Heads?
To answer this, let's consider where the principle travisrm89 invokes comes from. Let and be two mutually exclusive and exhaustive propositions, with meaning "the next observation is ." If our probability for updates to regardless of what we observe--that is, if for both --then
which says that our probability for should already be .
But the new information Beauty has after receiving one random bit doesn't fit the above pattern. In this special case of the problem her new information is
for some , where means "the first bit Beauty receives on day is y". Significantly, the two propositions and are not mutually exclusive if --both are true if the coin lands Tails and Beauty receives different first bits on Monday and Tuesday. Thus we have an exhaustive but not mutually exclusive set of possibilities, and the sum of their probabilities exceeds 1: .
Therefore,
which is why we can have even though .
It is only in the case of --Beauty's experiences on the two days are identical so the days are entirely indistinguishable--that the sum of observation probabilities is 1, and the prior and posterior probabilities are the same.
## Betting Arguments
Suppose that Beauty is offered a bet with a payoff of if the coin lands Heads, and a payoff of if the coin lands Tails. Either of these payoffs can be negative, in which case it is a loss. (The interesting cases are where one is positive and the other negative.) If the coin lands Tails she is offered the bet on both Monday and Tuesday. Since Beauty assesses the same probability of Heads on both Monday and Tuesday, she will either accept the bet both times or reject it both times. Thus the "objective" expected payout if she accepts the bet is
Beauty should accept the bet if, and only if, this quantity is positive. This is the case in the standard example where and .
The standard Thirder betting argument goes like this:
If Beauty assesses a probability for Heads after awakening on Monday/Tuesday, then she computes an expected payout of
and will accept the bet only iff this is positive. If then Beauty's expected payout is identical to the "objective" expected payout (up to a constant, positive factor) and so she will make the correct decision to accept or reject the bet, whatever the payoffs used.
The Halfer counterargument [reference?] goes like this:
Beauty knows full well that, if the coin lands Tails, she is going to compute the same probability of Heads on both Monday and Tuesday, and that she will therefore make identical decisions on those two days, and obtain identical outcomes. So in that case she is making a decision for two days, not just one. Therefore she should compute her expected payout as
and accept the bet only if this quantity is positive. If then her subjective expected payout is identical to the objective expected payout, and so she will make the correct decision to accept or reject the bet, whatever the payoffs used.
There are elements of truth to both positions. Any decision rule, including the rule that one should maximize expected gain, is a function that maps the available information to a recommended action. In the SB problem the available information is Beauty's background knowledge about the experimental setup, plus the stream of perceptions she has experienced since awakening. If the same stream of perceptions occurs on both days, Beauty's decision rule must give the same action both days, and so in this circumstance her payout for Tails is . However, if Beauty does not experience the stream of perceptions on the other day, then in principle her decision rule could give different actions for the two days, and her payout for Tails is just .
Given that Beauty experiences , we then have three cases for three different possible payouts:
• A: The coin lands Heads, and Beauty experiences on Monday. Payoff is . Using the notation of Part 1, the prior probability is | 2,063 | 10,111 | {"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} | 2.8125 | 3 | CC-MAIN-2021-31 | latest | en | 0.931498 |
https://www.thefreelibrary.com/The+evaluation+of+the+passive+samplers%27+performances-a0224712325 | 1,563,617,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526506.44/warc/CC-MAIN-20190720091347-20190720113347-00041.warc.gz | 860,865,877 | 15,429 | # The evaluation of the passive samplers' performances.
1. INTRODUCTION
Profiles of outdoor pollutant concentrations and indoor air quality can be evaluated by a multi-location measurement using the passive samplers. The passive sampler is an integrated and relatively inexpensive type monitor, providing personal exposure data for epidemiological research because it is light, small and easy to handle. An average concentration for a certain sampling period can be obtained with the passive samplers since it collects air pollutant of interest in its collection medium, but is not equipped with a detector. The major difference between an active and a passive sampler is the sampling process of the targeted air pollutant. The active sampler transports the pollutant to the medium or detector by drawing air, while the passive sampler relies on molecular diffusion (Zabiegala et al, 2005, Ott et al, 2007).
Based upon theoretical background of the diffusion process, the paper proposes an evaluation method for the performances of the passive sampler.
2. THEORETICAL BACKGROUND
The diffusion process described by Fick's law is analogized with electric resistance. The mass flow rate, N (mole/s) is proportional to a molecular diffusion coefficient, D ([cm.sub.2]/s), a cross sectional area for diffusion, A ([cm.sub.2]) and a concentration gradient of the interested air pollutant, cD (mole/[cm.sup.3])dx, where x (cm) is the one-dimensional position.
N = A x D x (dC/dx) (1)
An eddy diffusion coefficient in bulk air ranges from 100 to 10000 ([cm.sub.2]/s). Therefore, it is a reasonable assumption that the concentration gradient exists only around the collector of the passive sampler. Integration of equation (1) with boundary conditions of X=0, C=[C.sub.0], and X=L, C= [C.sub.a] gives:
N = (AD/L) ([C.sub.a]- [C.sub.0]) (2)
If N is thought of as a current, equation (2) is similar to Ohm's law. The difference in concentration ([C.sub.a] - [C.sub.0]) which is the driving force of the flow is comparable to the voltage difference.
N [right arrow] Current, ([C.sub.a] - [C.sub.0]) [right arrow] Voltage difference (3)
The proportional coefficient of current and voltage is resistance. In our case, resistance is expressed as follows
R = L / (AD) (4)
The unit of R for mass transfer is (s/[cm.sub.2]). The reciprocal of R is a sampling rate of the passive sampler, which is equivalent to the flow rate of the active sampler, F ([cm.sup.3]/s).
F = 1/R = (AD) / L (5)
There are usually three regions around the collection medium of the passive sampler where air pollutants can diffuse by molecular diffusion. This implies that there are three types of resistance during diffusion of the air pollutant from the bulk air to the collector. Air pollutants must pass a boundary layer around the interface of the air and the passive sampler, a stagnant region produced by a diffusion barrier and a boundary layer at the collector. These resistances connect in series. By analog with electrical resistance, the overall resistance of the diffusion becomes:
[R.sub.0] = [R.sub.1] + [R.sub.2] + [R.sub.3] (6)
If a small amount of the air pollutant of interest is absorbed, it is reasonable assumption that [C.sub.0] in equation (2) is nearly equal to zero. The amount of pollutant collected, M (mole), becomes:
M = (1/[R.sub.0]) * [C.sub.a] * t = F * [C.sub.a] * t (7)
This equation allows us to easily calculate the pollutant concentration during the exposure period of t.
The diffusion resistances are functions of various physical parameters. [R.sub.1] is described by the following equation using 5 as the thickness of the boundary layer.
[R.sub.1] = [delta]/D (8)
In the same manner [R.sub.2] becomes:
[R.sub.2] = L/D (9)
in which L is the length of the diffusion barrier.
If the reaction between the air pollutant of interest and the collection medium is first order, irreversible and rapid, with a reaction rate constant of k (1/s), the third resistance becomes:
[R.sub.3] = m/[(k x [D.sub.L]).sup.1/2] (10)
in which m is Henry's constant and [D.sub.L] is the diffusion coefficient in the liquid phase.
Environmental conditions can vary these parameters, that is, the environmental conditions determine the resistances. Surface wind velocity, temperature and relative humidity are the three major environmental factors affecting those (Godish, 2004). The surface wind velocity (WV) effects on the thickness of the boundary layer, [delta] and the diffusion barrier, L. At high WV, [delta] is thin. The diffusion coefficients and the chemical reaction rate constant are functions of temperature. Kinetic theory shows that the diffusion coefficient in an ideal gas is proportional to [T.sup.3/2] where T indicates absolute temperature. The reaction rate constant increases exponentially with elevation of temperature following Arrhenius' law. If the relative humidity affects the water content of the collection medium, the diffusion coefficient in the liquid phase and the reaction rate constant may be changed. The overall resistance to the diffusion process in the passive sampler could be described as functions of these environmental factors:
[R.sub.0] (WV, T, RH) = [delta](WV)/[A.sub.1]D(T) + L(WV)/[A.sub.2]D(T) + m(T)/[A.sub.3] [{k(T, RH) x [D.sub.L](T, RH)}.sup.1/2] (11)
Only the length of the diffusion barrier, L, can be controlled easily. For very large values of L, we can suppress the variation of the first and third terms due to change in the environmental conditions. However, if the second term is very large, the sensitivity of measurements, which are proportional to the reciprocal of [R.sub.0], is lowered. Therefore, we have to determine an appropriate length for the diffusion barrier.
In this derivation, the concentration of air pollutant is expressed by mole/[cm.sup.3]. If the concentration is presented by a volume to volume ratio, the temperature effect on concentration should be clarified. Since the passive sampler is designed to have large resistance at [R.sub.2] to eliminate the effect of surface wind velocity, the sampling rate is proportional to the molecular diffusion coefficient. Since the molecular diffusion coefficient theoretically depends on [T.sup.3/2], the amount of collected pollutant in the collection medium proportionally increases to [T.sup.3/2]. Whereas the amount of pollutant in a unit volume is reciprocally proportional to the square root of absolute temperature, [T.sup.1/2], as long as the concentration is expressed by the volume to volume ratio such as ppm.
Similar discussion to the temperature dependence can be done to the pressure effect. The molecular diffusion coefficient is proportional to the reciprocal of pressure, 1/P, while the amount of pollutant in a unit volume is proportional to P. The pressure dependence of measured concentration by the passive sampler is none if the concentration is indicated by ppm (Hill, 2000).
3. FACTORS TO BE EVALUATED
The passive samplers are provided with the sampling rate to get a concentration value as the volume to volume ratio. The sampling rate is determined under some representative conditions such as temperature of 20[degrees]C, relative humidity of 50%, surface wind velocity of 1 m/s and 1 atm. Environmental factors potentially change the sampling rate (Gillet et al, 2000).
