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https://eng.kakprosto.ru/how-17042-how-to-place-coefficients-in-chemistry
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Instruction
1
Before proceeding to the task, you need to understand that the figure that is put in front of a chemical element or formula is called a coefficient. And the figure standing after (and beneath) means index. In addition, you need to know that:
• ratio refers to all the chemical symbols standing after it in the formula
• coefficient is multiplied by the index (not sum!)
• the number of atoms of each element reacting substances must be the same number of atoms of these elements included in the composition of the reaction products.
For example, the formula 2H2SO4 means 4 atoms of H (hydrogen) 2 atoms of S (sulfur) and 8 atoms of O (oxygen).
2
1. Example No. 1. Consider the equation for the combustion of ethylene.
The combustion of organic substances are formed by carbon oxide (IV) (carbon dioxide) and water. Consistently try to place coefficients.
C2H4 + O2 => CO2+ H2O
Begin to analyze. In response to stepped 2 atoms of C (carbon), but it turned out only 1 atom, then put 2 CO2. Now their number is the same.
C2H4 + O2 => 2CO2+ H2O
Now look at H (hydrogen). The response entered the 4 atoms of hydrogen, and was the result of only 2 atoms, therefore H2O (water) put 2 – now turned also 4
C2H4 + O2 => 2CO2+ 2H2O
Consider all the atoms of O (oxygen), formed in the reaction (i.e., after the equal sign). 4 atoms in 2CO2 and 2 atoms in 2H2O – a total of 6 atoms. While reactions of 2 atoms, then a molecule of oxygen O2 set 3, which means that they became too 6.
C2H4 + 3O2 => 2CO2+ 2H2O
Thus, it turned out the same number of atoms of each element before and after the equal sign.
C2H4 + 3O2 => 2CO2+ 2H2O
3
2. Example No. 2. Consider the reaction of interaction of aluminum with dilute sulfuric acid.
Al + H2SO4 => Al2 (SO4) 3 + H2
Look at the S atoms included in the composition Al2 (SO4) 3 - their 3, and H2SO4 (sulfuric acid) only 1, therefore, before the sulfuric acid is also put 3.
Al + 3H2SO4 => Al2 (SO4) 3 + H2
But now turned to the reaction of 6 atoms of H (hydrogen), and after the reaction is only 2, so, before the molecule H2 (hydrogen) set is also 3, so overall, it was a 6.
Al + 3H2SO4 => Al2 (SO4) 3 + 3H2
Last look at the aluminum. As in Al2 (SO4) 3 (aluminium sulphate) only 2 of the aluminum atom, and to the reaction prior to Al (aluminum) put 2.
2Al + 3H2SO4 => Al2 (SO4) 3 + 3H2
Now the number of all atoms before and after reaction is the same. It turned out that arrange the coefficients in chemical equations is not so difficult. Enough to practice and succeed.
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# A 0.450 kg ice puck, moving east with a speed of 3.00 \frac{m}{s} has a head in collision with a 0.900 kg puck, initially at rest. Assuming a perfectly elastic collision what will be the speed and direction of each object after the collision?
A 0.450 kg ice puck, moving east with a speed of $3.00\frac{m}{s}$ has a head in collision with a 0.900 kg puck, initially at rest. Assuming a perfectly elastic collision what will be the speed and direction of each object after the collision?
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curwyrm
The final speeds of the two pucks are given by:
$v=\left[\frac{\left(M-m\right)}{\left(M+m\right)}\right]$
where v is the initial speed of the smaller puck
$V=\left[\frac{2m}{M+m}\right]$
The smaller puck will rebound and travel in the opposite direction of its original velocity at speed:
and the larger puck will continue in the same direction as the original velocity of the small puck:
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# Two equilibrium problems
1. May 1, 2005
### poodlefarm
1) I just want to check if this one is correct
reaction => CCl4(g) <==> C(s) = 2Cl2(g)
Kp= .76
find the initial partial pressure of CCl4 that will produce an equilibrium total pressure of 1.20 atm
p.total = p.CCl4 + p.Cl2
CCl4 <=> 2Cl2
i. press x 2x
change 1.20 - x 1.20 - 2x
kp = .76 = (p.Cl2^2)/(p.CCL4) = (1.2-2x)^2/(1.2-x)
x= .8557 or x= .1543
.8557 will not work so take x = .1543
p.CCl4 = 1.20- .1543 = 1.05 atm
the rub is that if I use x to find p.Cl2 I get .892 atm
1.05 + .892 =/= 1.2 ?
Q2.
Ok
constant temp =25 C constant volume
reaction; Nh4HS (g)<==> NH3(g) + H2S(g)
step 1; some NH4HS decomposed in an evacuated container to give a total pressure at equililbrium of .659 atm
step 2; extra NH3 is added. re-established equilibrium gives a partial pressure for NH3 of .750 atm
find Kp
I'm stuck on this one.
.
2. May 1, 2005
### Gokul43201
Staff Emeritus
Q1 ) CCl4(g) <==> C(s) + 2Cl2(g)
Let the dissociation constant of CCl4 be x. If you start with 1 mole of CCl4, then in equilibrium, how many moles of CCl4, Cl2 will you have ? Remember, CCl4 is being consumed and Cl2 is being produced. Now use the equation for Kp to arrive at a quadratic in x which you can solve to find x.
Next, if the container had some volume V, and there are n moles of CCl4 to start, what will the pressure be at some temperature T ? Write a similar equation for the equilibrium number of moles and call this pressure Po. From these two, you can eliminate V/T and find the value of Po.
3. May 1, 2005
### Gokul43201
Staff Emeritus
let the initial number of moles of NH4HS be n, and let the dissociation constant be x. Now proceed from there ...
4. May 1, 2005
### poodlefarm
Q1)
I'm a little lost.
assuming 1 mole of CCl4
initial ; CCl4 = 1 Cl2 = 0
concentration CCl4 = -x cl2 = +2x
equilibrium CCl4 1-x Cl2 2x
Kp = .76 = 4x^2/(1-x)
x = .35
not sure if I'm doing this right.
but here is where I really get lost.
I think you are refering to PV =nRT
P= 1.2 atm
so 1.2 = (nRT)/V
and Po = (.65RT)/V // .65= 1-x
I still don't see it
5. May 2, 2005
### Gokul43201
Staff Emeritus
So far, so good !
Umm. Let's try again.
Initially, you have n moles at a pressure Po. Finally, you have n(1 - x + 2x) = n(1+x) moles at a pressure of 1.2 atm, where the value of x is known from above. Make sure you understand where the n(1+x) comes from.
Initial : Po = nRT/V
Final :1.2 = n(1.35)RT/V
From this, find Po (as well as the final partial pressures to make sure they add to give 1.2 atm)
6. May 2, 2005
### GCT
The second one's a #\$%^&. If the initial pressure was known, you could calculate Kp simply by using the percent dissociation, no need for the second aspect of the information given. It'll need to be worked out mathematically, if a solution actually exist to the exact form of this problem. I'm able to deduce several meaningful equations so far, and the best one has been
$$P_{total,eq2}=.75atm + P_{0}$$ where $$P_{0}$$ is the initial pressure at the very beginning of the situation. If $$.75atm$$ were valid as the net change in pressure, than one could calculate the Kp by finding the moles of each compound in undergoing the reaction.
Nevertheless I'll wait for both of you to get to this part before proceeding.
7. May 2, 2005
### GCT
I was hoping for a more creative way to solve the second question, yet the only sure way to solve it seems to involve a messy equation at the end.
$$P_{total,eq1}=.659=P_{0}+x,~x=P_{NH3}=P_{H2S}$$
$$Kp= \frac{[x][x]}{[P_{0}-x]}=\frac{[x][x]}{[.659atm-2x]}$$
also
$$Kp= \frac{[.750atm][x- \Delta y]}{.659atm-2x+ \Delta y]}=$$
set the two equal to each other and solve for y in terms of x, plug back in the second Kp...set this Kp equal to the first Kp and solve for x.
perhaps a better way will involve the incorporating the temperature
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Divide Two Binary Numbers Online
Binary Calculator
Use the following calculators to perform the addition, subtraction, multiplication, or division of two binary values, as well as convert binary values to decimal values, and vice versa.
Convert Decimal Value to Binary Value
The binary system is a numerical system that functions virtually identically to the decimal number system that people are likely more familiar with.
While the decimal number system uses the number 10 as its base, the binary system uses 2.
Furthermore, although the decimal system uses the digits 0 through 9, the binary system uses only 0 and 1, and each digit is referred to as a bit. Apart from these differences, operations such as addition, subtraction, multiplication, and division are all computed following the same rules as the decimal system.
Almost all modern technology and computers use the binary system due to its ease of implementation in digital circuitry using logic gates. It is much simpler to design hardware that only needs to detect two states, on and off (or true/false, present/absent, etc.).
Using a decimal system would require hardware that can detect 10 states for the digits 0 through 9, and is more complicated.
Binary Division Logic
Below are some typical conversions between binary and decimal values:
Binary/Decimal Conversion
Decimal Binary 0 0 1 1 2 10 3 11 4 100 7 111 8 1000 10 1010 16 10000 20 10100
While working with binary may initially seem confusing, understanding that each binary place value represents 2n, just as each decimal place represents 10n, should help clarify.
Take the number 8 for example.
In the decimal number system, 8 is positioned in the first decimal place left of the decimal point, signifying the 100 place. Essentially this means:
8 × 100 = 8 × 1 = 8
Using the number 18 for comparison:
(1 × 101) + (8 × 100) = 10 + 8 = 18
In binary, 8 is represented as 1000.
Reading from right to left, the first 0 represents 20, the second 21, the third 22, and the fourth 23; just like the decimal system, except with a base of 2 rather than 10. Since 23 = 8, a 1 is entered in its position yielding 1000.
Using 18, or 10010 as an example:
18 = 16 + 2 = 24 + 21
10010 = (1 × 24) + (0 × 23) + (0 × 22) + (1 × 21) + (0 × 20) = 18
The step by step process to convert from the decimal to the binary system is:
1. Find the largest power of 2 that lies within the given number
2. Subtract that value from the given number
3. Find the largest power of 2 within the remainder found in step 2
4. Repeat until there is no remainder
5. Enter a 1 for each binary place value that was found, and a 0 for the rest
Using the target of 18 again as an example, below is another way to visualize this:
2n 24 23 22 21 20 Instances within 18 1 0 0 1 0 Target: 18 18 - 16 = 2 → 2 - 2 = 0
Converting from the binary to the decimal system is simpler.
Determine all of the place values where 1 occurs, and find the sum of the values.
EX: 10111 = (1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20) = 23
Hence: 16 + 4 + 2 + 1 = 23.
Binary addition follows the same rules as addition in the decimal system except that rather than carrying a 1 over when the values added equal 10, carry over occurs when the result of addition equals 2.
Refer to the example below for clarification.
Note that in the binary system:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0, carry over the 1, i.e. 10
EX:
10 11 11 10 1 + 1 0 1 1 1 = 1 0 0 1 0 0
The only real difference between binary and decimal addition is that the value 2 in the binary system is the equivalent of 10 in the decimal system.
Note that the superscripted 1's represent digits that are carried over.
Binary Division Calculator
A common mistake to watch out for when conducting binary addition is in the case where 1 + 1 = 0 also has a 1 carried over from the previous column to its right.
The value at the bottom should then be 1 from the carried over 1 rather than 0. This can be observed in the third column from the right in the above example.
Binary Subtraction
Similarly to binary addition, there is little difference between binary and decimal subtraction except those that arise from using only the digits 0 and 1.
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Borrowing occurs in any instance where the number that is subtracted is larger than the number it is being subtracted from. In binary subtraction, the only case where borrowing is necessary is when 1 is subtracted from 0.
When this occurs, the 0 in the borrowing column essentially becomes "2" (changing the 0-1 into 2-1 = 1) while reducing the 1 in the column being borrowed from by 1. If the following column is also 0, borrowing will have to occur from each subsequent column until a column with a value of 1 can be reduced to 0.
Convert Binary Value to Decimal Value
Refer to the example below for clarification.
Note that in the binary system:
0 - 0 = 0
0 - 1 = 1, borrow 1, resulting in -1 carried over
1 - 0 = 1
1 - 1 = 0
EX1:
-11 20 1 1 1 – 0 1 1 0 1 = 0 1 0 1 0
EX2:
-11 2-10 0 – 0 1 1 = 0 0 1
Note that the superscripts displayed are the changes that occur to each bit when borrowing.
The borrowing column essentially obtains 2 from borrowing, and the column that is borrowed from is reduced by 1.
Binary Multiplication
Binary multiplication is arguably simpler than its decimal counterpart.
Since the only values used are 0 and 1, the results that must be added are either the same as the first term, or 0. Note that in each subsequent row, placeholder 0's need to be added, and the value shifted to the left, just like in decimal multiplication. The complexity in binary multiplication arises from tedious binary addition dependent on how many bits are in each term.
Binary Numeral System
Refer to the example below for clarification.
Note that in the binary system:
0 × 0 = 0
0 × 1 = 0
1 × 0 = 0
1 × 1 = 1
EX:
1 0 1 1 1 × 1 1 1 0 1 1 1 + 1 0 1 1 1 0 = 1 0 0 0 1 0 1
As can be seen in the example above, the process of binary multiplication is the same as it is in decimal multiplication.
Note that the 0 placeholder is written in the second line.
Typically the 0 placeholder is not visually present in decimal multiplication. While the same can be done in this example (with the 0 placeholder being assumed rather than explicit), it is included in this example because the 0 is relevant for any binary addition / subtraction calculator, like the one provided on this page.
Without the 0 being shown, it would be possible to make the mistake of excluding the 0 when adding the binary values displayed above. Note again that in the binary system, any 0 to the right of a 1 is relevant, while any 0 to the left of the last 1 in the value is not.
EX:
1 0 1 0 1 1 0 0
= 0 0 1 0 1 0 1 1 0 0
≠ 1 0 1 0 1 1 0 0 0 0
Binary Division
The process of binary division is similar to long division in the decimal system.
How to construct the binary number system:
The dividend is still divided by the divisor in the same manner, with the only significant difference being the use of binary rather than decimal subtraction. Note that a good understanding of binary subtraction is important for conducting binary division.
Binary Division - How To Divide Two Binary Numbers
Refer to the example below, as well as to the binary subtraction section for clarification.
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This is an archived post. You won't be able to vote or comment.
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[–] 26 points27 points (34 children)
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I realize at 24 I would still get the same D- in physics as I did when I was 14.
[–] 26 points27 points (30 children)
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This is literally just plugging values into a simple equation.
[–] 27 points28 points (27 children)
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And remembering the equation. And a pencil.
[–] 6 points7 points (26 children)
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The equation is probably given to you.
[–] 26 points27 points (0 children)
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and my axe
[–] 5 points6 points (7 children)
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None of the physics teachers I've heard of give the equations because the equations are not given on the AP Physics Exam's multiple choice sections
[–] 0 points1 point (0 children)
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They do give it to you on written portions though, so he should have one at this stage in a test.
[–] -1 points0 points (5 children)
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I wouldn't know. I'm Canadian so I don't had anything to do with the AP system. But I am about to complete my bachelor's in engineering physics and the only equation I ever learned by heart was the Shrodinger's equation.
[–] 7 points8 points (2 children)
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so I don't had anything to do with
Spoken like a true engineer.
[–] 0 points1 point (1 child)
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English is my third language. I'm bound to make mistakes.
[–] 0 points1 point (0 children)
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'geers of every language are sloppy with grammar and spelling :)
[–] 0 points1 point (1 child)
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I live in Alberta and we had the AP system. Where did you go to high school that didn't?
[–] 0 points1 point (0 children)
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Québec.
[–] 1 point2 points (15 children)
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One of the best things my second college physics class did was not give us an equation sheet. At first I was annoyed (it was basically a class on electromagnetism, so tons of equations to memorize), but because my teacher actually explained how the equations came to be, rather than just telling us to plug and chug, I feel I developed a better understanding of the material. I think all science/math classes should take this approach from early on, maybe there'd be an increase in actually understanding the material. Any idiot can substitute values into an equation.
[–] 8 points9 points (8 children)
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I'm finishing off my degree in engineering physics, and you're wrong. First off, you did classical physics. Modern physics is far from being just plugging something into an equation, it's so much more then that. I can give you the Schrodinger's equation and tell you to modelize the harmonic oscillator, and you wouldn't know what the hell you're doing until you understand the idea behind the physics. Equations are tools that you use to describe something, nothing more. Second off, you took basic E&M. If you take E&M fields and waves you'll see that even if the Maxwell equations stay the same, working with them becomes much much harder, and you need to be able to use a decent amount of calculus or wave physics in the case of E&M waves. The difficulty lies there. Only when you do basic physics do you actually plug values into equations. Also, if you do something like crystallography, you will want to kill yourself if you don't have equations given to you, because there are thousands of equations and approximations and mathematical tools that might be useful, and not just 4 basic equations.
[–] 1 point2 points (2 children)
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I think you're missing the overall point of his post, which is not that rote memorization of equations is great. In all honesty, it seems like he basically said exactly what you did, in that an understanding of the math at hand is way more important than "plugging and chugging", since then your understanding of what you're looking at and trying to do is much more complete.
[–] 0 points1 point (1 child)
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He basically stated that giving equations leads to people not understanding the physics behind the problem, because they end up just plugging, while not understanding that the physics has very little to do with the actual equation. The equation is just a tool. It's like if he stated that by not giving the equation, people will understand what the equation is. The reality is, the understanding of physics must be completely independent from the equation.
[–] 0 points1 point (0 children)
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Not sure about that last sentence of yours. Once we get into advanced physics, the equations ARE the physics. Like...trying to describe what's physically going on in a lot of quantum mechanics and electromagnetism past like basic em (like, even just pre-relativity with the EM tensor) are gonna prob. be a bitch.
Sure, here's this vector, and we take its' curl, which is a cross product, which means we multiple these indices of three / four vectors and then we are gonna tensor that bitch.
Like, how are you gonna have someone without a math background really grasping the 'ideas' behind tensor analysis?
[–] 0 points1 point (3 children)
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Holy fuck, engineering physics? What are some jobs I could do with this?
[–] 1 point2 points (2 children)
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Basically, it's high-tech engineering, with an emphasis on R&D. Jobs you can get: nuclear engineer, optical engineer, biomedical engineer, nanotech engineer, semiconductor engineer. Any technology which has highly advanced physical concept to it is researched by someone who has an applied physics background, like in engineering physics.
[–] 0 points1 point (1 child)
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Yes this is what I want to do. It's sounds awesome. I knew I wanted to be an engineer just not what kind. Thanks for your help! What would you recommend my major and minor be?
[–] 1 point2 points (0 children)
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I don't know about minors since I'm at an engineering only school with only sciences and math courses. But for your major, just check out the different programs offered at the school you want to go to.
[–] 1 point2 points (0 children)
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I think all science/math classes should take this approach from early on, maybe there'd be an increase in actually understanding the material.
I can't possibly be in more agreement with this than I already am. Math in particular tends to be taught outside of the context in which it is useful, which makes it essentially unlearnable. This absolutely needs to change, at least in the US.
[–] 1 point2 points (0 children)
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We weren't given one, we had to make our own, but i've never heard of a class not allowing you to have one. I guess with enough studying you memorize them anyway, but it's kind of impractical. In any real life situation you have unending resources at your disposal. (I'm an E and M student right now)
[–] 0 points1 point (2 children)
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No formula sheet? Just progam all the equations into your calculator. Thats what I have always done.
[–] 0 points1 point (1 child)
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Our school issues us calculators specifically for that reason. No joke.
[–] 1 point2 points (0 children)
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Wow now I feel lucky to have my TI-83
[–] 1 point2 points (1 child)
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The difficulty of physics has never been the actual math. It's been analyzing a problem and knowing which equations to use, and how the parameters of a question can affect said equations (vectors, current direction, Emf direction, kirchoffs rules, etc).
Sincerely, Physics student.
[–] 0 points1 point (0 children)
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I know this.
Sincerely, Physics student as well.
[–] -1 points0 points (2 children)
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you took physics at 14?
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yeah freshmen year
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Probably just basic linear motion
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# Theoretical background¶
Below follows a brief overview of the theory underpinning the correlation functions supported by dynasor as well as their analysis. For a more detailed description and derivations of the correlations functions see for example .
## Dynamic structure factor¶
The particle density is denoted $$n(\boldsymbol{r},t)$$ and defined as
$n(\boldsymbol{r},t) = \sum _i ^N \delta (\boldsymbol{r} - \boldsymbol{r}_i(t)),$
where $$N$$ is the number of particles and $$\boldsymbol{r}_i(t)$$ is the position of particle $$i$$ at time $$t$$. The intermediate scattering function $$F(\boldsymbol{q},t)$$ is defined as the Fourier transform of the auto-correlation of the particle density
$F(\boldsymbol{q},t)=\frac{1}{N}\left<n(\boldsymbol{q},t)n(-\boldsymbol{q},0)\right> \,= \, \frac{1}{N}\sum _i ^N \sum _j ^N \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_j(0))\right] \right>,$
where $$\boldsymbol{q}$$ is a wave vector. The brackets denote an ensemble average, which in the case of dynasor is replaced by a time average. (The system under study should be ergodic in order for the time average to be a suitable substitute for the ensemble average.) Here, $$F(\boldsymbol{q},t)$$ is commonly referred to as the coherent part.
The incoherent intermediate scattering function, i.e., $$i=j$$, is defined as
$F_\mathrm{incoh}(\boldsymbol{q},t)=\frac{1}{N} \sum _i ^N \left < \mathrm{exp}\left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_i(0))\right] \right >$
and describes single particle motion, i.e., diffusive motion. $$F_\mathrm{incoh}(\boldsymbol{q},t)$$ is commonly referred to as the incoherent scattering function, but sometimes also referred to as the self-part of the intermediate scattering function.
The static structure factor is given by
$S(\boldsymbol{q}) = F(\boldsymbol{q},0),$
whereas the dynamic structure factor $$S(\boldsymbol{q},\omega)$$ is the Fourier transform of $$F(\boldsymbol{q},t)$$
$S(\boldsymbol{q},\omega) = \int _{-\infty} ^\infty F(\boldsymbol{q},t) \, e^{-iwt} \mathrm{d}t,$
where $$\omega$$ is the angular frequency. $$S(\boldsymbol{q},\omega)$$ exhibits peaks in the $$(\boldsymbol{q},\omega)$$ plane corresponding to the dispersion of lattice vibrations (phonons). The widths of these peaks are related to the phonon lifetimes. Computing $$S(\boldsymbol{q},\omega)$$ via molecular dynamics simulation has the advantage of fully including anharmonic effects and allowing one to study the temperature dependence of phonon dispersions.
