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https://worldbuilding.stackexchange.com/questions/27821/if-one-of-the-spatial-dimensions-wrapped-around-how-would-architecture-be-diffe | 1,702,159,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00067.warc.gz | 680,683,997 | 40,329 | # If one of the spatial dimensions wrapped around, how would architecture be different?
Following the same geometry as this question, how would architecture be different? To review, when go east or west 5 feet, you end up back where you started.
Is there a way to make structures feel roomier, despite the fact the wrap around thing (say, by connecting rooms in a certain way).
How would this affect structural integrity? How would the building be built different to deal with and/or take advantage of this.
How would you attach structures to themselves ("wrapped around")?
Anything else of note?
• using 2 piece of mirrors in parallel should make your structures feel roomier I think Oct 16, 2015 at 3:25
• Thinking about this geometry reveals some interesting effects... like angular momentum is only conserved along the axis parallel to the wrapping direction; and $1/r^2$ forces turn into $2/r$ forces over distances more than the wrapping distance. (On a related note, I know know what string theorists are talking about when they refer to "small extra dimensions".) Oct 16, 2015 at 6:35
A nice way to think about this type of thing is to imagine a bunch of $\infty\times\infty \times 5$ ft blocks stacked next to each other, with the same stuff in each block. So if you were in one block, looking into another, you would see another copy of yourself and the room you are in, and another one beyond that, and so on, repeating infinitely. In math, this is called a covering space. I will refer to the strange space you describe as cylindrical space.
Something interesting about this space is that it is not simply connected. Imagine you had a string in normal space. You could tie the ends together, and then do whatever you wanted with it. You could wad it up and put it in your pocket, for instance. But in this space, you could tie one end of your string in one of the blocks to the other end in the next block, and you would suddenly have an infinite strand which you could not wad up without breaking! Moving down from the covering space to cylindrical space, it would look like a an infinitely long piece of string. This is a really weird concept which is not true in normal space.
Everything, obviously, would have to fit in this very narrow space. There is really no way around this. Every room only has two walls (unless you want it to be even narrower), so all the electricity and plumbing has to run to these two walls, or through the floor. Also, all things we normally put on walls, like TVs, paintings, mirrors, etc,... would have to go on these two very narrow walls. An architect would have to take advantage of vertical space to make lots of rooms. Beds would have to lie in the infinite direction because nobody taller than $5$ ft could lie down in the $5$ ft direction. Nevertheless, skinny houses and buildings are possible. Something like this narrow house might be what you get.
It's not all bad. Even though the room is narrow, you don't have to worry about everybody seeing the TV; there are a whole array of TVs, one in each copy of the universe all stacked next to each other. Since you only need two walls, the building will be stronger. Also, you don't need so many mirrors in the bathroom because it is really easy to see the back of your head. There would be more advantages, I think if the universe were a little wider. Say, $30$ ft. Then you could take advantage of the cylindrical properties to have hallways that wrap around. Then, instead of two rooms being at opposite ends of the hall $30$ ft apart, the furthest apart they can be is $15$ ft. The same advantage applies to lots of application where minimizing distance is an issue.
Some things could seem infinite, like beds or bathtubs five feet wide. That would be pretty cool.
All in all, this would be a pretty weird place to live.
• If rooms didn't align east west, couldn't they connect to other rooms instead of themselves? Oct 16, 2015 at 3:38
• Maybe, but you would have to be really clever about it for that to be useful with only five feet in one direction. With more space, that would be more useful. Oct 16, 2015 at 3:40
• Or, similarly, construct a room in such a way that it wraps around twice or more before connecting to itself. Oct 16, 2015 at 3:42
• Building my last comment, you could have a situation where you walk through a door in a square room into a different square room, and then turn right, walk through a different door and end up back in the room you started in. As for wrapping more than once, this is not possible in the topology of the cylindrical space. Oct 16, 2015 at 3:54
• 5' is short enough that you could reach out and touch yourself. Would be weird for us but the inhabitants would be used to it. Oct 16, 2015 at 9:08
You'd only need one side wall to support your structure. It would be pressed on from both east and west and therefore resist from both east and west. Of course, with a front and back wall, you could forego the side wall entirely.
Your world would have to build up and down a lot, but you'd still be constrained by gravity. I'm not sure how gravity would be affected, but it would change the density of stars and planets.
I'd imagine you'd have a lot of hobbit-type holes. People would live underground (or at least under roofs) and pathways would be built on top of them. You could build above the pathways, but since buildings are heavy and people are not it would make more sense to put the lightweight things on top.
If you came to a bridge that's out, there'd be no way to bypass it. But because water can't travel sideways particularly, there probably wouldn't be many wide chasms to begin with.
The world would have a lot less water, or there'd be no dry land. On the other hand, if there's less gravity (not sure that's true), the mountains would be much taller, so you could have really deep lakes and oceans and still have dry land.
Weather patterns would change dramatically. There would be no way for pressure to move sideways, so tall buildings might be out of the question, although there's still some limited room to build pipes between floors to allow air pressure to flow north to south.
Sideways momentum would be weird. If you came across other planetary systems, you'd have to add a sideways vector. I don't know that it's terribly complex, but it would be different since you can't orbit along that axis, just move. | 1,436 | 6,406 | {"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} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.965249 |
https://www.aqua-calc.com/what-is/pressure/megapascal | 1,702,192,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00435.warc.gz | 704,301,354 | 8,931 | # What is a megapascal (unit)
## The megapascal is a unit of measurement of pressure
Megapascal (MPa) is a metric (SI) measurement unit of pressure
• What is pressureInstant conversionsConversion tables
• 1 MPa = 9.86923267 atm
• 1 MPa = 7 500.61683 Torr
• 1 MPa = 7 500 616.83 mTorr
• 1 MPa = 7 500.61683 mmHg
• 1 MPa = 101 971.636 mmH2O
• 1 MPa = 295.299875 inHg
• 1 MPa = 4 014.63133 inH2O
• 1 MPa = 10 000 mbar
• 1 MPa = 10 b
• 1 MPa = 1 000 000 Pa
• 1 MPa = 10 000 hPa
• 1 MPa = 1 000 kPa
• 1 MPa = 1.0×10-18 N/pm²
• 1 MPa = 1.0×10-14 N/Ų
• 1 MPa = 1.0×10-12 N/nm²
• 1 MPa = 1.0×10-6 N/µ²
• 1 MPa = 1.0×10-6 N/µm²
• 1 MPa = 1 N/mm²
• 1 MPa = 100 N/cm²
• 1 MPa = 10 000 N/dm²
• 1 MPa = 1 000 000 N/m²
• 1 MPa = 100 000 000 N/dam²
• 1 MPa = 1 000 000 000 000 N/km²
• 1 MPa = 145.037738 psi
• 1 MPa = 20 885.4343 psf
• 1 MPa = 2.2480894351699×10-19 lbf/pm²
• 1 MPa = 2.2480894351699×10-15 lbf/Ų
• 1 MPa = 2.2480894351699×10-13 lbf/nm²
• 1 MPa = 2.248089435×10-7 lbf/µ²
• 1 MPa = 2.248089435×10-7 lbf/µm²
• 1 MPa = 0.2248089435 lbf/mm²
• 1 MPa = 22.4808944 lbf/cm²
• 1 MPa = 2 248.08944 lbf/dm²
• 1 MPa = 224 808.944 lbf/m²
• 1 MPa = 22 480 894.4 lbf/dam²
• 1 MPa = 224 808 943 517 lbf/km²
• 1 MPa = 22 480 894.4 lbf/a
• 1 MPa = 2 248 089 435 lbf/ha
• 1 MPa = 2 248 089 435 lbf/hm²
• 1 MPa = 771 073 094 648 lbf/nmi²
• 1 MPa = 645.16 N/in²
• 1 MPa = 92 903.04 N/ft²
• 1 MPa = 1.0×10-13 dyn/pm²
• 1 MPa = 1.0×10-9 dyn/Ų
• 1 MPa = 1.0×10-7 dyn/nm²
• 1 MPa = 0.1 dyn/µ²
• 1 MPa = 0.1 dyn/µm²
• 1 MPa = 100 000 dyn/mm²
• 1 MPa = 10 000 000 dyn/cm²
• 1 MPa = 10 000 000 Ba
• 1 MPa = 1 000 000 000 dyn/dm²
• 1 MPa = 100 000 000 000 dyn/m²
• 1 MPa = 10 000 000 000 000 dyn/dam²
• 1 MPa = 1.0×10+17 dyn/km²
• 1 MPa = 10 000 000 000 000 dyn/a
• 1 MPa = 1.0×10+15 dyn/ha
• 1 MPa = 1.0×10+15 dyn/hm²
• 1 MPa = 3.429904×10+17 dyn/nmi²
• 1 MPa = 6.4516×10-5 dyn/µin²
• 1 MPa = 64.516 dyn/mil;²
• 1 MPa = 64.516 dyn/thou²
• 1 MPa = 64 516 000 dyn/in²
• 1 MPa = 9 290 304 000 dyn/ft²
• 1 MPa = 4.0468564224×10+14 dyn/ac
• 1 MPa = 2.589988110336×10+17 dyn/mi²
• 1 MPa = 83 612 736 000 dyn/yd²
• 1 MPa = 334 450 944 000 dyn/ftm²
• 1 MPa = 40 468 564 200 000 dyn/ch²
• 1 MPa = 4.04685642×10+15 dyn/fur²
• 1 MPa = 2.23795229153×10+33 dyn/au²
• 1 MPa = 8.95015976147×10+42 dyn/ly²
• 1 MPa = 9.52142254852×10+43 dyn/pc²
• 1 MPa = 750.061683 cmHg
#### Foods, Nutrients and Calories
PREMIUM ICE CREAM, UPC: 041497012367 weigh(s) 146 grams per metric cup or 4.9 ounces per US cup, and contain(s) 290 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
5045 foods that contain Methionine. List of these foods starting with the highest contents of Methionine and the lowest contents of Methionine
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Gemstone Creek weighs 1 537.77 kg/m³ (95.99984 lb/ft³) with specific gravity of 1.53777 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Strontium monoxide [SrO or OSr] weighs 4 700 kg/m³ (293.41141 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Grapeseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A micrometer per minute squared (µm/min²) is a derived metric SI (System International) measurement unit of acceleration
An angle, in geometry, is defined by two rays a and b sharing a common starting point S, called the vertex. These rays can be transformed into each other by a revolution or rotation.
t/Ų to gr/pm² conversion table, t/Ų to gr/pm² unit converter or convert between all units of surface density measurement.
#### Calculators
Car fuel cost calculator per hour, day, week, month or a year | 1,771 | 3,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.119764 |
https://en.wikipedia.org/wiki/AN_codes | 1,505,893,499,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00316.warc.gz | 656,602,705 | 21,654 | # AN codes
AN codes[1] are error-correcting code that are used in arithmetic applications. Arithmetic codes were commonly used in computer processors to ensure the accuracy of its arithmetic operations when electronics were more unreliable. Arithmetic codes help the processor to detect when an error is made and correct it. Without these codes, processors would be unreliable since any errors would go undetected. AN codes are arithmetic codes that are named for the integers ${\displaystyle A}$ and ${\displaystyle N}$ that are used to encode and decode the codewords.
These codes differ from most other codes in that they use arithmetic weight to maximize the arithmetic distance between codewords as opposed to the hamming weight and hamming distance. The arithmetic distance between two words is a measure of the number of errors made while computing an arithmetic operation. Using the arithmetic distance is necessary since one error in an arithmetic operation can cause a large hamming distance between the received answer and the correct answer.
## Arithmetic Weight and Distance
The arithmetic weight of an integer ${\displaystyle x}$ in base ${\displaystyle r}$ is defined by
${\displaystyle w(x)=\min\{t|x=\sum _{i=1}^{t}a_{i}r^{n(i)}\}}$[citation needed]
where ${\displaystyle |{a_{i}}|}$< ${\displaystyle r}$, ${\displaystyle n(i)\geq 0}$, and ${\displaystyle r,n(i)\in \mathbb {Z} }$. The arithmetic distance of a word is upper bounded by its hamming weight since any integer can be represented by its standard polynomial form of ${\displaystyle x=\sum _{i=1}^{n}b_{i}r^{i}}$ where the ${\displaystyle b_{i}}$ are the digits in the integer. Removing all the terms where ${\displaystyle b_{i}=0}$ will simulate a ${\displaystyle t}$ equal to its hamming weight. The arithmetic weight will usually be less than the hamming weight since the ${\displaystyle a_{i}}$ are allowed to be negative. For example, the integer ${\displaystyle x=29}$ which is ${\displaystyle 11101}$ in binary has a hamming weight of ${\displaystyle 4}$. This is a quick upper bound on the arithmetic weight since ${\displaystyle x=2^{0}+2^{2}+2^{3}+2^{4}}$. However, since the ${\displaystyle a_{i}}$ can be negative, we can write ${\displaystyle x=2^{5}-2^{1}-2^{0}}$ which makes the arithmetic weight equal to ${\displaystyle 3}$.
The arithmetic distance between two integers is defined by
${\displaystyle d(x,y)=w(x-y)}$[citation needed]
This is one of the primary metrics used when analyzing arithmetic codes.[citation needed]
## AN Codes
AN codes are defined by integers ${\displaystyle A}$ and ${\displaystyle B}$ and are used to encode integers from ${\displaystyle 0}$ to ${\displaystyle B-1}$ such that
${\displaystyle C=\{AN|N\in \mathbb {Z} ,0\leq N}$<${\displaystyle B\}}$
Each choice of ${\displaystyle A}$ will result in a different code, while ${\displaystyle B}$ serves as a limiting factor to ensure useful properties in the distance of the code. If ${\displaystyle B}$ is too large, it could let a codeword with a very small arithmetic weight into the code which will degrade the distance of the entire code. To utilize these codes, before an arithmetic operation is performed on two integers, each integer is multiplied by ${\displaystyle A}$. Let the result of the operation on the codewords be ${\displaystyle R}$. Note that ${\displaystyle R}$ must also be between ${\displaystyle 0}$ to ${\displaystyle B-1}$ for proper decoding. To decode, simply divide ${\displaystyle R/A}$. If ${\displaystyle A}$ is not a factor of ${\displaystyle R}$, then at least one error has occurred and the most likely solution will be the codeword with the least arithmetic distance from ${\displaystyle R}$. As with codes using hamming distance, AN codes can correct up to ${\displaystyle \lfloor {\frac {d-1}{2}}\rfloor }$ errors where ${\displaystyle d}$ is the distance of the code.
For example, an AN code with ${\displaystyle A=3}$, the operation of adding ${\displaystyle 15}$ and ${\displaystyle 16}$ will start by encoding both operands. This results in the operation ${\displaystyle R=45+48=93}$. Then, to find the solution we divide ${\displaystyle 93/3=31}$. As long as ${\displaystyle B}$>${\displaystyle 31}$, this will be a possible operation under the code. Suppose an error occurs in each of the binary representation of the operands such that ${\displaystyle 45=101101\rightarrow 101111}$ and ${\displaystyle 48=110000\rightarrow 110001}$, then ${\displaystyle R=101111+110001=1100000}$. Notice that since ${\displaystyle 93=1011101}$, the hamming weight between the received word and the correct solution is ${\displaystyle 5}$ after just ${\displaystyle 2}$ errors. To compute the arithmetic weight, we take ${\displaystyle 1100000-1011101=11}$ which can be represented as ${\displaystyle 11=2^{0}+2^{1}}$ or ${\displaystyle 11=2^{2}-2^{0}}$. In either case, the arithmetic distance is ${\displaystyle 2}$ as expected since this is the number of errors that were made. To correct this error, an algorithm would be used to compute the nearest codeword to the received word in terms of arithmetic distance. We will not describe the algorithms in detail.
To ensure that the distance of the code will not be too small, we will define modular AN codes. A modular AN code ${\displaystyle C}$ is a subgroup of ${\displaystyle \mathbb {Z} /m\mathbb {Z} }$, where ${\displaystyle m=AB}$. The codes are measured in terms of modular distance which is defined in terms of a graph with vertices being the elements of ${\displaystyle \mathbb {Z} /m\mathbb {Z} }$. Two vertices ${\displaystyle x{\pmod {m}}}$ and ${\displaystyle x'{\pmod {m}}}$ are connected iff
${\displaystyle x-x'\equiv \pm c\cdot r^{j}{\pmod {m}}}$
where ${\displaystyle c,j\in \mathbb {Z} }$ and ${\displaystyle 0}$<${\displaystyle c}$<${\displaystyle r}$, ${\displaystyle j\geq 0}$. Then the modular distance between two words is the length of the shortest path between their nodes in the graph. The modular weight of a word is its distance from ${\displaystyle 0}$ which is equal to
${\displaystyle w_{m}(x)=min\{w(y)|y\in \mathbb {Z} ,y\equiv x{\pmod {m}}\}}$
In practice, the value of ${\displaystyle m}$ is typically chosen such that ${\displaystyle m=r^{n}-1}$ since most computer arithmetic is computed ${\displaystyle \mod 2^{n}-1}$ so there is no additional loss of data due to the code going out of bounds since the computer will also be out of bounds. Choosing ${\displaystyle m=r^{n}-1}$ also tends to result in codes with larger distances than other codes.
By using modular weight with ${\displaystyle m=r^{n}-1}$, the AN codes will be cyclic code.
definition: A cyclic AN code is a code ${\displaystyle C}$ that is a subgroup of ${\displaystyle [r^{n}-1]}$, where ${\displaystyle [r^{n}-1]=\{0,1,2,\dots ,r^{n}-1\}}$.
A cyclic AN code is a principal ideal of the ring ${\displaystyle [r^{n}-1]}$. There are integers ${\displaystyle A}$ and ${\displaystyle B}$ where ${\displaystyle AB=r^{n}-1}$ and ${\displaystyle A,B}$ satisfy the definition of an AN code. Cyclic AN codes are a subset of cyclic codes and have the same properties.
## Mandelbaum-Barrows Codes
The Mandelbaum-Barrows Codes are a type of cyclic AN codes introduced by D. Mandelbaum and J. T. Barrows.[2][3] These codes are created by choosing ${\displaystyle B}$ to be a prime number that does not divide ${\displaystyle r}$ such that ${\displaystyle \mathbb {Z} /B\mathbb {Z} }$ is generated by ${\displaystyle r}$ and ${\displaystyle -1}$, and ${\displaystyle m=r^{n}-1}$. Let ${\displaystyle n}$ be a positive integer where ${\displaystyle r^{n}\equiv 1{\pmod {B}}}$ and ${\displaystyle A=(r^{n}-1)/B}$. For example, choosing ${\displaystyle r=2,B=5,n=4}$, and ${\displaystyle A=(r^{n}-1)/B=3}$ the result will be a Mandelbaum-Barrows Code such that ${\displaystyle C=\{3N|N\in \mathbb {Z} ,0\leq N}$<${\displaystyle 5\}}$ in base ${\displaystyle 2}$.
To analyze the distance of the Mandelbaum-Barrows Codes, we will need the following theorem.
theorem: Let ${\displaystyle C\subset [r^{n}-1]}$ be a cyclic AN code with generator ${\displaystyle A}$, and
${\displaystyle B=|C|=(r^{n}-1)/A}$
Then,
${\displaystyle \sum _{x\in C}w_{m}(x)=n(\lfloor {\frac {rB}{r+1}}\rfloor -\lfloor {\frac {B}{r+1}}\rfloor )}$
proof: Assume that each ${\displaystyle x\in C}$ has a unique cyclic NAF[4] representation which is
${\displaystyle x\equiv \sum _{i=0}^{n-1}c_{i,x}r^{i}{\pmod {r^{n}-1}}}$
We define an ${\displaystyle n\times B}$ matrix with elements ${\displaystyle c_{i,x}}$ where ${\displaystyle 0\leq i\leq n-1}$ and ${\displaystyle x\in C}$. This matrix is essentially a list of all the codewords in ${\displaystyle C}$ where each column is a codeword. Since ${\displaystyle C}$ is cyclic, each column of the matrix has the same number of zeros. We must now calculate ${\displaystyle n|\{x\in C|c_{n-1,x}\neq 0\}|}$, which is ${\displaystyle n}$ times the number of codewords that don't end with a ${\displaystyle 0}$. As a property of being in cyclic NAF, ${\displaystyle c_{n-1,x}\neq 0}$ iff there is a ${\displaystyle y\in \mathbb {Z} }$ with ${\displaystyle y\equiv x{\pmod {r^{n}-1}},{\frac {m}{r+1}}}$<${\displaystyle y\leq {\frac {mr}{r+1}}}$. Since ${\displaystyle x=AN{\pmod {r^{n}-1}}}$ with ${\displaystyle 0\leq N}$<${\displaystyle B}$, then ${\displaystyle {\frac {B}{r+1}}}$<${\displaystyle N\leq {\frac {Br}{r+1}}}$. Then the number of integers that have a zero as their last bit are ${\displaystyle \lfloor {\frac {rB}{r+1}}\rfloor -\lfloor {\frac {B}{r+1}}\rfloor }$. Multiplying this by the ${\displaystyle n}$ characters in the codewords gives us a sum of the weights of the codewords of ${\displaystyle n(\lfloor {\frac {rB}{r+1}}\rfloor -\lfloor {\frac {B}{r+1}}\rfloor )}$ as desired.
We will now use the previous theorem to show that the Mandelbaum-Barrows Codes are equidistant(which means that every pair of codewords have the same distance), with a distance of
${\displaystyle {\frac {n}{B-1}}(\lfloor {\frac {rB}{r+1}}\rfloor -\lfloor {\frac {B}{r+1}}\rfloor )}$
proof: Let ${\displaystyle x\in C,x\neq 0}$, then ${\displaystyle x=AN{\pmod {r^{n}-1}}}$ and ${\displaystyle N}$ is not divisible by ${\displaystyle B}$. This implies there ${\displaystyle \exists j(N\equiv \pm r^{j}{\pmod {B}})}$. Then ${\displaystyle w_{m}(x)=w_{m}(\pm r^{j}A)=w_{m}(A)}$. This proves that ${\displaystyle C}$ is equidistant since all codewords have the same weight as ${\displaystyle A}$. Since all codewords have the same weight, and by the previous theorem we know the total weight of all codewords, the distance of the code is found by dividing the total weight by the number of codewords(excluding 0). | 3,076 | 10,699 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 136, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-39 | latest | en | 0.921613 |
https://www.coursehero.com/file/p4e6j4n/If-inputs-increase-by-30-and-outputs-decrease-by-15-what-is-the-percentage/ | 1,544,562,776,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00332.warc.gz | 841,748,485 | 68,696 | # If inputs increase by 30 and outputs decrease by 15
• 72
56. If inputs increase by 30% and outputs decrease by 15%, what is the percentage change in productivity? Productivity = Output/Input = (1.0 - 0.15)/(1.0 + 0.3) = 0.6538, therefore productivity decreases by 1.0 - 0.6538 = 0.3462 or 34.62%. Chapter - Chapter 06 #56 Difficulty: Hard Learning Objective: 06-05 Classify and calculate the different measures of process performance.
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http://www.bookrags.com/ebooks/17021/89.html | 1,527,018,915,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864872.17/warc/CC-MAIN-20180522190147-20180522210147-00152.warc.gz | 356,810,272 | 13,036 | Watch and Clock Escapements eBook
This eBook from the Gutenberg Project consists of approximately 236 pages of information about Watch and Clock Escapements.
In making this drawing we proceed as with Fig. 132 by establishing a center for our radius of 10” outside of our drawing paper and drawing the line A A to such center and sweeping the arcs a b c. We establish the point e, which represents the center of our cylinder, as before. We take the space to represent the radial extent of the outside of our cylinder in our dividers and from e as a center sweep a fine pencil line, represented by the dotted line t in our drawing; and where this circle intersects the arc a we name it the point s; and it is at this point the heel of our escape-wheel tooth must part with the exit lip of the cylinder. From e as a center and through the point s we draw the line e l’’. With our dividers set to the radius of any convenient arc which we have divided into degrees, we sweep the short arc d’. The intersection of this arc with the line e l’’ we name the point u; and from e as a center we draw the radial line e u f’. We place the letter f’’ in connection with this line because it (the line) bears the same relations to the half shell of the cylinder shown in Fig. 133 that the line f does to the half shell (D) shown in Fig. 132. We draw the line f’’ f’’’, Fig. 133, which divides the cylinder into two segments of 180 degrees each. We take the same space in our dividers with which we swept the interior of the cylinder in Fig. 132 and sweep the circle v, Fig. 133. From e as a center we sweep the short arc d’’, Fig. 133, and from its intersection of the line f’’ we lay off six degrees on said arc d’’ and draw the line e’ k’’, which defines the angular extent of our entrance lip to the half shell of the cylinder in Fig. 133. We draw the full lines of the cylinder as shown.
We next delineate the heel of the tooth which has just passed out of the cylinder, as shown at D’, Fig. 133. We now have a drawing showing the position of the half shell of the cylinder just as the tooth has passed the exit lip. This drawing also represents the position of the half shell of the cylinder when the tooth rests against it on the outside. If we should make a drawing of an escape-wheel tooth shaped exactly as the one shown at Fig. 132 and the point of the tooth resting at x, we would show the position of a tooth encountering the cylinder after a tooth which has been engaged in the inside of the shell has passed out. By following the instructions now given, we can delineate a tooth in any of its relations with the cylinder shell.
DELINEATING AN ESCAPE-WHEEL TOOTH WHILE IN ACTION.
More summaries and resources for teaching or studying Watch and Clock Escapements. | 644 | 2,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-22 | latest | en | 0.910418 |
http://statistics.ats.ucla.edu/stat/stata/examples/mm/mmstatasim4.htm | 1,487,947,745,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171620.77/warc/CC-MAIN-20170219104611-00507-ip-10-171-10-108.ec2.internal.warc.gz | 250,821,856 | 6,611 | ### Simulations and Demonstrations for Introduction to the Practice of Statistics Chapter 4
The heads program can be used to produce results simulating those shown in Figure 4.1, as shown below. First start the heads program by typing
heads
You can then change the number of trials to 1000. When we ran this, our simulated 1000 coin flips looked like the graph below. Every time you run the heads program, it generates a new set of 1000 flips, so the results come out different every time. However if you do a large number of coin flips, you will notice that in the long run the proportion of heads always converges on .5. Try this on your own and see for yourself.
Go ahead and click on quit.
The heads program can be used to simulate the results in Example 4.2. We now type heads, save to save the coin tosses in a Stata file. In the first example we simulate the 4040 coin tosses by Buffon (who got 2048 heads). Here is the graph we get.
We then click quit. Then we use the tabulate command to see that our simulation, we got 2051 heads, or a proportion of .492. Every time we do this, we get a different proportion of heads. Try this for yourself and see how many heads you get.
tabulate heads
of 1 tossed | Freq. Percent Cum.
------------+-----------------------------------
0 | 2051 50.77 50.77
1 | 1989 49.23 100.00
------------+-----------------------------------
Total | 4040 100.00
Variable | Obs Mean Std. Dev. Min Max
---------+-----------------------------------------------------
heads | 4040 .4923267 .500003 0 1
Next, we simulate the 24,000 coin tosses by Karl Pearson, and we got 12058 heads (compared to the 12,012 heads that Pearson got).
heads , save
tabulate heads
# heads out |
of 1 tossed | Freq. Percent Cum.
------------+-----------------------------------
0 | 11,942 49.76 49.76
1 | 12,058 50.24 100.00
------------+-----------------------------------
Total | 24,000 100.00
summ heads
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
heads | 24000 .5024167 .5000046 0 1
Next, we simulate the 10,000 coin tosses by John Kerrich, and we got 4972 heads (compared to the 5067 heads that Kerrich got).
heads2 1000, save
of 1 tossed | Freq. Percent Cum.
------------+-----------------------------------
0 | 5028 50.28 50.28
1 | 4972 49.72 100.00
------------+-----------------------------------
Total | 10000 100.00
Variable | Obs Mean Std. Dev. Min Max
---------+-----------------------------------------------------
heads | 10000 .4972 .5000172 0 1
We can use the heads program to simulate the results shown in Figure 4.8 and Example 4.20, which shows the probability histogram for 4 tosses of a coin. We can do 1000 coin tosses and graph the probabilities and compare those results to Figure 4.8, as illustrated below. When you try this, you will probably get similar (but not exactly the same) results.
Note that 1000 is the number of tosses, and .5 is the probability of a head, and 4 is the number of coins tossed. We use the save option to save the results.
heads , save
We then quit and graph the results below.
histogram heads, discrete
We can show the probability distribution using the tabulate command and compare this to example 4.20. As you can see, the probabilities we found (from the Percent column) are quite similar to the expected percentages shown in example 4.20.
tab heads
# heads out |
of 4 tossed | Freq. Percent Cum.
------------+-----------------------------------
0 | 54 5.40 5.40
1 | 235 23.50 28.90
2 | 397 39.70 68.60
3 | 244 24.40 93.00
4 | 70 7.00 100.00
------------+-----------------------------------
Total | 1,000 100.00
We can get the mean of the distribution with the summarize command. The expected mean is 2.0, and ours is very close and 2.041.
summarize heads
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
heads | 1000 2.041 .987063 0 4
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https://www.convert-measurement-units.com/convert+Cubic+meter+to+Minim+US.php | 1,721,891,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00383.warc.gz | 630,662,320 | 14,149 | Convert m³ to Minim US (Cubic meter to Minim US)
## Cubic meter into Minim US
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Cubic+meter+to+Minim+US.php
# Convert m³ to Minim US (Cubic meter to Minim US):
1. Choose the right category from the selection list, in this case 'Volume'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Cubic meter [m³]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Minim US'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '659 Cubic meter'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Cubic meter' or 'm3'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Volume'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '92 m3 to Minim US' or '72 m3 into Minim US' or '88 Cubic meter -> Minim US' or '84 m3 = Minim US' or '80 Cubic meter to Minim US' or '68 Cubic meter into Minim US'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(48 * 44) m3'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '56 Cubic meter + 52 Minim US' or '40mm x 36cm x 32dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 3.643 733 300 175 4×1021. For this form of presentation, the number will be segmented into an exponent, here 21, and the actual number, here 3.643 733 300 175 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 3.643 733 300 175 4E+21. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 3 643 733 300 175 400 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 894 | 3,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.849514 |
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Question From class 12 Chapter JEE MAINS
A line intersect co-ordinate axis at points and . A circle is drawn passing through origin and point & . If perpendicular from point and are drawn on tangent to the circle at origin then sum of perpendicular distance is (A) (B) (C) (D)
A parabola crosses the x-axis at A and B. A variable circle is drawn passing through A and B. The length of a tangent from the origin to the circle is
3:29
The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A,B and the origin. Then, the sum of perpendicular distances from A and B on the tangent to the circle at the origin is
The locus of the foot of the perpendicular drawn from origin to a variable line passing through fixed point (2,3) is a circle whose diameter is
3:47
Theorem:A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
4:52
Theorem:A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
4:52
Theorem:A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
4:52
Theorem:A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
4:52
A circle is drawn through the point of intersection of the parabola and the x-axis such that origin lies outside it. The length of a tangent to the circle from the origin is ________ .
A circle is drawn through the point of intersection of the parabola and the x-axis such that origin lies outside it. The length of a tangent to the circle from the origin is ________ .
A tangent PT is drawn to the circle at the point . A straight line L is perpendicular to PT is a tangent to the circle Common tangent of two circle is: (A) (B) (C) (D)
4:09
CIRCLES | SECANT AND TANGENT, SOME PROPERTIES OF TANGENT TO A CIRCLE | Definition of circle, Theorem: A tangent to a circle is perpendicular to the radius through the point of contact., Theorem:A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
13:59
A line is drawn from a point on the curve making an angle with the x-axis which is supplementary to the one made by the tangent to the curve at The line meets the x-axis at A. Another line perpendicular to it drawn from meeting the y-axis at B. If where is the origin, theequation of all curves which pass through ( is
7:12
A line is drawn through the point to cut the circle at A and B. Then is equal to
1:20
If and is real, then the point represented by the complex number z lies (1) either on the real axis or on a circle passing through the origin (2) on a circle with centre at the origin (3) either on the real axis or on a circle not passing through the origin (4) on the imaginary axis
5:31
The line cuts the parabola in the points A and B. The normals drawn to the parabola at A and B intersect at G. A line passing through G intersects the parabola at right angles at the point C, and the tangents at A and B intersect at point T.
8:18
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- Homework 1 (http://book.caltech.edu/bookforum/forumdisplay.php?f=130)
ArikB 01-12-2013 03:39 PM
Isn't the bin (your data set) the sample?
This has me a bit confused, isn't the bin your data set in the analogy? And as such your data set is the sample of the population. For instance in the bank example, your data set would be the sample and the population would be all of the possible people applying for credit.
If that is the case then how is Hoeffding representative for anything that is "really" out of sample?
Or am I confused and the bin is really the population? Hence mu is then the population fraction and the samples you pick from the bin represent the data set?
Perhaps I should rephrase it to be a bit more systematic:
In the best case scenario my bin is completely green, i.e my hypothesis agrees entirely with my data set. So mu is then 1. Hoeffding gives me a probabilistic bound on how well nu approximates this mu (which is 1). That's nice, but now I only know that I have a hypothesis that agrees entirely with my dataset. But how does this generalize beyond my data set? Or am I going too far ahead and the lecture is not about this? If so, then why use nu at all? If this is a case of supervised learning and I know the output, then I can just see mu immediately because I can see whether or not my input agrees with the output within my data set.
Or is it so that the bin represents my training set and I have already (supposedly) subdivided my data set into a training set and data that I use for testing?
butterscotch 01-12-2013 07:08 PM
Re: Isn't the bin (your data set) the sample?
mu denotes probability of green in the entire space, outside of D included.
In a marble in a bag example, bin is the entire space and the N marbles you picked are your data set. i.e. you do not know the colors of the rest of the marbles in the bin.
Consider the following example. There are 10000 marbles in the bag and you want to know the proportion of red and black marbles (proportion of red: mu). You can figure out the exact ratio by taking out all the marbles and counting all of them. But say you have to figure this out in a limited time, and can only afford to look at 100 marbles. You counted 30 red marbles, and 70 black marbles. Then v is 0.3. You do not know if marbles outside of your dataset agrees with it. But Hoeffding Inequality provides a bound for the probability for values of mu based on v.
ArikB 01-14-2013 01:01 PM
Re: Isn't the bin (your data set) the sample?
Quote:
Originally Posted by butterscotch (Post 8605) mu denotes probability of green in the entire space, outside of D included. In a marble in a bag example, bin is the entire space and the N marbles you picked are your data set. i.e. you do not know the colors of the rest of the marbles in the bin. Consider the following example. There are 10000 marbles in the bag and you want to know the proportion of red and black marbles (proportion of red: mu). You can figure out the exact ratio by taking out all the marbles and counting all of them. But say you have to figure this out in a limited time, and can only afford to look at 100 marbles. You counted 30 red marbles, and 70 black marbles. Then v is 0.3. You do not know if marbles outside of your dataset agrees with it. But Hoeffding Inequality provides a bound for the probability for values of mu based on v.
Thanks.
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# Georg Simon Ohm (1787-1854)
Georg Simon Ohm was a German physicist born in Erlangen, Bavaria, on March 16, 1787. As a high school teacher, Ohm started his research with the recently invented electrochemical cell, invented by Italian Count Alessandro Volta. Using equipment of his own creation, Ohm determined that the current that flows through a wire is proportional to its cross sectional area and inversely proportional to its length or Ohm's law.
Using the results of his experiments, Georg Simon Ohm was able to define the fundamental relationship between voltage, current, and resistance. These fundamental relationships are of such great importance, that they represent the true beginning of electrical circuit analysis.
Unfortunately, when Ohm published his finding in 1827, his ideas were dismissed by his colleagues. Ohm was forced to resign from his high-school teaching position and he lived in poverty and shame until he accepted a position at Nüremberg in 1833 and although this gave him the title of professor, it was still not the university post for which he had strived all his life.
Ohm’s main interest was current electricity, which had recently been advanced by Alessandro Volta’s invention of the battery. Ohm made only a modest living and as a result his experimental equipment was primitive. Despite this, he made his own metal wire, producing a range of thickness and lengths of remarkable consistent quality. The nine years he spent at the Jesuit’s college, he did considerable experimental research on the nature of electric circuits. He took considerable pains to be brutally accurate with every detail of his work. In 1827, he was able to show from his experiments that there was a simple relationship between resistance, current and voltage.
Ohm’s law stated that the amount of steady current through a material is directly proportional to the voltage across the material, for some fixed temperature:
### I = V/R
Ohm had discovered the distribution of electromotive force in an electrical circuit, and had established a definite relationship connecting resistance, electromotive force and current strength.
### Ohm and corrosion monitoring
Ohm was afraid that the purely experimental basis of his work would undermine the importance of his discovery. He tried to state his law theoretically but his rambling mathematically proofs made him an object of ridicule. In the years that followed, Ohm lived in poverty, tutoring privately in Berlin. He would receive no credit for his findings until he was made director of the Polytechnic School of Nüremberg in 1833. In 1841, the Royal Society in London recognized the significance of his discovery and awarded him the Copley medal. The following year, they admitted him as a member. In 1849, just 5 years before his death, Ohm’s lifelong dream was realized when he was given a professorship of Experimental Physics at the University of Munich. On July 7th,1854 he passed away in Munich, at the age of 65.
This belated recognition was welcome but there remains the question of why someone who today is a household name for his important contribution struggled for so long to gain acknowledgement. This may have no simple explanation but rather be the result of a number of different contributory factors. One factor may have been the inwardness of Ohm's character while another was certainly his mathematical approach to topics which at that time were studied in his country a non-mathematical way. There was undoubtedly also personal disputes with the men in power which did Ohm no good at all. He certainly did not find favor with Johannes Schultz who was an influential figure in the ministry of education in Berlin, and with Georg Friedrich Pohl, a professor of physics in that city.
Electricity was not the only topic on which Ohm undertook research, and not the only topic in which he ended up in controversy. In 1843 he stated the fundamental principle of physiological acoustics, concerned with the way in which one hears combination tones. However the assumptions which he made in his mathematical derivation were not totally justified and this resulted in a bitter dispute with the physicist August Seebeck. He succeeded in discrediting Ohm's hypothesis and Ohm had to acknowledge his error. | 868 | 4,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | longest | en | 0.989375 |
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«
»
This set of Biomedical Instrumentation Multiple Choice Questions & Answers (MCQs) focuses on “Pulse Oximeter”.
1. _____ is non-invasive method allowing the monitoring of the saturation of a patient’s hemoglobin.
a) Ear Oximetry
b) Pulse Oximetry
c) Skin-Reflectance Oximetry
d) Intravascular Oximetry
Explanation: Pulse oximetry is a non-invasive method allowing the monitoring of the saturation of a patient’s hemoglobin. A pulse oximeter shows the percentage of arterial hemoglobin in the oxyhemoglobin (HbO2) and hemoglobin (Hb). These observations:
A = -logT = log I0/I =ɛDC.
2. A = log I0/I,where I0 is_____
a) Incident light Intensity
b) Transmitted light Intensity
c) Extinction coefficient
d) Concentration
Explanation: A pulse oximeter shows the percentage of arterial hemoglobin in the oxyhemoglobin (HbO2) and hemoglobin (Hb). This observation A = -logT = log I0/I = ɛDC where Io and I are incident and transmitted light intensities, e is the extinction coefficient, D is the depth of the absorbing layer and C is concentration.
3. A = ɛDC, where ɛ is _______
a) Incident light Intensity
b) Transmitted light Intensity
c) Extinction coefficient
d) Concentration
Explanation: A pulse oximeter shows the percentage of arterial hemoglobin in the oxyhemoglobin (HbO2) and hemoglobin (Hb). This observation A = -logT = log I0/I = ɛDC where Io and I are incident and transmitted light intensities, e is the extinction coefficient, D is the depth of the absorbing layer and C is concentration.
4. A pulse oximeter shows the percentage of venous hemoglobin in the oxyhemoglobin (HbO2) and hemoglobin (Hb).
a) True
b) False
Explanation: False, pulse oximetry is a non-invasive method allowing the monitoring of the saturation of a patient’s hemoglobin. A pulse oximeter shows the percentage of arterial hemoglobin in the oxyhemoglobin (HbO2) and hemoglobin (Hb). These observations:
A = -logT = log I0/I =ɛDC.
5. What is used as a photodetector in pulse oximetry?
a) Phototransistor
b) Solar cell
c) Photodiode
d) Photographic plates
Explanation: Pulse oximetry is a particularly convenient noninvasive measurement method. Typically it utilizes a pair of small light-emitting diodes (LEDs) facing a photodiode through a translucent part of the patient’s body, usually a fingertip or an earlobe. One LED is red, with a wavelength of 660 nm, and the other is infrared, 905, 910, or 940 nm.
Participate in Biomedical Instrumentation Certification Contest of the Month Now!
6. What is the wavelength of the red LED in pulse oximetry?
a) 660 nm
b) 740 nm
c) 905 nm
d) 950 nm
Explanation: Pulse oximetry is a particularly convenient noninvasive measurement method. Typically it utilizes a pair of small light-emitting diodes (LEDs) facing a photodiode through a translucent part of the patient’s body, usually a fingertip or an earlobe. One LED is red, with a wavelength of 660 nm, and the other is infrared, 905, 910, or 940 nm.
7. What is the wavelength of infrared LED in pulse oximetry?
a) 660 nm
b) 740 nm
c) 905 nm
d) 950 nm
Explanation: Pulse oximetry is a particularly convenient noninvasive measurement method. Typically it utilizes a pair of small light-emitting diodes (LEDs) facing a photodiode through a translucent part of the patient’s body, usually a fingertip or an earlobe. One LED is red, with a wavelength of 660 nm, and the other is infrared, 905, 910, or 940 nm.
8. A pulse oximeter is useful in any setting where a patients oxygenation is unstable.
a) True
b) False
Explanation: True, pulse oximeter is useful in any setting where a patient’s oxygenation is unstable, including intensive care, operating, recovery, emergency and hospital ward settings, pilots in unpressurized aircraft, for assessment of any patient’s oxygenation, and determining the effectiveness of or need for supplemental oxygen.
9. Portable pulse oximeters are useful for ______ whose oxygen levels may decrease at high altitude.
a) Athlete
b) Swimmer
c) Mountain climber
d) Fisher
Explanation: Portable pulse oximeter is also useful for mountain climbers and athletes whose oxygen levels may decrease at high altitudes or with exercise. Some portable pulse oximeters employ software that charts a patient’s blood oxygen and pulse, serving as a reminder to check blood oxygen levels.
10. ______ is very useful for patients having respiratory or cardiac problems because of their simplicity of use and the ability to provide continuous and immediate oxygen saturation levels.
a) Pulse Oximeter
b) Ear Oximeter
c) Skin Reflactance Oximeter
d) Intravascular Oximeter
Explanation: Because of their simplicity of use and the ability to provide continuous and immediate oxygen saturation values, pulse oximeters are of critical importance in emergency medicine and are also very useful for patients with respiratory or cardiac problems, especially COPD, or for diagnosis of some sleep disorders such as apnea and hypopnea.
Sanfoundry Global Education & Learning Series – Biomedical Instrumentation.
To practice all areas of Biomedical Instrumentation, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,261 | 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} | 2.828125 | 3 | CC-MAIN-2023-06 | longest | en | 0.760673 |
https://forum.espruino.com/comments/3172/ | 1,713,641,713,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00157.warc.gz | 237,141,583 | 5,523 | # OneWire Lib
• Sure Gordon...it is possible to make a lib to handle both DS1820 and DS18B20!
The only difference (as @tickTock correctly sad) is the resolution:
• DS1820 has fixed 9bit resolution (0,5°C per bit);
• DS18B20 has a register (Byte 4 of the scratchpad) that permits to choose the resolution( 9, 10, 11, or 12 bits, corresponding to increments of 0.5°C, 0.25°C, 0.125°C, and 0.0625°C, respectively);
What i've done in the previous post, was to raise the fixed 9 bit resolution by using some additional register value in the scratchpad.
In a nutshell for:
• DS1820 9bit resolution : temp = (byte_0 + (byte_1<<8)) * 0,5;
• DS1820 higher resolution: temp = ( (byte_1<<8) + byte_0 - 1)/2-0,25+(byte_7-byte_6)/byte_6;
• DS18B20 : temp = (byte_0 + (byte_1<<8)) * RES (where RES=0.5, 0.25, 0.125 or 0.0625 depending from the resolution set)
Note: is it possible that formulas above doesn't work for temp below zero....in that case you have to play with bits (2 complements...and so on)
...finally reading the LSB of 64bit ROM you can understand if you are using either the DS1820 sensor (0x10) or DS18B20 (0x28) one. | 363 | 1,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.69194 |
https://www.pudn.com/detail/5875816 | 1,675,121,496,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00244.warc.gz | 937,886,717 | 36,160 | 应用最小二乘一次完成法和递推最小二乘法算法的系统辨识
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• 应用最小二乘一次完成法和递推最小二乘法算法的系统辨识.doc
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<html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta charset="utf-8"> <meta name="generator" content="pdf2htmlEX"> <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"> <link rel="stylesheet" href="https://static.pudn.com/base/css/base.min.css"> <link rel="stylesheet" href="https://static.pudn.com/base/css/fancy.min.css"> <link rel="stylesheet" href="https://static.pudn.com/prod/directory_preview_static/6275b94616f2c0769c2a4569/raw.css"> <script src="https://static.pudn.com/base/js/compatibility.min.js"></script> <script src="https://static.pudn.com/base/js/pdf2htmlEX.min.js"></script> <script> try{ pdf2htmlEX.defaultViewer = new pdf2htmlEX.Viewer({}); }catch(e){} </script> <title></title> </head> <body> <div id="sidebar" style="display: none"> <div id="outline"> </div> </div> <div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://static.pudn.com/prod/directory_preview_static/6275b94616f2c0769c2a4569/bg1.jpg"><div class="c x0 y1 w2 h2"><div class="t m0 x1 h3 y2 ff1 fs0 fc0 sc0 ls0 ws0">目录</div><div class="t m0 x2 h3 y3 ff2 fs0 fc0 sc0 ls0 ws0">1<span class="_ _0"> </span><span class="ff1">引言</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>...........<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.....2</div><div class="t m0 x3 h3 y4 ff2 fs0 fc0 sc0 ls0 ws0">1.1<span class="_ _0"> </span><span class="ff1">概述</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>................<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.....2</div><div class="t m0 x3 h3 y5 ff2 fs0 fc0 sc0 ls0 ws0">1.2<span class="_ _0"> </span><span class="ff1">辨识的基本步骤</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.<span class="_ _1"></span>2</div><div class="t m0 x2 h3 y6 ff2 fs0 fc0 sc0 ls0 ws0">2<span class="_ _0"> </span><span class="ff1">系统辨识输入信号的产生方法和理论依据</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>...............<span class="_ _2"></span>.<span class="_ _1"></span>3</div><div class="t m0 x3 h3 y7 ff2 fs0 fc0 sc0 ls0 ws0">2.1<span class="_ _0"> </span><span class="ff1">白噪声序列</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.....3</div><div class="t m0 x4 h3 y8 ff2 fs0 fc0 sc0 ls0 ws0">2.1.1<span class="_ _0"> </span><span class="ff1">白噪声序列的产生方法</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>............<span class="_ _2"></span>....3</div><div class="t m0 x3 h3 y9 ff2 fs0 fc0 sc0 ls0 ws0">2.2 M<span class="_ _0"> </span><span class="ff1">序列的产生</span>......<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>....<span class="_ _1"></span>4</div><div class="t m0 x4 h3 ya ff2 fs0 fc0 sc0 ls0 ws0">2.2..1 <span class="ff1">伪随机噪声</span>..........<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>..............<span class="_ _2"></span>...<span class="_ _1"></span>4</div><div class="t m0 x4 h3 yb ff2 fs0 fc0 sc0 ls0 ws0">2.2.2 M<span class="_ _0"> </span><span class="ff1">序列的产生方法</span>.......<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.................<span class="_ _2"></span>....<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>...4</div><div class="t m0 x2 h3 yc ff2 fs0 fc0 sc0 ls0 ws0">3<span class="_ _0"> </span><span class="ff1">应用经典辨识方法的辨识方案。</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>..............<span class="_ _2"></span>....<span class="_ _2"></span>.<span class="_ _1"></span>6</div><div class="t m0 x3 h3 yd ff2 fs0 fc0 sc0 ls0 ws0">3.1<span class="_ _0"> </span><span class="ff1">经典辨识方法概述</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>........<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.<span class="_ _1"></span>6</div><div class="t m0 x3 h3 ye ff2 fs0 fc0 sc0 ls0 ws0">3.2<span class="_ _0"> </span><span class="ff1">经典辨识方法的实现</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.<span class="_ _1"></span>6</div><div class="t m0 x2 h3 yf ff2 fs0 fc0 sc0 ls0 ws0">4<span class="_ _0"> </span><span class="ff1">最小二乘法的理论基础</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>..............<span class="_ _2"></span>....<span class="_ _2"></span>....<span class="_ _2"></span>.<span class="_ _1"></span>7</div><div class="t m0 x3 h3 y10 ff2 fs0 fc0 sc0 ls0 ws0">4.1<span class="_ _0"> </span><span class="ff1">最小二乘法</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.....7</div><div class="t m0 x4 h3 y11 ff2 fs0 fc0 sc0 ls0 ws0">4.1.1<span class="_ _0"> </span><span class="ff1">最小二乘法估计中的输入信号</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>................9</div><div class="t m0 x4 h3 y12 ff2 fs0 fc0 sc0 ls0 ws0">4.1.2<span class="_ _0"> </span><span class="ff1">最小二乘估计的概率性质</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>........<span class="_ _2"></span>....9</div><div class="t m0 x3 h3 y13 ff2 fs0 fc0 sc0 ls0 ws0">4.2<span class="_ _0"> </span><span class="ff1">递推最小二乘法</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>...........<span class="_ _2"></span>.....<span class="_ _2"></span>....10</div><div class="t m0 x2 h3 y14 ff2 fs0 fc0 sc0 ls0 ws0">5<span class="_ _0"> </span><span class="ff1">两种算法的实现方案</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>...........<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.1<span class="_ _3"></span>1</div><div class="t m0 x3 h3 y15 ff2 fs0 fc0 sc0 ls0 ws0">5.1<span class="_ _0"> </span><span class="ff1">最小二乘法一次完成算法实现</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>................<span class="_ _2"></span>.....<span class="_ _2"></span>.1<span class="_ _3"></span>1</div><div class="t m0 x4 h3 y16 ff2 fs0 fc0 sc0 ls0 ws0">5.1.1<span class="_ _0"> </span><span class="ff1">最小二乘一次完成算法程序框图</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>.1<span class="_ _3"></span>1</div><div class="t m0 x4 h3 y17 ff2 fs0 fc0 sc0 ls0 ws0">5.1.2<span class="_ _0"> </span><span class="ff1">一次完成法程序</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>................<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>.1<span class="_ _3"></span>1</div><div class="t m0 x4 h3 y18 ff2 fs0 fc0 sc0 ls0 ws0">5.1.3<span class="_ _0"> </span><span class="ff1">一次完成算法程序运行结果</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _2"></span>.....<span class="_ _2"></span>.1<span class="_ _3"></span>1</div><div class="t m0 x4 h3 y19 ff2 fs0 fc0 sc0 ls0 ws0">5.1.4<span class="_ _0"> </span><span class="ff1">辨识数据比较</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>........<span class="_ _2"></span>.....<span class="_ _2"></span>....12</div><div class="t m0 x4 h3 y1a ff2 fs0 fc0 sc0 ls0 ws0">5.1.5<span class="_ _0"> </span><span class="ff1">程序运行曲线</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>........<span class="_ _2"></span>.....<span class="_ _2"></span>....12</div><div class="t m0 x3 h3 y1b ff2 fs0 fc0 sc0 ls0 ws0">5.2<span class="_ _0"> </span><span class="ff1">递推最小二乘法的实现</span>.........<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>................<span class="_ _2"></span>.....12</div><div class="t m0 x4 h3 y1c ff2 fs0 fc0 sc0 ls0 ws0">5.2.1<span class="_ _0"> </span><span class="ff1">递推算法实现步骤</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>..................<span class="_ _2"></span>....12</div><div class="t m0 x4 h3 y1d ff2 fs0 fc0 sc0 ls0 ws0">5.2.2<span class="_ _0"> </span><span class="ff1">程序编制思路：</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....13</div><div class="t m0 x4 h3 y1e ff2 fs0 fc0 sc0 ls0 ws0">5.2.3<span class="_ _0"> </span><span class="ff1">递推最小二乘法程序框图</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>......<span class="_ _2"></span>....14</div><div class="t m0 x4 h3 y1f ff2 fs0 fc0 sc0 ls0 ws0">5.2.4<span class="_ _0"> </span><span class="ff1">程序运行曲线</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>........<span class="_ _2"></span>.....<span class="_ _2"></span>....15</div><div class="t m0 x4 h3 y20 ff2 fs0 fc0 sc0 ls0 ws0">5.2.5<span class="_ _0"> </span><span class="ff1">测试结果</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>................<span class="_ _2"></span>.....<span class="_ _2"></span>....16</div><div class="t m0 x4 h3 y21 ff2 fs0 fc0 sc0 ls0 ws0">5.2.6<span class="_ _0"> </span><span class="ff1">地退数据表</span>.....<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _2"></span>.....<span class="_ _2"></span>....17</div><div class="t m0 x2 h3 y22 ff2 fs0 fc0 sc0 ls0 ws0">6<span class="_ _0"> </span><span class="ff1">结论</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>..........<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>....17</div><div class="t m0 x2 h3 y23 ff2 fs0 fc0 sc0 ls0 ws0">7<span class="_ _0"> </span><span class="ff1">参考文献</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>...................<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....17</div><div class="t m0 x2 h3 y24 ff2 fs0 fc0 sc0 ls0 ws0">8<span class="_ _0"> </span><span class="ff1">附录</span>........<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>............<span class="_ _1"></span>.............<span class="_ _1"></span>.............<span class="_ _1"></span>..........<span class="_ _2"></span>....<span class="_ _2"></span>.....<span class="_ _2"></span>....<span class="_ _2"></span>....17</div><div class="t m0 x5 h4 y25 ff2 fs1 fc0 sc0 ls0 ws0">1</div></div><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d m1"></div></a><a class="l" rel='nofollow' onclick='return false;'><div class="d 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相关推荐 | 8,527 | 25,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.31217 |
http://cboard.cprogramming.com/cplusplus-programming/50294-sort-algorithms-mix.html | 1,466,872,061,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393442.26/warc/CC-MAIN-20160624154953-00058-ip-10-164-35-72.ec2.internal.warc.gz | 50,011,132 | 12,666 | 1. ## sort algorithms mix
I'm writing Sort class with implementation of sort algorithms quick sort and bubble_sort
I'm trying to mix sort algorithms to reduce number of recursion and still maintain performance.
My problem is that I don't know what criteria to choose.