The range of temperature to be examined depends on conditions of the passive sampler application. If it is used in very cold weather, the sampling rate data at -5[degrees]C may be needed. The examination data of the sampling rate at 5, 20, 30 and 40[degrees]C can be enough to cover most cases.
Effects of relative humidity on the sampling rate are difficult to predict theoretically (Akutsu et al, 2000). It can change the reaction rate and diffusion coefficient in the collecting medium. Determination of sampling rates at 20, 40, 60, 80 and 95 % of RH can cover most of the applications.
Surface wind velocity affects the thickness of the boundary layer and the effective length of the diffusion barrier. At low wind velocity (indoor), the thickness of the boundary layer is thicker than at high wind velocity. If the wind velocity is high, eddies in bulk air can penetrate into the diffusion barrier and shorten its effective length. As a result, the sampling rate on the surface wind velocity is saturated at a certain high wind velocity. Determination of sampling rates at calm, 0.5, 1, 2, and 5 m/s is sufficient to evaluate the wind effects.
As concerning the factors to determine the precision and accuracy, pressure does not affect the concentration as long as it is expressed by the volume to volume ratio. There are two kinds of potential interferences by chemicals: the interference by coexisting pollutants on quantification and the impurity and/or instability of chemicals in the collecting medium. Sun light may also affect their stabilities.
The time constant or the response time determines minimum duration of the sampling. The time constant is estimated from the diffusion coefficient of the target air pollutant (D), sampling rate (F) and cross sectional area for diffusion (A) as time constant = D/(F/[A.sup.2]). The sampling time must be long enough comparing the time constant. The working range, lower and upper detection limits, is another factor to determine the sampling time. The range is indicated by concentration*exposure time. The lower limit is defined as three times of the standard deviation of blanks. Since the capacity of the collection medium is not infinite, the response of the sampler becomes nonlinear when the amount of the collected pollutant exceeds a certain level. The maximum amount within the linear region is defined as the upper detection limit.
Variability of blanks values is more important than the absolute blank value. The blank value is the amount of the target pollutant in the non-exposed sampler. The production blank indicates the extent of contamination due to the field operation. The variability of blanks determines the lower detection limit. Precision or variability of measurements conducted under identical conditions can be obtained by duplicate measurements (Yoshizumi, 1999).
4. CONCLUSIONS
The conditions described before are representative for quality of the environmental researches using passive samplers. The passive sampler is sensitive to environmental conditions such as wind velocity, air temperature, relative humidity. When designing the survey, the time constant and working range information are necessary. Precision and accuracy show the overall performance of the passive sampler.
5. REFERENCES
Akutsu, T.; Kumagai, K.; Uchiyama, S.;Tanabe, S.(2000) Development of measurement device for aldehyde emission rates using a diffusive sampler, Proceedings of Healthy Buildings, Seppanen & Sateri (ed), vol 1, pp.477-482, ISBN 952 5236 06 4, Helsinki, Finland, August 2000
Gillet, R.W.; Kreibich, H.; Ayers G.P. (2000) Measurement of indoor formaldehyde concentrations with a passive sampler, Environ.Sci.Technol., vol. 34 (10), pp2051-2056
Godish, T (2004) Air Quality, Lewis Publishers, ISBN 1 56670 586 X, USA
Hill, M; Gehrig, R.; Dorer, V.; Weber, A.; Hofer, P. (2000) Measurements of air change rates with the PFT method biassed by sink and temperature effect, Proceedings of Healthy Buildings, Seppanen & Sateri (ed), vol 2, pp.333-338, ISBN 952 5236 06 4, Helsinki, Finland, August 2000
Ott, W.R.; Steinemann, A.C. & Wallace L.A. (ed) (2007) Exposure Analysis, CRC Press, ISBN 1 56670 663 7, USA
Yoshizumi, K.; Ishibashi, Y.; Kudo, T.; Hedge, A.; Muramatsu K. (1999). Application of passive sampling methodology to the characterization of indoor air quality, Proceedings of Indoor Air '99, vol.2, pp. 870-875, ISBN 1 86081 295 3, Edinburgh, Scotland, August 1999, CRC Press, London
Zabiegala, B.; Partyka, M.; Namiesnik, J. (2005) Passive samplers in indoor air control, Air Pollution XIII, pp.195-204, ISBN 978-1-84564-014-9, WIT Press UK | 2,753 | 11,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.879867 |
https://cg-tower.com/how-many-milliseconds-ms-are-there-in-3-5-seconds-s/ | 1,638,823,152,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00465.warc.gz | 240,846,637 | 4,716 | Answer: The correct answer is 3500 miliseconds.Explanation:We are given a quantity of 3.5 seconds and also we require to convert it right into miliseconds.To transform it right into miliseconds, we will use the conversion factor:1 2nd = 1000 milisecondsNow, converting 3.5 seconds right into miliseconds:3,5 seconds = 3.5 × 1000 = 3500 milisecondsHence, the exactly answer is 3500 miliseconds.
You are watching: How many milliseconds (ms) are there in 3.5 seconds (s)?
3 main ago294.6 | 134 | 486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | longest | en | 0.788916 |
https://en.burgeronreport.com/gewinde-bestimmen-mit-der-schieblehre | 1,660,281,250,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00471.warc.gz | 241,897,355 | 12,534 | # Determine the thread with the caliper
To determine the size of an existing thread, two dimensions must be determined. It also depends on the type of thread. In Europe, metric threads are common, in Anglo-Saxon countries inch or inch threads.
The metric threads have round outside or inside diameters. If you measure numbers with decimal places when determining, it is an inch thread. Standard in Europe are metric threads with dimensional designations such as M18, M20, M25 or M30.
• Also read - The diameter of the thread is calculated metrically
If an inch or inch thread is determined, this can be converted to the metric system of measurement using a conversion table. One inch is 25.4 millimeters. In addition to the diameter, the so-called thread pitch is the decisive dimension when determining a thread.
## Metric or inch / inch
In simple terms, the thread pitch expresses the distance from one thread tip to the next. While the thread pitches are standardized for metric threads, they are used for inch or. Inch threads measured in a different way.
The number of thread crests over the length of one inch, which corresponds to 25.4 centimeters, is measured as the TPI value. In the case of metric threads, any deviations from the standardized standard pitch are indicated with an addition in the dimensions.
The permanently assigned pitch values can simply be added to the diameter of the thread. For example, if a diameter is 38.6 millimeters, 1.5 millimeters of thread pitch is the norm. This adds up to 40.1 millimeters and is rounded to the metric thread size M40.
This so-called nominal value determines the size of the thread. It is important to note that a screw or nut always needs a small deviation from the nominal value. Two to three tenths of a difference allow the play to permit screwing ability.
Tips & Tricks If you have a counterpart, nut or screw, available when determining a thread, select this for measuring. It is much easier to determine precisely. | 420 | 1,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.916487 |
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## Derivative of a function with vector arguments
I have the following function defined.
The intent if for this function to be very generic and allow any number of inputs. Therefore, Vin and Rin are vectors of the same length. Here's how those are setup.
The function itself works fine.
Now, I need to take the derivative of that function with respect to the individual components of Rin. The intent is to estimate how much effect the variation of each individual component has on the overall output. Ultimately the expression and number of parameters will be bigger than what I've shown here. I just scaled everything back to a simpler example to illustrate the actual problem. I was hoping the derivative with respect to Rin would give me a vector of the numerically evaluated derivative with respect to each individual resistance. Instead it tells me everything has to be scalar.
I found some old posts similar to this dating back to 2004 and 2005. They basically said this isn't possible. I was hoping something may have changed in 14 to 15 years.
Thanks,
Jon
4 REPLIES 4
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## Re: Derivative of a function with vector arguments
was hoping something may have changed in 14 to 15 years.
Unfortunately not (apart from the development of Prime which actually is a big step backwards).
You will have to find a workaround which sure is not as generic as you'd like it to be 😞
## Re: Derivative of a function with vector arguments
You can find your work-around on the internet.
## Re: Derivative of a function with vector arguments
Fred,
Thanks! Your approach didn't quite give me what I wanted but it got me thinking.
What I was trying to get was an approximate, linearized effect of each component on the output. Your solution gave me the sum of the derivatives for each component.
But, your approach got me thinking that I was being too lazy in trying to let Mathcad do all of the work. The partial derivative with respect to each resistor in my input vector just boils down to this:
Then, I can just pass my input vector to that function and get my individual results:
Thanks,
Jon
## Re: Derivative of a function with vector arguments
Mathcad is a great tool, but not the best one you have. NASA went to the moon with pencils and slide rules! | 515 | 2,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-05 | latest | en | 0.949064 |
https://deepnotes.io/sgd-momentum-adaptive | 1,632,638,500,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00190.warc.gz | 251,038,123 | 8,685 | # Solving the model - SGD, Momentum and Adaptive Learning Rate
## Thanks to active research, we are much better equipped with various optimization algorithms than just vanilla Gradient Descent. Lets discuss two more different approaches to Gradient Descent - Momentum and Adaptive Learning Rate.
Note: Complete source code can be found here https://github.com/parasdahal/deepnet
Once we have the model of our neural network, we need to find the best set of parameters to minimize the training/test loss and maximize the accuracy of the model. Model solver brings together training data, the model and the optimization algorithms to train the model. A model solver has:
1. Training and Test dataset
2. Reference to the model
3. Different type of optimizers like SGD, Adam
4. Record of loss and accuracy for the training phase for each epoch
5. Optimized parameters for the model
Let us start developing the ideas around building our model solver with a brief review of the backbone of all popular optimizers, Gradient Descent.