## Velocity correlation functions¶
It is often convenient to also consider current correlations based on the particle velocities. The current density $$\boldsymbol{j}(\boldsymbol{r},t)$$ is given by
$\begin{split}\boldsymbol{j}(\boldsymbol{r},t) &= \sum_i^N \boldsymbol{v}_i(t) \, \delta (\boldsymbol{r} - \boldsymbol{r}_i(t)) \\ \boldsymbol{j}(\boldsymbol{q},t) &= \sum_i^N \boldsymbol{v}_i(t) \, \mathrm{e}^{\mathrm{i}\boldsymbol{q} \cdot \boldsymbol{r}_i(t)},\end{split}$
where $$\boldsymbol{v}_i(t)$$ is the velocity of particle $$i$$ at time $$t$$. This can be split into a longitudinal and transverse part as
$\begin{split}\boldsymbol{j}_L(\boldsymbol{q},t) &= \sum_i ^N(\boldsymbol{v_i}(t) \cdot\hat{\boldsymbol{q}}) \, \hat{\boldsymbol{q}} \, \mathrm{e}^{\mathrm{i}\boldsymbol{q} \cdot \boldsymbol{r}_i(t)} \\ \boldsymbol{j}_T(\boldsymbol{q},t) &= \sum_i ^N \left[\boldsymbol{v_i}(t) - (\boldsymbol{v_i}(t) \cdot \hat{\boldsymbol{q}}) \, \hat{\boldsymbol{q}}\right] \, \mathrm{e}^{\mathrm{i}\boldsymbol{q} \cdot \boldsymbol{r}_i(t)}.\end{split}$
Now the correlation functions can be computed as
$\begin{split}C_L(\boldsymbol{q},t) &= \frac{1}{N}\left<\boldsymbol{j}_L(\boldsymbol{q},t)\cdot\boldsymbol{j}_L(-\boldsymbol{q},0)\right> \\ C_T(\boldsymbol{q},t) &= \frac{1}{N}\left<\boldsymbol{j}_T(\boldsymbol{q},t)\cdot\boldsymbol{j}_T(-\boldsymbol{q},0)\right>.\end{split}$
The longitudinal current can be related to the particle density as
$\begin{split}\frac{\partial }{\partial t}n(\boldsymbol{q}, t) &= \mathrm{i} \boldsymbol{q}\cdot \boldsymbol{j}(\boldsymbol{q}, t) \\ \omega^2 S(\boldsymbol{q},\omega) &= q^2 C_L(\boldsymbol{q},\omega),\end{split}$
which means some features can be easier to resolve in one function than the other. The current correlation functions can be thought of as spatially-dependent generalization of the velocity correlation function and are closely related to the phonon spectral energy density.
## Multi-component systems¶
In multi-component systems one can introduce partial correlations functions, which enables separation of the contributions by groups of particles, e.g., by type or site symmetry. For example, in the case of a binary system, one can define $$F_\mathrm{AA}(\boldsymbol{q},t)$$, $$F_\mathrm{BB}(\boldsymbol{q},t)$$, $$F_\mathrm{AB}(\boldsymbol{q},t)$$ and so on for all correlation functions introduced above. In dynasor we define the partial correlaion functions as
$F_\mathrm{AA}(\boldsymbol{q},t) = \frac{1}{N}\sum_{i\in A}^{N_\mathrm{A}} \sum_{j\in A}^{N_\mathrm{A}} \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_j(0))\right] \right>$
where the sums runs over all A atoms. Here, $$N_\mathrm{A}$$ refers to number of atoms of type A, $$N_\mathrm{B}$$ refers to number of atoms of type B, and $$N = N_\mathrm{A} + N_\mathrm{B}$$ refers to the total number of atoms in the system. For the cross-term, $$F_\mathrm{AB}(\boldsymbol{q},t)$$, we get
$\begin{split}F_\mathrm{AB}(\boldsymbol{q},t) = & \frac{1}{N}\sum_{i\in A}^{N_\mathrm{A}} \sum_{j\in B}^{N_\mathrm{B}} \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_j(0))\right] \right> + \frac{1}{N}\sum_{i\in B}^{N_\mathrm{B}} \sum_{j\in A}^{N_\mathrm{A}} \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_j(0))\right] \right> \\=& \frac{2}{N}\sum_{i\in A}^{N_\mathrm{A}} \sum_{j\in B}^{N_\mathrm{B}} \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_i(t)-\boldsymbol{r}_j(0))\right] \right>\end{split}$
Here, the factor two in the final expression is due to considering both the A-B and B-A terms. These normalization choices means the total intermediate scattering function $$F(\boldsymbol{q},t)$$ (as defined above) is given by
$F(\boldsymbol{q},t) = F_\mathrm{AA}(\boldsymbol{q},t) + F_\mathrm{AB}(\boldsymbol{q},t) + F_\mathrm{BB}(\boldsymbol{q},t)$
In some cases, instead of analyzing the partial functions directly, it is interesting to consider linear combinations of them. For example weighting them with relevant atomic form factors, masses or charges etc.
## Relation to scattering experiments¶
The correlation functions can be convoluted with atomic form factors and cross sections in order to obtain predictions for various types of neutron and X-ray scattering experiments.
In solids, it is frequently desirable to determine the above mentioned quantites along specific paths between high-symmetry $$\boldsymbol{q}$$-points. In isotropic samples, such as for example liquids, it is on the other hand usually preferable to compute these functions with respect to $$q=|\boldsymbol{q}|$$, i.e., a spherical average over wave vectors.
## Damped harmonic oscillator model¶
The correlation functions can be fitted to analytical expressions for damped harmonic oscillators in order to extract phonon frequencies and phonon lifetimes. While many different analytical forms can be considered for the damped harmonic oscillator, here we use the assumption that the correlation function of the position of the damped harmonic oscillator has a zero derivative at $$t=0$$. The position correlation function (corresponding to the intermediate scattering function) of a damped harmonic oscillator is given by
$\begin{split}F(t) &= A \mathrm{e}^{-\Gamma t/2} \big(\cos{ \omega_e t} + \frac{\Gamma}{2\omega_e}\sin{ \omega_e t}\big), \quad \omega_0 > \frac{\Gamma}{2} \\ F(t) &= A \mathrm{e}^{-\Gamma t/2} \big(\cosh{|\omega_e| t} + \frac{\Gamma}{2|\omega_e|}\sinh{|\omega_e| t}\big), \quad \omega_0 < \frac{\Gamma}{2}\end{split}$
where the fitting parameters are $$\omega_0$$, $$\Gamma$$ and $$A$$. The first case, $$\omega_0 > \Gamma/2$$ corresponds to an underdamped oscillator, and the second one to an overdamped oscillator. The eigenfrequency of the damped oscillator is $$\omega_e = \sqrt{\omega_0^2 - \Gamma^2/4}$$. The fitting can also be carried out in the frequency domain by fitting the dynamic structure factor to the following analytical function
$a(\omega) = A\frac{2\Gamma \omega_0^2}{(\omega^2 - \omega_0^2)^2 + (\Gamma\omega)^2}.$
The intermediate scattering function and dynamic structure factor can thus be analyzed by fitting them to the expressions above in order to extract frequency $$\omega_0$$ and damping $$\Gamma$$.
This analysis can be extended to the velocity correlation for the damped harmonic oscillator (corresponding to the current correlations) by considering the relation between the longitudinal current correlation and the dynamic structure factor, which gives the following solutions
$b(\omega) = B\frac{2\Gamma \omega^2}{(\omega^2 - \omega_0^2)^2 + (\Gamma\omega)^2}.$
In the time domain this function becomes
$\begin{split}b(t) &= B \mathrm{e}^{-\Gamma t/2} \big(\cos{ \omega_e t} - \frac{\Gamma}{2\omega_e}\sin{ \omega_e t} \big), \quad \omega_0 > \frac{\Gamma}{2} \\ b(t) &= B \mathrm{e}^{-\Gamma t/2} \big(\cosh{|\omega_e| t} - \frac{\Gamma}{2|\omega_e|}\sinh{|\omega_e| t} \big), \quad \omega_0 < \frac{\Gamma}{2}\end{split}$
Lastly, the damping $$\Gamma$$ is related to the inverse phonon lifetime $$\tau$$ according to
$\tau = \frac{2}{\Gamma}$
or in energy units as
$\tau = \frac{2\hbar}{\Gamma},$
where $$\hbar$$ appears since $$\Gamma$$ has units of rad/s.
If multiple modes are present in a correlation function they should be modeled as a sum of damped harmonic oscillators.
## Spectral energy density method¶
The spectral energy density (SED) is closely related to the phonon dispersion and thus also to the current correlations above. It is a measure of how the kinetic energy is partitioned over different wavelengths and frequencies in the system. The theoretical background for SED is quite simple and can be derived in several ways. The total, integrated kinetic energy of the infinite system can be written as
$E = \int dt \sum_n \sum_i \frac{1}{2} m_i v_i(n, t)^2.$
Here $$n$$ is the cell index and $$i$$ is the basis index, which together uniquely identify each atom in an infinite crystal. By using Parseval’s theorem for both the time integration and space summation the above is transformed to inverse spatial $$q$$-space and inverse temporal $$\omega$$-space
$E = \int d\omega \sum_q \sum_i \frac{1}{2} m_i v_i(q, \omega)^2.$
This is motivated by the Wiener–Khinchin theorem and the SED is simply defined as the quantity
$\text{SED}(q, \omega) = \sum_i \frac{1}{2} m_i v_i(q, \omega)^2$
For a more in-depth discussion see, e.g., .
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# INTRODUCING BIOTECHNOLOGY
Date of the Problem
October 25, 2021
Biotechnology is a field of engineering that uses living systems to produce a wide range of technologies and products used in medicine, energy, fuels and sensors. You can see biotechnology in action when, for example, you use yeast (a living organism) to create bread! Biotechnology involves collaboration with many other types of engineers, including microbiologists, chemical engineers, electrical engineers, data scientists and more.
Biotechnology can be used to monitor wastewater to determine what microorganisms are present in a human population. This is especially important when people with a virus may not show symptoms immediately; the presence of the virus can be detected in their wastewater before a person even knows they are sick.
Scientists can predict a spike of a virus in a population using a technique called PCR, which makes copies of DNA. Scientists take a small amount of DNA (in this case, found in wastewater) and amplify it in order to more easily detect a dangerous virus. If there is a very small amount of the virus in the wastewater sample, more PCR cycles must be conducted to get the dangerous virus to a detectable level. If there is a large amount of the virus, fewer cycles are needed to detect it. Scientists then use the results of the PCR cycle(s) to determine the number of copies of the dangerous virus per liter of wastewater. They can also compare the amount of the dangerous virus (Virus A in our example) to the amount of standard virus that is always detected in wastewater.
1. Using the information in the table, which location, Eastville or Westview, has a greater ratio of Virus A to standard virus in its wastewater?
For Eastville, there are (4 × 109)/(3 × 108) = 13.3 copies of Virus A per standard virus copy. For Westview, there are (2 × 107)/(9 × 105) = 22.2 copies of Virus A per standard virus copy. Westview has more of Virus A in its wastewater shed by its population.
2. When testing wastewater, scientists use a sample of 100 mL, rather than 1 liter. How many copies of Virus A would have been in Eastville’s wastewater sample? Express your answer in scientific notation.
In Eastville, there are 4 × 109 copies of Virus A per 1 liter (or 1 × 103 milliliters). So, we can set up and solve the following proportion to find the number of copies of Virus A in the original sample: (4 × 109)/(1 × 103) = x/(1 × 102) → (4 × 109)(1 × 102) = (1 × 103)x → 4 × 1011 = (1 × 103)x → x = (4 × 1011)/(1 × 103) = 4 × 108 copies. Alternatively, you could simply divide 4 × 109 by 10 (or 1 × 101).
3. The current population of Westview is 878,500 people. If 40% of the population is currently shedding Virus A into the wastewater, on average, how many copies of Virus A per liter of wastewater is shed by each person? Express your answer to the nearest whole number.
First, we need to find how many people are currently shedding Virus A. So, we’ll take 40% of 878,500, which is 0.4 × 878,500 = 351,400 = 3.514 × 105 people. So, on average, each person is shedding (2 × 107)/(3.514 × 105) = 0.5691519636 × 10257 copies.
4. The average number of PCR cycles needed in testing a wastewater sample is 35. If the entire process takes an average of 3 hours, what is the average number of minutes a single PCR cycle takes? Express your answer as a decimal to the nearest tenth.
The entire process takes 3 hours, which is 3 × 60 = 180 minutes, on average, for 35 PCR cycles. So, on average, 1 cycle takes 180/35 ≈ 5.1 minutes.
5. The first step of a PCR cycle lasts approximately 2 minutes, during which the DNA is at 96°C. The second step of a PCR cycle lasts approximately 2.25 minutes, during which the temperature is 60°C. Assuming an average total of 51 seconds of a PCR cycle is spent raising or lowering the temperature, and there are only these two steps in a PCR cycle, what percent of the time during a PCR cycle is spent with the temperature at 96°C? Express your answer to the nearest whole percent.
In order to get all given times into the same units, we must first convert 51 seconds to minutes: 51/60 = 0.85 minutes. So, the total time of a PCR cycle is 0.85 + 2 + 2.25 = 5.1 minutes. Thus, 2/5.1 ≈ 0.39, which is 39% of a cycle.
How is this relevant in the real world? Currently, scientists are monitoring wastewater to determine the level of SARS-CoV-2, the virus that causes COVID-19, in populations. A person with COVID-19 often does not show symptoms until 4-7 days after infection, but the virus can show up in their waste (a process called shedding) before they begin to show symptoms (which is called the asymptomatic stage). Spikes in SARS-CoV-2 in wastewater can be used to predict spikes in COVID-19 cases up to one week in advance.
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# Question S For the beam and loading shown, determinc (a) The equation of the elastic curve....
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Question S For the beam and loading shown, determinc (a) The equation of the elastic curve. (b) The slope equation. (c) The slope at end A (d) The deflection at the midpoint of the span.
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def traverseList(): for element in ['a', 2, 'word', ['1,2', 3]]: if type(element) == list: print('list of length:', len(element)) else: print(element) def traverseList2(l): for number in l: if number < 0: print("negative") else: print("not negative") # Example of building and returning a new list # # Returns a new list containing the negative numbers that appear in the # input list # def negativeListFrom(l): result = [] for number in l: if number < 0: print("before", result) result = result + [number] #result.append(number) print("after", result) return result # assume list1 and list2 are equal length lists of numbers # return a list where the ith item is the larger of the ith items # of list1 and list2 # def listOfBiggests(list1, list2): newList = [] index = 0 while index < len(list1): if list1[index] > list2[index]: newList = newList + [list1[index]] else: newList = newList + [list2[index]] index = index + 1 return newList def listOfBiggests2(list1, list2): newList = [] index = 0 for index in range(len(list1)): item1 = list1[index] item2 = list2[index] if item1 > item2: newList.append(item1) else: newList.append(item2) return newList # listOfLists is a list containing 0 or more lists of numbers # Return a new list that is the same length as listOfLists but where # the ith element of the result is the average of all the numbers # in the ith element of listOfLists # # When does this work/not work? Is the specification clear enough to know? # def getAverages(listOfLists): averages = [] for sublist in listOfLists: # compute average for sublist sublistSum = 0 for item in sublist: sublistSum = sublistSum + item # append sublist average to result averages.append(sublistSum/len(sublist)) return averages
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# 11.7: Global variables
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
In the previous example, known is created outside the function, so it belongs to the special frame called __main__. Variables in __main__ are sometimes called global because they can be accessed from any function. Unlike local variables, which disappear when their function ends, global variables persist from one function call to the next.
It is common to use global variables for flags; that is, boolean variables that indicate (“flag”) whether a condition is true. For example, some programs use a flag named verbose to control the level of detail in the output:
verbose = True
def example1():
if verbose:
print('Running example1')
If you try to reassign a global variable, you might be surprised. The following example is supposed to keep track of whether the function has been called:
been_called = False
def example2():
been_called = True # WRONG
But if you run it you will see that the value of been_called doesn’t change. The problem is that example2 creates a new local variable named been_called. The local variable goes away when the function ends, and has no effect on the global variable.
To reassign a global variable inside a function you have to declare the global variable before you use it:
been_called = False
def example2():
global been_called
been_called = True
The global statement tells the interpreter something like, “In this function, when I say been_called, I mean the global variable; don’t create a local one.”
Here’s an example that tries to update a global variable:
count = 0
def example3():
count = count + 1 # WRONG
If you run it you get:
UnboundLocalError: local variable 'count' referenced before assignment
Python assumes that count is local, and under that assumption you are reading it before writing it. The solution, again, is to declare count global.
def example3():
global count
count += 1
If a global variable refers to a mutable value, you can modify the value without declaring the variable:
known = {0:0, 1:1}
def example4():
known[2] = 1
So you can add, remove and replace elements of a global list or dictionary, but if you want to reassign the variable, you have to declare it:
def example5():
global known
known = dict()
Global variables can be useful, but if you have a lot of them, and you modify them frequently, they can make programs hard to debug.
This page titled 11.7: Global variables is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) .
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Commit ee13bc40 by David E. DeMarle Committed by Kitware Robot
### Merge topic 'Fix2DReflectionFilter'
```368b9586
Fix reflection for 2D cells, with valid results on nodes after triangulation
Acked-by: Kitware Robot <kwrobot@kitware.com>
Merge-request: !4788```
parents fab90595 368b9586
... ... @@ -209,7 +209,32 @@ vtkIdType vtkReflectionFilter::ReflectNon3DCell( } for (int j = numCellPts-1; j >= 0; j--) { newCellPts[numCellPts-1-j] = cellPts->GetId(j); // Let's take the connectivity of origin cell as follows: 0, 1, 2, 3. // Here, left : origin, right : symmetrized // 3 - 2 || 2' - 3' // | | || | | // 0 - 1 || 1' - 0' // Previous result of connectivity of symmetric cell gave: 3', 2', 1', 0' // Topologically connectivity is good but triangulation of this symmetric // cell is not simply the symmetry of the triangulation of the origin cell // if triangulation occurs between index nodes 1 and 3: // original nodes 1 and 3, symmetric nodes 0' and 2'. // 3 - 2 || 2' - 3' // | \ | || | \ | // 0 - 1 || 1' - 0' // Visually this is wrong for values on the nodes, result is not symmetric // as it should be. Our modification provides the following connectivity // for the symmetric cell: 0', 3', 2', 1'. // The triangulation behaves as expected now, as illustrated below: // original nodes 1 and 3, symmetric nodes 1' and 3'. // 3 - 2 || 2' - 3' // | \ | || | / | // 0 - 1 || 1' - 0' // The new symmetry behaves correctly on higher order as well: // 4 - 3 \ || / 3' - 4' // | \ | 2 || 2' | / | // 0 - 1 / || \ 1' - 0' newCellPts[(numCellPts-j)%numCellPts] = cellPts->GetId(j); } } } // end switch ... ...
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# Heston Monte Carlo or FFT Pricing
I am trying to better understand the Heston model and its implementation. It seems like a lot of people use the FFT method for calculating the call prices during the Heston calibration, but the Monte Carlo method is used to calculate the prices with the calibrated parameters. What is the point in this? Why not just use the FFT method for calculating both prices?
• FFT can be used to compute both prices. You would use MC simulations if you want to price exotic payoffs. Recall that FFT is limited to European-style options. Since these are often used for calibration, FFT is a fast way of finding your model parameters. There are, of course, further alternatives to pricing options under the Heston model (e.g. finite differences, trees and other Fourier methods). It all depends on your application Apr 30, 2020 at 15:12
• @KeSchn Ok, I think I understand what you are saying. So, the FFT method would work for calibration of parameters since the optimization can just use European call prices because there is no path dependency. And the Monte Carlo method would be needed to find values at specific points for exotic payoffs? Apr 30, 2020 at 16:01
• That's right Kevin. Heston (and many other models) give you a closed-form solution for European-style options. Implementing those with FFT gives you a very fast way of computing these option prices. Using liquid observed European option prices, you calibrate your model. Now you got $\kappa$, $\theta$ etc. Then, you can price (almost) every other derivative using either FFT, MC, FD or whatever. By the way, the COS method from Fang and Oosterlee is an even faster Fourier method and even simpler to implement, you may want to check it out :) Apr 30, 2020 at 16:33
• @KeSchn Great. That helps a lot. Thank you Apr 30, 2020 at 16:39
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# Multiple Kronecker Products
Looking at this paper here (see page 5, just above equation (7)) https://arxiv.org/pdf/1606.09225v1.pdf...
It shows an equation:
$$X_{3of4} = I \otimes I \otimes X \otimes I$$ Where $I$ is a 2x2 identity matrix, and X is the matrix: $$X = \left[ \begin{matrix} 0 & 1\\1 & 0\end{matrix} \right]$$
How would you work this out, as once you did $I \otimes I$, you would have a 4x4 matrix, so you couldn't then do it to the $X$ matrix...
Any help would be appreciatied
• I don't understand what your confusion here is. Certainly, you can compute $$((I\otimes I)\otimes X)\otimes I$$ Or, if there's an issue there, what is it? Nothing about the Kronecker product necessitates that the two matrices have the same shape or size. – Omnomnomnom Aug 20 '16 at 23:42
just for fun, \begin{align} I \otimes I \otimes X \otimes I = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} . \end{align}
As @Omnomnomnom points out, having computed one of the Kronecker products, you can continue with the others.
You can do this e.g. in Matlab/Octave with
I = eye(2); X = [0 1; 1 0]; kron(kron(kron(I, I), X), I).
• Thanks for that ;) did you just do them sequentially? The comment above said you don't actually need them to be the same size, that true? – Adam Kelly Aug 21 '16 at 0:48
• Ah yes! Thanks for that! – Adam Kelly Aug 21 '16 at 0:49
• @ user66081 , your equality $I\otimes I\otimes X\otimes I=\cdots$ is not quite correct. More precisely, $(E\otimes F)\otimes G$ and $E\otimes(F\otimes G)$ are not the same space but are isomorphic under a canonical isomorphism. Then, if you want explicitly write such a multiple tensor product, then you must choose some bracketing. Clearly, the explicit result depends on such a choice. – loup blanc Aug 29 '16 at 11:15
• @loupblanc: I understand; but we are talking about the Kronecker product of matrices here – user66081 Aug 29 '16 at 21:49
• @ user66081 , yes, you are right. – loup blanc Aug 29 '16 at 22:06
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# Directly Proportional
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Directly proportional means two variables that increase or decrease at the same time. If two variables are proportional if a change in one variable is accompanied by a change in another variable. We can also say that if two quantities are said to be in proportional then one quantity is a constant multiple of other quantity. Two quantities a and b are said to be directly proportional, if the relationship can be written as a = k b where k is a proportionality constant.