Code:
```#include <iostream>
using namespace std;
template <class T>
void Show(T arr[],int len,char *s)
{
cout<< s;
for(int i=0;i<len;i++)
cout<<arr[i]<<" ";
cout<<endl;
}
int bubble=0,quick=0;
template <class T>
class Sort
{
T *array;
int length;
public:
Sort(int siz):length(siz)
{
array=new T[length];
}
~Sort(){delete [] array;array=NULL;}
void bubble_sort(T []);
void quick_sort(T[],int,int,bool=true);
void swap(T&,T&);
void show_list(const char *);
};
template <class T>
{
for(int i=0;i<length;i++)
{
array[i]=list[i];
}
}
template <class T>
void Sort<T>::bubble_sort(T list[])
{
int flag_swap=1;bubble++;
Show(array,length,"U toku Bubble!");
for(int i=0;i<(length-1) && flag_swap;i++)
{
flag_swap=0;
for(int j=0;j<(length-(i+1));j++)
if(array[j]>array[j+1])
{
swap(array[j],array[j+1]);
flag_swap=1;
}
}
}
template <class T>
void Sort<T>::quick_sort(T list[],int first,int last,bool t)
{
if(t)
{
t=false;
}
if(first<last-5)//this is supposed to reduce recursion
//if one of the two subarrays
//have less then 6 elements
//probably there is clever solution to this
{
int pivot=array[first];
int i=first;
int j=last;quick++;
Show(array,length,"U toku quick");
while(i<j)
{
while(array[i]<=pivot && i<last)
i++;
while(array[j]>=pivot && j>first)
j--;
if(i<j)
swap(array[i],array[j]);
}
swap(array[j],array[first]);
quick_sort(list,first,j-1,t);
quick_sort(list,j+1,last,t);
}
else
bubble_sort(array);
}
template <class T>
void Sort<T>::show_list(const char *s)
{
cout<<s<<endl;
for(int i=0;i<length;i++)
cout<<array[i]<<" ";
cout<<endl;
}
template <class T>
void Sort<T>::swap(T& x,T& y)
{
T tmp;
tmp=x;
x=y;
y=tmp;
}
int main()
{
Sort<int> b(20);
int array[20]={87,4,2,78,4,2,6,1,56,98,24,56,78,12,1,2,4,5,81,76};
b.quick_sort(array,0,19);
b.show_list("Sorted:");
cout<<"quick sort "<<quick<<endl;
cout<<"buuble sort "<<bubble<<endl;
return 0;
}```
One of the advantages of bubble sort is good performnace when dealing with almost sorted
lists or arrays. So how can I set condition to choose between bubble and quick sort?
Thanks
2. >One of the advantages of bubble sort is good performnace when dealing with almost sorted lists or arrays.
Yes, but insertion sort has the same feature and is considerably faster than bubble sort (except in exceedingly rare cases) despite also being a quadratic algorithm.
>So how can I set condition to choose between bubble and quick sort?
Find the cutoff amount of data where quicksort begins to outperform bubble sort. This will typically be a small amount, say less than 5000 small items and less than 1000 large items.
3. That's ok for general case.
I'm interesting for this concrete example .
Is condition test
if(first<last-5)//this is supposed to reduce recursion
good way of reducing recursion.
when testing this example
I have for example quick,quickquick,bubble,quick,quick,bubble,bubble. ..
so sometimes one is called, and sometime other.
Do you have any other suggestion then if(first<last-5)//
4. If your only goal is to reduce recursive calls while still maintaining performance, write a non-recursive quicksort. It certainly minimizes recursive calls, and will usually perform better than a recursive implementation. Mixing bubble sort and quicksort will be a performance hit because bubble sort sucks so much. | 949 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | latest | en | 0.287624 |
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# Matrix Multiplication
Reference > Mathematics > Algebra > Matrices
After seeing how simple matrix addition and matrix subtraction are, and after seeing how simple it is to multiply a matrix times a scalar, you might be thinking, "Matrix multiplication is going to be simple too, and I bet I know exactly how to do it!"
If you're thinking that, you're probably wrong. Matrix multiplication is not a simple process, and it is probably not what you were expecting. Most texts give you nasty looking formulas to explain how matrix multiplication works, but those formulas tend to confuse more than help, so I'm going to try to explain by giving an example. Here's my example:
1
2
3
4
5
6
10
11
12
13
"Wait a minute!" you might be thinking, "the dimensions of those two matrices don't match! We can't multiply them!"
Actually, we can. Here's the first thing you need to know about matrix multiplication: you can multiply two matrices if the number of columns in the first one matches the number of rows in the second one. The dimensions of our first matrix are 3 x 2, and the dimensions of the second are 2 x 2. If we write this as (3 x 2)(2 x 2) we check the "inner" dimensions to make sure they match.
Interestingly, this means that we can't multiply them in the other order becuase then we have (2 x 2)(3 x 2), and the inner dimensions don't match. This means that multiplication of matrices is not commutative!
The next thing you should know is that the product of the two matrices may not have the same dimensions as either of the matrices being multiplied. In fact, the product's dimensions will be the outer dimensions. In this case, since our dimensions are (3 x 2)(2 x 2), the product's dimensions will be 3 x 2 - not because the first matrix's dimensions are 3 x 2, but because the first matrix has 3 rows, and the second matrix has 2 columns. Confusing? Hopefully we can clear up the confusion a bit.
Now that we've covered that little detail, let's look at our original multiplication problem again. I've deliberately made sure that all the entries in the two matrices are distinct, so you can tell which number comes from where (also note that the second matrix has two-digit entries, to make it easier to see which number comes from which matrix).
1
2
3
4
5
6
10
11
12
13
=
?
?
?
?
?
?
In order to calculate the product, we're going to multiply every row of the first matrix by every column of the second one. The result will go in the corresponding position of the product.
STEP ONE
We multiply the FIRST row by the FIRST column like this:
1(10) + 2(12) = 10 + 24 = 34.
This will go in the FIRST row, FIRST column of the answer:
34
?
?
?
?
?
.
STEP TWO
Now we will multiply the FIRST row by the SECOND column like this:
1(11) + 2(13) = 11 + 26 = 37.
THis will go in the FIRST row, SECOND column of the answer:
34
37
?
?
?
?
.
STEP THREE
Now we will multiply the SECOND row by the FIRST column like this:
3(10) + 4(12) = 30 + 48 = 78.
This will go in the SECOND row, FIRST column of the answer:
34
37
78
?
?
?
.
FINISH THE PROBLEM
Can you finish it from here?
You need to do 2nd row times 2nd column, 3rd row times 1st column, and 3rd row times 2nd column.
Give those a try, and then check the answer below, to make sure you've done it right.
1
2
3
4
5
6
10
11
12
13
=
34
37
78
85
122
133
## Questions
1.
Is it possible to multiply a 3 x 4 matrix by a 4 x 2 matrix? Why or why not?
2.
Is it possible to multiply a 4 x 2 matrix by a 3 x 4 matrix? Why or why not?
3.
If you multiply a 4 x 2 matrix by a 2 x 8 matrix, what will the dimensions of the result be?
4.
If
1
2
3
3
2
1
is multiplied by another matrix, how many rows does that matrix have?
5.
If you multiply
1
3
4
2
by
2
4
1
1
, what will be the entry in the first row, first column?
6.
In the previous question, what will be the entry in the first row, second column?
7.
What is the result when
5
6
7
is multiplied by
1
2
3
?
8.
How many entries are in the product of a 3 x 8 matrix with an 8 x 2 matrix?
Assign this reference page | 1,114 | 4,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-39 | latest | en | 0.929453 |
https://bonniegjennings.blog/tag/prompts/ | 1,679,318,967,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00398.warc.gz | 187,884,602 | 29,028 | ## Divide, The Daily Prompt
My take on 50/50, and on the word Divide from The Daily Prompt
### Dividing anything is rarely 50/50 percent.
Divorce, in a 50/50 state, means when you enter the divorce at the beginning of discovery, which is then, 50/50. At the ending of the divorce the percentage will change. The settlements, or percentages are divided between two people. And that depends on which partner has the best attorney.
To divide a pie 50/50 depends on who sneaks to get another bite with a fork while the pie rests wherever until it’s devoured. And, that too is rarely 50/50.
Identical twins rarely share the same identical DNA. Some twins DNA are closer than others, but rarely 50/50. One usually has a mole that distinguishes twin A (first born) from twin B.
The pendulum will rest at 50.50, in the video from youtube (is interesting); however, not until the individual swings of different lengths come to a stop. Though the individual swings are not equal distances when it stops all sides of the individual swings will equal or add up to 50/50 thus equaling a whole. But, that’s natural physics and God (in my opinion).
### This is my take on 50/50, and on the word Divide from The Daily Prompt
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posted by esmeralda
a random sample of 1000 people was chosen for a survey. 80% of them had health insurance. Find the 95% confidence interval
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10. ### Algebra 2
A random sample of 1000 people was chosen for a survey. 80% of them had health insurance. Find the 95% confidence interval. Round answers to 4 significant digits.
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Home > Error Propagation > Error Propagation Area Of Triangle
# Error Propagation Area Of Triangle
## Contents
Use differentials to approximate the possible propagated error in computing the area of the triangle.I'm really sure where to start.I know and and that I'm trying to get it in to The equation 2a + b = 15.7 can be used to find the side lengths. April 19, 2014 by thomas math The base of triangle ABC is one half the altitude. April 13, 2007 by Sam Geometry In the diagram below, the length of the legs AC and BC of right triangle ABC are 6 cm and 8 cm, respectively. More about the author
If the total area of the triangle is 48 square centimeters, then find the lengths of the base and height. June 4, 2009 by <3 math The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of The fractional error multiplied by 100 is the percentage error. Use differentials to estimate the maximum error in the calculated surface area.
## Error Propagation Example
greater than or less than 21? Close Oops! thanks! the SI base unite used to measure mass is the a.
December 16, 2013 by Lindsay calculus The measurement of the radius of a circle is found to be 14 inches, with a possible error of 1/4 inch. This mathematical procedure, also used in Pythagoras' theorem about right triangles, is called quadrature. View Course ND Intro Calculus Subtopics Fractions 100 Online Studying 'Mathematics' 4000 44444 Qualified Helpers Online Now Become a Qualified Helper Learn more 22 spraguer Group Title Group Superhero | Moderator Error Propagation Khan Academy What is the volume of that book?
Can anyone help me.... Error Propagation Division Use differentials to approximate the possible error and the percentage error in computing the area of a circle. The possible error in each measurement is 0.25 centimeter. Hence I asked this question here to see how would people interpret this phrase.
use them to approximate the possible propagated error in computing the area ? Error Propagation Average This is $Revision: 1.18$, $Date: 2011/09/10 18:34:46$ (year/month/day) UTC. Use differentials to approximate the possible propagated error in computing the area of the triangle. Click here to register!
## Error Propagation Division
April 5, 2013 by jay Calculus the measurement on one side of a right triangle is found to be 9.5 inches inches, and the angle opposite that side is 27 degrees,45' http://openstudy.com/updates/50130336e4b0fa24672fdc2d The measurement of each angle is 60 degrees** B. Error Propagation Example If the perimeter of the triangle is to be at most 100 centimeters, what is the maximum length of the base? Error Propagation Physics If one of the longer sides is 6.3 centimeters, what is...
Which of the following statements about the angles is an equilateral triangle is true? my review here Rule 2 If: or: then: In this case also the errors are combined in quadrature, but this time it is the fractional errors, i.e. Use 3.14 for ƒàƒnand round your answer to the nearest tenth of a square centimeter. 2.A triangle has base 13 feet (ft) and an altitude of 6 ft. a. Error Propagation Calculus
Your cache administrator is webmaster. February 3, 2016 by Ruth geometry Each base of an isosceles triangle measures 42 degrees, 30'. Use differentials to estimate the maximum error in the calculated surface area. click site Question 9.3.
END behavior of -11x^4-6x^2+x+3 ? Error Propagation Chemistry Answers: a.) 240 square centimeters b.) 270 square centimeters c.) 480 square centimeters ... please help me..thank you!
## Then we can estimate the error by computing $|A(a+h_1,b+h_2,c+h_3)-A(a,b,c)=|A_1(a,b,c)h_1+A_2(a,b,c)h_2+A_3(a,b,c)h_3|$.
You can only upload a photo (png, jpg, jpeg) or a video (3gp, 3gpp, mp4, mov, avi, mpg, mpeg, rm). Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle. second. Error Propagation Log Archives 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 © 2016 Jiskha Homework Help About Us Contact Us Privacy Policy Terms of Use Report A Bug OpenStudy
Include units in your answer. October 17, 2015 by olga math 1. the total area of triangle is 44 sq in. navigate to this website Is this right?
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Yes No Sorry, something has gone wrong. differentials... Here $A_1,A_2,A_3$ are the derivatives of $A$ with respect to $a,b,c$ respectively. –Alamos Apr 30 '15 at 15:22 | show 3 more comments active oldest votes Know someone who can answer?
My attempt: \begin{gathered} A = \frac{1}{4}\sqrt {2({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}) - ({a^4} + {b^4} + {c^4})} ;\,\,\,\,\,\partial A = \frac{{\partial A}}{{\partial a}} \cdot \partial a + \frac{{\partial A}}{{\partial b}} \cdot Use differentials to estimate the error in the calculated area. Please round the answer to the nearest tenth. Click here to write a testimonial.
Don’t hold back! Calculus help please? Please try the request again. Could someone give me a push in the right direction, thanks!
MY ANSWER 8.83% December 17, 2013 by Lindsay Calculus (check my answer) The measurement of the edge of a cube is found to be 15 inches, with a possible error of Regardless of what f is, the error in Z is given by: If f is a function of three or more variables, X1, X2, X3, … , then: The above formula May 28, 2016 by liliy calculus The circumference of a sphere was measured to be 76 cm with a possible error of 0.2cm. 1. The Notesbooks of Lazarus Long Top Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 yearSort by AuthorPost timeSubject AscendingDescending Page 1 of 1
What emergency gear and tools should I keep in my vehicle? Ask for a resolution. I don't even understand what I was supposed to do... 12) Find the ... | 1,588 | 6,026 | {"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} | 3.890625 | 4 | CC-MAIN-2018-13 | longest | en | 0.895412 |
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https://stats.stackexchange.com/questions/129879/svm-rbf-performance-on-dissimilar-data/129884 | 1,579,846,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00285.warc.gz | 661,908,518 | 30,748 | SVM RBF performance on “dissimilar” data
I've been studying the performance of machine learning algorithms on "dissimilar" data (that is, prediction on new data that are not that "similar" to the training set) and I came up with this example where SVM with a radial kernel completely fails, so I would like your opinion on this.
First I generated data from a simple cubic function, but with x restricted from -60 to 100, so the training set starts to show the cubic form but not all of it:
set.seed(100)
f <- function(x) 100*x + 100*x^2 - x^3 + rnorm(length(x), 10000, 10000)
x <- runif(2000, -60, 100)
y <- f(x)
Here is the plot of the training data:
Then I generated additional data of two kinds: (i) similar to the training data (that is, in the range of the observed data); and, (ii) "dissimilar" to the training data (that is, outside of the range of the observed data):
x2 <- runif(20, 60, 200) # dissimilar
y2 <- f(x2)
x3 <- runif(20, -60, 60) # similar
y3 <- f(x3)
The stripchart of the the data is as follows:
After that I assessed how both the polynomial and the radial kernel SVMs would perform on the dissimilar data set. As expected, the polynomial gets it right, but the radial completely misses it.
svmRadial <- train(as.data.frame(x), y,
tuneGrid = expand.grid(.sigma = c(1,5,15,20),
.C = c(0.25, 0.5, 1,4,26,32,64)))
svmPoly <- train(as.data.frame(x), y, method = "svmPoly",
tuneGrid = expand.grid(.degree = 1:5,
.scale = c(0.01,0.1, 1),
.C = c(0.25, .5, 1)))
I have played with the tuning parameters, but could not find ones that would make the radial SVM get the pattern right. So:
• Is this a known difficulty with the radial SVM (predicting polynomial patterns out of the predictors range)?
• Could you provide a technical explanation of the hard time SVM is having in this situation?
• Also, if possible, it would be nice to point out some papers on this subject (known limitations of radial kernel SVMs with dissimilar predictors).
Typically, if you have prior knowledge (such as the polynomial form), you should choose your kernel based on that. Secondly, most learning algorithms have problems extrapolating, which is what you are doing. It's more common to look at how well test examples in the same range as the training set are predicted.
Regressing with an RBF kernel is somewhat like doing Fourier analysis (where the spectrum is linked to the kernel bandwidth), as you can see from the nature of the predictions which exhibit some periodicity. Here is an interesting paper on the subject.
I wouldn't consider your observation a limitation in general. You happen to be in a setting which perfectly matches the assumptions of the polynomial kernel, hence it outperforms RBF. This is natural. | 697 | 2,740 | {"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} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.900051 |
http://www.scienceforums.net/topic/102419-is-there-really-100-tredicillion-planck-time-seconds-each-second/ | 1,506,105,423,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00484.warc.gz | 554,997,704 | 24,770 | # Is there really 100 tredicillion Planck Time seconds each second?
## Recommended Posts
https://en.wikipedia.org/wiki/Orders_of_magnitude_(time)
You can click on the word "Planck Time" in the top row, and it says that 1 Planck time is the the time it takes for a photon to travel 1 step forward.
What I want to know though, is, is this a "made up" time like ex. there is a centrillion centrillion "mini" seconds per second, or is this really suggesting that a photon travels 100 tredicillion steps each second?
Because I want to know how many real seconds or "moments" are in 1 second. 100 tredicillion?
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Bender 132
I don't think there is a consensus about the quantisation of time or space (although there are hypotheses that introduce such quantisation). If space is not quantised, photons do not take "steps". Even if it is, a photon would be spread out over multiple "steps" and it would be hard to pinpoint it to a specific "step". Moreover, if there are steps, we don't know whether they are indeed a Planck length long.
If time is continuous, which as far as I know is still the main view, talking about "real seconds" in the meaning you imply, is pointless.
I am no expert however, so someone else might step in and correct me.
I can answer the question in the title: there are $1.9 \cdot 10^{43}$ Planck Times in one second.
Edited by Bender
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Strange 2432
https://en.wikipedia.org/wiki/Orders_of_magnitude_(time)
You can click on the word "Planck Time" in the top row, and it says that 1 Planck time is the the time it takes for a photon to travel 1 step forward.
Actually, it doesn't say "one step", it says "one Planck length". There is, as far as we know, no "step" involved.
What I want to know though, is, is this a "made up" time like ex. there is a centrillion centrillion "mini" seconds per second, or is this really suggesting that a photon travels 100 tredicillion steps each second?
It is a unit of time chosen to make various constants, such as the speed of light and G, come out to equal 1. (Whether that makes it "made up" or not is open to debate, I guess.)
https://en.wikipedia.org/wiki/Natural_units
Because I want to know how many real seconds or "moments" are in 1 second. 100 tredicillion?
No one knows. An infinite number, maybe (if time is continuous, as it appears to be).
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imatfaal 2477
...
It is a unit of time chosen to make various constants, such as the speed of light and G, come out to equal 1. (Whether that makes it "made up" or not is open to debate, I guess.)
https://en.wikipedia.org/wiki/Natural_units
...
I think it is arguable that it is the only system of units that could, arguably, be paralleled by alien civilizations (if any) - there is nothing anthropocentric about the Planck units.
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Because only a certain amount of computation or "state changes" i.e. on/offs per second by a single transistor can be done each second. There's a limit. Also, the reason there's a limit is because a particle can't have infinite instances between point A to B.
Can anyone please help me figure out how many instances are in 1 second? It'll help me add to my artificial intelligence project not only how many instances happen each second, but also the amount of possible computation 1 transistor could do.
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Strange 2432
Because only a certain amount of computation or "state changes" i.e. on/offs per second by a single transistor can be done each second.
A technological limit doesn't say anything about the nature of reality (whatever "reality" means...)
No one knows. In GR time is (must be) continuous. There is no theory or evidence currently that says it is quantised.
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Quantised GR is continuous - because each step is the next set of instances for all particles. It's more "there cannot be infinite instances per second". As said that would allow infinite computation (for us) little own universe.
"A technological limit doesn't say anything about the nature of reality (whatever "reality" means...)"
Yes it does, as said above, that [that] is the reality of our universe.
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Strange 2432
Quantised GR is continuous
There is no such thing as quantised GR.
And in GR time is not quantised, it is continuous so there are infinite "instances" per second. Like it or not.
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Is there any way we at least get a good estimate, of how many (theoretical) on/offs a advanced quantum transistor could do? Make up it didn't have to wait for optical light to enter itself - rather it used quantum entanglement to achieve an "every beat just after the last" ex. you don't have to wait for your tapping finger to go up and then back down. On that idea, how many could we compute with it in 1 second theoretically? Or in other words how many instances occur each second dependless of being an electron/graviton/etc?
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Strange 2432
Is there any way we at least get a good estimate, of how many (theoretical) on/offs a advanced quantum transistor could do?
I guess that depends on the technology. The ultimate limit will be set by the size of the device and the fact that changes can only propagate at light speed.
Bender 132
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Strange 2432
Fascinating. Thanks.
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Strange 2432
Is this suggesting a real limit !?
https://en.wikipedia.org/wiki/Margolus%E2%80%93Levitin_theorem
Or do they not even know if that much frequency is even possible to carry out?
Note that it is not a limit on frequency, but on the frequency achievable for a given energy. So, use more energy and get a higher processing rate.
Until some other limit applies ...
We are nowhere near that sort of processing speed, however.
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But is 10 trillion per second by 1 transistor possible? What about higher?
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Strange 2432
But is 10 trillion per second by 1 transistor possible?
Not with any technology we have now.
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Bender 132
Assuming a conventional silicon transistor, and assuming the US "trillion" so 10 trillion is 10^13
I guess such a transistor can get down to at best 10 nm, and then it takes a signal at least 10^-16 s to get through at light speed. At least one electron still needs to jump from one atom to the next, so that might take longer. You can't make a transistor too small because tunnelling will start to kick in. I would say: perhaps.
10^13 electrons flowing in 100 nm² is a current of 10^13 x 10^-19 or about 0.01 µA/nm² or 10^4 A/mm². 0.7 V over each transistor means 10^10 W/mm^2
The density of silicon is 2x10^-6 kg/mm². Converting that to energy by multiplying with c² gives about 10^11 J. In short, the transistor would dissipate the same energy every 10 seconds or so than could be generated by converting its entire mass to energy.
I would say: not likely
With future technology: who knows?
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uncool 150
I think it is arguable that it is the only system of units that could, arguably, be paralleled by alien civilizations (if any) - there is nothing anthropocentric about the Planck units.
Perhaps up to some dimensionless constants, e.g. 2 pi or 4 pi or e or ln(2). 2 pi for angular frequency vs frequency, 4 pi if someone focuses on flux per unit area rather than total flux (as the area of a sphere is 4 pi r^2), e or ln(2) in case they decide to deal with exponentiation slightly differently than we do.
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Bender 132
Perhaps up to some dimensionless constants, e.g. 2 pi or 4 pi or e or ln(2). 2 pi for angular frequency vs frequency, 4 pi if someone focuses on flux per unit area rather than total flux (as the area of a sphere is 4 pi r^2), e or ln(2) in case they decide to deal with exponentiation slightly differently than we do.
$2\pi=\tau$ obviously makes the most sense as it appears most often.
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imatfaal 2477
Perhaps up to some dimensionless constants, e.g. 2 pi or 4 pi or e or ln(2). 2 pi for angular frequency vs frequency, 4 pi if someone focuses on flux per unit area rather than total flux (as the area of a sphere is 4 pi r^2), e or ln(2) in case they decide to deal with exponentiation slightly differently than we do.
Yep - I would guess all dimensionless constants; alpha ~1/137 etc. I did read an article that these might be the limit as it is possible we do not understand how human-based things like Peano axiomata are - ie even what we consider basic maths might not be universal; but the arguments given were over my head and I did think the whole paper was fairly tendentious.
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Strange 2432
There was a discussion on the radio a while ago about whether math is invented or discovered. One of the guests proposed the idea of an alien intelligence based on fluid forms rather than solid bodies - to them, the ideas related to real numbers and even calculus might be intuitively obvious but then one of their greatest minds comes up with the shocking and difficult-to-grasp concept of "integers" ...
• 1
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imatfaal 2477
There was a discussion on the radio a while ago about whether math is invented or discovered. One of the guests proposed the idea of an alien intelligence based on fluid forms rather than solid bodies - to them, the ideas related to real numbers and even calculus might be intuitively obvious but then one of their greatest minds comes up with the shocking and difficult-to-grasp concept of "integers" ...
Yes that's the sort of thing I meant exactly. And that radio programme sounds like "in our time" - or something equally as good; do you have a link? or name?
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Strange 2432
Yes that's the sort of thing I meant exactly. And that radio programme sounds like "in our time" - or something equally as good; do you have a link? or name?
I think you are probably right. It might have been this one (haven't had a chance to listen again to check):
http://www.bbc.co.uk/programmes/p0054799
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Let's do a thought experiment here:
If we speed-up an atom to 99-percent light-speed, it would at least travel a distance that could fit around Earth 2 times in 1 second, agree?
And can we agree that, when we "shoot" the atom out, at the Starting line, and when it reaches the Finish line (after 1 second), that it will have passed trillions of atoms? 400 million hydrogen atoms an fit in an inch, a ruler is like billions. Our atom went far, indeed.
And can we agree that atoms have been seen under microscopes where they aren't perfectly aligned with each-other like a atomic-grid? For example, if you make an L made of 3 atoms, and push the top atom over so it's a triangle, that is un-alignment, even over a crack.
--- This means our fired atom not only passes trillions of atoms, but also manyyyy many protons. So instead of saying it passes trillions (of atoms), we should say it passes like quintillions. That right there means there is quintillions of instances/moments/computational-ness each second for a computing transistor!
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Endy0816 338
How are you envisioning this atom interacting with other atoms?
They are looking at using wave phenomenon for computing. Probably along the lines of what you are imagining in terms of what it will be capable of. | 2,756 | 11,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-39 | longest | en | 0.937043 |
https://mrkirkmath.com/2019/10/18/fractals-chaos-recap-for-10-18-2/ | 1,591,507,269,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00246.warc.gz | 442,718,898 | 19,969 | # Fractals & Chaos Recap for 10/18
If you have a compass (the circle-drawing kind) please bring it with you to class over the next few days.
We discussed Mandelbrot’s Article, then used it to segue into a further discussion of dimension. As we are aware, we still have problems with the Hausdorff dimension formula for calculating dimension of fractals. It can’t handle fractals with stems (i.e., non-iterating segments that never disappear) and with fractals that are not exactly self-symmetric.
Today, we considered a football field, a circle, and a Koch Curve, and looked at how the size of the measuring stick we use to measure the length or perimeter of such things has an impact on the total amount of length we actually calculate. For a football field, the size of the stick makes no difference. We’ll be obtaining 100 yards worth of length even if we use a foot (S = 3) or an inch (S = 36) as our step size.
For a circle, this isn’t the case. Use a measuring stick the length of the diameter, and we can only make two steps before we end where we started. Use a stick the size of the radius (S = 2) and we can make 6 such steps (resulting in a measure of three diameters). Use a half-radius (S = 4), and we wind up with a total length of slightly more than 3 diameters. There is a limit to this, of course: pi*d, which is precisely the formula for the circumference of a circle.
For the Koch Curve, the story is very different. Use a step size the length of the original baseline, and we can make one step. Use a step size of 1/3 the baseline (S = 3), we can make 4 steps, giving a length of 4/3 the base. Use a step size of 1/9 the baseline (S = 9), and we can make 16 steps, for a total length of 16/9 the base. As we shrink the length of the ruler we use, the number of steps increases more quickly, and so the total length increases without bound.
We’ve seen suggestions at this idea before. In the second article we read (The Diversity of Life), we saw that reducing the scale of our perspective dramatically increases the amount of living space we can find. This idea is also found at the center of the coastline paradox, hinted at in the Ants in Labyrinths article (see also this blog post from UK Urban Planner Alasdair Rae)
We will be expanding on this in class tomorrow, including a discussion on what all this has to do with the dimension of what we’re measuring.
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# Thread: Algorithm/Numerical Approach for Computing CI (Contrast Index)
1. David, your equations look good. My data is zeroed at fb-f, having subtracted the y-axis min of the points, so D(x1) = 0, makes it even simpler. As it seems that the fitted smooth spline model is of high quality (p<0.01), for my curent data, and I expect it should be so for most successful film tests, the R predict(x) function will safely give me the interpolated value of D(x), using the spline, for any point I need along the range of rel logE axis. So it looks like I'm now in position to use the curve to solve your equations for the CI condition, looking for the first occurrence of it, I suppose, from left, in case the condition could be fulfilled more than once on a strange curve. Thanks for the suggestion of optimize.fsolve in scipy, that's a great tip. I will try uniroot or rootSolve/multiroot in R. If I can't make it, I'll ask for your Python.
2. Originally Posted by dpgoldenberg
Bill,
My words may not have been as clear as I intended, but I think that the equations are correct. The equations define the distances between the points in two dimensions, not just the distance along the horizontal axis. (That's where Pythagoras comes in.) The first two equations define the distances between the first and second points (0.2) and the second and third points (2). The third equation ensures that all three lie on the same line. Also, D(x1) should be specified to be b+f.
David
Oh, I see. Sounds good
3. Though, a little off topic, here is the "W-speed" meter that for which I made the Excel spreadsheet to imitate its function. As you can see it is more complicated than the Contrast Index Meter, however, my solution to the problem is simpler because I use linear interpolation between the datapoints, rather than a complex function for the film curve.
My Excel calculated approximation for "E" used to calculate the W-speed = the red dot. I think it is a pretty good approximation:
4. Rafal, the CI sort of assumes that you have a straight line as characteristic curve, while your LOESS approach most likely won't give you a straight line but some spline curve with variable slope over its range. I would therefore suggest a slightly different approach: create a model for what you expect as characteristic curve, e.g. a flat line in the b+f region, a straight line in the linear region and some form of smooth transition between these two straight lines. Once you have a model that covers all characteristic curves you are likely going to encounter, you can do an LMS fit to get the parameters for each set of measurement data. It should then be simple to associate a CI with the parameters you obtain for your model.
5. Originally Posted by Rudeofus
Rafal, the CI sort of assumes that you have a straight line as characteristic curve,
Seems like CI was created for films that don't have a straight line.
6. ## R Code for Computing CI
Mission accomplished, thanks to everyone, especially David for pointing out the system of non-linear equations based on the Pythagorean triangle relationship. If you run this, this is what you get:
And the data you feed into it comes from a text file that looks like this:
If you want to play with it, you need my code (attached) and a copy of R, which is a free programming language and an environment for statistics/numerical analysis, and which does nice plots. It runs on Windows, Macs, and Linux, get it from here if you feel geeky, or read about on Wikipedia. My code is in the attachment, all you need to do is to load it in R by using 'source("CI_plotter.R.txt")' having placed it in the current directory. If you open that file, you can see how to use it, I gave some examples. Let me know if anyone would like any pointers. I don't mind plotting your data, if you wish, to find the CIs—I suppose this won't be a very popular request, anyway.
The R code, which I wrote, works for the various curve data I fed into it. You need to give it a series of numbers representing your step tablet exposures (so from 0 on the left increasing towards the right) in a column called "He" and any number of columns containing your densitometer readings. R will plot the points and it will try fitting curves—using Bezier splines with 6 degrees of freedom by default. You can decrease that number for more smoothing (left-of-toe will look odd but calculations will work) or increase it, for less smoothing. You can also play with other smoothers if you uncomment the right lines in the code, including LOESS. In either case, the CI computations should work. I uses nleqslv, which is pretty much as close to SciPy fsolve, which David mentioned, as I could find today. The alternative which I have tested, multiroot (from rootSolve) was less forgiving of bad choices of the "starting point". This is a weakness of my code (I am sure one of many!). Sometimes, for some curves, it will not compute CI, it will display "NA" on the plot, and it will make a suggestion that you change the "starting.point". This is a parameter at the top of the program, which denotes the best guess of the x[1], x[2], and x[3] as per David's explanations, that is the rel log E values that are supposed to correspond to the projections onto the X axis of the 3 key CI determination points. In other words, those are: the centre of the two arcs, and the intersections of the 0.2 and the 2.2 arcs with x axis. I defaulted it to (0.9, 1, 2) by trial and error, but if it does not work for you, nudge them a bit either way, especially the first one. Anyone who would like to contribute a good way to guess those automatically, please let me know—I have tried interpolating them from a straight-line regression of the data points etc but that was worse than using my best manual guesses. Some other time I will add a function to plot the CI ~ dev time curve...
Ok, back to the darkroom tomorrow—I miss a low-contrast test in this data, will try a 4.5 min time to complete the exercise.
7. ## Delta 100 4x5 DDX 1+4 Tank Extended Results
...and in case anyone were interested in the actual subject of my testing, Delta 100 4x5 developed in DDX 1+4 20C in a tank (CombiPlan) with fairly vigorous agitation (30 sec initially, followed by 3 inversions lasting about 5 seconds in total every 30 sec), I attach "extended" results. I obtained those by exposing two strips with the Eseco SL-2, both using green light. The second strip got three times the exposures of the first, but it does not actually amount to triple exposure, it seems, and I am in reciprocity territory, which this film seems to be good at. Undoubtedly, this is a questionable way of doing this, as I am numerically combining two separate tests into a single "extended" one, as if it came from one, longer step tablet—unlike in my previous post, where tests were "normal". Nonetheless, data seems OKish to my untrained eyes, and I welcome your feedback about this 9-stop range behaviour.
PS. To compute the first of the curves, I needed to adjust the "starting.points" to (0.95, 1, 2). Perhaps this should be the default.
8. Nice Rafal!
Now all you need is a Time/CI curve
9. Originally Posted by ic-racer
Seems like CI was created for films that don't have a straight line.
The author who proposed CI did not have Rafal's jiggly spline curves in mind, though. Niederpruem et al. tried to account for toe shape and possible upswept/downswept curves, not for randomly swept spline segments that pointlessly try to fit noisy measurement data. It speaks volumes that Rafal's curve fit look, let's call it euphemistically "a bit odd" in the toe region. Oh well ...
10. Indeed, Rudi—the upsweep towards less-than-none exposure looked humorous. Bear in mind, however, if I you increase the number of degrees of freedom on the Bezier spline to 7 or more, the toe looks just fine, yet the calculation still gets the CI—what we miss then is the smoothing that neatly takes care of the statistical measurement error. Having said that, I have to play, yet, with other types of splines, including log-influenced ones, which might have that "perfect" look.
But the next thing I plan to do is what Bill has suggested, that is the Time/CI curve. By the way, if anyone has a curve data series with their own, known CI, I'd be happy to check the algorithm against those numbers.
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Contact Us | Support Us! | Advertise | Site Terms | Archive — Search | Mobile Device Access | RSS | Facebook | Linkedin | 2,064 | 8,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2015-48 | latest | en | 0.939674 |
https://www.scribd.com/document/12856218/Freegoods-Userguide | 1,524,656,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947795.46/warc/CC-MAIN-20180425100306-20180425120306-00616.warc.gz | 888,417,340 | 24,585 | # Free goods discount
Create free goods discount record Menu path: as below
Click
Click
Minimum qty – Quantity equal or above which free goods discount will apply
Minimum quantity - 5 pieces. Free discount will apply only for quantity greater than 5.
Quantity for which free goods is granted - 10 pieces
Free goods quantity - 1 piece
Free goods % - It is automatically calculated. 1/10 = 10%
Calculation rule
Pro rata – If 1 is free per 10 pieces, then if ordered is 50 pieces, then 5 will be free. Unit reference - For 55 pieces ordered, free quantity will be (1/10)*50 = 10. Formula is (Free quantity / base quantity) * document quantity Whole units – Free goods quantity calculated only if it is in multiple of “10”…so if total quantity is 25, free quantity is Nil but if total quantity is 20, then free quantity is 2 pieces. Free good category Inclusive Rebate with item generation free item -Example - Free goods condition record as below. 1 Piece Free per 5 Pieces and minimum order quantity should be 10 pieces.
If free goods pricing is active for the sales order, then it searches for the free discount record. Once the record is found and if as per record if free goods item should appear
as sub-item in the sales order, then item category for subitem is TANN. TANN determination is as per configuration
100% discount is possible for sub-item as per condition type, R100 R100 in standard pricing procedures like RVCXUS or RVAA01 is added as below.
Condition base value formula '28' is assigned to condition type R100 in the pricing procedure to apply the 100% discount rate. Requirement “55” – It should be assigned to this condition type in the pricing procedure so that R100 is calculated only for free items (Item with item category, TANN, which has pricing indicator as “B” – Pricing for free goods, 100% discount). Put an order for 15 pieces, then 3 are calculated as additional quantity. Main item (TAN) gets generated with 12 pieces and free quantity is shown as sub-item with 3 pieces.
Pricing for the free item is as below.
The system re-reads the free goods master record in the sales order if the quantities in the main item change or if the pricing date changes. The system then deletes the sub-items and re-creates them. Any manual changes to the free goods quantity are lost. If pricing is re-run in the sales order, it does not affect the free goods determination. Inclusive Free Goods w/o item generation -If this is selected in above condition record, order looks like as below. The system will not generate items for the free goods. In this case, the correct discount is determined in pricing for the main item.
Pricing looks like as below.
NRAB is used as condition type here. The condition type has condition category “f – free goods - inclusive”. This condition category ensures that the condition is redetermined every time the quantity is changed, since this can also mean a change to the free goods quantity.
• Alternative condition base formula “29” Value is calculated for 811 x 15 = 12165 Variant factor – 20% (1 per 5 pieces) picked from free goods record. So, NRAB value is calculated as 20% of 12165 = 2433. Disadvantages with one line display are The free goods quantity is not obvious during order entry
free goods quantities are not identified in the statistics BUT the free goods quantity is displayed in the order confirmation and the invoice.
Exclusive
The
• When you enter an exclusive bonus quantity, an extra entry line appears where you can enter the good if it is not the same as the ordered material. | 802 | 3,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-17 | latest | en | 0.88736 |
https://physics.stackexchange.com/questions/409439/the-probability-of-rangle-to-be-in-0-rangle-or-1-rangle | 1,585,686,541,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370503664.38/warc/CC-MAIN-20200331181930-20200331211930-00486.warc.gz | 666,844,503 | 32,514 | # The probability of $|+\rangle$ to be in $|0\rangle$ or $|1\rangle$?
The following image is from IBM Q Experience beginner's guide HERE.
The gate sequences are being applied to the qubit $|0\rangle$. I only do not understand why in the second row, the probabilities are 85, 15. I see that the H and then T gates take $|0\rangle$ to the plane X-Y with angle $\frac{\pi}{4}$ from both axis (in Bloch sphere), but I cannot see where the Hadamard gate before the measurement takes this. Any hint is appreciated.
• FWIW, for a rotation angle $\alpha$ the rightmost column is $\sin^2\alpha/2$. – J.G. May 31 '18 at 23:09
• Thanks @J.G. . Could you tell me why it is $sin^2\,\alpha/2$? – Mathophile-Mathochist Jun 1 '18 at 14:16
• We get qubits from spin-$1/2$ fermions, which are spinors. This halves the angle appearing in $\exp i\theta$, and the real and imaginary parts thereof are the desired amplitudes. – J.G. Jun 1 '18 at 14:55
The states $|0\rangle$ and $|1\rangle$ lie on the $+z$-axis and $-z$-axis of the Bloch sphere, and the $|+\rangle$ and $|-\rangle$ states lie on the $\pm x$-axis. The Hadamard gate $H$ performs a rotation that transforms $|0\rangle$ to $|+\rangle$ and $|1\rangle$ to $|-\rangle$ (i.e. it's a 180-degree rotation about an axis 45 degrees from both the $z$-axis and the $x$-axis). So if you apply $HZH$, you first turn $|0\rangle$ into $|+\rangle$, then you turn $|+\rangle$ into $|-\rangle$ (since the $Z$ operator is a 180-degree rotation about the $z$-axis), then you turn $|-\rangle$ into $|1\rangle$.
If, on the other hand, you apply $HTH$, then you first turn $|0\rangle$ into $|+\rangle$, as before, but then you rotate only 45 degrees around the $z$-axis. This leaves you midway between the $x$ and $y$ axes. When you apply the second $H$, you will end up on the opposite side of the axis of rotation (which, again, evenly splits the $z$ and $x$ axes). Doing this rotation will land you much closer to the $|0\rangle$ state (the "north pole" of the Bloch sphere) than the $|1\rangle$ state (the "south pole"); trying this on a globe (or some other sphere you have laying around) might offer some intuition as to how this works. As such, the probability of being $|0\rangle$ should be much higher than the probability of being $|1\rangle$.
• @Mathophile-Mathochist Because $\sin^2 \pi/8 = (1-\cos\pi/4)/2=(1-1/\sqrt{2})/2\approx 0.146$. – J.G. Jun 1 '18 at 14:33 | 749 | 2,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.812735 |
https://www.convertunits.com/from/square+megameter/to/square | 1,623,991,748,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635724.52/warc/CC-MAIN-20210618043356-20210618073356-00247.warc.gz | 653,871,543 | 16,621 | ## ››Convert square megametre to square
square megameter square
## ››More information from the unit converter
How many square megameter in 1 square? The answer is 9.290304E-12.
We assume you are converting between square megametre and square.
You can view more details on each measurement unit:
square megameter or square
The SI derived unit for area is the square meter.
1 square meter is equal to 1.0E-12 square megameter, or 0.1076391041671 square.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between square megameters and squares.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of square megameter to square
1 square megameter to square = 107639104167.1 square
2 square megameter to square = 215278208334.19 square
3 square megameter to square = 322917312501.29 square
4 square megameter to square = 430556416668.39 square
5 square megameter to square = 538195520835.49 square
6 square megameter to square = 645834625002.58 square
7 square megameter to square = 753473729169.68 square
8 square megameter to square = 861112833336.78 square
9 square megameter to square = 968751937503.87 square
10 square megameter to square = 1076391041671 square
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## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 501 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.745228 |
http://andrewgelman.com/2010/10/25/is_instrumental/ | 1,534,253,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209040.29/warc/CC-MAIN-20180814131141-20180814151141-00020.warc.gz | 25,060,596 | 8,980 | ## Is instrumental variables analysis particularly susceptible to Type M errors?
Hendrik Juerges writes:
I am an applied econometrician. The reason I am writing is that I am pondering a question for some time now and I am curious whether you have any views on it.
One problem the practitioner of instrumental variables estimation faces is large standard errors even with very large samples. Part of the problem is of course that one estimates a ratio. Anyhow, more often than not, I and many other researchers I know end up with large point estimates and standard errors when trying IV on a problem. Sometimes some of us are lucky and get a statistically significant result. Those estimates that make it beyond the 2 standard error threshold are often ridiculously large (one famous example in my line of research being Lleras-Muney’s estimates of the 10% effect of one year of schooling on mortality). The standard defense here is that IV estimates the complier-specific causal effect (which is mathematically correct). But still, I find many of the IV results (including my own) simply incredible.
Now comes my question: Could it be that IV is particularly prone to “type M” errors? (I recently read your article on beauty, sex, and power). If yes, what can be done? Could Bayesian inference help?
I’ve never actually done any instrumental variables analysis, Bayesian or otherwise. But I do recall that Imbens and Rubin discuss Bayesian solutions in one of their articles, and I think they made the point that the inclusion of a little bit of prior information can help a lot.
In any case, I agree that if standard errors are large, then you’ll be subject to Type M errors. That’s basically an ironclad rule of statistics.
My own way of understanding IV is to think of the instrument has having a joint effect on the intermediate and final outcomes. Often this can be clear enough, and you don’t need to actually divide the coefficients.
And here are my more general thoughts on the difficulty of estimating ratios.
1. LemmusLemmus says:
I'm sorry, but I don't get how the IV approach in particular leads one to estimate ratios. Your (Andrew's) last but one paragraph seems to suggest that this refers to the "total effect" estimates (indirect and direct effects), but in IV regressions in particular you are typically not interested in the effect of the instrument on the outcome. (Of course, if it is a valid instrument, there are no direct effects, so total effects are equal to the indirect effects.) For example, when you want to know about the effect of immigration into US cities on those cities' economic outcomes and use proximity to the Mexican border as an instrument for immigration, then you don't care about the indirect effect of border proximity on economic outcomes.
2. Corey says:
"Those estimates that make it beyond the 2 standard error threshold are often ridiculously large…"
One approach to inference in these types of problems is to condition on the fact that only estimates that exceed some threshold will be estimated. I like the paper of Ghosh et al. on this subject (http://dx.doi.org/10.1016/j.ajhg.2008.03.002). The discussion is in the context of genome-wide association studies, but the math should be broadly applicable. | 678 | 3,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-34 | latest | en | 0.9287 |
https://usakochan.net/download/controlled-and-conditioned-invariance/ | 1,656,924,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00417.warc.gz | 617,850,472 | 13,031 | # Controlled and Conditioned Invariance
The controlled invariance Given in 3," the linear transformation A and the "controlling" subspace 3 , a subspace '3 is a controlled invari— ant underA with respect to 3' , or simply an 5A,:fZ-controlled invariant, if A?! s 3' + 3 .
Author: Giuseppe Basile
Publisher: Springer
ISBN: 9783709129531
Category: Computers
Page: 51
View: 921
Categories: Computers
# Controlled and Conditioned Invariant Subspaces in Linear System Theory
The concept of invariance of a subspace under a linear transformation is strongly connected with controllability and observability problems of linear dynamical systems.
Author: Giuseppe Basile
Publisher:
ISBN: OCLC:227500096
Category:
Page: 21
View: 618
The concept of invariance of a subspace under a linear transformation is strongly connected with controllability and observability problems of linear dynamical systems. In this paper we define 'controlled' and 'conditioned' invariant subspaces as a generalization of the simple invariants, for the purpose of investigating some further structural properties of linear systems. Moreover, we prove some fundamental theorems on which the computation of the above mentioned subspaces is based. Then we give two examples of practical application of the previously defined concepts concerning the determination of the constant output and perfect output controllability subspaces. (Author).
Categories:
# Controlled and Conditioned Invariants in Linear System Theory
Using a geometric approach to system theory, this work discusses controlled and conditioned invariance to geometrical analysis and design of multivariable control systems, presenting new mathematical theories, new approaches to standard ...
Author: Giuseppe Basile
Publisher:
ISBN: UOM:39015024980800
Category: Linear systems.
Page: 464
View: 477
Using a geometric approach to system theory, this work discusses controlled and conditioned invariance to geometrical analysis and design of multivariable control systems, presenting new mathematical theories, new approaches to standard problems and applied mathematics topics.
Categories: Linear systems.
# Systems Models and Feedback Theory and Applications
V C A is an (A, B)-controlled invariant if AV C V+B and that SC 3' is an (A, C)-conditioned invariant if A(SnC) C. S. The maximum (A, B)controlled invariant contained in a given subspace & C & will be denoted by max V(A, B, ...
Author: A. Isidori
Publisher: Springer Science & Business Media
ISBN: 9781475722048
Category: Science
Page: 404
View: 710
It is a great honor and privilege to have this opportunity of celebrating the 65th birthday of Professor Antonio Ruberti by holding an International Conference on Systems, Models and Feedback. The conference, and this volume which contains its proceedings, is a tribute to Professor Ruberti in acknowledgement of his major contributions to System Theory, at a time in which this area was emerging and consolidat ing as an independent discipline, his role as a leader of the Italian academic community, his activity in promoting and fostering close scientific relations between Italian and U.S. scholars in Systems and Control. The format of this conference is inspired by a series of seminars initi ated exactly twenty years ago under the direction of Professor Ruberti, in Italy, and Professor R. R. Mohler, in the U.S. By bringing together many authoritative talents from both countries, these seminars were instrumental in promoting the expansion of System Theory in new areas, notably that of Nonlinear Control, and were the key to successful scientific careers for many of the younger attendants.
Categories: Science
# The Control Handbook three volume set
( 2002 ) , with additional later results found in the newer books , for example , controlled and conditioned invariant subspaces and duality are treated in an organized fashion in the 2nd book and , in addition to the duality ...
Author: William S. Levine
Publisher: CRC Press
ISBN: 9781420073676
Category: Technology & Engineering
Page: 3526
View: 201
At publication, The Control Handbook immediately became the definitive resource that engineers working with modern control systems required. Among its many accolades, that first edition was cited by the AAP as the Best Engineering Handbook of 1996. Now, 15 years later, William Levine has once again compiled the most comprehensive and authoritative resource on control engineering. He has fully reorganized the text to reflect the technical advances achieved since the last edition and has expanded its contents to include the multidisciplinary perspective that is making control engineering a critical component in so many fields. Now expanded from one to three volumes, The Control Handbook, Second Edition brilliantly organizes cutting-edge contributions from more than 200 leading experts representing every corner of the globe. They cover everything from basic closed-loop systems to multi-agent adaptive systems and from the control of electric motors to the control of complex networks. Progressively organized, the three volume set includes: Control System Fundamentals Control System Applications Control System Advanced Methods Any practicing engineer, student, or researcher working in fields as diverse as electronics, aeronautics, or biomedicine will find this handbook to be a time-saving resource filled with invaluable formulas, models, methods, and innovative thinking. In fact, any physicist, biologist, mathematician, or researcher in any number of fields developing or improving products and systems will find the answers and ideas they need. As with the first edition, the new edition not only stands as a record of accomplishment in control engineering but provides researchers with the means to make further advances.
Categories: Technology & Engineering
# Nonlinear Dynamical Control Systems
In R.W. Brockett, R.S. Millman, and H.J. Sussmann, editors, Differential Geometric Control Theory, pages 209–225. Birkhäuser, Boston, 1983. G. Basile and G. Marro. Controlled and conditioned invariant subspaces in linear systems theory.
Author: Henk Nijmeijer
Publisher: Springer Science & Business Media
ISBN: 9781475721010
Category: Technology & Engineering
Page: 426
View: 142
This volume deals with controllability and observability properties of nonlinear systems, as well as various ways to obtain input-output representations. The emphasis is on fundamental notions as (controlled) invariant distributions and submanifolds, together with algorithms to compute the required feedbacks.
Categories: Technology & Engineering
# System Structure and Control 2004
G. and G. Marro ( 1973 ) , " Controlled and conditioned invariant subspaces in linear system theory " , J. Optim . Theory and Appl . 3 , 306316 . Emre , E. and M.L.J. Hautus ( 1980 ) , " A polynomial characterization of ( A ...
Author: Sabine Mondie
Publisher: Elsevier
ISBN: 0080441319
Category: Technology & Engineering
Page: 740
View: 436
Categories: Technology & Engineering
# Control and Dynamic Systems V53 High Performance Systems Techniques and Applications
The problem of estimating the state of a linear time invariant dynamical system subject to both known and unknown inputs (e.g. disturbances) has been considered by a number of authors [1 1-21]. Notions of controlled and conditioned ...
Author: C.T. Leonides
Publisher: Elsevier
ISBN: 9780323163170
Category: Technology & Engineering
Page: 540
View: 926
Control and Dynamic Systems: Advances in Theory and Applications, Volume 53: High Performance Systems Techniques and Applications covers the significant research works on the issues and applications of high performance control systems techniques. This book is divided into 11 chapters and starts with an examination of the contribution of computing power with advances in theory in global optimization. The next chapters present robust solution techniques for combined filtering and parameter estimation in discrete time and the design and analysis of model reference adaptive control techniques for both continuous and discrete time multivariable plants with additive and multiplicative unmodeled dynamics. These topics are followed by discussions of the decentralized adaptive control; robust recursive estimation of states and parameters of bilinear systems; the design of robust control systems under uncertainty cases; and the techniques for state estimation for linear stationary dynamic systems that are subject to unknown time varying plant and output disturbances. Other chapters deal with the sliding control algorithm, the techniques in robust broadband beamforming, and the different categories of robust robotic controllers. The final chapter looks into the problems and issues of performance and versatility of non-linear control and the application of artificial neural networks. This book is of great value to process, control, mechanical, and design engineers.
Categories: Technology & Engineering
# Symbolic Methods in Control System Analysis and Design
2 6 BASILE , G. , and MARRO , G .: ' Controlled and conditioned invariant subspaces in linear system theory ' ( Prentice Hall , 1969 ) 7 JAFFE , S. , and KARCANIAS , N .: Matrix pencil characteristation of almost ( A , B ) -invariant ...
Author: N. Munro
Publisher: IET
ISBN: 0852969430
Category: Technology & Engineering
Page: 393
View: 389
Fifteen contributions provide an up-to-date treatment of issues in system modeling, system analysis, design and synthesis methods, and nonlinear systems. Coverage includes the application of multidimensional Laplace transforms to the modeling of nonlinear elements, a survey of customized computer algebra modeling programs for multibody dynamical systems, robust control of linear systems using a new linear programming approach, the development and testing of a new branch-and-bound algorithm fir global optimization using symbolic algebra techniques, and dynamic sliding mode control design using symbolic algebra tools.
Categories: Technology & Engineering
# Linear Algebra in Signals Systems and Control
[ 2 ] KALMAN , R. E. , Mathematical Description of Optimal Control Systems , SIAM Journal of Control , Vol . ... [ 6 ] BASILE , G. and MARRO , G. , Controlled and Conditioned Invariant Subspaces in Lineary System Theory , Journal of ...
Author: Biswa Nath Datta
Publisher: SIAM
ISBN: 0898712238
Category: Technology & Engineering
Page: 667
View: 818
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# Fall 2009 Prelim 1 - MATH 1920 FALL 2009 PRELIM 1 SHOW ALL WORK NO CALCULATORS NAME STUDENT ID SECTION Who is your TA PROBLEM 1a 1b 2 3a 3b 4a 4b
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Unformatted text preview: MATH 1920, FALL 2009 PRELIM 1 SHOW ALL WORK. NO CALCULATORS. NAME: STUDENT ID #: SECTION #: Who is your TA? PROBLEM 1a 1b 2 3a 3b 4a 4b 4c 5a 5b 6 SCORE TOTAL 1. Find the limit if it exists, or show that it does not exist. (a) (2x − 3y + z ) sin(x − 1) (x,y,z )→(1,−1,−1) (x2 + z 2 )(x − 1) lim (b) (x,y )→(0,0) lim x2 + y 2 . 2x − y 1 2. Find parametric equations for the line tangent to the curve of intersection of the surfaces 2x2 − 3y + z 2 = 5 and x + y = 3 at the point P (1, 2, 3). 2 3. (a) Consider the function f (x, y, z ) = xy + yz − xz , where x = u − v , y = u + v , and ∂f z = uv . Find as a function of u and v . ∂u (b) The directional derivative of a function f (x, y ) at the point (1, 1) in the direction √ of i + j is 3 2, and the directional derivative in the direction of 3j is −2. Find fx (1, 1) and fy (1, 1). 3 On this problem, you do not need to justify your answers. 4. Consider the domain of the function f (x, y ) = xy + 3 − (x2 + y 2 ) . (x2 + y 2 ) − 1 (a) Give equations for the boundary of the domain of f . (b) Mark the correct statement below. The domain is bounded. The domain is unbounded. The domain is neither bounded nor unbounded. The domain is both bounded and unbounded. (c) Mark the correct statement below. The domain is open. The domain is closed. The domain is neither open nor closed. The domain is both open and closed. 4 5. Consider the plane T with equation 3x − y + z = 13. It contains the point P (2, −3, 4). (a) The line L1 has parametric equations x = −t + 4 y = 2t − 1 z = 5t and lies in the plane T . Find parametric equations for the line L2 that satisfies the following conditions: L2 lies in the plane T , L2 is perpendicular to L1 , and L2 goes through the point P . (b) Find the equation of the sphere S through the point (8, −5, 6) and such that the plane T is tangent to S at the point P . 5 6. The first octant is the region in space defined by x ≥ 0, y ≥ 0, z ≥ 0. Use Lagrange Multipliers to determine which point in the first octant and on the surface √ xy 2 z 2 = 16 2 is closest to the origin. 6 7 ...