Gradient Descent is the most common optimization algorithm used in Machine Learning. It uses gradient of loss function to find the global minima by taking one step at a time toward the negative of the gradient (as we wish to minimize the loss function).
If $$\Delta w$$ and $$\Delta b$$ are the changes we need to find in direction of $$w$$ and $$b$$ respectively of our Loss Vs. Parameters curve, then the gradient of the loss function L is defined as,
$\Delta L = \left(\frac{\partial L}{\partial w},\frac{\partial L}{\partial b}\right)$
The way gradient descent works is to now repeatedly compute the gradient $$\Delta L$$ and then move in the opposite direction, “falling down” the slope of the valley.
\begin{align} w_k \leftarrow w_k - \alpha \frac{\partial L}{\partial w_k} \\ b_k \leftarrow b_k - \alpha \frac{\partial L}{\partial b_k} \\ \end{align}
where $$\alpha$$ is known as Learning Rate, which represents the size of the step taken at each iteration by Gradient Descent.
Gradient Descent along with Backpropagation algorithm has become the de-facto learning algorithm of neural networks.
In other optimizing algorithm like Newton’s Method and BFGS, we need to calculate second order partial derivative matrix or Hessian matrix. Although the algorithm makes efficient updates and doesn’t require learning rate parameter, we have to calculate second order partial derivatives matrix for every parameter with respect to every other parameter, which makes it very computationally costly and highly ineffective in terms of memory.
Gradient Descent requires only first order derivatives of parameters with respect to the loss function, which is efficiently calculated by Backpropagation, and so it shines above these techniques because of its simplicity and efficiency.
There are other variations of Gradient Descent with few ideas added to allow faster convergence to the optimum. The most popular algorithms are:
2. SGD with Momentum
5. RMSprop
When training input is very large, gradient descent is quite slow to converge. Stochastic Gradient Descent is the preferred variation of gradient descent which estimates the gradient from a small sample of randomly chosen training input in each iteration called minibatches.
Minibatches
Minibatches are generated by shuffling the training data and randomly selecting certain number of training samples. This number of samples is called minibatch size and is a parameter to SGD. Here is the code:
def get_minibatches(X,y,minibatch_size):
m = X.shape[0]
minibatches = []
X,y = shuffle(X,y)
for i in range (0,m,minibatch_size):
X_batch = X[i:i+minibatch_size,:,:,:]
y_batch = y[i:i+minibatch_size,]
minibatches.append((X_batch,y_batch))
return minibatches
Update Rule
SGD uses a very simple update rule to change the parameters along the negative gradient. Assume we have a list learnable parameters for each layer in order params and a similar list for gradients grads calculated by backward pass with tuples of gradients for each learnable parameter, our simple update rule would be:
def vanilla_update(params,grads,learning_rate=0.01):
param[i] += - learning_rate * grad[i]
SGD
Every complete exposure of the training dataset is called epoch. The SGD algorithm iterates for a given number of epochs. It uses the above get_minibatches and vanilla_update functions to brings together the requirements for our model solver.
def sgd(nnet,X_train,y_train,minibatch_size,epoch,learning_rate,verbose=True,\
X_test=None,y_test=None):
minibatches = get_minibatches(X_train,y_train,minibatch_size)
for i in range(epoch):
loss = 0
if verbose:
print("Epoch {0}".format(i+1))
for X_mini, y_mini in minibatches:
if verbose:
train_acc = accuracy(y_train,nnet.predict(X_train))
test_acc = accuracy(y_test,nnet.predict(X_test))
print("Loss = {0} | Training Accuracy = {1} | Test Accuracy = {2}".\
format(loss,train_acc,test_acc))
return nnet
### Momentum
Momentum technique is an approach which provides an update rule that is motivated from the physical perspective of optimization. Imagine a ball in a hilly terrain is trying to reach the deepest valley. When the slope of the hill is very high, the ball gains a lot of momentum and is able to pass through slight hills in its way. As the slope decreases the momentum and speed of the ball decreases, eventually coming to rest in the deepest position of valley.
This technique modifies the standard SGD by introducing velocity $$v$$ , which is the parameter we are trying to optimize, and friction $$\mu$$, which tries to control the velocity and prevents overshooting the valley while allowing faster descent. The gradient only has direct influence on the velocity, which in turn has an effect on the position. Mathematically,
\begin{align} v &= \mu v - \alpha \Delta L \\ w &= w + v \end{align}
Which translates to code as
def momentum_update(velocity,params,grads,learning_rate=0.01,mu=0.9):
v[i] = mu*v[i] + learning_rate * grad[i]
param[i] -= v[i]
The advantage of momentum is that it makes very small change to SGD but provides a big boost to speed of learning. We need to store the velocity for all the parameters, and use this velocity for making the updates. Here is the modified function for SGD which uses the above momentum update rule.
def sgd_momentum(nnet,X_train,y_train,minibatch_size,epoch,learning_rate,mu = 0.9,\
verbose=True,X_test=None,y_test=None):
minibatches = get_minibatches(X_train,y_train,minibatch_size)
for i in range(epoch):
loss = 0
velocity = []
for param_layer in nnet.params:
p = [np.zeros_like(param) for param in list(param_layer)]
velocity.append(p)
if verbose:
print("Epoch {0}".format(i+1))
for X_mini, y_mini in minibatches:
if verbose:
train_acc = accuracy(y_train,nnet.predict(X_train))
test_acc = accuracy(y_test,nnet.predict(X_test))
print("Loss = {0} | Training Accuracy = {1} | Test Accuracy = {2}".\
format(loss,train_acc,test_acc))
return nnet
Nesterov’s Accelerated Gradient is a clever variation of momentum that works slightly better than standard momentum. The idea behind Nesterov’s momentum is that instead of calculating the gradient at the current position, we calculate the gradient at a position that we know our momentum is about to take us, called as “look ahead” position. From physical perspective, it makes sense to make judgements about our final position based on the position that we know we are going to be in a short while.
The implementation makes a slight modification to standard SGD Momentum by nudging our parameters slightly in the direction of the velocity and calculating the gradients there. Here is the code:
def sgd_momentum(nnet,X_train,y_train,minibatch_size,epoch,learning_rate,mu = 0.9,\
verbose=True,X_test=None,y_test=None,nesterov = False):
minibatches = get_minibatches(X_train,y_train,minibatch_size)
for i in range(epoch):
loss = 0
velocity = []
for param_layer in nnet.params:
p = [np.zeros_like(param) for param in list(param_layer)]
velocity.append(p)
if verbose:
print("Epoch {0}".format(i+1))
for X_mini, y_mini in minibatches:
# if nesterov is enabled, nudge the params forward by momentum
if nesterov:
for param,ve in zip(nnet.params,velocity):
for i in range(len(param)):
param[i] += mu*ve[i]
if verbose:
train_acc = accuracy(y_train,nnet.predict(X_train))
test_acc = accuracy(y_test,nnet.predict(X_test))
print("Loss = {0} | Training Accuracy = {1} | Test Accuracy = {2}".format(loss,train_acc,test_acc))
return nnet
Until now we have used a global and equal learning rate for all our parameters. So all of our parameters are being updated with constant factor. But what if we could speed up or slow down this factor, even for each parameter, as the training progresses? We could adaptively tune the learning throughout the training phases and know which direction to accelerate and which to decelerate. Several methods that use such adaptive learning rates have been proposed, most notably AdaGrad, RMSprop and ADAM.
AdaGrad (original paper) keeps track of per parameter sum of squared gradient and normalizes parameter update step. The idea is that parameters which receive big updates will have their effective learning rate reduced, while parameters which receive small updates will have their effective learning rate increased. This way we can accelerate the convergence by accelerating per parameter learning.
def adagrad_update(cache,params,grads,learning_rate=0.01):
param[i] += - learning_rate * grad[i] / (np.sqrt(cache[i])+1e-8) # for preventing divide by 0
RMSprop
A disadvantage of AdaGrad is that cache[i] += grad[i]**2 part of the update is monotonically increasing. This can pose problems because the learning rate can steadily decrease to the point where it stops the learning altogether. RMSprop (unpublished, citation here) combats this problem by decaying the past squared gradient by a factor decay_rate to control the aggressive learning rates. Here decay_rate is a hyperparameter with typical values like 0.9,0.99 and so on.
def rmsprop_update(cache,params,grads,learning_rate=0.01,decay_rate=0.9):
cache[i] = decay_rate * cache[i] + (1-decay_rate) * grad[i]**2
param[i] += - learning_rate * grad[i] / (np.sqrt(cache[i])+1e-4)
Adam (original paper) is a recently proposed and currently state of the art first order optimization algorithm. It is an improvement upon RMSprop by adding momentum to the update rule, combining best of the both momentum and adaptive learning worlds. We introduce two more parameters beta1 and beta2 with recommended values 0.9 and 0.999 respectively.
Another thing to note is that Adam includes bias correction mechanism, which compensates for first few iterations when both cache and velocity are biased at zero as they are initialized to zero.