Problem 1: The term A is directly proportional to x. And when A is 12, x is 4. Find the value of A when x is 10.
Solution: Since A is directly proportional to x.
=> This can be written as A = k x, where k is proportionality constant.
=> Given When A is12, x is 4
=> Find out constant from the known values A = k x
=> 12 = k * 4
=> By dividing 4 on both sides, we get k = 3
=> When x is 10 then A = k x = 3 * 10 = 30
=> Therefore, when x is 10 the value of A is 30.
Problem 2: A term Y is directly proportional to the square of x. And when Y is 24, x is 2. Find the value of Y when x is 5.
Solution: Given Y is directly proportional to x^2.
=> So, Y = k x^2
=> Substitute the given values (Y= 24, x = 2)
=> Y = k x^2
=> 24 = k* 2^2
=> 24 = k * 4
=> Dividing by 4 on both sides we get k = 6
=> When x = 5 then Y = k x^2 = 6 * 5^2 = 6 * 25 = 150
=> Therefore, when x is 5 then the value of y is 150.
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# Calculate the y component of the electric field at a generic point(x,y)
## Homework Statement
an electric dipole is placed in the x-y plane, with the positive charge +q placed in position (+a, 0) and the negative charge -q placed in position (-a, 0)
define r(subscript +) and r(subscript -) the vectors connecting the point defined by r and the two points (+a, 0) and (-a, 0), and r(subscript+) and r(subscript -) their magnitudes; calculate the y component of the electric field at a generic point (x, y) of the plane as a function of r(subscript +) and r(subscript -) (it is quicker to use directly the field equation from a point charge than using the potential)
## The Attempt at a Solution
E=q/(4pi epsilon0) [(1/([r(+)]^2 +(x+a)^2))+(1/([r(-)]^2 + (x-a)^2))]
Related Introductory Physics Homework Help News on Phys.org
Spinnor
Gold Member
In the above was a hint:
"... (it is quicker to use directly the field equation from a point charge than using the potential) ... "
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Angular momentum: why can't I answer this problem using linear momentum rules?
• mm2424
In summary, a 1.0 g bullet is fired into a 0.5 kg block attached to the end of a 0.6 m nonuniform rod of mass 0.5 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A (the figure shows a vertical rod labeled A at its top. A block is attached to its bottom end. The bullet flies into the block). The rotational inertia of the rod alone about the axis at A is 0.060 (kg)m^2. Treat the block as a particle. If the angular speed (ω) of the system about A just after impact is 4.5 rad/s, what is the bullet's
Homework Statement
A 1.0 g bullet is fired into a 0.5 kg block attached to the end of a 0.6 m nonuniform rod of mass 0.5 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A (the figure shows a vertical rod labeled A at its top. A block is attached to its bottom end. The bullet flies into the block). The rotational inertia of the rod alone about the axis at A is 0.060 (kg)m^2. Treat the block as a particle. If the angular speed (ω) of the system about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact.
Homework Equations
I have the worked out solution and see that we need to use L = Iω and L = (linear momentum) x r.
The Attempt at a Solution
When I first tried this problem, I tried to do it as a linear momentum problem. I set it up as (mass of the bullet)(initial velocity of bullet) = (mass of the block, bullet and rod system)(velocity of the block, bullet, rod system). I then found the velocity of the block, bullet, rod system using V=ωR.
I understand that this is an angular momentum problem and that I should have used angular momentum equations to solve it. However, I tried to turn it into a linear momentum problem. My strategy was to use the w given to find v of the block/rod/bullet system after impact (using V=ωR). Since my strategy didn't yield the correct answer, I now see that my reasoning was flawed :).
One thing in particular that confuses me about all of this is that the actual solution requires that we take the linear momentum of the bullet and multiply it by the length of the rod...which seems like taking linear momentum and turning it into angular momentum. I therefore thought/hoped/prayed there was a way to turn angular momentum into linear momentum (by finding V from the w given)...in part because the L = r x p and L = Iω equations seem like magic whereas linear momentum equations seem to make sense.
Can someone please explain the error in my logic? While I'm sure that a completely adequate explanation requires an explanation of the cross product, I think I'm clinically incapable of understanding what it is...so if there's a conceptual way of explaining why I can't tackle this problem with linear momentum...and why the actual method of multiplying the bullet's linear momentum by R and equating it to Iω works, I'd be super grateful!
Thanks!
Welcome to PF .
One end of the rod is fixed, so the center of mass of the rod has only 1/2 of the velocity of the rest of the system.
Oh, I see. I have a quick follow up, if that's ok.
I think what confuses me about this problem is the fact that angular momentum can be written as both r x p and Iω.
The solution to this problem seems to require that we write the angular momentum of the bullet before it strikes as r x p while writing the angular momentum of the system after the strike as Iω. What's conceptually hard for me to reconcile is that it doesn't seem possible to write the angular momentum of the bullet in the Iω form before the strike, since it has no rotational velocity. It seems like ω for the bullet prior to the strike is 0, which would make the angular momentum before the strike 0...but this conflicts with the fact that r x p for the bullet is not 0.
Right now, the decision of whether to express angular momentum as r x p or Iω seems arbitrary, which makes the concept seem a bit mysterious. Is there some way of thinking about these problems that I'm missing? Thanks a million for any advice!
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https://mathematicalremarks.wordpress.com/2012/12/16/uniqueness-of-hyperbolic-geodesics/
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## Uniqueness of hyperbolic geodesics
The Poincare’ model of hyperbolic geometry consists of the unit disk $\mathbb{D}=\{z\in\mathbb{C}: |z|<1\}$ where the geodesics are arcs of circles perpendicular to the unit circle $\partial\mathbb{D}=\{|z|=1\}$. It turns out that given two points $z,w\in\mathbb{D}$ there is a unique circle orthogonal to $\partial\mathbb{D}$ passing through $z$ and $w$.
A nice way to prove the uniqueness of geodesics is to use the stereographic projection. Let $\mathcal{S}$ be the unit sphere in $\mathbb{R}^3$ and $N$ its north pole. The stereographic projection puts in one-to-one correspondence each point $z\in\mathbb{C}$ with the unique point $z^*\in \mathcal{S}\setminus\{N\}$ that lies on the line through $N$ and $z$. For example, the unit disk $\mathbb{D}$ corresponds to the southern hemisphere on the sphere $\mathcal{S}$.
One of the main properties of the stereographic projection is that it is a conformal map, i.e., it preserves angles between curves. Proving this is a nice exercise in geometry.
Another property is that the stereographic projection maps circles to “circles”, in the sense that circles on $\mathcal{S}$ through the north pole are mapped to lines in $\mathbb{C}$ and those that don’t go through $N$ are mapped to actual circles in $\mathbb{C}$.
Armed with these two facts we can now easily establish the uniqueness of hyperbolic geodesics. Given two points $z,w\in\mathbb{D}$ consider their images $z^*, w^*$ on the southern hemisphere of $\mathcal{S}$. It is clear that $z^*$ and $w^*$ do not lie on a vertical line. Therefore, there is a unique plane through $z^*$ and $w^*$ perpendicular to the $(x,y)$-plane in $\mathbb{R}^3$, i.e. $\mathbb{C}$. The intersection of this plane with $\mathcal{S}$ is a circle that is perpendicular to the equator of $\mathcal{S}$. Under the stereographic projection it will be mapped to a “circle” through the points $z,w$ and, by conformality, its image will be orthogonal to the image of the equator, i.e., $\partial\mathbb{D}$ (the equator is fixed by the projection).
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https://mymathtutors.com/factors-of-quadratic/radical-expressions/basics-of-algebra-negatives.html
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Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
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https://community.fabric.microsoft.com/t5/Desktop/SUMIFs-CALCULATE-not-doing-what-I-need-it-to-do/m-p/622845
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Continued Contributor
## SUMIFs/CALCULATE not doing what I need it to do
Power BI sample here (page 4) https://www.dropbox.com/s/rbjoqspdjdmtquq/Power%20BI%20Forum%20Sample.pbix?dl=0
As shown in column BY of the Excel Data, I have used the following excel formula: '=SUMIFS([Cumulative IRR - Upside],[Asset ID],[@[Asset ID]],[Count],[Project Term (mos)]+1) to generate flattened data of an output on an [Asset ID] by [Asset ID] level
My attempt to create the same in Power BI is the following programming:
CALCULATE(
SUM('_Bronn Portfolio Analysis Table'[Cumulative IRR - Upside]),
ALLSELECTED('_Bronn Portfolio Analysis Table'[Asset ID]),
FIlter('_Bronn Portfolio Analysis Table','_Bronn Portfolio Analysis Table'[Count]='_Bronn Portfolio Analysis Table'[Project Term (mos)]+1)
)
The problem I have with this is that some [Project]s have one [Asset ID] and some have multiple. As an example, if you select "Alpha" in the BI slicer, the output from the above Measure is 62.4%. Alternatively, if you select "Charlie" in the BI slicer, the output is 21.1%. The only difference between these is that there are 3 [Asset ID]s in "Alpha" and only one in "Charlie". Because there may be different purchase prices on an [Asset ID] by [Asset ID] level, I don't think the right answer is dividing by number of [Assets ID]s., so essentially, I'm trying to get a weighted average of the output, weighted in this case by the [Equity Contribution] column.
Help please
1 ACCEPTED SOLUTION
Super User
@mrothschild
To solve the immediate problem with the data as it is currently structured, try this measure:
```IRR Forecast Upside Weighted Average =
CALCULATE (
DIVIDE (
SUMX (
'_Bronn Portfolio Analysis Table',
'_Bronn Portfolio Analysis Table'[Cumulative IRR - Upside] * '_Bronn Portfolio Analysis Table'[Equity Contribution]
),
SUM ( '_Bronn Portfolio Analysis Table'[Equity Contribution] )
),
ALLSELECTED ( '_Bronn Portfolio Analysis Table'[Asset ID] ),
FILTER (
SUMMARIZE (
'_Bronn Portfolio Analysis Table',
'_Bronn Portfolio Analysis Table'[Count],
'_Bronn Portfolio Analysis Table'[Project Term (mos)]
),
'_Bronn Portfolio Analysis Table'[Count] = '_Bronn Portfolio Analysis Table'[Project Term (mos)] + 1
)
)```
• The average of Cumulative IRR - Upside weighted by Equity Contribution is in green.
• This expression makes sense as long as the final FILTER results in one row per Asset ID (appears to be the case)
• I also rewrote the FILTER as FILTER ( SUMMARIZE (...) ), in order to apply a filter just on the required columns.
• I'm not sure the ALLSELECTED is required, but it depends how you are using this measure so left it there.
As a side note, I would consider restructuring the data model so that you have an Asset table (one row per asset) related to the '_Bronn Portfolio Analysis Table'. This would avoid repeating common values across every row of a given Asset ID, and may simplify some of the DAX.
Regards,
Owen
Owen Auger
Did I answer your question? Mark my post as a solution!
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2 REPLIES 2
Super User
@mrothschild
To solve the immediate problem with the data as it is currently structured, try this measure:
```IRR Forecast Upside Weighted Average =
CALCULATE (
DIVIDE (
SUMX (
'_Bronn Portfolio Analysis Table',
'_Bronn Portfolio Analysis Table'[Cumulative IRR - Upside] * '_Bronn Portfolio Analysis Table'[Equity Contribution]
),
SUM ( '_Bronn Portfolio Analysis Table'[Equity Contribution] )
),
ALLSELECTED ( '_Bronn Portfolio Analysis Table'[Asset ID] ),
FILTER (
SUMMARIZE (
'_Bronn Portfolio Analysis Table',
'_Bronn Portfolio Analysis Table'[Count],
'_Bronn Portfolio Analysis Table'[Project Term (mos)]
),
'_Bronn Portfolio Analysis Table'[Count] = '_Bronn Portfolio Analysis Table'[Project Term (mos)] + 1
)
)```
• The average of Cumulative IRR - Upside weighted by Equity Contribution is in green.
• This expression makes sense as long as the final FILTER results in one row per Asset ID (appears to be the case)
• I also rewrote the FILTER as FILTER ( SUMMARIZE (...) ), in order to apply a filter just on the required columns.
• I'm not sure the ALLSELECTED is required, but it depends how you are using this measure so left it there.
As a side note, I would consider restructuring the data model so that you have an Asset table (one row per asset) related to the '_Bronn Portfolio Analysis Table'. This would avoid repeating common values across every row of a given Asset ID, and may simplify some of the DAX.
Regards,
Owen
Owen Auger
Did I answer your question? Mark my post as a solution!
Blog
Twitter
LinkedIn
Continued Contributor
Thanks so much for your help - much appreciated.
Regarding restructuring the data, ironically, for years, I've avoided using pivot tables in excel, prefering INDEX to present pivoted data. I just started using/learning BI a few weeks ago and flattened my matrix data to start the process. I'm certain you're right about a better way to access the data and make programming and auditing more functional and elegant, but I'm not quite understanding the relationship functionality across different tables yet.
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http://genie-from-klein-bottle.com/topicsPages/number/negative.htm
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|
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| 31,088,664
| 3,619
|
### home | courses | topics | theorems | starters | worksheets | timeline | KS3 | KS4 | KS5
There is a bit longer history to the negative numbers than what you will find here - so investigate links on the side if you want to find more.
People thought about the negative numbers for a long time before they accepted them as 'real' numbers. Sometimes they would get negative numbers as solutions to some equations, as well as use them to represent them as 'debts' when there was not enough to be covered by a positive value (for example when subtracting a larger from a smaller number).
But it was only in the 16 th century that they became famous. Girolamo Cardano, a famous Italian mathematician and medical doctor (or physician - that is how they were then called) for the first time described negative numbers as 'fictitious' and accepted the possibility of them being the solutions to equations. He described this in his famous book Ars Magna which was published in 1545.
Descartes studied further the possible solutions of equations and called the negative solutions 'false' and other solutions, which contained the square root of negative numbers, 'imaginary'. This is where we get our name for imaginary numbers.
The thing is that you can show (or imagine) all real numbers on a number line of a kind - even the irrational numbers can be imagined to slide down the number line, albeit their position is very difficult to pin!
Back to negative numbers - they rose to become accepted members of the numbers community sometime in the 17th century. Now we use them freely and without any confusion.
But the irrational, imaginary, and complex numbers remained something of an enigma for the greatest part of the 18 th century, and even today you don't learn about them until you are about 16. Complex numbers are also very important to understand if you want to understand the Fundamental Theorem of Algebra.
Girolamo Cardano (1501-1576) - a famous Italian mathematician.
Some other pages on negative numbers:
Leo Rogers' page on the history of negative numbers
Anne Boye's page for teachers
and another page - nicely giving the timeline of developments
artefacts | numerals | concepts | people | places | pythagoreans | egyptians | babylonians
Search WWW Search www.mathsisgoodforyou.com
_____________________________________________________________________________________________________________________
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| 0.971369
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https://tsfa.co/what-is-the-mad-in-arithmetic-23
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| 656,986,410
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What is the mad in math
What is Mean Absolute Deviation (MAD) Formula? The mean deviation or absolute deviation is
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Mean Absolute Deviation: Definition, Finding & Formula
Mean absolute deviation (MAD) is a measure of the average absolute distance
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Mean absolute deviation (MAD) review (article)
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Mean Absolute Deviation
Mean absolute deviation (MAD) review Mean absolute deviation The mean absolute deviation
Get Started
| 567
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|
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| 3.1875
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CC-MAIN-2023-14
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http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=212143
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|
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```Question 294017
1 x+4
------ - ----
x^2-16 x^2-3x-4
========================
Factor the denominators:
=============================
1/[(x-4)(x+4)] - (x+4)/[(x-4)(x+1)]
----
lcd = (x+1)(x-4)(x+4)
====
Rewrite each fraction with the lcd as its denominator:
(x+1)/lcd - (x+4)^2/lcd
============================
Combine the numerators over the lcd:
[x+1 - (x+4)^2]/lcd
-----
Simplify the numerator:
[x+1 -x^2 - 8x - 16]/lcd
---
Rearrange:
[-(x^2 +7x + 15)]/[(x+1)(x^2-16)]
------------------------------------------------
Cheers,
Stan H.
```
| 193
| 550
|
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| 3.515625
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https://testbook.com/question-answer/identify-the-following-sequential-component--6086a83ded132e437ef0ee0f
| 1,638,691,432,000,000,000
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crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00110.warc.gz
| 617,715,541
| 31,411
|
# Identify the following sequential component.
This question was previously asked in
DFCCIL Executive S&T 2016 Official Paper
View all DFCCIL Executive Papers >
1. Master-slave flip flop
2. Clocked flip flop
3. J-K flip flop
4. R-S flip flop
Option 4 : R-S flip flop
## Detailed Solution
Explanation:
The given sequential component is of RS Flip Flop.
Here A = R and B = S
The truth table for the circuit is shown:
Inputs (SR) Output (Qn+1) Action 0 0 Q No change 0 1 0 Reset 1 0 1 Set 1 1 0 Forbidden (Undefined)
The difference between latches and flip flops is shown
Latches Flip Flops Latches are building block of sequential circuits and they are built using logic gates Flip flops are also building blocks of sequential circuits but they are made using latches Latches continuously change input and output changes correspondingly Flip flop output changes only when the clock is applied Latches are level sensitive Flip flops are edge sensitive
| 233
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| 2.890625
| 3
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| 0.879484
|
https://www.codea.io/talk/discussion/comment/31039/
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| 769,536,520
| 7,632
|
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# How to make walls
edited October 2013 Posts: 11
I am fairly new to programming and have been working on a small game which involves an ellipse moving around the screen and capturing smaller ones. The only problem I am having with it is keeping the ball on the screen. It does not involve gravity. Is there any way i can make invisible walls on the side of the screen to set a boundary.
``````function setup()
-- this creates the circle
c1 = physics.body(CIRCLE,70)
c1.x=400
c1.y=400
c1.type=STATIC
end
function draw()
background(147, 119, 170, 255)
--this is what is shown as the circle
ellipse(c1.x, c1.y, 100, 100)
strokeWidth(10)
-- Do your drawing here
end
function touched(touch)
--makes sure the circle only moves where you drag it
if touch.state == MOVING then
c1.x = touch.x
c1.y = touch.y
end
end
--not finished
``````
Tagged:
• Posts: 577
This entire thing seems weird, how are you having trouble with it going off screen if it can only follow your finger?
• Posts: 11
I just want the circle to remain on screen and not have half of it not showing. I just want it to look more professional
• edited October 2013 Posts: 2,051
`c1.x = math.max(math.min(touch.x, WIDTH - c1.radius), c1.radius)`
`c1.y = math.max(math.min(touch.y, HEIGHT - c1.radius), c1.radius)`
That should do it.
• Mod
Posts: 10,049
@newow Try this.
``````function setup()
-- this creates the circle
c1 = physics.body(CIRCLE,70)
c1.x=400
c1.y=400
c1.type=STATIC
end
function draw()
background(147, 119, 170, 255)
--this is what is shown as the circle
ellipse(c1.x, c1.y, 100, 100)
strokeWidth(10)
-- Do your drawing here
end
function touched(touch)
--makes sure the circle only moves where you drag it
if touch.state == MOVING then
if touch.x>50 and touch.x<WIDTH-50 then
c1.x = touch.x
end
if touch.y>50 and touch.y<HEIGHT-50 then
c1.y = touch.y
end
end
end
--not finished
``````
• Posts: 577
I see, well you will have to get rid of the
``````c1.x = touch.x
c1.y = touch.y
``````
Because that would allow it to go through walls, for example, lets say you have a brick wall in the middle of the screen, and the ball is on one side of it. If you tap on the other side of the wall, the ball will magically teleport over there. So you want to "add force" to the physic body, here is a example:
``````function touched(t)
if t.state ~= ENDED then
dx = t.x-c1.x
dy = t.y-c1.y
if(dx < 0) then
dx = -speed
elseif(dx > 0) then
dx = speed
end
if(dy < 0) then
dy = -speed
elseif(dy > 0) then
dy = speed
end
force = vec2(dx, dy)
c1.linearVelocity = force
end
end
end
``````
Basically, we are testing where the finger is around the ball, and adding a little bit of force to make it slowly move. Now to answer your question, to make the walls you would do something like this:
``````floor = physics.body(EDGE,vec2(0,0),vec2(WIDTH,0))
left = physics.body(EDGE,vec2(0,0),vec2(0,HEIGHT))
right = physics.body(EDGE, vec2(WIDTH, 0), vec2(WIDTH, HEIGHT))
top = physics.body(EDGE, vec2(0, HEIGHT),vec2(WIDTH,HEIGHT))
``````
• Posts: 11
Thanks
• Posts: 11
Hopefully I don't have anymore problems with this
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# 93.42 kg to lbs - 93.42 kilograms to pounds
Do you need to know how much is 93.42 kg equal to lbs and how to convert 93.42 kg to lbs? Here it is. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to highlight that all this article is devoted to one amount of kilograms - this is one kilogram. So if you want to learn more about 93.42 kg to pound conversion - keep reading.
Before we go to the more practical part - this is 93.42 kg how much lbs calculation - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 93.42 kg to lbs? 93.42 kilograms it is equal 205.9558451604 pounds, so 93.42 kg is equal 205.9558451604 lbs.
## 93.42 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, in formal International System of Units (in abbreviated form SI).
Sometimes the kilogram could be written as kilogramme. The symbol of the kilogram is kg.
Firstly the kilogram was defined in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but difficult to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was replaced by another definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 93.42 kilogram to pounds
You know a little bit about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to underline that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Of course, this unit is in use also in another systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is just equal 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 93.42 kg?
93.42 kilogram is equal to 205.9558451604 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 93.42 kg in lbs
The most theoretical section is already behind us. In next section we will tell you how much is 93.42 kg to lbs. Now you learned that 93.42 kg = x lbs. So it is time to know the answer. Have a look:
93.42 kilogram = 205.9558451604 pounds.