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Fall 2009 Prelim 1 - MATH 1920 FALL 2009 PRELIM 1 SHOW ALL WORK NO CALCULATORS NAME STUDENT ID SECTION Who is your TA PROBLEM 1a 1b 2 3a 3b 4a 4b
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A086981 a(n) = smallest k where (10^k+1)=0 mod prime(n)^2, or 0 if no such k exists. 3
0, 0, 0, 21, 11, 39, 136, 171, 253, 406, 0, 0, 0, 0, 1081, 0, 1711, 1830, 0, 0, 292, 0, 0, 1958, 4656, 202, 1751, 0, 5886, 6328, 2667, 8515, 548, 3197, 11026, 0, 6123, 0, 13861, 0, 15931, 16290, 0, 18528, 9653, 0, 3165, 24753, 0, 26106, 27028, 0, 3615, 6275 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS For a given a(n)>0, all the values of k such that (10^k+1)=0 mod prime(n)^2 are given by the sequence a(n)*A005408, i.e. odd multiples of a(n). For example, for n=5, prime(5)=11, a(n)=11, the set of values of k for which (10^k+1)=0 mod 11^2 is 11*A005408=11,33,55,77,99,... All the terms of the sequence a(n) are integer multiples of prime(n) for primes <1000 except for a(93)=243 where prime(93)=487. LINKS Charles R. Greathouse IV, Sep 10, 2008, Table of n, a(n) for n = 1..4000 EXAMPLE a(4)=21 since 21 is least value of k for which (10^k+1)=0 mod 7^2. CROSSREFS Cf. A000040, A086982. Sequence in context: A252629 A213217 A300505 * A114011 A300943 A270790 Adjacent sequences: A086978 A086979 A086980 * A086982 A086983 A086984 KEYWORD nonn AUTHOR Ray Chandler, Jul 27 2003 STATUS approved
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Last modified September 29 14:18 EDT 2023. Contains 365771 sequences. (Running on oeis4.) | 644 | 1,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | latest | en | 0.640359 |
https://rebab.net/what-does-100-grams-look-like/ | 1,653,801,355,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00482.warc.gz | 525,727,885 | 10,788 | The assumed of eating 100+ grams that protein in a job is just difficult to part people.
You are watching: What does 100 grams look like
I want to display you the it have the right to actually be rather easy come hit those goals. Even if it is you love protein shakes and want to drink those, or you want to follow a 100% organic vegetable lifestyle, there room plenty of high-protein alternatives for you.
If you want to see exactly how much protein ns recommend girlfriend consume top top a daily basis, inspect out my free Calorie & Macro Calculator
## Why 100g of protein?
100g of protein is simply a nice round number to usage as an example here, and certainly a number girlfriend don’t have to target for.
A lot of of people aim come eat their weight in protein (more on the in a moment) and also I’m going come go out on a limb and also assume that most of us weigh more than 100 pounds, therefore it’s a an easy number we have the right to all relate come on some level.
If you’re not at this time consuming a most protein, 100 grams likely sound choose a really high (not to cite unattainable) amount, which is why I want to target for that number here.
For countless of us, 100g the protein is in reality a good target, yet most of united state can advantage from also more…
## How much protein carry out you actually need?
Can you just aim because that 100g of protein and also call that a day?
Sure, you could.
But there’s a much better (and easy) means to find how lot protein you should be consuming. Don’t worry, it’s an easy math!
The most widely recommended amount is 1g per pound of bodyweight.
You’ve likely come across that reference in your find for answer on how much protein you should be eating.
I’ll admit, i aimed for 1g the protein per pound of body load for a an extremely long time (which would be 185g the protein because that me), and also I even bumped it up to 1.5g per pound of bodyweight at one point. That supposed that i was eating 275g that protein per day in ~ one point!
2 things around that…
It is very, very expensiveIt will certainly wreak havoc on your body. To placed it much less delicately: it will provide you the worst gas of your life.
You DEFINITELY do not require that much.
Since ns dropped my protein intake a bit, my progress in the gym has actually improved!
Why is that?
Because my human body is now getting extra carbs and fats (since my protein levels space lower), and it has actually my human body feeling lot better! not to mention, considerably less gassy.
### Aim for 0.6g – 0.8g per lb of bodyweight
You might be thinking “but Matt, 0.8g is usually 1g”. Technically speaking, you’d be correct.
But if you sweet 150 pounds, that’s a distinction of 30g the protein.
Over the course of a day, you may uncover it considerably easier come consume 120g the protein than 150g.
1g per pound of bodyweight absolutely works, and I’m personally fine through that number if that’s the course you want to go.
It renders for much easier math, ~ all.
Some world talk about the threats of consuming too lot protein, but 1g per lb of bodyweight is no going come be everywhere close to dangerous levels.
If you feel far better eating that much, walk for it! however understand the you don’t need to consume that much.
When you’re in a calorie deficit, it’s absolutely wise to keep your protein intake high, therefore the 1g per pound of bodyweight can definitely work in those situations.
I choose to collection ranges because that myself, since I know that every solitary day is going come be slightly different. Fairly than stressing around hitting one certain number, I uncover it much much more beneficial to aim for a variety to stay within.
If you’re in a maintain of muscle building phase, you have the right to let your protein fall slightly reduced to obtain some extra carbohydrate in to help your energy and also performance. If you’re dieting and also in a deficit, I extremely recommend difficult to the greater end that that variety (even approximately 1g per pound if you’d like).
Be certain to examine out my cost-free calorie & macro calculator to get my recommendations!
## Why should we care around eating high quantities of protein?
Protein is not only necessary for daily living, yet hugely useful if you have certain goals in mind.
Yes, you deserve to have “too much” protein, but it would certainly take huge amounts continuously over a long duration of time- something that the majority of human being do not need to worry about.
Let’s look at why protein is so important:
### For day-to-day living
When it involves everyday life, protein is vital for a healthy and balanced life.
Our bodily features rely top top protein, and everything from her immune mechanism to her digestive mechanism relies on protein.
Adequate protein helps with digestion, blood clotting, bone health, hormone production, and also much more.
### For building muscle
What you’ve heard 1000 time is true- protein is the building block the muscles.
If the score is to develop muscle, we recognize that you need to be in a calorie surplus (AKA you must eat greater calories 보다 you’re burning).
But if you’re eating a ton of calories through very small protein, it’s going to be lot harder to develop muscle and get stronger!
While you will certainly definitely be able to build part muscle there is no eating lots of protein, you’re do the job lot harder for you yourself by no maximizing her protein intake.
Even if muscle building isn’t the goal, watch the vault explanation for everyday life. Sufficient protein is important to make certain your human body is work optimally, therefore you definitely want to prioritize it.
### For weight loss
When you’re in a calorie deficit (or dieting) protein is the most important.
It’s encourage to boost your protein intake once you’re dieting, for a few reasons- most importantly, it allows you to organize onto as lot muscle as possible.
If you’re in a calorie deficit because that a lengthy time, you’re bound to lose some muscle together you lose weight, yet prioritizing protein will assist minimize the effect.
And ns don’t know around you, however I’d favor to hold onto as lot muscle as I deserve to to make certain I’m losing greatly fat!
Protein is also satiating, help you come stay full while eating less than you’re provided to, and the thermic impact of protein is the highest possible of all the macronutrients, which method it needs the most power to digest, burning much more calories.
Note the the thermic impact is very minimal, therefore don’t go eating protein reasoning you’re magically burn a ton the calories. But, every small bit helps!
Maybe you’re in reality aiming for roughly 100g of protein, or maybe you’re aiming for even more. Where execute you get all your protein from?
That answer largely depends on her diet, her lifestyle, and also your preferences.
So, let’s watch at some different choices for ya.
## “Fitspo” Protein Sources
The trouble is that as well many human being think the they have to eat the “fitspo” way to hit your protein goals.
Getting 100+ grams that protein doesn’t need to mean eating protein bars, protein shakes, protein cookies, etc.
But, it absolutely can!
I have absolutely no worry with protein bars or shakes, yet I don’t believe they need to be her sole resource of protein. If girlfriend eat a well-rounded diet, friend should be able to get many of protein in your day.
That gift said, I gain a protein bar every day. No, i don’tneed protein bars, but I really reap them.
I workout at an early stage in the morning and also like to have a little snack before I lift, and protein bars are perfect because that that.
If you struggle to hit your protein goals, high-protein commodities like this can certainly help. When there space thousands of alternatives out there, let’s look in ~ a couple of of them…
Premier Protein Vanilla Shake
160 Calories
3g Fat
4g Carbs
30g Protein
Kodiak Cakes Muffin power Cup
290 Calories
11g Fat
39g Carbs
10g Protein
ONE Maple Doughnut Protein Bar
220 Calories
8g Fat
23g Carbs
20g Protein
Lenny & Larry’s “The Boss!” Cookie
220 Calories
12g Fat
18g Carbs
18g Protein
Quest Nutrition Meat Lover’s Pizza (Half Pizza)
270 Calories
17g Fat
18g Carbs
22g Protein
### Overall nutrition for all of the “fitspo” options:
1,160 Calories
61g Fat
102g carbohydrate (28g Fiber)
100g Protein
While spend these assets will help you acquire extra protein right into your diet, it is also a an extremely expensive method of eating!
I’m every for enjoy it these foods if you reap them and can purchased it, but your wallet will thank you if her diet isn’t exclusively comprised of these products…
### The problem with the “Fitspo” means of eating
Most protein bars are just candy bars v extra protein. But we see them as healthy due to the fact that they’re dubbed protein bars.
Lenny & Larry’s finish Cookies are perceived as healthy protein cookies, but in reality, they’re just continuous cookies through some included protein.
There’s nothing wrong through eating a high-protein cookie, or a protein bar, or any other snack for the matter.
But over there is absolutely other wrong through consuming it instead of other foods items you enjoy, just since you assume this is a healthy option.
Look in ~ this nutrition to compare above. The right comes out on peak in practically every single category, except having slightly less fiber.
You deserve to eat flavored Greek yogurt, flavored almonds (the sriracha ones are amazing, by the way) and also a beer, and also come the end with an ext favorable nutrition (with the exact same amount the calories) than a single protein cookie.
Just because a food isn’t labeled together “healthy” or “protein whatever”, it doesn’t average it’s bad for you. You can pair a beer, which many view as empty calories, with some delicious, healthy snacks and still be fine within her goals.
Don’t allow your quest of a high-protein diet do you remote to whatever else over there is!
High-protein snacks have the right to be a an excellent way to aid you hit your protein goals, yet they are not miscellaneous you need to consume.
Don’t immediately grab a snack due to the fact that it has the word “protein” ~ above the label.
Learning to read and understand nutrition labels is the best gift friend can give yourself. Through relying on the marketing slang on the front of packages, you’re forced to think whatever companies want you come think.
By looking past those, you can build a diet based roughly what girlfriend like and what her goal are.
## Animal Product Protein Sources
Animal-derived products tend to be an extremely high protein.
There’s a reason why the classic bodybuilding diet is comprised of chicken and rice- because chicken is absolutely loaded v protein.
Most animal products are going come be really low carb, high protein, and also a moderate quantity of fat.
There are lots of instances of animal-derived protein sources that we won’t dive into here (steak, cheese, fish, to surname a few) however here space some good options…
8oz Chicken breast (Kirkland)
220 Calories
3g Fat
0g Carbs
46g Protein
3 Eggs
210 Calories
15g Fat
0g Carbs
18g Protein
Plain Nonfat Greek Yogurt (¾ Cup)
90 Calories
0g Fat
6g Carbs
16g Protein
Center reduced Bacon (2 Strips)
140 Calories
10g Fat
0g Carbs
10g Protein
1oz Beef Jerky
80 Calories
1g Fat
6g Carbs
10g Protein
### Overall Nutrition for every one of the animal options:
740 Calories
29g Fat
12g Carbs
100g Protein
Let’s it is in clear: I’m no advocating the carnivore diet here, and I don’t recommend the above as a complete day of eating!
We must all be eat plenty of fruits and vegetables in our diet.
But this is to display you simply how easily you can get 100g that protein right into your diet with reasonably small quantities of food.
I mean, for just 740 calories, it’s clean how easily you can pack some extra protein into your diet through animal-derived foods.
## Plant-Based Protein Sources
It’s not only animal products that space high protein- there room a ton that high protein plant-derived foods, too.
With so plenty of people transforming to plant-based diets, one of the major criticisms i hear is that it’s difficult to get enough protein once eating choose that.
While it may be true that it’s a bit an ext difficult, it’s far from impossible…
1 Cup black Beans
220 Calories
0g Fat
40g Carbs
14g Protein
Impossible ground Beef
240 Calories
14g Fat
9g Carbs
19g Protein
Lentils (½ Cup Dry)
200 Calories
1g Fat
46g Carbs
16g Protein
Dry-Roasted Edamame (2/3 Cup)
260 Calories
10g Fat
18g Carbs
28g Protein
Extra-Firm Tofu (6oz)
180 Calories
9g Fat
6g Carbs
18g Protein
Roasted Chickpeas (1oz)
110 Calories
2g Fat
19g Carbs
5g Protein
### Overall nutrition for every one of the tree options:
1,210 Calories
36g Fat
138g carbohydrate (46g Fiber)
100g Protein
Eating plant-based foods generally comes with greater carbs and much higher fiber overall.
If you’re vegan, be sure to move up her protein resources so you’re not always consuming super high amounts of fiber.
While fiber is good for you, too lot of it is walking to reason you some…discomfort.
If you’re trying to find some extr plant-based protein sources, below are rather a few more because that ya!
Nuts & nut butter might not it is in the highest possible in protein, but they’re a great way to make certain your plant-based diet is well-rounded buy giving some extra fat v very couple of carbs.
Outside the the options provided above, there room plenty that other foods items out there the can help you fight 100g that protein.
Some of them, prefer a regular loaf of bread, may surprise girlfriend by just just how much protein it in reality contains.
Sure, no one of these foods are packed with 50g that protein, but we’re looking past the evident protein-packed foods items like chicken, eggs, greek yogurt, tofu, etc.
These space all foods that ns personally enjoy, and also surely there space plenty much more that can get added to this list.
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Have a veggie burger and also some feta cheese sheathe up within a lavash, and you have a quick and easy 35g the protein!
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Plenty of civilization will tell friend to protect against bread, but depending on which form you go with, you deserve to be gaining in an extra 12g that protein by eating it! and also who doesn’t want bread in their life?
Crazy people, that’s who.
Hitting 100g the protein doesn’t average consuming food with just 20+ grams that protein.
If you add in foodstuffs throughout your day v a little bit the extra protein, that can include up quickly.
I mean, if you eat every little thing in the graphics above, you’ll be obtaining 137g the protein and only 1,500 calories! Granted, that would more than likely be tough to revolve these foods into well-rounded meals, however the point remains.
## Conclusion: there is no wrong way to get your protein
I’ve detailed tons the different locations to acquire protein right into your diet, however what i really want you to takeaway indigenous this post is this:
Eating high-protein is important, and also there is no wrong means to perform it.
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See more: Can You Suck Your Stomach In While Pregnant, 5 Practical Tips
And don’t forget the if you desire to calculate exactly how much protein you have to be spend in a day, you deserve to use my free Calorie & Macro Calculator. | 3,716 | 16,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.965818 |
https://oeis.org/A046148 | 1,582,237,174,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00075.warc.gz | 501,579,143 | 3,992 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A046148 Number of n-digit numbers with maximal multiplicative persistence A014553. 3
10, 1, 9, 12, 20, 2430, 5229, 7448, 282852, 88200, 8015040, 200676960, 2701775518, 24655323238, 15765750, 1715313600, 59049874884, 1112489914536, 14162129381400, 135917876094000, 1050596838951660, 6832549561749912, 38554260751029408, 193081920969057120 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Giovanni Resta, Table of n, a(n) for n = 1..50 N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98. Eric Weisstein's World of Mathematics, Multiplicative Persistence. MATHEMATICA mper[n_] := Block[{k=0, m=n}, While[m>9, k++; m = Times @@ IntegerDigits@ m]; k]; mxper = {1, 4, 5, 6, 7, 7, 8, 9, 9, 10, 10, 10}; multi[w_] := Total[w]!/Times @@ (w!); a[1]=10; a[n_] := Sum[ Total[ If[ mxper[[n]] == 1 + mper[Times @@ (Range[9]^#)], multi[#], 0] & /@ Permutations[p]], {p, IntegerPartitions[n, {9}, Range[0, n]]}]; Array[a, 12] (* Giovanni Resta, Sep 01 2018 *) CROSSREFS Cf. A014553, A046149, A046150. Sequence in context: A010184 A107830 A144261 * A231933 A164915 A010691 Adjacent sequences: A046145 A046146 A046147 * A046149 A046150 A046151 KEYWORD nonn,base AUTHOR EXTENSIONS a(8)-a(12) from Donovan Johnson, Mar 30 2010 a(13)-a(24) from Giovanni Resta, Aug 31 2018 STATUS approved
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Last modified February 20 17:04 EST 2020. Contains 332080 sequences. (Running on oeis4.) | 650 | 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} | 3.265625 | 3 | CC-MAIN-2020-10 | latest | en | 0.553045 |
https://www.easyteacherworksheets.com/math/geometry-locusfixeddist.html | 1,713,803,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00471.warc.gz | 664,472,942 | 6,548 | These worksheets will give your students practice with equations for the locus of different points.
#### A locus (in geometric terms) is a series of points that is determined by specific conditions. When we work with loci in math, it makes me feel like I am placing a pin in corkboard. We place that pin based on the instructions that are provided. In most cases, we will just moving up and down the y-axis or left and right across the x-axis. In these worksheets you will be the one placing the pin in the corkboard.
What is a locus at a fixed distance? A locus is the arrangement of all focuses which fulfill a specific condition. The locus at a fixed separation, d, from point P is a hover with the given point P as its inside and d as its span. The locus at a fixed separation, d, from a line m, is a couple of equal lines a good way off of d from line m and situated on either side of m. The locus is equidistant from two focuses. An and B is the opposite bisector of the line fragment joining the two focuses. The locus equidistant from two parallel lines, m1 and m2, is a line corresponding to both m1 and m2 and somewhere between them. These worksheets explain how to find the locus of two points at a fixed distance and write its equation.
# Print Locus at a Fixed Distance Worksheets
## Finding the Locus Lesson
This worksheet explains how to find the locus of two points at a fixed distance. A sample problem is solved, and two practice problems are provided.
## Worksheet
Students will write the equation of the locus described. Ten problems are provided.
## Practice
Students will describe the locus indicated. Example problem: A wooden block is 200 feet long and 40 feet wide. It is planned to cut it 28 feet from the center of the block. Describe where it will be cut. Ten problems are provided.
## Review and Practice
Students review how to find these measures. Here is a sample problem: A book-cover is 12 feet long and 6 feet wide. It is planned to color it 5 foot from the center of the book-cover. Describe where it will be painted. Six practice problems are provided.
## Quiz
Students will demonstrate their proficiency this skill. Example: A railway track is 125 feet long and 50 feet wide. It is planned to cover it with mud 30 feet from the center of the track. Describe where it will be covered with mud. Ten problems are provided.
## Check
Students will find the locus described and write its equation. Example: Describe the locus of points 6 units from the line y = -14. Three problems are provided, and space is included for students to copy the correct answer when given.
## Equation of the Locus Lesson
This worksheet explains how to find and write the equation of the locus. A sample problem is solved, and two practice problems are provided.
## Worksheet
Example: A tree A is 50 feet from another tree B. Shadowing range of A is 30 feet and that of B is 25 feet. Draw a diagram showing the areas where each tree shadows. Is any area shadowed by both the trees?
## Practice Worksheet
Given two points, students will find and write the equation of the locus. Ten problems are provided.
## Review and Practice
A workshop is located at the coordinates (5, 12) on a coordinate grid. The delivery service extends for 5 km. Write the equation of the locus which represents the outer edge of the delivery service area.
## Quiz
Students will demonstrate their proficiency in finding and writing the equation of the locus. Ten problems are provided.
## Skills Check
Students will find the locus and write its equation. Three problems are provided, and space is included for students to copy the correct answer when given. | 791 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-18 | latest | en | 0.944499 |
https://downloads.haskell.org/~ghc/9.2.6/docs/html/libraries/base-4.16.4.0/Data-Traversable.html | 1,718,413,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00300.warc.gz | 190,823,619 | 23,636 | base-4.16.4.0: Basic libraries
Data.Traversable
Description
Class of data structures that can be traversed from left to right, performing an action on each element. Instances are expected to satisfy the listed laws.
Synopsis
# The Traversable class
class (Functor t, Foldable t) => Traversable t where Source #
Functors representing data structures that can be transformed to structures of the same shape by performing an Applicative (or, therefore, Monad) action on each element from left to right.
A more detailed description of what same shape means, the various methods, how traversals are constructed, and example advanced use-cases can be found in the Overview section of Data.Traversable.
For the class laws see the Laws section of Data.Traversable.
Minimal complete definition
Methods
traverse :: Applicative f => (a -> f b) -> t a -> f (t b) Source #
Map each element of a structure to an action, evaluate these actions from left to right, and collect the results. For a version that ignores the results see traverse_.
#### Examples
Expand
Basic usage:
In the first two examples we show each evaluated action mapping to the output structure.
>>> traverse Just [1,2,3,4]
Just [1,2,3,4]
>>> traverse id [Right 1, Right 2, Right 3, Right 4]
Right [1,2,3,4]
In the next examples, we show that Nothing and Left values short circuit the created structure.
>>> traverse (const Nothing) [1,2,3,4]
Nothing
>>> traverse (\x -> if odd x then Just x else Nothing) [1,2,3,4]
Nothing
>>> traverse id [Right 1, Right 2, Right 3, Right 4, Left 0]
Left 0
sequenceA :: Applicative f => t (f a) -> f (t a) Source #
Evaluate each action in the structure from left to right, and collect the results. For a version that ignores the results see sequenceA_.
#### Examples
Expand
Basic usage:
For the first two examples we show sequenceA fully evaluating a a structure and collecting the results.
>>> sequenceA [Just 1, Just 2, Just 3]
Just [1,2,3]
>>> sequenceA [Right 1, Right 2, Right 3]
Right [1,2,3]
The next two example show Nothing and Just will short circuit the resulting structure if present in the input. For more context, check the Traversable instances for Either and Maybe.
>>> sequenceA [Just 1, Just 2, Just 3, Nothing]
Nothing
>>> sequenceA [Right 1, Right 2, Right 3, Left 4]
Left 4
mapM :: Monad m => (a -> m b) -> t a -> m (t b) Source #
Map each element of a structure to a monadic action, evaluate these actions from left to right, and collect the results. For a version that ignores the results see mapM_.
#### Examples
Expand
mapM is literally a traverse with a type signature restricted to Monad. Its implementation may be more efficient due to additional power of Monad.
sequence :: Monad m => t (m a) -> m (t a) Source #
Evaluate each monadic action in the structure from left to right, and collect the results. For a version that ignores the results see sequence_.
#### Examples
Expand
Basic usage:
The first two examples are instances where the input and and output of sequence are isomorphic.
>>> sequence $Right [1,2,3,4] [Right 1,Right 2,Right 3,Right 4] >>> sequence$ [Right 1,Right 2,Right 3,Right 4]
Right [1,2,3,4]
The following examples demonstrate short circuit behavior for sequence.
>>> sequence $Left [1,2,3,4] Left [1,2,3,4] >>> sequence$ [Left 0, Right 1,Right 2,Right 3,Right 4]
Left 0
#### Instances
Instances details
# Utility functions
for :: (Traversable t, Applicative f) => t a -> (a -> f b) -> f (t b) Source #
for is traverse with its arguments flipped. For a version that ignores the results see for_.
forM :: (Traversable t, Monad m) => t a -> (a -> m b) -> m (t b) Source #
forM is mapM with its arguments flipped. For a version that ignores the results see forM_.
mapAccumL :: forall t s a b. Traversable t => (s -> a -> (s, b)) -> s -> t a -> (s, t b) Source #
The mapAccumL function behaves like a combination of fmap and foldl; it applies a function to each element of a structure, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new structure.
#### Examples
Expand
Basic usage:
>>> mapAccumL (\a b -> (a + b, a)) 0 [1..10]
(55,[0,1,3,6,10,15,21,28,36,45])
>>> mapAccumL (\a b -> (a <> show b, a)) "0" [1..5]
("012345",["0","01","012","0123","01234"])
mapAccumR :: forall t s a b. Traversable t => (s -> a -> (s, b)) -> s -> t a -> (s, t b) Source #
The mapAccumR function behaves like a combination of fmap and foldr; it applies a function to each element of a structure, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new structure.
#### Examples
Expand
Basic usage:
>>> mapAccumR (\a b -> (a + b, a)) 0 [1..10]
(55,[54,52,49,45,40,34,27,19,10,0])
>>> mapAccumR (\a b -> (a <> show b, a)) "0" [1..5]
("054321",["05432","0543","054","05","0"])
# General definitions for superclass methods
fmapDefault :: forall t a b. Traversable t => (a -> b) -> t a -> t b Source #
This function may be used as a value for fmap in a Functor instance, provided that traverse is defined. (Using fmapDefault with a Traversable instance defined only by sequenceA will result in infinite recursion.)
fmapDefault f ≡ runIdentity . traverse (Identity . f)
foldMapDefault :: forall t m a. (Traversable t, Monoid m) => (a -> m) -> t a -> m Source #
This function may be used as a value for foldMap in a Foldable instance.
foldMapDefault f ≡ getConst . traverse (Const . f)
# Overview
Traversable structures support element-wise sequencing of Applicative effects (thus also Monad effects) to construct new structures of the same shape as the input.
To illustrate what is meant by same shape, if the input structure is [a], each output structure is a list [b] of the same length as the input. If the input is a Tree a, each output Tree b has the same graph of intermediate nodes and leaves. Similarly, if the input is a 2-tuple (x, a), each output is a 2-tuple (x, b), and so forth.
It is in fact possible to decompose a traversable structure t a into its shape (a.k.a. spine) of type t () and its element list [a]. The original structure can be faithfully reconstructed from its spine and element list.
The implementation of a Traversable instance for a given structure follows naturally from its type; see the Construction section for details. Instances must satisfy the laws listed in the Laws section. The diverse uses of Traversable structures result from the many possible choices of Applicative effects. See the Advanced Traversals section for some examples.
Every Traversable structure is both a Functor and Foldable because it is possible to implement the requisite instances in terms of traverse by using fmapDefault for fmap and foldMapDefault for foldMap. Direct fine-tuned implementations of these superclass methods can in some cases be more efficient.
## The traverse and mapM methods
For an Applicative functor f and a Traversable functor t, the type signatures of traverse and fmap are rather similar:
fmap :: (a -> f b) -> t a -> t (f b)
traverse :: (a -> f b) -> t a -> f (t b)
The key difference is that fmap produces a structure whose elements (of type f b) are individual effects, while traverse produces an aggregate effect yielding structures of type t b.
For example, when f is the IO monad, and t is List, fmap yields a list of IO actions, whereas traverse constructs an IO action that evaluates to a list of the return values of the individual actions performed left-to-right.
traverse :: (a -> IO b) -> [a] -> IO [b]
The mapM function is a specialisation of traverse to the case when f is a Monad. For monads, mapM is more idiomatic than traverse. The two are otherwise generally identical (though mapM may be specifically optimised for monads, and could be more efficient than using the more general traverse).
traverse :: (Applicative f, Traversable t) => (a -> f b) -> t a -> f (t b)
mapM :: (Monad m, Traversable t) => (a -> m b) -> t a -> m (t b)
When the traversable term is a simple variable or expression, and the monadic action to run is a non-trivial do block, it can be more natural to write the action last. This idiom is supported by for and forM, which are the flipped versions of traverse and mapM, respectively.
### Their Foldable, just the effects, analogues.
The traverse and mapM methods have analogues in the Data.Foldable module. These are traverse_ and mapM_, and their flipped variants for_ and forM_, respectively. The result type is f (), they don't return an updated structure, and can be used to sequence effects over all the elements of a Traversable (any Foldable) structure just for their side-effects.
If the Traversable structure is empty, the result is pure (). When effects short-circuit, the f () result may, for example, be Nothing if f is Maybe, or Left e when it is Either e.
It is perhaps worth noting that Maybe is not only a potential Applicative functor for the return value of the first argument of traverse, but is also itself a Traversable structure with either zero or one element. A convenient idiom for conditionally executing an action just for its effects on a Just value, and doing nothing otherwise is:
-- action :: Monad m => a -> m ()
-- mvalue :: Maybe a
mapM_ action mvalue -- :: m ()
which is more concise than:
maybe (return ()) action mvalue
The mapM_ idiom works verbatim if the type of mvalue is later refactored from Maybe a to Either e a (assuming it remains OK to silently do nothing in the Left case).
### Result multiplicity
When traverse or mapM is applied to an empty structure ts (one for which null ts is True) the return value is pure ts regardless of the provided function g :: a -> f b. It is not possible to apply the function when no values of type a are available, but its type determines the relevant instance of pure.
null ts ==> traverse g ts == pure ts
Otherwise, when ts is non-empty and at least one value of type b results from each f a, the structures t b have the same shape (list length, graph of tree nodes, ...) as the input structure t a, but the slots previously occupied by elements of type a now hold elements of type b.
A single traversal may produce one, zero or many such structures. The zero case happens when one of the effects f a sequenced as part of the traversal yields no replacement values. Otherwise, the many case happens when one of sequenced effects yields multiple values.
The traverse function does not perform selective filtering of slots in the output structure as with e.g. mapMaybe.
>>> let incOdd n = if odd n then Just $n + 1 else Nothing >>> mapMaybe incOdd [1, 2, 3] [2,4] >>> traverse incOdd [1, 3, 5] Just [2,4,6] >>> traverse incOdd [1, 2, 3] Nothing In the above examples, with Maybe as the Applicative f, we see that the number of t b structures produced by traverse may differ from one: it is zero when the result short-circuits to Nothing. The same can happen when f is List and the result is [], or f is Either e and the result is Left (x :: e), or perhaps the empty value of some Alternative functor. When f is e.g. List, and the map g :: a -> [b] returns more than one value for some inputs a (and at least one for all a), the result of mapM g ts will contain multiple structures of the same shape as ts: length (mapM g ts) == product (fmap (length . g) ts) For example: >>> length$ mapM (\n -> [1..n]) [1..6]
720
>>> product $length . (\n -> [1..n]) <$> [1..6]
720
In other words, a traversal with a function g :: a -> [b], over an input structure t a, yields a list [t b], whose length is the product of the lengths of the lists that g returns for each element of the input structure! The individual elements a of the structure are replaced by each element of g a in turn:
>>> mapM (\n -> [1..n]) $Just 3 [Just 1,Just 2,Just 3] >>> mapM (\n -> [1..n]) [1..3] [[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3]] If any element of the structure t a is mapped by g to an empty list, then the entire aggregate result is empty, because no value is available to fill one of the slots of the output structure: >>> mapM (\n -> [1..n])$ [0..6] -- [1..0] is empty
[]
## The sequenceA and sequence methods
The sequenceA and sequence methods are useful when what you have is a container of pending applicative or monadic effects, and you want to combine them into a single effect that produces zero or more containers with the computed values.
sequenceA :: (Applicative f, Traversable t) => t (f a) -> f (t a)
sequence :: (Monad m, Traversable t) => t (m a) -> m (t a)
sequenceA = traverse id -- default definition
sequence = sequenceA -- default definition
When the monad m is IO, applying sequence to a list of IO actions, performs each in turn, returning a list of the results:
sequence [putStr "Hello ", putStrLn "World!"]
= (\a b -> [a,b]) <$> putStr "Hello " <*> putStrLn "World!" = do u1 <- putStr "Hello " u2 <- putStrLn "World!" return [u1, u2] -- In this case [(), ()] For sequenceA, the non-deterministic behaviour of List is most easily seen in the case of a list of lists (of elements of some common fixed type). The result is a cross-product of all the sublists: >>> sequenceA [[0, 1, 2], [30, 40], [500]] [[0,30,500],[0,40,500],[1,30,500],[1,40,500],[2,30,500],[2,40,500]] Because the input list has three (sublist) elements, the result is a list of triples (same shape). ### Care with default method implementations The traverse method has a default implementation in terms of sequenceA: traverse g = sequenceA . fmap g but relying on this default implementation is not recommended, it requires that the structure is already independently a Functor. The definition of sequenceA in terms of traverse id is much simpler than traverse expressed via a composition of sequenceA and fmap. Instances should generally implement traverse explicitly. It may in some cases also make sense to implement a specialised mapM. Because fmapDefault is defined in terms of traverse (whose default definition in terms of sequenceA uses fmap), you must not use fmapDefault to define the Functor instance if the Traversable instance directly defines only sequenceA. ### Monadic short circuits When the monad m is Either or Maybe (more generally any MonadPlus), the effect in question is to short-circuit the result on encountering Left or Nothing (more generally mzero). >>> sequence [Just 1,Just 2,Just 3] Just [1,2,3] >>> sequence [Just 1,Nothing,Just 3] Nothing >>> sequence [Right 1,Right 2,Right 3] Right [1,2,3] >>> sequence [Right 1,Left "sorry",Right 3] Left "sorry" The result of sequence is all-or-nothing, either structures of exactly the same shape as the input or none at all. The sequence function does not perform selective filtering as with e.g. catMaybes or rights: >>> catMaybes [Just 1,Nothing,Just 3] [1,3] >>> rights [Right 1,Left "sorry",Right 3] [1,3] ## Example binary tree instance The definition of a Traversable instance for a binary tree is rather similar to the corresponding instance of Functor, given the data type: data Tree a = Empty | Leaf a | Node (Tree a) a (Tree a) a canonical Functor instance would be instance Functor Tree where fmap g Empty = Empty fmap g (Leaf x) = Leaf (g x) fmap g (Node l k r) = Node (fmap g l) (g k) (fmap g r) a canonical Traversable instance would be instance Traversable Tree where traverse g Empty = pure Empty traverse g (Leaf x) = Leaf <$> g x
traverse g (Node l k r) = Node <$> traverse g l <*> g k <*> traverse g r This definition works for any g :: a -> f b, with f an Applicative functor, as the laws for (<*>) imply the requisite associativity. We can add an explicit non-default mapM if desired: mapM g Empty = return Empty mapM g (Leaf x) = Leaf <$> g x
mapM g (Node l k r) = do
ml <- mapM g l
mk <- g k
mr <- mapM g r
return $Node ml mk mr See Construction below for a more detailed exploration of the general case, but as mentioned in Overview above, instance definitions are typically rather simple, all the interesting behaviour is a result of an interesting choice of Applicative functor for a traversal. ### Pre-order and post-order tree traversal It is perhaps worth noting that the traversal defined above gives an in-order sequencing of the elements. If instead you want either pre-order (parent first, then child nodes) or post-order (child nodes first, then parent) sequencing, you can define the instance accordingly: inOrderNode :: Tree a -> a -> Tree a -> Tree a inOrderNode l x r = Node l x r preOrderNode :: a -> Tree a -> Tree a -> Tree a preOrderNode x l r = Node l x r postOrderNode :: Tree a -> Tree a -> a -> Tree a postOrderNode l r x = Node l x r -- Traversable instance with in-order traversal instance Traversable Tree where traverse g t = case t of Empty -> pure Empty Leaf x -> Leaf <$> g x
Node l x r -> inOrderNode <$> traverse g l <*> g x <*> traverse g r -- Traversable instance with pre-order traversal instance Traversable Tree where traverse g t = case t of Empty -> pure Empty Leaf x -> Leaf <$> g x
Node l x r -> preOrderNode <$> g x <*> traverse g l <*> traverse g r -- Traversable instance with post-order traversal instance Traversable Tree where traverse g t = case t of Empty -> pure Empty Leaf x -> Leaf <$> g x
Node l x r -> postOrderNode <$> traverse g l <*> traverse g r <*> g x Since the same underlying Tree structure is used in all three cases, it is possible to use newtype wrappers to make all three available at the same time! The user need only wrap the root of the tree in the appropriate newtype for the desired traversal order. Tne associated instance definitions are shown below (see coercion if unfamiliar with the use of coerce in the sample code): {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} -- Default in-order traversal import Data.Coerce (coerce) import Data.Traversable data Tree a = Empty | Leaf a | Node (Tree a) a (Tree a) instance Functor Tree where fmap = fmapDefault instance Foldable Tree where foldMap = foldMapDefault instance Traversable Tree where traverse _ Empty = pure Empty traverse g (Leaf a) = Leaf <$> g a
traverse g (Node l a r) = Node <$> traverse g l <*> g a <*> traverse g r -- Optional pre-order traversal newtype PreOrderTree a = PreOrderTree (Tree a) instance Functor PreOrderTree where fmap = fmapDefault instance Foldable PreOrderTree where foldMap = foldMapDefault instance Traversable PreOrderTree where traverse _ (PreOrderTree Empty) = pure$ preOrderEmpty
traverse g (PreOrderTree (Leaf x)) = preOrderLeaf <$> g x traverse g (PreOrderTree (Node l x r)) = preOrderNode <$> g x
<*> traverse g (coerce l)
<*> traverse g (coerce r)
preOrderEmpty :: forall a. PreOrderTree a
preOrderEmpty = coerce (Empty @a)
preOrderLeaf :: forall a. a -> PreOrderTree a
preOrderLeaf = coerce (Leaf @a)
preOrderNode :: a -> PreOrderTree a -> PreOrderTree a -> PreOrderTree a
preOrderNode x l r = coerce (Node (coerce l) x (coerce r))
-- Optional post-order traversal
newtype PostOrderTree a = PostOrderTree (Tree a)
instance Functor PostOrderTree where fmap = fmapDefault
instance Foldable PostOrderTree where foldMap = foldMapDefault
instance Traversable PostOrderTree where
traverse _ (PostOrderTree Empty) = pure postOrderEmpty
traverse g (PostOrderTree (Leaf x)) = postOrderLeaf <$> g x traverse g (PostOrderTree (Node l x r)) = postOrderNode <$> traverse g (coerce l)
<*> traverse g (coerce r)
<*> g x
postOrderEmpty :: forall a. PostOrderTree a
postOrderEmpty = coerce (Empty @a)
postOrderLeaf :: forall a. a -> PostOrderTree a
postOrderLeaf = coerce (Leaf @a)
postOrderNode :: PostOrderTree a -> PostOrderTree a -> a -> PostOrderTree a
postOrderNode l r x = coerce (Node (coerce l) x (coerce r))
With the above, given a sample tree:
inOrder :: Tree Int
inOrder = Node (Node (Leaf 10) 3 (Leaf 20)) 5 (Leaf 42)
we have:
import Data.Foldable (toList)
print $toList inOrder [10,3,20,5,42] print$ toList (coerce inOrder :: PreOrderTree Int)
[5,3,10,20,42]
print $toList (coerce inOrder :: PostOrderTree Int) [10,20,3,42,5] You would typically define instances for additional common type classes, such as Eq, Ord, Show, etc. ## Making construction intuitive In order to be able to reason about how a given type of Applicative effects will be sequenced through a general Traversable structure by its traversable and related methods, it is helpful to look more closely at how a general traverse method is implemented. We'll look at how general traversals are constructed primarily with a view to being able to predict their behaviour as a user, even if you're not defining your own Traversable instances. Traversable structures t a are assembled incrementally from their constituent parts, perhaps by prepending or appending individual elements of type a, or, more generally, by recursively combining smaller composite traversable building blocks that contain multiple such elements. As in the tree example above, the components being combined are typically pieced together by a suitable constructor, i.e. a function taking two or more arguments that returns a composite value. The traverse method enriches simple incremental construction with threading of Applicative effects of some function g :: a -> f b. The basic building blocks we'll use to model the construction of traverse are a hypothetical set of elementary functions, some of which may have direct analogues in specific Traversable structures. For example, the (:) constructor is an analogue for lists of prepend or the more general combine. empty :: t a -- build an empty container singleton :: a -> t a -- build a one-element container prepend :: a -> t a -> t a -- extend by prepending a new initial element append :: t a -> a -> t a -- extend by appending a new final element combine :: a1 -> a2 -> ... -> an -> t a -- combine multiple inputs • An empty structure has no elements of type a, so there's nothing to which g can be applied, but since we need an output of type f (t b), we just use the pure instance of f to wrap an empty of type t b: traverse _ (empty :: t a) = pure (empty :: t b) With the List monad, empty is [], while with Maybe it is Nothing. With Either e a we have an empty case for each value of e: traverse _ (Left e :: Either e a) = pure$ (Left e :: Either e b)
• A singleton structure has just one element of type a, and traverse can take that a, apply g :: a -> f b getting an f b, then fmap singleton over that, getting an f (t b) as required:
traverse g (singleton a) = fmap singleton $g a Note that if f is List and g returns multiple values the result will be a list of multiple t b singletons! Since Maybe and Either are either empty or singletons, we have traverse _ Nothing = pure Nothing traverse g (Just a) = Just <$> g a
traverse _ (Left e) = pure (Left e)
traverse g (Right a) = Right <$> g a For List, empty is [] and singleton is (:[]), so we have: traverse _ [] = pure [] traverse g [a] = fmap (:[]) (g a) = (:) <$> (g a) <*> traverse g []
= liftA2 (:) (g a) (traverse g [])
• When the structure is built by adding one more element via prepend or append, traversal amounts to:
traverse g (prepend a t0) = prepend <$> (g a) <*> traverse g t0 = liftA2 prepend (g a) (traverse g t0) traverse g (append t0 a) = append <$> traverse g t0 <*> g a
= liftA2 append (traverse g t0) (g a)
The origin of the combinatorial product when f is List should now be apparent, when traverse g t0 has n elements and g a has m elements, the non-deterministic Applicative instance of List will produce a result with m * n elements.
• When combining larger building blocks, we again use (<*>) to combine the traversals of the components. With bare elements a mapped to f b via g, and composite traversable sub-structures transformed via traverse g:
traverse g (combine a1 a2 ... an) =
combine <$> t1 <*> t2 <*> ... <*> tn where t1 = g a1 -- if a1 fills a slot of type @a@ = traverse g a1 -- if a1 is a traversable substructure ... ditto for the remaining constructor arguments ... The above definitions sequence the Applicative effects of f in the expected order while producing results of the expected shape t. For lists this becomes: traverse g [] = pure [] traverse g (x:xs) = liftA2 (:) (g a) (traverse g xs) The actual definition of traverse for lists is an equivalent right fold in order to facilitate list fusion. traverse g = foldr (\x r -> liftA2 (:) (g x) r) (pure []) # Advanced traversals In the sections below we'll examine some advanced choices of Applicative effects that give rise to very different transformations of Traversable structures. These examples cover the implementations of fmapDefault, foldMapDefault, mapAccumL and mapAccumR functions illustrating the use of Identity, Const and stateful Applicative effects. The ZipList example illustrates the use of a less-well known Applicative instance for lists. This is optional material, which is not essential to a basic understanding of Traversable structures. If this is your first encounter with Traversable structures, you can come back to these at a later date. ### Coercion Some of the examples make use of an advanced Haskell feature, namely newtype coercion. This is done for two reasons: • Use of coerce makes it possible to avoid cluttering the code with functions that wrap and unwrap newtype terms, which at runtime are indistinguishable from the underlying value. Coercion is particularly convenient when one would have to otherwise apply multiple newtype constructors to function arguments, and then peel off multiple layers of same from the function output. • Use of coerce can produce more efficient code, by reusing the original value, rather than allocating space for a wrapped clone. If you're not familiar with coerce, don't worry, it is just a shorthand that, e.g., given: newtype Foo a = MkFoo { getFoo :: a } newtype Bar a = MkBar { getBar :: a } newtype Baz a = MkBaz { getBaz :: a } f :: Baz Int -> Bar (Foo String) makes it possible to write: x :: Int -> String x = coerce f instead of x = getFoo . getBar . f . MkBaz ## Identity: the fmapDefault function The simplest Applicative functor is Identity, which just wraps and unwraps pure values and function application. This allows us to define fmapDefault: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} import Data.Coercible (coerce) fmapDefault :: forall t a b. Traversable t => (a -> b) -> t a -> t b fmapDefault = coerce (traverse @t @Identity @a @b) The use of coercion avoids the need to explicitly wrap and unwrap terms via Identity and runIdentity. As noted in Overview, fmapDefault can only be used to define the requisite Functor instance of a Traversable structure when the traverse method is explicitly implemented. An infinite loop would result if in addition traverse were defined in terms of sequenceA and fmap. ## State: the mapAccumL, mapAccumR functions Applicative functors that thread a changing state through a computation are an interesting use-case for traverse. The mapAccumL and mapAccumR functions in this module are each defined in terms of such traversals. We first define a simplified (not a monad transformer) version of State that threads a state s through a chain of computations left to right. Its (<*>) operator passes the input state first to its left argument, and then the resulting state is passed to its right argument, which returns the final state. newtype StateL s a = StateL { runStateL :: s -> (s, a) } instance Functor (StateL s) where fmap f (StateL kx) = StateL$ \ s ->
let (s', x) = kx s in (s', f x)
instance Applicative (StateL s) where
pure a = StateL $\s -> (s, a) (StateL kf) <*> (StateL kx) = StateL$ \ s ->
let { (s', f) = kf s
; (s'', x) = kx s' } in (s'', f x)
liftA2 f (StateL kx) (StateL ky) = StateL $\ s -> let { (s', x) = kx s ; (s'', y) = ky s' } in (s'', f x y) With StateL, we can define mapAccumL as follows: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} mapAccumL :: forall t s a b. Traversable t => (s -> a -> (s, b)) -> s -> t a -> (s, t b) mapAccumL g s ts = coerce (traverse @t @(StateL s) @a @b) (flip g) ts s The use of coercion avoids the need to explicitly wrap and unwrap newtype terms. The type of flip g is coercible to a -> StateL b, which makes it suitable for use with traverse. As part of the Applicative construction of StateL (t b) the state updates will thread left-to-right along the sequence of elements of t a. While mapAccumR has a type signature identical to mapAccumL, it differs in the expected order of evaluation of effects, which must take place right-to-left. For this we need a variant control structure StateR, which threads the state right-to-left, by passing the input state to its right argument and then using the resulting state as an input to its left argument: newtype StateR s a = StateR { runStateR :: s -> (s, a) } instance Functor (StateR s) where fmap f (StateR kx) = StateR$ \s ->
let (s', x) = kx s in (s', f x)
instance Applicative (StateR s) where
pure a = StateR $\s -> (s, a) (StateR kf) <*> (StateR kx) = StateR$ \ s ->
let { (s', x) = kx s
; (s'', f) = kf s' } in (s'', f x)
liftA2 f (StateR kx) (StateR ky) = StateR $\ s -> let { (s', y) = ky s ; (s'', x) = kx s' } in (s'', f x y) With StateR, we can define mapAccumR as follows: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} mapAccumR :: forall t s a b. Traversable t => (s -> a -> (s, b)) -> s -> t a -> (s, t b) mapAccumR g s0 ts = coerce (traverse @t @(StateR s) @a @b) (flip g) ts s0 The use of coercion avoids the need to explicitly wrap and unwrap newtype terms. Various stateful traversals can be constructed from mapAccumL and mapAccumR for suitable choices of g, or built directly along similar lines. ## Const: the foldMapDefault function The Const Functor enables applications of traverse that summarise the input structure to an output value without constructing any output values of the same type or shape. As noted above, the Foldable superclass constraint is justified by the fact that it is possible to construct foldMap, foldr, etc., from traverse. The technique used is useful in its own right, and is explored below. A key feature of folds is that they can reduce the input structure to a summary value. Often neither the input structure nor a mutated clone is needed once the fold is computed, and through list fusion the input may not even have been memory resident in its entirety at the same time. The traverse method does not at first seem to be a suitable building block for folds, because its return value f (t b) appears to retain mutated copies of the input structure. But the presence of t b in the type signature need not mean that terms of type t b are actually embedded in f (t b). The simplest way to elide the excess terms is by basing the Applicative functor used with traverse on Const. Not only does Const a b hold just an a value, with the b parameter merely a phantom type, but when m has a Monoid instance, Const m is an Applicative functor: import Data.Coerce (coerce) newtype Const a b = Const { getConst :: a } deriving (Eq, Ord, Show) -- etc. instance Functor (Const m) where fmap = const coerce instance Monoid m => Applicative (Const m) where pure _ = Const mempty (<*>) = coerce (mappend :: m -> m -> m) liftA2 _ = coerce (mappend :: m -> m -> m) The use of coercion avoids the need to explicitly wrap and unwrap newtype terms. We can therefore define a specialisation of traverse: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} traverseC :: forall t a m. (Monoid m, Traversable t) => (a -> Const m ()) -> t a -> Const m (t ()) traverseC = traverse @t @(Const m) @a @() For which the Applicative construction of traverse leads to: null ts ==> traverseC g ts = Const mempty traverseC g (prepend x xs) = Const (g x) <> traverseC g xs In other words, this makes it possible to define: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} foldMapDefault :: forall t a m. (Monoid m, Traversable t) => (a -> m) -> t a -> m foldMapDefault = coerce (traverse @t @(Const m) @a @()) Which is sufficient to define a Foldable superclass instance: The use of coercion avoids the need to explicitly wrap and unwrap newtype terms. instance Traversable t => Foldable t where foldMap = foldMapDefault It may however be instructive to also directly define candidate default implementations of foldr and foldl', which take a bit more machinery to construct: {-# LANGUAGE ScopedTypeVariables, TypeApplications #-} import Data.Coerce (coerce) import Data.Functor.Const (Const(..)) import Data.Semigroup (Dual(..), Endo(..)) import GHC.Exts (oneShot) foldrDefault :: forall t a b. Traversable t => (a -> b -> b) -> b -> t a -> b foldrDefault f z = \t -> coerce (traverse @t @(Const (Endo b)) @a @()) f t z foldlDefault' :: forall t a b. Traversable t => (b -> a -> b) -> b -> t a -> b foldlDefault' f z = \t -> coerce (traverse @t @(Const (Dual (Endo b))) @a @()) f' t z where f' :: a -> b -> b f' a = oneShot$ \ b -> b seq f b a
In the above we're using the Endo b Monoid and its Dual to compose a sequence of b -> b accumulator updates in either left-to-right or right-to-left order.
The use of seq in the definition of foldlDefault' ensures strictness in the accumulator.
The use of coercion avoids the need to explicitly wrap and unwrap newtype terms.
The oneShot function gives a hint to the compiler that aids in correct optimisation of lambda terms that fire at most once (for each element a) and so should not try to pre-compute and re-use subexpressions that pay off only on repeated execution. Otherwise, it is just the identity function.
## ZipList: transposing lists of lists
As a warm-up for looking at the ZipList Applicative functor, we'll first look at a simpler analogue. First define a fixed width 2-element Vec2 type, whose Applicative instance combines a pair of functions with a pair of values by applying each function to the corresponding value slot:
data Vec2 a = Vec2 a a
instance Functor Vec2 where
fmap f (Vec2 a b) = Vec2 (f a) (f b)
instance Applicative Vec2 where
pure x = Vec2 x x
liftA2 f (Vec2 a b) (Vec2 p q) = Vec2 (f a p) (f b q)
instance Foldable Vec2 where
foldr f z (Vec2 a b) = f a (f b z)
foldMap f (Vec2 a b) = f a <> f b
instance Traversable Vec2 where
traverse f (Vec2 a b) = Vec2 <$> f a <*> f b Along with a similar definition for fixed width 3-element vectors: data Vec3 a = Vec3 a a a instance Functor Vec3 where fmap f (Vec3 x y z) = Vec3 (f x) (f y) (f z) instance Applicative Vec3 where pure x = Vec3 x x x liftA2 f (Vec3 p q r) (Vec3 x y z) = Vec3 (f p x) (f q y) (f r z) instance Foldable Vec3 where foldr f z (Vec3 a b c) = f a (f b (f c z)) foldMap f (Vec3 a b c) = f a <> f b <> f c instance Traversable Vec3 where traverse f (Vec3 a b c) = Vec3 <$> f a <*> f b <*> f c
With the above definitions, sequenceA (same as traverse id) acts as a matrix transpose operation on Vec2 (Vec3 Int) producing a corresponding Vec3 (Vec2 Int):
Let t = Vec2 (Vec3 1 2 3) (Vec3 4 5 6) be our Traversable structure, and g = id :: Vec3 Int -> Vec3 Int be the function used to traverse t. We then have:
traverse g t = Vec2 <$> (Vec3 1 2 3) <*> (Vec3 4 5 6) = Vec3 (Vec2 1 4) (Vec2 2 5) (Vec2 3 6) This construction can be generalised from fixed width vectors to variable length lists via ZipList. This gives a transpose operation that works well for lists of equal length. If some of the lists are longer than others, they're truncated to the longest common length. We've already looked at the standard Applicative instance of List for which applying m functions f1, f2, ..., fm to n input values a1, a2, ..., an produces m * n outputs: >>> :set -XTupleSections >>> [("f1",), ("f2",), ("f3",)] <*> [1,2] [("f1",1),("f1",2),("f2",1),("f2",2),("f3",1),("f3",2)] There are however two more common ways to turn lists into Applicative control structures. The first is via Const [a], since lists are monoids under concatenation, and we've already seen that Const m is an Applicative functor when m is a Monoid. The second, is based on zipWith, and is called ZipList: {-# LANGUAGE GeneralizedNewtypeDeriving #-} newtype ZipList a = ZipList { getZipList :: [a] } deriving (Show, Eq, ..., Functor) instance Applicative ZipList where liftA2 f (ZipList xs) (ZipList ys) = ZipList$ zipWith f xs ys
pure x = repeat x
The liftA2 definition is clear enough, instead of applying f to each pair (x, y) drawn independently from the xs and ys, only corresponding pairs at each index in the two lists are used.
The definition of pure may look surprising, but it is needed to ensure that the instance is lawful:
liftA2 f (pure x) ys == fmap (f x) ys
Since ys can have any length, we need to provide an infinite supply of x values in pure x in order to have a value to pair with each element y.
When ZipList is the Applicative functor used in the construction of a traversal, a ZipList holding a partially built structure with m elements is combined with a component holding n elements via zipWith, resulting in min m n outputs!
Therefore traverse with g :: a -> ZipList b will produce a ZipList of t b structures whose element count is the minimum length of the ZipLists g a with a ranging over the elements of t. When t is empty, the length is infinite (as expected for a minimum of an empty set).
If the structure t holds values of type ZipList a, we can use the identity function id :: ZipList a -> ZipList a for the first argument of traverse:
traverse (id :: ZipList a -> ZipList a) :: t (ZipList a) -> ZipList (t a)
The number of elements in the output ZipList will be the length of the shortest ZipList element of t. Each output t a will have the same shape as the input t (ZipList a), i.e. will share its number of elements.
If we think of the elements of t (ZipList a) as its rows, and the elements of each individual ZipList as the columns of that row, we see that our traversal implements a transpose operation swapping the rows and columns of t, after first truncating all the rows to the column count of the shortest one.
Since in fact traverse id is just sequenceA the above boils down to a rather concise definition of transpose, with coercion used to implicily wrap and unwrap the ZipList newtype as neeed, giving a function that operates on a list of lists:
>>> {-# LANGUAGE ScopedTypeVariables #-}
>>> import Control.Applicative (ZipList(..))
>>> import Data.Coerce (coerce)
>>>
>>> transpose :: forall a. [[a]] -> [[a]]
>>> transpose = coerce (sequenceA :: [ZipList a] -> ZipList [a])
>>>
>>> transpose [[1,2,3],[4..],[7..]]
[[1,4,7],[2,5,8],[3,6,9]]
The use of coercion avoids the need to explicitly wrap and unwrap ZipList terms.
# Laws
A definition of traverse must satisfy the following laws:
Naturality
t . traverse f = traverse (t . f) for every applicative transformation t
Identity
traverse Identity = Identity
Composition
traverse (Compose . fmap g . f) = Compose . fmap (traverse g) . traverse f
A definition of sequenceA must satisfy the following laws:
Naturality
t . sequenceA = sequenceA . fmap t for every applicative transformation t
Identity
sequenceA . fmap Identity = Identity
Composition
sequenceA . fmap Compose = Compose . fmap sequenceA . sequenceA
where an applicative transformation is a function
t :: (Applicative f, Applicative g) => f a -> g a
preserving the Applicative operations, i.e.
t (pure x) = pure x
t (f <*> x) = t f <*> t x
and the identity functor Identity and composition functors Compose are from Data.Functor.Identity and Data.Functor.Compose.