Here is the full implementation of Adam:
X_test=None,y_test=None,beta1=0.9,beta2=0.999):
minibatches = get_minibatches(X_train,y_train,minibatch_size)
for i in range(epoch):
loss = 0
velocity,cache = [],[]
for param_layer in nnet.params:
p = [np.zeros_like(param) for param in list(param_layer)]
velocity.append(p)
cache.append(p)
if verbose:
print("Epoch {0}".format(i+1))
t = 1
for X_mini, y_mini in minibatches:
c[i] = beta1 * c[i] + (1-beta1) * grad[i]
mt = c[i] / (1 - beta1**t)
v[i] = beta2 * v[i] + (1-beta2) * (grad[i]**2)
vt = v[i] / (1 - beta2**t)
print(vt)
param[i] += - learning_rate * mt / (np.sqrt(vt) + 1e-4)
t+=1
if verbose:
train_acc = accuracy(y_train,nnet.predict(X_train))
test_acc = accuracy(y_test,nnet.predict(X_test))
print("Loss = {0} | Training Accuracy = {1} | Test Accuracy = {2}".\
format(loss,train_acc,test_acc))
return nnet | 2,707 | 11,634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-39 | longest | en | 0.840986 |
https://www.1ju.org/py-data-structure/python-sorting-algorithms | 1,723,471,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00611.warc.gz | 494,597,080 | 65,564 | # Python排序算法
• 冒泡排序
• 合併排序
• 插入排序
• 希爾排序
• 選擇排序
## 冒泡排序
``````def bubblesort(list):
# Swap the elements to arrange in order
for iter_num in range(len(list)-1,0,-1):
for idx in range(iter_num):
if list[idx]>list[idx+1]:
temp = list[idx]
list[idx] = list[idx+1]
list[idx+1] = temp
list = [19,2,31,45,6,11,121,27]
bubblesort(list)
print(list)``````
``[2, 6, 11, 19, 27, 31, 45, 121]``
## 合併排序
``````def merge_sort(unsorted_list):
if len(unsorted_list) <= 1:
return unsorted_list
# Find the middle point and devide it
middle = len(unsorted_list) // 2
left_list = unsorted_list[:middle]
right_list = unsorted_list[middle:]
left_list = merge_sort(left_list)
right_list = merge_sort(right_list)
return list(merge(left_list, right_list))
# Merge the sorted halves
def merge(left_half,right_half):
res = []
while len(left_half) != 0 and len(right_half) != 0:
if left_half[0] < right_half[0]:
res.append(left_half[0])
left_half.remove(left_half[0])
else:
res.append(right_half[0])
right_half.remove(right_half[0])
if len(left_half) == 0:
res = res + right_half
else:
res = res + left_half
return res
unsorted_list = [64, 34, 25, 12, 22, 11, 90]
print(merge_sort(unsorted_list))``````
``[11, 12, 22, 25, 34, 64, 90]``
## 插入排序
``````def insertion_sort(InputList):
for i in range(1, len(InputList)):
j = i-1
nxt_element = InputList[i]
# Compare the current element with next one
while (InputList[j] > nxt_element) and (j >= 0):
InputList[j+1] = InputList[j]
j=j-1
InputList[j+1] = nxt_element
list = [19,2,31,45,30,11,121,27]
insertion_sort(list)
print(list)``````
``[2, 11, 19, 27, 30, 31, 45, 121]``
## 希爾排序
``````def shellSort(input_list):
gap = len(input_list) / 2
while gap > 0:
for i in range(gap, len(input_list)):
temp = input_list[i]
j = i
# Sort the sub list for this gap
while j >= gap and input_list[j - gap] > temp:
input_list[j] = input_list[j - gap]
j = j-gap
input_list[j] = temp
# Reduce the gap for the next element
gap = gap/2
list = [19,2,31,45,30,11,121,27]
shellSort(list)
print(list)``````
``[2, 11, 19, 27, 30, 31, 45, 121]``
## 選擇排序
``````def selection_sort(input_list):
for idx in range(len(input_list)):
min_idx = idx
for j in range( idx +1, len(input_list)):
if input_list[min_idx] > input_list[j]:
min_idx = j
# Swap the minimum value with the compared value
input_list[idx], input_list[min_idx] = input_list[min_idx], input_list[idx]
l = [19,2,31,45,30,11,121,27]
selection_sort(l)
print(l)``````
``[2, 11, 19, 27, 30, 31, 45, 121]`` | 895 | 2,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.445309 |
https://math.stackexchange.com/questions/2937405/a-strange-inductive-proof-induction-on-n-for-all-positive-integers-n-n-ge1/2937411 | 1,628,162,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00246.warc.gz | 390,083,717 | 35,830 | # A strange inductive proof: Induction on $n$, for all positive integers $n,n\ge1$
Prove by induction on $$n$$ that, for all positive integers $$n, n\ge1$$.
My Try:
Base case is true for $$n=1$$.
Inductive step:
$$P(k)$$ is true. $$\implies k\ge1$$
We need to show that $$(k+1)\ge1$$
From here how should I proceed.
Can anyone explain this strange inductive proof.
• Hint: $k\ge 1 \Rightarrow k+1\ge 2$. – rogerl Oct 1 '18 at 1:01
$$k\ge1\implies k+1\ge1+1=2\gt1$$. | 172 | 475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-31 | latest | en | 0.730911 |
https://www.enjoymathematics.com/blog/monty-hall-problem/ | 1,638,391,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00357.warc.gz | 794,566,062 | 18,362 | # Monty Hall Problem
### Problem Statement
There are 3 doors behind which are two goats and a car. You pick door 1 hoping for the car but don’t open it right away. Monty Hall, the game show host who knows what's behind the doors, opens door 3, which has a goat. Here's the game: do you want to pick door No. 2? Is it to your advantage to switch your choice?
Key takeaway: The Monty Hall problem is one of the exciting puzzles to learn the application of probability in algorithmic problem-solving. This helps us to understand that: sometimes our statistical intuition or assumptions are incorrect!
### Solution Idea: Monty Hall Problem
Here most people will assume that both doors are equally like to have the prize, and there is no reason to change. So, most of them stick with their initial choice. But the solution is just the opposite: if we switch doors, we double our probability of winning! But the critical question is — how? Let’s think.
If our initial choice is incorrect, then Monty will always open any remaining doors that do not have the prize. In this situation, the only closed door will have the award, and changing the choice will be lead us to the award. So the conclusion is: if our initial choice is incorrect, then switching the door will ensure 100% chances of success! Think!
• The probability of choosing the incorrect door in the first choice = 2/3, and the “Don’t Switch” strategy always leads to failure. Or in other words, the “Switch” strategy always leads to success.
• Similarly, The probability of choosing the correct door in the first choice = 1/3, and the “Switch” strategy always leads to failure.
• In different words, the probability of choosing the incorrect door is high in the first choice. So our best thought would be: we should "Switch" because our initial choice can be wrong with 2/3 probability! Think!
We will get a clear picture by analyzing and preparing the list of different choices and outcomes: there are only nine different combinations! In the following table, each row shows a different combination of initial door choice, prize door, and the results for the two options — “Don’t Switch” and “Switch.” There could be three doors for the first choice, and for each door, there could be three possibilities: either prize will be present in the same door or any one of the other two doors.
The above table shows all possible situations, and we need to count the number of wins for each door strategy. It is visible that out of 9 choices, if we “switch,” then we win 6 times (2/3 probability), and if we “Don’t switch,” then we win 3 times(1/3 probability). I hope this table convinces you that the probability of winning doubles when you switch doors. But again, the critical question is: why this happens? Let's think!
To understand the solution, we first need to know why your brain gets the incorrect solution that is 50/50. Actually, we are using false assumptions, and that’s why we can’t trust our answers.
Mostly we think of probabilities for independent, random events. Flipping a coin is a good example. The probability of heads is 0.5, and we obtain that simply by dividing the specific outcome by the total number of outcomes. That’s why it feels so right that the final two doors each have a probability of 0.5. However, to produce the correct answer for this method, the process must be random and have probabilities that do not change. Unfortunately, the Monty Hall problem does not satisfy either requirement.
The only random portion of the process is your first choice. When you pick one of the three doors, you truly have a 0.33 probability of picking the correct door. The “Don’t Switch” column in the table verifies this by showing you’ll win 33% of the time if you stick with your initial random choice.
The process stops being random when Monty Hall uses his knowledge about the prize’s location. It’s easiest to understand if you think about it from Monty’s point of view. When it’s time for him to open a door, there are two doors he can open. If he chose the door using a random process, he’d do something like flip a coin. However, Monty is constrained because he doesn’t want to reveal the prize. Monty very carefully opens only a door that does not contain the award. The end result is that the door he doesn’t show you, and lets you switch to, has a higher probability of containing the prize. That’s how the process is neither random nor has constant probabilities.
### Here’s how the solution to the Monty Hall Problem works?
The probability that your initial door choice is wrong is 0.66. The following sequence is deterministic when you choose the wrong door. Therefore, it happens 66% of the time:
1. You pick the incorrect door by random chance. The prize is behind one of the other two doors.
2. Monty knows the prize location. He opens the only door available to him that does not have the award.
3. By the process of elimination, the award must be behind the door that he does not open.
Because this process occurs 66% of the time and always ends with the prize behind the door that Monty allows you to switch to, the “Switch To” door must have the prize 66% of the time. That matches the table!
Our mental assumptions were based on independent, random events. However, Monty knows the prize location and uses this knowledge to affect the outcomes in a non-random fashion. Once you understand how Monty uses his knowledge to pick a door, the results make sense.
Enjoy learning, Enjoy thinking! | 1,206 | 5,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-49 | latest | en | 0.938052 |
https://tackk.com/00zkgb | 1,480,726,231,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00354-ip-10-31-129-80.ec2.internal.warc.gz | 886,088,504 | 6,975 | # Newtons first law
definition- An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.This law is often called "the law of inertia".
in my own words- its when an object is moving stay at the same speed and direction intill friction stops it
i could let a ball roll then stop it
# second law
Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
in my own words- the greater mass the more force
i could push a ball into many things | 156 | 711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.94087 |
https://meta.stackexchange.com/questions/228213/why-was-the-minimum-comment-length-of-15-chosen | 1,721,013,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00855.warc.gz | 358,705,564 | 29,962 | Why was the minimum comment length of 15 chosen?
Note: I am not asking why is there a minimum limit.
Previously closed as dup, but neither of the "duplicates" answer this question!!!
The suggested dups were:
Both of which ask why the limit is there. I know why it's there. I'm happy it's there.
This question is asking why it is 15 (and not 14, or 16, etc)?
Was any thought or research put into the choice, or was it just based on feel?