It is an accurate result of how much 93.42 kg to pound. You can also round it off. After rounding off your outcome will be as following: 93.42 kg = 205.524 lbs.
You learned 93.42 kg is how many lbs, so have a look how many kg 93.42 lbs: 93.42 pound = 0.45359237 kilograms.
Naturally, this time you can also round off the result. After rounding off your outcome will be as following: 93.42 lb = 0.45 kgs.
We also want to show you 93.42 kg to how many pounds and 93.42 pound how many kg outcomes in charts. Look:
We want to begin with a table for how much is 93.42 kg equal to pound.
### 93.42 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
93.42 205.9558451604 205.5240
Now see a table for how many kilograms 93.42 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
93.42 0.45359237 0.45
Now you learned how many 93.42 kg to lbs and how many kilograms 93.42 pound, so it is time to go to the 93.42 kg to lbs formula.
### 93.42 kg to pounds
To convert 93.42 kg to us lbs a formula is needed. We are going to show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 205.9558451604 outcome in pounds
The first version of a formula give you the most accurate outcome. Sometimes even the smallest difference could be considerable. So if you need an accurate result - first formula will be the best for you/option to convert how many pounds are equivalent to 93.42 kilogram.
So move on to the second version of a formula, which also enables calculations to know how much 93.42 kilogram in pounds.
The second formula is as following, look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second version is simpler. It can be the best choice if you need to make a conversion of 93.42 kilogram to pounds in easy way, for example, during shopping. You only have to remember that final outcome will be not so correct.
Now we want to show you these two formulas in practice. But before we are going to make a conversion of 93.42 kg to lbs we want to show you easier way to know 93.42 kg to how many lbs without any effort.
### 93.42 kg to lbs converter
An easier way to learn what is 93.42 kilogram equal to in pounds is to use 93.42 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on longer formula which we gave you above. Due to 93.42 kg pound calculator you can effortless convert 93.42 kg to lbs. Just enter number of kilograms which you need to calculate and click ‘convert’ button. The result will be shown in a flash.
So try to calculate 93.42 kg into lbs with use of 93.42 kg vs pound converter. We entered 93.42 as an amount of kilograms. Here is the outcome: 93.42 kilogram = 205.9558451604 pounds.
As you see, this 93.42 kg vs lbs converter is intuitive.
Now we are going to our main issue - how to convert 93.42 kilograms to pounds on your own.
#### 93.42 kg to lbs conversion
We are going to start 93.42 kilogram equals to how many pounds conversion with the first formula to get the most correct result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 205.9558451604 the outcome in pounds
So what have you do to know how many pounds equal to 93.42 kilogram? Just multiply amount of kilograms, in this case 93.42, by 2.20462262. It gives 205.9558451604. So 93.42 kilogram is exactly 205.9558451604.
It is also possible to round it off, for instance, to two decimal places. It gives 2.20. So 93.42 kilogram = 205.5240 pounds.
It is time for an example from everyday life. Let’s convert 93.42 kg gold in pounds. So 93.42 kg equal to how many lbs? As in the previous example - multiply 93.42 by 2.20462262. It is exactly 205.9558451604. So equivalent of 93.42 kilograms to pounds, when it comes to gold, is equal 205.9558451604.
In this example you can also round off the result. This is the outcome after rounding off, this time to one decimal place - 93.42 kilogram 205.524 pounds.
Now we can move on to examples calculated using short formula.
#### How many 93.42 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 205.524 the result in pounds
So 93.42 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, this time 93.42, by 2.2. Let’s see: 93.42 * 2.2 = 205.524. So 93.42 kilogram is equal 2.2 pounds.
Make another conversion using this formula. Now convert something from everyday life, for example, 93.42 kg to lbs weight of strawberries.
So calculate - 93.42 kilogram of strawberries * 2.2 = 205.524 pounds of strawberries. So 93.42 kg to pound mass is exactly 205.524.
If you learned how much is 93.42 kilogram weight in pounds and can calculate it using two different formulas, let’s move on. Now we are going to show you all outcomes in tables.
#### Convert 93.42 kilogram to pounds
We are aware that results presented in charts are so much clearer for most of you. We understand it, so we gathered all these results in tables for your convenience. Thanks to this you can quickly make a comparison 93.42 kg equivalent to lbs outcomes.
Let’s start with a 93.42 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
93.42 205.9558451604 205.5240
And now let’s see 93.42 kg equal pound chart for the second formula:
Kilograms Pounds
93.42 205.524
As you see, after rounding off, if it comes to how much 93.42 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Remember it when you want to do bigger number than 93.42 kilograms pounds conversion.
#### How many kilograms 93.42 pound
Now you learned how to calculate 93.42 kilograms how much pounds but we will show you something more. Are you interested what it is? What do you say about 93.42 kilogram to pounds and ounces conversion?
We are going to show you how you can convert it step by step. Begin. How much is 93.42 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 93.42, by 2.20462262. So 93.42 * 2.20462262 = 205.9558451604. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 93.42 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your outcome will be equal 2 pounds and 33 ounces.
As you can see, conversion 93.42 kilogram in pounds and ounces simply.
The last conversion which we want to show you is conversion of 93.42 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we give you it, see:
• 93.42 kilograms meters = 7.23301385 foot pounds,
• 93.42 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 93.42 foot pounds to kilograms meters you have to multiply 93.42 by 0.13825495. It is 0.13825495. So 93.42 foot pounds is exactly 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 93.42 foot pounds will be equal 0.14 kilogram meters.
We hope that this conversion was as easy as 93.42 kilogram into pounds conversions.
This article is a huge compendium about kilogram, pound and 93.42 kg to lbs in calculation. Thanks to this conversion you learned 93.42 kilogram is equivalent to how many pounds.
We showed you not only how to make a conversion 93.42 kilogram to metric pounds but also two another conversions - to check how many 93.42 kg in pounds and ounces and how many 93.42 foot pounds to kilograms meters.
We showed you also another solution to do 93.42 kilogram how many pounds conversions, this is using 93.42 kg en pound converter. It is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you are able to do 93.42 kilogram equal to how many pounds conversion - on your own or with use of our 93.42 kgs to pounds calculator.
So what are you waiting for? Let’s calculate 93.42 kilogram mass to pounds in the best way for you.
Do you want to make other than 93.42 kilogram as pounds calculation? For example, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so simply as for 93.42 kilogram equal many pounds.
### How much is 93.42 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 93.42 kg in pounds , we prepared for you an additional section. Here you can see all you need to know about how much is 93.42 kg equal to lbs and how to convert 93.42 kg to lbs . Let’s see.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 93.42 kg to pound conversion formula look? . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 93.42 kilogram to pounds? The correct answer is 205.9558451604 lbs.
There is also another way to calculate how much 93.42 kilogram is equal to pounds with another, easier version of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So in this case, 93.42 kg equal to how much lbs ? The result is 205.9558451604 pounds.
How to convert 93.42 kg to lbs in just a moment? It is possible to use the 93.42 kg to lbs converter , which will do the rest for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
93.01 kg to lbs = 205.052 93.02 kg to lbs = 205.074 93.03 kg to lbs = 205.096 93.04 kg to lbs = 205.118 93.05 kg to lbs = 205.14 93.06 kg to lbs = 205.162 93.07 kg to lbs = 205.184 93.08 kg to lbs = 205.206 93.09 kg to lbs = 205.228 93.1 kg to lbs = 205.25 93.11 kg to lbs = 205.272 93.12 kg to lbs = 205.294 93.13 kg to lbs = 205.316 93.14 kg to lbs = 205.339 93.15 kg to lbs = 205.361 93.16 kg to lbs = 205.383 93.17 kg to lbs = 205.405 93.18 kg to lbs = 205.427 93.19 kg to lbs = 205.449 93.2 kg to lbs = 205.471 93.21 kg to lbs = 205.493 93.22 kg to lbs = 205.515 93.23 kg to lbs = 205.537 93.24 kg to lbs = 205.559 93.25 kg to lbs = 205.581
93.26 kg to lbs = 205.603 93.27 kg to lbs = 205.625 93.28 kg to lbs = 205.647 93.29 kg to lbs = 205.669 93.3 kg to lbs = 205.691 93.31 kg to lbs = 205.713 93.32 kg to lbs = 205.735 93.33 kg to lbs = 205.757 93.34 kg to lbs = 205.779 93.35 kg to lbs = 205.802 93.36 kg to lbs = 205.824 93.37 kg to lbs = 205.846 93.38 kg to lbs = 205.868 93.39 kg to lbs = 205.89 93.4 kg to lbs = 205.912 93.41 kg to lbs = 205.934 93.42 kg to lbs = 205.956 93.43 kg to lbs = 205.978 93.44 kg to lbs = 206 93.45 kg to lbs = 206.022 93.46 kg to lbs = 206.044 93.47 kg to lbs = 206.066 93.48 kg to lbs = 206.088 93.49 kg to lbs = 206.11 93.5 kg to lbs = 206.132
93.51 kg to lbs = 206.154 93.52 kg to lbs = 206.176 93.53 kg to lbs = 206.198 93.54 kg to lbs = 206.22 93.55 kg to lbs = 206.242 93.56 kg to lbs = 206.264 93.57 kg to lbs = 206.287 93.58 kg to lbs = 206.309 93.59 kg to lbs = 206.331 93.6 kg to lbs = 206.353 93.61 kg to lbs = 206.375 93.62 kg to lbs = 206.397 93.63 kg to lbs = 206.419 93.64 kg to lbs = 206.441 93.65 kg to lbs = 206.463 93.66 kg to lbs = 206.485 93.67 kg to lbs = 206.507 93.68 kg to lbs = 206.529 93.69 kg to lbs = 206.551 93.7 kg to lbs = 206.573 93.71 kg to lbs = 206.595 93.72 kg to lbs = 206.617 93.73 kg to lbs = 206.639 93.74 kg to lbs = 206.661 93.75 kg to lbs = 206.683
93.76 kg to lbs = 206.705 93.77 kg to lbs = 206.727 93.78 kg to lbs = 206.75 93.79 kg to lbs = 206.772 93.8 kg to lbs = 206.794 93.81 kg to lbs = 206.816 93.82 kg to lbs = 206.838 93.83 kg to lbs = 206.86 93.84 kg to lbs = 206.882 93.85 kg to lbs = 206.904 93.86 kg to lbs = 206.926 93.87 kg to lbs = 206.948 93.88 kg to lbs = 206.97 93.89 kg to lbs = 206.992 93.9 kg to lbs = 207.014 93.91 kg to lbs = 207.036 93.92 kg to lbs = 207.058 93.93 kg to lbs = 207.08 93.94 kg to lbs = 207.102 93.95 kg to lbs = 207.124 93.96 kg to lbs = 207.146 93.97 kg to lbs = 207.168 93.98 kg to lbs = 207.19 93.99 kg to lbs = 207.212 94 kg to lbs = 207.235
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Overview of the Student[MultivariateCalculus] Subpackage - Maple Programming Help
Home : Support : Online Help : Education : Student Packages : Multivariate Calculus : Student/MultivariateCalculus
Overview of the Student[MultivariateCalculus] Subpackage
Calling Sequence Student[MultivariateCalculus][command](arguments) command(arguments)
Description
• The Student[MultivariateCalculus] subpackage is designed to help teachers present and students understand the basic material of a standard course in multivariate calculus. There are two principal components to the subpackage: interactive and visualization. These components are described in the following sections.
• Each command in the Student[MultivariateCalculus] subpackage can be accessed by using either the long form or the short form of the command name in the command calling sequence.
As the underlying implementation of the Student[MultivariateCalculus] subpackage is a module, it is also possible to use the form Student[MultivariateCalculus]:-command or Student:-MultivariateCalculus:-command to access a command. For more information, see Module Members.
• The Maple Command Completion facility is helpful for entering the names of Student package commands.
• Many of the commands and tutors in the Student[MultivariateCalculus] package can be accessed through the context-menu. These commands are consolidated under the Student[MultivariateCalculus] name.
• Note: Though some of the commands in this package may return complex values, it is assumed that the user is working with the calculus of real valued functions of real variables.
Visualization
• The visualization routines are designed to assist in the understanding of basic multivariate calculus concepts, theorems, and computations. These routines normally produce a Maple plot and most can optionally return one or more symbolic representations of the studied quantity.
• You have considerable control over the presentation of plots produced by the visualization routines. The display of each object included in the plot can be adjusted by using a corresponding option in the calling sequence. For details, see the individual command help pages.
• For more information on this functionality, see Student[MultivariateCalculus][VisualizationOverview].
The visualization commands are:
Interactive
• The interactive routines use Maple's Maplet technology to assist you to work through the standard problems of multivariate calculus in a visually directed manner. These commands display one or more dialog boxes allowing you to plot a function and change the various plot options.
For more information on this functionality, see Student[MultivariateCalculus][InteractiveOverview].
The interactive commands are:
Computation
The computation routines provide tools that perform standard multivariate calculus operations.
For more information on this functionality, see Student[MultivariateCalculus][ComputationOverview].
The computation commands are:
Lines and Planes
The Student[MultivariateCalculus] package also contains routines for working with lines and planes in two and three dimensions. For more information on this functionality, see Student[MultivariateCalculus][Line] and Student[MultivariateCalculus][Plane].
The commands for lines and planes are:
Getting Help with a Command in the Package
To display the help page for a particular Student[MultivariateCalculus] command, see Getting Help with a Command in a Package.
Example Worksheet
For introductory examples, see MultivariateCalculus Example Worksheet.
Applications
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Engineering Courses
Chapter 2: Engineering Mathematics Exam Tests
Engineering Mathematics MCQs - Chapter 2
# First Order Ordinary Differential Equations Multiple Choice Questions (MCQs) PDF Download - 1
Books:
Apps:
The First Order Ordinary Differential Equations Multiple Choice Questions (MCQs) with Answers PDF, First Order Ordinary Differential Equations MCQs PDF Download e-Book Ch. 2-1 to study Engineering Mathematics Course. Practice Homogeneous And InHomogeneous Differential Equations MCQs, First Order Ordinary Differential Equations trivia questions and answers PDF for online engineering programs. The First Order Ordinary Differential Equations MCQs App Download: Free learning app for concepts of solution, seperation of variables, number types career test to learn online courses.
The Multiple Choice Question (MCQ Quiz): Linear differential equation of any order if there is no constant term is present is known as; "First Order Ordinary Differential Equations" App Download (Free) with answers: Non homogeneous ODE; Homogeneous ODE; Partial ODE; Nonlinear ODE; for online engineering programs. Solve General Laplace transform Examples Quiz Questions, download Google eBook (Free Sample) for job assessment test.
## First Order Ordinary Differential Equations Questions and Answers : MCQ Quiz 1
MCQ 1:
Linear differential equation of any order if there is no constant term is present is known as
1. Homogeneous ODE
2. Non homogeneous ODE
3. Partial ODE
4. nonlinear ODE
MCQ 2:
Interval used to indicate the interval of all integers between a and b is called
1. infinite interval
2. closed interval
3. integer interval
4. open interval
MCQ 3:
Separation of variables method is also termed as
1. Laplace method
2. Fourier method
3. Len method
4. Gauss method
MCQ 4:
F(x)=-c/b is a
1. Homogeneous solution
2. Non homogeneous solution
3. Partial solution
4. nonlinear solution
MCQ 5:
A value that represents a quantity along a line is
1. natural number
2. whole number
3. real number
4. transfer number
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The Devil 30-gon (Posted on 2018-04-17)
If A1, A2, A3, . . . , A30 are the vertices of a regular 30-gon inscribed in a unit circle, then find
|A1A2|2 + |A1A3|2 + |A1A4|2 + ... + |A1A30|2
No Solution Yet Submitted by Danish Ahmed Khan No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
solution Comment 3 of 3 |
My trig is pretty sketchy, pun intended, but here goes.
Using the law of cosines and assuming a unit circle with 30-gon side s, s^2 = 2-2cos(x) where x is angle formed by the two radii/legs.
Each of the 30 distance squares can be paired with one 180 degrees away. Since cos(x) = -cos(x+180), all the cosine sums cancel and we're left with 30*2 = 60 as the desired answer.
Posted by xdog on 2018-04-17 19:03:54
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# 49 Yard to Decimeters (49 yd to dm)
Convert 49 Yard to Decimeters (yd to dm) with our conversion calculator and conversion tables. To convert 49 yd to dm use direct conversion formula below.
49 yd = 448.056 dm.
You also can convert 49 Yard to other Length (popular) units.
49 YARD
=
448.056 DECIMETERS
Direct conversion formula: 1 Yard / 9.144 = 1 Decimeters
Opposite conversion: 49 Decimeters to Yard
Check out conversion of 49 yd to most popular length units:
49 yd to Kilometers
49 yd to Millimeters
49 yd to Miles
49 yd to Centimeter
49 yd to Inches
## Conversion table: Yard to Decimeters
YARD DECIMETERS
1 = 9.144
2 = 18.288
3 = 27.432
4 = 36.576
5 = 45.72
7 = 64.008
8 = 73.152
9 = 82.296
10 = 91.44
DECIMETERS YARD
1 = 0.10936132983377
2 = 0.21872265966754
3 = 0.32808398950131
4 = 0.43744531933508
5 = 0.54680664916885
7 = 0.7655293088364
8 = 0.87489063867017
9 = 0.98425196850394
10 = 1.0936132983377
## Nearest numbers for 49 Yard
YARD DECIMETERS
96.25 yd = 880.11 dm
145 yd = 1325.88 dm
150 yd = 1371.6 dm
182.9 yd = 1672.4376 dm
230 yd = 2103.12 dm
250 yd = 2286 dm
256 yd = 2340.864 dm
302 yd = 2761.488 dm
322.7 yd = 2950.7688 dm
328 yd = 2999.232 dm
350 yd = 3200.4 dm
368 yd = 3364.992 dm
548.3 yd = 5013.6552 dm
726.14 yd = 6639.82416 dm
865 yd = 7909.56 dm
1000 yd = 9144 dm
1582 yd = 14465.808 dm
3600 yd = 32918.4 dm
4413 yd = 40352.472 dm
4928 yd = 45061.632 dm
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# Doom Engine Vertical Stretch Formula
## Recommended Posts
The Doom engine stretches everything on screen vertically. What i'm looking for is some kind of percentage that it is stretched.
I need to make some kind of formula to figure out how much stretch (or vertical squish rather) I need for my 3d Models so i don't have to eyeball it every time I'm putting in a new model.
I have squash/stretch control over "depth, height, width". So "1, 1, 1" will end up looking stretched because the doom (and any source port) engine is vertically challenged and to fix this I would do something close to "1, 0.8, 1". Simple, but i'm not mathemagical so the moment I need to change the depth or width I have to spend a painful amount of time eyeballing it to find an approximate height that looks right.
Is there a formula out there for this already?
##### Link to post
Not a formula or a proper answer, but this might be of some slight use to you
##### Link to post
The Doom engine originally displayed 16:10 content (320 / 200) on a 4:3 display. Try .833333 for your height scale.
##### Link to post
Or 1.2 if you're moving the other direction.
Here are some examples. You make a 10x10 graphic and then ask the Doom engine to display that. It does so dutifully but while transforming the internal 16:10 logical framebuffer into a 4:3 display. The vertical scale-up is a factor of 1.2 so you end up displaying what looks like 10x12 instead. 10 x 1.2 == 12.
So now, we want to pre-correct for this so that, say, a round circle actually looks round and not like an oval. The number Bloodshedder gave was the inverse of 1.2 (0.8333...)
10 x 0.8333.... = 8.3333... - so your graphic will need to be 10x8 to come out looking close to round. Getting it exact in this case is impossible since you can't have a third of a pixel.
##### Link to post
Ok so if i need to change the overall scale and keep the ratio, how do I figure the vertical squish/stretch
For example, if i needed to squish the width and depth to 0.52 (.52, .?, .52), how would I figure the vertical squish? Sorry, i'm mathematically challenged.
##### Link to post
Since you're using models, I figure you're using an advanced port. You might be able to specify the vertical scaling to apply to the actor, which will then compensate aspect ratio correction. I know it's possible in GZDoom; I'm not familiar enough to tell you about Doomsday/Risen3D, but if it's not possible already I'm sure it could be requested from the developers.
##### Link to post
I am talking about scaling the actor within the Doomsday Engine. Sorry, I neglected to mention that.
The formula i'm trying to come up with (or have other people come up with rather) is one i would use to scale my models within the Doomsday Engine Definitions.
I didn't include that info because it is more of a math question than anything else.
##### Link to post
.52 x
---- = -------
1 0.83333
Solve for x (cross multiply)
##### Link to post
Excellent! Thank you.
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# Older blog entries for sab39 (starting at number 32)
Privacy
I often find myself working on my laptop on the train, or in another public place. Sometimes I'd like to spend that time writing things that I'd prefer not to be read over my shoulder by random fellow travelers. It occurred to me that this goal could probably be attained with a utility that would rot13 all text on a tty (my laptop is text-mode-only due to being a 486 with 4Mb RAM :) ).
However, my searches for software to do this have come up empty so far. I'd even settle for a VIM mode that would accept keystrokes normally, but display everything rot13'd - everything I'm concerned about right now would be in VIM. Is it really true that nobody's come up with a "conceal what you're doing" rot13 program?
Chicago: You can do much better than the three-fold increase in filesize that your solution implies. Off the top of my head, I can see an immediate way to increase the filesize by just twofold-plus-a-small-constant with almost the same benefit as your solution. Since this is entirely thought up in a couple of minutes by someone who's not well-versed in error correction mathematics, I guarantee it's possible to do much better than my solution, too. So don't go ahead and implement my solution - rather take the existence of my solution as a heads-up that there are a lot of ways to improve, and read some literature on the subject to find out what the state of the art is. I'd be interested to know myself how close I was able to get to that!
First, represent a as 10 and b as 01. By using this technique, you can identify any single-bit error, but not correct it - if you get 11, you know that's wrong, but you can't tell whether it's supposed to be a or b. (Choosing 01 and 10 rather than 00 and 11 protects against the kind of error where the channel is dead and always reports zeros, or ones). After reading this full stream, the recipient will have a result that contains three states: definitely a, definitely b, or unknown (we deliberately ignore the possibility of getting errors in two consecutive bits; we can catch that case later).
Then, after transmitting your entire stream, send a checksum of the correct data. Use a well-known checksumming algorithm like md5 that is known to have good properties, rather than rolling your own, which almost certainly won't work as well. Transmit the checksum using the same 10 / 01 encoding as the rest of the content.