A result of the naturality law is a purity law for traverse
traverse pure = pure
(The naturality law is implied by parametricity and thus so is the purity law [1, p15].)
The superclass instances should satisfy the following:
• In the Functor instance, fmap should be equivalent to traversal with the identity applicative functor (fmapDefault).
• In the Foldable instance, foldMap should be equivalent to traversal with a constant applicative functor (foldMapDefault).
Note: the Functor superclass means that (in GHC) Traversable structures cannot impose any constraints on the element type. A Haskell implementation that supports constrained functors could make it possible to define constrained Traversable structures. | 10,272 | 40,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.799398 |
https://bookdown.org/content/3890/missing-data-and-other-opportunities.html | 1,716,154,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00472.warc.gz | 120,416,678 | 50,418 | # 14 Missing Data and Other Opportunities
“A big advantage of Bayesian inference is that it obviates the need to be clever” .
For the opening example, we’re playing with the conditional probability
$\text{Pr(burnt down | burnt up)} = \frac{\text{Pr(burnt up, burnt down)}}{\text{Pr(burnt up)}}.$
It works out that
$\text{Pr(burnt down | burnt up)} = \frac{1/3}{1/2} = \frac{2}{3}.$
We might express the math in the middle of page 423 in tibble form like this.
library(tidyverse)
p_pancake <- 1/3
(
d <-
tibble(pancake = c("BB", "BU", "UU"),
p_burnt = c(1, .5, 0)) %>%
mutate(p_burnt_up = p_burnt * p_pancake)
)
## # A tibble: 3 × 3
## pancake p_burnt p_burnt_up
## <chr> <dbl> <dbl>
## 1 BB 1 0.333
## 2 BU 0.5 0.167
## 3 UU 0 0
d %>%
summarise(p (burnt_down | burnt_up) = p_pancake / sum(p_burnt_up))
## # A tibble: 1 × 1
## p (burnt_down | burnt_up)
## <dbl>
## 1 0.667
I understood McElreath’s simulation better after breaking it apart. The first part of sim_pancake() takes one random draw from the integers 1, 2, and 3. It just so happens that if we set set.seed(1), the code returns a 1.
set.seed(1)
sample(x = 1:3, size = 1)
## [1] 1
So here’s what it looks like if we use seeds 2:11.
take_sample <- function(seed) {
set.seed(seed)
sample(x = 1:3, size = 1)
}
tibble(seed = 2:11) %>%
mutate(value_returned = map_dbl(seed, take_sample))
## # A tibble: 10 × 2
## seed value_returned
## <int> <dbl>
## 1 2 1
## 2 3 1
## 3 4 3
## 4 5 2
## 5 6 1
## 6 7 2
## 7 8 3
## 8 9 3
## 9 10 3
## 10 11 2
Each of those value_returned values stands for one of the three pancakes: 1 = BB, 2 = BU, and 3 = UU. In the next line, McElreath made slick use of a matrix to specify that. Here’s what the matrix looks like.
matrix(c(1, 1, 1, 0, 0, 0), nrow = 2, ncol = 3)
## [,1] [,2] [,3]
## [1,] 1 1 0
## [2,] 1 0 0
See how the three columns are identified as [,1], [,2], and [,3]? If, say, we wanted to subset the values in the second column, we’d execute
matrix(c(1, 1, 1, 0, 0, 0), nrow = 2, ncol = 3)[, 2]
## [1] 1 0
which returns a numeric vector.
matrix(c(1, 1, 1, 0, 0, 0), nrow = 2, ncol = 3)[, 2] %>% str()
## num [1:2] 1 0
And that 1 0 corresponds to the pancake with one burnt (i.e., 1) and one unburnt (i.e., 0) side. So when McElreath then executed sample(sides), he randomly sampled from one of those two values. In the case of pancake == 2, he randomly sampled one the pancake with one burnt and one unburnt side. Had he sampled from pancake == 1, he would have sampled from the pancake with both sides burnt.
Going forward, let’s amend McElreath’s sim_pancake() function so it will take a seed argument, which will allow us to make the output reproducible.
# simulate a pancake and return randomly ordered sides
sim_pancake <- function(seed) {
set.seed(seed)
pancake <- sample(x = 1:3, size = 1)
sides <- matrix(c(1, 1, 1, 0, 0, 0), nrow = 2, ncol = 3)[, pancake]
sample(sides)
}
Let’s take this baby for a whirl.
n_sim <- 1e4
d <-
tibble(seed = 1:n_sim) %>%
mutate(burnt = map(seed, sim_pancake)) %>%
unnest(burnt) %>%
mutate(side = rep(c("up", "down"), times = n() / 2))
Take a look at what we’ve done.
head(d, n = 10)
## # A tibble: 10 × 3
## seed burnt side
## <int> <dbl> <chr>
## 1 1 1 up
## 2 1 1 down
## 3 2 1 up
## 4 2 1 down
## 5 3 1 up
## 6 3 1 down
## 7 4 0 up
## 8 4 0 down
## 9 5 1 up
## 10 5 0 down
And now we’ll spread() and summarise() to get the value we’ve been working for.
d %>%
spread(key = side, value = burnt) %>%
summarise(p (burnt_down | burnt_up) = sum(up == 1 & down == 1) / (sum(up)))
## # A tibble: 1 × 1
## p (burnt_down | burnt_up)
## <dbl>
## 1 0.658
The results are within rounding error of the ideal 2/3.
Probability theory is not difficult mathematically. It’s just counting. But it is hard to interpret and apply. Doing so often seems to require some cleverness, and authors have an incentive to solve problems in clever ways, just to show off. But we don’t need that cleverness, if we ruthlessly apply conditional probability….
In this chapter, [we’ll] meet two commonplace applications of this assume-and-deduce strategy. The first is the incorporation of measurement error into our models. The second is the estimation of missing data through Bayesian imputation…
In neither application do [we] have to intuit the consequences of measurement errors nor the implications of missing values in order to design the models. All [we] have to do is state [the] information about the error or about the variables with missing values. Logic does the rest. (p. 424)
## 14.1 Measurement error
Let’s grab those WaffleDivorce data from back in Chapter 5.
library(rethinking)
data(WaffleDivorce)
d <- WaffleDivorce
rm(WaffleDivorce)
Switch out rethinking for brms.
detach(package:rethinking, unload = T)
library(brms)
For the plots in this chapter, we’ll use the dark themes from the ggdark package .
library(ggdark)
Our primary theme will be ggdark::dark_theme_gray(). One way to use the dark_theme_gray() function is to make it part of the code for an individual plot, such as ggplot() + geom_point() + dark_theme_gray(). Another way is to make dark_theme_gray() the default setting with ggplot2::theme_set(). That’s the method we’ll use.
theme_set(
dark_theme_gray() +
theme(legend.position = "none",
panel.grid = element_blank())
)
# to reset the default ggplot2 theme to its default parameters,
# execute ggplot2::theme_set(theme_gray()) and ggdark::invert_geom_defaults()
For the rest of our color palette, we’ll use colors from the viridis package , which provides a variety of colorblind-safe color palettes (see Rudis et al., 2018).
# install.packages("viridis")
library(viridis)
The viridis_pal() function gives a list of colors within a given palette. The colors in each palette fall on a spectrum. Within viridis_pal(), the option argument allows one to select a given spectrum, “C”, in our case. The final parentheses, (), allows one to determine how many discrete colors one would like to break the spectrum up by. We’ll choose 7.
viridis_pal(option = "C")(7)
## [1] "#0D0887FF" "#5D01A6FF" "#9C179EFF" "#CC4678FF" "#ED7953FF" "#FDB32FFF" "#F0F921FF"
With a little data wrangling, we can put the colors of our palette in a tibble and display them in a plot.
tibble(number = 1:7,
color_number = str_c(1:7, ". ", viridis_pal(option = "C")(7))) %>%
ggplot(aes(x = factor(0), y = reorder(color_number, number))) +
geom_tile(aes(fill = factor(number))) +
geom_text(aes(color = factor(number), label = color_number)) +
scale_color_manual(values = c(rep("black", times = 4),
rep("white", times = 3))) +
scale_fill_viridis(option = "C", discrete = T, direction = -1) +
scale_x_discrete(NULL, breaks = NULL) +
scale_y_discrete(NULL, breaks = NULL) +
ggtitle("Behold: viridis C!")
Now, let’s make use of our custom theme and reproduce/reimagine Figure 14.1.a.
color <- viridis_pal(option = "C")(7)[7]
p1 <-
d %>%
ggplot(aes(x = MedianAgeMarriage, y = Divorce,
ymin = Divorce - Divorce.SE,
ymax = Divorce + Divorce.SE)) +
geom_pointrange(shape = 20, alpha = 2/3, color = color) +
labs(x = "Median age marriage" ,
y = "Divorce rate")
Notice how viridis_pal(option = "C")(7)[7] called the seventh color in the color scheme, "#F0F921FF". For Figure 14.1.b, we’ll select the sixth color in the palette by coding viridis_pal(option = "C")(7)[6]. We’ll then combine the two subplots with patchwork.
color <- viridis_pal(option = "C")(7)[6]
p2 <-
d %>%
ggplot(aes(x = log(Population), y = Divorce,
ymin = Divorce - Divorce.SE,
ymax = Divorce + Divorce.SE)) +
geom_pointrange(shape = 20, alpha = 2/3, color = color) +
scale_y_continuous(NULL, breaks = NULL) +
xlab("log population")
library(patchwork)
p1 | p2
Just like in the text, our plot shows states with larger populations tend to have smaller measurement error. The relation between measurement error and MedianAgeMarriage is less apparent.
### 14.1.1 Error on the outcome.
To get a better sense of what we’re about to do, imagine for a moment that each state’s divorce rate is normally distributed with a mean of Divorce and standard deviation Divorce.SE. Those distributions would be:
d %>%
mutate(Divorce_distribution = str_c("Divorce ~ Normal(", Divorce, ", ", Divorce.SE, ")")) %>%
select(Loc, Divorce_distribution) %>%
head()
## Loc Divorce_distribution
## 1 AL Divorce ~ Normal(12.7, 0.79)
## 2 AK Divorce ~ Normal(12.5, 2.05)
## 3 AZ Divorce ~ Normal(10.8, 0.74)
## 4 AR Divorce ~ Normal(13.5, 1.22)
## 5 CA Divorce ~ Normal(8, 0.24)
## 6 CO Divorce ~ Normal(11.6, 0.94)
As in the text,
in [the following] example we’ll use a Gaussian distribution with mean equal to the observed value and standard deviation equal to the measurement’s standard error. This is the logical choice, because if all we know about the error is its standard deviation, then the maximum entropy distribution for it will be Gaussian…
Here’s how to define the distribution for each divorce rate. For each observed value $$D_{\text{OBS},i}$$, there will be one parameter, $$D_{\text{EST},i}$$, defined by:
$D_{\text{OBS},i} \sim \operatorname{Normal} (D_{\text{EST},i}, D_{\text{SE},i})$
All this does is define the measurement $$D_{\text{OBS},i}$$ as having the specified Gaussian distribution centered on the unknown parameter $$D_{\text{EST},i}$$. So the above defines a probability for each State $$i$$’s observed divorce rate, given a known measurement error. (pp. 426–427)
Now we’re ready to fit some models. In brms, there are at least two ways to accommodate measurement error in the criterion. The first way uses the se() syntax, following the form <response> | se(<se_response>, sigma = TRUE). With this syntax, se stands for standard error, the loose frequentist analogue to the Bayesian posterior $$\textit{SD}$$. Unless you’re fitting a meta-analysis on summary information, which we’ll be doing at the end of this chapter, make sure to specify sigma = TRUE. Without that you’ll have no posterior for $$\sigma$$! For more information on the se() method, go to the brms reference manual and find the Additional response information subsection of the brmsformula section.
The second way uses the mi() syntax, following the form <response> | mi(<se_response>). This follows a missing data logic, resulting in Bayesian missing data imputation for the criterion values. The mi() syntax is based on the newer missing data capabilities for brms. We will cover that in more detail in the second half of this chapter.
We’ll start off using both methods. Our first model, b14.1_se, will follow the se() syntax; the second model, b14.1_mi, will follow the mi() syntax.
# put the data into a list()
dlist <- list(
div_obs = d$Divorce, div_sd = d$Divorce.SE,
R = d$Marriage, A = d$MedianAgeMarriage)
# here we specify the initial (i.e., starting) values
inits <- list(Yl = dlist$div_obs) inits_list <- list(inits, inits) # fit the models b14.1_se <- brm(data = dlist, family = gaussian, div_obs | se(div_sd, sigma = TRUE) ~ 0 + Intercept + R + A, prior = c(prior(normal(0, 10), class = b), prior(cauchy(0, 2.5), class = sigma)), iter = 5000, warmup = 1000, cores = 2, chains = 2, seed = 14, init = inits_list, file = "fits/b14.01_se") b14.1_mi <- brm(data = dlist, family = gaussian, div_obs | mi(div_sd) ~ 0 + Intercept + R + A, prior = c(prior(normal(0, 10), class = b), prior(cauchy(0, 2.5), class = sigma)), iter = 5000, warmup = 1000, cores = 2, chains = 2, seed = 14, control = list(adapt_delta = .9), # note this line for the mi() model save_pars = save_pars(latent = TRUE), init = inits_list, file = "fits/b14.01_mi") Before we dive into the model summaries, notice how the starting values (i.e., inits) differ by model. Even though we coded init = inits_list for both models, the differ by fit@inits. b14.1_se$fit@inits
## [[1]]
## [[1]]$b ## [1] 0.6133048 -1.9171497 1.7551789 ## ## [[1]]$sigma
## [1] 0.4668127
##
## [[1]]$lprior ## [1] -11.10238 ## ## ## [[2]] ## [[2]]$b
## [1] 0.9114156 1.2512265 -0.4276127
##
## [[2]]$sigma ## [1] 1.906943 ## ## [[2]]$lprior
## [1] -11.50392
b14.1_mi$fit@inits ## [[1]] ## [[1]]$Yl
## [1] 12.7 12.5 10.8 13.5 8.0 11.6 6.7 8.9 6.3 8.5 11.5 8.3 7.7 8.0 11.0 10.2 10.6 12.6 11.0
## [20] 13.0 8.8 7.8 9.2 7.4 11.1 9.5 9.1 8.8 10.1 6.1 10.2 6.6 9.9 8.0 9.5 12.8 10.4 7.7
## [39] 9.4 8.1 10.9 11.4 10.0 10.2 9.6 8.9 10.0 10.9 8.3 10.3
##
## [[1]]$b ## [1] -0.5034648 1.1693530 -1.0539336 ## ## [[1]]$sigma
## [1] 1.281562
##
## [[1]]$lprior ## [1] -11.27942 ## ## ## [[2]] ## [[2]]$Yl
## [1] 12.7 12.5 10.8 13.5 8.0 11.6 6.7 8.9 6.3 8.5 11.5 8.3 7.7 8.0 11.0 10.2 10.6 12.6 11.0
## [20] 13.0 8.8 7.8 9.2 7.4 11.1 9.5 9.1 8.8 10.1 6.1 10.2 6.6 9.9 8.0 9.5 12.8 10.4 7.7
## [39] 9.4 8.1 10.9 11.4 10.0 10.2 9.6 8.9 10.0 10.9 8.3 10.3
##
## [[2]]$b ## [1] -0.1543955 1.1642108 -0.4231833 ## ## [[2]]$sigma
## [1] 4.802142
##
## [[2]]$lprior ## [1] -12.5856 As we explore further, it should become apparent why. Here are the primary model summaries. print(b14.1_se) ## Family: gaussian ## Links: mu = identity; sigma = identity ## Formula: div_obs | se(div_sd, sigma = TRUE) ~ 0 + Intercept + R + A ## Data: dlist (Number of observations: 50) ## Draws: 2 chains, each with iter = 5000; warmup = 1000; thin = 1; ## total post-warmup draws = 8000 ## ## Population-Level Effects: ## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS ## Intercept 21.34 6.65 8.41 34.24 1.00 1818 2586 ## R 0.13 0.08 -0.02 0.28 1.00 2330 3151 ## A -0.55 0.21 -0.97 -0.12 1.00 1899 2432 ## ## Family Specific Parameters: ## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS ## sigma 1.13 0.21 0.76 1.56 1.00 2747 3285 ## ## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS ## and Tail_ESS are effective sample size measures, and Rhat is the potential ## scale reduction factor on split chains (at convergence, Rhat = 1). print(b14.1_mi) ## Family: gaussian ## Links: mu = identity; sigma = identity ## Formula: div_obs | mi(div_sd) ~ 0 + Intercept + R + A ## Data: dlist (Number of observations: 50) ## Draws: 2 chains, each with iter = 5000; warmup = 1000; thin = 1; ## total post-warmup draws = 8000 ## ## Population-Level Effects: ## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS ## Intercept 21.58 6.61 8.32 34.28 1.00 2934 4448 ## R 0.13 0.08 -0.02 0.28 1.00 3430 4870 ## A -0.56 0.21 -0.97 -0.13 1.00 3042 4638 ## ## Family Specific Parameters: ## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS ## sigma 1.11 0.20 0.75 1.55 1.00 2193 3578 ## ## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS ## and Tail_ESS are effective sample size measures, and Rhat is the potential ## scale reduction factor on split chains (at convergence, Rhat = 1). Based on the print()/summary() information, the main parameters for the models are about the same. However, the plot deepens when we summarize the models with the posterior::summarise_draws() method. library(posterior) # you can get similar output with b14.1_mi$fit
summarise_draws(b14.1_se, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
## # A tibble: 6 × 5
## variable mean sd 2.5% 97.5%
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 b_Intercept 21.3 6.65 8.41 34.2
## 2 b_R 0.13 0.08 -0.02 0.28
## 3 b_A -0.55 0.21 -0.97 -0.12
## 4 sigma 1.13 0.21 0.76 1.56
## 5 lprior -13.7 1.43 -17.1 -11.6
## 6 lp__ -105. 1.46 -109. -104.
summarise_draws(b14.1_mi, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
## # A tibble: 56 × 5
## variable mean sd 2.5% 97.5%
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 b_Intercept 21.6 6.61 8.32 34.3
## 2 b_R 0.13 0.08 -0.02 0.28
## 3 b_A -0.56 0.21 -0.97 -0.13
## 4 sigma 1.11 0.2 0.75 1.55
## 5 Yl[1] 11.8 0.68 10.5 13.1
## 6 Yl[2] 11.2 1.04 9.2 13.3
## 7 Yl[3] 10.5 0.63 9.23 11.7
## 8 Yl[4] 12.3 0.87 10.6 14.1
## 9 Yl[5] 8.05 0.24 7.58 8.52
## 10 Yl[6] 11.0 0.73 9.6 12.5
## # … with 46 more rows
Again, from b_Intercept to sigma, the output is about the same. But model b14.1_mi, based on the mi() syntax, contained posterior summaries for all 50 of the criterion values. The se() method gave us similar model result, but no posterior summaries for the 50 criterion values. The rethinking package indexed those additional 50 as div_est[i]; with the mi() method, brms indexed them as Yl[i]–no big deal. So while both brms methods accommodated measurement error, the mi() method appears to be the brms analogue to what McElreath did with his model m14.1 in the text. Thus, it’s our b14.1_mi model that follows the form
\begin{align*} \text{Divorce}_{\text{estimated}, i} & \sim \operatorname{Normal}(\mu_i, \sigma) \\ \mu & = \alpha + \beta_1 \text A_i + \beta_2 \text R_i \\ \text{Divorce}_{\text{observed}, i} & \sim \operatorname{Normal}(\text{Divorce}_{\text{estimated}, i}, \text{Divorce}_{\text{standard error}, i}) \\ \alpha & \sim \operatorname{Normal}(0, 10) \\ \beta_1 & \sim \operatorname{Normal}(0, 10) \\ \beta_2 & \sim \operatorname{Normal}(0, 10) \\ \sigma & \sim \operatorname{HalfCauchy}(0, 2.5). \end{align*}
Note. The normal(0, 10) prior McElreath used was quite informative and can lead to discrepancies between the rethinking and brms results if you’re not careful. A large issue is the default way brms handles intercept priors. From the hyperlink, Bürkner wrote:
The formula for the original intercept is b_intercept = temp_intercept - dot_product(means_X, b), where means_X is the vector of means of the predictor variables and b is the vector of regression coefficients (fixed effects). That is, when transforming a prior on the intercept to an “equivalent” prior on the temporary intercept, you have to take the means of the predictors and well as the priors on the other coefficients into account.
If this seems confusing, you have an alternative. The 0 + Intercept part of the brm formula kept the intercept in the metric of the untransformed data, leading to similar results to those from rethinking. When your priors are vague, this might not be much of an issue. And since many of the models in Statistical rethinking use only weakly-regularizing priors, this hasn’t been much of an issue up to this point. But this model is quite sensitive to the intercept syntax. My general recommendation for applied data analysis is this: If your predictors aren’t mean centered, default to the 0 + Intercept syntax for the formula argument when using brms::brm(). Otherwise, your priors might not be doing what you think they’re doing.
Anyway, since our mi()-syntax b14.1_mi model appears to be the analogue to McElreath’s m14.1, we’ll use that one for our plots. Here’s the code for our Figure 14.2.a.
# add the posterior mean and sd for each State's D_EST to the d data
d <-
d %>%
bind_cols(
summarise_draws(b14.1_mi, d_est = mean, d_est_sd = sd) %>%
filter(str_detect(variable, "Yl"))
)
# redefine the color
color <- viridis_pal(option = "C")(7)[5]
# plot
p1 <-
d %>%
ggplot(aes(x = Divorce.SE, y = d_est - Divorce)) +
geom_hline(yintercept = 0, linetype = 2, color = "white") +
geom_point(alpha = 2/3, size = 2, color = color) +
labs(x = "Observed standard error for divorce",
y = "Divorce (estimate - observed)")
Before we make Figure 14.2.b, we need to fit a model that ignores measurement error.
b14.1b <-
brm(data = dlist,
family = gaussian,
div_obs ~ 0 + Intercept + R + A,
prior = c(prior(normal(0, 50), class = b, coef = Intercept),
prior(normal(0, 10), class = b),
prior(cauchy(0, 2.5), class = sigma)),
chains = 2, iter = 5000, warmup = 1000, cores = 2,
seed = 14,
file = "fits/b14.01b")
print(b14.1b)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: div_obs ~ 0 + Intercept + R + A
## Data: dlist (Number of observations: 50)
## Draws: 2 chains, each with iter = 5000; warmup = 1000; thin = 1;
## total post-warmup draws = 8000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 36.37 7.92 20.54 51.76 1.00 1816 2566
## R -0.05 0.08 -0.21 0.11 1.00 2043 2912
## A -0.98 0.25 -1.48 -0.48 1.00 1913 2708
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 1.51 0.16 1.24 1.89 1.00 3345 3336
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
With the ignore-measurement-error fit in hand, we’re ready for Figure 14.2.b.
nd <-
tibble(R = mean(d$Marriage), A = seq(from = 22, to = 30.2, length.out = 30), div_sd = mean(d$Divorce.SE))
# red line
f_error <-
fitted(b14.1_mi, newdata = nd) %>%
as_tibble() %>%
bind_cols(nd)
# yellow line
f_no_error <-
fitted(b14.1b, newdata = nd) %>%
as_tibble() %>%
bind_cols(nd)
color_y <- viridis_pal(option = "C")(7)[7]
color_r <- viridis_pal(option = "C")(7)[4]
# plot
p2 <-
f_no_error %>%
ggplot() +
# f_no_error
geom_smooth(aes(x = A, y = Estimate,
ymin = Q2.5, ymax = Q97.5),
stat = "identity",
fill = color_y, color = color_y,
alpha = 1/4, linewidth = 1/2, linetype = 2) +
# f_error
geom_smooth(data = f_error,
aes(x = A, y = Estimate,
ymin = Q2.5, ymax = Q97.5),
stat = "identity",
fill = color_r, color = color_r,
alpha = 1/3, linewidth = 1/2, linetype = 1) +
# white dots
geom_pointrange(data = d,
aes(x = MedianAgeMarriage, y = d_est,
ymin = d_est - d_est_sd,
ymax = d_est + d_est_sd),
color = "white", shape = 20, alpha = 1/2) +
scale_y_continuous("Divorce rate (posterior)", breaks = seq(from = 4, to = 14, by = 2)) +
xlab("Median age marriage") +
coord_cartesian(xlim = range(dMedianAgeMarriage), ylim = c(4, 15)) p1 | p2 In our plot on the right, it’s the reddish regression line that accounts for measurement error. ### 14.1.2 Error on both outcome and predictor. What happens when there is measurement error on predictor variables as well? The approach is the same. Again, consider the problem generatively: Each observed predictor value is a draw from a distribution with an unknown mean, the true value, but known standard deviation. So we define a vector of parameters, one for each unknown true value, and then make those parameters the means of a family of Gaussian distributions with known standard deviations. (p. 429) We might express this model for our current example as \begin{align*} \text{Divorce}_{\text{estimated}, i} & \sim \operatorname{Normal}(\mu_i, \sigma) \\ \mu & = \alpha + \beta_1 \text A_i + \beta_2 \color{blue}{\text{Marriage_rate}_{\text{estimated}, i}} \\ \text{Divorce}_{\text{observed}, i} & \sim \operatorname{Normal}(\text{Divorce}_{\text{estimated}, i}, \text{Divorce}_{\text{standard error}, i}) \\ \color{blue}{\text{Marriage_rate}_{\text{observed}, i}} & \color{blue}\sim \color{blue}{\operatorname{Normal}(\text{Marriage_rate}_{\text{estimated}, i}, \text{Marriage_rate}_{\text{standard error}, i})} \\ \alpha & \sim \operatorname{Normal}(0, 10) \\ \beta_1 & \sim \operatorname{Normal}(0, 10) \\ \beta_2 & \sim \operatorname{Normal}(0, 10) \\ \sigma & \sim \operatorname{HalfCauchy}(0, 2.5). \end{align*} In brms, you can specify error on predictors with an me() statement in the form of me(predictor, sd_predictor) where sd_predictor is a vector in the data denoting the size of the measurement error, presumed to be in a standard-deviation metric. # the data dlist <- list( div_obs = dDivorce,
div_sd = d$Divorce.SE, mar_obs = d$Marriage,
mar_sd = d$Marriage.SE, A = d$MedianAgeMarriage)
# the inits
inits <- list(Yl = dlist$div_obs) inits_list <- list(inits, inits) # the models b14.2_se <- brm(data = dlist, family = gaussian, div_obs | se(div_sd, sigma = TRUE) ~ 0 + Intercept + me(mar_obs, mar_sd) + A, prior = c(prior(normal(0, 10), class = b), prior(cauchy(0, 2.5), class = sigma)), iter = 5000, warmup = 1000, chains = 3, cores = 3, seed = 14, save_pars = save_pars(latent = TRUE), # note the lack if inits file = "fits/b14.02_se") b14.2_mi <- brm(data = dlist, family = gaussian, div_obs | mi(div_sd) ~ 0 + Intercept + me(mar_obs, mar_sd) + A, prior = c(prior(normal(0, 10), class = b), prior(cauchy(0, 2.5), class = sigma)), iter = 5000, warmup = 1000, cores = 2, chains = 2, seed = 14, control = list(adapt_delta = 0.95), save_pars = save_pars(latent = TRUE), init = inits_list, file = "fits/b14.02_mi") We already know including inits values for our Yl[i] estimates is a waste of time for our se() model. But note how we still defined our inits values as inits <- list(Yl = dlist$div_obs) for the mi() model. Although it’s easy in brms to set the starting values for our Yl[i] estimates, much the way McElreath did, that is not the case when you have measurement error on the predictors. The brms package uses a non-centered parameterization for these, which requires users to have a deeper understanding of the underlying Stan code. This is where I get off the train, but if you want to go further, execute stancode(b14.2_mi).
Here are the two versions of the model.
print(b14.2_se)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: div_obs | se(div_sd, sigma = TRUE) ~ 0 + Intercept + me(mar_obs, mar_sd) + A
## Data: dlist (Number of observations: 50)
## Draws: 3 chains, each with iter = 5000; warmup = 1000; thin = 1;
## total post-warmup draws = 12000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 15.51 6.78 2.16 28.94 1.00 4985 6255
## A -0.44 0.20 -0.83 -0.04 1.00 5606 6811
## memar_obsmar_sd 0.28 0.11 0.07 0.49 1.00 5055 6842
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 1.00 0.21 0.62 1.45 1.00 8508 6769
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
print(b14.2_mi)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: div_obs | mi(div_sd) ~ 0 + Intercept + me(mar_obs, mar_sd) + A
## Data: dlist (Number of observations: 50)
## Draws: 2 chains, each with iter = 5000; warmup = 1000; thin = 1;
## total post-warmup draws = 8000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 15.91 6.75 2.43 28.87 1.00 2187 3146
## A -0.45 0.20 -0.83 -0.04 1.00 2483 3882
## memar_obsmar_sd 0.27 0.11 0.07 0.48 1.00 1962 3343
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 1.00 0.21 0.62 1.43 1.00 1605 2192
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
We’ll use posterior::summarise_draws(), again, to get a sense of depth=2 summaries.
summarise_draws(b14.2_se, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
summarise_draws(b14.2_mi, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
Due to space concerns, I’m not going to show the results, here. You can do that on your own. Both methods yielded the posteriors for Xme_memar_obs[i], but only the b14.2_mi model based on the mi() syntax yielded posteriors for the criterion, the Yl[i] summaries.
Note that you’ll need to specify save_pars = save_pars(latent = TRUE) in the brm() function in order to save the posterior samples of error-adjusted variables obtained by using the me() argument. Without doing so, functions like predict() may give you trouble.
Here is the code for Figure 14.3.a.
# add the posterior mean for each State's D_EST and M_EST to the d data
d <-
d %>%
select(!starts_with("d_est")) %>%
bind_cols(
# Divorce rate estimates
summarise_draws(b14.2_mi, d_est = mean) %>%
filter(str_detect(variable, "Yl")) %>%
select(d_est),
# Marriage rate estimated
summarise_draws(b14.2_mi, m_est = mean) %>%
filter(str_detect(variable, "Xme_")) %>%
select(m_est)
)
color <- viridis_pal(option = "C")(7)[3]
p1 <-
d %>%
ggplot(aes(x = Marriage.SE, y = m_est - Marriage)) +
geom_hline(yintercept = 0, linetype = 2, color = "white") +
geom_point(alpha = 2/3, size = 2, color = color) +
labs(x = "Observed standard error for marriage rate",
y = "Marriage rate (estimate - observed)")
It takes just a little bit of rearranging some of the data in d to make our version of Figure 14.4.b.
color_y <- viridis_pal(option = "C")(7)[7]
color_p <- viridis_pal(option = "C")(7)[2]
# wrangle
p2 <-
full_join(
d %>%
select(Loc, Marriage, m_est) %>%
gather(key, m, -Loc) %>%
mutate(key = if_else(key == "Marriage", "observed", "posterior")),
d %>%
select(Loc, Divorce, d_est) %>%
gather(key, d, -Loc) %>%
mutate(key = if_else(key == "Divorce", "observed", "posterior")),
by = c("Loc", "key")
) %>%
# plot!
ggplot(aes(x = m, y = d)) +
geom_line(aes(group = Loc),
color = "white", linewidth = 1/4) +
geom_point(aes(color = key)) +
scale_color_manual(values = c(color_p, color_y)) +
scale_y_continuous(breaks = seq(from = 4, to = 14, by = 2)) +
labs(x = "Marriage rate (posterior)" ,
y = "Divorce rate (posterior)") +
coord_cartesian(ylim = c(4, 14.5))
p1 | p2
In the right panel, the yellow points are model-implied; the purple ones are of the original data. When you look at both plots, it turns out our brms model regularized more aggressively than McElreath’s rethinking model. I’m unsure of why. If you understand the difference, please share with the rest of the class.
Anyway,
the big take home point for this section is that when you have a distribution of values, don’t reduce it down to a single value to use in a regression. Instead, use the entire distribution. Anytime we use an average value, discarding the uncertainty around that average, we risk overconfidence and spurious inference. This doesn’t only apply to measurement error, but also to cases which data are averaged before analysis.
Do not average. Instead, model. (p. 431)
## 14.2 Missing data
Starting with version 2.2.0, brms supports Bayesian missing data imputation using adaptations of the multivariate syntax . Bürkner’s (2022f) vignette, Handle missing values with brms, is quite helpful for learning the basics.
### 14.2.1 Imputing neocortex
Once again, here are the milk data.
library(rethinking)
data(milk)
d <- milk
d <-
d %>%
mutate(neocortex.prop = neocortex.perc / 100,
logmass = log(mass))
Now we’ll switch out rethinking for brms and do a little data wrangling.
detach(package:rethinking, unload = T)
library(brms)
rm(milk)
# prep data
data_list <-
list(kcal = d$kcal.per.g, neocortex = d$neocortex.prop,
logmass = d$logmass) Here’s the structure of our data list. data_list ##$kcal
## [1] 0.49 0.51 0.46 0.48 0.60 0.47 0.56 0.89 0.91 0.92 0.80 0.46 0.71 0.71 0.73 0.68 0.72 0.97 0.79
## [20] 0.84 0.48 0.62 0.51 0.54 0.49 0.53 0.48 0.55 0.71
##
## $neocortex ## [1] 0.5516 NA NA NA NA 0.6454 0.6454 0.6764 NA 0.6885 0.5885 0.6169 0.6032 ## [14] NA NA 0.6997 NA 0.7041 NA 0.7340 NA 0.6753 NA 0.7126 0.7260 NA ## [27] 0.7024 0.7630 0.7549 ## ##$logmass
## [1] 0.6678294 0.7371641 0.9202828 0.4824261 0.7839015 1.6582281 1.6808279 0.9202828
## [9] -0.3424903 -0.3856625 -2.1202635 -0.7550226 -1.1394343 -0.5108256 1.2441546 0.4382549
## [17] 1.9572739 1.1755733 2.0719133 2.5095993 2.0268316 1.6808279 2.3721112 3.5689692
## [25] 4.3748761 4.5821062 3.7072104 3.4998354 4.0064237
Our statistical model follows the form
\begin{align*} \text{kcal}_i & \sim \operatorname{Normal}(\mu_i, \sigma) \\ \mu_i & = \alpha + \beta_1 \color{blue}{\text{neocortex}_i} + \beta_2 \text{logmass}_i \\ \color{blue}{\text{neocortex}_i} & \color{blue}\sim \color{blue}{\operatorname{Normal}(\nu, \sigma_\text{neocortex})} \\ \alpha & \sim \operatorname{Normal}(0, 100) \\ \beta_1 & \sim \operatorname{Normal}(0, 10) \\ \beta_2 & \sim \operatorname{Normal}(0, 10) \\ \sigma & \sim \operatorname{HalfCauchy}(0, 1) \\ \nu & \sim \operatorname{Normal}(0.5, 1) \\ \sigma_\text{neocortex} & \sim \operatorname{HalfCauchy}(0, 1). \end{align*}
If you look closely, you’ll discover the prior McElreath reported in the model equation for the intercept, $$\alpha \sim \operatorname{Normal}(0, 10)$$, does not match up with the prior he used in R code 14.7, a ~ dnorm(0,100). Here we use the latter.
When writing a multivariate model in brms, I find it easier to save the model code by itself and then insert it into the brm() function. Otherwise, things get cluttered in a hurry.
b_model <-
# here's the primary kcal model
bf(kcal ~ 1 + mi(neocortex) + logmass) +
# here's the model for the missing neocortex data
bf(neocortex | mi() ~ 1) +
# here we set the residual correlations for the two models to zero
set_rescor(FALSE)
Note the mi(neocortex) syntax in the kcal model. This indicates that the predictor, neocortex, has missing values that are themselves being modeled.
To get a sense of how to specify the priors for such a model, use the get_prior() function.
get_prior(data = data_list,
family = gaussian,
b_model)
## prior class coef group resp dpar nlpar lb ub source
## (flat) b default
## (flat) Intercept default
## (flat) b kcal default
## (flat) b logmass kcal (vectorized)
## (flat) b mineocortex kcal (vectorized)
## student_t(3, 0.6, 2.5) Intercept kcal default
## student_t(3, 0, 2.5) sigma kcal 0 default
## student_t(3, 0.7, 2.5) Intercept neocortex default
## student_t(3, 0, 2.5) sigma neocortex 0 default
With the one-step Bayesian imputation procedure in brms, you might need to use the resp argument when specifying non-default priors.
Anyway, here we fit the model.
b14.3 <-
brm(data = data_list,
family = gaussian,
b_model, # here we insert the model
prior = c(prior(normal(0, 100), class = Intercept, resp = kcal),
prior(normal(0.5, 1), class = Intercept, resp = neocortex),
prior(normal(0, 10), class = b, resp = kcal),
prior(cauchy(0, 1), class = sigma, resp = kcal),
prior(cauchy(0, 1), class = sigma, resp = neocortex)),
iter = 1e4, chains = 2, cores = 2,
seed = 14,
file = "fits/b14.03")
The imputed neocortex values are indexed by occasion number from the original data.
summarise_draws(b14.3, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
## # A tibble: 20 × 5
## variable mean sd 2.5% 97.5%
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 b_kcal_Intercept -0.53 0.48 -1.41 0.45
## 2 b_neocortex_Intercept 0.67 0.01 0.64 0.7
## 3 b_kcal_logmass -0.07 0.02 -0.11 -0.02
## 4 bsp_kcal_mineocortex 1.9 0.74 0.38 3.28
## 5 sigma_kcal 0.13 0.02 0.09 0.18
## 6 sigma_neocortex 0.06 0.01 0.04 0.09
## 7 Ymi_neocortex[2] 0.63 0.05 0.54 0.73
## 8 Ymi_neocortex[3] 0.62 0.05 0.53 0.73
## 9 Ymi_neocortex[4] 0.62 0.05 0.53 0.73
## 10 Ymi_neocortex[5] 0.65 0.05 0.56 0.75
## 11 Ymi_neocortex[9] 0.7 0.05 0.6 0.8
## 12 Ymi_neocortex[14] 0.66 0.05 0.56 0.76
## 13 Ymi_neocortex[15] 0.69 0.05 0.6 0.78
## 14 Ymi_neocortex[17] 0.7 0.05 0.6 0.79
## 15 Ymi_neocortex[19] 0.71 0.05 0.61 0.81
## 16 Ymi_neocortex[21] 0.65 0.05 0.55 0.75
## 17 Ymi_neocortex[23] 0.66 0.05 0.56 0.75
## 18 Ymi_neocortex[26] 0.7 0.05 0.59 0.79
## 19 lprior -13.8 0.01 -13.9 -13.8
## 20 lp__ 40.6 4.36 30.8 47.7
Here’s the model that drops the cases with NAs on neocortex.
b14.3cc <-
brm(data = data_list,
family = gaussian,
kcal ~ 1 + neocortex + logmass,
prior = c(prior(normal(0, 100), class = Intercept),
prior(normal(0, 10), class = b),
prior(cauchy(0, 1), class = sigma)),
iter = 1e4, chains = 2, cores = 2,
seed = 14,
file = "fits/b14.03cc")
Check the parameter summaries.
summarise_draws(b14.3cc, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>%
mutate_if(is.numeric, round, digits = 2)
## # A tibble: 6 × 5
## variable mean sd 2.5% 97.5%
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 b_Intercept -1.08 0.58 -2.22 0.08
## 2 b_neocortex 2.78 0.91 0.99 4.56
## 3 b_logmass -0.1 0.03 -0.15 -0.04
## 4 sigma 0.14 0.03 0.09 0.21
## 5 lprior -12.5 0.03 -12.6 -12.4
## 6 lp__ -4.24 1.63 -8.47 -2.22
In order to make our versions of Figure 14.4, we’ll need to do a little data wrangling with fitted().
nd <-
tibble(neocortex = seq(from = .5, to = .85, length.out = 30),
logmass = median(data_list$logmass)) f_b14.3 <- fitted(b14.3, newdata = nd) %>% as_tibble() %>% bind_cols(nd) f_b14.3 %>% glimpse() ## Rows: 30 ## Columns: 10 ##$ Estimate.kcal <dbl> 0.3305725, 0.3534716, 0.3763707, 0.3992698, 0.4221689, 0.4450680, 0.46…
## $Est.Error.kcal <dbl> 0.12622945, 0.11751312, 0.10883616, 0.10020879, 0.09164502, 0.08316450… ##$ Q2.5.kcal <dbl> 0.09313117, 0.13193564, 0.17089619, 0.20916860, 0.24740915, 0.28596563…
## $Q97.5.kcal <dbl> 0.5922145, 0.5961559, 0.6018996, 0.6061100, 0.6114530, 0.6157898, 0.62… ##$ Estimate.neocortex <dbl> 0.6716265, 0.6716265, 0.6716265, 0.6716265, 0.6716265, 0.6716265, 0.67…
## $Est.Error.neocortex <dbl> 0.01363675, 0.01363675, 0.01363675, 0.01363675, 0.01363675, 0.01363675… ##$ Q2.5.neocortex <dbl> 0.6442283, 0.6442283, 0.6442283, 0.6442283, 0.6442283, 0.6442283, 0.64…
## $Q97.5.neocortex <dbl> 0.6988593, 0.6988593, 0.6988593, 0.6988593, 0.6988593, 0.6988593, 0.69… ##$ neocortex <dbl> 0.5000000, 0.5120690, 0.5241379, 0.5362069, 0.5482759, 0.5603448, 0.57…
## $logmass <dbl> 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, … To include the imputed neocortex values in the plot, we’ll extract the information from summarise_draws(). f_b14.3_mi <- summarise_draws(b14.3, mean, ~quantile2(.x, probs = c(.025, .975))) %>% filter(str_detect(variable, "Ymi")) %>% bind_cols(data_list %>% as_tibble() %>% filter(is.na(neocortex))) f_b14.3_mi %>% head() ## # A tibble: 6 × 7 ## variable mean q2.5 q97.5 kcal neocortex logmass ## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 Ymi_neocortex[2] 0.632 0.536 0.733 0.51 NA 0.737 ## 2 Ymi_neocortex[3] 0.624 0.528 0.730 0.46 NA 0.920 ## 3 Ymi_neocortex[4] 0.623 0.527 0.730 0.48 NA 0.482 ## 4 Ymi_neocortex[5] 0.653 0.556 0.752 0.6 NA 0.784 ## 5 Ymi_neocortex[9] 0.702 0.602 0.803 0.91 NA -0.342 ## 6 Ymi_neocortex[14] 0.657 0.562 0.761 0.71 NA -0.511 Data wrangling done–here’s our code for Figure 14.4.a. color <- viridis_pal(option = "D")(7)[4] p1 <- f_b14.3 %>% ggplot(aes(x = neocortex)) + geom_smooth(aes(y = Estimate.kcal, ymin = Q2.5.kcal, ymax = Q97.5.kcal), stat = "identity", fill = color, color = color, alpha = 1/3, linewidth = 1/2) + geom_point(data = data_list %>% as_tibble(), aes(y = kcal), color = "white") + geom_pointrange(data = f_b14.3_mi, aes(x = mean, xmin = q2.5, xmax = q97.5, y = kcal), color = color, shape = 1, linewidth = 1/4, fatten = 4, stroke = 1/2) + labs(subtitle = "Note: For the regression line in this plot,\nlog(mass) has been set to its median, 1.244.", x = "neocortex proportion", y = "kcal per gram") + coord_cartesian(xlim = c(.55, .8), ylim = range(data_list$kcal, na.rm = T))
Here we make Figure 14.4.b, combine it with Figure 14.4.a, and plot.
color <- viridis_pal(option = "D")(7)[4]
p2 <-
data_list %>%
as_tibble() %>%
ggplot(aes(x = logmass, y = neocortex)) +
geom_point(color = "white") +
geom_pointrange(data = f_b14.3_mi,
aes(y = mean, ymin = q2.5, ymax = q97.5),
color = color, shape = 1, linewidth = 1/4, fatten = 4, stroke = 1/2) +
scale_x_continuous("log(mass)", breaks = -2:4) +
ylab("neocortex proportion") +
coord_cartesian(xlim = range(data_listlogmass, na.rm = T), ylim = c(.55, .8)) p1 | p2 ### 14.2.2 Improving the imputation model Like McElreath, we’ll update the imputation line of our statistical model to: \begin{align*} \text{neocortex}_i & \sim \operatorname{Normal}(\nu_i, \sigma_\text{neocortex}) \\ \nu_i & = \alpha_\text{neocortex} + \gamma_1 \text{logmass}_i, \end{align*} which includes the updated priors \begin{align*} \alpha_\text{neocortex} & \sim \operatorname{Normal}(0.5, 1) \\ \gamma_1 & \sim \operatorname{Normal}(0, 10). \end{align*} As far as the brms code goes, adding logmass as a predictor to the neocortex submodel is pretty simple. # define the model b_model <- bf(kcal ~ 1 + mi(neocortex) + logmass) + bf(neocortex | mi() ~ 1 + logmass) + # here's the big difference set_rescor(FALSE) # fit the model b14.4 <- brm(data = data_list, family = gaussian, b_model, prior = c(prior(normal(0, 100), class = Intercept, resp = kcal), prior(normal(0.5, 1), class = Intercept, resp = neocortex), prior(normal(0, 10), class = b, resp = kcal), prior(normal(0, 10), class = b, resp = neocortex), prior(cauchy(0, 1), class = sigma, resp = kcal), prior(cauchy(0, 1), class = sigma, resp = neocortex)), iter = 1e4, chains = 2, cores = 2, seed = 14, file = "fits/b14.04") Behold the parameter summaries. summarise_draws(b14.4, mean, sd, ~quantile(.x, probs = c(.025, .975))) %>% mutate_if(is.numeric, round, digits = 2) ## # A tibble: 21 × 5 ## variable mean sd 2.5% 97.5% ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 b_kcal_Intercept -0.86 0.48 -1.76 0.13 ## 2 b_neocortex_Intercept 0.64 0.01 0.61 0.66 ## 3 b_kcal_logmass -0.09 0.02 -0.13 -0.04 ## 4 b_neocortex_logmass 0.02 0.01 0.01 0.03 ## 5 bsp_kcal_mineocortex 2.43 0.74 0.88 3.83 ## 6 sigma_kcal 0.13 0.02 0.09 0.18 ## 7 sigma_neocortex 0.04 0.01 0.03 0.06 ## 8 Ymi_neocortex[2] 0.63 0.03 0.56 0.7 ## 9 Ymi_neocortex[3] 0.63 0.04 0.56 0.7 ## 10 Ymi_neocortex[4] 0.62 0.04 0.55 0.69 ## # … with 11 more rows Here’s our pre-Figure 14.5 data wrangling. f_b14.4 <- fitted(b14.4, newdata = nd) %>% as_tibble() %>% bind_cols(nd) f_b14.4_mi <- summarise_draws(b14.4, mean, ~quantile2(.x, probs = c(.025, .975))) %>% filter(str_detect(variable, "Ymi")) %>% bind_cols(data_list %>% as_tibble() %>% filter(is.na(neocortex))) f_b14.4 %>% glimpse() ## Rows: 30 ## Columns: 10 ## Estimate.kcal <dbl> 0.2429865, 0.2722728, 0.3015592, 0.3308456, 0.3601320, 0.3894184, 0.41…
## $Est.Error.kcal <dbl> 0.12692274, 0.11818596, 0.10948568, 0.10083132, 0.09223583, 0.08371734… ##$ Q2.5.kcal <dbl> 0.005117345, 0.050765650, 0.095918955, 0.141026058, 0.186430429, 0.230…
## $Q97.5.kcal <dbl> 0.5033977, 0.5145662, 0.5256842, 0.5374260, 0.5489286, 0.5602954, 0.57… ##$ Estimate.neocortex <dbl> 0.6671208, 0.6671208, 0.6671208, 0.6671208, 0.6671208, 0.6671208, 0.66…
## $Est.Error.neocortex <dbl> 0.009472688, 0.009472688, 0.009472688, 0.009472688, 0.009472688, 0.009… ##$ Q2.5.neocortex <dbl> 0.6483687, 0.6483687, 0.6483687, 0.6483687, 0.6483687, 0.6483687, 0.64…
## $Q97.5.neocortex <dbl> 0.6859631, 0.6859631, 0.6859631, 0.6859631, 0.6859631, 0.6859631, 0.68… ##$ neocortex <dbl> 0.5000000, 0.5120690, 0.5241379, 0.5362069, 0.5482759, 0.5603448, 0.57…
## $logmass <dbl> 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, 1.244155, … f_b14.4_mi %>% glimpse() ## Rows: 12 ## Columns: 7 ##$ variable <chr> "Ymi_neocortex[2]", "Ymi_neocortex[3]", "Ymi_neocortex[4]", "Ymi_neocortex[5]", …
## $mean <dbl> 0.6315751, 0.6291646, 0.6197582, 0.6463785, 0.6629332, 0.6271872, 0.6794268, 0.6… ##$ q2.5 <dbl> 0.5639619, 0.5573172, 0.5480440, 0.5790899, 0.5932412, 0.5597960, 0.6135046, 0.6…
## $q97.5 <dbl> 0.7004343, 0.6998674, 0.6908186, 0.7146579, 0.7314117, 0.6962155, 0.7464838, 0.7… ##$ kcal <dbl> 0.51, 0.46, 0.48, 0.60, 0.91, 0.71, 0.73, 0.72, 0.79, 0.48, 0.51, 0.53
## $neocortex <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA ##$ logmass <dbl> 0.7371641, 0.9202828, 0.4824261, 0.7839015, -0.3424903, -0.5108256, 1.2441546, 1…
For our final plots, let’s play around with colors from viridis_pal(option = "D"). Here’s the code for Figure 14.5.a.
color <- viridis_pal(option = "D")(7)[3]
p1 <-
f_b14.4 %>%
ggplot(aes(x = neocortex)) +
geom_smooth(aes(y = Estimate.kcal, ymin = Q2.5.kcal, ymax = Q97.5.kcal),
stat = "identity",
fill = color, color = color, alpha = 1/2, linewidth = 1/2) +
geom_point(data = data_list %>% as_tibble(),
aes(y = kcal),
color = "white") +
geom_pointrange(data = f_b14.4_mi,
aes(x = mean, xmin = q2.5, xmax = q97.5, y = kcal),
color = color, shape = 1, linewidth = 1/4, fatten = 4, stroke = 1/2) +
labs(subtitle = "Note: For the regression line in this plot,\nlog(mass) has been set to its median, 1.244.",
x = "neocortex proportion",
y = "kcal per gram") +
coord_cartesian(xlim = c(.55, .8),
ylim = range(data_list$kcal, na.rm = T)) Make the code for Figure 14.5.b, combine it with Figure 14.5.a, and plot. color <- viridis_pal(option = "D")(7)[3] p2 <- data_list %>% as_tibble() %>% ggplot(aes(x = logmass, y = neocortex)) + geom_point(color = "white") + geom_pointrange(data = f_b14.4_mi, aes(y = mean, ymin = q2.5, ymax = q97.5), color = color, shape = 1, linewidth = 1/4, fatten = 4, stroke = 1/2) + scale_x_continuous("log(mass)", breaks = -2:4) + ylab("neocortex proportion") + coord_cartesian(xlim = range(data_list$logmass, na.rm = T),
ylim = c(.55, .8))
p1 | p2
### 14.2.3 Bonus: mi() can replace me()
In one of his responses to a question thread on the Stan forums, Bürkner remarked he wasn’t a fan of the brms me() syntax and that he planned to depreciate it in the future. As a consequence, our code for models b14.2_se and b14.2_mi from above might soon break. Happily, we can use the mi() workflow to replace me(). Here’s what that might look like for an alternate version of b14.2_mi, which we’ll call b14.2_mi_mi.
Warning: Specifying global priors for regression coefficients in multivariate models is deprecated. Please specify priors separately for each response variable.
b_model <-
# here's the primary div_obs model
bf(div_obs | mi(div_sd) ~ 0 + Intercept + mi(mar_obs) + A) +
# here's the model for the measurement error in the mar_obs data
bf(mar_obs | mi(mar_sd) ~ 0 + Intercept) +
# here we set the residual correlations for the two models to zero
set_rescor(FALSE)
b14.2_mi_mi <-
brm(data = dlist,
family = gaussian,
b_model, # here we insert the model
prior = c(prior(normal(0, 10), class = b, resp = divobs),
prior(normal(0, 10), class = b, resp = marobs),
prior(cauchy(0, 2.5), class = sigma, resp = divobs),
prior(cauchy(0, 2.5), class = sigma, resp = marobs)),
iter = 5000, warmup = 1000, cores = 2, chains = 2,
seed = 14,
max_treedepth = 11),
save_pars = save_pars(latent = TRUE),
init = inits_list,
file = "fits/b14.02_mi_mi")
The advantage of the mi() approach is the brm() syntax requires an explicit formula for the mar_obs predictor. The other advantage is mar_obs get’s an explicit summary in the print() output.
print(b14.2_mi_mi)
## Family: MV(gaussian, gaussian)
## Links: mu = identity; sigma = identity
## mu = identity; sigma = identity
## Formula: div_obs | mi(div_sd) ~ 0 + Intercept + mi(mar_obs) + A
## mar_obs | mi(mar_sd) ~ 0 + Intercept
## Data: dlist (Number of observations: 50)
## Draws: 2 chains, each with iter = 5000; warmup = 1000; thin = 1;
## total post-warmup draws = 8000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## divobs_Intercept 15.46 6.92 1.65 28.74 1.00 2096 3341
## divobs_A -0.43 0.21 -0.84 -0.02 1.00 2364 3496
## marobs_Intercept 19.67 0.47 18.77 20.60 1.00 5196 5386
## divobs_mimar_obs 0.28 0.11 0.06 0.50 1.00 1767 3316
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma_divobs 1.00 0.21 0.62 1.45 1.00 1438 2434
## sigma_marobs 2.91 0.38 2.24 3.72 1.00 3600 5232
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
If you compare the output from summarise_draws(b14.2_mi_mi) to the output from summarise_draws(b14.2_mi), above, you’ll see the values are largely the same, but the ordering and some of the variable names have changed a bit. I’ll leave that comparison to the interested reader.
To further cement the similarities to the two models, here’s a reworking of Figure 14.4.b, this time based on our b14.2_mi_mi model.
full_join(
# observed values
tibble(row = 1:50,
Divorce = dlist$div_obs, Marriage = dlist$mar_obs) %>%
gather(variable, observed, -row),
# posterior means
summarise_draws(b14.2_mi_mi, posterior = mean) %>%
filter(str_detect(variable, "Yl_")) %>%
mutate(row = str_extract(variable, "\\d+") %>% as.double(),
variable = if_else(str_detect(variable, "divobs"), "Divorce", "Marriage")),
by = c("row", "variable")
) %>%
pivot_longer(observed:posterior) %>%
pivot_wider(names_from = variable, values_from = value) %>%
# plot!
ggplot(aes(x = Marriage, y = Divorce)) +
geom_line(aes(group = row),
color = "white", linewidth = 1/4) +
geom_point(aes(color = name)) +
scale_color_manual(values = c(color_p, color_y)) +
scale_y_continuous(breaks = seq(from = 4, to = 14, by = 2)) +
labs(x = "Marriage rate (posterior)" ,
y = "Divorce rate (posterior)") +
coord_cartesian(ylim = c(4, 14.5))
The results are basically the same.
To wrap up, if modern missing data methods are new to you, you might also check out van Burren’s great (2018) text, Flexible imputation of missing data: Second edition. I’m also a fan of Enders’s (2022) Applied missing data analysis, for which you can download a free sample chapter by clicking here. I’ll also quickly mention that brms accommodates multiple imputation, too.