Some examples of reasons for choosing 15:
• Was some statistical analysis done on comment sizes?
• Was the limit borrowed from an external academic or commercial source?
• Was it decided on by committee?
• Is it because it's the largest length that can be represented by a single hex digit?
Also note: I don't think it's a bad limit or needs changing. I am just curious to know why 15 specifically was chosen.
The reason I'm asking is if the reason has firm basis it, that basis could be valuable info for someone researching lower thresholds of meaningful communication.
• Unicorns fly in groups of 15. Thought it was obvious :) Commented Apr 3, 2014 at 9:38
• `"spolsky atwood"` is 15 bytes long (including the `NUL` character). Commented Apr 3, 2014 at 9:38
• @Łukasz, [questioner is] not asking why is there a minimum limit. Commented Apr 3, 2014 at 9:47
• @FrédéricHamidi but the answer to that questions gives an explanation: because 15 is mentioned in some holy quote from Jeff. Quotes from Jeff are considered holy because unicorns talk to him when he sleeps or something that way... Commented Apr 3, 2014 at 9:50
• why is the max comment length 600
– Bolu
Commented Apr 3, 2014 at 10:39
• Guys, I'm asking why was 15 chosen. Those dups do not ask or answer that Commented Apr 3, 2014 at 10:45
• It's the median age of emotional maturity of users on the site. Not sure if that's why it was picked, but just sayin'
– user50049
Commented Apr 3, 2014 at 11:37
• @TimPost Maybe it's the median IQ of users on the site. Commented Apr 3, 2014 at 11:48
• @TimPost ...Is not! Commented Apr 3, 2014 at 11:49
• I have a feeling the only serious answer here is "no special reason". Just a moment decision by Jeff when he wrote the code. Commented Apr 3, 2014 at 11:55
• @ShadowWizard et al., the answer to "Why this particular number?" is almost always "Because we had to pick one and this one felt about right." The answer to "Why a minimum at all?" is more interesting and addressed in other places. Commented Apr 3, 2014 at 20:13
• @DavidFullerton Sure, but I am curious if any thought or research went into it. Commented Apr 3, 2014 at 20:46
• @ShadowWizard you seem to doubt that Jeff really spoke with unicorns in his dreams ;) Commented Apr 3, 2014 at 20:56
• @ShadowWizard I don't really talk to Jeff on a regular basis, and it seems a little silly to send him an email about this :) If there was research, it was probably of the form "What are some crappy comments we don't want to allow?" and set it just longer than that. But probably he just picked a number. Commented Apr 4, 2014 at 15:29
• Thanks @David, your reply is more than what most here expected, I'm pretty sure of that. As for email asking "Hey Jeff, why did you set the minimum comment length to 15?" he'll go after you with some giant letter (maybe T for trivial? ;)) so better not push your luck. Commented Apr 4, 2014 at 20:14
Perhaps because `15` is the magic constant of the unique order-3 normal magic square. | 930 | 3,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.961799 |
https://www.nagwa.com/en/videos/306138234837/ | 1,726,583,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00547.warc.gz | 825,280,738 | 33,728 | Question Video: Evaluating Algebraic Expressions | Nagwa Question Video: Evaluating Algebraic Expressions | Nagwa
# Question Video: Evaluating Algebraic Expressions
Evaluate 𝑥 + 𝑦 + 𝑧 + 6 for 𝑥 = −5, 𝑦 = −6, and 𝑧 = −9.
01:25
### Video Transcript
Evaluate 𝑥 plus 𝑦 plus 𝑧 plus six for 𝑥 equals negative five, 𝑦 equals negative six, and 𝑧 equals negative nine.
We are told in the question that 𝑥 equals negative five, 𝑦 is equal to negative six, and, finally, 𝑧 is equal to negative nine. To evaluate 𝑥 plus 𝑦 plus 𝑧 plus six, we need to add negative five, negative six, negative nine, and positive six. Adding a negative number is the same as subtracting that number.
This leaves us with negative five minus six minus nine plus six. Negative five minus six gives us negative 11. Subtracting nine from this gives us negative 20. Negative 20 plus six is equal to negative 14. The value of 𝑥 plus 𝑦 plus 𝑧 plus six when 𝑥 equals negative five, 𝑦 equals negative six, and 𝑧 equals negative nine is negative 14.
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• Realistic Exam Questions | 331 | 1,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.784858 |
https://tradesmith.com/bayesian-priors-and-the-trouble-with-antibody-tests/ | 1,718,714,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00462.warc.gz | 523,382,871 | 12,134 | # Bayesian Priors and the Trouble with Antibody Tests
By John Banks
On the COVID-19 medical front, the hot new thing is antibody tests.
The idea is that, if a person tests positive for antibodies, it would show they have already contracted COVID-19. That would imply (though not guarantee) they are safe from getting infected again.
Antibody tests could also indicate the overall percentage of a population that has already had COVID-19.
In the state of New York, for example, 21% of New York City residents may have been infected (as reported by the New York Times).
Because the COVID-19 caseload was highly concentrated in New York City — one of the most densely populated urban centers in the country — it drops off quickly for surrounding areas. Numbers for Long Island suggest the antibody rate was 17%; for Westchester and Rockland counties it was 12%; and the overall rate was 4% for the rest of New York state.
These antibody tests have a big drawback, though. For most of the antibody tests in circulation, which vary widely in origin and quality, the accuracy is terrible. This renders them useless, or even dangerous.
To show why accuracy is so important, let’s walk through a quick probability exercise.
• Imagine the actual rate of COVID-19 cases in your area is 4%, meaning that 4 out of every 100 people in your city or county have actually been infected.
• You take an antibody test with a 96% accuracy rate and a 4% false-positive rate. Lab work has shown that, 4% of the time, this test returns a “false positive” diagnosis (identifying antibodies where they don’t exist).
• Your test results come back positive.
• Given what you know about the actual rate of infection in your area — and the 96% accuracy rate of the test — what is the actual probability you have antibodies (and already had COVID-19)?
Most people would assume something like this: “The test is shown to be 96% accurate. That means a 96% chance I have antibodies and have already had COVID-19.”
Given the data provided, the odds of having COVID-19 antibodies are coin-flip. Even with a 96% accuracy rate, the test has told you nothing much at all.
Even worse, it may have created false confidence by supporting the notion you already had the virus and are safe (when the real chance of this is just 50/50).
To understand why this is so, we have to start with the prior probability distribution, a concept known as the “”prior” or “Bayesian prior.”
We explained that, in reality, 4% of the local population, or 4 people out of 100, had actual antibodies. We can represent that percentage with a 10×10 grid as shown below.
In our grid example, the 96 blue squares represent people with no antibodies. The four red squares are the people who do have antibodies, representing 4% of the population.
Now let’s show the results of the antibody test.
Remember we stipulated that the test has a 96% accuracy rate and a 4% false positive rate. The grid below represents what it would look if this test were given to the same population of 100 people, where 4% actually do have antibodies.
The red squares with a “T” represent “True” positive results. The orange squares with an “F” represent “False” positive results — the 4% of the time where the test gets it wrong.
As the visual implies, anybody who tests positive cannot know if they occupy a red square (true result) or an orange square (false positive result).
What’s more, the number of red squares and orange squares are equal, which shows why a test with 96% accuracy can deliver a 50% verdict (no better than a coin flip). The test reliably identifies the four true positives, but also identifies four false ones. Hence, if your results come back positive, the odds of being red (true) are 4/8; but the odds of being orange (false) are also 4/8.
A quicker, though less intuitive, way to explain what is happening here is to point out that the prior probability distribution (the Bayesian prior) and the testing error rate have a mathematical relationship. The smaller the first number in comparison to the second, the more useless the test result becomes.
This also matters a great deal when it comes to assuming the rate of antibodies (and thus COVID-19 cases) within a general population.
Say, for example, an antibody test assumes a 5% antibody rate within a population but the test is only 90% accurate. One might think “90%, that’s pretty good” — but if the error percentage (10%) is double the output percentage (5%), the test is worse than useless (in terms of providing false confidence).
A team of more than 50 scientists from University of California San Francisco, University of California Berkeley, the Chan Zuckerberg Biohub, and the Innovative Genomics Institute recently tested the accuracy of 14 different COVID-19 antibody tests. Their initiative is called the COVID-19 Testing Project.)
“Of the 14 tests, only three delivered consistently reliable results,” the New York Times reports. “Even the best had some flaws.”
In consistency terms, three out of 14 is a terrible hit rate.
As for the others — they are worse than useless, per the New York Times again: “Four of the tests produced false-positive rates ranging from 11% to 16%; many of the rest hovered around 5%.”
These are not numbers to be trusted.
Hopefully the quality of these antibody tests will improve dramatically before they are rolled out at scale — and especially if the test results are used to shape policy decisions.
As it stands, far too many of these tests could deliver dangerous misinformation: Imagine taking a test with a touted 95% accuracy rate, assuming you are safe (because you already had COVID-19), and then finding out the tragic way you were wrong. | 1,247 | 5,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.957555 |
https://www.lmfdb.org/EllipticCurve/Q/46389k/ | 1,597,279,012,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00249.warc.gz | 740,613,603 | 43,291 | # Properties
Label 46389k Number of curves 6 Conductor 46389 CM no Rank 1 Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("46389.d1")
sage: E.isogeny_class()
## Elliptic curves in class 46389k
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
46389.d6 46389k1 [1, 0, 0, 2163, 8712] [2] 52992 $$\Gamma_0(N)$$-optimal
46389.d5 46389k2 [1, 0, 0, -8882, 68355] [2, 2] 105984
46389.d3 46389k3 [1, 0, 0, -86197, -9688798] [2] 211968
46389.d2 46389k4 [1, 0, 0, -108287, 13686840] [2, 2] 211968
46389.d4 46389k5 [1, 0, 0, -75152, 22229043] [2] 423936
46389.d1 46389k6 [1, 0, 0, -1731902, 877125297] [2] 423936
## Rank
sage: E.rank()
The elliptic curves in class 46389k have rank $$1$$.