Here's how the recipient of the data can figure out the corrected data from the original stream in conjunction with the checksum: Assume each "definitely known" bit is correct, but try both values for each unknown bit (in both the data and the checksum). Calculate the checksum of the data and see if it matches the checksum that was sent. This algorithm is O(2^n) in the number of "unknowns" found, but that number is expected to be low. If the number is too high to calculate in a reasonable amount of time, the transmission can be flagged as corrupt and the user can try again. If all has gone well, though, there should be exactly one combination of values for the unknown bits that leads to the checksum matching. Voila, you've now got your correct sequence of 'a's and 'b's.
There are a couple of ways that this can fail to give a corrected stream, but even those cases can be identified. One is mentioned above - too many "unknowns". Another is if there were errors in two consecutive bits, turning a b into a "definite a". That situation will be flagged because no combination of values for the "unknowns" will result in a matching checksum. The other possibility is the astronomical possibility that an incorrect sequence of 'a's and 'b's would generate a correct checksum. With a good checksumming algorithm, that should be absolutely impossible without also producing a too-large number of errors (and hence "unknowns"), so the transfer would be flagged as corrupt anyway.
As an Englishman living in the US, I'm double-gutted today... I felt that if an England-US final was still possible by today, then it would be the most likely outcome. So much for that idea.
Sing along with me:
We're going home, we're going -- England's going home
We're going home, we're going home, we're going -- England's going home
We're going home, we're going home, we're going -- England's going home
We're going home, we're going home, we're going -- England's going home
Everyone seems to know the line
They say it every ti-i-ime
They just know
They're so sure
That this time we could really make it
We could be a big hit
But I know that we're shit
Cause I remember
Three lions on the shirt
Awful refereeing
Let the Hand of God
Send us home all seething
So many hopes so many dreams
But all those England tea-ea-eams
Couldn't break
With the theme:
Cause I still see those penalties missed
And the refs that we hissed
And how we were all pissed
At England losing
Three lions on the shirt
Another fricking shootout
Beckam lying hurt
Had to get the boot out
Yeah I know you believe
But don't be naive...
Three lions on the shirt
Two Brazilians scoring
When the final comes
I will still be snoring
We're going home, we're going home, we're going -- England's going home
25 Apr 2002 (updated 25 Apr 2002 at 18:16 UTC) »
bjf: See my past diary entry for an idea for a conceptually simple but not-yet-written application. At least, I've never been able to find an existing implementation...
This might also be interesting to habes because it sounds like something that Audacity might be able to implement extremely easily. Or be done using a plugin to Audacity, if such a mechanism exists. It's basically a problem in user-interface design with respect to sounds, and it seems like Audacity has the bulk of that problem solved.
[updated to split paragraphs and to make a link to the relevant past diary entry]
D'oh!
Forgot to grab my laptop on the way out of the door. That's two train-traveling hours of potential japitools hacking that turned into sitting around doing nothing.
Life
dyork: Congratulations! My first son or daughter (uncooperative little thing wouldn't tell us which) is due June 4th. Been busily painting the room and assembling cribs, etc...
Gateway supports my right to enjoy digital music legally. Isn't that nice of them? Anyone know if their computers are any good? I'd like to recommend an ethical company to friends who ask about buying computers, but I don't want to reccommend a crappy computer...
I guess I'll just keep doing my part by listening to the dude and the cow singing their song...
Umm...
That seems to be everything I can think of to say right now. Huh.
Nothing
Nothing much of note has happened with regard to free software, which is why I haven't posted any diary entries despite having advodiary to make it painless. Of course, "nothing to do with free software" doesn't mean nothing at all. I've been busy painting and assembling furniture for the baby's room, and with my "real" job. I'm not sure how appropriate that kind of discussion is for this forum; I know that I don't mind seeing it in other people's diaries, but I do tend to just skim over it. But it's more important to my life than japitools is, so I guess posting a little bit about it even though it's technically offtopic is okay.
japitools
I've been working mostly on revamping the first pass of japitools, which is the Japize java program. I've revamped the algorithms for iterating over the classes to ensure that the output is sorted in strict alphabetical order with an exception for java.lang.Object. I also rethought the commandline options and added the capability for Japize to dump it's output directly into a .gz file by using Java's GZIPOutputStream class. Finally, I wrote a perl wrapper called japize to (a) allow running the program as japize rather than as "java net.wuffies.japi.Japize", and (b) check the program's arguments for invalid syntax early, so that you don't have to wait for a JVM to start up just to get told that you misspelled "packages".
In the process I discovered a few issues in the jode.bytecode library that Japize relies on for loading classes from zipfiles. I mailed the Jode author about them, and it turns out that the latest Jode in CVS already solves all my problems. Although this code hasn't been publically released yet, I think I'll definitely be using it once it is: it will allow me to vastly simplify several places where I'm currently fighting against the Jode API rather than working with it.
japitools - mind-numbing details
I just need to make a few more tweaks to Japize, and then it will be time to move on to pass 2: the japicompat perl script that tests APIs for compatibility with each other. Aside from making this use a constant and bounded amount of memory (instead of O(size-of-the-JDK-measured-in-methods), which is pretty huge and growing exponentially every release, it seems) I want to add some more options, like comparing bidirectionally instead of unidirectionally, more generalized filtering of the output, and machine-readable error output instead of human-readable. This last change is in order to support the new planned third pass - japifilter.
japifilter will perform the part of the japicompat process that couldn't be done as part of the second pass without using O(N) memory, optionally translate the errors to humanly-readable, and add new options like the ability to exclude errors that appear in another file. That will enable me to say "Show all errors between kaffe and jdk1.1 except for those that also appear between jdk1.2 and jdk1.1" - so I can evaluate kaffe's progress to full 1.1 compatibility without false positives for parts of kaffe that are already jdk1.2 compatible instead.
4 Apr 2002 (updated 4 Apr 2002 at 18:27 UTC) »
cmiller: A nice enhancement to advodiary would be to use a temporary filename that ends in .html, to trigger the correct vim syntax hilighting mode. :set syntax=html works well in the meantime, though.
japitools
A recent post on the GNU Classpath mailing list set me thinking about japitools again. The problem with it has been that when faced with an API as mind-bogglingly huge as JDK1.2 and above, it starts swapping to hell even on a well-endowed system. And of course dies entirely on my trusty 486DX-4MB laptop, which is the only place I can take the time to actually develop it. This problem had led me to get discouraged and demotivated with it, so I responded to the Classpath post advocating its use but also angling for a possible new maintainer.
It turns out that after 2 years of demotivation this got me thinking about the problem again, and I figured out some algorithms that would make it possible to operate on gargantuan APIs without any use of O(N) memory. It came down to a few key observations:
• Comparing two lists can be done in O(1) space if you can guarantee that both lists are sorted identically. By imposing a strict ordering requirement on .japi files, I can make a minimal japicompat that uses O(1) space and straight O(N) time.
• You can iterate through a hierarchical structure in order in O(Depth*NumItemsInAnyGivenDirectory) space, by loading the list of subdirectories, sorting it, and then recursively processing each one in the same way. This means that Japize, the tool that produces .japi files, can satisfy the strict ordering requirement without requiring O(N) space itself. Also, this nice feature applies even when combining multiple sources (directories and zip files, in my case) of hierarchy information. It does cause multiple redundant scans through the same zipfile index, but that doesn't take any memory. And on systems that do have memory to spare, the disk cache will ensure that the physical disk only gets read once anyway.
• There's no need to use the same ordering through every stage of the processing. The japicompat first pass requires an alphabetical ordering because hierarchical information might be inconsistent between the two files. However, a filtering pass to eliminate duplicate errors requires a hierarchical sort to keep memory usage in O(NonDuplicateErrors) as opposed to O(ErrorsIncludingDuplicates) which is a much bigger number and can be O(N) in some common situations (exactly what we're trying to avoid). These conflicting needs can both be met by simply applying a sorting pass in between, rather than (as was previously done) keeping all errors in memory and filtering from that (possibly O(N) sized) data structure at the end.
• Conventional wisdom about sorting doesn't apply to a partial order. Conventional wisdom says that sorting a list of length N requires at least O(N) space anyway. But the only requirement for the filtering pass is that all superclasses appear prior to their subclasses. This sort can be accomplished by simply counting the number of superclasses, and performing multiple passes through the data writing out the classes with 0 superclasses, then 1, then 2, etc. This can be optimized even further through creating a temporary file for each superclass-count, and then concatenating them all together. As above, I can rely on the disk cache to help out systems where all N items really can fit into memory. This optimization uses O(N) space again, but on disk rather than in memory. And in nicely ordered writes, rather than random-access page swapping.
• Domain-specific knowledge can help in optimization. I happen to know that the only class with zero superclasses is java.lang.Object. Knowing this, if I can arrange that java.lang.Object is already first, I can save myself a pass. Since the requirement for the first pass was only "I must know for sure that both lists are sorted the same", I can alter the original sort to enforce this rule.
Armed with these tricks, it seems that I can completely avoid O(N) memory usage in every stage of japitools. I still have to actually implement it, but it seems that my call for a new maintainer was premature: I've found myself some motivation again. Always nice when that happens.
3 Apr 2002 (updated 3 Apr 2002 at 15:45 UTC) »
Maybe having advodiary to use will help me update more frequently... I've wanted to make more timely entries for a while...
"Killer App"
tk: I had read your last diary entry out of context, in that it mentioned wanting a "killer app" but didn't mention Delitalk explicitly. So the suggestion was just a "killer app" in general, not a killer app for Delitalk explicitly. Sorry.
Just for posterity's sake, I'll post the idea here also...
I have a number of cassette tapes that I don't have CD equivalents to, and most of these I simply cannot get CD equivalents to for any amount of money - either because the tapes pre-dated CDs being widely available, the band has split up and CDs are no longer available, or the tape was made by a "friend of a friend of a friend"'s band who I am no longer able to get in touch with (and who probably don't have the resources to make a CD anyway).
But I'd still like to be able to rip and encode these cassettes into digital form (Ogg, of course). It's relatively easy to use sox or equivalent to record the whole of one side of a cassette into a single wav file (using a walkman and a headphone-out-to-line-in cable). But the difficult part is then going through the cassette and identifying where one track starts and the next ends, and encoding those parts into ogg individually (and, of course, ID3 tagging them and auto-creating a playlist in the same way that abcde does for CDs).
My theory is that it should be possible to hook into xmms or something and let the user use xmms's position control to jump to the end of each track, hit a button indicating "track 1 ends and track 2 starts here" and, after repeating that process for all the tracks and typing in the track (and artist for multi-artist tapes) name, generate an ogg file for each track and a playlist.
Bonus points for making it easy to process two sides of a cassette into a single playlist, applying some kind of filter to the sound to eliminate tape hiss, and for coming up with an inventive UI for noting the exact point in the sound where one track ends and another begins. You could also support cases where tracks overlap slightly (eg the band starts playing the next song before the last chord of the previous one has died down) or where there's some dead space between tracks that you want to skip entirely.
Personal
Life is good...
mjg59: Interesting the way cam.ac.uk people are instantly recognizable by the format of their preferred nicks :)
Shame the RH/AOL merger rumours are apparently false, because I just thought of this AOLinux one-liner:
You've got mail(1)!
23 older entries...
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Short Quiz in Math 6 performs two or more different operations on whole numbers with or without exponents and grouping symbols
Mathematics
Feel free to use or edit a copy
includes Teacher and Student dashboards
M6NS-IIf-149
Track each student's skills and progress in your Mastery dashboards
• edit the questions
• save a copy for later
• start a class game
• automatically assign follow-up activities based on students’ scores
• assign as homework
• share a link with colleagues
• print as a bubble sheet
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
5 questions
• Q1
Solve 2 x [5 + (3 x 2)]
26
22
60
5
30s
M6NS-IIf-149
• Q2
Where would you insert parentheses to make the equation true? 7 + 8 x 3 + 45
7 + 8 x (3 + 45)
(7 + 8) x 3 + 45
7 + (8 x 3 + 45)
7 + (8 x 3) + 45
30s
M6NS-IIf-149
• Q3
Solve: 100 ÷ (10 - 8) + 12
14
10
62
30s
M6NS-IIf-149
• Q4
Write the numerical expression for the sum of 7 and 5 divided by 3.
(7 + 5) ÷ 3
3 ÷ 7 - 5
3 ÷ 7 + 5
(7 - 5) ÷ 3
30s
M6NS-IIf-149
• Q5
Write a statement for this numerical expression: 6 + 24 ÷ 4
6 minus 24 divided by 4
four divided by the difference of 6 and 24
6 added to the quotient of 24 and 4
30s
M6NS-IIf-149
Teachers give this quiz to your class
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# Dog Age Calculator: Dog Years to Human Years
## Dog Years: How Old is Your Dog?
Pixabay
How old is your dog in human years? Multiplying your dog’s age by seven is easy, but isn’t very accurate.
This carefully graded system piles the equivalent human years onto a dog’s life more quickly during the dog’s rapid growth to maturity.
If we think like a dog, here’s how a dog’s age compares to a human’s age! There may be some differences between breeds, but this should give you a good sense of where your dog is in the development/aging process.
### Dog Years to Human Years Chart
Dog Years
Human Years
#### (dog’s age in equivalent human years, based on stage of development/aging)
6 months 10 years
1 year 15
2 years 24
3 28
4 32
5 36
6 40
7 44
8 48
9 52
10 56
11 60
12 64
13 68
14 72
15 74 ½
16 77
17 79 ½
18 82
19 84 ½
20 87
21 89 ½
22 92
23 94 ½
24 97
25 99 ½
26 102
27 104 ½
28 107
29 109 ½
30 112
Is your dog starting to enter its elderly years? Get some tips for caring for an old pup. Also, check our home remedies for your pet’s arthritis.
Ready to congratulate your dog for their birthday? Try out these yummy pet treat recipes.
## What do you want to read next?
### You have the terminology wrong
I don't understand why you have dog years and human years the way you do. A human year equates one year. When we use the terminology of dog year it implies the equivalent in dog years, not "human years" as you state.
### dog years
We understand your point, and it can be confusing. “Dog Years” in this case is the actual dog’s age, such as would be on a veterinarian chart (tied into Gregorian calendar years). “Human Years” is the equivalent age in human years that matches the stage in development of the dog at the specified dog age (the one on the veterinarian chart). So, for example, because a dog ages faster than a human, a 1-year-old dog according to the Gregorian calendar (dog years) would be the equivalent in age and development of a 15-year-old human (human years). We’ll think about the headings and description to try to make this clearer. Thank you for your feedback!
### PUPPIES
I have a adorable new puppy her name is Bubbles. I love her so much. She is 2.5 months human years but I cant work out how old she is in dog years. Please help me. Thanks.
### puppy age
Unfortunately, our chart does not cover dogs that are less than 6 months in age. Puppies mature rapidly in their first months of development; just how much will depend on the dog breed (size) and other factors. For a very rough estimate, you might use the formula: 6 months dog age/10 years human age = 2.5 months dog age/x years human age, which yields a little over 4 years in human terms. Hope this helps–and congratulations on your new puppy!
### Cats
Does the dog age chart apply to CATS too??
### Cats
No it doesn't, but I sure wish they'd publish a chart that does.
### cat ages
Traditionally, it was said that each year a cat lived was equivalent to having lived 7 human years (so, a cat 3 years old would be 21 years in human terms). However, this turns out to be incorrect. Cats mature faster during their first 2 years. Actually, looking at the dog chart above, it is very similar. The first 2 years of a cat’s life, it is estimated that they have matured to about 24 or 25 years of a human’s life. Then from there on, the cat develops more slowly, every year being the equivalent to about 4 years of a human. Maturity rate, aging, and lifespan of an individual cat will depend on environment, health, breed, etc., but this is a general guideline.
### Dog age
I have a female rat terrier that is 17/79 1\2 years old. Very loyal and loving dog. My new puppy is 4mths old so she isn't quite 15 human years, my Res Boston Terrier is 8 dogs years & 52 human years. All are family dogs. Love them all.
### DOG AGE CALCULATOR: DOG YEARS TO HUMAN YEARS
I've also seen where a pup's first year was equivalent to 15 years old, and subsequent years equaled 7 years...
### My American Pit Bull Lived to
My American Pit Bull Lived to be 15 or 16 years old and died from health problems
### Horses
how do horses age?
### horse ages
The age of a horse equivalent to human age will vary upon many factors, such as breed, etc. In general, though, it is thought that horses age about 6.5 years for each human year until age 4, when it is a young adult of about 20 in human years. After age 4, they age about 2.5 years for every human year.
### My boys
Bruno made it to 48 or more....
Levi's currently 64....
Luv dawgies like they the only real true good in a bad bad world....
Bye
And bye
Thanks
aE
### age
I have two questions
1. If a dog is born 12-5 how old is she??
2. If a dog is born 2-10 how old is she?? Both of this year
### Ginger
Gingersnap was 13.5 yrs approx 110 in her years. She was a shelter dog mix of German shepherd, Chelsea collie, beagle, and husky.
Little on the heavy side about 110-125
One of the best dogs I had, smart, loyal, protective when need in her mind, loved kids, and would kill if you forced your way in the house with out a ok given. She was great miss her after 2001.
### Our dog Lindsay lived to 98
Our dog Lindsay lived to 98 using this formula, not bad for a dog with only 3 legs. We miss you girl.
### Like my dog throws up idk how
Like my dog throws up idk how long he is going to live how do u know when your dog going to die
### Mia Harris-how do you know when your dog is going to die
I realize that it's been almost 2yrs since you posted that your dog was throwing up & you didn't know how much longer he was going to live...I hope you don't mind me asking, but were you ever able to get him to your Vet's office? Did he stop throwing up? If he has passed on, I am VERY sorry for your loss. I pray that he didn't suffer. Please forgive me if I open old wounds. I was merely trying to educate myself. As I too have a fur-son who seems to have a stomach issue. He's always eating grass. But not just any grass...he seeks out certain weeds & very carefully bites off the top of each stalk individually. Do you have any advice. Please & thank you.
Be blessed, Kriss VanOrnum✌☺
### Dog Eating Grass
Hi Kriss,
Whatever grass your dog may be eating, there are a couple of reasons he may be doing it. He could be have stomach aches and trying to relieve the distress by attempting to vomit, eating large quantities of grass to do so. If your dog has not vomited, it could just be regular dog behavior. They explore the world around them with their mouths, eating and chewing things as they go. If vomiting does occur, take your pup to the vet right away.
### Dog throwing up
My 13 year old black lab mix has a problem with throwing up. I took him to the vet and she said that he was having pancreatic problems and gave me pills which helped immediately, but then they stopped working as well. The pills are lifetime, named pannekare.
Good luck.
### My german shepherd is 56yrs
My german shepherd is 56yrs old..Fiona passed away on 21.5.15..will miss her..
RIP MY BABY
### April 26 my pekinesee/cocker
April 26 my pekinesee/cocker spaniel mix will be 83 1/2 . I've had him since he was 5weeks old. Had check up recently vet said he had strong heart that of a puppy. A little arthritis in one hip. For 10 yrs ge ran with my boys for miles through woods. He went every where they did. Now they're grown it's just us 2
### My American Cocker is 60 yrs
My American Cocker is 60 yrs old and still loves to play with his toys.And has a very good appetite.
### My Jack a poo is the exact
My Jack a poo is the exact same age as me! 32 years young!
### My pinkinnise was 4 years old
My pinkinnise was 4 years old are years he would be 32 years Dogie years that a shame the grow so fast I want him to stay a baby . That growing old fast.
### That makes my Collie mix
That makes my Collie mix right around 72. Good life!
### Mines 1 years old.But in
Mines 1 years old.But in human years 15 year old
### Y dog is 36 she is a pary of
Y dog is 36 she is a pary of our family I was there ecacly when she was born so if she ever died I would be so so so so so so so so so sad
### Cool! My pitbull mix is 32
Cool! My pitbull mix is 32 and here i am calling her an old lady ^^. Im 17 and my sog is 32 XD
### My goodness, a mini pin we
My goodness, a mini pin we had was almost 87 when we lost him, We lost him 4 yrs. ago and still miss him
### My dog died after my
My dog died after my birthday. So sad, he is 82 years old.
### we just got a new maltese
we just got a new maltese pomeranian mix puppy we cant decide if we should name it macy or chloe.
### Awww..what a coincidence!!
Awww..what a coincidence!! Our yorkie/miniature schnauzer mix i gave to my niece they named her Mazy..and our Yorkie/Shih tzu mix we got last week we named Chloe.
### My 2 yr. old Goldadore is the
My 2 yr. old Goldadore is the puppiest 24 year old there can be!
### Jus got a pup she is 6 weeks
Jus got a pup she is 6 weeks how old she in human. Yrs
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# Showing that a linear map $\mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3$ is an isomorphism.
Let $t_1,t_2,t_3$ be distinct real numbers and consider the linear map
$$T : \mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3, \quad \quad p(x) \mapsto \begin{pmatrix}p(t_1)\\p(t_2)\\p(t_3)\end{pmatrix}$$
I want to show that $T$ is an isomorphism. Since it is given that $T$ is linear, I just need to show that $T$ is bijective. My initial approach was to solve the system
$$\begin{pmatrix} a + bt_1 + ct_1^2\\ a + bt_2 + ct_2^2\\ a + bt_3 + ct_3^2 \end{pmatrix} = \begin{pmatrix} r_1\\ r_2\\ r_3 \end{pmatrix}$$
for $a,b,c$ in terms of $r_1,r_2,r_3$, thus determining the inverse map $T^{-1}$ mapping a point in $\mathbb{R}^3$ to a polynomial of degree $\leq 2$.
I solved this system using Maple, and got a solution which was defined when $t_1,t_2,t_3$ where not all equal, which is fine by their definition, but I am wondering if there is a nicer argument, especially since the solution is rather ugly.
(this is not homework)
-
Personally, I find the answer by Yuki most intuitive, and understandable. By considering the current votes, it seems like the majority disagrees (though only slightly). I can't find a question on meta about this "dilemma". – utdiscant Aug 14 '12 at 3:02
You should accept the answer you consider most helpful. If anyone stumbles into this question, he will be able to see the one with the top score (or next-to-top score, if the accepted answer is the top-scoring one) right below the accepted one anyway. There are even badges here for having nonaccepted answers with higher scores than accepted ones. ;) Personally, I find Jacob Schlather's suggestion to be the most elegant. – tomasz Aug 14 '12 at 4:47
Hint:
$T$ is onto: Langrange Polynomial.
Injectivity: if $p$ is an polynomial of degree $\le2$, knowing values in 3 distinct points, you know $p$.