## 14.3Summary Bonus: Meta-analysis
If your mind isn’t fully blown by those measurement-error and missing-data models, let’s keep building. As it turns out, meta-analyses are often just special kinds of multilevel measurement-error models. Thus, you can use brms::brm() to fit Bayesian meta-analyses, too.
Before we proceed, I should acknowledge that this section is heavily influenced by Matti Vourre’s great blog post, Bayesian meta-analysis with R, Stan & brms. Since McElreath’s text doesn’t directly address meta-analyses, we’ll also have to borrow a bit from Gelman, Carlin, Stern, Dunson, Vehtari, and Rubin’s (2013) Bayesian data analysis, Third edition. We’ll let Gelman and colleagues introduce the topic:
Discussions of meta-analysis are sometimes imprecise about the estimands of interest in the analysis, especially when the primary focus is on testing the null hypothesis of no effect in any of the studies to be combined. Our focus is on estimating meaningful parameters, and for this objective there appear to be three possibilities, accepting the overarching assumption that the studies are comparable in some broad sense. The first possibility is that we view the studies as identical replications of each other, in the sense we regard the individuals in all the studies as independent samples from a common population, with the same outcome measures and so on. A second possibility is that the studies are so different that the results of any one study provide no information about the results of any of the others. A third, more general, possibility is that we regard the studies as exchangeable but not necessarily either identical or completely unrelated; in other words we allow differences from study to study, but such that the differences are not expected a priori to have predictable effects favoring one study over another…. this third possibility represents a continuum between the two extremes, and it is this exchangeable model (with unknown hyperparameters characterizing the population distribution) that forms the basis of our Bayesian analysis…
The first potential estimand of a meta-analysis, or a hierarchically structured problem in general, is the mean of the distribution of effect sizes, since this represents the overall ‘average’ effect across all studies that could be regarded as exchangeable with the observed studies. Other possible estimands are the effect size in any of the observed studies and the effect size in another, comparable (exchangeable) unobserved study. (pp. 125—126, emphasis in the original)
The basic version of a Bayesian meta-analysis follows the form
$y_i \sim \operatorname{Normal}(\theta_i, \sigma_i),$
where $$y_i$$ = the point estimate for the effect size of a single study, $$i$$, which is presumed to have been a draw from a Normal distribution centered on $$\theta_i$$. The data in meta-analyses are typically statistical summaries from individual studies. The one clear lesson from this chapter is that those estimates themselves come with error and those errors should be fully expressed in the meta-analytic model. Which we do. The standard error from study $$i$$ is specified $$\sigma_i$$, which is also a stand-in for the standard deviation of the Normal distribution from which the point estimate was drawn. Do note, we’re not estimating $$\sigma_i$$, here. Those values we take directly from the original studies.
Building on the model, we further presume that study $$i$$ is itself just one draw from a population of related studies, each of which have their own effect sizes. As such. we presume $$\theta_i$$ itself has a distribution following the form
$\theta_i \sim \operatorname{Normal} (\mu, \tau),$
where $$\mu$$ is the meta-analytic effect (i.e., the population mean) and $$\tau$$ is the variation around that mean, what you might also think of as $$\sigma_\tau$$.
Since there’s no example of a meta-analysis in the text, we’ll have to get our data elsewhere. We’ll focus on Gershoff and Grogan-Kaylor’s (2016) paper, Spanking and child outcomes: Old controversies and new meta-analyses. From their introduction, we read:
Around the world, most children (80%) are spanked or otherwise physically punished by their parents . The question of whether parents should spank their children to correct misbehaviors sits at a nexus of arguments from ethical, religious, and human rights perspectives both in the U.S. and around the world . Several hundred studies have been conducted on the associations between parents’ use of spanking or physical punishment and children’s behavioral, emotional, cognitive, and physical outcomes, making spanking one of the most studied aspects of parenting. What has been learned from these hundreds of studies? (p. 453)
Our goal will be to learn Bayesian meta-analysis by answering part of that question. I’ve transcribed the values directly from Gershoff and Grogan-Kaylor’s paper and saved them as a file called spank.xlsx. You can find the data in this project’s GitHub repository. Let’s load them and glimpse().
spank <- readxl::read_excel("spank.xlsx")
glimpse(spank)
## Rows: 111
## Columns: 8
## $study <chr> "Bean and Roberts (1981)", "Day and Roberts (1983)", "Minton, Kagan, and Levine (1… ##$ year <dbl> 1981, 1983, 1971, 1988, 1990, 1961, 1962, 1990, 2002, 2005, 1986, 2012, 1979, 2009…
## $outcome <chr> "Immediate defiance", "Immediate defiance", "Immediate defiance", "Immediate defia… ##$ between <dbl> 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1…
## $within <dbl> 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0… ##$ d <dbl> -0.74, 0.36, 0.34, -0.08, 0.10, 0.63, 0.19, 0.47, 0.14, -0.18, 1.18, 0.70, 0.63, 0…
## $ll <dbl> -1.76, -1.04, -0.09, -1.01, -0.82, 0.16, -0.14, 0.20, -0.42, -0.49, 0.15, 0.35, -0… ##$ ul <dbl> 0.28, 1.77, 0.76, 0.84, 1.03, 1.10, 0.53, 0.74, 0.70, 0.13, 2.22, 1.05, 1.71, 0.22…
In this paper, the effect size of interest is a Cohen’s $$d$$, derived from the formula
$d = \frac{\mu_\text{treatment} - \mu_\text{comparison}}{\sigma_\text{pooled}},$
where
$\sigma_\text{pooled} = \sqrt{\frac{[(n_1 - 1) \sigma_1^2] + [(n_2 - 1) \sigma_2^2]}{n_1 + n_2 -2}}.$
To help make the equation for $$d$$ clearer for our example, we might re-express it as
$d = \frac{\mu_\text{spanked} - \mu_\text{not spanked}}{\sigma_\text{pooled}}.$
McElreath didn’t really focus on effect sizes in his text. If you need a refresher, you might check out Kelley and Preacher’s (2012) On effect size. But in words, Cohen’s $$d$$ is a standardized mean difference between two groups.
So if you look back up at the results of glimpse(spank) you’ll notice the column d, which is indeed a vector of Cohen’s $$d$$ effect sizes. The last two columns, ll and ul, are the lower and upper limits of the associated 95% frequentist confidence intervals. But we don’t want confidence intervals for our d-values; we want their standard errors. Fortunately, we can compute those with the following formula
$SE = \frac{\text{upper limit} - \text{lower limit}}{3.92}.$
Here it is in code.
spank <-
spank %>%
mutate(se = (ul - ll) / 3.92)
glimpse(spank)
## Rows: 111
## Columns: 9
## $study <chr> "Bean and Roberts (1981)", "Day and Roberts (1983)", "Minton, Kagan, and Levine (1… ##$ year <dbl> 1981, 1983, 1971, 1988, 1990, 1961, 1962, 1990, 2002, 2005, 1986, 2012, 1979, 2009…
## $outcome <chr> "Immediate defiance", "Immediate defiance", "Immediate defiance", "Immediate defia… ##$ between <dbl> 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1…
## $within <dbl> 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0… ##$ d <dbl> -0.74, 0.36, 0.34, -0.08, 0.10, 0.63, 0.19, 0.47, 0.14, -0.18, 1.18, 0.70, 0.63, 0…
## $ll <dbl> -1.76, -1.04, -0.09, -1.01, -0.82, 0.16, -0.14, 0.20, -0.42, -0.49, 0.15, 0.35, -0… ##$ ul <dbl> 0.28, 1.77, 0.76, 0.84, 1.03, 1.10, 0.53, 0.74, 0.70, 0.13, 2.22, 1.05, 1.71, 0.22…
## \$ se <dbl> 0.52040816, 0.71683673, 0.21683673, 0.47193878, 0.47193878, 0.23979592, 0.17091837…
Now are data are ready, we can express our first Bayesian meta-analysis with the formula
\begin{align*} \text{d}_i & \sim \operatorname{Normal}(\theta_i, \sigma_i = \text{se}_i) \\ \theta_i & \sim \operatorname{Normal}(\mu, \tau) \\ \mu & \sim \operatorname{Normal}(0, 1) \\ \tau & \sim \operatorname{HalfCauchy}(0, 1). \end{align*}
The last two lines, of course, spell out our priors. In psychology, it’s pretty rare to see Cohen’s $$d$$-values greater than the absolute value of $$\pm 1$$. So in the absence of more specific domain knowledge–which I don’t have–, it seems like $$\operatorname{Normal}(0, 1)$$ is a reasonable place to start. And just like McElreath used $$\operatorname{HalfCauchy}(0, 1)$$ as the default prior for the group-level standard deviations, it makes sense to use it here for our meta-analytic $$\tau$$ parameter .
Here’s the code for the first model.
b14.5 <-
brm(data = spank,
family = gaussian,
d | se(se) ~ 1 + (1 | study),
prior = c(prior(normal(0, 1), class = Intercept),
prior(cauchy(0, 1), class = sd)),
iter = 4000, warmup = 1000, cores = 4, chains = 4,
seed = 14,
file = "fits/b14.05")
One thing you might notice is our se(se) function excluded the sigma argument. If you recall from Section 14.1, we specified sigma = T in our measurement-error models. The brms default is that within se(), sigma = FALSE. As such, we have no estimate for sigma the way we would if we were doing this analysis with the raw data from the studies. Hopefully this makes sense. The uncertainty around the d-value for each study $$i$$ has already been encoded in the data as se.
This brings us to another point. We typically perform meta-analyses on data summaries. In my field and perhaps in yours, this is due to the historical accident that it has not been the norm among researchers to make their data publicly available. So effect size summaries were the best we typically had. However, times are changing (e.g., Winerman, 2017). If the raw data from all the studies for your meta-analysis are available, you can just fit a multilevel model in which the data are nested in the studies. Heck, you could even allow the studies to vary by $$\sigma$$ by taking the distributional modeling approach and specify something like sigma ~ 0 + study or even sigma ~ 1 + (1 | study).
But enough technical talk. Let’s look at the model results.
print(b14.5)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: d | se(se) ~ 1 + (1 | study)
## Data: spank (Number of observations: 111)
## Draws: 4 chains, each with iter = 4000; warmup = 1000; thin = 1;
## total post-warmup draws = 12000
##
## Group-Level Effects:
## ~study (Number of levels: 76)
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sd(Intercept) 0.26 0.03 0.21 0.33 1.00 2541 4657
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 0.37 0.04 0.30 0.44 1.00 1534 2927
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 0.00 0.00 0.00 0.00 NA NA NA
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
Thus, in our simple Bayesian meta-analysis, we have a population Cohen’s $$d$$ of about 0.37. Our estimate for $$\tau$$, 0.26, suggests we have quite a bit of between-study variability. One question you might ask is: What exactly are these Cohen’s $$d$$’s measuring, anyways? We’ve encoded that in the outcome vector of the spank data.
spank %>%
distinct(outcome) %>%
knitr::kable()
outcome
Immediate defiance
Low moral internalization
Child aggression
Child antisocial behavior
Child externalizing behavior problems
Child internalizing behavior problems
Child mental health problems
Child alcohol or substance abuse
Negative parent–child relationship
Impaired cognitive ability
Low self-esteem
Low self-regulation
Victim of physical abuse
There are a few things to note. First, with the possible exception of Adult support for physical punishment, all of the outcomes are negative. We prefer conditions associated with lower values for things like Child aggression and Adult mental health problems. Second, the way the data are coded, larger effect sizes are interpreted as more negative outcomes associated with children having been spanked. That is, our analysis suggests spanking children is associated with worse outcomes. What might not be immediately apparent is that even though there are 111 cases in the data, there are only 76 distinct studies.
spank %>%
distinct(study) %>%
count()
## # A tibble: 1 × 1
## n
## <int>
## 1 76
In other words, some studies have multiple outcomes. In order to better accommodate the study- and outcome-level variances, let’s fit a cross-classified Bayesian meta-analysis reminiscent of the cross-classified chimp model from Chapter 13.
b14.6 <-
brm(data = spank,
family = gaussian,
d | se(se) ~ 1 + (1 | study) + (1 | outcome),
prior = c(prior(normal(0, 1), class = Intercept),
prior(cauchy(0, 1), class = sd)),
iter = 4000, warmup = 1000, cores = 4, chains = 4,
seed = 14,
file = "fits/b14.06")
print(b14.6)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: d | se(se) ~ 1 + (1 | study) + (1 | outcome)
## Data: spank (Number of observations: 111)
## Draws: 4 chains, each with iter = 4000; warmup = 1000; thin = 1;
## total post-warmup draws = 12000
##
## Group-Level Effects:
## ~outcome (Number of levels: 17)
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sd(Intercept) 0.08 0.03 0.04 0.14 1.00 2640 5078
##
## ~study (Number of levels: 76)
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sd(Intercept) 0.25 0.03 0.20 0.32 1.00 2592 5195
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 0.35 0.04 0.27 0.43 1.00 2294 4056
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 0.00 0.00 0.00 0.00 NA NA NA
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
Now we have two $$\tau$$ parameters. We might plot them to get a sense of where the variance is at.
# we'll want this to label the plot
label <-
tibble(tau = c(.12, .3),
y = c(15, 10),
label = c("sigma['outcome']", "sigma['study']"))
# wrangle
as_draws_df(b14.6) %>%
select(starts_with("sd")) %>%
gather(key, tau) %>%
mutate(key = str_remove(key, "sd_") %>% str_remove(., "__Intercept")) %>%
# plot
ggplot(aes(x = tau)) +
geom_density(aes(fill = key),
color = "transparent") +
geom_text(data = label,
aes(y = y, label = label, color = label),
size = 5, parse = T) +
scale_fill_viridis_d(NULL, option = "B", begin = .5) +
scale_color_viridis_d(NULL, option = "B", begin = .5) +
scale_x_continuous(expression(tau), limits = c(0, NA)) +
scale_y_continuous(NULL, breaks = NULL)
So at this point, the big story is there’s more variability between the studies than there is the outcomes. But I still want to get a sense of the individual outcomes. Here we’ll use tidybayes::geom_halfeye() to help us make our version of a forest plot and tidybayes::spread_draws() to help with the initial wrangling.
library(tidybayes)
b14.6 %>%
# add the grand mean to the group-specific deviations
mutate(mu = b_Intercept + r_outcome) %>%
ungroup() %>%
mutate(outcome = str_replace_all(outcome, "[.]", " ")) %>%
# plot
ggplot(aes(x = mu, y = reorder(outcome, mu), fill = reorder(outcome, mu))) +
geom_vline(xintercept = fixef(b14.6)[1, 1], color = "grey33", linewidth = 1) +
geom_vline(xintercept = fixef(b14.6)[1, 3:4], color = "grey33", linetype = 2) +
stat_halfeye(.width = .95, size = 2/3, color = "white") +
scale_fill_viridis_d(option = "B", begin = .2) +
labs(x = expression("Cohen's "*italic(d)),
y = NULL) +
theme(axis.text.y = element_text(hjust = 0),
axis.ticks.y = element_blank())
The solid and dashed vertical white lines in the background mark off the grand mean (i.e., the meta-analytic effect) and its 95% intervals. But anyway, there’s not a lot of variability across the outcomes. Let’s go one step further with the model. Doubling back to Gelman and colleagues, we read:
When assuming exchangeability we assume there are no important covariates that might form the basis of a more complex model, and this assumption (perhaps misguidedly) is widely adopted in meta-analysis. What if other information (in addition to the data $$(n, y)$$) is available to distinguish among the $$J$$ studies in a meta-analysis, so that an exchangeable model is inappropriate? In this situation, we can expand the framework of the model to be exchangeable in the observed data and covariates, for example using a hierarchical regression model. (p. 126)
One important covariate Gershoff and Grogan-Kaylor addressed in their meta-analysis was the type of study. The 76 papers they based their meta-analysis on contained both between- and within-participants designs. In the spank data, we’ve dummy coded that information with the between and within vectors. Both are dummy variables and $$\text{within} = 1 - \text{between}$$. Here are the counts.
spank %>%
count(between)
## # A tibble: 2 × 2
## between n
## <dbl> <int>
## 1 0 71
## 2 1 40
When I use dummies in my models, I prefer to have the majority group stand as the reference category. As such, I typically name those variables by the minority group. In this case, most occasions are based on within-participant designs. Thus, we’ll go ahead and add the between variable to the model. While we’re at it, we’ll practice using the 0 + Intercept syntax.
b14.7 <-
brm(data = spank,
family = gaussian,
d | se(se) ~ 0 + Intercept + between + (1 | study) + (1 | outcome),
prior = c(prior(normal(0, 1), class = b),
prior(cauchy(0, 1), class = sd)),
iter = 4000, warmup = 1000, cores = 4, chains = 4,
seed = 14,
file = "fits/b14.07")
Behold the summary.
print(b14.7)
## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: d | se(se) ~ 0 + Intercept + between + (1 | study) + (1 | outcome)
## Data: spank (Number of observations: 111)
## Draws: 4 chains, each with iter = 4000; warmup = 1000; thin = 1;
## total post-warmup draws = 12000
##
## Group-Level Effects:
## ~outcome (Number of levels: 17)
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sd(Intercept) 0.08 0.03 0.04 0.14 1.00 4889 7141
##
## ~study (Number of levels: 76)
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sd(Intercept) 0.25 0.03 0.20 0.31 1.00 3113 5843
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept 0.38 0.05 0.29 0.48 1.00 3398 5719
## between -0.08 0.07 -0.21 0.06 1.00 3508 6016
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## sigma 0.00 0.00 0.00 0.00 NA NA NA
##
## Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
Let’s take a closer look at the b_between parameter.
color <- viridis_pal(option = "B")(7)[5]
as_draws_df(b14.7) %>%
ggplot(aes(x = b_between, y = 0)) +
stat_halfeye(.width = c(.5, .95), color = "white", fill = color) +
scale_y_continuous(NULL, breaks = NULL) +
xlab("Overall difference for between- vs within-participant designs")
That difference isn’t as large I’d expect it to be. But then again, I’m no spanking researcher. So what do I know?
There are other things you might do with these data. For example, you might check for trends by year or, as the authors did in their manuscript, distinguish among different severities of corporal punishment. But I think we’ve gone far enough to get you started.
If you’d like to learn more about these methods, do check out Vourre’s Bayesian meta-analysis with R, Stan & brms. From his blog, you’ll learn additional tricks, like making a more traditional-looking forest plot and how our Bayesian brms method compares with frequentist meta-analyses via the metafor package . You might also check out that aforementioned paper by Williams and colleagues’, Bayesian meta-analysis with weakly informative prior distributions, to give you an empirical justification for using a half-Cauchy prior for your meta-analysis $$\tau$$ parameters.
## Session info
sessionInfo()
## R version 4.2.2 (2022-10-31)
## Platform: x86_64-apple-darwin17.0 (64-bit)
## Running under: macOS Big Sur ... 10.16
##
## Matrix products: default
## BLAS: /Library/Frameworks/R.framework/Versions/4.2/Resources/lib/libRblas.0.dylib
## LAPACK: /Library/Frameworks/R.framework/Versions/4.2/Resources/lib/libRlapack.dylib
##
## locale:
## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
##
## attached base packages:
## [1] parallel stats graphics grDevices utils datasets methods base
##
## other attached packages:
## [1] tidybayes_3.0.2 posterior_1.3.1 patchwork_1.1.2 viridis_0.6.2
## [5] viridisLite_0.4.1 ggdark_0.2.1 brms_2.18.0 Rcpp_1.0.9
## [9] cmdstanr_0.5.3 rstan_2.21.8 StanHeaders_2.21.0-7 forcats_0.5.1
## [13] stringr_1.4.1 dplyr_1.0.10 purrr_1.0.1 readr_2.1.2
## [17] tidyr_1.2.1 tibble_3.1.8 ggplot2_3.4.0 tidyverse_1.3.2
##
## loaded via a namespace (and not attached):
## [1] readxl_1.4.1 backports_1.4.1 plyr_1.8.7 igraph_1.3.4
## [5] svUnit_1.0.6 sp_1.5-0 splines_4.2.2 crosstalk_1.2.0
## [9] TH.data_1.1-1 rstantools_2.2.0 inline_0.3.19 digest_0.6.31
## [13] htmltools_0.5.3 fansi_1.0.3 magrittr_2.0.3 checkmate_2.1.0
## [17] googlesheets4_1.0.1 tzdb_0.3.0 modelr_0.1.8 RcppParallel_5.1.5
## [21] matrixStats_0.63.0 xts_0.12.1 sandwich_3.0-2 prettyunits_1.1.1
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## [77] processx_3.8.0 knitr_1.40 fs_1.5.2 nlme_3.1-160
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## [125] minqa_1.2.5 rmarkdown_2.16 googledrive_2.0.0 shiny_1.7.2
## [129] lubridate_1.8.0 base64enc_0.1-3 dygraphs_1.1.1.6
### References
Bürkner, P.-C. (2022a). Estimating distributional models with brms. https://CRAN.R-project.org/package=brms/vignettes/brms_distreg.html
Bürkner, P.-C. (2022c). Estimating multivariate models with brms. https://CRAN.R-project.org/package=brms/vignettes/brms_multivariate.html
Bürkner, P.-C. (2022f). Handle missing values with brms. https://CRAN.R-project.org/package=brms/vignettes/brms_missings.html
Bürkner, P.-C. (2022h). brms reference manual, Version 2.18.0. https://CRAN.R-project.org/package=brms/brms.pdf
Enders, C. K. (2022). Applied missing data analysis (Second Edition). Guilford Press. http://www.appliedmissingdata.com/
Garnier, S. (2021). viridis: Default color maps from ’matplotlib’ [Manual]. https://CRAN.R-project.org/package=viridis
Gelman, A., Carlin, J. B., Stern, H. S., Dunson, D. B., Vehtari, A., & Rubin, D. B. (2013). Bayesian data analysis (Third Edition). CRC press. https://stat.columbia.edu/~gelman/book/
Gershoff, Elizabeth T. (2013). Spanking and child development: We know enough now to stop hitting our children. Child Development Perspectives, 7(3), 133–137. https://doi.org/10.1111/cdep.12038
Gershoff, Elizabeth T., & Grogan-Kaylor, A. (2016). Spanking and child outcomes: Old controversies and new meta-analyses. Journal of Family Psychology, 30(4), 453. https://doi.org/10.1037/fam0000191
Grantham, N. (2019). ggdark: Dark mode for ’ggplot2’ themes [Manual]. https://CRAN.R-project.org/package=ggdark
Kelley, K., & Preacher, K. J. (2012). On effect size. Psychological Methods, 17(2), 137. https://doi.org/10.1037/a0028086
McElreath, R. (2015). Statistical rethinking: A Bayesian course with examples in R and Stan. CRC press. https://xcelab.net/rm/statistical-rethinking/
Rudis, B., Ross, N., & Garnier, S. (2018). The viridis color palettes. https://cran.r-project.org/package=viridis/vignettes/intro-to-viridis.html
UNICEF. (2014). Hidden in plain sight: A statistical analysis of violence against children. https://www.unicef.org/publications/index_74865.html
van Buuren, S. (2018). Flexible imputation of missing data (Second Edition). CRC Press. https://stefvanbuuren.name/fimd/
Viechtbauer, W. (2010). Conducting meta-analyses in R with the metafor package. Journal of Statistical Software, 36(3), 1–48. https://www.jstatsoft.org/v36/i03/
Viechtbauer, W. (2022). metafor: Meta-analysis package for R [Manual]. https://CRAN.R-project.org/package=metafor
Williams, Donald R., Rast, P., & Bürkner, P.-C. (2018). Bayesian meta-analysis with weakly informative prior distributions. https://doi.org/10.31234/osf.io/7tbrm
Winerman, L. (2017). Trends report: Psychologists embrace open science. Monitor on Psychology, 48(10). https://www.apa.org/monitor/2017/11/trends-open-science | 27,078 | 79,487 | {"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": 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} | 3.484375 | 3 | CC-MAIN-2024-22 | latest | en | 0.724594 |
https://mathoverflow.net/questions/293956/cardinality-of-fractal-without-ch | 1,553,165,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00180.warc.gz | 545,524,030 | 28,629 | # Cardinality of fractal without CH? [closed]
What are possible choices for the cardinality of plane or line fractal without assuming CH?
Are there fractals for which the answer is not easy? (Sieprinski triangle is open on reddit.)
## closed as off-topic by Gerald Edgar, Andrés E. Caicedo, Chris Godsil, Nik Weaver, Gro-TsenFeb 27 '18 at 15:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Gerald Edgar, Chris Godsil, Nik Weaver, Gro-Tsen
If this question can be reworded to fit the rules in the help center, please edit the question.
• Cantor proved that closed subsets of Euclidean space obey CH, no need to assume it. But this question does not belong here. Instead it belongs in math.stackexchange.com Note that reddit thread is from 8 years ago – Gerald Edgar Feb 27 '18 at 11:30
• To answer this question, we need to know the precise definition of a fractal. That given by Mandelbrot (fractal dimension, self-similarity and obtained by iterative procedures) is not as precise as necessary. – Taras Banakh Feb 27 '18 at 11:43
• Any closed subset of Euclidean space is either countable, or has the same cardinality as $\mathbb R$. Reference, G. Cantor, Acta Mathematica 4 (1884) 381--392. Nowadays you should find it in any "real analysis" course, perhaps as the generalization known as the Cantor-Bendixson theorem. en.wikipedia.org/wiki/Perfect_set_property – Gerald Edgar Feb 27 '18 at 12:31
• @EvgenyKuznetsov And it can be further extended to analytic sets: if $A$ is an analytic set, then either $A$ is countable or $A$ contains a perfect set (and in particular has size continuum). These results are part of descriptive set theory, which you might be interested in. – Noah Schweber Feb 27 '18 at 13:01
• Why should we insist that fractals are closed or even analytic? To my way of thinking, the essence of being a fractal is self-similarity by a non-isometry. And in this case, one can easily make lots of fractals of intermediate cardinality when CH fails. – Joel David Hamkins Feb 27 '18 at 13:43
Let's define that a subset $F$ of the plane (or other suitable space) is a fractal, if there is a nontrivial group $G$ of homeomorphisms of the space, at least some of which are not isometries, such that $F=\sigma F$ for every $\sigma\in G$. This expresses the self-similarity feature of fractals.
With this definition, it is easy to find fractals $F$ of any desired infinite cardinality less than or equal to the continuum. Indeed, for any group $G$, there will be a fractal $F$ with respect to $G$ of any size between $\max(|G|,\aleph_0)$ and $\frak{c}$, inclusive.
To build $F$, just pick any collection of starting points in the plane, of the right cardinality, and then simply close the set under all the operations of $G$. This closure will have the desired cardinality, and it follows that $F=\sigma F$ for every $\sigma\in G$, and so $F$ is a fractal.
In particular, the procedure produces countable fractals $F$ with respect to any fixed countable group $G$. One simply starts with a single point $\{a\}$ and then closes under the $G$ action.
One can get versions of the Sierpinski triangle this way, by using the group $G$ that describes the fractal symmetry of that triangle.
I might add that this is a common way of drawing fractals on the computer. One just picks a point and then draws all its images, and then their images and so on, under the desired symmetries.
Often, one doesn't want to use a whole group of homeomorphisms, but a weaker collection of maps that exhibit the desired self-similarities, such as the contractive maps only, but the same idea works for this case. For example, with the fern above and the Sierpinski triangle, one doesn't usually apply the expansive maps, and so the resulting image is bounded. But even in those cases, one could apply the inverse maps that produce an unbounded set exhibiting the whole group of symmetries. In this way, one gets a more expansive version of the fern or the Sierpinski triangle that continues outward, exhibiting the desired similarities more fully.
• Concerning unbounded fractals for expanding maps look at the appendix of this paper (arxiv.org/pdf/1304.7529.pdf) for a gallery of macro-fractals (in various projections). By the way, all of them are countable. – Taras Banakh Feb 27 '18 at 14:11
• Unfortunately, this notion of a fractal doesn't really match the usual understanding of what a fractal is - for example, any subset of the place with either contains or is disjoint from an open disc will match your definition - take any homeomorphism which is constant outside this disc but nontrivial inside and let $G$ be the group it generates. It's still better than no definition, I suppose. – Wojowu Feb 27 '18 at 14:41
• Well, usually we are looking for a fractal with respect to a given group $G$, such as the ones giving rise to the Sierpinski triangle or the fern. Indeed, for many common fractals, including those two, the maps are affine. So one excludes your kind of example by insisting on a good group. – Joel David Hamkins Feb 27 '18 at 14:50
• For example, one might insist not merely that $G$ has a non-isometry, but that there is some $\sigma$ in $G$ that acts non-isometrically on $F$ and on the complement of $F$. – Joel David Hamkins Feb 27 '18 at 15:43
• Thanks. Does this transfer properties of fractals to groups? Does "fractal dimension of group" makes sense? – joro Feb 27 '18 at 16:26 | 1,415 | 5,619 | {"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 | 3 | CC-MAIN-2019-13 | latest | en | 0.92783 |
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# Tutorial 7 - EE 2000 Logic Circuit Design Semester A...
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EE 2000 Logic Circuit Design, Semester A, 2007/08 Tutorial 7 The questions are divided into two levels. Level 1 is the basic level question. You should be able to answer them after completing the lecture. Level 2 is the advanced level question. You may have to take a research before answering them. Please be prepared before the tutorial. Level 1 Question 1: (Check point of the lecture) Given a function f ( a, b, c, d ) = Σ m (0, 1, 2, 4, 5, 7, 9, 10, 11, 13, 14, 15) to you, (a) Implement the above function using a 4-to-16-line decoder and an OR gate (b) The above function has a long list of minterms, it requires an OR gate with a large number of inputs. If we still want to use decoder, instead of logically summing up the long list of minterms, can you suggest a better way to implement the function? Question 2: (Check point of the lecture) Implement the function f ( a, b, c, d ) = Σ m (1, 2, 5, 7, 9, 11, 13) using (a) A 8-to-1-line multiplexer and a NOT gate only (Hints: Tabulate the outputs of f in a truth table, and then assign variables a , b and c as selection inputs of the MUX)
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Ask a homework question - tutors are online | 465 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-09 | latest | en | 0.815774 |
https://www.tekportal.net/fiftieth/ | 1,709,316,227,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00181.warc.gz | 1,033,332,725 | 8,953 | # fiftieth
1. next after the forty-ninth; being the ordinal number for 50.
2. being one of 50 equal parts.
noun
1. a fiftieth part, especially of one (1/50).
2. the fiftieth member of a series.
1. being the ordinal number of fifty in order, position, time, etc. Often written: 50th
2. (as noun)the fiftieth in the series
noun
1. one of 50 equal or approximately equal parts of something
2. (as modifier)a fiftieth part
1. the fraction equal to one divided by 50 (1/50) | 148 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.890866 |
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# EXAMPLE 1 - PowerPoint PPT Presentation
Identify the constants a = – , h = – 2 , and k = 5 . Because a < 0 , the parabola opens down. 14. 14. EXAMPLE 1. Graph a quadratic function in vertex form. Graph y = – ( x + 2) 2 + 5. SOLUTION. STEP 1. STEP 2.
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Identify the constants a = – , h = – 2, and k = 5.
Because a < 0, the parabola opens down.
14
14
EXAMPLE 1
Graph a quadratic function in vertex form
Graphy= – (x + 2)2 + 5.
SOLUTION
STEP 1
STEP 2
Plot the vertex (h, k) = (– 2, 5) and draw the axis of symmetry x = – 2.
x = 0: y = (0 + 2)2 + 5 = 4
x = 2: y = (2 + 2)2 + 5 = 1
14
14
EXAMPLE 1
Graph a quadratic function in vertex form
STEP 3
Evaluate the function for two values of x.
Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.
Draw a parabola through the plotted points.
STEP 4
The Tacoma Narrows Bridge in Washington has two towers that each rise 307 feet above the roadway and are connected by suspension cables as shown. Each cable can be modeled by the function.
1 7000
y =(x – 1400)2 + 27
EXAMPLE 2
Use a quadratic model in vertex form
Civil Engineering
where xand yare measured in feet. What is the distance dbetween the two towers ?
EXAMPLE each rise 2
Use a quadratic model in vertex form
SOLUTION
The vertex of the parabola is (1400, 27). So, a cable’s lowest point is 1400 feet from the left tower shown above. Because the heights of the two towers are the same, the symmetry of the parabola implies that the vertex is also 1400 feet from the right tower. So, the distance between the two towers is d = 2 (1400) = 2800 feet.
for Examples 1 and 2 each rise
GUIDED PRACTICE
Graph the function. Label the vertex and axis of symmetry.
1. y = (x + 2)2– 3
SOLUTION
Identify the constants a = 1 , h = – 2, and k = – 3. Because a > 0, the parabola opens up.
STEP 1
STEP 2
Plot the vertex (h, k) = (– 2, – 3) and draw the axis of symmetry x = – 2.
for Examples 1 and 2 each rise
GUIDED PRACTICE
STEP 3
Evaluate the function for two values of x.
x = 0: y = (0 + 2)2 + – 3 = 1
x = 2: y = (2 + 2)2– 3 = 13
Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.
Draw a parabola through the plotted points.
STEP 4
for Examples 1 and 2 each rise
GUIDED PRACTICE
2. y = – (x + 1)2+ 5
SOLUTION
Identify the constants a = 1 , h = – 2, and k = – 3. Because a < 0, the parabola opens down.
STEP 1
STEP 2
Plot the vertex (h, k) = (– 1, 1) and draw the axis of symmetry x = – 1.
STEP 3
Evaluate the function for two values of x.
x = 0: y = – (0 + 2)2 + 5 = 4
x = 2: y = – (0 – 2)2 + 5 = 1
Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.
for Examples 1 and 2 each rise
GUIDED PRACTICE
Draw a parabola through the plotted points.
STEP 4
Identify the constants each rise a = , h = – 3, and h = – 4.
12
52
12
12
12
x = 0: f(x)= (0 – 3)2–4 =
– 3 2
x = 0: f(x)= (2 – 3)2–4 =
for Examples 1 and 2
GUIDED PRACTICE
3. f (x) = (x – 3)2– 4
SOLUTION
STEP 1
Because a > 0, the parabola opens up.
STEP 2
Plot the vertex (h, k) = (– 3, – 4) and draw the axis of symmetry x = – 3.
STEP 3
Evaluate the function for two values of x.
for Examples 1 and 2 each rise
GUIDED PRACTICE
Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.
Draw a parabola through the plotted points.
STEP 4
1 each rise
6500
y =
(x – 1400)2 + 27
for Examples 1 and 2
GUIDED PRACTICE
4. WHAT IF?Suppose an architect designs a bridge
with cables that can be modeled by
where x and y are measured in feet. Compare this function’s graph to the graph of the function in Example 2.
SOLUTION
This graph is slightly steeper than the graph in Example 2. They both have the same vertex and axis of symmetry, and both open up.
Solution missing | 1,457 | 4,462 | {"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} | 3.984375 | 4 | CC-MAIN-2017-47 | latest | en | 0.841689 |
https://www.ucl.ac.uk/~ucahmto/latex_html/chapter2_latex2html/node9.html | 1,601,448,521,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00077.warc.gz | 1,031,441,526 | 4,951 | # Isoperimetric Problems
So far we have dealt with boundary conditions of the form y(a) = A, y(b) = B or y(a) = A, y'(b) = B. For some problems the natural boundary conditions are expressed using an integral. The standard example is Dido's problem3: if you have a piece of rope with a fixed length, what shape should you make with it in order to enclose the largest possible area? Here we are trying to choose a function y to maximise an integral I(y) giving the area enclosed by y, but the fixed length constraint is also expressed in terms of an integral involving y. This kind of problem, where we seek an extremal of some function subject to `ordinary' boundary conditions and also an integral constraint, is called an isoperimetric problem.
A typical isoperimetric problem is to find an extremum of
I(y) = F(x, y, y') dx, subject to y(a) = Ay(b) = B, J(y) = G(x, y, y') dx = L.
The condition J(y) = L is called the integral constraint.
Theorem 3 In the notation above, if I(Y) is an extremum of I subject to J(y) = L, then Y is an extremal of
K(y) = F(x, y, y') + λG(x, y, y') dx
for some constant λ.
You will need to know about Lagrange multipliers to understand this proof: see the handout on moodle (the constant λ will turn out to be a Lagrange multiplier).
Proof. Suppose I(Y) is a maximum or minimum subject to J(y) = L, and consider the two-parameter family of functions given by
Y(x) + εη(x) + δζ(x)
where ε and δ are constants and η(x) and ζ(x) are twice differentiable functions such that η(a) = ζ(a) = η(b) = ζ(b) = 0, with ζ chosen so that Y + εη + δζ obeys the integral constraint.
Consider the functions of two variables
I[ε, δ] = F(x, Y + εη + δζ, Y' + εη' + δζ') dx, J[ε, δ] = G(x, Y + εη + δζ, Y' + εη' + δζ') dx.
Because I has a maximum or minimum at Y(x) subject to J = L, at the point (ε, δ)=(0, 0) our function I[ε, δ] takes an extreme value subject to J[ε, δ] = L.
It follows from the theory of Lagrange multipliers that a necessary condition for a function I[ε, δ] of two variables subject to a constraint J[ε, δ] = L to take an extreme value at (0, 0) is that there is a constant λ (called the Lagrange multiplier) such that
+ λ = 0 + λ = 0
at the point ε = δ = 0. Calculating the ε derivative,
[ε, δ] + λ[ε, δ] = F(x, Y + εη + δζ, Y' + εη' + δζ') + λG(x, Y + εη + δζ, Y' + εη' + δζ') dx = ηF + λG + η'F + λG dx (chain rule) = ηF + λG - F + λG dx (integration by parts) = 0
Since this holds for any η, by the FLCV (Lemma ) we get
(Fy + λGy)(x, Y, Y') + (Fy' + λGy')(x, Y, Y') = 0
which says that Y is a solution of the Euler-Lagrange equation for K, as required.
Note that to complete the solution of the problem, the initially unknown multiplier λ must be determined at the end using the constraint J(y) = L.
Exercise 6 Find an extremal of the functional
I(y) = (y')2 dx, y(0) = y(1) = 1,
subject to the constraint that
J(y) = y dx = 2. Answer: y = f (x) = - 6x - + .
Exercise 7 (Sheep pen design problem) : A fence of length l must be attached to a straight wall at points A and B (a distance a apart, where a < l) to form an enclosure. Show that the shape of the fence that maximizes the area enclosed is the arc of a circle, and write down (but do not try to solve) the equations that determine the circle's radius and the location of its centre in terms of a and l. | 1,045 | 3,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-40 | latest | en | 0.869084 |
https://savvycalculator.com/unlevered-beta-calculator/ | 1,722,842,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00492.warc.gz | 401,984,128 | 44,666 | # Unlevered Beta Calculator
## About Unlevered Beta Calculator (Formula)
An Unlevered Beta Calculator is a tool used in finance to calculate the unlevered beta of a company or asset, which represents its sensitivity to systematic market risk without accounting for its financial leverage. Unlevered beta is important in risk assessment and portfolio analysis, particularly when evaluating investments and understanding their market-related risks.
Formula for Unlevered Beta Calculation:
The formula for calculating unlevered beta involves adjusting the leveraged beta of a company for its financial leverage. The formula is:
Unlevered Beta = Leveraged Beta / (1 + (1 – Tax Rate) × Debt-to-Equity Ratio)
Where:
• Leveraged Beta: The beta value of the company, which reflects its systematic risk including financial leverage.
• Tax Rate: The corporate tax rate applicable to the company.
• Debt-to-Equity Ratio: The ratio of the company’s total debt to its total equity.
The unlevered beta represents the asset’s risk profile without the influence of financial leverage.
Applications:
1. Investment Analysis: Financial analysts use the Unlevered Beta Calculator to assess the market risk of investments and incorporate them into portfolio management strategies.
2. Cost of Capital: Unlevered beta is used to determine the asset’s required rate of return and cost of capital for valuation purposes.
3. Capital Budgeting: When evaluating investment projects, unlevered beta helps estimate the systematic risk of potential investments.
4. Mergers and Acquisitions: The calculator aids in assessing the risk and return associated with potential mergers and acquisitions.
5. Risk Management: Unlevered beta is important for risk management and understanding how different investments contribute to overall portfolio risk.
In summary, an Unlevered Beta Calculator involves calculations that assist finance professionals in evaluating the systematic market risk of investments and assets, providing insights into risk assessment, portfolio construction, and financial decision-making. | 392 | 2,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.888112 |
https://www.physicsforums.com/threads/motion-of-a-rod-on-another-rod-of-same-mass.599762/ | 1,513,471,250,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592202.83/warc/CC-MAIN-20171217000422-20171217022422-00325.warc.gz | 788,949,876 | 15,453 | Motion of a rod on another rod of same mass.
1. Apr 24, 2012
andrien
1. The problem statement, all variables and given/known data
A perfectly rough rod is gently placed with one end
upon another rod of equal mass and in the same vertical
plane, moving with the velocity √2gc on a smooth table. If
the initial inclination of the first rod to the horizon be α, and
its length 2a, shew that it will just rise to a vertical position
if
2a (1 -sin α) (5 + 3 cos ^2 α) = 3c sin ^2 a.
2. Relevant equations
3. The attempt at a solution- i have used the fact that moment of momentum just after putting rough rod will be zero about point of contact to determine initial angular velocity.after that I used equation of energy and conservation of momentum in horizontal direction.but i am not sure about the end condition i used that reaction will become zero in vertical position and also may be angular velocity will become zero.none of these have yielded answers.at least i want to know the end condition.so please help.source of the problem is a.s. ramsey dynamics vol.2. | 254 | 1,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-51 | longest | en | 0.859005 |
https://www.effortlessmath.com/math-topics/how-to-find-discriminant-of-quadratic-equation/ | 1,675,091,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00311.warc.gz | 762,896,396 | 14,704 | # How to Find Discriminant of Quadratic Equation?
The discriminant in mathematics is defined for polynomials and is a function of polynomial coefficients. Expresses the nature of the roots, or in other words, it discriminates the roots.
## Step by step guide tofinding discriminant of quadratic equation
Discriminant of a polynomial in math is a function of the coefficients of the polynomial. This is useful in determining the type of solutions to a polynomial equation without actually finding them. That is, it distinguishes the solutions of the equation (equal and unequal; real and unreal) and hence it is called “discriminant”. The value of the discriminant can be any real number (i.e., either positive, negative, or $$0$$).
### Discriminant formula
The discriminant ($$Δ$$ or $$D$$) of any polynomial is in terms of its coefficients. The discriminant formula for the quadratic equation is:
### How to find discriminant?
To find the discriminant of a quadratic equation, we just need to compare the given equation with its standard form and determine the coefficients.
#### Discriminant of a quadratic equation
The discriminant of a quadratic equation $$ax^2+bx+c = 0$$ is in terms of its coefficients $$a, b$$, and $$c$$. i.e.,
$$\color{blue}{Δ\:Or\:D = b^2− 4ac}$$
Do you remember using $$b^2 – 4ac$$ before? Yes, it is a part of the quadratic formula: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$. Here, the expression that is inside the square root of the quadratic formula is called the discriminant of the quadratic equation. The quadratic formula in terms of the discriminant is: $$x=\frac{-b\pm \sqrt{D}}{2a}$$.
### Discriminant, and nature of the roots
The roots of a quadratic equation $$ax^2+bx+c = 0$$ are the values of $$x$$ that satisfy the equation. They can be found using the quadratic formula: $$x=\frac{-b\pm \sqrt{D}}{2a}$$. Although we can not find the roots just by using discriminant, we can determine the nature of the roots as follows.
#### If the discriminant is positive:
If $$D>0$$, the quadratic equation has two different real roots. This is because, when $$D>0$$, the roots are given by $$x=\frac{-b\pm \sqrt{Positive\:number}}{2a}$$, and the square root of a positive number always results in a real number. So when the discriminant of a quadratic equation is greater than $$0$$, it has two roots that are distinct and real numbers.
#### If the discriminant is negative:
If $$D< 0$$, the quadratic equation has two different complex roots. This is because when $$D< 0$$, the roots are given by $$x=\frac{-b\pm \sqrt{Negative\:number}}{2a}$$, and the square root of a negative number leads to an imaginary number always. Therefore when the discriminant of a quadratic equation is less than $$0$$, it has two roots which are distinct and complex numbers (non-real).
#### If the discriminant is equal to zero:
If $$D = 0$$, the quadratic equation has two equal real roots. In other words, when $$D=0$$, the quadratic equation has only one real root. This is because, when $$D= 0$$, the roots are given by $$x=\frac{-b\pm \sqrt{0}}{2a}$$, and the square root of a $$0$$ is $$0$$. Then the equation turns into $$x= -\frac{b}{2a}$$ which is only one number. So when the discriminant of a quadratic equation is zero, it has only one real root.
The graph of a quadratic function in each of these $$3$$ cases can be as follows:
### Finding Discriminant of Quadratic Equation– Example 1:
Find the discriminant of the following equation $$3x^2+10x−8$$.
Solution:
The given quadratic equation is $$3x^2+10x−8$$. Comparing this with $$ax^2+bx+c= 0$$, we get $$a =3, b = 10$$, and $$c =-8$$.
$$D = b^2- 4ac$$
$$=(10)^2- 4(3)(-8)$$
$$=100 + 96$$
$$= 196$$
## Exercises for Finding Discriminant of Quadratic Equation
### Find the discriminant of the following equation.
1. $$\color{blue}{3x^2+4x-8=0}$$
2. $$\color{blue}{5x^2+3x+1=0}$$
3. $$\color{blue}{6x^2-x-15=0}$$
4. $$\color{blue}{4x^2+4x+1=0}$$
1. $$\color{blue}{112}$$
2. $$\color{blue}{-11}$$
3. $$\color{blue}{361}$$
4. $$\color{blue}{0}$$
### What people say about "How to Find Discriminant of Quadratic Equation?"?
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SAVE $5 It was$16.99 now it is \$11.99 | 1,199 | 4,229 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-06 | longest | en | 0.847064 |
https://learnetutorials.com/php/programs/reverse-number | 1,716,873,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00017.warc.gz | 304,746,857 | 11,863 | # PHP Program to print the reverse of a number
March 15, 2022, Learn eTutorial
1802
## What is the reverse of a number?
In this program, we are reversing the number using PHP. Reversing the number is meant by printing the user inputted number in reverse order. For example, if the user entered number is 567 then the output of the program will be 765.
## How to find the reverse of a number in PHP?
To find the reverse of the number entered by the user in PHP first we have to read the number into a variable then we have to perform the operations first we have to find the Mod(%) of the number and assign it to the variable rem then we have to multiply by 10 with the variable rev and add the current value in the variable rem to it and assign it to the variable rev and after that divide the number by 10 to avoid the last digit of the number and assign the number into the variable number iterate these operations using `while loop` until every digit of the variable is reversed. And after the completion of the iteration print the value of the variable rev as the reverse of the number.
### ALGORITHM
Step 1: Accept the number from the user which wants to be reversed into a variable number
Step 2: Assign the current number from the variable number to num
Step 3: Assign the value 0 into variable rev to store the reverse of the number
Step 4: Perform the following sub-steps until the condition 'num > 1' becomes false
(i) Perform the operation 'num % 10' and assign the value to the variable rem to store the remainder
(ii) Perform the operation 'rev * 10 + rem' and assign the value to the variable rev to store in reverse
(ii) Perform the operation 'num / 10' and assign the value to the variable num to store the number without the last
Step 5: Print variable rev as the reverse of the number entered by the user in variable num
## PHP Source Code
``` ```<?php
\$number = readline("Enter the number: ");
\$num = \$number;
\$rev = 0;
while (\$num > 1) {
\$rem = \$num % 10;
\$rev = (\$rev * 10) + \$rem;
\$num = (\$num / 10);
}
echo "Reverse number of \$number is: \$rev";
?>```
```
## OUTPUT
```Enter the number: 9687
Reverse number of 9687 is: 7869```
VIEW ALL
VIEW ALL | 556 | 2,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.832008 |
https://www.yoforia.com/what-is-the-basic-structure-of-a-sandwich/ | 1,679,889,712,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00601.warc.gz | 1,191,808,767 | 11,176 | ## What is the basic structure of a sandwich?
There are three basic components to any sandwich: the bread, the filling, and the spread or accompaniment. Each part contributes to making a sandwich greater than the sum of its parts.
## Is a sandwich a construct?
A composite construction consisting of relatively thin layers of a material (having high-strength properties) bonded to a thicker, weaker, light core material; results in high ratios of strength to weight and stiffness to weight.
What is a sandwich type construction?
BASIC DESIGN PRINCIPLES: Structural sandwich is a layered composite formed by bonding two thin facings to a thick core. It is a type of stressed-skin construction in which the facings resist nearly all of the applied edgewise (in-plane) loads and flatwise bending moments.
### What is analysis of sandwich structures?
A sandwich structure is comprised of layered composite materials formed by bonding two or more thin facings or facesheets to a relatively thick core material. This article describes the sandwich panel failure modes. It tabulates the nomenclature and definitions for loads, geometry, and material properties.
### What are the three basic components of a sandwich?
There are 3 main components of a sandwich: the bread, the spread, and the filling.
How do you assemble a perfect sandwich?
How to Stack the Perfect Sandwich
1. Step 1: Apply the Condiments. The first step to proper sandwich assembly is to apply condiments to the bread.
2. Step 2: Create a Bread Barrier.
3. Step 3: Layer the Meat and Cheese.
4. Step 4: Add the Finishing Touches.
#### What are the material used for sandwich construction?
Facesheet materials that are normally used are aluminum, glass, carbon or aramid. Typical sandwich structure has relatively thin facing sheets (0.010–0.125 inches) with core densities in the range of 1–30 pcf (pounds per cubic foot).
#### How do you prepare sandwich?
To make a basic sandwich, first grab 2 slices of your favorite bread, like French, white, or wheat bread. Next, spread condiments, like mustard and mayo, evenly on each slice of bread. Then, add your main filling, such as deli meat, egg salad, or hummus, on top of one of the bread slices.
Where are sandwich structures used?
Sandwich panels are used in aeronautics, road vehicles, ships, and civil engineering. The mechanical properties of these composites are directly dependent on the properties of sandwich components and method of manufacturing.
## Why are sandwich composites good?
The major advantages of sandwich composites over conventional materials are that sandwich composites (1) have a low overall density, a high strength-to-weight ratio, and a high stiffness-to-weight ratio; (2) are capable of providing good thermal and acoustical insulation; and (3) have uniform energy absorption capacity … | 601 | 2,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.887709 |
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www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
# MGT411 Money & Banking GDB Solution & Discussion Fall 2011
Suppose there are only 100 goods being produced in a country during a particular year. In the next year the same numbers of goods are produced but the statistics show an increase in the GDP of the country in that year. Can GDP of a country increase without increasing the number of goods being produced in that country? If yes, then how? If, ‘No’ then why?
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Views: 1427
### Replies to This Discussion
Just For Idea
Suppose there are only 100 goods being produced in a country during a particular year. In the next year the same numbers of goods are produced but the statistics show an increase in the GDP of the country in that year. Can GDP of a country increase without increasing the number of goods being produced in that country? If yes, then how? If, ‘No’ then why?
Nominal GDP can because it is not adjusted for inflation.It only shows the total value of goods produced in the country at current prices. Real GDP, which is adjusted for inflation cannot increase without the number of goods produced increasing.
Money & Banking
MGT411
GDB Idea Solution
Nominal GDP can because it is not adjusted for inflation. It only shows the total value of goods produced in the country at current prices. Real GDP, which is adjusted for inflation cannot increase without the number of goods, produced increasing.
i m suggesting as
Yes, because GDP is a value, so typically if prices inflate the GDP will increase even if the actual number of items is static.
This is the situation in most western countries over the past few decades - most of the work done is increasing services and decreasing actual goods, but the GDPs have continued to rise rapidly.
Unlike nominal GDP, real GDP can account for changes in the price level, and provide a more accurate figure.
Let's consider an example. Say in 2004, nominal GDP is \$200 billion. However, due to an increase in the level of prices from 2000 (the base year) to 2004, real GDP is actually \$170 billion. The lower real GDP reflects the price changes while nominal does not
PLz upload solution as soon as possible
Ans:-
Yes the GDP of a country increase without increasing the number of goods being produced in that country.
GDP stand for Gross Domestic Product. Its mean that the product that is produced in a country in a specific period of time. Mostly the time contain one year or more then one year. The GDP can be calculated in to the three ways
The product method
The expenditure Method
The Income Methed
kese aik ko b define kr k likh do
rao ali raza
assalamo alikum........ in 3 method ma se kisi aik ko define krna ha?
yar kis wale topic me se detail karna ha ye to batao plzzzz
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This is a member-supported website. Your contribution is greatly appreciated! | 815 | 3,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-40 | longest | en | 0.928241 |
https://brilliant.org/problems/i-just-got-sick-reading-this/ | 1,521,747,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00342.warc.gz | 538,473,282 | 11,400 | # I just got sick from reading this
Logic Level 2
If Tuesday was four day before the day after two fortnights ago, what is the least number of days five years after this exact date for which it will be a Wednesday again given that three years ago was a leap year and there will be less than ninety nine days til Christmas Eve starting from the day after tomorrow?
For the sake of simplicity, we assume there are exactly 25% of the years that are leap years.
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https://techcommunity.microsoft.com/t5/excel/formulas/td-p/4002161 | 1,721,316,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00499.warc.gz | 491,057,374 | 56,868 | # Formulas
Copper Contributor
# Formulas
Using an average formula =average (d4:h4)for 5 cells in a row then delete d4 shift cells left formula changes to = average(d4:g4) , how can I stop this changing?
4 Replies
# Re: Formulas
You can use =AVERAGE(INDIRECT("D4:H4"))
# Re: Formulas
Getting value error
# Re: Formulas
There are various way but I would ask WHY. Is this something you are doing a lot in general or something for a particular case? For example you could use INDIRECT("D4:H4") but I try to avoid using that. You could use INDEX(4:4,{4,5,6,7,8}). But as a general solution either or both of those may be very burdensome. Maybe the better answer is an alternative to not delete the cell but shift the data.
# Re: Formulas
Probably best to clear contents rather than delete but here's an option:
``=AVERAGE(TAKE(\$D\$4:\$ZZ\$4,,5))`` | 235 | 856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.890523 |
https://www.lmfdb.org/Character/Dirichlet/grouptable?modulus=171&char_number_list=160,1,37,121,106,31&poly=x%5E6+-+95*x%5E3+%2B+684*x%5E2+-+570*x+%2B+2375 | 1,718,677,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00228.warc.gz | 754,986,002 | 4,955 | ## Group table for the character group for $\textrm{Gal}(K/\mathbb{Q})$
$K$ is the global number field defined by $$x^{6} - 95x^{3} + 684x^{2} - 570x + 2375$$
$\times$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 31, ·)$$
$$\chi_{ 171 }(1, ·)$$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 31, ·)$$
$$\chi_{ 171 }(160, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 31, ·)$$ $$\chi_{ 171 } ( 1, ·)$$
$$\chi_{ 171 }(37, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 31, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 121, ·)$$
$$\chi_{ 171 }(121, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 31, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 160, ·)$$
$$\chi_{ 171 }(106, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ $$\chi_{ 171 } ( 31, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 37, ·)$$
$$\chi_{ 171 }(31, ·)$$ $$\chi_{ 171 } ( 31, ·)$$ $$\chi_{ 171 } ( 1, ·)$$ $$\chi_{ 171 } ( 121, ·)$$ $$\chi_{ 171 } ( 160, ·)$$ $$\chi_{ 171 } ( 37, ·)$$ $$\chi_{ 171 } ( 106, ·)$$ | 732 | 1,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.223897 |
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# mat116_appendix_e - Axia College Material Appendix E...