## Modular form 46389.2.a.d
sage: E.q_eigenform(10)
$$q - q^{2} + q^{3} - q^{4} + 2q^{5} - q^{6} - q^{7} + 3q^{8} + q^{9} - 2q^{10} - 4q^{11} - q^{12} + 2q^{13} + q^{14} + 2q^{15} - q^{16} - 6q^{17} - q^{18} - 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 4 & 4 & 8 & 8 \\ 2 & 1 & 2 & 2 & 4 & 4 \\ 4 & 2 & 1 & 4 & 8 & 8 \\ 4 & 2 & 4 & 1 & 2 & 2 \\ 8 & 4 & 8 & 2 & 1 & 4 \\ 8 & 4 & 8 & 2 & 4 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with Cremona labels. | 702 | 1,543 | {"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 | 3 | CC-MAIN-2020-34 | latest | en | 0.435635 |
http://earthmeasured.com/annual-parallax-doesnt-exist-since-flat-earth-doesnt-move/ | 1,561,492,857,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999946.25/warc/CC-MAIN-20190625192953-20190625214953-00261.warc.gz | 53,021,386 | 10,290 | # Annual Parallax doesn’t exist since Flat Earth doesn’t move.
We could say that the annual parallax of a star is the angle defined by the sun, the star and the Earth. (considering the star perpendicular to the line that unites the Sun and the Earth).
π is the angle SÂE in the picture below. (annual parallax angle).
The star considered to be nearer to the Earth is Proxima Centauri that has an official parallax angle of 0.75 seconds of degree. The parallax effect due to the movement of the observer on the Earth orbit around the sun means a periodical movement of the star on the celestial sphere. The ellipse thus projected by this movement on the celestial sphere by is called parallax ellipse and has a periodicity of one year.
# Annual parallax is a hoax
Warning: once again the so said “aligned astronomers” don’t consider the movement of the sun in the galaxy, but only the rotation of the Earth around the sun. The parallax thus shouldn’t originate an ellipse but a spiral during the year.
How do astronomers determinate the parallax angle? This determination is one of the most difficult but most important key point of sidereal astronomy.
This issue is so important because when knowing the parallax of one star, its distance from the Earth can be determined . We can thus understand the efforts of astronomers in their attempts to be absolutely accurate in determining the parallax angles of stars.
The first parallax to be determined was 61 Cygni. It was calculated by Bessel in Konigsberg in 1837-38.
There is more than one method to determine the parallax. Here you will read about the trigonometric method. To determine the parallax of a star S the astronomers chose two stars A and B with parallax almost equal to zero because they are very far. A and B must be aligned on a parallel to the ecliptic one on a side and one on the other side of the star S. During the year A and B will remain fixed in the celestial sphere while S, nearer to the Earth, will move toward A for six months and toward B for the rest of the year. By measuring for one year the amount of these movements it is possible to determine the parallax. These very little angles where measured by using an eliometer. Today the preferred method is photography that “consents much precision”. The idea is simple: when the star S is at one extreme of the ellipse, one picture is made, another when the star is on the other side of the ellipse, after six month, and another picture of control is made after one year. Pictures are checked and from the movements of S in respect of all the other stars the parallax is determined. Parallax is considered to be a strong evidence of the rotation of the Earth around the sun. If Earth were motionless this phenomenon wouldn’t exist.
A consideration I have to do is that the parallax angle is really small, always smaller than one second of degree.
Consider a circle, divide it in 360°. Then take one single segment and divide it 3600 times. Well, the parallax angle of the nearest star is even smaller.
This angle is even smaller than the aberration angle (you certainly remember it was calculated as 20”,45). But both these angles are smaller of the refraction angle. So we have three ellipses (the parallax, the aberration and the refraction ellipses) that superpose one over the other. The refraction ellipse, the greater one, is very changeable during the year depending on temperature and pressure of the air. Also the aberration depends on the air temperature, since light speed depends on the dielectric constant of the mean and consequently on temperature.
So how is it possible to evaluate with a photograph the contribution of the aberration ellipse, and even more difficult, the contribution of refraction when it could be sufficient a slight hot current of air in the moment the picture is made to change all the results?
The conclusion is that the annual parallax doesn’t exist, cannot be measured and absolutely cannot be used to determine distances of the stars or of planets.
## Real accurate maethods to measure the height of planets and stars
This doesn’t mean that it is not possible to measure the distance of a star from the Earth and in this blog we have already shown some example when we have calculated the height of the sun or the height of Polaris. We have not used the annual parallax obviously but a parallax calculation that we call triangulation. We can remember these two cases:
Height of the sun. To make a triangulation we need two observers that from two distant places (parallax) on the Earth can measure the angle of the sun. With these two angles we can track two lines that define the height of the sun. If you are alone you can consider that on solstice the sun will be vertical over the Capricorn and make thus the triangulation with that point. Height of Polaris. Also with Polaris triangulation is easy. Polaris is in fact exactly on the North Pole. When we have measured the angle from our latitude, since we know the distance from the pole, we can easily calculate the height of Polaris.
Next post will be about planets: we can use tha magical square to know something more about their height and trajectory . | 1,128 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-26 | latest | en | 0.925179 |
https://byjus.com/question-answer/a-stream-of-water-flowing-horizontally-with-a-speed-of-15-m-s-2-gushes/ | 1,643,221,233,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00640.warc.gz | 210,258,537 | 22,207 | Question
# A stream of water flowing horizontally with a speed of 15 m/s2 gushes out of a tube of cross - sectional area 10−2 m2, and hits at a vertical wall nearby. What is the force (in newtons) exerted on the wall by the impact of water, assuming it does not rebound?
Solution
## Figure shows a column of water of length 15 m and cross - section area 10−2 m2 at rest. Let us calculate the mass of water flowing out per second and hitting the wall. When the stream of water has a speed of 15 m/s, the particle of water at section B would have travelled a distance of 15 m in one second and arrived at the section A. The water contained in this column would have gone out in one second and hit the wall. Thus, the volume of water coming out of A in one second = volume of water contained in a cylinder of length 15 m and area of cross - section 0.01 m2 = Length×Area of cross - section =15×10−2=0.15 m3 Mass of water gushing out per second = Volume of water×Density of water =0.15×1000=150 kg [density of water =1000 kg/m3] |Change in momentum| =2250−0=2250 N Since change in momentum per second is the force, hence force exerted on the wall will be 2250 N.Physics
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https://www.physicsforums.com/threads/use-excel-to-find-the-similarity-between-functions.832809/ | 1,571,289,917,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00077.warc.gz | 1,044,720,450 | 16,240 | # Use Excel to find the similarity between functions.
#### 24forChromium
I am writing a paper and I came up with a function that uses a hypothetical relationship to predict the value of one variable at different points in time, I graphed it, and then graphed the actual readings from an experiment. How can quantitatively describe how close the two trends are? In other words, how can I quantitatively describe the amount of support the hypothetical relationship get from the experiment? See image for more ideas, predictive function gives values in blue, actual readings are orange.
Related Computing and Technology News on Phys.org
#### DrClaude
Mentor
You could perform a χ2 test. You can also calculate the R2 for your model, since you already have the value for the linear regression.
If you don't know what any of that means, you can start with https://en.wikipedia.org/wiki/Goodness_of_fit
#### FactChecker
Gold Member
2018 Award
You can get the confidence intervals for the slope and intercept of the regression line. For a given confidence level, your theoretical parameters will be in or out of the confidence interval. Just eyeballing the data, the regression line, and your theoretical line, I think there is something missing from your theory. I think it is likely to be statistically rejected.
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• Solo and co-op problem solving | 341 | 1,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | longest | en | 0.874947 |
https://aufdercouch.net/one-fourth-plus-one-fourth-equals/ | 1,652,995,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00300.warc.gz | 172,550,918 | 5,080 | The calculator performs an easy and progressed operations through fractions, expressions with fractions merged with integers, decimals, and mixed numbers. It additionally shows detailed step-by-step information around the fraction calculation procedure. Solve difficulties with two, three, or much more fractions and numbers in one expression.
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Rules for expressions v fractions: Fractions - merely use a front slash between the numerator and denominator, i.e., for five-hundredths, get in 5/100. If you room using mixed numbers, be sure to leaving a solitary space in between the whole and portion part.The cut separates the numerator (number over a portion line) and also denominator (number below).Mixed numerals (mixed fractions or blended numbers) compose as integer separated by one space and fraction i.e., 12/3 (having the exact same sign). An example of a negative mixed fraction: -5 1/2.Because slash is both indicators for fraction line and division, us recommended usage colon (:) together the operator of department fractions i.e., 1/2 : 3.Decimals (decimal numbers) go into with a decimal suggest . and also they are instantly converted to fountain - i.e. 1.45.The colon : and also slash / is the prize of division. Can be offered to divide mixed numbers 12/3 : 43/8 or can be offered for write complicated fractions i.e. 1/2 : 1/3.An asterisk * or × is the symbol because that multiplication.Plus + is addition, minus sign - is subtraction and ()<> is mathematics parentheses.The exponentiation/power price is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
Examples: • adding fractions: 2/4 + 3/4• individually fractions: 2/3 - 1/2• multiply fractions: 7/8 * 3/9• separating Fractions: 1/2 : 3/4• indices of fraction: 3/5^3• fractional exponents: 16 ^ 1/2• including fractions and mixed numbers: 8/5 + 6 2/7• dividing integer and also fraction: 5 ÷ 1/2• complex fractions: 5/8 : 2 2/3• decimal to fraction: 0.625• fraction to Decimal: 1/4• portion to Percent: 1/8 %• comparing fractions: 1/4 2/3• multiplying a portion by a entirety number: 6 * 3/4• square root of a fraction: sqrt(1/16)• reduce or simplifying the fraction (simplification) - dividing the numerator and also denominator the a fraction by the same non-zero number - equivalent fraction: 4/22• expression v brackets: 1/3 * (1/2 - 3 3/8)• compound fraction: 3/4 the 5/7• fountain multiple: 2/3 of 3/5• divide to uncover the quotient: 3/5 ÷ 2/3The calculator follows popular rules because that order that operations. The most common mnemonics because that remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, that or Order, Division, Multiplication, Addition, Subtraction.