-
Just take $1,x,x^2$. They transform into $\begin{pmatrix}1\\ 1 \\ 1\end{pmatrix},\begin{pmatrix}t_1\\ t_2 \\ t_3\end{pmatrix},\begin{pmatrix}t_1^2\\ t_2^2 \\ t_3^2\end{pmatrix}$. To show that they are linearly independent, calculate the determinant of $\begin{pmatrix}1&t_1&t_1^2\\1&t_2& t_2^2 \\1&t_3& t_3^2\end{pmatrix}$, which is, incidentally, Vandermonde matrix.
-
The result heavily used here, and a rather important one, is: an operator $\,T:V\to V\,\,\,,\,\,V\,$ a vector space, is an isomorphism iff the image of some basis is again a basis (or more general: the image of any linearly independent is again lin. indep.) +1 – DonAntonio Aug 14 '12 at 2:30
@DonAntonio: Not really: rather than using the result, I've explicitly shown that it is of rank 3. :) – tomasz Aug 14 '12 at 2:31
Hint: What is in the kernel of your map?
-
I think it may be easier to show that the associated matrix always has non-zero determinant. This is a known result.
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# Re: an old bug on the 85
Yeah, this turns out not to be a bug at all, but has to do with Hermitian
matrices. The Hermitian is a transpose where each element is also replaced
with its complex conjugate.
geoffh
stephen soltesz wrote:
> > I recall less than a year ago someone reporting a bug on the 85
> > involving complex numbers and (this is what I've forgotten) either the
> > dot() or cross() product. I played with both using complex numbers and
> > didn't see any problem.. but need to know what it was or what caused it,
> > so as not to mess any calculations up..
> >
> > So, if anyone knows to what I am referring, or if it was a different
> > function, let me know. I remember it relating to EE majors, for some
> > reason.
> >
> > thankyou.
> > stephen.
> >
>
> As an interesting follow-up to this bug is another manifestation I
> discovered while trying to find an easy workaround.
> I wanted to make two 1x2 matrixes take the first, then transpose the
> second and multiply the two.. 1x2 * 2x1 and get a 1x1 answer.
>
> As it turns out, if you transpose a complex matrix, say [[(a,b),(c,d)]]
> you will get
> [[(a,-b)]
> [(c,-d)]]
> This occurs for any dimension complex matrix MxN.
> Based on this, I'd guess that the vector functions are a subset of the
> matrix operations already in ROM. More specifically the transpose
> function. When someone calls the dot() function the ti pretends you
> have two 1xN matrixes, transposes the second and multiplies the two.
> This is consistent with what I've observed, but I may be confusing the
> correlation with simple causality.
>
> I now have two questions as a result of this. I found four FAQ's on the
> internet.
> 1. maintained by ticalc.org 2. By TI itself 3. by Ray Kremer 4. by
> Stephan Bird.
> was on none of them... Why?
>
> The second question is, does this problem manifest itself in the current
> ti-86 rom.. and if so WHY!!??
>
> Please reguard these questions as merely rhetorical.. The fact that I
> have a buggy calc has gotten me a little pissed.
>
> --
> "I will take my hands and weave them to a little house,
> and there you shall keep a dream--"
> --Miriamne. "Winterset"
Follow-Ups:
References:
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Re: t-test question
• To: mathgroup at smc.vnet.net
• Subject: [mg64163] Re: t-test question
• Date: Fri, 3 Feb 2006 01:03:50 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```On 2/2/06 at 7:09 PM, attila at biking.taiiku.tsukuba.ac.jp (Csukas
Attila) wrote:
>I have found an equation for comparing two means.
>Ue = (na - 1)*Ua + (nb - 1)Ub/na + nb - 2 This is for the pooled
>value of distribution and
>t0 = |xa-xb|/ Root Ue*(1/na+1/nb) This is for the t value.
This would be done in Mathematica as:
t0= Abs[xa-xb]/Sqrt[Ue (1/na+1/nb)]
>Sample values na=342 nb=170 Ua=6.39 Ub=6.07 xa=177.19 xb=170.37
with these values
In[18]:=Ue = (na - 1)*Ua + (nb - 1)*(Ub/na) + nb - 2
Out[18]=2349.9895029239765
and
In[23]:=t0=Abs[xa-xb]/Sqrt[Ue (1/na+1/nb)]
Out[23]=1.49918
--
To reply via email subtract one hundred and four
```
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Select Page
Do not proceed unless you have read at least Chapter 32, Spherical Time. If you are ready a whole new world awaits. The Journey begins here if you missed it.
Here is where the sinuous tunnels of the Torah take another turn and lead us to even more wondrous revelations, pathways, and connections. In congruence with the sequential 13 42 55 occurrences of specific initials in the Shema, we note that there are 13 initials Hei (ה) in those same 248 words, connecting the letters Alefs(א), Lamed (ל), Vav(ו), and Hei (ה) through their intersection of 13, 42, and 55 respectively. All 4 letters intersect as one: Lamed (ל) with its value of 30 and ordinal value of 12 for a total of 42; Alef (א) through the 42 occurrences in the main Shema, Vav(ו) though its (12 + 30) = 42 and (42 + 13) = 55 occurrences schematic within the Shema; and Hei (ה) with its 13 occurrences.
### A Concealed Name of 42
The 4 connected initials in the Shema have the numerical value 42 and spell out the Name Eloah (אלוה), G-d on the ceiling above us.
It is not a commonly used Name of G-d. In fact, it is used only once in the Torah, in Devarim 32:15 in Hazeinu, Moses’ Song. My Rav always said, “If you want to know what is important, look at what is least.” A permutation of the Name (אלוה) is also the first word of the Book of Shmot (ואלה), as in “These are the Names…”
There are 60 occurrences of permutations of the Name Eloah (אלוה) in the Torah and the Name Eloah (אלוה) itself of numerical value 42 is found 60 times in the entire Tanakh. Nothing is coincidental. Everything coincidental is an entrance to discovery.
The Book of Job is the only Book of 42 chapters, and the Name of G-d (אלוה) of numerical value 42 is found there 41 of those 60 times. From our otherworldly perspective we see that it is by design that Job (איוב) has a numerical value of 19 and that (19 + 41) = 60.
Moreover, those 41 times the Name of G-d (אלוה) appears in the 42-chapter Job plus the 1 time in the Torah, make 42 times between them, establishing a specific and potent luminous bond between them. A still small voice tells us that the connection of (18 + 42) = 60 is not coincidence either and that it is written into the cosmos, sealed in the 6th planet. We can imagine all sorts of connections but have no idea what that means.
A warm, no, a hot dry wind blows through and we are engulfed in a sea of sand and numbers, a forest of numbers. From within the wilderness, a glowing grove appears and as we swiftly make our way to it, we see that the word Bamidbar (במדבר) is found 60 times in the Torah. In the center of the grove, we see that its numerical value is 248 and that its ordinal value is 41, exactly like Abraham (אברהם). Surrounded by the desert wilderness, we are reminded that the 4th Book of the Torah, Bamidbar is synonymous with the 42-Letter Name and the 42 journeys. The 248 words of the Shema are not only associated with Abraham, but they represent the concept of Bamidbar (במדבר) every which way. The center of the grove expands to form a proper desert, where we see that Bamidbar Sinai (במדבר־סיני), the Sinai desert, has the numerical value of (248 + 130) = 378, that of the 27 positions of the Alef-bet and Essential Cube of Creation. Is it a cube?
The desert wind further tells us that the initials (בס) correspond to the 62 Yuds(י) in the 10 commandments and thus 620, Keter. In the same breath we are told that the final letters (רי) correspond to 210, the mountain (ההר).
As a mountain arises in that desert, we learn that (ההר־סיני) has the value of (210 + 130) = 340, the same as Sefer (ספר), Book, carved out of the two central letters (סר) in Mount Sinai (ההר־סיני). The Book (הספר) = 345, Moses (מהש).
The triplet sefer (ספר) is found 90 times in the Torah, as in the ordinal value of Bamidbar Sinai (במדבר־סיני). On deeper reflection, when the 42 journeys or letters are separated out of Bamidbar Sinai (במדבר־סיני), we are left with (דבר־סיני) and (במ) or 42 + 336, as in the 336 letters of the 112 Triplets.
Is the referred to Book The 5 Books of Moses, or the Book of Mt Sinai, or is it referring to the Book of Bamidbar, or all 3?
Following the path back out of the wilderness we take shelter in a cool cave and find ourselves in the chamber of the conjunction of the Shema and the Book of Job. They are indelibly written together with the Name of 42, as is Bamidbar (במדבר), whose small gematria adds up to 14. It must be understood that the 15 Triplets of the Shema conceal the 14 Triplets of the 42 letters of the Name of G-d, and that the 4 initials (אלוה) are woven into the 248 words of the Shema precisely 148 times.
As it says in Job 42:16, “and Job lived 140 years after this [the renewed blessing of Hashem], which we see is connected with the concealed 14 Triplets. Indeed, the 41 mentions of (אלוה) in Job are a mirrored reflection of the number 14. Moreover, in Job 42:12, we see that it says, “So the LORD blessed the latter end of Job more than his beginning; and he had fourteen thousand [14,000] sheep … and 8000 [other animals].” Once again there is a connection between Job and the Shema through the 148 occurrences of the initials in (אלוה), the Name of 42 that pervades the 42-chapter Book of Job. It would seem that the scholars that think that the Book of Job is a pieced together fable are wrong as usual about the Tanakh, and that there is a powerful message and connection for us in Job.
Following the intricate tunnels that branch off from G-d’s blessing in Job 42:12, we come to a conjunction where Abraham of numerical value 248, whose name embodies the 248 words of the Shema, is told by G-d to leave home, his birthplace. He is shown the promised land and receives the greatest blessing by G-d. This is all in Genesis verse 42:12. The trails and passages merge for all sorts of different reasons, yet each time they do they form a chamber, sometimes large sometimes small. Each chamber is filled will all sorts of treasures and relationships.
It is left to us to understand the linkages of these twin blessings by G-d. Job went on to have 10 children, as did Abraham. In both cases it must relate to being blessed with the 10 sefirot.
Nonetheless, we are back in the Cavern of the Shema’s initials. One clear hidden message is found in the splitting of the initials. The 248 initials are split into 148 and 100, with 100 being exactly 1/10th or a tithing of all 1000 letters in the Shema, and 148 being connected to the Name of G-d (אלוה) of 42. Four (4) letters form 148 and connect to the Name (אלוה) of 42, and 13 letters connect to 100.
As we have seen, there are multiple ways that the Shema connects to the 9th sefira of Chochma of numerical value 73. Through a new network of tunnels and imagery we learn that of the 13 letters that combine to total 100 initials, the lowest 9 in terms of occurrences (1 ,1, 2, 2, 3, 4, 4, 5, 5) total 27, as in the design of the 9 planes and 27 positions of the Essential Cube of Creation, leaving (10027) = 73 for the remaining 4 (עבמי). Those 4 (עבמי) have a numerical value of 122, the number of words in the middle paragraph of the Shema. Moreover, their ordinal value, 41, is reflective of the 41 occurrences of (אלוה) in Job and the ordinal values of Abraham and Bamidbar.
Then, if we separate out the last of those 4, the Yud (י) of numerical value 10, as in the 10 sefirot, etc. we are left with (עבמ). We do not see any flying saucers, but those first 3 letters (עבמ) of the four are the same as in UFO (עבמ), and they include the two sets (במ) and (עב) of the 4 Essential Elements. Moreover those 3 last initials (עבמ) total 112, as in the full 112 Triplets. Tempted to take, just for fun, another passage leading off the UFO initials (עבמ), where their collective (13 + 22 + 16) occurrences are found 51 times in the Shema, as in area 51, or Edom. We choose not to and instead stay the course.
In dawns on us that the 248 initials and the 5 elements of the Shema, are structurally designed to equal 42, 112 and 27, a perfect representation of the 5 Essentials Elements of Creation, which are comprised of the 112 Triplets surrounding the central Cube of 27 positions that total 42 throughout, and whose directional flow forms the 10 loops of the 10 sefirot (dimensions).
It is not just the initials in the Shema that form connecting passages, the words do as well. For example, we just stated that there are 13 initial letters not connected to the Name of G-d (אלוה) of 42. The value 13 is that of the prominent word echad (אחד) in the Shema’s first verse, but with a visual sleight of hand on the ceiling above us, it is shown that there are also 13 YHVH (יהוה) in the Shema. As it says in the first verse, “The Lord is One (אחד).” There are 13 (יהוה) encircling (אלוה).
Suddenly, we realize that metaphysically, every initial in the Shema is connected to a Name of G-d, either (אלוה) or (יהוה). Then add to that the paragraph or 42, and the paragraph of 72, and the hidden 42 Letters of the Name of G-d in the first verse, and this seems like a pretty powerful portal indeed.
Initials are what we connect to first, the seed of what the final outcome will be. It is through the connections of the Shema’s first verse and the 248 initials where we can focus our intent on G-d. Job lost his faith in his fellow man and in the relevance of being righteous, but he never lost his faith in G-d, who he knew was truly the seed of all things.
The Lord Giveth, the Lord taketh away. In Hazeinu, Moses’ Song that hints at the time of Moshiach, Moses tells us of all the great abundance G-d gave his chosen people, until in Devarim 32:15, the lone verse where we find the Name (אלוה). It is then that the people are called Jeshurun for the first time. They became fat, they spurned G-d, their source, and they worshipped alien practices, sacrificing to demons. G-d was ready to take everything from them as he had Job. He was prepared to exterminate them, yet instead of letting His enemies win because of the shortcomings of His people, He destroyed their mutual enemies, wiping them away. It forces us to ask ourselves if we are at that point that Moshe sang about, and if that is the event horizon in the Spherical Time bubble points to.
At that moment, a gigantic shadow passes over head and we wonder if the connection to Job is meant to draw our attention to the Leviathan that Job spoke of at length? Or did we release it by traversing his passages.
Rather than give into the fear of what might be out there or what was coming, we follow a bright, short, enclosed trail from the 17 letters used as initials and from the Name of 42 (אלוה) to the Spherical Time arena where we see that (37602018) = 1742. What we also find at trail’s end is that the 5 letters not utilized as initials in the Shema (גחטסר) equal 280, as in the 5 final letters in the alef-bet, which obviously are not included either, making a total of (280 + 3500) = 3780 or 378 x 10 for the 10 letters.
While there, we see that those 17 letters tied to Abraham and the 248 words/initials form yet another equation: (1700 + 248) = 1948, with 1948 being one of the most critical years marked on both calendars, including Abraham’s birth in 1948 and the birth of Israel as a nation. Only with the perspective of hyperspace vision are we able to further see that while the addition of the two time paradigm radii (3760 + 2018) = 5778, the subtraction of them from one another is (37602018) = 1742, the distance that needs to be breached by the speed-of-light. Moreover, while (3760 + 2018) = 5778, the event horizon at the far end of the radii, the equation (37602018) represents its origin point. Think of it as the two equal radii subtracted from one another. Equal in distance, not length. Therefore, 1742, or the metaphysical fusion of the specific unique initials (letters) and the Name of 42, may be a way to nullify time, especially since the P/S cipher of 1742 is 56/14 = 4, as in (אלוה).
### The Exile
With the enlarged shadow once again swooping over us, our thoughts return to the Leviathan (לִוְיָתָֽן) that has many ominous forms—snake, serpent, whale, crocodile, dragon—and is also known as the giant crocodiles (תַּנִּינִם), which is found later in that same Devarim chapter 32 within Moses’ Song of the End. Another pseudonym for the Leviathan is also found in Job 26:13 and also twice in Isaiah 27. It is the eclipse serpent Nachash Barak (נָחָשׁ בָּרִחַ).
On its own barak (בָּרִחַ) means to escape, to flee, and is a permutation of “sword” (חרב), both of numerical value 210, which is significant. Yes, it is the height of the Pyramid in cubits, but more important it is the 210 years of exile in Egypt. It is as if it were predetermined that it would take 210 years to escape, similar to the predetermined 5778 years necessary to escape physicality. The clear mirror shows us that it is so similar that within days, the time interval proportion 210/5778 is equal to.363424 and/or .3634470, as in the Spherical Time quotient (36304.24470).
Thanks to the clear mirror we realize 210 was not just a cool number for the Torah to choose for the Israelite exile in Egypt, it was entirely predicated on the Spherical Time bubble, and as we have already seen it had to end 3330 years before 5778. We further realize that that year, 2448 HC, also had to be (-1313 BCE) to align the giving of the Torah with the giving of the 33 Hebrew letters, whose exponent coefficient xn is 1.313. If not, the exponential scale would not have assigned the correct letter values so that barak (בָּרִחַ), meaning escape and to flee, and sword (חרב), and mountain (ההר) would equal 210 or (5 x 42), the 5 Names of 42.
It occurs to us that those 210 years in Goshen might have been a bubble within the bubble. Just as we have those thoughts a giant dreidel appears in front of us, with the 4 letters (גשנה), Goshna, of numerical value 358, Moshiach, engraved into its 4 sides. That is not a new concept to us, but what is, is that the dreidel is a spinning cube sitting on top of a pyramid. What we also realized is that (358210) = 148, as in the 148 initials in the Shema that connect with the Name of 42 (אלוה).
As our consciousness expands, the more connections we can make and hold.
### Male and Female Letters
The dreidel gone and replaced by a distant circling shadow, and we realize that the Leviathan (לִוְיָתָֽן) is Malchut. Both words have the same gematria, 496, and the same ordinal value, 64. Together, their complete values each represent twice the final judgement of the 5 final letters, or (280 x 2). The clear mirror gives us further insight. The 5 final letters (ךםןףץ) can total 280 and can also total 3500, or together the 10 duplicate letters equal 3780, which is 10 x 378, the sum of the 27 letters/positions of the Essential Cube of Creation, just as we saw with the initials of the Shema. Does this mean the 5 final letters balance out the entire 27 letters? As we try to picture this balancing seesaw, we realize that 27 less the 10 duplicate letters (כמנפצךםןףץ) equal 17, as in the numerical value of tov(טוב), meaning good, and as in the 17 letters of the Shema used as initials. We are constantly reminded how important that portal entrance is.
We do not know why this is important, but the ordinal value of the 10 duplicate final letters (כמנפצךםןףץ) is 73 and 125, making the ordinal value of the other 17 letters, 180, which is 18 less than (73 + 125).
Clearly those differentials are important, though we do not know if they are as important as what the clear mirror is now showing us. Since the sum of the 27 letter values is 4995 and the 10 duplicate final letters (כמנפצךםןףץ) is 3780, the value of the 17 standard letters is (49953780) = 1215, the dawn of time.
This also means that the 17 letters utilized as initials to words in the Shema also equal 1215 and thus reflect the dawn of creation too.
If that was not mind twisting enough, it is then revealed that the complete value of those 17 (אבגדהוזחטילסעקרשת) standard letters (1215 + 180) = 1395 matches the gematria value of the 6 Matriarchs—Sarah, Leah, Rivka, Eve, Miriam, and Rachel—whose names are embedded in the 14 Triplets of the 42-Letter Name matrix, (505, 36, 307, 19, 290, and 238) = 1395.
At once, we ask ourselves of the alef-bet is divided into male/female energy archetypes: 10 final letters versus or complementary to 17 standard letters, a 26.0 to 74.0 female to male ratio in terms of their complete value. Is this why the letters (עד) of numerical value 74 in the Shema are enlarged?
Drawn into the 17 (אבגדהוזחטילסעקרשת) female letters, we see that they break down sequentially into the first 10 (אבגדהוזחטי) of numerical value 55, the last 4 (קרשת) of numerical value 1000 and the middle 3 (לסע), whose complete value is 203, as in the first Triplet in the Torah (ברא). Besides 55 being reflective of an entrance and 1000 of Binah, we are not sure what this means for the inner cosmos, only that the 10 and 17 proportion matches that of the initials in the Shema. And as the clear mirror suggests, 10/27 = the 42-Letter Name plus 1/42-Letter Name, in other words, (.3701 + 1/3701) = (.3701 + .2701) = .3703701…, the merging of the 42-Letter Name with the first verse of the Torah.
### A New World Order
Just as it seems like we have already learned enough for a lifetime, the shadow settles over us again, and we are reminded that the Zohar talks about Job, and independently about the Leviathan, in reference to the same portion, Bo, the portion of the final 3 plagues and the fleeing from Pharaoh. In Exodus paragraph 33, in 12:41 it says, “At the end of 430 years, all of God’s armies left Egypt in broad daylight.” The year was 2448 so (2448430) is 2018, when G-d made his Covenant with Abraham. “All of God’s armies”
In Spherical Time that also means 2018 CE.
The large shadowy cloud lifts and is replaced by the thin shadow of the tip of the Tower of Truth. It lengthens and casts a shadow over our feet. Once it does, we see that 430 is the numerical value of Shekel (שקל) and that the Torah tells us specifically that 1 Shekel is equivalent to 20 gerah, which metaphysically means that the 430 years was the same as the 210 years, since the height of the Tower, and of the 20 stacked cubes (1 +2 + 3…20) = 210. The value 20 is a reference to Keter, the completion of the process.
Only then does it dawn on us that the year 1776 HC when the Tower of Truth was created was 242 years earlier than the 430-year countdown from Abraham to Exodus began, matching the 242 years from 1776 CE to 2018 CE, the physical time radius, marking both dates (HC and CE) 1776 and 2018 as the same two radii. On the same 1776 radius, yet on opposite extremes, a new world order was established. On one end, they came together to build a city and a tower to circumvent Spiritual Time and control physicality, and on the other end they formed the Illuminati and established a nation they could physically control.
Only now does the 430-year time span make sense. The “120 years of man” corresponds to the 120 years from the Flood in 1656 to 1776, and also to the last 120 years from 1898 to 2018 CE. To view the Torah in any way other than hyperspace is like viewing the Mona Lisa through a kaleidoscope.
As for the last 120 years, on the surface that year was like many other: the US fought and won the Spanish-American war, yet another war initiated by a mysterious false flag event; they annexed Hawaii to complete the 50 States, and Brooklyn (ב) joined Manhattan (מ) to create New York City. Those 120 years would see an exponential rise in technology and the transfer from a human civilization to an AI-aided society. It would mimic the rise from one man to one society able to construct a city and a tower. There was, though, an inconspicuous time marker planted back in 1898. On April 18th, the Boston Marathon (במ) or 42, was won for the 2nd time, as in 2:42, in a record 2:42:00 time. Markers are clues left for us during the recycling of time to remind us to wake up.