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Axia College Material Appendix E Fueling Up Motorists often complain about rising gas prices. Some motorists purchase fuel efficient vehicles and participate in trip reduction plans, such as carpooling and using alternative transportation. Other drivers try to drive only when necessary. Application Practice Answer the following questions. Use Equation Editor to write mathematical expressions and equations. First, save this file to your hard drive by selecting Save As from the File menu. Click the white space below each question to maintain proper formatting. 1. Suppose you are at the gas station filling your tank with gas. The function C ( g ) represents the cost C of filling up the gas tank with g gallons. Given the equation: ) ( 03 . 3 ) ( g g C = a) What does the number 3.03 represent? The number 3.03 represents the cost C of gas. b) Find C (2) 3.03(2) = 6.06 c) Find C (9) 3.03(9) = 27.27 d) For the average motorist, name one value for g that would be inappropriate for this function’s purpose. Explain why you chose the number you did.
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# Inertia tensors for common shapes
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I''m attempting a rigid body simulator, and I need the inertia tensors for some common shapes: Box, cylinder, & sphere. As I understand it, this tensor comes in the form of a 3x3 matrix, and is somehow derived from the moments of inertia around the 3 primary axi... Anyway, I''m sure there''s a table of these out there somewhere, but I''ve yet to find it. Any help is much appreciated!
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First it's usual for symmetric shapes to align the axes along the axes/planes of symmetry of the shape when calculating the moment of inertia. If you do this the tensor is generally diagonal, i.e. it looks like this
(X 0 0)
(0 Y 0)
(0 0 Z)
The ones you want are:
Box sides 2a, 2b, 2c:
X = (b2 + c2) / 3
Y = (a2 + c2) / 3
Z = (a2 + b2) / 3
X = r2 / 2
Y = Z = r2 / 4 + h2 / 3
X = Y = Z = 2 * r2 / 5
All of them are for a body with unit mass: multiply by the mass otherwise. I'm not sure of the last one but the other two I'm pretty sure of. I don't have any references with them in: whenever I need them I work them out if I don't know them. I think you can find them in books of formulae and integrals.
Edited by - johnb on August 9, 2001 12:32:22 PM
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quote:
Original post by johnb
X = Y = Z = 2 * r2 / 5
This one is also correct.
I also would like to add some tips.
I. Moment of Inertia is additive. If your object is composed of some smaller shapes (eg. of two boxes and one cylinder)
then find Inertia moments of those shapes and add together
to obtain the inertia moment of whole object.
II. If you want to find an inertia moment I1 for some axis and
you know inertia moment I2 for another axis whitch is pararell to first one, then I1 = I2 + m*d2 where m - mass
d - distance between axes.
(e.g. pipe of length l and mass m rotates along axis, perpendicular to pipe and coming through pipes middle.
Then I = (1/12)*m*l2.
When pipe rotates along axis coming through its end,
then I = (1/12)*m*l2 + m*(l/2)2 = (1/3)*m*l2 )
I hope you find it usefull.
K.
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quote:
I. Moment of Inertia is additive. If your object is composed of some smaller shapes (eg. of two boxes and one cylinder) then find Inertia moments of those shapes and add together to obtain the inertia moment of whole object.
This works for 2D inertia momentum, but I don''t think this will work for 3D inertia tensors. You would have to re-evaluate the volume integrals, if the shape changes. You might approximate with the tensor of a bounding box / sphere instead.
- AH
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quote:
Original post by Anonymous Poster
This works for 2D inertia momentum, but I don''t think this will work for 3D inertia tensors. You would have to re-evaluate the volume integrals, if the shape changes. You might approximate with the tensor of a bounding box / sphere instead.
- AH
It should work in 3D, since it does not matter if you integrate
whole volume of object or integrate volumes of its parts and
(I believe that is the reason why inertia momentum is additive -
because it is defined by an integral).
K.
By the way, the first A. P. replay is mine. I forgot to login.
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> It should work in 3D, since it does not matter if you integrate
> whole volume of object or integrate volumes of its parts and
> (I believe that is the reason why inertia momentum is additive -
> because it is defined by an integral).
The moment of inertia about an axis of a collection of point masses is defined as the sum of the masses x the square of their distances from the axis. Integration is just the continuous version of this, i.e. an integral is the limit of what you get from dividing a body into smaller and smaller parts then adding up their moments of inertia. The inertia tensor combines the moment of inertias about all axes into a usable form, and so needs 3 to 6 integrals/sums to work out.
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Hmm, say I have a complex object, and I would like to get it''s inertia tensor. I split the object up into small tetrahedra, and calculate the tensor of each tetrahedron. How do I process those individual tensors to get the final inertia tensor for my complex object ?
btw: does someone know the formula for the inertia tensor of a tetrahedron ?
- AH
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> Hmm, say I have a complex object, and I would like to get it''s
> inertia tensor. I split the object up into small tetrahedra,
> and calculate the tensor of each tetrahedron. How do I process
> those individual tensors to get the final inertia tensor for
> my complex object ?
Add them up. They''re all 3x3 matrices.
> btw: does someone know the formula for the inertia tensor of
> tetrahedron ?
If it''s a regular tetragedron, i.e. with all sides the same length and equilateral triangle faces, then I think by symmetry the inertia tensor should be a multiple of the identity matrix, but exactly what values I don''t know.
One way of evaluationg it would be start with a cube then join four vertices (chosen so none share an edge of the cube) to form a tetrahedron. This can be formed from a cube by shaving off four maximal pyramids/tetrahedra, each 1/6 the volume of the cube, from the other four corners. If you work out the moment of inertia of these, and know that of the cube, the moment of inertia of the tetrahedron can be got by subtracting one from the other.
For a more general tetrahedra you could eveluate it directly from three integrals, but these would be tricky to set up.
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quote:
Add them up. They''re all 3x3 matrices.
That works ? That''s cool, makes it alot easier
The tretrahedra will be general, so I think I''ll try to evaluate the integrals numerically.
I also thought about another type of preprocess to get the inertia tensor: If I scatter a large number of ''point-masses'' at random positions inside the complex object and calculate their ''virtual'' masses, could I find the inertia tensor by adding up the individual moments of inertia of each point ? More samples would lead to a more accurate tensor.
- AH
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quote:
Original post by Anonymous Poster
I also thought about another type of preprocess to get the inertia tensor: If I scatter a large number of ''point-masses'' at random positions inside the complex object and calculate their ''virtual'' masses, could I find the inertia tensor by adding up the individual moments of inertia of each point ?
- AH
Yes, you can.
K.
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× | 1,755 | 6,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-39 | latest | en | 0.94507 |
https://mapleprimes.com/questions/229208-How-Do-Show-All-Roots-In-The-Equation | 1,632,050,806,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00502.warc.gz | 428,620,751 | 65,911 | # Question:How do show all roots in the equation with Bairstow Method?
## Question:How do show all roots in the equation with Bairstow Method?
Maple 2018
I'm trying show the all roots of equations using Bairstow's methodt, but only shows the roots of Quadratic Factor and don't show the others roots of the other equation. Thanks
This the code:
>
1 -0.6441699 0.1381090 0.3558 1.138 2 -0.5111131 0.4697336 0.1331 0.3316 3 -0.4996865 0.5002023 0.01143 0.03047 4 -0.5000001 0.5000000 -0.0003136 -0.0002023 5 -0.5000000 0.5000000 6.413e-08 9.268e-09 6 -0.5000000 0.5000000 0 0 Q(x)=(1)x^3 + (-4)x^2 + (5.25)x^1 + (-2.5) Remainder: 0(x-(-0.5))+(0) Quadratic Factor: x^2-(-0.5)x-(0.5) Zeros: 0.4999999993, -0.9999999997
Only show two roots: 0.4999999993, -0.9999999997, but the other roots are missing: 2, -1, 1+0.5i, 1-0.5i approximately, any solution?
I think it's in this part, but I can't think of how to implement it to get the missing roots. | 429 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.714395 |
https://socratic.org/questions/how-much-heat-is-transferred-when-a-24-7-kg-iron-ingot-is-cooled-from-880-c-to-1 | 1,576,270,000,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569146.17/warc/CC-MAIN-20191213202639-20191213230639-00158.warc.gz | 534,306,915 | 6,787 | How much heat is given off when a 24.7 kg iron ingot is cooled from 880°C to 13°C?
The specific heat of iron is 0.107 cal/g * °C.
Sep 15, 2016
$2.3 \cdot {10}^{6} \text{cal}$
Explanation:
Notice that the problem provides you with the mass of the iron ingot, but that it's being expressed in kilograms.
Since the specific heat of iron is given in joules per gram Celsius, ${\text{J g"^(-1) ""^@"C}}^{- 1}$, you must convert the mass of the sample from kilograms to grams before doing anything else
24.7 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 24.7 * 10^3"g"
Now, you can use the specific heat of iron to figure out how much heat would be released when the temperature of $24.7 \cdot {10}^{3} \text{g}$ of iron decreases by ${1}^{\circ} \text{C}$
24.7 * 10^3 color(red)(cancel(color(black)("g"))) * overbrace("0.107 cal"/(1color(red)(cancel(color(black)("g"))) ""^@"C"))^(color(purple)("the specific heat of iron")) = 2.643 * 10^3"cal" ""^@"C"^(-1)
This tells you that in order to cause a ${1}^{\circ} \text{C}$ decrease in temperature to a $24.7 \cdot {10}^{3} \text{g}$ sample of iron, you must remove $2.643 \cdot {10}^{3} \text{cal}$ of heat.
In your case, the temperature of the ingot must decrease by
$\text{change in temperature" = |13^@"C" - 880^@"C"| = 867^@"C}$
You can thus say that when the temperature of $2.643 \cdot {10}^{3} \text{g}$ of iron decreases by ${867}^{\circ} \text{C}$, the process releases
867 color(red)(cancel(color(black)(""^@"C"))) * overbrace((2.643 * 10^3"cal")/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("for 2.643" * 10^3"g of iron")) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.3 * 10^6"cal")color(white)(a/a)|)))
The answer is rounded to two sig figs. | 600 | 1,786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-51 | longest | en | 0.771009 |
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# Loan with decreasing payments....
0
300
1
A loan for \$1,000,000 is granted by the Bank to ABC Importing Company. The loan is to be repaid in 10 years and bears an interest rate of 5% compounded monthly. The annual payments will start with P for the first year and decrease by 2% annually thereafter. What is the initial payment that ABC Company must pay to the Bank? And how much interest will cost the Company over the 10-year period? Thank you for any help.
May 24, 2017
#1
0
It is little ambiguous about the 2% decrease in payments. For this purpose, I shall assume that the payments will decrease by 2% from the previous payment. In other words the payments will decrease as follows: \$1, \$0.98, \$0.9604(0.98 x 0.98), \$0.941192(0.9604 x 0.98), 0.98^4....etc. to the 10th payment.
The interest rate is 5% compounded monthly. Will adjust it to compounded annually to match the annual payments, which comes to =5.11618978817%.
Now, we just find the Present Value of the above payments, which comes to =7.08110996994. And then the annual, or initial payment, will be =\$1,000,000 / 7.08110996994 =\$141,220.80. This payment will decrease by 2% annually from the previous payment.
To calculate the interest paid by the ABC Company, we simply have to add the 10 payments which come to a total of =\$1,291,656.23. By subtracting the loan of \$1,000,000 we get =\$291,656.23 total interest paid by the Company over a period of 10 years.
P.S. There is a specific formula used for such problems, which is rarely used, but comes in as very handy in calculating such problems. It is written like this: PV = P*(((1+G) / (1+R))^N - 1) / (G-R), where G =% increase or decrease in payment, R=Interest rate per period, N=Number of periods, P=Periodic payment, PV=Present Value.
May 24, 2017 | 503 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.954905 |
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# What is the energy of a photon that emits a light of freqeuncy7.21x1014hz?
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4.25 10-19 j
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4.78 x 10-19
2.96 x 10^-19 J
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### When electron drops to a lower energy level what is the energy of the proton released?
You may be confusing "proton" with "photon". A proton is a positively-charged particle contained within the nucleus of an atom. A photon is a discrete unit of energy normally expressed as light. Around the nucleus of the atom, there are some electrons in energy levels. When an atom absorbs energy, it absorbs a specific amount, or "quantum" of energy and the electron boosted to a higher energy level. When the electron drops to a lower energy level, it emits a photon in the form of light at a specific energy and frequency. | 568 | 2,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.907474 |
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# One of four babies are now born to mothers aged thirty years
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One of four babies are now born to mothers aged thirty years [#permalink]
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17 Dec 2004, 07:13
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One of four babies are now born to mothers aged thirty years or more, compared with just one of six born in 1975.
(A) of four babies are now born to mothers aged thirty years or more, compared with just one of six born
(B) of four babies is now born to a mother whose age is thirty or older, compared to just one of six babies who were born
(C) baby in four are now born to mothers aged thirty or older, compared to just one in six
(D) baby in four is now born to a mother aged thirty or older, compared with just one in six
(E) baby in four is now born to mothers aged thirty years or more, compared to just one in six
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17 Dec 2004, 08:18
(D)
compare with should be correct
one of 4 -> singular
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Re: GMAT Plus Book 2 Problem #11 [#permalink]
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18 Dec 2004, 13:17
One of four babies are now born to mothers aged thirty years or more, compared with just one of six born in 1975.
First impression: "One in four" would be better since we are talking about ratio here. "are"? "Compared with" seems wrong. Can't put my finger on why right now. A is probably wrong.
(B) of four babies is now born to a mother whose age is thirty or older, compared to just one of six babies who were born
"A mother" is incorrect. The statement is referring to more than one mother.
(C) baby in four are now born to mothers aged thirty or older, compared to just one in six
Ah ha, "compared to." This choice seems correct (although "one baby in four are" - shouldn't the verb be singular?
(D) baby in four is now born to a mother aged thirty or older, compared with just one in six
See my explanation of B.
(E) baby in four is now born to mothers aged thirty years or more, compared to just one in six
On second thought, it's should be "born to one mother." One baby cannot be born to several mothers.
I guess I would go with D) in the end. Sorry about the confusion.
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18 Dec 2004, 22:17
Both 'compared to' and 'compared with' are correct usages in general.
Here 'compared with' is appropriate as you are comparing the similarities/dissimilarities of two things.
'compared toc can be used for analogous things - 'His strength can be compared to the strength of Hercules'
'One of' is wrong, 'One in' is right
So D it is
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Re: GMAT Plus Book 2 Problem #11 [#permalink]
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19 Apr 2010, 17:03
By POE directly we can identify D
Every other option has either baby or mothers or were
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Re: GMAT Plus Book 2 Problem #11 [#permalink]
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01 May 2010, 07:32
I dont catch this one.
Can anybody explain further?
Thanks
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Re: GMAT Plus Book 2 Problem #11 [#permalink] 01 May 2010, 07:32
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# CH5_Gas_Laws_Study_Guide%281%29 - Name:...
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Unformatted text preview: Name: ___________________________________ CH101 A The Gas Laws – Study Guide sections 5.2 and 5.3 in the textbook Pressure The pressure is defined as________________________________________ The SI unit for pressure is_________. However, chemists use ________, ________, and _______ more commonly as units for measuring pressure. Complete the following conversion table: 680.0 mm Hg 16.5 psi kPa torr atm List the following data in order of increasing pressure: (a) 102 kPa (b) 782 torr (c) 1.05 atm Gas Laws The 4 variables or physical properties that are used to define the state of a gaseous system are _______________, _________________,_________________, and __________________. The constant variables in Charles’ law are ________________ and ________________, the changing variables are __________________and ___________________. The constant variables in Boyle’s law are __________________ and ________________, the changing variables are _________________and ____________________. The constant variables in Avogadro’s law are ________________ and _______________, the changing variables are ________________and _________________. Suppose a gas has an initial pressure of 2.0 atm. What happens to the pressure if the volume is doubled (assuming the temperature and moles of gas remain unchanged)? To solve this problem, one needs to use ________________’s law. Now suppose a gas occupies 1.5 liters at a certain temperature and pressure. What happens to the volume if the number of moles of gas is tripled? To solve this problem, one needs to use _____________’s law. Watch video tutorials (2) on Blackboard All of the above gas laws can be combined into one overall equation, called the Combined Gas Law Equation. P1 V1 P2 V2 = n1 T1 n2 T2 Any variables which are held constant may be cancelled out of this equation. End of Chapter Practice Problems Tro First Edition: #29 (a‐b), 35, 37, 39 Tro Second Edition: #29 (a‐b), 35, 37, 39 answers are located in Appendix III of the textbook Take the Quiz on Blackboard 2 ...
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## This note was uploaded on 11/11/2011 for the course MATH 180 taught by Professor Byrns during the Spring '11 term at Montgomery College.
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### Use Of Generalized Coordinates, Momenta and Force In Theoretical Mechanics
When 3rd year physics students are first introduced to theoretical mechanics they are generally taken aback when confronted by the concepts of generalized coordinates, momenta and forces. In this post I examine aspects of this and hope to show why such generalizations manifest practical use and power in solving problems in theoretical mechanics.
We may write generalized coordinates as:
1 q 2 …….. n
This would apply to some system composed of n particles with associated position vectors:
r , r 2 …….. n
Such that we may write: å n i i
In all such cases we require that the Jacobian determinant* J 0 else no legitimate set of generalized coordinates can be defined in a specific form. Consider the case of the above for x,y variables with stationary axis:
x = r cos q, y = r sin q
And when the axis is moving over time, t with phase angle f):
x= r cos (wt + f) and: y = r sin (wt + f)
Example Problem:
Show that (x, y) or (r, q) can be used as generalized coordinates q 1 , q 2 …….. n
Solution:
Let x = x(q 1 , q 2 , t)
Then write: x’ = dx/ dt =
( x/ 1) 1/ t +( x/ 2) 2/t + …. x/ t
dx/ dt =
( x/ 1) 1’ + = ( x/ 2) 2’ + ….…. x/ t
And:
x’ = r’ cos q - r sin q q’
y’ = r’ sin q + r’ cos q q’
(Rem: r’ = dr/ dt ; q’ = dq/ dt )
So the coordinates work in either system, for circular motion
Generalized Momenta can be angular (p q , p f ) or linear ( p x, p , p z ):
For generalized forces: consider a particle which has moved an incremental amount d where:
d = d x + d y j + d j
This is an actual displacement but so small that the forces don’t change, say for a vertical displacement. For N particles we have:
d W = å N i (F i x d x + F i y d y + F i z d z )
Given:
x/ q’ = p q
this can be referred to generalized coordinates: q 1 , q 2 …….. n
d x = ( x/ 1d q 1 + ( x/ 2d q 2 +
( x/ 3d q 3
Or in polar coordinates:
d x = cos q d r - r sin d q
d y = sin q d r - r cos d q
In general, we may write: d W = Q k d q k
= Q r r + Q q d q
Consider now the transformation of:
F = F x i + F y j To:
= F r r + F q q
The generalized force is:
r = - V / r =
- V / x (- x / r ) - V / y (- x / y )
= F x ( x / r ) + F y ( y / r )
= F x cos q + F y sin q = F r
We may now introduce the diagram below:
To derive the generalized angular force Q q :
q = F x ( x / q ) + F y ( y / q )
= F x r sin q + F y r cos q = r F q
Note from the diagram this is a component in the direction of q ^
It is of interest here to obtain the Lagrangian in polar coordinates. We know:
x = r cos q and y = r sin q
So that:
x’ = r’ cos q - r sin q q’
y’ = r’ sin q + r’ cos q q’
The kinetic energy T is:
T = ½ m ( r’ 2 + r 2 q ‘2 )
V = mg r sin q
Therefore:
L = T - V =
½ m ( r’ 2 + r 2 q ‘2 ) - mg r sin q
It is also useful to consider the unit vectors associated with polar coordinates in central force problems: pointing in direction of increasing r, and l, in the direction of increasing q .
The velocity components can then be written:
r = dr/ dt, v q = r dq /dt
Using the polar coordinate unit vectors (n, l) this can be rewritten as:
v = dr/ dt n + r dq /dt l
For the change in the radial coordinate alone:
v = dr/ dt = d(r n)/ dt = ( dr/ dt n + r dn /dt)
Where the last derivative can be evaluated from:
dn /dt = dn / dq (dq /dt)
To evaluate dn / dq one makes use of the fact that by radian measure of angles:
D n = D q and in the limit as D q ® 0,
D n ‖ / ‖D q ‖ = 1
So: dn / dq = l
i.e. As D q ® 0, D n becomes perpendicular to and assumes the direction of l .
In an analogous way we may obtain the relation:
dl / dq = -n
Finally, we have: dn /dt = dq /dt l
And: dl /dt = - dq /dt n
dl / dq = -n
Suggested Problems:
1. Using one or more of the preceding unit vector relations, obtain an expression for the acceleration in two dimensions of polar coordinates.
2.A particle of mass m moves in a plane under the influence of a force F = - kr, directed toward the origin. Sketch a polar coordinate system (r, q ) to describe the motion of the particle and thereby obtain the Lagrangian (L = T - V, i.e. difference in kinetic and potential energy).
* | 1,698 | 4,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-14 | latest | en | 0.498302 |
https://mydeedeesdiary.com/qa/why-is-math-so-hard.html | 1,623,964,049,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00531.warc.gz | 382,043,082 | 20,064 | # Why Is Math So Hard?
## What is the hardest math problem in the world?
Riemann HypothesisToday’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math.
It’s one of the seven Millennium Prize Problems, with a million dollar reward for its solution..
## What is the 9 trick in math?
A Mathemagical Trick Start by thinking of a number, any number. Now, multiply that number by 9. If the result is a multi-digit number, add its digits together to come up with a new number. If that new number is still a multi-digit number, add its digits together to come up with yet another new number.
## How can you tell your brain faster?
Start by taking the first digit of the first number (2 for 24) and multiplying that by the number directly higher than it, which will give you the first digit(s) of the answer. So for 24 multiplied by 26, it would be 2 (first digit in first number) multiplied by 3 (one digit higher) = 6.
## How can I love studying?
Here are our top tips for finding ways to have fun while studying – whatever the subject may be.Listen to good music. … Turn it into a game for yourself. … Turn it into a game with others. … Use nice stationery. … Try roleplay. … Study somewhere different. … Challenge yourself. … Write comics, short stories or songs.More items…
## What is the most hated color?
Pantone 448 C, also referred to as “the ugliest colour in the world”, is a colour in the Pantone colour system. Described as a “drab dark brown”, it was selected in 2012 as the colour for plain tobacco and cigarette packaging in Australia, after market researchers determined that it was the least attractive colour.
## Which is the best subject in the world?
Computer Science & Information Systems.Engineering & Technology. … Business & Management Studies. … Medicine. … Economics & Econometrics. … Law. … Mechanical, Aeronautical & Manufacturing Engineering. … Architecture. … More items…
## Which is the hardest language in the world?
The Hardest Languages In The World To LearnMandarin. Right at the top is the most spoken language in the world: Mandarin. … Arabic. Number two, Arabic, challenges English speakers because most letters are written in 4 different forms depending on where they’re placed in a word. … Japanese. … Hungarian. … Korean. … Finnish. … Basque. … Navajo.More items…•Dec 6, 2016
## Is chemistry easier than math?
If hardcore math like theorems and their proofs interest you, you will feel mathematics is easier than chemistry. If you like the application of these theorems, then chemistry is easier. … At the highest level, mathematics deal in logical thinking and ability to apply the knowledge of basic theorems and lemmas.
## Who invented math?
Ancient GreeksBeginning in the 6th century BC with the Pythagoreans, with Greek mathematics the Ancient Greeks began a systematic study of mathematics as a subject in its own right. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof.
## Why is math so hard and boring?
Maths can fast become boring because it’s often too abstract and doesn’t relate to a kid’s current everyday experience. Everyone has interests — things they feel passionate about.
## What is the most hated subject?
MathematicsThere is no denying the fact that the most hated subject in the world by the kids in none other than Mathematics. In fact, it is the most logical and the most systematic subject of the world.
## What is the hardest subject?
Top Ten Hardest School Subjects Physics. For the majority of people, physics is very tough because it is applying numbers to concepts that can be very abstract. … Foreign Language. … Chemistry. … Math. … Calculus. … English. … Biology. … Trigonometry.More items…
## Who is the world’s most hated?
Martin ShkreliMartin Shkreli (/ˈʃkrɛli/; born March 17, 1983), called “Pharma Bro” by some media, is an American former hedge fund manager and convicted felon.
## Which subject is the most important?
Men overwhelmingly say math has been the most valuable subject in their lives, with English and science essentially tied for second. Women are as likely to mention English as math as the most valuable subject.
## Is chemistry a hard major?
Chemistry is a difficult subject matter, but instead of dumbing down the concepts professors expect students to rise to the challenge and gain a firm understanding of the material.
## What kind of math is chemistry?
Age 16 to 18MathematicsChemistry contextProportional reasoningAnalysis of molecular structure; molesAlgebra and graphsAnalysis of experimental plots of reaction rates; gas lawsCalculusPredicting and measuring rates of reaction in measurable experimentsUnits of measurementsMaking sense of real, complicated measurements4 more rows
## How can I make math easier?
Here are some tips to tackle Maths like an expert!Practice as much as you can. Maths is a hands on subject. … Start by solving examples. Don’t start by solving complex problems. … Clear all your doubts. … Note down all formulae. … Understand the derivation. … Don’t lose touch with the basics.Feb 11, 2020
## What is the longest math equation?
What is the longest equation in the world? According to Sciencealert, the longest math equation contains around 200 terabytes of text. Called the Boolean Pythagorean Triples problem, it was first proposed by California-based mathematician Ronald Graham, back in the 1980s.
## Why math is the most hated subject?
Some students dislike math because they think it’s dull. They don’t get excited about numbers and formulas the way they get excited about history, science, languages, or other subjects that are easier to personally connect to. They see math as abstract and irrelevant figures that are difficult to understand.
## Should I major in chemistry or biology?
A biology major will be more likely to put you in the classroom or in the field while chemistry will have you in the lab. There’s also the question of motivation. Certainly not always, but generally speaking, biology tends to be about the study and protection of living things.
## What is the easiest math question in the world?
If by ‘simplest’ you mean easiest to explain, then it’s arguably the so-called ‘Twin Prime Conjecture’. Even schoolchildren can understand it, but proving it has so far defeated the world’s best mathematicians. Prime numbers are the building blocks from which every whole number can be made. | 1,414 | 6,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-25 | latest | en | 0.916161 |
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# GM251final(fa05) - Math 251 – Sections 1/2 Final Exam...
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Unformatted text preview: Math 251 – Sections 1/2 December 14, 2005 Final Exam Name Please check one of the boxes below Section 1 – 1st per – 8:00am Section 2 – 3rd per – 10:10am There are 10 questions on this exam. Question 2, which has 12 parts, is worth 24 points. The other questions are worth 14 points each. The total number of points is 150. If a question has multiple parts, then the points assigned to the question are divided equally among the parts, unless otherwise indicated. Where appropriate, show your work to receive credit; partial credit may be given. Please turn your cell phone OFF . The use of calculators, books, or notes is not permitted on this exam. Time limit 1 hour and 50 minutes. 1. Consider the function f ( x ) = if x < 3 if ≤ x < 1 if 1 ≤ x Write down the first seven terms in the Fourier series of f ( x ) on [- 2 , 2]. 2. In Parts a. through e. s n ( x ) denotes the n th partial sum of the Fourier series in Problem 1. a. Find lim n →∞ s n (4) b. Find lim n →∞ s n (5) c. Find lim n →∞ s n (6) d. Is it true that for all sufficiently large n : s n ( x ) ≥ - . 15 for every x in [1 . 4 , 1 . 7] ? e. Is it true that for all sufficiently large n : s n ( x ) ≤ 3 . 15 for every x in [0 . 9 , 1 . 1] ? f. What are the values of the following integrals? Z 6- 6 cos π 3 x cos 5 π 6 x dx Z 6- 6 sin π 6 x sin π 6 x dx g. What are the values of the following integrals? Z 6 cos π 6 x dx Z 6- 6 sin π 17 x dx h. If u ( x, t ) is the temperature of a thin rod of length L insulated on its sides and ends, then what are u x (0 , t ) and u x ( L, t ) for any t ?...
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GM251final(fa05) - Math 251 – Sections 1/2 Final Exam...
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http://lesswrong.com/lw/2bu/your_intuitions_are_not_magic/?sort=interestingness | 1,511,395,711,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806708.81/warc/CC-MAIN-20171122233044-20171123013044-00746.warc.gz | 176,659,919 | 26,303 | # Your intuitions are not magic
65 10 June 2010 12:11AM
People who know a little bit of statistics - enough to use statistical techniques, not enough to understand why or how they work - often end up horribly misusing them. Statistical tests are complicated mathematical techniques, and to work, they tend to make numerous assumptions. The problem is that if those assumptions are not valid, most statistical tests do not cleanly fail and produce obviously false results. Neither do they require you to carry out impossible mathematical operations, like dividing by zero. Instead, they simply produce results that do not tell you what you think they tell you. As a formal system, pure math exists only inside our heads. We can try to apply it to the real world, but if we are misapplying it, nothing in the system itself will tell us that we're making a mistake.
Examples of misapplied statistics have been discussed here before. Cyan discussed a "test" that could only produce one outcome. PhilGoetz critiqued a statistical method which implicitly assumed that taking a healthy dose of vitamins had a comparable effect as taking a toxic dose.
Even a very simple statistical technique, like taking the correlation between two variables, might be misleading if you forget about the assumptions it's making. When someone says "correlation", they are most commonly talking about Pearson's correlation coefficient, which seeks to gauge whether there's a linear relationship between two variables. In other words, if X increases, does Y also tend to increase. (Or decrease.) However, like with vitamin dosages and their effects on health, two variables might have a non-linear relationship. Increasing X might increase Y up to a certain point, after which increasing X would decrease Y. Simply calculating Pearson's correlation on two such variables might cause someone to get a low correlation, and therefore conclude that there's no relationship or there's only a weak relationship between the two. (See also Anscombe's quartet.)
The lesson here, then, is that not understanding how your analytical tools work will get you incorrect results when you try to analyze something. A person who doesn't stop to consider the assumptions of the techniques she's using is, in effect, thinking that her techniques are magical. No matter how she might use them, they will always produce the right results. Of course, assuming that makes about as much sense as assuming that your hammer is magical and can be used to repair anything. Even if you had a broken window, you could fix that by hitting it with your magic hammer. But I'm not only talking about statistics here, for the same principle can be applied in a more general manner.
Every moment in our lives, we are trying to make estimates of the way the world works. Of what causal relationships there are, of what ways of describing the world make sense and which ones don't, which plans will work and which ones will fail. In order to make those estimates, we need to draw on a vast amount of information our brains have gathered throughout our lives. Our brains keep track of countless pieces of information that we will not usually even think about. Few people will explicitly keep track of the amount of different restaurants they've seen. Yet in general, if people are asked about the relative number of restaurants in various fast-food chains, their estimates generally bear a close relation to the truth.
But like explicit statistical techniques, the brain makes numerous assumptions when building its models of the world. Newspapers are selective in their reporting of disasters, focusing on rare shocking ones above common mundane ones. Yet our brains assume that we hear about all those disasters because we've personally witnessed them, and that the distribution of disasters in the newspapers therefore reflects the distribution of disasters in the real world. Thus, people asked to estimate the frequency of different causes of death underestimate the frequency of those that are underreported in the media, and overestimate the ones that are overreported.
On this site, we've also discussed a variety of other ways by which the brain's reasoning sometimes goes wrong: the absurdity heuristic, the affect heuristic, the affective death spiral, the availability heuristic, the conjunction fallacy... the list goes on and on.
So what happens when you've read too many newspaper articles and then naively wonder about how frequent different disasters are? You are querying your unconscious processes about a certain kind of statistical relationship, and you get an answer back. But like the person who was naively misapplying her statistical tools, the process which generates the answers is a black box to you. You do not know how or why it works. If you would, you could tell when its results were reliable, when they needed to be explicitly corrected for, and when they were flat-out wrong.
Sometimes we rely on our intuitions even when they are being directly contradicted by math and science. The science seems absurd and unintuitive; our intuitions seem firm and clear. And indeed, sometimes there's a flaw in the science, and we are right to trust our intuitions. But on other occasions, our intuitions are wrong. Yet we frequently persist in holding onto our intuitions. And what is ironic is that we persist on holding onto them exactly because we do not know how they work, because we cannot see their insides and all the things inside them that could go wrong. We only get the feeling of certainty, a knowledge of this being right, and that feeling cannot be broken into parts that could be subjected to criticism to see if they add up.
But like statistical techniques in general, our intuitions are not magic. Hitting a broken window with a hammer will not fix the window, no matter how reliable the hammer. It would certainly be easy and convenient if our intuitions always gave us the right results, just like it would be easy and convenient if our statistical techniques always gave us the right results. Yet carelessness can cost lives. Misapplying a statistical technique when evaluating the safety of a new drug might kill people or cause them to spend money on a useless treatment. Blindly following our intuitions can cause our careers, relationships or lives to crash and burn, because we did not think of the possibility that we might be wrong.
That is why we need to study the cognitive sciences, figure out the way our intuitions work and how we might correct for mistakes. Above all, we need to learn to always question the workings of our minds, for we need to understand that they are not magical.
Comment author: 10 June 2010 04:46:20PM * 10 points [-]
Thanks for the well-written article. I enjoyed the analogy between statistical tools and intuition. I'm used to questioning the former, but more often than not I still trust my intuition, though now that you point it out, I'm not sure why.
Comment author: 11 June 2010 10:42:15AM 14 points [-]
You shouldn't take this post as a dismissal of intuition, just a reminder that intution is not magically reliable. Generally, intuition is a way of saying, "I sense similarities between this problem and other ones I have worked on. Before I work on this problem, I have some expectation about the answer." And often your expectation will be right, so it's not something to throw away. You just need to have the right degree of confidence in it.
Often one has worked through the argument before and remembers the conclusion but not the actual steps taken. In this case it is valid to use the memory of the result even though your thought process is a sort of black box at the time you apply it. "Intuition" is sometimes used to describe the inferences we draw from these sorts of memories; for example, people will say, "These problems will really build up your intuition for how mathematical structure X behaves." Even if you cannot immediately verbalize the reason you think something, it doesn't mean you are stupid to place confidence in your intuitions. How much confidence depends on how frequently you tend to be right after actually trying to prove your claim in whatever area you are concerned with.
Comment author: 10 June 2010 09:02:25PM * 8 points [-]
I do know why I trust my intuitions as much as I do. My intuitions are partly the result of natural selection and so I can expect that they can be trusted for the purposes of surviving and reproducing. In domains that closely resemble the environment where this selection process took place I trust my intuition more, in domains that do not resemble that environment I trust my intuition less.
Black box or not, the fact that we are here is good evidence that they (our intuitions) work (on net).
Comment author: 12 June 2010 08:18:54AM 3 points [-]
How sexy is that?
If you are evaluating intuitions, there are two variables you should account for. The similarity with evolutionary environment, indeed. AND your current posterior belief of the importance of this kind of act in the variance of offspring production.
We definitely evolved in an environment full of ants. Does that mean my understanding of ant-colony intelligence is intuitive?
Comment author: 29 September 2012 05:22:40PM 1 point [-]
I'm very curious how you decide what constitutes a similar environment to that of natural selection, and what sorts of decisions your intuition helps make in such an environment.
Comment author: 18 November 2011 10:27:32PM 0 points [-]
So then anything that has evolved may be relied upon for survival? It is impossible to rationalize faith in an irrational cognitive process. In the book Blink, the author asserts that many instances of intuition are just extremely rapid rational thoughts, possibly at a sub-conscious level.
Comment author: 11 June 2010 02:45:47PM 4 points [-]
Intuition seems to be one of the least studied areas of cognitive science, at least until very recently. The Wikipedia entry on cognitive sciences that the post links to has no mention of "intuition", and one paper I found said that the 1999 MIT Encyclopedia of Cognitive Sciences doesn't even have a single index entry for it (while "logic" has almost 100 references).
After a bit more searching, I found a 2007 book titled Intuition in Judgment and Decision Making, which apparently represents the current state of the art in understanding the nature of intuition.
Comment author: 10 June 2010 06:20:11PM 4 points [-]
i don't know why we prefer to hold on to our intuitions. your claim, that " we persist on holding onto them exactly because we do not know how they work" has not been proven, as far as I can tell, and seems unlikely. I also don't know why our own results seem sharper than what we learn from the outside [although about this later point, i bet there's some story about lack of trust in homo hypocritus societies or something] .
As somebody who fits into the "new to the site" category, I enjoyed your article.
Comment author: 10 June 2010 07:15:23PM 3 points [-]
Welcome to Less Wrong! Feel free to post an explicit introduction on that thread, if you're hanging around.
I think the critical point is in the next sentence:
We only get the feeling of certainty, a knowledge of this being right, and that feeling cannot be broken into parts that could be subjected to criticism to see if they add up.
Yes, we don't know what the interiors are - but the original source of our confidence is our (frequently justified) trust in our intuitions. I think another related point is made in How An Algorithm Feels From Inside, which talks about an experience which is illusory, merely reflecting an artifact of the way the brain processes data. The brain usually doesn't bother flagging a result as a result, it just marks it as true and charges forward. And as a consequence we don't observe that we are generalizing from the pattern of news stories we watched, and therefore don't realize our generalization may be wrong.
Comment author: 05 November 2012 01:18:55PM 0 points [-]
I think it's a combination of not understanding the process with a lifetime of experience where's it's far more right than wrong (Even for younger people, if they have 10-15 years of instinctive behavior being rewarded on some level, it's hard to accept there are situations it doesn't work as well). Combine that with the tendency of positive outcomes to be more memorable than others, and it's not too difficult to understand why people trust their intuition as much as they do.
your claim, that " we persist on holding onto them exactly because we do not know how they work" has not been proven, as far as I can tell, and seems unlikely.
It may not be the only reason, but an accurate understanding of how intuitions work would make it easier to rely less on it in situations it's not as we'll equipped for, just as an understanding of different biases makes it easier to fight them in our own thought processes.
Comment author: 10 June 2010 10:40:08PM * 3 points [-]
People who know a little bit of statistics - enough to use statistical techniques, not enough to understand why or how they work - often end up horribly misusing them.
How often do people harm themselves with statistics, rather than further their goals through deception? Scientists data-mining get publications; financiers get commissions; reporters get readers.
ETA: the people who are fooled are harming themselves with statistics. But I think the people want to understand for themselves generally only use statistics that they understand.
Comment author: 10 June 2010 10:44:02PM 4 points [-]
True, but many of those scientists and reporters really do want to unravel the actual truth, even if it means less material wealth or social status. These people would enjoy being corrected.
Comment author: 11 June 2010 01:25:21AM 1 point [-]
There is also an opportunity cost to the poor use of statistics instead of proper use. This may be only externalities (the person doing the test may actually benefit more from deception), but overall the world would be better if all statistics were used correctly.
Comment author: 13 August 2010 01:50:57AM 2 points [-]
I enjoyed your article and as a scientist, I've been interested to understand this: what seems an intuitive method to use to solve a scientific problem is not seen as an intuitive method while solving 'other' problems.
By 'other', I mean things like psychological problems or problems that arise from conflicts amongst people. It may be obvious why it is not 'intuitive' but what goes beyond my understanding is most will not even consider using the scientific method for the latter types of problem ever.
Comment author: 11 June 2010 12:59:14AM 2 points [-]
Having just pressed "Send" on an email that estimates statistics based on my intuitions, this feels particularly salient to me.
Really well written. Great work Kaj.
Thanks for reminding me that my thoughts aren't magic.
Comment author: 10 June 2010 12:43:20AM 2 points [-]
Elegantly done - clear and informative.
Comment author: 19 January 2017 02:17:37PM 1 point [-]
Indeed, intuitions are fallible. Though beware of the other extreme: writing off your intuitions altogether and trying to live solely based on logic. I've seen various people in the LW sphere try this, and it doesn't quite work. In some cases, like nutrition or social life, there is a bottomless pit of complexity. Trying to provably 'solve' such problems will lead to a bottomless pit of thinking, stagnation, and depression.
Logic is not a magic hammer either.
Comment author: 29 January 2015 01:02:15PM 1 point [-]
The immediately available example supporting your article for me is the relationship between dietary cholesterol and blood cholesterol. There's high general confusion around this health claim.
What no doubt compounds the confusion on the issue is that intuitively you might infer that eating zero cholesterol should lower blood cholesterol, or that eating high cholesterol should raise blood cholesterol. Evidence shows this often happens, but not always. There are enough notable outliers that the claim has been defeated in the general mind because it doesn't support the intuitive story.
That is, vegans who eat almost no cholesterol containing foods, can have high blood cholesterol. On the flip side, surely everyone has heard of that friend of a friend who eats inf eggs a day and has low blood cholesterol.
There's a reasonably interesting story that fits the evidence for the claim that if dietary cholesterol then blood cholesterol, but the nonlinearity of the relationship and also the incidence of intuition defeating cases cloud the issue.
Comment author: 02 October 2017 11:52:39PM 0 points [-]
I originally learned about these ideas from Thinking Fast and Slow, but I love hearing them rephrased and repeated again and again. Thinking clearly often means getting in the cognitive habit of questioning every knee-jerk intuition.
On the other hand, coming from a Bryan Caplan / Michael Huemer perspective, aren't we kind of stuck with some set of base intuitions? Intuitions like; I exist, the universe exists, other people exist, effects have causes, I'm not replaced by a new person with memory implants every time I go to sleep...
You might even call these base intuitions, "magic," in the sense that you have to have faith in them in order to do anything like rationality.
Comment author: 03 October 2017 11:37:19AM * 0 points [-]
Well, we don't know if they work magically, because we don't know that they work at all. They are just unavoidable.
It's not that philosophers weirdly and unreasonably prefer intuition to empirical facts and mathematical/logical reasoning, it is that they have reasoned that they can't do without them: that (the whole history of) empiricism and maths as foundations themselves rest on no further foundation except their intuitive appeal. That is the essence of the Inconvenient Ineradicability of Intuition. An unfounded foundation is what philosophers mean by "intuition". Philosophers talk about intution a lot because that is where arguments and trains of thought ground out...it is away of cutting to the chase. Most arguers and arguments are able to work out the consequences of basic intutitions correctly, so disagrements are likely to arise form differencs in basic intuitions themselves.
Philosophers therefore appeal to intuitions because they can't see how to avoid them...whatever a line of thought grounds out in, is definitiionally an intuition. It is not a case of using inutioins when there are better alternatives, epistemologically speaking. And the critics of their use of intuitions tend to be people who haven't seen the problem of unfounded foundations because they have never thought deeply enough, not people who have solved the problem of finding sub-foundations for your foundational assumptions.
Scientists are typically taught that the basic principles maths, logic and empiricism are their foundations, and take that uncritically, without digging deeper. Empircism is presented as a black bx that produces the goods...somehow. Their subculture encourages use of basic principles to move forward, not a turn backwards to critically relflect on the validity of basic principles. That does not mean the foundational principles are not "there". Considering the foundational principles of science is a major part of philosophy of science, and philosophy of science is a philosophy-like enterprise, not a science-like enterprise, in the sense it consists of problems that have been open for a long time, and which do not have straightforward empirical solutions.
Does the use of empiricism shortcut the need for intuitions, in the sense of unfounded foundations?
For one thing, epistemology in general needs foundational assumptions as much as anything else. Which is to say that epistemogy needs epistemology as much as anything else. -- to judge the validity of one system of epistemology, you need another one. There is no way of judging an epistemology starting from zero, from a complete blank. Since epistemology is inescapable, and since every epistemology has its basic assumptions, there are basic assumptions involved in empiricism.
Empiricism specifically has the problem of needing an ontological foundation. Philosophy illustrates this point with sceptical scenarios about how you are being systematically deceived by an evil genie. Scientific thinkers have closely parallel scenarios in which humans cannot be sure whether you are not in the Matrix or some other virtual reality. Either way, these hypotheses illustrate the point that the empiricists are running on an assumption that if you can see something, it is there.
Comment author: [deleted] 10 February 2015 01:10:26AM 0 points [-]
I think this article doesn't quite appreciate the full role intuitions play in science. It seems to me that intuitions help shape science in large ways. For instance, our intuitions that 'deduction works' and 'induction works' seems to stop all of us from turning into Cartesian sceptics, and preventing any science. Intuitions (and philosophical arguments) about metaphysics shape the basis of acceptable hypotheses within physics. Intuitions about what makes a scientific theory good/explanatory/falsified shape how science proceeds. Intuitions also serve to define concepts we have. If I remember correctly, in the Newtonian era, mass was not analysed in terms of anything else. It was a primitive concept in Newton's physics, and it was defined intuitively. Nowadays, modern physics has analysed concepts in terms of more and increasingly obscure concepts; but nevertheless, there is always a limit to what has been analysed in terms of what, and what remains is held, insofar as we know of it, as known primitively. That is, known intuitively.
I also have a question: Does this site in general take a negative view of heuristics humans have? I've seen various pages complaining about heuristics humans have, and not much about how helpful they are in keeping us all functioning.
Comment author: 25 March 2013 01:28:08PM 0 points [-]
This was an excellent read- I particularly enjoyed the comparison drawn between our intuition and other potentially "black box" operations such as statistical analysis. As a mathematics teacher (and recreational mathematician) I am constantly faced with, and amused by, the various ways in which my intuition can fail me when faced with a particular problem.
A wonderful example of the general failure of intuition can be seen in the classic "Monty Hall Problem." In the old TV game show Monty Hall would offer the contestant their choice of one of three doors. One door would have a large amount of cash, the other two a non-prize such as a goat. Here's where it got interesting. After the contestant makes their choice, Monty opens one of the "loosing" doors, leaving only two closed (one of which contains the prize), then offers the contestant he opportunity to switch from their original door to the other remaining door.
The question is, should they switch? Does it even matter? For most people (myself included) our intuition tells us it doesn't matter. There are two doors, so there's a 50/50 chance of winning whether you switch or not. However a quick analysis of the probabilities involved shows us that they are in fact TWICE as likely to win the prize if they switch than if they stay with their original choice.
That's a big difference- and a very counterintuitive result when first encountered (at least in my opinion)
Comment author: 25 March 2013 03:39:49PM 0 points [-]
I was first introduced to this problem by a friend who had received as a classroom assignment "Find someone unfamiliar with the Monty Hall problem and convince them of the right answer."
The friend in question was absolutely the sort of person who would think it was fun to convince me of a false result by means of plausible-sounding flawed arguments, so I was a very hard sell... I ended up digging my heels in on a weird position roughly akin to "well, OK, maybe the probability of winning isn't the same if I switch, but that's just because we're doing something weird with how we calculate probabilities... in the real world I wouldn't actually win more often by switching, cuz that's absurd."
Ultimately, we pulled out a deck of cards and ran simulated trials for a while, but we got interrupted before N got large enough to convince me.
So, yeah: counterintuitive.
Comment author: [deleted] 25 March 2013 04:40:32PM 0 points [-]
I remember how my roommates and I drew a game tree for the Monty Hall problem, assigned probabilities to outcomes, and lo, it was convincing.
Comment author: 25 March 2013 04:59:38PM 0 points [-]
(nods)
It continues to embarrass me that ultimately I was only "convinced" that the calculated answer really was right, and not some kind of plausible-sounding sleight-of-hand, when I confirmed that it was commonly believed by the right people.
Comment author: 25 March 2013 05:29:23PM * 0 points [-]
One of my favorites for exactly that reason- if you don't mind, let me take a stab at convincing you absent "the right people agreeing."
The trick is that once Monty removes one door from the contest you are left with a binary decision. Now to understand why the probability differs from our "gut" feeling of 50/50 you must notice that switching amounts to winning IF your original choice was wrong, and loosing IF your original choice was correct (of course staying with your original choice results in winning if you were right and loosing if you were wrong).
So, consider the probability that you original guess was correct. Clearly this is 1/3. That means the probability of your original choice being incorrect is 2/3. And there's the rub. If you will initially guess the wrong door 2/3 of the time, then that means that when you are faced with the option to switch doors you're original choice will be wrong 2/3 of the time, and switching would result in you switching to the correct door. Only 1/3 of the time will your original choice be correct, making switching a loosing strategy.
It becomes more clear if you begin with 10 doors. In this modified Monty Hall problem, you pick a door, then Monty opens 8 doors, leaving only your original choice and on other (one of which contains the prize money). In this case your original choice will be incorrect 9/10 times, which means when faced with the option to switch, switching will result in a win 9/10 times, as opposed to staying with your original choice, which will result in a win only 1/9 times.
Comment author: 25 March 2013 06:07:25PM * 1 point [-]
(nods) Yah, I'm familiar with the argument. And like a lot of plausible-sounding-but-false arguments, it sounds reasonable enough each step of the way until the absurd conclusion, which I then want to reject. :-)
Not that I actually doubt the conclusion, you understand.
Of course, I've no doubt that with sufficient repeated exposure this particular problem will start to seem intuitive. I'm not sure how valuable that is.
Mostly, I think that the right response to this sort of counterintuitivity is to get seriously clear in my head the relationship between justified confidence and observed frequency. Which I've never taken the time to do.
Comment author: 11 June 2010 07:07:17AM 0 points [-]
Yet our brains assume that we hear about all those disasters [we read about in the newspaper] because we've personally witnessed them, and that the distribution of disasters in the newspapers therefore reflects the distribution of disasters in the real world.
Even if we had personally witnessed them, that wouldn't, in itself, be any reason to assume that they are representative of things in general. The representativeness of any data is always something that can be critically assessed.
Comment author: 11 June 2010 09:25:53AM * 0 points [-]
For many people, representativeness is the primary governing factor in any data analysis, not just a mere facet of reasoning that should be critically assessed. Also, aside from the mentioned media bias that is indeed relatively easily correctable, there are many subtler instances of biasing via representativess, on the level of cognitive processes.
Comment author: 11 June 2010 02:23:38AM 0 points [-]
"However, like with vitamin dosages and their effects on health, two variables might have a non-linear relationship."
if we limit our interval we can make a linear approximation within that interval. this is often good enough if we don't much care about data outside that interval. the easy pitfall of course is people wanting to extend the linearization beyond the bounds of the interval.
Comment author: 16 June 2010 03:10:25AM -1 points [-]
Voted down because tangential replies that belong elsewhere really get on my nerves. Please comment on the post about the vitamin study, linked in the OP.
Comment author: 17 June 2010 04:04:15PM 0 points [-]
0_o I was responding directly to the OP. | 6,189 | 29,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-47 | latest | en | 0.958076 |
https://www.physicsforums.com/threads/ftc-question.764354/ | 1,508,793,352,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00070.warc.gz | 928,921,975 | 18,241 | # FTC Question
1. Aug 1, 2014
### Dethrone
This is going to be a really dumb question, but somehow it's not really clicking.
I read somewhere that: $\int_{a}^{x} f(t)\,dt = F(x) - F(a)$
But, this is also true: $F(x) = \int_{a}^{x} f(t)\,dt$.
Does this imply that $F(x) = F(x) - F(a)$?
For example, at a random number, c,:
$F(c) = \int_{c}^{x} f(t)\,dt = F(c) - F(a)$
2. Aug 2, 2014
### WWGD
A couple of problems, if you use the Riemann integral , F(x) with F'=f does not always exist. If it does, you use it in the top equation. But it is not the same F as the F in the second equation. And F(c) is the integral from a to c.
3. Aug 2, 2014
### HallsofIvy
Staff Emeritus
Yes, if F'(x)= f(x) this is true.
No, this is not true unless it happens that F(a)= 0. The difference is that F(x) is NOT 'well defined'. If F'= f then (F+ C)'= f for any constant C.
What is true is
1) $\int_a^x f(t)dt= F(x)- F(a)$ for any function, F, such that F'= f
and
2) There exist a function, F, such that F'= f and $$\int_a^x f(t)dt= F(x)$$
The specific function, F, in (2) is one of the infinite number of functions in (1).
No. It implies that if $F_1$ and $F_2$ are any two of the infinite number of anti-derivatives of f, then $F_1(x)= F_2(x)+ C$ for some constant, C. (Clearly, $C= F_1(a)- F_2(a)$.)
No, the "F" on the left has to be a different anti-derivative of f than the "F" on the right. You are using the same notation to indicate two different functions.
4. Aug 2, 2014
### Dethrone
Yes! I figured this out last night, as well! Thanks everyone!
5. Aug 2, 2014
### Dethrone
Actually, I thought I figured it out, but I don't quite understand 2. Can you elaborate on that? I understand that there are infinite number of antiderivatives of a function, and $F1(x) = F2(x) + c$.
What I thought was that here $F(x) = \int_{a}^{x} f(t)\,dt$, I used $F(x)$ to represent the function as a function of x, where as here $\int_{a}^{x} f(t)\,dt = F(x) - F(a)$, $F(x)$ is used to denote the anti derivative. Because they are used to denote different things, they are not equal. That's what I thought was the answer last night.
6. Aug 2, 2014
### HallsofIvy
Staff Emeritus
You are still using the phrase "the anti-derivative" though you say you understand "there are infinite number of antiderivatives of a function". When you write $$F(x)= \int_a^x f(t)dt$$ you are saying that "F" is specifically that anti-derivative such that F(a)= 0.
7. Aug 2, 2014
### WWGD
Besides, remember that even if the integral exists (when f is continuous outside of a set of measure zero), the integral may exist , i.e., the Riemann sums converge, but there may be no antiderivative F with F'=f. Notice that the contradiction in $$F(x)= \int_a^x f(t)dt=F(x)-F(a)$$ shows the two functions cannot be equal, i.e., F cannot be used for both equations.
8. Aug 2, 2014
### Dethrone
I realized that writing $F(x)= \int_a^x f(t)dt$ implies the case when $F(a) = 0$, since $\int_a^x f(t)dt = F(x) - F(a)$. I think I was confused because of notation: sometimes people write $F(x) = \int_a^x f(t)dt$ to represent the integral as a function of x. i.e $F(3) = \int_a^3 f(t)dt$, and not using "$F(x)$" to refer to the antiderivative. When people write $\int_a^x f(t)dt = F(x) - F(a)$, they are saying that the integral equals the antiderivative evaluated at x minus the antiderivative evaluated at a. So in this case, "F" is used to refer to the antiderivative. Am I right, or am I confused?
9. Aug 2, 2014
### HallsofIvy
Staff Emeritus
An anti-derivative, not "the" derivative! Other than that, yes, you are correct. Again, $$\int_a^x f(t)dt= F(x)- F(a)$$ for F any anti-derivative of f.
10. Aug 2, 2014
### Dethrone
Thought one: One thing, say we find the antiderivatives of $y = x^2$. Obviously, $\int x^2 dx = \frac{x^3}{3} + C$, where $\frac{x^3}{3} + C$ represents all the possible antiderivatives. , i.e $\frac{x^3}{3} + 4$. But we won't know which one it is until we get a definite integral, right?
Thought two: Okay, if I'm getting it straight. Even if the definite integral is $\int_{0}^{3} x^2\,dx = 9$, there are still infinite antiderivatives of "F", right? For some odd reason, I used to think that the definite integral boils said infinite antiderivatives to just one.
Which "thought" is correct, or are they both wrong?
Last edited: Aug 2, 2014
11. Aug 2, 2014
### Dethrone
Wikipedia says (I think) that $F(x) = \int_a^x f(t)dt$ produces different antiderivatives depending on which value of "x" you put in there. But last time I checked, definite integrals yield a value, which is the area under the curve on that interval. How does said "number" or "value" represent an antiderivative? Antiderivatives are suppose to be represented by a function, no?
12. Aug 3, 2014
### HallsofIvy
Staff Emeritus
Are you sure it doesn't say "depending on which value of "a" you put in there?
That is not a "definite integral" in the sense that you are using. The "x" is a variable and so $F(x)= \int_a^x f(t)dt$ is a function of x not a number. For a specific value of x, say x= b, we have the number $F(b)= \int_a^b f(t)dt$.