See more: How Fast Is A Lacrosse Shot ? What Is The Average Speed Of Lacrosse Shots
GEMDAS - Grouping icons - brackets (), Exponents, Multiplication, Division, Addition, Subtraction. it is in careful, constantly do multiplication and division prior to addition and also subtraction. Some operators (+ and -) and (* and also /) has the exact same priority and also then have to evaluate from left come right.
## Fractions in indigenous problems:
next math difficulties » | 846 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-21 | latest | en | 0.860049 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/11%3A_Basic_Science_of_Nanomaterials/11.02%3A_Physics_and_Length_Scales-_Cavity_Laser_Coulomb_Blockade_Nanoscale_Magnets | 1,723,674,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00091.warc.gz | 131,672,703 | 34,387 | # 11.2: Physics and Length Scales- Cavity Laser, Coulomb Blockade, Nanoscale Magnets
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The special properties of nanomaterials do not derive from different laws of physics, which are the same for objects large and small. For example, Newton's second law ($$F = ma$$), Coulomb's law ($$E= \frac{q_{1}q_{2}}{4 \pi \varepsilon_{0}r}$$), and the laws of energy and momentum conservation are the same for buckyballs (C60) and full-size soccer balls. Nevertheless, the physics of electrons, atoms, and photons naturally produce characteristic length scales, some of which we have already seen. For example, in Chapter 6 we discovered that the mean free path of an electron in a good metal is about 40 nm. In Chapter 10, we learned that the Bohr radius of an electron or hole in doped Si is about 4 nm, and that the coherence length of Cooper pairs in semiconductors is somewhere between a few nm and 1 µm. When objects become small relative to these characteristic lengths, their physical properties change in interesting ways. Materials that exist at the relevant length scale are called mesoscopic (meso = "between," scopic = "size") meaning that they cross over from one kind of behavior - the bulk behavior of large objects - to another. This length scale is different for different kinds of properties, but for many it happens between 1 and 100 nm. We illustrate this point with a few examples.
## The cavity laser
A vertical cavity surface-emitting laser (or VCSEL) is a semiconductor-based device that emits light in the vertical direction relative to the plane of the chip. These devices are being developed and used for high power applications such as laser surgery, infrared illumination for military surveillance, and laser cutting tools. The basic design of a VCSEL is shown at the right. The VCSEL is basically a light-emitting diode, incorporating p-type and n-type regions of the III-V semiconductor (Al,Ga)As. However, it has two special features. First, the refractive index of the semiconductor is modulated above and below the junction to make Bragg mirrors. These mirrors reflect light emitted in the junction, so that the photon density becomes very high there, a necessary condition for stimulated emission and lasing. The Bragg mirror stack is asymmetric (thinner on the bottom), so that light can escape from the junction in one direction only. Second, the junction itself consists of a thin "quantum well" layer of (In,Ga)As, a III-V semiconductor with a smaller band gap than the surrounding (Al,Ga)As layers.
A VCSEL device structure. This is a bottom-emitting multiple-quantum-well VCSEL.
The quantum well structure, the band diagram of which is illustrated at the left, is the nanoscale part of the laser. The energies of the conduction and valence bands of (Al,Ga)As flank those of the thin (In,Ga)As layer. Therefore, electrons and holes injected into that layer cannot escape: electrons in (In,Ga)As do not have enough energy to climb "up" to the conduction band of (Al,Ga)As and holes cannot climb "down." Electrons confined to such a small well behave as a particle-in-a-box (as we learned in the context of electrides in Chapter 9).
The electron has a kinetic energy defined by the equation:
$KE= \frac{h^{2}n^{2}}{8mL^{2}}$
We can calculate the energy difference between the lowest (n=1) and next lowest (n=2) levels, which is inversely proportional to the square of the thickness (L) of the (In,Ga)As layer. In this calculation we need to use the effective mass of the electron in (In,Ga)As, which is about 7% of the electron rest mass[1]. With an 8 nm thick layer, this energy is:
$E = \frac{(2^{2}-1^{2})(6.626 \cdot 10^{-34} Js)^{2}}{8 \cdot 0.07 \cdot (9.1 \cdot 10^{-31} kg)(8 \cdot 10^{-9} m)^{2}} = 4.0 \cdot 10^{-20} J = 0.25eV$
The VCSEL cavity will thus have a resonant energy of 0.25 eV and emit photons at this energy in the infrared (λ ≈ 5000 nm). Note that because of the inverse square dependence of the cavity energy on layer thickness, lasers based on this design can only function at nanoscale dimensions. When the cavity is three times thicker, its resonant energy becomes comparable to the thermal energy at room temperature (kT = 0.026 eV), and the lasing effect is thermally "washed out."
A capacitor is a (macroscopic) device that stores electrical charge. The basic structure of a capacitor is shown at the right. When a voltage is applied to such a device, it develops a charge (± Q) on the two plates that is proportional to the voltage:
$C= \frac{Q}{V}$
Basic design of a parallel plate capacitor of area A and with a dielectric thickness of d
The magnitude of the capacitance C is determined by the permittivity ε and the dimensions of the dielectric layer, A and d.
$C= \frac{\varepsilon A}{d}$
We can also calculate the work done in charging the capacitor up (i.e., the energy stored by charging the capacitor) by integrating the voltage times the charge:
$E = int^{Q}_{0} V(q)dq = \int^{Q}_{0} \frac{q}{C} dq = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2}CV^{2} = \frac{1}{2}VQ$
Now it is interesting to ask, what happens to a capacitor when we make it very small? This is of particular interest in a device called a single electron transistor, a schematic diagram of which is shown at the right. The metallic gate lead is separated from a "quantum dot," which can be a metal or semiconductor particle, by a thin dielectric layer. This metal-dielectric-metal sandwich acts as a capacitor, and from the equation above, the energy needed to charge it by a single electron (Q = e) is:
$E= \frac{e^{2}}{2C}$
where e is the charge of the electron, 1.602 x 10-19 Coulomb. If the gate width and lateral dimensions are very small - say 2 nm as is readily achievable in self-assembled Coulomb blockade devices[2][3] - then for a typical insulating dielectric, a voltage of about 200 mV is needed to charge the quantum dot by a single electron. Again, this effect is unique to the nanoscale, because a 10 times larger device area would make the single-electron charging voltage about 20 mV, which is smaller than the thermal energy kT (26 meV). Thus for devices larger than about 5-6 nm, individual electron charging events are washed out at room temperature by thermal fluctuations.
Schematic of a single-electron transistor.
How can a nanoscale capacitor like this act as a transistor, which functions as a switch in an electrical circuit? The effect comes from the mutual repulsion of electrons. An electron on the quantum dot repels any other electron that would be forced onto it by applying a small voltage between the source and the drain. Hence the conductance of the quantum dot is very low at a gate bias of zero volts, or at any gate bias (200, 400, 600 mV...) that places an integer number (1, 2, 3,...) of electrons on the dot. But halfway in between these voltages (e.g., at 100, 300, 500 mV) the energy is the same whether there are n or n+1 electrons on the dot. This means that electrons can hop on and off without changing their energy, i.e., they can tunnel through the dot from source to drain. This effect gives peaks in the conductance of the dot at regular steps in the gate voltage. In effect, the gate can act as a switch, as in a conventional field-effect transistor. Single-electron transistors are being researched as ultra-sensitive electrometers and single-molecule chemical sensors, since a tiny change in the electrostatic environment of the dot can switch the device on or off.
## Nanoscale magnets
Ferro- and ferrimagnetic materials such as iron and chromium oxide are used for digitial storage of information in hard disks. The individual memory bits, which can be oriented perpendicular or parallel to the plane of the disk as shown at the right, store a logical "0" or "1" depending on the orientation of their magnetic dipole. To be useful, this information must be non-volatile, i.e., the magnetic bit must retain its polarization in the absence of an applied field from the read/write head.
Longitudinal and perpendicular recording, two types of writing heads on a hard disk.
The storage density of such magnetic memories is impressive. A 2.5" hard drive can now store 1 TB of information, using rod-shaped magnetic grains that are approximately 0.5 µm long. We now have good synthetic methods for making these same materials as crystals with dimensions of only a few nanometers. Why aren't those nanocrystals used to make even more dense memory disks?
The reason is that the energy needed to flip the magnetization (i.e., to turn a "0" into a "1" and vice-versa) is strongly size-dependent. For a ferro- or ferrimagnet this energy is equal to Mr3, where M is the magnetic energy per unit volume and r is the characteristic dimension (e.g., the length of the edge of a cube, or the diameter of a sphere) of the magnetic grain. For typical materials such as iron, this energy becomes comparable to kT when r is about 3-5 nm. Such small particles are superparamagnetic, meaning that they still have a large magnetic moment because of the ordering of their spins, but they do not retain a permanent polarization in the absence of an applied magnetic field. Superparamagnetic particles are thus not useful for magnetic memories, but they are interesting and practical in other ways, for example in ferrofluids, magnetic resonance imaging (MRI), and some emerging medical diagnostic and therapeutic applications.
A ferrofluid containing superparamagnetic Fe3O4 nanoparticles, which are coated with oleic acid and suspended in oil, in the field of a strong permanent magnet.