The shadow of the Tower is creeping across our bodies and looming larger. As we combine the Hebrew word for City Yir (עיר), of numerical value 280 like the 5 final letters and the 46th (ערי) of the 72 Triplets with the name Enoch (חנוך), as the city named by Cain, we see the words “was building a city,” BNehנה) Yir (עיר) before us.
The words of Cain permute before our eyes to (הר־בעני) or mountain in New York City (NYC). Moreover, the name Cain (קין) forms the final letters of New York (ניו־יורק). We flash ahead to the Freedom Tower’s height of 1776 ft. Then back to the Torah text where we learn that the city was to be found in the valley of Shinar (שענר) of numerical value 620, as in Keter, with the middle letters having the value 120. Were they able to build their city so fast—within 120 years of the Flood—because Cain (קין) had laid the groundwork for it? Did he attach Enoch’s (חנוך) name to it to download the necessary spiritual blueprints? Does harnessing Enoch’s (חנוך) name with the proper intentions help us to access Spiritual Time? Was the Torah advising us that the same people who tried in 1776 HC would try again in NYC 4000 years later, and once again name it “One World?” They made it 541 meters high, like the 541 rooms in the Capitol Building in Washington D.C. and like the numerical value of Israel (541). Why? Why are both the Capitol Building and the World Trade Center Plaza also the same 16 acres, as in Yud-Vav (יו)?
The Capitol Building began construction back in 1793, 27 years after 1776 CE and 155 years before Israel became a nation, as was promised to Abraham in 2018 HC. Abraham was born in 1948 HC and on the opposite side of the physicality radius Israel became a nation in 1948 CE. It would take another 70 years for the 45th US President to declare Jerusalem the capital of Israel by moving its embassy there in 2018 CE, fulfilling the promise on the opposite side of the radius.
For most of these cosmic symmetries to happen Spherical Time would need to exist.
To better understand this time connection with the Exodus, we tunnel deeper into the Torah, and see that “All of God’s armies” “went out [left] (יָצְאוּ)” Egypt with the Israelites. The word “went out [left] (יָצְאוּ)” has a numerical value 107, as in the 107th triangular number and the sum of the integers from 1 – 107 that sum to 5778.
Is 107 and the 10 and 17 proportion of the letters related?
We do not need to understand all of the technology and terminology embedded in this verse, only that G-d’s intentions were tied to Spherical Time and a possible warning that All of God’s armies marched alongside us in 2448 after 430 years from 2018 HC and that would need to happen at the edge of physical time (2018 CE). The same 430 years from the edge of Spiritual Time would be 3330 years from the center, creating a time circle with a diameter of 6660 years.
The Hebrew word for soul, nefesh (נפש) has a numerical value of 430, so are we talking about the maturation of the Israelite soul from Abraham to Moses, who were born 420 years apart?
It was 430 years since Abraham set out on his Journey and Moses brought the Israelites out of the womb of Egypt. The Hebrew word womb (רחם) and Abraham have the same gematria value, 248 and the same ordinal value, and on Abraham’s trip to Egypt, the Zohar makes a point of telling us that G-d sealed up all the wombs in Egypt. Tying the two sojourns to Egypt together, the Torah makes a point of telling us that G-d rained plagues down on Pharaoh and Egypt to free Sarah, and that Abraham left with treasures and riches. We can understand that all of God’s armies accompanied the Jewish Soul (Bnei Israel) on its growth journey from its conception with Abraham to its birth through the waters of the Red Sea. We can further understand the role of Sarah as an allusion to the Shechinah.
The gestation period for humans is 40 weeks and the Torah also mention a 400-year period for the Egyptian exile, which corresponds to the period from the miraculous birth of Isaac to Moses’ crossing of the Red Sea. Those 40 weeks are mimicked in the 40 days on Sinai, and the 40 years in the desert, both connected to Bamidbar, and the 40 days and 40 nights of rain.
When we convert those physical years of gestation and exile, 210 and 430, to spiritual ones, by multiplying by the speed-of-light (1.86), they become 390.6 and 799.8 respectively, as in the 390,625 words, letters, and verses in the Torah, and the 79,975 words in the Torah—converting the physical to the spiritual.
The deeper the Torah tunnels get, the brighter the light. This is because the tunnels are not underground; they are channels within the upper realms, passages on the path of ascension.
Another abstract branching tunnel from (37602018) = 1742 passage shows us that the diameter of the 3760 radius spiritual sphere is (3760 x 2) = 7520 and that (5778 + 1742, the nullification point) also equals 7520. At first, we question why we are being shown this, as it is just mathematical slight-of-hand. Then it dawns on us that it is an entirely different perspective: The right injection at the nullification point can replace physical time with spiritual time, elevating us to a pure conscious state. The full span across spiritual time is 7520 spiritual years. The limit or diameter to the event horizon in 5778 years, an admixture of the two radii, physical and spiritual time. When 1742 years are added, it matches pure Spiritual Time. The gap that needs to be breached is 1742 years.
Next, we see that the first time a sequence of words in the Torah totaled 5778 in gematria starts off with the words, “It Happened,” a separate phrase at the end of Genesis 1:11 and then it continues with the entirety of verse 1:12 about the spreading of the trees, and that each tree will spread its own fruit depending on its seeds. It ends with, “it was morning, it was night, third day.” This can be understood as a reference to the third time the tree-of-life reality was available to us, in 5778. “It Happened.” It is also an obvious connection to the 112 Triplets and the moment they would be revealed.
From there, we see that the phrase “Bnei Adam ‘Sons of Adam (אדם בני)” of numerical value 107 represents the lineage of Adam that stretches 5778 years from Adam until now, as in the 107th triangular number and the sum of the integers from 1 – 107.
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# Christmas Around the World Activities | Christmas Around the World Math
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(22 MB|4 different problem solving and pictures pages with 3 levels of each picture)
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1. This Winter Holidays Around the World activity BUNDLE is a festive and fun way to practice place value standards. Your students will love solving the problems and coloring the pictures in these holidays and Christmas around the world math worksheets, while you enjoy all of the extra standard-based p
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These Christmas Around the World activities are a festive and fun way to practice place value standards. Your students will love solving the problems and coloring the pictures in these Christmas Around the World math worksheets, while you enjoy all of the extra standard-based practice they will be receiving!
This easy, no-prep activity includes 3 LEVELS of each problem solving and color picture to meet a variety of class and student needs.
Be sure to look at the preview to see what is included in this festive and engaging activity.
Would you like to save some money? Click here to SAVE 20% today by purchasing this holiday color by number in my Winter Holidays Around the World Math Color by Number BUNDLE.
This diverse Christmas standard-based activity includes...
-Christmas in Germany (3 levels)
• Read and compare numbers to 1,000
• Read and compare numbers to 1,000,000
• Read and compare numbers to the thousandths place
-Christmas in Mexico
• Round to the nearest 10 and 100
• Round whole numbers to any place value
• Round numbers with decimals
-Christmas in Sweden
• Add and subtract numbers to 1,000
• Add and subtract numbers to 1,000,000
• Add and subtract numbers with decimals to the hundredths
-Christmas in Australia
• Multiply and divide math facts
• Multiply up to 4x1 digits and divide with 1-digit divisors and remainders
• Multiply multi-digit numbers and divide with 2-digit divisors and remainders
-Answer Keys for ALL VERSIONS of the problem solving and color by number
pages
This activity is perfect for math centers, early finishers, homework, morning work, or for any needed extra practice!
DON'T MISS WINTER HOLIDAYS AROUND THE WORLD COLOR BY NUMBER!
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You may also be interested in these fun and engaging activities...
Winter Holidays Math Project 3rd Grade
Winter Holidays Math Project 4th Grade
Winter Holidays Math Project 5th Grade
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Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Fluently multiply multi-digit whole numbers using the standard algorithm.
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Analyse the bicycle frame design sketched in Figure 1. Use a vertical load of 70.xx kg at the... 1 answer below »
Analyse the bicycle frame design sketched in Figure 1. Use a vertical load of 70.xx kg at the seat
location and 12.xx kg at the handle bar location. Assume for the first analysis that all the members are
tubular steel with a 1 in. outside diameter and 0.062 in. wall thickness. xx is the sum of the last digits
of the group members student ID number. From the first analysis:
A) Determine if any yield failures are likely if the material is a high-carbon steel. Find the density,
Young’s elastic modulus and yield strength of the material.
B) If yielding will occur, refine the design by replacement of highly stressed members with a more
substantial section or by altering the design layout to eliminate yield failures. If the frame is
overdesigned, refine the design to reduce weight.
C) Do the deflections seem excessive? Is there a specific location that seems to be too flexible. This
As the design engineers, you are asked to build the FE model of the bicycle frame. To accomplish this,
you are required to:
1) Develop a Finite Element Model of the original frame design and manually calculate the nodal
displacements and the stresses in each members (elements) (20 marks);
2) Develop FE MATLAB programs to calculate the frame displacements and stresses. Use these
programs to answer (A), (B) and (C). The main program should also be able to display the
3) Build and analyse the frame in ANSYS WORKBENCH and compare the displacements and the
stresses obtained in (2) with the ANSYS result in (2) (20 marks);
4) Provide comment on the results of your FE analysis e.g. the use of FE method for the analysis,
discussion on (A), (B) and (C) (20 marks); State any assumptions you make in the analysis and
justify them.
5) Present you analysis as professional engineering report (
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University of Wollongong Faculty of Engineering and Information Sciences MECH419/928 Assignment 1 Design of bicycle frame Autumn Session – 2019 Due Friday week 7, 19 April (Group project of max 2 members) Rules: 1. The assignment may be completed individually or by a group (of maximum 2 members). The group formation is your own responsibility. Members may be from the same or different tutorial groups. 2. The assignment is due in week 7 (Friday 19April) and submitted to EEC using the barcoded sheet (see subject outline under assignment submission). Late submission will incur penalty as described in the subject outline. Students should make themselves aware of the university academic integrity policies (see below types of non-tolerated misconduct). Copying is cheating and will result in significant penalty and will be referred to the school assessment committee who will determine the consequences. 3. As this assignment is considered as engineering project undertaken by group of engineers, group synergy and team work are part of the learning. So if the assignment is completed by group, include statement indicating the effort or contribution to the assignment by each member and describe how you group work together to solve the problem. The report is expected to be not more than 2000 words long. 4. The MATLAB script files and function files must be included as appendix in the hard copy of the report. You must have the ANSYS file handy in case we would like to review them. Do not include pages and pages of ANSYS Output in the report. Analyse the bicycle frame design sketched in Figure 1. Use a vertical load of 70.xx kg at the seat location and 12.xx kg at the handle bar location. Assume for the first analysis that all the members are tubular steel with a 1 in. outside diameter and 0.062 in. wall thickness. xx is the sum of the last digits of the group members student ID number. From the first analysis: A) Determine if any yield...
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https://isabelle.in.tum.de/repos/isabelle/file/87ad6e8a5f2c/src/HOL/ex/LocaleTest2.thy
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src/HOL/ex/LocaleTest2.thy
author wenzelm Sun, 03 Jun 2007 23:16:47 +0200 changeset 23219 87ad6e8a5f2c parent 22757 d3298d63b7b6 child 23919 af871d13e320 permissions -rw-r--r--
tuned document;
```
(* Title: HOL/ex/LocaleTest2.thy
ID: \$Id\$
Author: Clemens Ballarin
Copyright (c) 2007 by Clemens Ballarin
More regression tests for locales.
Definitions are less natural in FOL, since there is no selection operator.
Hence we do them in HOL, not in the main test suite for locales:
FOL/ex/LocaleTest.thy
*)
header {* Test of Locale Interpretation *}
theory LocaleTest2
imports Main
begin
ML {* set quick_and_dirty *} (* allow for thm command in batch mode *)
ML {* set show_hyps *}
ML {* set show_sorts *}
subsection {* Interpretation of Defined Concepts *}
text {* Naming convention for global objects: prefixes D and d *}
(* Group example with defined operation inv *)
locale Dsemi =
fixes prod (infixl "**" 65)
assumes assoc: "(x ** y) ** z = x ** (y ** z)"
locale Dmonoid = Dsemi +
fixes one
assumes lone: "one ** x = x"
and rone: "x ** one = x"
definition (in Dmonoid)
inv where "inv(x) == THE y. x ** y = one & y ** x = one"
lemma (in Dmonoid) inv_unique:
assumes eq: "y ** x = one" "x ** y' = one"
shows "y = y'"
proof -
from eq have "y = y ** (x ** y')" by (simp add: rone)
also have "... = (y ** x) ** y'" by (simp add: assoc)
also from eq have "... = y'" by (simp add: lone)
finally show ?thesis .
qed
locale Dgrp = Dmonoid +
assumes linv_ex: "EX y. y ** x = one"
and rinv_ex: "EX y. x ** y = one"
lemma (in Dgrp) linv:
"inv x ** x = one"
apply (unfold inv_def)
apply (insert rinv_ex [where x = x])
apply (insert linv_ex [where x = x])
apply (erule exE) apply (erule exE)
apply (rule theI2)
apply rule apply assumption
apply (frule inv_unique, assumption)
apply simp
apply (rule inv_unique)
apply fast+
done
lemma (in Dgrp) rinv:
"x ** inv x = one"
apply (unfold inv_def)
apply (insert rinv_ex [where x = x])
apply (insert linv_ex [where x = x])
apply (erule exE) apply (erule exE)
apply (rule theI2)
apply rule apply assumption
apply (frule inv_unique, assumption)
apply simp
apply (rule inv_unique)
apply fast+
done
lemma (in Dgrp) lcancel:
"x ** y = x ** z <-> y = z"
proof
assume "x ** y = x ** z"
then have "inv(x) ** x ** y = inv(x) ** x ** z" by (simp add: assoc)
then show "y = z" by (simp add: lone linv)
qed simp
interpretation Dint: Dmonoid ["op +" "0::int"]
where "Dmonoid.inv (op +) (0::int)" = "uminus"
proof -
show "Dmonoid (op +) (0::int)" by unfold_locales auto
note Dint = this (* should have this as an assumption in further goals *)
{
fix x
have "Dmonoid.inv (op +) (0::int) x = - x"
by (auto simp: Dmonoid.inv_def [OF Dint])
}
then show "Dmonoid.inv (op +) (0::int) == uminus"
by (rule_tac eq_reflection) (fast intro: ext)
qed
thm Dmonoid.inv_def Dint.inv_def
lemma "- x \<equiv> THE y. x + y = 0 \<and> y + x = (0::int)"
apply (rule Dint.inv_def) done
interpretation Dclass: Dmonoid ["op +" "0::'a::ab_group_add"]
where "Dmonoid.inv (op +) (0::'a::ab_group_add)" = "uminus"
proof -
show "Dmonoid (op +) (0::'a::ab_group_add)" by unfold_locales auto
note Dclass = this (* should have this as an assumption in further goals *)
{
fix x
have "Dmonoid.inv (op +) (0::'a::ab_group_add) x = - x"
by (auto simp: Dmonoid.inv_def [OF Dclass] equals_zero_I [symmetric])
}
then show "Dmonoid.inv (op +) (0::'a::ab_group_add) == uminus"
by (rule_tac eq_reflection) (fast intro: ext)
qed
interpretation Dclass: Dgrp ["op +" "0::'a::ring"]
apply unfold_locales
apply (rule_tac x="-x" in exI) apply simp
apply (rule_tac x="-x" in exI) apply simp
done
(* Equation for inverse from Dclass: Dmonoid effective. *)
thm Dclass.linv
lemma "-x + x = (0::'a::ring)" apply (rule Dclass.linv) done
(* Equations have effect in "subscriptions" *)
(* In the same module *)
lemma (in Dmonoid) trivial:
"inv one = inv one"
by rule
thm Dclass.trivial
(* Inherited: interpretation *)
lemma (in Dgrp) inv_inv:
"inv (inv x) = x"
proof -
have "inv x ** inv (inv x) = inv x ** x"
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# Comparing two collections for equality irrespective of the order of items in them
I would like to compare two collections (in C#), but I'm not sure of the best way to implement this efficiently.
In my case, two collections would be equal if they both contain the same items (no matter the order).
Example:
``````collection1 = {1, 2, 3, 4};
collection2 = {2, 4, 1, 3};
collection1 == collection2; // true
``````
What I usually do is to loop through each item of one collection and see if it exists in the other collection, then loop through each item of the other collection and see if it exists in the first collection. (I start by comparing the lengths).
``````if (collection1.Count != collection2.Count)
return false; // the collections are not equal
foreach (Item item in collection1)
{
if (!collection2.Contains(item))
return false; // the collections are not equal
}
foreach (Item item in collection2)
{
if (!collection1.Contains(item))
return false; // the collections are not equal
}
return true; // the collections are equal
``````
However, this is not entirely correct, and it's probably not the most efficient way to do compare two collections for equality.
An example I can think of that would be wrong is:
``````collection1 = {1, 2, 3, 3, 4}
collection2 = {1, 2, 2, 3, 4}
``````
Which would be equal with my implementation. Should I just count the number of times each item is found and make sure the counts are equal in both collections?
The examples are in some sort of C# (let's call it pseudo-C#), but give your answer in whatever language you wish, it does not matter.
Note: I used integers in the examples for simplicity, but I want to be able to use reference-type objects too (they do not behave correctly as keys because only the reference of the object is compared, not the content).
• How about algorithm? All answer related by compare something, generic lists compare linq etc. Really did we promised to someone that we will never use algorithm as an old fashioned programmer? Jun 8, 2012 at 12:09
• You are not checking for Equality you are checking for Equivalence. It's nitpicky but an important distinction. And a long time ago. This is a good Q+A. Mar 23, 2015 at 23:51
• You may be interested in this post, which discusses a tuned version of the dictionary-based method described below. One issue with most simple dictionary approaches is that they don't handle nulls properly because .NET's Dictionary class doesn't permit null keys. Jan 3, 2016 at 18:10
It turns out Microsoft already has this covered in its testing framework: CollectionAssert.AreEquivalent
Remarks
Two collections are equivalent if they have the same elements in the same quantity, but in any order. Elements are equal if their values are equal, not if they refer to the same object.
Using reflector, I modified the code behind AreEquivalent() to create a corresponding equality comparer. It is more complete than existing answers, since it takes nulls into account, implements IEqualityComparer and has some efficiency and edge case checks. plus, it's Microsoft :)
``````public class MultiSetComparer<T> : IEqualityComparer<IEnumerable<T>>
{
public MultiSetComparer(IEqualityComparer<T> comparer = null)
{
m_comparer = comparer ?? EqualityComparer<T>.Default;
}
public bool Equals(IEnumerable<T> first, IEnumerable<T> second)
{
if (first == null)
return second == null;
if (second == null)
return false;
if (ReferenceEquals(first, second))
return true;
if (first is ICollection<T> firstCollection && second is ICollection<T> secondCollection)
{
if (firstCollection.Count != secondCollection.Count)
return false;
if (firstCollection.Count == 0)
return true;
}
return !HaveMismatchedElement(first, second);
}
private bool HaveMismatchedElement(IEnumerable<T> first, IEnumerable<T> second)
{
int firstNullCount;
int secondNullCount;
var firstElementCounts = GetElementCounts(first, out firstNullCount);
var secondElementCounts = GetElementCounts(second, out secondNullCount);
if (firstNullCount != secondNullCount || firstElementCounts.Count != secondElementCounts.Count)
return true;
foreach (var kvp in firstElementCounts)
{
var firstElementCount = kvp.Value;
int secondElementCount;
secondElementCounts.TryGetValue(kvp.Key, out secondElementCount);
if (firstElementCount != secondElementCount)
return true;
}
return false;
}
private Dictionary<T, int> GetElementCounts(IEnumerable<T> enumerable, out int nullCount)
{
var dictionary = new Dictionary<T, int>(m_comparer);
nullCount = 0;
foreach (T element in enumerable)
{
if (element == null)
{
nullCount++;
}
else
{
int num;
dictionary.TryGetValue(element, out num);
num++;
dictionary[element] = num;
}
}
return dictionary;
}
public int GetHashCode(IEnumerable<T> enumerable)
{
if (enumerable == null) throw new
ArgumentNullException(nameof(enumerable));
int hash = 17;
foreach (T val in enumerable)
hash ^= (val == null ? 42 : m_comparer.GetHashCode(val));
return hash;
}
}
``````
Sample usage:
``````var set = new HashSet<IEnumerable<int>>(new[] {new[]{1,2,3}}, new MultiSetComparer<int>());
Console.WriteLine(set.Contains(new [] {3,2,1})); //true
Console.WriteLine(set.Contains(new [] {1, 2, 3, 3})); //false
``````
Or if you just want to compare two collections directly:
``````var comp = new MultiSetComparer<string>();
Console.WriteLine(comp.Equals(new[] {"a","b","c"}, new[] {"a","c","b"})); //true
Console.WriteLine(comp.Equals(new[] {"a","b","c"}, new[] {"a","b"})); //false
``````
``````var strcomp = new MultiSetComparer<string>(StringComparer.OrdinalIgnoreCase);
Console.WriteLine(strcomp.Equals(new[] {"a", "b"}, new []{"B", "A"})); //true
``````
• Hello Ohad, Please read the following long debate in the topic , stackoverflow.com/questions/371328/… If you change object hashcode , while its in a hashset it will interrupt with the hashset proper action and might cause an exception . The rule is as following : If two objects are equals - they must have same hash code. If two objects have same hashcode - it isnt a must for them to be equal. Hashcode must stay same for entire object's lifetime! Thats why you impelment ICompareable and IEqualrity . Sep 1, 2013 at 23:54
• @JamesRoeiter Perhaps my comment was misleading. When a dictionary encounters a hashcode it already contains, it checks for actual equality with an `EqualityComparer` (either the one you supplied or `EqualityComparer.Default`, you can check Reflector or the reference source to verify this). True, if objects change (and specifically thier hashcode changes) while this method is running then the results are unexpected, but that just means this method is not thread safe in this context. Sep 2, 2013 at 9:04
• @JamesRoeiter Suppose x and y are two objects we want to compare. If they have different hashcodes, we know they are different (because equal items have equal hashcodes), and the above implementation is correct. If they have the same hashcode, the dictionary implementation will check for actual equality using the specified `EqualityComparer` (or `EqualityComparer.Default` if none was specified) and again the implementation is correct. Sep 11, 2013 at 15:32
• @CADbloke the method has to be named `Equals` because of the `IEqualityComparer<T>` interface. What you should be looking at is the name of the comparer itself. In this case it's `MultiSetComparer` which makes sense. Mar 24, 2015 at 8:25
A simple and fairly efficient solution is to sort both collections and then compare them for equality:
``````bool equal = collection1.OrderBy(i => i).SequenceEqual(
collection2.OrderBy(i => i));
``````
This algorithm is O(N*logN), while your solution above is O(N^2).