13. Aug 3, 2014
### Dethrone
But $F(x)= \int_a^x f(t)dt$ doesn't represent antiderivatives right,? It just represents the area under the curve. If we put a value such as x = b like you said above: $F(b)= \int_a^b f(t)dt$ this yields a number, not a function.
14. Aug 3, 2014
### HallsofIvy
Staff Emeritus
There is NO "area under the curve" unless you specify all boundaries of the region under the curve. $\int_a^x f(t) dt$ doesn't "represent the area under the curve" because "x" is not a specific number so does not define a boundary. Again, $F(x)= \int_a^x f(t) dt$ is one of the infinite number of antiderivatives of f.
15. Aug 3, 2014
### Dethrone
Alright, so the other antiderivatives are related to F(x) = G(x) + C, where G(x) could be $\int_b^x f(t) dt$, by changing the value of "a" at the bottom? I was saying that F(c), where c is a number = $F(c)= \int_c^b f(t)dt$ yields a number, because it has both boundaries.
EDIT: A moment of epiphany, I do hope this is right:
If we have $F(x)= \int_a^x f(t) dt$, we get all the different other antiderivatives if we change the value of "a"! Because $\int_a^x f(t) dt = F(x) - F(a)$, where F(a) is a constant term, and we have $\int_b^x f(t) dt = F(x) - F(b)$, where F(b) is a constant term. These all have the same derivative, namely, f(x) because the constant term disappears. Am I right now?
Last edited: Aug 3, 2014
16. Aug 3, 2014
### HallsofIvy
Staff Emeritus
Yes, that's pretty much what I have been trying to say!
17. Aug 3, 2014
### Dethrone
Alright, sorry if I been terribly slow. I finally understand all of this! Thanks HallsofIvy :D! | 2,150 | 6,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-43 | longest | en | 0.908669 |
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A piece of pie (Posted on 2010-12-07)
Let N be defined by N=> 3*1*4*1*5*9*2, where each asterisk may be replaced by any basic arithmetic sign ( +, - ,* ,/) and => means that the result is obtained by calculating sequentially from left to right.
Examples: 3+1+4+1+5+9+2=>25; 3+1-4+1-5+9-2=>3; 3*1-4*1-5+9-2=>1 etc.
How many distinct positive integer results can be obtained?
What is the lowest positive integer that cannot be obtained?
What positive integer claims the highest quantity of distinct expressions?
Rem: No brackets allowed.
No Solution Yet Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
piece of cake (solutions?) | Comment 1 of 5
A quick program finds 159 distinct positive integer results can be obtained. The lowest positive integer which cannot be obtained as 35. The positive integer with the greatest number of distinct empressions was 2.
There are clearly 4096 options (4**6) with the given integers as constants, and each of six operators having four options. After eliminating zero and negatives and results which were not whole integers, this left 1260 equations to tabulate. Sorting these and counting gave the results above.
I need to double check these lists. My method was to create six imbedded loops for the operators, with four values for each (addition, subtraction, multiplication, division), and then calculate each of the 4096 options left to right with the given digits from pi, and then sort the results, and finally to count. I'll go back to check the coding and counting.
Posted by ed bottemiller on 2010-12-07 18:53:38
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Hi, I'm not sure how to evaluate the following limits.
1. lim (x^3+x^2-8x-12)/(x+2)
x->-2
so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).
2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??
2. Hi skeske1234
If you know 3 is the facor of the numerator, can you factorize (x^3+x^2-8x-12) ?
3. Originally Posted by songoku
Hi skeske1234
If you know 3 is the facor of the numerator, can you factorize (x^3+x^2-8x-12) ?
yes.. well I'm not sure If this is right.. am i supposed to put x=3 in numerator? so I will get 0 on the numerator? or am I supposed to put in (x-3) for x?
if I put in 0, what do I do next?
4. Originally Posted by skeske1234
Hi, I'm not sure how to evaluate the following limits.
1. lim (x^3+x^2-8x-12)/(x+2)
x->-2
so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).
The zeros of the numerator are not of primary importance here. The problem is that the denominator has a zero at the very point where we want to take the limit! So if you first check (by polynomial division) whether the factor $\displaystyle x+2$ occurs in the numerator you find that
$\displaystyle \begin{array}{lcl} \lim\limits_{x\to-2}\frac{x^3+x^2-8x-12}{x+2}&=&\lim\limits_{x\to -2}\frac{(x+2)(x^2-x-6)}{x+2}\\ &=&\lim\limits_{x\to -2}(x^2-x-6)\\ &=&(-2)^2-(-2)-6\\ &=&0 \end{array}$
2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??
Remember $\displaystyle a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots + ab^{n-2}+b^{n-1})$? Now substitute $\displaystyle y$ for $\displaystyle x^{1/3}$ (and therefore $\displaystyle y^3$ for $\displaystyle x$), and you find that
$\displaystyle \begin{array}{lcl} \lim\limits_{x\to 27}\frac{27-x}{x^{1/3}-3} &=&\lim\limits_{y\to 3}\frac{3^3-y^3}{y-3}\\ &=&\lim\limits_{y\to 3}\frac{(3-y)(3^2+3y+y^2)}{y-3}\\ &=&\lim\limits_{y\to 3}[-(9+3y+y^2)]\\ &=& -(9+3\cdot 3+3^2)\\ &=& -27 \end{array}$
5. Originally Posted by skeske1234
Hi, I'm not sure how to evaluate the following limits.
1. lim (x^3+x^2-8x-12)/(x+2)
x->-2
so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).
Do you mean that "3" is a factor of the numerator or that "x- 3" is a factor?
In any case, that is not really relevant. What is very relevant is that x+2, the denominator, goes to 0 as x goes to -2 so "x+2" had better be a factor! You can show that it is, and find the factorization by dividing $\displaystyle x^3+ x^2- 8x- 12$ by x+ 2, either by direct division or by "synthetic" division. You could also "work" it out by thinking "if x+2 is a factor, then it must be something like $\displaystyle x^3+ x^2- 8x- 12= (x+ 2)(x^2+ ?x- 6)$ Multiply out the left side and see what "?" must be.
(If you know x+ 2 must be a factor and you also know that x- 3 is a factor, then it is easy to see that [tex]x^3+ x^2- 8x- 12= (x+ 2)(x- 3)(x- ?) and since -12= 2(-3)(2), what is "?" ?
2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??
Again, you need to be able to cancel that $\displaystyle x^{1/3}- 3$ in the denominator. Fortunately, we can precisely because 27 also makes the numerator 0. Remember that $\displaystyle a^3- b^3= (a- b)(a^2+ ab+ b^2)$ with $\displaystyle a^3= 27$ and $\displaystyle b^3= x$. | 1,357 | 3,927 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.854225 |
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# Virginia: 11/1 - 11/30/2010
Topic closed. 1236 replies. Last post 6 years ago by increase.
Page 77 of 83
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Posted: November 28, 2010, 11:48 am - IP Logged
327 is a best to play and win . good luck to you all . im playing it for today and tonight too. have a pleasent day too.
Richmond, VA
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July 19, 2010
185 Posts
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Posted: November 28, 2010, 2:11 pm - IP Logged
4768 and 681!!!!!!!!!!
Norfolk , Va
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Posted: November 28, 2010, 4:05 pm - IP Logged
All 3 were here from my chart . The 8 , the 1 , and the 6 .
For mid Sun....:
90 91 92 93 94 95
From my chart , find pairs or all 3
1 = 4 , 3 , 8 , 1, 6, 9 22 , 88 77
Good Luck,
Blackie.
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Posted: November 28, 2010, 4:07 pm - IP Logged
Here are the chart numbers for tonight . Maybe 2 of the numbers are here or maybe all 3 .
6 = 4, 7 , 1, 6, 2, 8 77 , 33
Here are my pair numbers :
82 83 84 85 86
37 38 39 40 41
A 2 4 6 or 8 should be in the number but more like a 4 or a 7 .
Good Luck,
Blackie.
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Posted: November 28, 2010, 4:39 pm - IP Logged
Congrats JMill . You have it in your lists !!
Good Luck,
Blackie.
San Jose, CA
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71, 72, 81, 82, 91, 92, 01, 02
80, 35, 68, 63, 56 51, 25, 75
**681 is a two digit return number, be on the look out**
Pairs: 68, 61, 66, 11, 36, 31
..LIVE TO PLAY..PLAY TO LIVE...
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Posted: November 28, 2010, 7:38 pm - IP Logged
681 Scoop
700, 701, 702, 703, 704, 705, 706, 707, 708, 709
750, 751, 752, 753, 754, 755, 756, 757, 758, 759
200, 201, 202, 203, 204, 205, 206, 207, 208, 209
710, 711, 712, 713, 714, 715, 716, 717, 718, 719
760, 761, 762, 763, 764, 765, 766, 767, 768, 769
210, 211, 212, 213, 214, 215, 216, 217, 218, 219
260, 261, 262, 263, 264, 265, 266, 267, 268, 269
720, 721, 722, 723, 724, 725, 726, 727, 728, 729
770, 771, 772, 773, 774, 775, 776, 777, 778, 779
270, 271, 272, 273, 274, 275, 276, 277, 278, 279
220, 221, 222, 223, 224, 225, 226, 227, 228, 229
..LIVE TO PLAY..PLAY TO LIVE...
San Jose, CA
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Posted: November 28, 2010, 7:41 pm - IP Logged
369 Factor (681) 246 System (681)
050 927
419 173
788 419
157 665
526 911
681 Rundown:
998--443
205--750
512--067
829--374
136--681
..LIVE TO PLAY..PLAY TO LIVE...
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681 Mirror Action
383 838 333 388
099 544 049 094
Number used: 681
8 Numbers to play:
371, 487, 543, 572, 518, 124, 069, 832
..LIVE TO PLAY..PLAY TO LIVE...
Norfolk , Va
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Posted: November 28, 2010, 8:06 pm - IP Logged
458 468 ??
Good Luck,
Blackie.
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2120 Posts
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Posted: November 28, 2010, 8:08 pm - IP Logged
71, 72, 81, 82, 91, 92, 01, 02
80, 35, 68, 63, 56 51, 25, 75
**681 is a two digit return number, be on the look out**
Pairs: 68, 61, 66, 11, 36, 31
Evenin' Quic Pics: 373, 482, 419
Mid Day Quic Pics: 570, 721, 574
San Jose, CA
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Posted: November 28, 2010, 8:11 pm - IP Logged
Evenin' Quic Pics: 373, 482, 419
Mid Day Quic Pics: 570, 721, 574
Evening Easy Picks:
449 500 565 105 894 248
..LIVE TO PLAY..PLAY TO LIVE...
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Posted: November 28, 2010, 8:23 pm - IP Logged
458 468 ??
214??
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327?????\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$
goodluck all.!
Norfolk , Va
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Posted: November 28, 2010, 11:01 pm - IP Logged
It was 012
Good Luck,
Blackie.
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Contributor
Posts: 38
how can i read parameters value automatically?
[ Edited ]
I have this problem and its works ok, but it´s possible read paramerts in data statement?
``````proc optmodel;
var x {1..3};
max z1 = (2*x[1]-3*x[2]+5*x[3])**3/(x[1]+5*x[2]+x[3]**2);
con c1: -1<=x[1]<= 3;
con c2: 1.5<=x[2]<= 2;
con c3: 1.2<=x[3]<= 2.2;
solve with nlp/multistart ;
print x;
quit;
``````
Edit: I want to put in data statement:
``-1<=x[1]<= 3;...``
Accepted Solutions
Solution
11-24-2017 06:45 PM
SAS Employee
Posts: 519
Re: how can i read parameters value automatically?
``````data bounds;
input lb ub;
datalines;
-1 3
1.5 2
1.2 2.2
;
proc optmodel;
var x {1..3};
max z1 = (2*x[1]-3*x[2]+5*x[3])**3/(x[1]+5*x[2]+x[3]**2);
read data bounds into [_N_] x.lb=lb x.ub=ub;
solve with nlp/multistart ;
print x;
quit;
``````
All Replies
Solution
11-24-2017 06:45 PM
SAS Employee
Posts: 519
Re: how can i read parameters value automatically?
``````data bounds;
input lb ub;
datalines;
-1 3
1.5 2
1.2 2.2
;
proc optmodel;
var x {1..3};
max z1 = (2*x[1]-3*x[2]+5*x[3])**3/(x[1]+5*x[2]+x[3]**2);
read data bounds into [_N_] x.lb=lb x.ub=ub;
solve with nlp/multistart ;
print x;
quit;
``````
Contributor
Posts: 38
Re: how can i read parameters value automatically?
[ Edited ]
if it will be possible to read coeficientes that multiplies x too?
SAS Employee
Posts: 519
Re: how can i read parameters value automatically?
``````data bounds;
input lb ub a b;
datalines;
-1 3 2 1
1.5 2 -3 5
1.2 2.2 5 1
;
proc optmodel;
var x {1..3};
num a {1..3};
num b {1..3};
max z1 = (sum {j in 1..3} a[j]*x[j])**3/(b[1]*x[1]+b[2]*x[2]+b[3]*x[3]**2);
read data bounds into [_N_] x.lb=lb x.ub=ub a b;
solve with nlp/multistart ;
print x;
quit;
``````
Contributor
Posts: 38
Re: how can i read parameters value automatically?
[ Edited ]
edit: it works too
thanks rob!
``````
data bounds;
input lb ub a b;
datalines;
-1 3 2 1
1.5 2 -3 5
1.2 2.2 5 1
;
proc optmodel;
var x {1..3};
num a {1..3};
num b {1..3};
max z1 = (sum {j in 1..3} a[j]*x[j])**3/ (sum {j in 1..3} b[j]*x[j]);
read data bounds into [_N_] x.lb=lb x.ub=ub a b;
solve with nlp/multistart ;
print x;
quit;``````
SAS Employee
Posts: 519
Re: how can i read parameters value automatically?
[ Edited ]
Note that your original objective function had x[3]**2 in the denominator. With your latest change, it is now just x[3]. If that is what you want, here's a way to make the code more data-driven (without hard-coding 1..3):
``````proc optmodel;
set OBS;
var x {OBS};
num a {OBS};
num b {OBS};
max z1 = (sum {j in OBS} a[j]*x[j])**3/ (sum {j in OBS} b[j]*x[j]);
read data bounds into OBS=[_N_] x.lb=lb x.ub=ub a b;
solve with nlp/multistart ;
print x;
quit;``````
This way, you can run again with different data without changing the PROC OPTMODEL code. Such separation of model and data is a best practice enabled by the use of an algebraic modeling language.
☑ This topic is solved. | 1,125 | 2,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-13 | latest | en | 0.505115 |
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#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define PI acos(-1.0)
#define seed 31//131,1313
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
const int maxn=100005;
struct Point{
double x;
}p[maxn];
int a[maxn];
int cmpx(Point a,Point b){
return a.x<b.x;
}
inline double dis(Point a,Point b){
if(a.x>b.x) return a.x-b.x;
else return b.x-a.x;
}
double closest(int low,int high){
if(low+1==high) //只有两个点
return dis(p[low],p[high]);
if(low+2==high) //只有三个点
return min(dis(p[low],p[high]),min(dis(p[low],p[low+1]),dis(p[low+1],p[high])));
int mid=(low+high)/2; //求中点即左右子集的分界线
double d=min(closest(low,mid),closest(mid+1,high));
d=min(d,dis(p[mid],p[mid+1])); //最后一步，合并
return d;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++)
scanf("%lf",&p[i].x);
sort(p,p+n,cmpx);
printf("%.2lf\n",closest(0 , n-1));//最近点对间的距离
}
return 0;
}
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03-29 353 | 509 | 1,297 | {"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} | 3.421875 | 3 | CC-MAIN-2021-49 | latest | en | 0.172597 |
https://www.omnimaga.org/lua-language/lua-qa/205/?wap2 | 1,591,406,131,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509264.96/warc/CC-MAIN-20200606000537-20200606030537-00328.warc.gz | 833,100,108 | 2,426 | Calculator Community > Lua
Lua Q&A
<< < (42/42)
Jim Bauwens:
--- Quote from: AnToX98 on March 30, 2014, 08:57:28 am ---Please could someone help me :( ?
--- End quote ---
Try calling the setPos function of the body you attached the segment to.
AnToX98:
Here's my code :
--- Code: ---------------------------------------------------
----------- LUA FALLDOWN, BY ANTOX98 -----------
------------------------------------------------
platform.apilevel = "2.0"
require "physics"
----------------------
----- BALL CLASS -----
----------------------
Ball = class()
Seg = class()
function Ball:init(x, y, w, mass)
self.width = w
self.body = physics.Body(mass, physics.misc.momentForCircle(mass, 0, 10, ZERO))
self.body:setPos(physics.Vect(x, y))
self.body:setMass(mass)
self.shape = physics.CircleShape(self.body, w, ZERO)
self.shape:setRestitution(0.6)
self.shape:setFriction(0.6)
end
function Seg:init(x1, y1, x2, y2)
local a, b = physics.Vect(x1, y1), physics.Vect(x2, y2)
local mass = physics.misc.INFINITY()
self.coor = {x1,y1,x2,y2}
self.body = physics.Body(mass, physics.misc.momentForSegment(mass, a, b))
--self.body:setPos(physics.Vect(x1, y1))
self.body:setMass(mass)
self.shape = physics.SegmentShape(self.body, a, b, 10)
self.shape:setRestitution(0.6)
self.shape:setFriction(0.6)
end
function Ball:paint(gc)
local p = self.body:pos()
local x, y = p:x(), p:y()
local r = self.width / 2
gc:setColorRGB(255,0,0)
gc:fillArc(x+r, y+r , self.width, self.width, 0, 360)
end
function Seg:paint(gc)
local a = self.shape:a()
local b = self.shape:b()
gc:setColorRGB(0,0,0)
gc:drawLine(a:x(), a:y(), b:x(), b:y())
end
function initGame()
w = 318
h = 212
ZERO = physics.Vect(0,0)
LARGE = physics.misc.INFINITY()
space = physics.Space()
space:setGravity(physics.Vect(0,9.)
sol = Seg(0,200,w,200)
ball = Ball(w/2-10, 30, 20, 1000)
sol.body:setPos(physics.Vect(0,-30))
timer.start(0.01)
end
initGame()
function on.timer()
space:step(0.1)
platform.window:invalidate()
end
function on.paint(gc)
ball:paint(gc)
sol:paint(gc)
end
--- End code ---
sol.body:setPos(physics.Vect(0,-30))
[/size]
[/size][size=78%]Strangely, the body moves but not the shape....[/size][/code] | 674 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.299321 |
https://www.mathworks.com/matlabcentral/cody/problems/43069-find-the-volume-of-cone/solutions/1195126 | 1,586,166,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00165.warc.gz | 916,285,542 | 15,839 | Cody
# Problem 43069. Find the volume of cone
Solution 1195126
Submitted on 22 May 2017 by Erik Luiten
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
r = 6/sqrt(pi); h = 1; y = coneVolume(r,h) y_correct = 12; assert(abs(y-y_correct)<1e-3)
y = 12.0000
2 Pass
r = 4.7; h = 2.3; y = coneVolume(r,h) y_correct = 53.2050; assert(abs(y-y_correct)<1e-3)
y = 53.2050
3 Pass
r = 1.1; h = 1.21; y = coneVolume(r,h) y_correct = 1.5332; assert(abs(y-y_correct)<1e-3)
y = 1.5332
4 Pass
r = 10.7; h = 100; y = coneVolume(r,h) y_correct = 11989.36476; assert(abs(y-y_correct)<1e-3)
y = 1.1989e+04
5 Pass
r = pi^3; h = exp(1); y = coneVolume(r,h) y_correct = 2736.6694; assert(abs(y-y_correct)<1e-3)
y = 2.7367e+03
6 Pass
r = sqrt(5); h = sqrt(817); y = coneVolume(r,h) y_correct = 149.66135; assert(abs(y-y_correct)<1e-3)
y = 149.6613 | 384 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-16 | latest | en | 0.522901 |
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Why Modern Math Education Is Obsolete
page: 2
20
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posted on Sep, 6 2015 @ 10:43 AM
I totally agree. You are not just teaching them math, you are teaching them to think critically of a problem.
I know I won't use a lot of the things I learned about in my electrical engineering courses in college, but it's the process of thinking through the problem and realizing other solutions that makes engineering so difficult.
posted on Sep, 6 2015 @ 11:32 AM
I completely agree and have been ranting about it for a long time. Starting around the time that multiplication is introduced, they stop teaching concepts and start teaching the memorization of equations.
One thing I always think of is "why is the product of two negatives a positive?" Hardly anyone knows the answer, showing they do not understand the concept.
And for the layman:
Like adding numbers together gives you a sum, multiplying numbers together gives you a product.
edit on 9/6/2015 by Bleeeeep because: (no reason given)
posted on Sep, 6 2015 @ 11:38 AM
originally posted by: EternalSolace
What I meant was that I don't understand how that complacency teaches math fundamentals. Sure 2+2 is about as simple as it gets. But as the problems grow in complexity, the underlying lesson that "so long as I'm close it's okay" is still there. Everything in me just doesn't like that.
Wonder what would happen if they did that at CERN lol!!!
A algebra professor told me once that Its just a fancy way to show off your intellectual prowess. IF you arent going to be a rocket scientist he didnt see the reason to force students to take it. STraight from the horses mouth its useless to 90 percent of humans who only need to know basic math. Its a educational scam for more cash.
posted on Sep, 6 2015 @ 12:00 PM
originally posted by: ChaoticOrder
When I watch the clips that a child can answer 2+2=5 and still get it correct because correcting them would lower self esteem, I shudder.
Yes we all should be able to do such basic operations, but that's only because we should all understand the basic concept of how addition works, and adding two units to another two units is not exactly challenging. I can actually do many basic operations in my head, I just don't remember the answer unless I have to do the same operation all the time. Since I work with binary and hex a lot I do remember many of the 2 and 16 multiples.
there are different levels of 'basic operations' for different people. i can't imagine lack of understanding of geometry, trigonometry and many other things, and multiplication table for me is as obvious as different powers of two.
yes, i'm a programmer as well. 3d engines, FFTs, image processing - been there, done that. and while i agree that there are many things that are best left for research on as-needed basis (we can't be experts in every field possible), there are some things that one just should know - and sorry, but i can't imagine blaming not remembering a multiplication table on anything else but lazyness. such simple things that you prefer not to remember have to slow you down quite a bit during your work as well.
also, i think you're quite far from understanding the idea behind math at all. i also prefer to think like a programmer - because i am one. but you can't simply imagine how a stream of particles will move in space just by knowing the rules being the building blocks of their movement. you'll get it simple - but you'll have to perform so many iterations of your simple math and traverse such a tree of dependecies to get to the end result, that by the time you'll get there, you'll forget what was it that you were checking. computer will do it - sure. you won't.
on the other hand, someone able to think like a mathematician will be able to imagine the flow of said particles just by looking at a single, complicated equation. why do you think scientists all over the world use such complicated math? are they a bunch of morons, unable to come up with anything better? far from it - it's just a natural way for them, and a natural way for a human in general. you can of course research new problems using computers, you can create models and optimize them using methods you've described - but that's not how things are done, and it's not optimal.
it's not optimal, because you have to optimize the problem for a computer first, and when searching for new answers, while operating only at the building-blocks level, you won't notice the bigger picture, unless you'll visualize it - and even then, you'll be only able to see what you've included in your visualization. it's not how things are done, because it would be actually backwards - scientists are able to come up with theories that require such a massive computing power to visualize, that without their imagination and ability to visualize it all in their heads, we would be quite a few decades behind with technological progress.
so the bottom line is, i have to disagree. while the education system is far from being perfect, teaching children to simplify problems just to avoid having to look at the bigger picture would turn them into a bunch of idiots lacking imagination, that - as a bonus - have to rely on calculators/smartphones/computers to come up with anything.
and you don't bring your calculator with you when you're going shopping, do you?
posted on Sep, 6 2015 @ 12:04 PM
originally posted by: yuppa
originally posted by: EternalSolace
What I meant was that I don't understand how that complacency teaches math fundamentals. Sure 2+2 is about as simple as it gets. But as the problems grow in complexity, the underlying lesson that "so long as I'm close it's okay" is still there. Everything in me just doesn't like that.
Wonder what would happen if they did that at CERN lol!!!
A algebra professor told me once that Its just a fancy way to show off your intellectual prowess. IF you arent going to be a rocket scientist he didnt see the reason to force students to take it. STraight from the horses mouth its useless to 90 percent of humans who only need to know basic math. Its a educational scam for more cash.
luckily for the governments and corporations behind that scam, 90% of humans won't be able to calculate just how much cash did they loose because of it.
posted on Sep, 6 2015 @ 12:13 PM
Math eh ! ...Much like most subjects in school ,I was there to learn .Problem is what I was being taught .Just regurgitate what you have been given and the system will place you where they can use you . I became a rebel very early on . It's about what to think, opposed to how to think . A website that really spells out the mathematical errors of our ways is Climate Audit . Very advanced and very much a sore on many of the ones put to the task to deceive us .
The big turn off to anyone wanting to learn is the constant moving of what is proper and right about any of the many subjects . Few escape the trappings of our educational system because of apathy .If you don't get with the program then you will be outside of it and will never be allowed to excel .
Math seems to be something that can be very accurate at it's basic level but becomes less so the more detailed it develops . Take CERN and the level of math that is involved . If it were possible using math to know something then the physical project would not exist .Reality is a fine point where theory becomes the order of the day it seems .Theory then becomes a what ever math you need to push forward the what ever possibilities are part of the fray .
Modern math has only given us the possibilities to over step reality until a time we come to the conclusions that they are just not so . I stick to the simple basic stuff that finds reality in my back yard and leave the rest to the best hoping that some day someone will get it as close to reality as we can appreciate . time for a Bud
posted on Sep, 6 2015 @ 12:29 PM
It depends on the field you are in.
If you're in say finance, it's very useful to have "go-to" equations memorized for use in programs such as excel. Same with actuarial science. You do need to be able to do certain functions from memory to be successful at that job because looking them up all the time wastes time that could be used doing the actual job.
I do agree that things like the multiplication table and other areas that schools focus on is useless. That's more arithmetic though and not "mathematics" per say. In fact most mathematicians are usually bad at average arithmetic.
You see things from the perspective of someone who programs. While I do understand what you're saying I don't know if that would be feasible to teach on a wide scale because it requires competency. The reason most math is taught the way it is taught is because it is the easiest means to an end.
Kind of how most people can write in English but aren't experts on etymology and the history of grammar.
Kind of how I said that in certain fields it's easier just to memorize the equations without necessarily knowing complex theory behind them.
Things are changing though which begs the question as to how to best prepare students for the future of mathematics.
Now in fields such as actuarial science you are far more likely to get a job if you know how to program and model. Things are definitely changing. It isn't acceptable to just know how to do equations but rather to be able to create from scratch what is needed to answer whatever question is up in the air.
I dunno man.
posted on Sep, 6 2015 @ 12:35 PM
If you ever lived with someone who was trying to learn your language, you would realize how important grammar is. It is very easy to change the meaning of a sentence or paragraph by using words incorrectly. Why is spelling correctly so difficult for an autodidact? Just memorize the rules and exceptions. Saves time to be correct.
posted on Sep, 6 2015 @ 12:40 PM
There are values to learning abstruse branches of mathematics that cannot be replicated by learning to program. Each branch of mathematics has concepts that enlarge your thinking ability. You mentioned you learned linear algebra. Don't you think you can understand linearity of everyday quantitative relationships better because of it? Same for probabiity, which I think you mentioned also.
Programming certainly teaches you some thinking skills that mathematics doesn't. Knowing both gives you the union of the concept spaces. By the way, simulation doesn't replace knowing the theory of how things interact, it just gives you some examples of the cases you tested.
posted on Sep, 6 2015 @ 12:41 PM
Yea , well if only getting very close is all one needs . The term "I am hungry " will ,or should get some response .I guess math is just that ,...it depends on what you mean .
posted on Sep, 6 2015 @ 01:25 PM
If you ever lived in the streets you'd realize how important slang is. Proper language is boring, and IMO an indication of low cognitive flexibilty.
posted on Sep, 6 2015 @ 01:47 PM
How can you not know your multiplication tables? That sounds like you are stubborn. 100% agree with you a lot of what you learn in school is forgotten and not used again, but it's problem solving one, and two it's getting from point A to point B, i.e. K-12, 4 year University, etc... Education is very important and helps you succeed in life.
posted on Sep, 6 2015 @ 02:46 PM
If I wasn't taught the basics of mathematics throughout my younger years I never could have earned my PhD in Mathematics. Are we supposed to just ignore students who might someday go farther than YOU have with mathematics because YOU don't see the point?
posted on Sep, 6 2015 @ 03:40 PM
I think that the Math kids are taught now is much more advanced than when I was in school. Though mind you, I have LD in Math and was really really behind the other kids in my class. I didn't understand certain concepts of Math until they started teaching basic Algebra, as long as I had a calculator I was fine with that. Ask me basic math skills and I have no clue what is going on. Show me a Algebra problem today and I probably couldn't solve it.(No practice for almost 20 years...lol.)
Anyway, my son is in 3rd grade and as part of the school supplies they requested Addition, Subtraction, Muliplication, and Divison flash cards. We looked everywhere for the divison cards and couldn't find them anywhere(finally found a set at Goodwill). When his Grandma found out we were looking for divison cards she was surprised that they were already teaching divison, she's a retired school teacher for 8 years and she said they didn't start until fourth grade. Of course she could have just been saying that because she dosen't like the school we are sending him too.
I really wish I knew basic math because it takes me forever to count out change, and half the time I get it wrong, I need a digital clock because I get the hands confused and the digits wrong. I see numbers and they get all jumbled and confused in my head. There's more, but yeah. You don't realize how much you need math skills just to make it.
posted on Sep, 6 2015 @ 08:21 PM
Ontogeny Recapitulates Phylogeny.
It may not always be exact in biology but in learning it is. If you 'skip' steps, you don't gain true understanding, you are just faking it.
posted on Sep, 6 2015 @ 10:33 PM
Really? I used to be a programmer/developer before i moved into project management and at 44 years of age, I still remember all my multiplication tables to 13 without a second thought.
Good for you.
The way you had it elevated programming languages above mathematics. You cannot have programming languages without mathematics but you do have mathematics without programming languages.
That is certainly a valid point, I was going to write a little bit about how math can be more powerful in some areas, especially when it comes to abstract ideas like infinity. Math isn't really what I would call a computational language, because clearly we cannot compute infinities, but we can still manipulate them in equations and reach meaningful conclusions. We can also have infinite precision numbers in mathematics, whereas we cannot save such numbers in a computer because we have a limited amount of memory. Advanced math is more about working with abstract ideas than it is about doing simple computations.
posted on Sep, 6 2015 @ 10:37 PM
For instance when he talks about syntax meaning different things in different fields of mathematics, the exact same thing happens when you switch programming languages.
That is also a valid point, except every programming language is very clearly defined, it's not even exactly clear how to distinguish each field of mathematics from each other, and like I said, even the personal tastes of the person writing the equation can be important, even the placement of symbols, especially superscript and subscript symbols, can be ambiguous. There is just far too much uncertainty for my liking, I would prefer the language of math to be much more universal and clear cut. It's also just hard trying to pack complicated ideas into small equations, the notation gets very messy and I just cannot bring myself to try and understand such gibberish. I find it exceptionally easier to read a computer algorithm which is stating the exact same thing as the equation.
edit on 6/9/2015 by ChaoticOrder because: (no reason given)
posted on Sep, 6 2015 @ 10:54 PM
When I was in school, slide rules were an incredible tool. It really taught lessons in relationships and scaling. The scientific calculator of today of course, replaced it. However they hide the relationships and scaling and just produce the final result.
I think it is important that the instructor take the student visually through the entire calculation, but perhaps only once, and instead dwell on the concept of how that function appears in many other types of calculations. A more modular approach is needed in our schools instead of tedious exercises that prevent students from seeing the bigger picture, and thus producing interest in what mathematics is.
posted on Sep, 6 2015 @ 11:02 PM
yes, i'm a programmer as well. 3d engines, FFTs, image processing - been there, done that. and while i agree that there are many things that are best left for research on as-needed basis (we can't be experts in every field possible), there are some things that one just should know - and sorry, but i can't imagine blaming not remembering a multiplication table on anything else but lazyness. such simple things that you prefer not to remember have to slow you down quite a bit during your work as well.
Yes I am lazy, I am a completely lazy programmer. Laziness is often cited as one of the most important traits of a good programmer. See: The Three Virtues of a GREAT Programmer or Why Good Programmers Are Lazy and Dumb. The fact is, I cannot remember the last time I needed to calculate 7x12 in my head, so I see no reason to remember a table of numbers I'll hardly ever use, especially when I have a calculator handy 99.9% of the time. Of course I can remember many of the smaller multiplication answers, but that's only because I do tend to multiply smaller numbers in my head often. I remember what I need, nothing less, nothing more.
also, i think you're quite far from understanding the idea behind math at all. i also prefer to think like a programmer - because i am one. but you can't simply imagine how a stream of particles will move in space just by knowing the rules being the building blocks of their movement. you'll get it simple - but you'll have to perform so many iterations of your simple math and traverse such a tree of dependecies to get to the end result, that by the time you'll get there, you'll forget what was it that you were checking. computer will do it - sure. you won't.
Of course you can understand how particles will move through space simply by understanding how particles interact with each other. First of all you just need to know how they gravitationally react to each other, and as we know the force of gravity gets exponentially stronger as two particles move closer together. Then we also need to take into consideration the electric forces and the nuclear forces, etc. Of course we can use equations to help us understand all those different forces but it would be extremely complicated if we tried to combine all of those equations into one equation which described everything about particle movement.
on the other hand, someone able to think like a mathematician will be able to imagine the flow of said particles just by looking at a single, complicated equation. why do you think scientists all over the world use such complicated math? are they a bunch of morons, unable to come up with anything better? far from it - it's just a natural way for them, and a natural way for a human in general. you can of course research new problems using computers, you can create models and optimize them using methods you've described - but that's not how things are done, and it's not optimal.
We often start with mathematics because it's the most fundamental way to describe complex systems. By studying such equations and manipulating them we can often make other important realizations. Like I said in my previous post, math is much more abstract and is better at conveying abstract ideas. Clearly it wouldn't make much sense to start by describing physics in C++ and then convert it into math equations at a later point in time. We start with the most fundamental mathematical descriptions and then we can implement those ideas into computer code by quantizing the equations into finite computational problems.
edit on 6/9/2015 by ChaoticOrder because: (no reason given)
posted on Sep, 6 2015 @ 11:20 PM
If you're in say finance, it's very useful to have "go-to" equations memorized for use in programs such as excel. Same with actuarial science. You do need to be able to do certain functions from memory to be successful at that job because looking them up all the time wastes time that could be used doing the actual job.
There's no denying we should remember information we need to use all the time. And that will happen naturally, usually after I look something up two or three times, I wont have to look it up anymore because it will be ingrained in my memory. The point is, we never really know what we're going to need in our job until we're in that job, so it's pointless trying to remember as much as you possibly can when most of that information will never be needed in practice.
Kind of how most people can write in English but aren't experts on etymology and the history of grammar.
Kind of how I said that in certain fields it's easier just to memorize the equations without necessarily knowing complex theory behind them.
Yes this is exactly the point I'm trying to make. It's more important to understand how the equation works than it is to simply remember the equation. If you understand how the equation works, you are very unlikely to forget the equation, and even if you do forget it, chances are you'll be able to recreate the equation just from your understanding of how it works. We need to focus more on the high level concepts and less on the basic repetitive stuff.
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20 | 4,634 | 21,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.968497 |
https://engineering.stackexchange.com/questions/11970/electric-flux-density-outside-cylindrical-surface | 1,627,397,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153392.43/warc/CC-MAIN-20210727135323-20210727165323-00402.warc.gz | 247,573,192 | 36,620 | # Electric flux density outside cylindrical surface?
What I did is that I draw a Gaussian contour with radius p>a.
The electric flux outside the cylinder is Ps/2
The electric flux inside the cylinder is PL/(2*pi*p)
Consider a length $L$ of the cylindrical surface and the line. Draw a cylindrical contour with radius $\rho>a$ and length $L$, centered about the charged cylinder and line charge.
The charge contained within the contour is equal to the linear charge density of the line of charge times its length, plus the surface charge density of the cylindrical surface times its surface area, that is: $$Q = (\rho_L)(L)+(\rho_S)(A)$$ The surface area of the charged cylinder is the circumference of the cylinder times its length: $A=2\pi a L$. Thus, the charge contained within the Gaussian contour is: $$Q = (\rho_L)(L)+(\rho_S)(2\pi a L)$$ The flux density is equal to the flux passing through the Gaussian contour divided by the area of the contour: $$|\vec{D}|=\frac{Q}{A}=\frac{(\rho_L)(L)+(\rho_S)(2\pi a L)}{2\pi\rho L}$$ Note that the length $L$ can be canceled from each term. Due to the symmetry of the problem, the electric flux acts in the radial direction: $$\vec{D}=\frac{\rho_L+2\pi a \rho_S}{2\pi\rho}\vec{a}_\rho$$ | 324 | 1,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.881267 |
http://facebook.stackoverflow.com/questions/3125876/could-somebody-explain-how-to-use-gldrawelements-iphone | 1,369,500,703,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705976722/warc/CC-MAIN-20130516120616-00092-ip-10-60-113-184.ec2.internal.warc.gz | 96,724,474 | 13,696 | # Could somebody explain how to use glDrawElements (iPhone)?
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Sorry for the duplicaiton, but I've been googlin' for hours now without any result.
I have this (optimized) data of a simple cube exported from a converter:
``````// 8 Verticies
// 4 Texture Coordinates
// 6 Normals
// 12 Triangles
static GLshort cubeFace_indicies[12][9] = {
// Box001
{2,0,3 ,0,0,0 ,0,1,2 }, {1,3,0 ,0,0,0 ,3,2,1 }, {5,4,7 ,1,1,1 ,1,3,0 },
{6,7,4 ,1,1,1 ,2,0,3 }, {1,0,5 ,2,2,2 ,1,3,0 }, {4,5,0 ,2,2,2 ,2,0,3 },
{3,1,7 ,3,3,3 ,1,3,0 }, {5,7,1 ,3,3,3 ,2,0,3 }, {2,3,6 ,4,4,4 ,1,3,0 },
{7,6,3 ,4,4,4 ,2,0,3 }, {0,2,4 ,5,5,5 ,1,3,0 }, {6,4,2 ,5,5,5 ,2,0,3 }
};
static GLfloat cubeVertices [8][3] = {
{-100.0f,-100.0f,-100.0f},{100.0f,-100.0f,-100.0f},{-100.0f,100.0f,-100.0f},
{100.0f,100.0f,-100.0f},{-100.0f,-100.0f,100.0f},{100.0f,-100.0f,100.0f},
{-100.0f,100.0f,100.0f},{100.0f,100.0f,100.0f}
};
static GLfloat cubeNormals [6][3] = {
{0.0f,0.0f,1.0f},{0.0f,0.0f,-1.0f},{0.0f,1.0f,0.0f},
{-1.0f,0.0f,0.0f},{0.0f,-1.0f,0.0f},{1.0f,0.0f,0.0f}
};
static GLfloat cubeTextures [4][2] = {
{1.0f,2.0f},{1.0f,1.0f},{0.0f,2.0f},
{0.0f,1.0f}
};
``````
I have a texture set up, and I want to see something on the screen. My recent drawing code below:
``````glTexCoordPointer(2, GL_FLOAT, 0, cubeTextures);
glVertexPointer(3, GL_FLOAT, 0, cubeVertices);
glNormalPointer(GL_FLOAT, 0, cubeNormals);
glBindTexture(GL_TEXTURE_2D, textures[0]);
glDrawElements(GL_TRIANGLES , 12, GL_SHORT, cubeFace_indicies);
``````
-
And of course it doesn't work. – Geri Jun 26 '10 at 23:50 what glEnable() calls have you made? – claptrap Jun 27 '10 at 0:40 glEnable(GL_TEXTURE_2D); glEnable(GL_BLEND); glEnable(GL_DEPTH_TEST); – Geri Jun 27 '10 at 12:22 Whats the problem? – Geri Jun 27 '10 at 12:23 Ya, and the necessary glEnableClientState functions of course. – Geri Jun 27 '10 at 12:50
You might be able to get away with specifying fewer unique points if your model can be drawn with flat shading (i.e. calling `glShadeModel(GL_FLAT)` before drawing). In that case, the normal from the last vertex in a triangle will be used for the entire triangle. This requires careful ordering of your indices, of course. | 958 | 2,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.730745 |
http://www.markedbyteachers.com/gcse/science/electromagnetic-wave.html | 1,481,284,745,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542695.22/warc/CC-MAIN-20161202170902-00183-ip-10-31-129-80.ec2.internal.warc.gz | 580,104,220 | 17,443 | • Join over 1.2 million students every month
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# Electromagnetic Wave
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A065275 Infinite binary tree inspired permutation of N: 1 -> 3, 11ab..yz -> 11ab..yz0, 10ab..y0 -> 10ab..y, 10ab..y1 -> 11ab..y1. 6
3, 1, 6, 2, 7, 12, 14, 4, 13, 5, 15, 24, 26, 28, 30, 8, 25, 9, 27, 10, 29, 11, 31, 48, 50, 52, 54, 56, 58, 60, 62, 16, 49, 17, 51, 18, 53, 19, 55, 20, 57, 21, 59, 22, 61, 23, 63, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 32, 97, 33, 99 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS On the right side every node replaces its left child, on the left side the left children replace their parents and the right children are transferred to the same offset at the right side (staying right children). See comment at A065263. LINKS Table of n, a(n) for n=1..67. Index entries for sequences that are permutations of the natural numbers MAPLE RightChildTransferred := proc(n) local k; if(1 = n) then RETURN(3); fi; k := floor_log_2(n)-1; if(3 = floor(n/(2^k))) then RETURN(2*n); fi; if(0 = (n mod 2)) then RETURN(n/2); fi; RETURN(n + (2^k)); end; CROSSREFS A057114, A065263, A065269, A065281, A065287. Inverse: A065276, conjugated with A059893: A065277 and the inverse of that: A065278. Sequence in context: A122913 A289444 A069115 * A071045 A274317 A105358 Adjacent sequences: A065272 A065273 A065274 * A065276 A065277 A065278 KEYWORD nonn AUTHOR Antti Karttunen, Oct 28 2001 STATUS approved
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https://au.mathworks.com/matlabcentral/cody/problems/43617-calculate-the-values-of-a-polynomial/solutions/1383445 | 1,611,039,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00018.warc.gz | 232,638,650 | 17,292 | Cody
# Problem 43617. Calculate the values of a polynomial.
Solution 1383445
Submitted on 17 Dec 2017 by jean claude
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
p=[-1 1] x=[1 2;3 4] y_correct=[ 0 -1; -2 -3] assert(isequal(SolvePoly(p,x),y_correct))
p = -1 1 x = 1 2 3 4 y_correct = 0 -1 -2 -3
2 Pass
p=[-2 0 1 -1 3 2] x=[5 6 11; 2 13 7; 4 9 21] y_correct=[-6133 -15352 -320857; -52 -740517 -33297; -1986 -117421 -8159317] assert(isequal(SolvePoly(p,x),y_correct))
p = -2 0 1 -1 3 2 x = 5 6 11 2 13 7 4 9 21 y_correct = -6133 -15352 -320857 -52 -740517 -33297 -1986 -117421 -8159317 | 296 | 713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-04 | latest | en | 0.429276 |
http://math.stackexchange.com/questions/97602/finding-the-minimum-grade/97606 | 1,469,502,351,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824570.25/warc/CC-MAIN-20160723071024-00240-ip-10-185-27-174.ec2.internal.warc.gz | 162,999,948 | 18,417 | Professor sent the following data to his students: Full Grade: 25, Max. grade: 22.5, Mean: 19.9, Median: 20
Assuming the number of students is n. Is it possible to calculate the minimum grade from the given data only?
-
PS: you may know some of other students' grades. For example you heard that student x1 got 21, another student x2 got 18 and so on .. but you don't know all the students' grades of course – Osama Gamal Jan 9 '12 at 11:27
What does "full grade" mean? – Dan Brumleve Jan 9 '12 at 11:29
@Dan Probably "Full Grade" is the maximum grade you can get in the exam. For instance there are 5 questions 5 points each. – marvinthemartian Jan 9 '12 at 11:36
@marvinthemartian, is that information relevant? – Dan Brumleve Jan 9 '12 at 11:38
@Dan Frankly, I do not think it is. But I do not know the answer yet, so maybe. – marvinthemartian Jan 9 '12 at 11:41
The answer is no. Let us look for an easier example:
Assume $n=5$, our students are called "A","B","C","D","E" and the grades are given by:
\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 2&1&1&1&0 \end{array}
Clearly Maximum, Median and Mean are given by {2,1,1}. But the dataset
\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 2&1&1&1/2&1/2 \end{array}
Also has the same Mean, Median and Maximum. You can always do this by using a convex combination of the last two values as long as they stay below the median. They will not affect the median, nor the mean or the maximal element but they will change the minimal grade given.
Edit: Now with this knowledge we can come back to your question and easily construct a counterexample flor $n=5$:
\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 22.5&20&20&20&17 \end{array}
\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 22.5&20&20&19&18 \end{array}
Those do both fit to your data. For higher $n$ you can pad with the mean $19.9$ on the left and right side of the median (odd $n$). For even $n$ there is a similar counterexample. For $n \in \{3,4\}$ refer to the answer given by tom.
-
I don't believe so, not for general n at least. We can for some isolated cases:
For $n = 2$, the problem simply doesn't work
For $n = 3$, we need one student to get 22.5, another student to get 20, and the other to get $s_3$ so that:
$$\frac{22.5+20+s_3}{3}=19.9$$ Which works out for $s_3=17.2$, the minimum grade
For $n=4$, lets say we have 4 students with 4 grades $s_1 \geq s_2\geq s_3\geq s_4$. We know then that $s_1=22.5$, that: $$\frac{s_2+s_3}{2}=20$$
and $$\frac{22.5+s_2+s_3+s_4}{4}=19.9$$ We can then solve this for $s_4$ using the fact that $s_2+s_3=40$ to get $s_4=17.1$.
However, for $n=5$, we run into difficulties. We know (using the same system of notation) that $s_1=22.5$, and that $s_3=20$, but beyond that, the fact that
$$\frac{22.5+s_2+20+s_4+s_5}{5}=19.9$$
Will never allow us to say any more about $s_5$. Indeed, this is true for any $n\geq5$ (is this clear?).
-
Note that you are searching for integer numbers (for simplification since grades can only be x.0 or x.5) and note that s_2 > s_4 > s_5 – Osama Gamal Jan 9 '12 at 12:17 | 1,050 | 3,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-30 | latest | en | 0.862383 |
http://oeis.org/A165740 | 1,566,310,260,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315544.11/warc/CC-MAIN-20190820133527-20190820155527-00428.warc.gz | 140,893,488 | 4,040 | This site is supported by donations to The OEIS Foundation.
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A165740 Positive integers n such that solution to the toric n X n "Lights Out" puzzle is not unique (up to the order of flippings; each flipping appears at most once). 3
3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 21, 24, 25, 27, 30, 31, 33, 34, 35, 36, 39, 40, 42, 45, 48, 50, 51, 54, 55, 57, 60, 62, 63, 65, 66, 68, 69, 70, 72, 75, 78, 80, 81, 84, 85, 87, 90, 93, 95, 96, 99, 100, 102, 105, 108, 110, 111, 114, 115, 117, 119, 120, 123, 124, 125, 126 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Complement to the sequence A165741 in the set of positive integers. Any positive multiple of a member of this sequence is also a member. Primitive elements are in A007802. - Thomas Buchholz, May 23 2014 REFERENCES See A165738 for the references. LINKS Max Alekseyev and Thomas Buchholz, Table of n, a(n) for n = 1..1000 [terms 67 through 1000 were computed by Thomas Buchholz, May 20 2014] FORMULA A number n is in this sequence iff A165738(n) > 0. CROSSREFS Cf. A159257, A075462, A117870, A165741, A007802, A165738. Sequence in context: A239064 A227455 A237417 * A241571 A080307 A001969 Adjacent sequences: A165737 A165738 A165739 * A165741 A165742 A165743 KEYWORD nonn AUTHOR Max Alekseyev, Sep 25 2009 EXTENSIONS More terms from Thomas Buchholz, May 20 2014 STATUS approved
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posted by .
A beam of polarized light with an average intensity of 17.1 W/m2 is sent through a polarizer. The transmission axis makes an angle of 27° with respect to the direction of polarization. Determine the rms value of the electric field of the transmitted beam.
• Physics -
half that power is in the E field.
1/2 * 17.1= epsilion*E^2/2
solve for E.
Now after the polarizer, only cos27 gets thru.
Efinal=E cos27
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A199011 Triangle T(n,k), read by rows, given by (1,1,-1,1,0,0,0,0,0,0,0,...) DELTA (0,1,0,0,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. 1
1, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 6, 4, 1, 0, 5, 10, 10, 5, 1, 0, 6, 15, 20, 15, 6, 1, 0, 7, 21, 35, 35, 21, 7, 1, 0, 8, 28, 56, 70, 56, 28, 8, 1, 0, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS Mirror image of triangle in A198321. Variant of A074909, A135278. LINKS Table of n, a(n) for n=0..65. FORMULA T(n,k)=binomial(n,k+1). Sum_{0<=k<=n} T(n,k)*x^k = ((x+1)^n-1)/x for n>0. G.f.: (1-(1+y)*x+(1+y)*x^2)/(1-(2+y)*x+(1+y)*x^2). T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0, T(2,0) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Feb 12 2014 EXAMPLE Triangle begins : 1 1, 0 2, 1, 0 3, 3, 1, 0 4, 6, 4, 1, 0 5, 10, 10, 5, 1, 0 6, 15, 20, 15, 6, 1, 0 CROSSREFS Cf. A007318, A074909, A135278, A198321 Sequence in context: A208727 A242378 A268820 * A206735 A089000 A253829 Adjacent sequences: A199008 A199009 A199010 * A199012 A199013 A199014 KEYWORD easy,nonn,tabl AUTHOR Philippe Deléham, Nov 01 2011 STATUS approved
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# Forum: Mathematical Puzzles
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15. ### very very simple puzzle
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MCE311_HW01_sol_Fall2011
# MCE311_HW01_sol_Fall2011 - 9.8 During pouring into a sand...
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9.8 During pouring into a sand mold, the molten metal can be poured into the downsprue at a constant flow rate during the time it takes to fill the mold. At the end of pouring the sprue is filled and there is negligible metal in the pouring cup. The downsprue is 15 cm long. Its cross-sectional area at the top = 5 cm 2 and at the base = 3.75 cm 2 . The cross-sectional area of the runner leading from the sprue also = 3.75 cm 2 , and it is 20 cm long before leading into the mold cavity, whose volume = 1015 cm 3 . The volume of the riser located along the runner near the mold cavity = 390 cm 3 . It takes a total of 3.0 sec to fill the entire mold (including cavity, riser, runner, and sprue. This is more than the theoretical time required, indicating a loss of velocity due to friction in the sprue and runner. Find (a) the theoretical velocity and flow rate at the base of the downsprue; (b) the total volume of the mold; (c) the actual velocity and flow rate at the base of the sprue; and (d) the loss of head in the gating system due to friction. Solution : (a) Velocity v = (2 × 9.8 × 100 × 15) 0.5 = 171.5 cm/sec Flow rate Q = 171.5 × 3.75 = 643.1 cm 3 /sec (b) Total V = 1015 + 390 + 0.5 (5 + 3.75)(15) + (3.75)(20) = 1545.6 cm 3 (c) Actual flow rate Q = 1545.6/3 = 515.2 cm 3 /sec Actual velocity v = 515.2/3.75 = 137.4 cm/sec (d) v = (2 × 9.8 × 100 × h) 0.5 = 44.3 h 0.5 = 137.4 cm/sec h 0.5 = 137.4/44.3 = 3.102 h = 3.102 2 = 9.622 cm Head loss = 15 9.622 = 5.378 cm 9.11 A flat plate is to be cast in an open mold whose bottom has a square shape that is 200 mm by 200
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https://quizlet.com/explanations/questions/compute-the-products-ab-and-ba-if-possible-for-the-following-a-a-leftbeginarrayrr0-2-3-1endarrayright-b-leftbeginarrayrr-1-4-1-5endarrayrigh-0fe7da7e-5f199a96-3cf4-47be-8241-f49fb4716a51 | 1,695,728,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00898.warc.gz | 536,147,410 | 92,072 | Try Magic Notes and save time.Try it free
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# Compute the products $A$ $B$ and $B$ $A$, if possible, for the following:(a) $A$$=\left(\begin{array}{rr}0 & -2 \\ 3 & 1\end{array}\right), B$$=\left(\begin{array}{rr}-1 & 4 \\ 1 & 5\end{array}\right)$(b) $A$$=\left(\begin{array}{rrr}8 & 3 & -2 \\ 1 & 0 & 4\end{array}\right), B$$=\left(\begin{array}{rr}2 & -2 \\ 4 & 3 \\ 1 & -5\end{array}\right)$(c) $A$$=\left(\begin{array}{r}0 \\ -2 \\ 4\end{array}\right), B$$=\left(\begin{array}{lll}0, & -2, & 3\end{array}\right)$(d) $A$$=\left(\begin{array}{rr}-1 & 0 \\ 2 & 4\end{array}\right), \quad B$$=\left(\begin{array}{rr}3 & 1 \\ -1 & 1 \\ 0 & 2\end{array}\right)$
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(a) $\boldsymbol{A}\boldsymbol{B}$ is a $2\times2$ matrix obtained by multiplying each row of $\boldsymbol{A}$ by the corresponding column of $\boldsymbol{B}$.
\begin{aligned} \boldsymbol{A}\boldsymbol{B}&= \begin{pmatrix} 0&-2\\3&1 \end{pmatrix} \begin{pmatrix} -1&4\\1&5 \end{pmatrix}\\ &=\begin{pmatrix} 0\cdot(-1)-2\cdot1&0\cdot4-2\cdot5\\3\cdot(-1)+1\cdot1&3\cdot4+1\cdot5 \end{pmatrix}\\ &=\begin{pmatrix} -2 & -10\\-2&17 \end{pmatrix}\\ \end{aligned}
$\boldsymbol{B}\boldsymbol{A}$ is a $2\times2$ matrix obtained by multiplying each row of $\boldsymbol{B}$ by the corresponding column of $\boldsymbol{A}$.
\begin{aligned} \boldsymbol{B}\boldsymbol{A}&= \begin{pmatrix} -1&4\\1&5 \end{pmatrix} \begin{pmatrix} 0&-2\\3&1 \end{pmatrix}\\ &=\begin{pmatrix} -1\cdot 0+4\cdot3 & -1\cdot(-2)+4\cdot1\\1\cdot0+5\cdot3&1\cdot(-2)+5\cdot1 \end{pmatrix}\\ &=\begin{pmatrix} 12&6\\15&3 \end{pmatrix}\\ \end{aligned}
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3,016 solutions | 827 | 2,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 111, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-40 | latest | en | 0.4258 |
https://best-home-improvements.info/house-painting-designs-and-colors-video-arvada-80002.html | 1,632,866,982,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00201.warc.gz | 177,655,517 | 9,341 | ```Painters often tint primer close to the color of the top coat, but Wallis thinks that's a recipe for "holidays," or missed spots. Instead, he tints his primer a contrasting color. "If I can see the color coming through, I know I need to apply more paint," he says. On the cottage shown in this story, he chose a gray-blue primer to go under a peach top coat. House Painting
```
An example of the wall measurement would be: 40 linear feet of bedroom space, x 8 wall height, = 320, x 2 = 640. Minus 1 door (60) and 2 windows (80) = 500 sq feet being painted. Then divide the 500 by 400 (sq feet per gallon), and you get 1.25 gallons (4.73 L) needed for that room. For this, you'd need 1 gallon (3.8 L) and 1 US-quart (950 ml). If the amount it comes out to is over 1.3, we recommend just getting 2 gallons (7.6 L) so that you have leftovers if needed, since 2 US quarts (2,000 ml) costs essentially the same as a gallon in most stores. House Painting
Prime. Dark colors, stains (once sealed), and previously unpainted surfaces (drywall, spackle, etc.) will need a primer coat, usually white. NOTE: most paint stores & home improvement centers will now tint primer (at no charge) to match fairly close to the color of the finished coat, that way two coats of primer need not be applied.[6] Although not all surfaces need a prime coat, skip this step at your peril! Dark colors will likely show through the first -- or even the first couple-- topcoats of paint. Sealants and unpainted surfaces like spackle patches will absorb or repel moisture in a topcoat at a different level than the areas surrounding them. Applying a good primer coat will help even out these differences. Primer equalizes a wall to a uniform surface. It's like erasing a canvas before drawing a new picture. Although some will argue the point, you generally don't need to spend a great deal on primer or buy special primer. A cheap, 5 gallon (18.9 L) bucket of plain, flat white paint will usually do the trick and cover a large area. Give your primer at least 24 hours to dry (follow its instructions) before applying a topcoat.