In these three illustrative examples (involving light emission, electronic conduction, and magnetic behavior), the transition to new properties involves a crossover in which the characteristic energy of the system is comparable to the thermal energy kT. It just so happens that for many physical phenomena, this crossover occurs on the length scale of nanometers.
This page titled 11.2: Physics and Length Scales- Cavity Laser, Coulomb Blockade, Nanoscale Magnets is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Chemistry 310 (Wikibook) via source content that was edited to the style and standards of the LibreTexts platform. | 4,179 | 14,605 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.203623 |
http://mathhelpforum.com/business-math/29104-inventory-model-print.html | 1,526,911,019,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864186.38/warc/CC-MAIN-20180521122245-20180521142245-00495.warc.gz | 182,433,022 | 2,744 | # Inventory Model
• Feb 25th 2008, 11:24 AM
DooBeeDoo
Inventory Model
assuming no shortages permitted and an order lead time of zero, calculate the economic order quantity and economic order time for the firm:
if:
x(demand) = 1000 units/month
h (holding cost) = 1% of the value of stock/month
k(fixed cost)=$180/delivery c (variable cost) =$4/unit
the economic order quantity is (2kx/h) square rooted.
any help appreciated!
• Feb 25th 2008, 12:13 PM
colby2152
Quote:
Originally Posted by DooBeeDoo
assuming no shortages permitted and an order lead time of zero, calculate the economic order quantity and economic order time for the firm:
if:
x(demand) = 1000 units/month
h (holding cost) = 1% of the value of stock/month
k(fixed cost)=$180/delivery c (variable cost) =$4/unit
the economic order quantity is (2kx/h) square rooted.
any help appreciated!
$\displaystyle EOQ=\sqrt{\frac{2kx}{h}}\Rightarrow \sqrt{\frac{2*180*1000}{0.1}} = ???$
I will leave you and your calculate to complete the problem. | 291 | 1,010 | {"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} | 3.328125 | 3 | CC-MAIN-2018-22 | latest | en | 0.798825 |
http://www.jiskha.com/display.cgi?id=1354421175 | 1,462,439,072,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00083-ip-10-239-7-51.ec2.internal.warc.gz | 603,215,515 | 3,689 | Thursday
May 5, 2016
# Homework Help: PHYSICS
Posted by HELP on Saturday, December 1, 2012 at 11:06pm.
After 6.00 kg of water at 80.4 oC is mixed in a perfect thermos with 3.00 kg of ice at 0.0 oC, the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be ÄS = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs. | 155 | 565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-18 | longest | en | 0.913928 |
http://www.thescienceforum.com/physics/15171-thermal-value-print.html | 1,591,430,399,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00082.warc.gz | 208,670,074 | 4,253 | # Thermal value
• July 10th, 2009, 06:35 PM
marcusclayman
Thermal value
I need some direction. Please excuse my lack of understanding on the subject.
I am not preoccupied with the efficiency of this project, it's just an idea that I'd like to do some research on and draw up some designs.
The idea is this, cooling using Ice from winter, stored in a thermal building so that the ice lasts throughout the summer.
To state the obvious variables
x temperatures outside
y thermal value of the building
z thermal mass of the ice
is this correct? are there other things I need to take into account?
Now I'm assuming that the most efficient shape is a sphere, and for the sake of this thought experiment will be using a thermal dome for the building.
The idea is this.
During the winter, a big block of ice is manufactured along side of a water source. The block is made as big as possible and shaped roughly as a dome. As spring approaches, the block is made into the shape of a dome. A thermal dome is then built around the pile of ice.
for those who don't know how to make a thermal dome
1: spray closed cell Styrofoam around the form, in this case, on the outside of the ice
2: stick steel re-bar into the Styrofoam
3: spray cement onto the steel re-bar and Styrofoam
A vent would be needed, or a vacuum would be created as the ice melts. BUT a vacuum keeps things cooler right? If so, pressure of a certain amount could be tolerated.
I want to figure out how long the ice would last.
Where should I start?
• July 10th, 2009, 06:52 PM
Bunbury
Presumably you don't want to just sit there and watch it melt; you want to do something useful with it - right? So the place to start is to decide what the purpose of your giant ice block is. To do anything useful you will have to exchange heat with some other medium - maybe air if you want to cool a building like the early days of air conditioning, or maybe run water pipes through it to produce cold water, or maybe just store food. The rate of melting will depend on what you are exchanging heat with.
Once you've decide what you're going to do with it, the heat transfer through your well-insulated dome will probably be much smaller than the heat exchange with your process whatever it is. If it isn't you are pretty much wasting your time.
• July 10th, 2009, 07:45 PM
marcusclayman
I don't have anything planned for this. I just want to learn about thermal dynamics, and investigate whether or not there is a practical use of this.
Whether it is to run pipes through for cooling, which I doubt would be worth it; or if it's merely to use as a giant walk in refrigerator/freezer.
• July 10th, 2009, 08:39 PM
Bunbury
Then you would start by calculating the thermal resistance of the wall of your building, which is the sum of resistances of the styrofoam and the concrete, The rersistance is calculated from the thickness divided by the thermal conductivity. You should correct for the spherical shape, but id the radius is large you can treat it as a flat wall. Then add in a film resistance for air outside the dome.
Then it starts to get complicated because you now have to look at the mass of ice itself. If it is really cold then you can warm it up without melting it until its surface reaches 0 deg C and it starts to melt. You will have to involve the Biot number for the mass of ice, and if ther ice is in close contact with th dome you should include the dome in your Biot calculation. I''m sure you can google how to do this.
Then it gets really complicated because once the ice starts to melt you will have an air space between the ice mass and the dome, and you now have to consider the film resitances between the doem and air and the ice and air. The melt water will increase the rate of heat transfer, while the mass of ice decreases so your Biot number is no longer a constant.
Since this is a transient case, not steady state, you would probably do best to model it in a spreadsheet with a little convergence routine. I did this once for an autoclave sublimating uranium hexafluoride.
Ice storage is not a new idea by the way. There are ice houses in most US cities dating from the days before mechanical refrigeration.
http://query.nytimes.com/mem/archive...669D946797D6CF
Maybe you should start with a simpler steady state model if you are just trying to learn some heat transfer.
• July 11th, 2009, 09:20 AM
marcusclayman
Indeed, thank you, a little direction goes a long way when you don't know where to start.
I got stuck trying to figure out how to "reciprocate" to determine R value, or U value, don't remember which.
I guess it would be easier to find a calculator, and figure out what it's doing.
• July 11th, 2009, 11:07 AM
Bunbury
Resistances in series are simply additive (as in electrical analogy). Once you've added up all your resistances the U value is the reciprocal.
• July 11th, 2009, 02:57 PM
marcusclayman
I got that far, but in trying to figure out what "reciprocal" means, I found dozens of things, a few of which were mathematical, all of which were beyond my comprehension.
• July 11th, 2009, 05:17 PM
Bunbury
R = r1 + r2 + r3 + r4
U = 1/R
1/R is the reciprocal of R
There is most likely a 1/x function on your calculator.
• July 11th, 2009, 07:19 PM
marcusclayman
Ah, that's simple enough. I can do that in my head, lol. The formula I found was much more complicated than that. | 1,326 | 5,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.96711 |
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### Hard Logic Puzzle
Hard Logic Puzzle - 17 April
A gardener planted the mangoes trees in a mathematical ordered way (i.e. he planted tree in rows).
He created total of 19 rows of trees.
Note: There should be more than 2 trees to count as a row.
What is the minimum number of trees required by the gardener to have 19 rows of trees ?
1. 3 19 =57
3. By constructing all diagonals of octagon and counting all points of intersection,we get 57,as answer.
4. 34 trees
he will make 17 rows of two trees each = 17 rows (34 tress)
now if we conside the first trees in each it can be considered as 1 row
|||ly for the second tree = 2 rows (0 tree req.)
5. This comment has been removed by the author.
6. 18 Trees
Two blocks of 3x3 trees offset diagonally.
T.T.T
.T.T.T
T.T.T
.T.T.T
T.T.T
.T.T.T
7. 18 trees
8. I also get 18 trees in 2 (3x3) blocks diagonally adjacent (longest row being 6 trees)
1. How? I get only 17 rows like this.
T.T.T.
T.T.T.
T.T.T.
.........T.T.T.
.........T.T.T.
.........T.T.T.
Each square gives you 7 rows (3 vertical, 3 horizontal, 1 diagonal). That is 14 rows. Plus you get 1 long diagonal, which now makes total of 15 rows.
Now we need 4 more rows, but there can be found only 2 more - 2 rows with 4 trees on each, making the total of 17 rows. Still missing 2 rows?
9. A B C
D E F G
H I J K
L M N O
Found 20 using this layout with 15 trees.
Only 14 rows you have there. Where do you take the needed 5 more?
10. 37 trees, it is due to the gravity of the jupiter!!
ATTRACTING THE TREES!
11. The answer is 2, the question was "what is the MINIMUM number of trees required by the gardener to have 19 rows of trees" The statement before the question gave away the answer suggesting that there should be more that 2 trees to make a row.
1. 2 (minimum) * 19 = 38
38 is the minimum number of trees to make 19 rows
12. Ashok RajpurohitJune 13, 2013 at 3:34 PM
208
2+3+4+5+...+19+19 = 208
13. This is not a hard logic puzzle.. this is what you call a fucking dumb one
14. Just need 5 trees in one main row, since there are 25 rows within it
T T T T T
1 row of 5
5 rows of 4
10 rows of 3 (5x4x3/6)
10 rows of 2 (5x4/2)
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https://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=yard&To=fl+head | 1,620,934,056,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00392.warc.gz | 212,444,040 | 3,840 | Partner with ConvertIt.com
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 458 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.698101 |
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