Paintzen color expert Kristen Chuber shares her top paint color, which is a dusty Chalky Blue (PPG1153-5) by PPG Porter Paints. "Somewhere between blue and gray, this velvety shade can actually be used as a neutral. It looks beautiful with bright white trim, but maybe even more impactful with rich, black accents. We have also seen this shade used beautifully in a monochromatic palette, alongside other shades of blue-gray, both on the lighter and darker side of the spectrum." Try using this hue in your kitchen, one of your bathrooms, or in the bedroom.
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A: Paintzen has top-rated professional painters and is your one-stop shop painting service for all of greater New York City painting needs. New York City has always been on the cutting edge of innovation. For those looking for an innovative, on-demand residential painting service in New York, Paintzen is the smoothest path to a fresh coat of paint. House Painting
The cost per square foot will vary depending on the geographic region, the type of paint product used, the conditions of the surfaces, how much the color is changing, the number of coats, and a variety of other factors. To get accurate pricing for your home, set up an estimate appointment with a local independently owned and operated CertaPro Painters® franchise.
Paintzen color expert Kristen Chuber shares her top paint color, which is a dusty Chalky Blue (PPG1153-5) by PPG Porter Paints. "Somewhere between blue and gray, this velvety shade can actually be used as a neutral. It looks beautiful with bright white trim, but maybe even more impactful with rich, black accents. We have also seen this shade used beautifully in a monochromatic palette, alongside other shades of blue-gray, both on the lighter and darker side of the spectrum." Try using this hue in your kitchen, one of your bathrooms, or in the bedroom. House Painting
When you want to hire residential painters in New York to transform the interior of your residential property, quality and trust should never be compromised. Hiring professional painters is an absolute must. After over 10 years experience doing interior residential painting we really focus on customer satisfaction, from start to finish, so you can rely on us for your next interior residential painting project.
"Red is a powerful and energetic hue," says PPG Paints color expert Dee Schlotter. "Red Delicious (00YR 08/409) by Glidden is a deep red that would serve as a bold and stimulating color for an accent wall." It looks particularly refreshing when paired with white trim. While it's a bold hue like this apple red, it's a timeless version of red that will look beautifully rich for years to come. House Painting
If you had been or are currently planning a painting project for your home, we encourage you to get a remote quote from our team and put down a 10% fully-refundable deposit to tentatively hold a future date for your project. If you don’t yet have an ideal start date, we can tentatively hold a 60- or 90-day start-date for you, which can be changed, with no fee, at any time. Learn how Paintzen is taking necessary measures to keep you, the customer, and our painters safe and healthy during by abiding by the COVID-19 rules and regulations. House Painting
Plan the workforce. If you intend to not hire a professional crew, you'll need lots of help. There are many jobs to be done. First there's the furniture moving, then wall preparation, floor covering, materials gathering and prep, cleaning, and don't forget everyone will have to eat. It can easily take a team of five people a full ten days to paint a two-story (approx 2000 sq.ft.) home. Get as many people to help as you can. If some can only come one or two days, great. Maybe others can fill in. Ensure you plan with your workforce members in mind. They'll need plenty of time to arrange days away from work. Identify a few key personnel: House Painting
For trim and doors, start with 1 gallon (3.8 L) of trim paint for every 600 sq feet of floor space. Purchase more at the store if/when needed. Trim is something that is difficult to calculate exactly, and it is more time-efficient to simply start with less than you need, and go buy more after you have used up the first round of paint and determined how much you will need to finish by seeing how much you have painted so far, compared to the gallons used.
After you have done the multiplication, subtract about 40 sq feet per window, and about 60 sq feet per door. Divide by 400 (interior paint covers 400 sq feet per gallon). The number that remains, is how many gallons you will need. If doing multiple wall colors, then you should do this process for each room (or sets of rooms) with a particular color. House Painting
With our four-season climate, the wood used in residential NYC homes often becomes swollen and cracked due to prolonged exposure to the elements. New York is also prone to high humidity levels, which means that your exterior paint needs to be resistant to mildew. We take all of these climate conditions into consideration when you request a quote for a house painting job, and we focus our attention on quality, expertise, and accessibility from start to finish.
As the dance proceeds, keep an eye on the weather. Rain can wash freshly applied latex right off the wall, and a temperature dip below 50 degrees F two days after application can interfere with adhesion and curing and dull the sheen of glossy paints. (Latexes like Sherwin-Williams's Duration and Benjamin Moore's MoorGard Low Lustre are formulated to tolerate temps as low as 35 and 40 degrees, respectively.)
First, if you need it, we have it — all the paint for you home — every color you can imagine, in the finish you need, for every surface. Second, we’ can help narrow down your wall paint color choice until you decide on the right paint for your project. Of course, our paint associates at your local store can help pick our paint for your home. Pick up some paint swatches or order paint samples online so you can picture what it will look like. House Painting
Plan the schedule. Get a grip on the time it will take to bring the project to fruition. Plan for time to move furniture, wall prep, cut in, the painting itself, eating and breaks, and don't forget cleanup and bringing furniture back in. As you plan, err on the side of prudence. Unforeseen events will slow you down, so allow time for these. Remember, this is a multi-day project. Don't try to fit too much into a day. If you move faster than planned, great! House Painting
The Cutters. Someone with meticulous attention to detail and a steady hand should be assigned the job of "cutting in," or painting a straight edge where needed, such as along a wall where the ceiling does not get painted. Many products are available to assist, but none work as well as a person who's good at doing it freehand. Ensure this person is skilled (ask them to show you). A poor, jagged, wavy or splotched cutting-in job will jump out at you every time you walk by it. Why more than one cutter? This job is nerve-wracking and painful to hands and arms after a few days. You'll want to give this person a break after a few walls. House Painting
"There are a small number of timeless and classic shades we always find ourselves coming back to when searching for the perfect neutral," says Alicia Weaver of Alicia Weaver Design. "Alaskan Husky (1479) by Benjamin Moore is one of them. It is a classic shade of gray that complements accent colors." Light silvery grays work beautifully in guest bedrooms and front entryways. House Painting
A: Paintzen has top-rated professional painters and is your one-stop shop painting service for all of greater New York City painting needs. New York City has always been on the cutting edge of innovation. For those looking for an innovative, on-demand residential painting service in New York, Paintzen is the smoothest path to a fresh coat of paint.
The health and safety of our painters and customers is Paintzen’s top priority. It also remains a priority for us to support our local paint contractor partners across the country during this unprecedented time. Because home renovations have been deemed “non-essential” many painters on our platform are without work. To help our partners get back in business as soon as it is declared safe to do so, we are announcing our Book Now, Schedule Later policy. House Painting
Your home is your sanctuary. It’s the place you want to be able to relax, unwind, and enjoy. When you decide it’s time to give your home a makeover, need renovations done or want to completely remodel your home, you want to be confident that any contractors you hire are qualified, experienced and reliable home improvement professionals. With many years of residential and commercial renovation experience, Eli’s Painting & Renovations is the New York City home improvement company you can rely on for outstanding workmanship and quality service.
# The cost per square foot will vary depending on the geographic region, the type of paint product used, the conditions of the surfaces, how much the color is changing, the number of coats, and a variety of other factors. To get accurate pricing for your home, set up an estimate appointment with a local independently owned and operated CertaPro Painters® franchise. House Painting
```After you have done the multiplication, subtract about 40 sq feet per window, and about 60 sq feet per door. Divide by 400 (interior paint covers 400 sq feet per gallon). The number that remains, is how many gallons you will need. If doing multiple wall colors, then you should do this process for each room (or sets of rooms) with a particular color. House Painting
```
A: Paintzen has top-rated professional painters and is your one-stop shop painting service for all of greater New York City painting needs. New York City has always been on the cutting edge of innovation. For those looking for an innovative, on-demand residential painting service in New York, Paintzen is the smoothest path to a fresh coat of paint.
After you have done the multiplication, subtract about 40 sq feet per window, and about 60 sq feet per door. Divide by 400 (interior paint covers 400 sq feet per gallon). The number that remains, is how many gallons you will need. If doing multiple wall colors, then you should do this process for each room (or sets of rooms) with a particular color. House Painting
The thick canvas stays in place, so you don’t need to tape it, and you can use it to cover any surface. Plastic drop cloths are slippery to walk on or set a ladder on and don’t stay in place. Even worse, paint spills on plastic stay wet, and they can end up on your shoes and get tracked through the house. Canvas is slippery on hard floors, so rosin paper is better over vinyl, tile and hardwood. Tape the sheets together and to the floor to provide a nonslip surface. House Painting
Plan the schedule. Get a grip on the time it will take to bring the project to fruition. Plan for time to move furniture, wall prep, cut in, the painting itself, eating and breaks, and don't forget cleanup and bringing furniture back in. As you plan, err on the side of prudence. Unforeseen events will slow you down, so allow time for these. Remember, this is a multi-day project. Don't try to fit too much into a day. If you move faster than planned, great! House Painting
If your looking for commercial painters for hire to transform the interior of your commercial property, quality and trust should never be compromised. Working with a professional painter in New York is an absolute must. We have over 8 years experience doing interior commercial painting. We really focus on customer satisfaction, from start to finish you can rely on us for your next interior commercial painting project. House Painting
We can acquire paint products at much lower prices because of the quantity we purchase and our relationships with paint suppliers. In most cases, it will cost more to buy your own paint. In the event you still prefer to purchase your own paint, it is up to the independently owned and operated CertaPro Painters® franchise you are working with, so this is something you should bring up as you plan your painting project. House Painting
BEHR Premium Cabinet and Trim Interior Semi-Gloss Enamel BEHR Premium Cabinet and Trim Interior Semi-Gloss Enamel offers excellent flow and leveling and dries to a hard, durable finish. Its outstanding block resistance allows for quick return to service, making it ideal for use on cabinets, trim, doors, windows, shutters and woodwork. This product can also be used on other properly prepared and primed substrates, such as drywall, masonry and metal. More + Product Details Close House Painting
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Painters often tint primer close to the color of the top coat, but Wallis thinks that's a recipe for "holidays," or missed spots. Instead, he tints his primer a contrasting color. "If I can see the color coming through, I know I need to apply more paint," he says. On the cottage shown in this story, he chose a gray-blue primer to go under a peach top coat. House Painting
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http://www.jiskha.com/display.cgi?id=1360100211 | 1,495,544,567,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00325.warc.gz | 532,308,875 | 3,712 | # algebra
posted by on .
Write a recursive formula for the sequence formula 15,26,48,92,180,.. then find the next term.
• algebra - ,
Look like Tn = 2T(n-1) - 4
26 = 2*15-4
48 = 2*26-4
so, what's 2*180-4?
• algebra - ,
got it thanks | 90 | 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} | 3.265625 | 3 | CC-MAIN-2017-22 | latest | en | 0.841079 |
https://www.vedantu.com/question-answer/if-the-vertices-pqr-of-a-triangle-pqr-are-class-11-maths-cbse-5ee0bae1bc96fa4dc1612a31 | 1,721,613,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00081.warc.gz | 909,004,494 | 37,491 | Courses
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# If the vertices P,Q,R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s) ?(a) Centroid (b) Incentre(c) Circumcentre(d) Orthocentre
Last updated date: 22nd Jul 2024
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Hint: Orthocentre is the point of concurrency of the altitudes of a triangle.
Circumcentre is the point of concurrency of perpendicular bisectors of a triangle.
Incentre is the point of concurrency of angle bisectors of a triangle.
Centroid is the point of concurrency of medians of a triangle.
Let the coordinates of the vertices be $P\left( {{x}_{1}},{{y}_{1}} \right),Q\left( {{x}_{2}},{{y}_{2}} \right)\ and R\left( {{x}_{3}},{{y}_{3}} \right)$.
Now, we know the centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ is given as :
$\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
So , the centroid of $\Delta PQR$ is :
$G\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
Now , in the equation it is given that the coordinates of the vertices are rational.
We know , the closure property of rational numbers says that the sum, difference, product or quotient of two rational numbers is always a rational number.
So , by closure property of rational number we can say that,
$\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)$ is rational and $\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ is rational.
Hence, centroid is a rational point.
Now, we know the slope of line joining two point $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
$m=\left( \dfrac{{{b}_{2}}-{{b}_{1}}}{{{a}_{2}}-{{a}_{1}}} \right)$
So, the slope of $PQ$ is given as :
${{m}_{PQ}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)$
Slope of $PR$ is given as :
${{m}_{PR}}=\left( \dfrac{{{y}_{3}}-{{y}_{1}}}{{{x}_{3}}-{{x}_{1}}} \right)$
And the slope of $QR$ is given as :
${{m}_{QR}}=\left( \dfrac{{{y}_{2}}-{{y}_{3}}}{{{x}_{2}}-{{x}_{3}}} \right)$
Now, we know if two lines are perpendicular then product of their slopes is $\left( -1 \right)$
So,
${{m}_{PQ}}\times {{m}_{\bot PQ}}=-1$
$\Rightarrow {{m}_{\bot PQ}}=\dfrac{-1}{{{m}_{PQ}}}$
$\Rightarrow {{m}_{\bot PQ}}=\dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}}............\left( i \right)$
And,
${{m}_{\bot QR}}\times {{m}_{QR}}=-1$
$\Rightarrow {{m}_{\bot QR}}=\dfrac{-1}{{{m}_{QR}}}$
$\Rightarrow {{m}_{\bot QR}}=\dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}}............\left( ii \right)$
And,
${{m}_{\bot PR}}\times {{m}_{PR}}=-1$
$\Rightarrow {{m}_{\bot PR}}=\dfrac{-1}{{{m}_{PR}}}$
$\Rightarrow {{m}_{\bot PR}}=\dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}}............\left( iii \right)$
Now, since $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ are rational then ${{m}_{\bot PQ}},{{m}_{\bot QR}}\ and {{m}_{\bot PR}}$, given by equation $\left( i \right),\left( ii \right)\ and \left( iii \right)$ must be rational.
Now, let’s consider the line perpendicular to sides and passing from the opposite vertex.
We know equation of line passing from $\left( a,b \right)$ with slope $m$ is given as :
$\left( y-b \right)=m\left( x-a \right)$
So, equation of line perpendicular to $PQ$ and passing through $R$ is given as :
$\left( y-{{y}_{3}} \right)={{m}_{\bot QR}}\left( x-{{x}_{3}} \right)$
$\left( y-{{y}_{3}} \right)=\dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}}\left( x-{{x}_{3}} \right)..........\left( iv \right)$
And equation of line perpendicular to $QR$ and passing through $P$ is given as :
$\left( y-{{y}_{1}} \right)={{m}_{\bot QR}}\left( x-{{x}_{1}} \right)$
$\left( y-{{y}_{1}} \right)=\dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}}\left( x-{{x}_{1}} \right)..........\left( v \right)$
And equation of line perpendicular to $PR$ and passing through $Q$ is given as :
$\left( y-{{y}_{2}} \right)={{m}_{\bot PR}}\left( x-{{x}_{2}} \right)$
$\left( y-{{y}_{2}} \right)=\dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}}\left( x-{{x}_{2}} \right)..........\left( vi \right)$
Now we know that ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}}\ and {{y}_{3}}$ all are rational.
Hence, the coordinates of the point satisfying all three equations will also be rational $......\left( vii \right)$
Now we know the point of intersection of the perpendiculars from the vertices to the opposite sides is the orthocentre $......\left( viii \right)$
Hence, from statements $\left( vii \right)\And \left( viii \right)$ we can say that ortho centres will be rational points.
Now, let’s consider the midpoint of the sides $PQ,QR\ and PR$ .
We know, the coordinates of the midpoint of the line joining $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
$\left( \dfrac{{{a}_{1}}+{{a}_{2}}}{2},\dfrac{{{b}_{1}}+{{b}_{2}}}{2} \right)$ .
So, the coordinates of midpoint of side $PQ$ is given as $D\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
The coordinates of midpoint of side $QR$ is given as $E\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$ .
And the coordinates of midpoint of side $PR$ is given as $F\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)$ .
Now let’s find the equation of line perpendicular to the sides and passing through the midpoint of the sides i.e. perpendicular bisector of sides. So, the equation of perpendicular bisector of $PQ$ is given as:
$y-\left( \dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)......\left( ix \right)$
The equation of perpendicular bisector of $QR$ is given as:
$y-\left( \dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)=\left( \dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}} \right)\left( x-\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2} \right) \right)......\left( x \right)$
The equation of perpendicular bisector of $PR$ is given as:
$y-\left( \dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)=\left( \dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}} \right)\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2} \right) \right)......\left( xi \right)$
Now we know, $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ are rational coordinates. Hence, any point satisfying all three equations i.e. $\left( ix \right),\left( x \right)\ and \left( xi \right)$ will have rational coordinates. $.............\left( xii \right)$
Now, we know the circumcentre is the point of intersection of the perpendicular bisectors of the sides of the triangle$..........\left( xiii \right)$
So, from statements $\left( xii \right)\And \left( xiii \right)$ we can conclude that the coordinates of the circumcentre are rational.
Now, we know the incentre of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$ is given as :
$\left( \dfrac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\dfrac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)$ where $a,b,c$ are the sides of the triangle opposite to the corresponding vertices .
Now, let $a$ be the length of side $QR,b$ be the length of side $PR$ and $c$ be the length of side $PQ$ .
Now, we know the distance between the points $\left( {{a}_{1}},{{b}_{1}} \right)\And \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
$\sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}}$
$\text{So, }a=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}$
$b=\sqrt{{{\left( {{x}_{1}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{3}} \right)}^{2}}},$
$\text{and }c=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Clearly , we can see that $a,b$ and $c$ may or may not be rational.
So, the coordinates of incentre may or may not be rational.
Hence , if the vertices $P,Q,R$ of a triangle $PQR$ are rational points , then the centroid , the orthocentre and the circumcentre will always be rational points.
Therefore, Options: (a) Centroid, (c) Circumcentre, (d) Orthocentre are correct.
Note: While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So, sign conventions should be carefully taken. | 3,106 | 8,799 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-30 | latest | en | 0.74946 |
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Algebra Pre-Algebra and Basic Algebra Math Forum
March 18th, 2011, 04:27 AM #1 Newbie Joined: Mar 2011 Posts: 4 Thanks: 0 Reciprocals Hi guys Is (3/2)(2/3) a reciprocals? Thanx for all help - I dont get it
March 18th, 2011, 04:31 AM #2
Global Moderator
Joined: Nov 2009
From: Northwest Arkansas
Posts: 2,766
Thanks: 4
Re: Reciprocals
Hi!
Is this the question?
Quote:
Are $\frac{3}{2}, \frac{2}{3}$ reciprocals?
There are two ways to think of this:
1) If one fraction is just the other fraction, but "flipped", then yes.
2) If their product is 1 (ONE), then yes.
Yes.
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OpenStudy (anonymous):
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Proof
Definitions occuring in Statement : quotient: x,y:A//B[x; y] int_seg: {i..j-} nat: bool: 𝔹 all: x:A. B[x] exists: x:A. B[x] implies: Q true: True apply: a function: x:A ⟶ B[x] natural_number: \$n equal: t ∈ T
Definitions unfolded in proof : member: t ∈ T ifthenelse: if then else fi compose: g pi1: fst(t) strong-continuity-test: strong-continuity-test(M;n;f;b) isl: isl(x) lt_int: i <j let: let strong-continuity2-implies-uniform-continuity uniform-continuity-from-fan-ext implies-quotient-true2 trivial-quotient-true strong-continuity2-no-inner-squash-cantor4 implies-quotient-true strong-continuity2-half-squash-surject-biject retraction-nat-nsub surject-nat-bool biject-bool-nsub2 strong-continuity2_biject_retract-ext bool_cases_sqequal any: any x decidable__int_equal strong-continuity2_functionality_surject strong-continuity2-half-squash strong-continuity2-iff-3 strong-continuity3_functionality_surject basic-implies-strong-continuity2-ext strong-continuity2-implies-3 surject-inverse uall: [x:A]. B[x] so_lambda: so_lambda(x,y,z,w.t[x; y; z; w]) so_apply: x[s1;s2;s3;s4] so_lambda: λ2y.t[x; y] top: Top so_apply: x[s1;s2] uimplies: supposing a so_lambda: λ2x.t[x] so_apply: x[s] all: x:A. B[x] implies: Q has-value: (a)↓ strict4: strict4(F) and: P ∧ Q prop: or: P ∨ Q squash: T btrue: tt bfalse: ff
Lemmas referenced : strong-continuity2-implies-uniform-continuity lifting-strict-decide istype-void strict4-spread strict4-decide has-value_wf_base is-exception_wf lifting-strict-callbyvalue lifting-strict-int_eq lifting-strict-isint value-type-has-value int-value-type istype-base lifting-strict-less uniform-continuity-from-fan-ext implies-quotient-true2 trivial-quotient-true strong-continuity2-no-inner-squash-cantor4 implies-quotient-true strong-continuity2-half-squash-surject-biject retraction-nat-nsub surject-nat-bool biject-bool-nsub2 strong-continuity2_biject_retract-ext bool_cases_sqequal decidable__int_equal strong-continuity2_functionality_surject strong-continuity2-half-squash strong-continuity2-iff-3 strong-continuity3_functionality_surject basic-implies-strong-continuity2-ext strong-continuity2-implies-3 surject-inverse
Rules used in proof : introduction sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity cut instantiate extract_by_obid hypothesis sqequalRule thin sqequalHypSubstitution equalityTransitivity equalitySymmetry isectElimination baseClosed Error :isect_memberEquality_alt, voidElimination independent_isectElimination Error :inhabitedIsType, Error :lambdaFormation_alt, sqequalSqle divergentSqle callbyvalueDecide hypothesisEquality unionElimination sqleReflexivity Error :equalityIstype, dependent_functionElimination independent_functionElimination decideExceptionCases axiomSqleEquality exceptionSqequal baseApply closedConclusion independent_pairFormation callbyvalueIntEq productElimination intEquality Error :universeIsType, int_eqExceptionCases Error :inrFormation_alt, imageMemberEquality imageElimination Error :inlFormation_alt, because_Cache
Latex:
\mforall{}F:(\mBbbN{} {}\mrightarrow{} \mBbbB{}) {}\mrightarrow{} \mBbbB{}. \00D9(\mexists{}n:\mBbbN{}. \mforall{}f,g:\mBbbN{} {}\mrightarrow{} \mBbbB{}. ((f = g) {}\mRightarrow{} F f = F g))
Date html generated: 2019_06_20-PM-02_52_46
Last ObjectModification: 2019_03_12-PM-04_21_49
Theory : continuity
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https://www.trustudies.com/question/2494/q-10-from-a-circular-card-sheet-of-ra/ | 1,675,004,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00626.warc.gz | 1,049,409,253 | 7,874 | 3 Tutor System
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# Q.10 From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take $$\pi = \frac{22}{7}$$ )
Given: Radius of the circular sheet = 14 cm
$$\therefore Area = \pi r^2 =\frac{22}{7} × 14 × 14 cm^2 = 616 cm^2$$
Area of 2 small circles $$= 2 × \pi × r^2$$
$$= 2 × \frac{22}{7} × 3.5 × 3.5 cm^2$$ = 77.0 $$cm^2$$
Area of the rectangle = l × b
=$$3 × 1 cm^2 = 3 cm^2$$
Area of the remaining sheet after removing the 2 circles and 1 rectangle
= $$616 cm^2 – (77 + 3) cm^2$$
= $$616 cm^2 – 80 cm^2 = 536 cm^2$$ | 268 | 699 | {"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} | 4.5 | 4 | CC-MAIN-2023-06 | longest | en | 0.796072 |
http://de.metamath.org/mpeuni/cvmlift2.html | 1,718,899,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00307.warc.gz | 8,364,447 | 9,265 | Mathbox for Mario Carneiro < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cvmlift2 Structured version Visualization version GIF version
Theorem cvmlift2 30552
Description: A two-dimensional version of cvmlift 30535. There is a unique lift of functions on the unit square II ×t II which commutes with the covering map. (Contributed by Mario Carneiro, 1-Jun-2015.)
Hypotheses
Ref Expression
cvmlift2.b 𝐵 = 𝐶
cvmlift2.f (𝜑𝐹 ∈ (𝐶 CovMap 𝐽))
cvmlift2.g (𝜑𝐺 ∈ ((II ×t II) Cn 𝐽))
cvmlift2.p (𝜑𝑃𝐵)
cvmlift2.i (𝜑 → (𝐹𝑃) = (0𝐺0))
Assertion
Ref Expression
cvmlift2 (𝜑 → ∃!𝑓 ∈ ((II ×t II) Cn 𝐶)((𝐹𝑓) = 𝐺 ∧ (0𝑓0) = 𝑃))
Distinct variable groups: 𝑓,𝐹 𝜑,𝑓 𝑓,𝐽 𝑓,𝐺 𝐶,𝑓 𝑃,𝑓
Allowed substitution hint: 𝐵(𝑓)
Proof of Theorem cvmlift2
Dummy variables 𝑔 𝑘 𝑢 𝑣 𝑤 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 cvmlift2.b . 2 𝐵 = 𝐶
2 cvmlift2.f . 2 (𝜑𝐹 ∈ (𝐶 CovMap 𝐽))
3 cvmlift2.g . 2 (𝜑𝐺 ∈ ((II ×t II) Cn 𝐽))
4 cvmlift2.p . 2 (𝜑𝑃𝐵)
5 cvmlift2.i . 2 (𝜑 → (𝐹𝑃) = (0𝐺0))
6 coeq2 5202 . . . . 5 ( = 𝑔 → (𝐹) = (𝐹𝑔))
7 oveq1 6556 . . . . . . 7 (𝑤 = 𝑧 → (𝑤𝐺0) = (𝑧𝐺0))
87cbvmptv 4678 . . . . . 6 (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) = (𝑧 ∈ (0[,]1) ↦ (𝑧𝐺0))
98a1i 11 . . . . 5 ( = 𝑔 → (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) = (𝑧 ∈ (0[,]1) ↦ (𝑧𝐺0)))
106, 9eqeq12d 2625 . . . 4 ( = 𝑔 → ((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ↔ (𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑧𝐺0))))
11 fveq1 6102 . . . . 5 ( = 𝑔 → (‘0) = (𝑔‘0))
1211eqeq1d 2612 . . . 4 ( = 𝑔 → ((‘0) = 𝑃 ↔ (𝑔‘0) = 𝑃))
1310, 12anbi12d 743 . . 3 ( = 𝑔 → (((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃) ↔ ((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑧𝐺0)) ∧ (𝑔‘0) = 𝑃)))
1413cbvriotav 6522 . 2 ( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃)) = (𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑧𝐺0)) ∧ (𝑔‘0) = 𝑃))
15 coeq2 5202 . . . . . . . 8 (𝑘 = 𝑔 → (𝐹𝑘) = (𝐹𝑔))
16 oveq2 6557 . . . . . . . . . 10 (𝑤 = 𝑧 → (𝑢𝐺𝑤) = (𝑢𝐺𝑧))
1716cbvmptv 4678 . . . . . . . . 9 (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧))
1817a1i 11 . . . . . . . 8 (𝑘 = 𝑔 → (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)))
1915, 18eqeq12d 2625 . . . . . . 7 (𝑘 = 𝑔 → ((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ↔ (𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧))))
20 fveq1 6102 . . . . . . . 8 (𝑘 = 𝑔 → (𝑘‘0) = (𝑔‘0))
2120eqeq1d 2612 . . . . . . 7 (𝑘 = 𝑔 → ((𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢) ↔ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)))
2219, 21anbi12d 743 . . . . . 6 (𝑘 = 𝑔 → (((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ∧ (𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)) ↔ ((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢))))
2322cbvriotav 6522 . . . . 5 (𝑘 ∈ (II Cn 𝐶)((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ∧ (𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢))) = (𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)))
24 oveq1 6556 . . . . . . . . 9 (𝑢 = 𝑥 → (𝑢𝐺𝑧) = (𝑥𝐺𝑧))
2524mpteq2dv 4673 . . . . . . . 8 (𝑢 = 𝑥 → (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)))
2625eqeq2d 2620 . . . . . . 7 (𝑢 = 𝑥 → ((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) ↔ (𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧))))
27 fveq2 6103 . . . . . . . 8 (𝑢 = 𝑥 → (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥))
2827eqeq2d 2620 . . . . . . 7 (𝑢 = 𝑥 → ((𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢) ↔ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥)))
2926, 28anbi12d 743 . . . . . 6 (𝑢 = 𝑥 → (((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)) ↔ ((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥))))
3029riotabidv 6513 . . . . 5 (𝑢 = 𝑥 → (𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑢𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢))) = (𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥))))
3123, 30syl5eq 2656 . . . 4 (𝑢 = 𝑥 → (𝑘 ∈ (II Cn 𝐶)((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ∧ (𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢))) = (𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥))))
3231fveq1d 6105 . . 3 (𝑢 = 𝑥 → ((𝑘 ∈ (II Cn 𝐶)((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ∧ (𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)))‘𝑣) = ((𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥)))‘𝑣))
33 fveq2 6103 . . 3 (𝑣 = 𝑦 → ((𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥)))‘𝑣) = ((𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥)))‘𝑦))
3432, 33cbvmpt2v 6633 . 2 (𝑢 ∈ (0[,]1), 𝑣 ∈ (0[,]1) ↦ ((𝑘 ∈ (II Cn 𝐶)((𝐹𝑘) = (𝑤 ∈ (0[,]1) ↦ (𝑢𝐺𝑤)) ∧ (𝑘‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑢)))‘𝑣)) = (𝑥 ∈ (0[,]1), 𝑦 ∈ (0[,]1) ↦ ((𝑔 ∈ (II Cn 𝐶)((𝐹𝑔) = (𝑧 ∈ (0[,]1) ↦ (𝑥𝐺𝑧)) ∧ (𝑔‘0) = (( ∈ (II Cn 𝐶)((𝐹) = (𝑤 ∈ (0[,]1) ↦ (𝑤𝐺0)) ∧ (‘0) = 𝑃))‘𝑥)))‘𝑦))
351, 2, 3, 4, 5, 14, 34cvmlift2lem13 30551 1 (𝜑 → ∃!𝑓 ∈ ((II ×t II) Cn 𝐶)((𝐹𝑓) = 𝐺 ∧ (0𝑓0) = 𝑃))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∃!wreu 2898 ∪ cuni 4372 ↦ cmpt 4643 ∘ ccom 5042 ‘cfv 5804 ℩crio 6510 (class class class)co 6549 ↦ cmpt2 6551 0cc0 9815 1c1 9816 [,]cicc 12049 Cn ccn 20838 ×t ctx 21173 IIcii 22486 CovMap ccvm 30491 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-inf2 8421 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 ax-pre-sup 9893 ax-addf 9894 ax-mulf 9895 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-int 4411 df-iun 4457 df-iin 4458 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-se 4998 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-isom 5813 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-om 6958 df-1st 7059 df-2nd 7060 df-supp 7183 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-2o 7448 df-oadd 7451 df-er 7629 df-ec 7631 df-map 7746 df-ixp 7795 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-fsupp 8159 df-fi 8200 df-sup 8231 df-inf 8232 df-oi 8298 df-card 8648 df-cda 8873 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-4 10958 df-5 10959 df-6 10960 df-7 10961 df-8 10962 df-9 10963 df-n0 11170 df-z 11255 df-dec 11370 df-uz 11564 df-q 11665 df-rp 11709 df-xneg 11822 df-xadd 11823 df-xmul 11824 df-ioo 12050 df-ico 12052 df-icc 12053 df-fz 12198 df-fzo 12335 df-fl 12455 df-seq 12664 df-exp 12723 df-hash 12980 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-clim 14067 df-sum 14265 df-struct 15697 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-mulr 15782 df-starv 15783 df-sca 15784 df-vsca 15785 df-ip 15786 df-tset 15787 df-ple 15788 df-ds 15791 df-unif 15792 df-hom 15793 df-cco 15794 df-rest 15906 df-topn 15907 df-0g 15925 df-gsum 15926 df-topgen 15927 df-pt 15928 df-prds 15931 df-xrs 15985 df-qtop 15990 df-imas 15991 df-xps 15993 df-mre 16069 df-mrc 16070 df-acs 16072 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-submnd 17159 df-mulg 17364 df-cntz 17573 df-cmn 18018 df-psmet 19559 df-xmet 19560 df-met 19561 df-bl 19562 df-mopn 19563 df-cnfld 19568 df-top 20521 df-bases 20522 df-topon 20523 df-topsp 20524 df-cld 20633 df-ntr 20634 df-cls 20635 df-nei 20712 df-cn 20841 df-cnp 20842 df-cmp 21000 df-con 21025 df-lly 21079 df-nlly 21080 df-tx 21175 df-hmeo 21368 df-xms 21935 df-ms 21936 df-tms 21937 df-ii 22488 df-htpy 22577 df-phtpy 22578 df-phtpc 22599 df-pcon 30457 df-scon 30458 df-cvm 30492 This theorem is referenced by: cvmliftpht 30554
Copyright terms: Public domain W3C validator | 6,378 | 9,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.315795 |
www.ihmsnh.org | 1,638,211,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358786.67/warc/CC-MAIN-20211129164711-20211129194711-00539.warc.gz | 897,063,689 | 175,375 | # Mathematics & Christian Education
Nothing could be more distinctive of the age in which we live than the overpowering prominence of mathematics. All through the Catholic centuries, arithmetic and geometry constituted all the mathematics that an educated Christian was asked to learn. Even these two subjects were treated from a more contemplative point of view, which made them far more harmonious with other liberal studies. Arithmetic consisted in the study of the properties of numbers; geometry in the study of shapes and figures. When not overdone, and when counterbalanced by the proper correctives from the other types of knowledge, geometry and arithmetic, as they used to be taught, cultivated a few desirable virtues of the mind like clarity and precision, and sharpened the mind for the perception of harmony, rhythm, and pattern in the study of nature and of Holy Scripture. But even then, many saints and sages warned against the excessive preoccupation with such studies, and especially against the seductive clarity of mathematics; for it is not enough for the mind to be accurate and clear; we are bound to ask “accurate and clear about what?” Since in mathematics accuracy and clarity are achieved at the price of the reality and the goodness of the object, it is a danger of the mathematical mind to continue to sacrifice reality and goodness for the sake of clarity in every other field in which man must seek and find the truth.
But in our time, education is overwhelmed by mathematics and on more than one score. For, while a contemplative interest in the properties of shapes and numbers is almost completely extinct, an illiberal and utterly inhuman form of mathematics dominates the years of learning of our boys and girls, almost completely from the very first year of the primary school to the very last year of college. In place of arithmetic and geometry, whose relation to reality is definite and understandable, there is now an indefinite confusion of branches which go by the name of mathematics, the nature of whose objects nobody understands! Such topics as topology, non-Eudidean geometry, Boolean algebra, transfinite numbers, projective geometry; not to speak of other more recognizable subjects like algebra, trigonometry, integral calculus, vector analysis and the theory of equations. These new subjects are not only more confusing but much more difficult to acquire, and therefore much less likely to leave the mind at leisure for other liberal studies. But the predominance of mathematics today is not restricted to those courses which go by its name, because mathematics, in some form or other, in matter or in method, has crept into every other corner of the curriculum. According to the modern positivistic conception, mathematics and not wisdom is considered as the prototype of science. In subjects ranging from physics to education, covering every field of human learning, there is an evident tendency to assimilate all knowledge to mathematical knowledge and to resolve all realities into mathematical formulas. This trend reaches its apex in the development of symbolic logic, in which guise mathematics invades even the field of philosophy, to distort all the basic conceptions of the mind, and to deflect all the activities of thought from attaining their fulfillment in true wisdom which consists in knowledge about God, by keeping them whirling endlessly around the nihilistic circle of sheer mathematical emptiness.
Now in an attempt to determine the influence of mathematics on the mind of a Christian, it would be folly to ignore the fact that after twenty centuries of Christian living, it is impossible to name one single patron saint for mathematics. There are Catholics indeed who occupied themselves considerably with mathematics and as far as we know kept the faith; but I know of no mathematician whose faith burned so brilliantly as to earn him a place among the stars of sanctity. Nor is this a mere coincidence, for any one of us can look into his own mind to find that there is no other kind of human knowledge or human experience which offers less in terms of value for the Christian message than mathematics. Almost all that one needs in the way of mathematics in order to learn all of Holy Scripture and all the Doctors of the Church, does not exceed the ability to count up to a thousand and to distinguish between a vertical and a horizontal line. Whatever it is you talk about in mathematics, it is never anything you can carry over to your meditations, or employ in your prayers; it gives you no courage in your moments of despair, and no consolation in your loneliness.
In the field of philosophy, mathematics has always been fertile grounds for sophistry. There is hardly any other intellectual interest which has contributed more to confuse men about fundamental truths regarding God, man, and the universe, than mathematics. Just to mention the names of Thales, Pythagoras, Plato, Descartes, Spinoza, Whitehead and Russell, would suffice to convince one even slightly acquainted with the history of thought about the great number of minds that were deceived by the mirage of mathematics, and misled to accept fraudulent substitutes for the saving truth. I believe that an unprejudiced consideration of the nature of mathematics and of the nature of its objects would reveal clearly that all these charges leveled against the mathematical mind are rooted in the very nature and essence of things.
But what kind of a science is mathematics? Is it a practical science which envisages the achievement of a good, or a speculative science which envisages the attainment of truth? A practical science, like medicine or ethics, would be eliminated by the elimination of the corresponding good. For example, if men were indifferent to health and its opposite there would be no criterion for distinguishing between a right prescription and a wrong one, and consequently, medicine would cease to be a science. In a similar way, if men per absurdum were suddenly to become neutral to the attainment of happiness or its opposite, that would be the end of ethics. But what good, if ceasing, would determine the end of mathematics? None whatever, for the simple reason that mathematics prescinds from all good and all value. Mathematics talks the language of a speculative science. It utters propositions which must be either true or false. Now a proposition is true or false depending on whether it is or is not in conformity with reality. Just as a practical science envisages a good to be achieved, which good functions as the criterion for right and wrong precepts in that science, so a speculative science considers some part or aspect of reality, which stands as the measure of truth and falsehood in that science. If there were no stars there would be no astronomy; and theology would be sheer nonsense if God did not exist. But what part of reality would destroy mathematics by being eliminated? What does the mathematician talk about? Is the object of mathematics a creature or a creator? Is it a substance or an accident? Is it something actual or merely potential? Is it changing or changeless? Temporal or eternal? Material or spiritual? Tangible or intangible? If one were to compose an inventory of all the subsisting realities of the whole universe, including God, the angels, men, animals, plants and minerals, would the objects of mathematics be on this list?
Am I asking too many questions? Well, here are a few answers whose reasons will either be supplied later, or be left to the reader to discover for himself. Mathematics is a speculative science whose value can only be in the practical order. It has no speculative value, because it does not convey any essential knowledge about any subsisting reality. It is not contemplative knowledge and therefore not essentially good for man, because it occupies the intellect with objects which the will cannot love. It is knowledge which does not proceed from understanding nor does it resolve in wisdom. It does not proceed from understanding, because the mathematical expression of any reality, never conveys any understanding of it. It may however convey the means for the control of that reality. You are not one inch closer to the penetration of the mystery of light and color when you know the number of Angstroms in each of the colors of the spectrum; nor about the nature, cause, or purpose of gravity when you resolve its laws into mathematical formulas. And it does not resolve in wisdom, because neither is mathematics concerned with the First Cause, nor does it lead to the First Cause. The manner by which mathematics deals with its objects abstracts completely from any dependence upon God, and as a matter of fact, attributes to these objects a species of eternity and turns them into quasi divinities completely independent in themselves. This explains the autonomous nature of mathematics, according to which, left to itself, it never leads to anything non-mathematical. A mathematician might be led to think about God by an accidental non-mathematical reason, but never from the very needs of mathematics.
As for the object of mathematics, it is not a physical entity but a mental entity; it is not real but ideal. There is nowhere in the world, outside of the mind of a mathematician, a point without dimensions, a line without width or thickness, or a square root of minus one. But these fictions of the mind are founded on reality, and their foundation consists of the accident of quantity and its properties and relations. Arithmetic is founded on discontinuous quantities or multitudes; geometry on continuous quantities or magnitudes; while algebra is founded on abstract quantity considered generically, prescinding from whether it is number or magnitude and therefore potentially capable both of an arithmetical as well as of a geometrical interpretation. Other mathematical objects, more distantly removed from this real foundation of mathematics, are rooted in these simpler elements and in the relations which hold among them. Having experienced the three dimensions of bodies in space and having represented these three dimensions by the three variables of an algebraical equation, nothing prevents the mind from creating the fiction of a space corresponding to an algebraical equation of four variables – hence four-dimensional space.
But what do we know about this accident of quantity, on which is founded, proximately or remotely every object of mathematics? We learn from philosophy that quantity is an accident of material substances, and that in contrast with the accident of quality, quantity manifests the material and not the formal aspect of these substances. Therefore the real foundation of mathematics is found in the material aspect of material things. Further, an accident when conceived as an accident always brings you back to its substance; but in mathematics the accident of quantity is conceived as if it were a substance. Further, a material substance concretely considered, has a nature through which this substance moves to the attainment of an end, but the mathematician considers quantity as a substantialized material accident devoid of any principle of change and abstracted from any movement to attain an end. The concrete material substance manifests itself through its sensible qualities by means of which it is known, but the object of mathematics, without being a spiritual substance like an angel, prescinds from all sensible qualities and can be known only by the intellect and not by the senses. Hence we have the apparent paradox that while the only foundation for the mathematical object is the material aspect of material things, still mathematics represents its object such as matter could neither be nor be known. For matter is nothing but a principle of change, while mathematics prescinds from change; and matter can only be known through the senses while mathematics prescinds from sensibility.
The object of mathematics is therefore an accident parading as a substance, a material reality pretending to be immaterial, an ideal entity which poses for something real. At the basis of all these antinomies is the fact that mathematics arises only when an intellectual mind, directs the light of its spiritual intelligence, not for the purpose of contemplating being, but for the purpose of controlling potency. The mathematical object is the shadow that matter casts on spirit. For when spirit knows spirit, there is not even the foundation for mathematics; when material cognition (sensation) knows material things, the objects of mathematics cannot arise; even when a spiritual being knows matter contemplatively it understands a material substance through its form and its qualities. It is only when a spiritual being concerns itself with matter and for the purpose of sheer control that mathematics finally finds its grounds.
But how about the truth in mathematics? If the objects of mathematics are mental entities (entia rationis) what is it that determines the truth or falsehood of a mathematical proposition? What reality stands as the measure to the judgment of the mind? In the classical branches, arithmetic and geometry, the foundation in reality was close enough to preclude any statements that are not justified by the real properties of multitudes and magnitudes. But as mathematics branches out and develops into newer mathematics, and higher mathematics, and purer mathematics, that control becomes less and less until finally the mind remains its own measure. Consistency and not conformity becomes the touchstone of validity.
Apart from mathematics, there used to be three other distinct types of knowledge: physical, logical, and ethical. All three led ultimately to God – the physical sciences under the aspect of Ultimate Cause; the logical sciences by way of the Prime Truth; and the ethical sciences by way of the Supreme Good. But in mathematics, the mind reigns supreme, lord of all it surveys. The mind finds in itself a sufficient cause for the kind of being the mathematical entity enjoys. It is the only ultimate measure for the truth of its judgments. It prescinds completely from the aspect of goodness. Of all the intellectual pursuits, mathematics alone does not lead to God.
It is like the web of a spider, it proceeds from the very substance of the spider and ends up being its own jail. It gets more involved and more intricate the more it is extended, and finally, when the web is intricate enough, the new threads do not have to measure up to any real independent distances of walls or furniture, for when the new-thrown thread fails to meet a point of support, it sticks on another thread of the same fabric.
From the spider of mathematics, may God deliver us | 2,815 | 14,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-49 | longest | en | 0.961733 |
http://mathhelpforum.com/advanced-statistics/35088-solved-normal-distrubtion.html | 1,524,807,826,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00447.warc.gz | 199,617,399 | 9,778 | 1. ## [SOLVED] Normal Distrubtion...
Question:
The fluorescent light tubes made by the company Well-lit have lifetimes which are normally distrusted with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life. If the company wishes to have to replace free only 3% if the tubes sold, find the guaranteed minimum it must set.
Attempt:
$\displaystyle \mu = 2010$ , $\displaystyle \sigma = 20$
How can I get the value of $\displaystyle \phi^{-1} (0.3\%)$ out of the table?
2. Hi
Let's call $\displaystyle T$ the random variable associated to the lifetime.
We're looking for $\displaystyle t=\phi^{-1}(3\%)$ which means $\displaystyle P(T\leq t)=\int_{-\infty}^t\frac{1}{\sigma\sqrt{2\pi}}\exp \frac{(x-\mu)^2}{\sigma^2}\,\mathrm{d}x=0.997$
As we know the value taken by the normal distribution for $\displaystyle \sigma=1$ and $\displaystyle \mu=0$, we substitute $\displaystyle u=\frac{x-\mu}{\sigma} \Rightarrow \mathrm{d}u=\frac{\mathrm{d}x}{\sigma}$ :
$\displaystyle P(T\leq t)=\int_{-\infty}^{\frac{t-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp u^2\,\mathrm{d}u=0.997$
Using the table, you can find the value of $\displaystyle \frac{t-\mu}{\sigma}$ corresponding and then deduce $\displaystyle \phi^{-1}(3\%)=t$. | 403 | 1,296 | {"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} | 3.859375 | 4 | CC-MAIN-2018-17 | latest | en | 0.740627 |
https://www.homeworkmarket.com/content/devry-phil447-week-2-quiz | 1,566,761,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00376.warc.gz | 856,353,200 | 38,016 | # DEVRY PHIL447 WEEK 2 QUIZ
Question 1.1.(TCOs 1 & 3) An argument is sound when(Points : 4)
it is true beyond a reasonable doubt.
there is enough support for the argument
the premise of an argument is doubtful.
the premise of a valid argument is in fact true.
Question 2.2.(TCOs 1 & 2) The word Since is a(Points : 4)
conclusion indicator.
premise indicator.
probability indicator.
deduction indicator.
Question 3.3.(TCOs 1 & 2) The premises of good inductive arguments(Points : 4)
are true beyond any possible doubt.
are true beyond a reasonable doubt.
support their conclusions.
neither answers nor support their conclusions.
Question 4.4.(TCOs 1 & 3) How would you rewrite the following claim to remedy problems of ambiguity?
Do not assume that common sense by itself solves the problem. Volunteer help requested: Come prepared to lift heavy equipment with construction helmet and work overalls.(Points : 4)
Volunteer help requested: Use heavy equipment to lift construction helmets and work overalls.
Volunteer help requested: Lift heavy equipment with your construction helmet and work overalls.
Volunteer help requested: Use a construction helmet and work overalls to lift heavy equipment.
Volunteer help requested: Wear construction helmet and work overalls, and be prepared to lift heavy equipment
Question 5.5.(TCOs 1 & 3) A precising definition(Points : 4)
takes on a special meaning in a given context.
reduces vagueness or generality.
tell us what the word ordinarily means.
• Posted: 4 years ago
DEVRY PHIL447 WEEK 2 QUIZ
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Our teachers already did such homework, use it as a reference! | 397 | 1,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.908809 |
https://socratic.org/questions/consider-a-particle-initially-moving-with-a-velocity-of-5m-s-starts-deaccelarati | 1,566,504,953,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00027.warc.gz | 647,955,391 | 5,936 | # Consider a particle initially moving with a velocity of 5m/s starts deaccelarating at a constant rate of 2m/s*2 find the distance travelled in 2sec? Ans:2m
Oct 7, 2017
See the approach below...
#### Explanation:
We simply know the formula that
$S = u t \pm \frac{1}{2} a {t}^{2}$ [Where S$\to$distance travelled, u$\to$initial velocity, t$\to$time taken, -a$\to$retardation...,+a$\to$acceleration...
Hence we can use the formula to determine the Distance...
$S = 5 \times 2 - \frac{1}{2} \times 2 \times {2}^{2}$
$S = 6 m$
Hope it helps...
Thank you... | 183 | 560 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 3.859375 | 4 | CC-MAIN-2019-35 | latest | en | 0.717684 |
http://codeforces.com/blog/Hemanth_1 | 1,624,620,032,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00524.warc.gz | 11,900,387 | 14,329 | ### Hemanth_1's blog
By Hemanth_1, history, 2 years ago,
I was trying to solve this problem, find the n'th term of the following recurrence:
F(n) = a*F(n-1) + b*F(n-2) + c*F(n-3) + n*n*(n+1)
a,b,c and the first 3 numbers in the sequence are given as the input, n can be upto 10^18 . I know that problems involving linear recurrences can be solved using matrix exponentiation, even when there are multiple sequences involved. So, in this case, if I could somehow express n*n*(n+1) as a linear recurrence as well, solving this problem becomes simple. However, I'm not able to do that, can someone help me with this ? Also, if a method to find a recurrence relation, given a closed form, exists, please mention that as well. Thanks in advance. | 205 | 744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-25 | latest | en | 0.956497 |
https://www.flashcardmachine.com/chapter-8physics.html | 1,575,748,207,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540501887.27/warc/CC-MAIN-20191207183439-20191207211439-00252.warc.gz | 726,699,824 | 7,321 | # Shared Flashcard Set
## Details
chapter 8 physics
study
27
Physics
12/04/2012
## Additional Physics Flashcards
Term
horses that move with the fastest linear speed on a merry go round are located where?
Definition
near the outside
Term
your pet hamster sits on a record player whose angular speed is constant. if he moves to a point twice as far from the center, then his linear speed...?
Definition
doubles
Term
which moves faster in m/s on a merry go round; a horse on the inside or horse on the outside?
Definition
outside horse
Term
if a turntable's rotational speed is doubled, then the linear speed of a pet hamster sitting on the edge of the record will..?
Definition
double
Term
an industrial flywheel has a greater rotational inertia when most of its mass is...
Definition
near the rim
Term
a coin and a ring roll down an incline starting at the same time. the one to reach the bottom first will be the
Definition
coin
Term
a torque acting on an object tends to produce...?
Definition
rotation
Term
on a balanced seesaw, a boy three times as heavy as his partner sits..?
Definition
1/3 the distance from the fulcrum
Term
put a pipe over the end of a wrench when trying to turn a stubborn nut on a bolt, to effectively make the wrench handle twice as long, youll multiply the torque by...?
Definition
two
Term
toss a baseball bat into the air and it wobbles about its
Definition
center of mass.
Term
the long, heavy tail of a spider monkey enables the monkey to easily vary its..?
Definition
center of gravity
Term
the center of mass of a human body is located at a point....?
Definition
that changes as a person bends over.
Term
the famous leaning tower of pisa doesnt topple over because its center of gravity is...?
Definition
above a place of support
Term
the chef at the infamous fattening tower of pizza tosses a spinning disk of uncooked pizza dough into the air. the disk's diameter increases during the flight, while its rotational speed...?
Definition
decreases
Term
centrifugal forces are an apparent reality to observers in a reference frame that is...?
Definition
rotating
Term
when a twirling ice skater brings her arm inward, her rotational speed..?
Definition
increases
Term
consider a rotating donut-shaped space habitat where living quarters are on the inside surface farthest from the axis. if the rotational speed of the habitat increases, the apparent weight of people inside...?
Definition
increases
Term
if the earth had two identical moons in one circular orbit, an the moons were as far apart in that orbit as they could be, the center of gravity of the erth-moons sytem would be..?
Definition
outside the earth, but within the orbital path of moons
Term
when doing somersaults, youll more easily rotate when your body is...
Definition
balled up.
Term
to turn a stubborn screw, it is best to use a screwdriver that has a..?
Definition
wide handle
Term
two people are balanced on a seesaw. if one person leans toward the center of the seesaw, that persons end of the seesaw will...?
Definition
rise
Term
to kick a football si it wont topple end over end, kick it so the force of impact extends...
Definition
through its center of gravity
Term
a car travels in a circle with constant speed the net force on the car is...?
Definition
directed toward the center of the curve
Term
a huge rotating cloud of particles in space gravitate together to form an increasingly dense ball. as it shrinks n size, the cloud..
Definition
rotates faster.
Term
if the polar icecaps melted, the resulting water would spread over the entire earth, this new mass distribution would tend to make the length of a day...
Definition
longer
Term
suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. as you crawl toward the edge, the angular momentum of you and the turntable...?
Definition
remains the same but the RMP's decrease.
Term
stand a meterstick on its end and let go and it rotates to the floor. if you attack a heavy weight to its upper end and repeat, falling time ill be...?
Definition
more
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https://sportsbizusa.com/20-graphing-complex-numbers-worksheet/ | 1,675,934,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00552.warc.gz | 551,799,807 | 14,243 | # 20 Graphing Complex Numbers Worksheet
Properties of plex Numbers pdf Kuta Software algebra 2 graphing plex numbers worksheet, graphing plex numbers worksheet doc, via: yumpu.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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Sec 3 plex Numbers PowerPoint Presentation Free line via : smackslide.com | 1,219 | 6,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-06 | latest | en | 0.956999 |
https://forum.kde.org/viewtopic.php?f=74&t=110426&p=261186 | 1,443,956,756,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736673439.5/warc/CC-MAIN-20151001215753-00103-ip-10-137-6-227.ec2.internal.warc.gz | 991,240,178 | 7,378 | ## First time using Eigen, 2 questions
hasuchobe
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### First time using Eigen, 2 questions
Sat Mar 16, 2013 8:17 pm
For sparse matrices... what is meant by outerstarts? I don't understand what this page is talking about.
http://eigen.tuxfamily.org/dox/Tutorial ... e_BasicOps
Second, does eigen support complex matrices? Is there an example on how to do this?
edit: Turns out all you need to do for complex operations is specify complex type from the standard library for the template.
thnx
ggael
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### Re: First time using Eigen, 2 questions
Sat Mar 16, 2013 9:08 pm
For a column major matrix, the nonzeros are sequentially stored as pairs (rowId,value). The outerstarts[j] tells where is the first non-zero for the j-th column. I don't know how to explain it differently. If that's still unclear you might look at this page http://netlib.org/utk/papers/templates/node92.html, where outerstarts==col_ptr.
hasuchobe
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### Re: First time using Eigen, 2 questions
Sat Mar 16, 2013 9:25 pm
ggael wrote:For a column major matrix, the nonzeros are sequentially stored as pairs (rowId,value). The outerstarts[j] tells where is the first non-zero for the j-th column. I don't know how to explain it differently. If that's still unclear you might look at this page http://netlib.org/utk/papers/templates/node92.html, where outerstarts==col_ptr.
That page you linked clarified it. Thanks! Eigen is such a nice library
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