If the collections have certain properties, you may be able to implement a faster solution. For example, if both of your collections are hash sets, they cannot contain duplicates. Also, checking whether a hash set contains some element is very fast. In that case an algorithm similar to yours would likely be fastest.
• You just have to add a using System.Linq; first to make it work May 21, 2010 at 16:44
• if this code is within a loop and collection1 gets updated and collection2 remains untouched, notice even when both collections have the same object, debugger would show false for this "equal" variable. May 21, 2010 at 17:11
• @Chaulky - I believe the OrderBy is needed. See: dotnetfiddle.net/jA8iwE May 28, 2014 at 20:34
• Which was the other answer referred to as "above"? Possibly stackoverflow.com/a/50465/3195477 ? Dec 11, 2019 at 19:16
• Take care to chose the comparison carefully. Elements may not be re-ordered at all if e.g. two elements are comparably the same (but are different objects). May 25 at 9:35
Create a Dictionary "dict" and then for each member in the first collection, do dict[member]++;
Then, loop over the second collection in the same way, but for each member do dict[member]--.
At the end, loop over all of the members in the dictionary:
`````` private bool SetEqual (List<int> left, List<int> right) {
if (left.Count != right.Count)
return false;
Dictionary<int, int> dict = new Dictionary<int, int>();
foreach (int member in left) {
if (dict.ContainsKey(member) == false)
dict[member] = 1;
else
dict[member]++;
}
foreach (int member in right) {
if (dict.ContainsKey(member) == false)
return false;
else
dict[member]--;
}
foreach (KeyValuePair<int, int> kvp in dict) {
if (kvp.Value != 0)
return false;
}
return true;
}
``````
Edit: As far as I can tell this is on the same order as the most efficient algorithm. This algorithm is O(N), assuming that the Dictionary uses O(1) lookups.
• This is almost what I want. However, I'd like to be able to do this even if I am not using integers. I'd like to use reference objects, but they do not behave properly as keys in dictionaries. Sep 8, 2008 at 19:09
• Mono, your question is moot if your Items are not comparable. If they cannot be used as keys in Dictionary, there is no solution available. Sep 16, 2008 at 15:50
• I think Mono meant the keys are not sortable. But Daniel's solution is clearly intended to be implemented with a hashtable, not a tree, and will work as long as there's an equivalence test and a hash function. Oct 1, 2008 at 15:29
• Upvoted of course for the help, but not accepted since it's missing an important point (which I cover in my answer). Jan 6, 2009 at 16:41
• FWIW, you can simplify your last foreach loop and return statement with this: `return dict.All(kvp => kvp.Value == 0);` Feb 26, 2016 at 12:29
This is my (heavily influenced by D.Jennings) generic implementation of the comparison method (in C#):
``````/// <summary>
/// Represents a service used to compare two collections for equality.
/// </summary>
/// <typeparam name="T">The type of the items in the collections.</typeparam>
public class CollectionComparer<T>
{
/// <summary>
/// Compares the content of two collections for equality.
/// </summary>
/// <param name="foo">The first collection.</param>
/// <param name="bar">The second collection.</param>
/// <returns>True if both collections have the same content, false otherwise.</returns>
public bool Execute(ICollection<T> foo, ICollection<T> bar)
{
// Declare a dictionary to count the occurence of the items in the collection
Dictionary<T, int> itemCounts = new Dictionary<T,int>();
// Increase the count for each occurence of the item in the first collection
foreach (T item in foo)
{
if (itemCounts.ContainsKey(item))
{
itemCounts[item]++;
}
else
{
itemCounts[item] = 1;
}
}
// Wrap the keys in a searchable list
List<T> keys = new List<T>(itemCounts.Keys);
// Decrease the count for each occurence of the item in the second collection
foreach (T item in bar)
{
// Try to find a key for the item
// The keys of a dictionary are compared by reference, so we have to
// find the original key that is equivalent to the "item"
// You may want to override ".Equals" to define what it means for
// two "T" objects to be equal
T key = keys.Find(
delegate(T listKey)
{
return listKey.Equals(item);
});
// Check if a key was found
if(key != null)
{
itemCounts[key]--;
}
else
{
// There was no occurence of this item in the first collection, thus the collections are not equal
return false;
}
}
// The count of each item should be 0 if the contents of the collections are equal
foreach (int value in itemCounts.Values)
{
if (value != 0)
{
return false;
}
}
// The collections are equal
return true;
}
}
``````
• Nice job, but Note: 1. In contrast to Daniel Jennings solution, This is not O(N) but rather O(N^2), because of the find function inside the foreach loop on the bar collection; 2. You can generalize the method to accept IEnumerable<T> instead of ICollection<T> with no further modification to the code Apr 1, 2010 at 23:20
• `The keys of a dictionary are compared by reference, so we have to find the original key that is equivalent to the "item"` - this is not true. The algorithm is based on wrong assumptions and while works, it is terribly inefficient. Dec 21, 2018 at 6:55
You could use a Hashset. Look at the SetEquals method.
• of course, using a HashSet assumes no duplicates but if so HashSet is the best way to go Oct 4, 2008 at 4:40
• As `ToHashSet()` is now built into Linq, the `SetEquals()` solution can be written as a very compact, efficient one-liner: `collection1.ToHashSet().SetEquals(collection2)`. That doesn't support duplicates, but it's easily the shortest answer, it requires no external libraries, and it even runs in amortized O(n) time. Sep 5 at 14:44
``````static bool SetsContainSameElements<T>(IEnumerable<T> set1, IEnumerable<T> set2) {
var setXOR = new HashSet<T>(set1);
setXOR.SymmetricExceptWith(set2);
return (setXOR.Count == 0);
}
``````
Solution requires .NET 3.5 and the `System.Collections.Generic` namespace. According to Microsoft, `SymmetricExceptWith` is an O(n + m) operation, with n representing the number of elements in the first set and m representing the number of elements in the second. You could always add an equality comparer to this function if necessary.
• Interesting and rare fact. Thanks for the knowledge Mar 24, 2021 at 21:18
• Best answer here, concise, correct and fast. Should be upvoted. Aug 16, 2021 at 4:51
If you use Shouldly, you can use ShouldAllBe with Contains.
``````collection1 = {1, 2, 3, 4};
collection2 = {2, 4, 1, 3};
collection1.ShouldAllBe(item=>collection2.Contains(item)); // true
``````
And finally, you can write an extension.
``````public static class ShouldlyIEnumerableExtensions
{
public static void ShouldEquivalentTo<T>(this IEnumerable<T> list, IEnumerable<T> equivalent)
{
list.ShouldAllBe(l => equivalent.Contains(l));
}
}
``````
UPDATE
A optional parameter exists on ShouldBe method.
``````collection1.ShouldBe(collection2, ignoreOrder: true); // true
``````
• I've just found on latest version that there is a parameter `bool ignoreOrder` on ShouldBe method. Nov 14, 2017 at 15:28
• Fantastic reference to Shouldly. Nov 23, 2021 at 19:47
EDIT: I realized as soon as I posed that this really only works for sets -- it will not properly deal with collections that have duplicate items. For example { 1, 1, 2 } and { 2, 2, 1 } will be considered equal from this algorithm's perspective. If your collections are sets (or their equality can be measured that way), however, I hope you find the below useful.
The solution I use is:
``````return c1.Count == c2.Count && c1.Intersect(c2).Count() == c1.Count;
``````
Linq does the dictionary thing under the covers, so this is also O(N). (Note, it's O(1) if the collections aren't the same size).
I did a sanity check using the "SetEqual" method suggested by Daniel, the OrderBy/SequenceEquals method suggested by Igor, and my suggestion. The results are below, showing O(N*LogN) for Igor and O(N) for mine and Daniel's.
I think the simplicity of the Linq intersect code makes it the preferable solution.
``````__Test Latency(ms)__
N, SetEquals, OrderBy, Intersect
1024, 0, 0, 0
2048, 0, 0, 0
4096, 31.2468, 0, 0
8192, 62.4936, 0, 0
16384, 156.234, 15.6234, 0
32768, 312.468, 15.6234, 46.8702
65536, 640.5594, 46.8702, 31.2468
131072, 1312.3656, 93.7404, 203.1042
262144, 3765.2394, 187.4808, 187.4808
524288, 5718.1644, 374.9616, 406.2084
1048576, 11420.7054, 734.2998, 718.6764
2097152, 35090.1564, 1515.4698, 1484.223
``````
• The only issue with this code is that it only works when comparing value types or comparing the pointers to reference types. I could have two different instances of the same object in the collections, so I need to be able to specify how to compare each. Can you pass a comparison delegate to the intersect method? Jun 19, 2009 at 12:59
• Sure, you can pass a comparer delegate. But, note the above limitation regarding sets that I added, which puts a significant limit on its applicability.
– Schmidty
Jun 19, 2009 at 14:39
• The Intersect method returns a distinct collection. Given a = {1,1,2} and b ={2,2,1}, a.Intersect(b).Count() != a.Count, which causes your expression to correctly return false. {1,2}.Count != {1,1,2}.Count See link[/link] (Note that both sides are made distinct before comparison.) Aug 12, 2016 at 18:29
In the case of no repeats and no order, the following EqualityComparer can be used to allow collections as dictionary keys:
``````public class SetComparer<T> : IEqualityComparer<IEnumerable<T>>
where T:IComparable<T>
{
public bool Equals(IEnumerable<T> first, IEnumerable<T> second)
{
if (first == second)
return true;
if ((first == null) || (second == null))
return false;
return first.ToHashSet().SetEquals(second);
}
public int GetHashCode(IEnumerable<T> enumerable)
{
int hash = 17;
foreach (T val in enumerable.OrderBy(x => x))
hash = hash * 23 + val.GetHashCode();
return hash;
}
}
``````
Here is the ToHashSet() implementation I used. The hash code algorithm comes from Effective Java (by way of Jon Skeet).
• What is the point of Serializable for Comparer class? :o Also you can change the input to `ISet<T>` to express it is meant for sets (ie no duplicates). Jun 24, 2015 at 17:29
• @nawfal thanks, don't know what I was thinking when I marked it Serializable... As for `ISet`, the idea here was to treat the `IEnumerable` as a set (because you got an `IEnumerable` to begin with), though considering the 0 upvotes in over 5 years that may not have been the sharpest idea :P Jun 24, 2015 at 21:24
Why not use .Except()
``````// Create the IEnumerable data sources.
// Create the query. Note that method syntax must be used here.
IEnumerable<string> differenceQuery = names1.Except(names2);
// Execute the query.
Console.WriteLine("The following lines are in names1.txt but not names2.txt");
foreach (string s in differenceQuery)
Console.WriteLine(s);
``````
http://msdn.microsoft.com/en-us/library/bb397894.aspx
• `Except` won't work for counting duplicate items. It will return true for sets {1,2,2} and {1,1,2}. Jan 31, 2013 at 9:07
• @CristiDiaconescu you could do a ".Distinct()" first to remove any duplicates Apr 12, 2018 at 14:04
• The OP is asking for `[1, 1, 2] != [1, 2, 2]` . Using `Distinct` would make them look equal. Apr 14, 2018 at 22:04
A duplicate post of sorts, but check out my solution for comparing collections. It's pretty simple:
This will perform an equality comparison regardless of order:
``````var list1 = new[] { "Bill", "Bob", "Sally" };
var list2 = new[] { "Bob", "Bill", "Sally" };
bool isequal = list1.Compare(list2).IsSame;
``````
This will check to see if items were added / removed:
``````var list1 = new[] { "Billy", "Bob" };
var list2 = new[] { "Bob", "Sally" };
var diff = list1.Compare(list2);
var onlyinlist1 = diff.Removed; //Billy
var inbothlists = diff.Equal; //Bob
``````
This will see what items in the dictionary changed:
``````var original = new Dictionary<int, string>() { { 1, "a" }, { 2, "b" } };
var changed = new Dictionary<int, string>() { { 1, "aaa" }, { 2, "b" } };
var diff = original.Compare(changed, (x, y) => x.Value == y.Value, (x, y) => x.Value == y.Value);
foreach (var item in diff.Different)
Console.Write("{0} changed to {1}", item.Key.Value, item.Value.Value);
//Will output: a changed to aaa
``````
Original post here.
This simple solution forces the `IEnumerable`'s generic type to implement `IComparable`. Because of `OrderBy`'s definition.
If you don't want to make such an assumption but still want use this solution, you can use the following piece of code :
``````bool equal = collection1.OrderBy(i => i?.GetHashCode())
.SequenceEqual(collection2.OrderBy(i => i?.GetHashCode()));
``````
Here's my extension method variant of ohadsc's answer, in case it's useful to someone
``````static public class EnumerableExtensions
{
static public bool IsEquivalentTo<T>(this IEnumerable<T> first, IEnumerable<T> second)
{
if ((first == null) != (second == null))
return false;
if (!object.ReferenceEquals(first, second) && (first != null))
{
if (first.Count() != second.Count())
return false;
if ((first.Count() != 0) && HaveMismatchedElement<T>(first, second))
return false;
}
return true;
}
private static bool HaveMismatchedElement<T>(IEnumerable<T> first, IEnumerable<T> second)
{
int firstCount;
int secondCount;
var firstElementCounts = GetElementCounts<T>(first, out firstCount);
var secondElementCounts = GetElementCounts<T>(second, out secondCount);
if (firstCount != secondCount)
return true;
foreach (var kvp in firstElementCounts)
{
firstCount = kvp.Value;
secondElementCounts.TryGetValue(kvp.Key, out secondCount);
if (firstCount != secondCount)
return true;
}
return false;
}
private static Dictionary<T, int> GetElementCounts<T>(IEnumerable<T> enumerable, out int nullCount)
{
var dictionary = new Dictionary<T, int>();
nullCount = 0;
foreach (T element in enumerable)
{
if (element == null)
{
nullCount++;
}
else
{
int num;
dictionary.TryGetValue(element, out num);
num++;
dictionary[element] = num;
}
}
return dictionary;
}
static private int GetHashCode<T>(IEnumerable<T> enumerable)
{
int hash = 17;
foreach (T val in enumerable.OrderBy(x => x))
hash = hash * 23 + val.GetHashCode();
return hash;
}
}
``````
• How well does this perform, any ideas? Nov 8, 2013 at 20:04
• I only use this for small collections, so have not thought about Big-O complexity or done benchmarking. HaveMismatchedElements alone is O(M*N) so it may not perform well for large collections. Nov 8, 2013 at 21:22
• If `IEnumerable<T>`s are queries then calling `Count()` is not a good idea. Ohad's original answer's approach of checking if they are `ICollection<T>` is the better idea. Jun 28, 2015 at 10:22
Here is a solution which is an improvement over this one.
``````public static bool HasSameElementsAs<T>(
this IEnumerable<T> first,
IEnumerable<T> second,
IEqualityComparer<T> comparer = null)
{
var firstMap = first
.GroupBy(x => x, comparer)
.ToDictionary(x => x.Key, x => x.Count(), comparer);
var secondMap = second
.GroupBy(x => x, comparer)
.ToDictionary(x => x.Key, x => x.Count(), comparer);
if (firstMap.Keys.Count != secondMap.Keys.Count)
return false;
if (firstMap.Keys.Any(k1 => !secondMap.ContainsKey(k1)))
return false;
return firstMap.Keys.All(x => firstMap[x] == secondMap[x]);
}
``````
`````` public static bool AreEquivalentIgnoringDuplicates<T>(this IEnumerable<T> items, IEnumerable<T> otherItems)
{
var itemList = items.ToList();
var otherItemList = otherItems.ToList();
var except = itemList.Except(otherItemList);
return itemList.Count == otherItemList.Count && except.IsEmpty();
}
public static bool AreEquivalent<T>(this IEnumerable<T> items, IEnumerable<T> otherItems)
{
var itemList = items.ToList();
var otherItemList = otherItems.ToList();
var except = itemList.Except(otherItemList);
return itemList.Distinct().Count() == otherItemList.Count && except.IsEmpty();
}
``````
Tests for these two:
`````` [Test]
public void collection_with_duplicates_are_equivalent()
{
var a = new[] {1, 5, 5};
var b = new[] {1, 1, 5};
a.AreEquivalentIgnoringDuplicates(b).ShouldBe(true);
}
[Test]
public void collection_with_duplicates_are_not_equivalent()
{
var a = new[] {1, 5, 5};
var b = new[] {1, 1, 5};
a.AreEquivalent(b).ShouldBe(false);
}
``````
erickson is almost right: since you want to match on counts of duplicates, you want a Bag. In Java, this looks something like:
``````(new HashBag(collection1)).equals(new HashBag(collection2))
``````
I'm sure C# has a built-in Set implementation. I would use that first; if performance is a problem, you could always use a different Set implementation, but use the same Set interface.
There are many solutions to this problem. If you don't care about duplicates, you don't have to sort both. First make sure that they have the same number of items. After that sort one of the collections. Then binsearch each item from the second collection in the sorted collection. If you don't find a given item stop and return false. The complexity of this: - sorting the first collection: NLog(N) - searching each item from second into the first: NLOG(N) so you end up with 2*N*LOG(N) assuming that they match and you look up everything. This is similar to the complexity of sorting both. Also this gives you the benefit to stop earlier if there's a difference. However, keep in mind that if both are sorted before you step into this comparison and you try sorting by use something like a qsort, the sorting will be more expensive. There are optimizations for this. Another alternative, which is great for small collections where you know the range of the elements is to use a bitmask index. This will give you a O(n) performance. Another alternative is to use a hash and look it up. For small collections it is usually a lot better to do the sorting or the bitmask index. Hashtable have the disadvantage of worse locality so keep that in mind. Again, that's only if you don't care about duplicates. If you want to account for duplicates go with sorting both.
In many cases the only suitable answer is the one of Igor Ostrovsky , other answers are based on objects hash code. But when you generate an hash code for an object you do so only based on his IMMUTABLE fields - such as object Id field (in case of a database entity) - Why is it important to override GetHashCode when Equals method is overridden?
This means , that if you compare two collections , the result might be true of the compare method even though the fields of the different items are non-equal . To deep compare collections , you need to use Igor's method and implement IEqualirity .
James
Allowing for duplicates in the `IEnumerable<T>` (if sets are not desirable\possible) and "ignoring order" you should be able to use a `.GroupBy()`.
I'm not an expert on the complexity measurements, but my rudimentary understanding is that this should be O(n). I understand O(n^2) as coming from performing an O(n) operation inside another O(n) operation like `ListA.Where(a => ListB.Contains(a)).ToList()`. Every item in ListB is evaluated for equality against each item in ListA.
Like I said, my understanding on complexity is limited, so correct me on this if I'm wrong.
``````public static bool IsSameAs<T, TKey>(this IEnumerable<T> source, IEnumerable<T> target, Expression<Func<T, TKey>> keySelectorExpression)
{
// check the object
if (source == null && target == null) return true;
if (source == null || target == null) return false;
var sourceList = source.ToList();
var targetList = target.ToList();
// check the list count :: { 1,1,1 } != { 1,1,1,1 }
if (sourceList.Count != targetList.Count) return false;
var keySelector = keySelectorExpression.Compile();
var groupedSourceList = sourceList.GroupBy(keySelector).ToList();
var groupedTargetList = targetList.GroupBy(keySelector).ToList();
// check that the number of grouptings match :: { 1,1,2,3,4 } != { 1,1,2,3,4,5 }
var groupCountIsSame = groupedSourceList.Count == groupedTargetList.Count;
if (!groupCountIsSame) return false;
// check that the count of each group in source has the same count in target :: for values { 1,1,2,3,4 } & { 1,1,1,2,3,4 }
// key:count
// { 1:2, 2:1, 3:1, 4:1 } != { 1:3, 2:1, 3:1, 4:1 }
var countsMissmatch = groupedSourceList.Any(sourceGroup =>
{
var targetGroup = groupedTargetList.Single(y => y.Key.Equals(sourceGroup.Key));
return sourceGroup.Count() != targetGroup.Count();
});
return !countsMissmatch;
}
``````
If comparing for the purpose of Unit Testing Assertions, it may make sense to throw some efficiency out the window and simply convert each list to a string representation (csv) before doing the comparison. That way, the default test Assertion message will display the differences within the error message.
Usage:
``````using Microsoft.VisualStudio.TestTools.UnitTesting;
// define collection1, collection2, ...
Assert.Equal(collection1.OrderBy(c=>c).ToCsv(), collection2.OrderBy(c=>c).ToCsv());
``````
Helper Extension Method:
``````public static string ToCsv<T>(
this IEnumerable<T> values,
Func<T, string> selector,
string joinSeparator = ",")
{
if (selector == null)
{
if (typeof(T) == typeof(Int16) ||
typeof(T) == typeof(Int32) ||
typeof(T) == typeof(Int64))
{
selector = (v) => Convert.ToInt64(v).ToStringInvariant();
}
else if (typeof(T) == typeof(decimal))
{
selector = (v) => Convert.ToDecimal(v).ToStringInvariant();
}
else if (typeof(T) == typeof(float) ||
typeof(T) == typeof(double))
{
selector = (v) => Convert.ToDouble(v).ToString(CultureInfo.InvariantCulture);
}
else
{
selector = (v) => v.ToString();
}
}
return String.Join(joinSeparator, values.Select(v => selector(v)));
}
``````
Here's my stab at the problem. It's based on this strategy but also borrows some ideas from the accepted answer.
``````public static class EnumerableExtensions
{
public static bool SequenceEqualUnordered<TSource>(this IEnumerable<TSource> source, IEnumerable<TSource> second)
{
return SequenceEqualUnordered(source, second, EqualityComparer<TSource>.Default);
}
public static bool SequenceEqualUnordered<TSource>(this IEnumerable<TSource> source, IEnumerable<TSource> second, IEqualityComparer<TSource> comparer)
{
if (source == null)
throw new ArgumentNullException(nameof(source));
if (second == null)
throw new ArgumentNullException(nameof(second));
if (source.TryGetCount(out int firstCount) && second.TryGetCount(out int secondCount))
{
if (firstCount != secondCount)
return false;
if (firstCount == 0)
return true;
}
IEqualityComparer<ValueTuple<TSource>> wrapperComparer = comparer != null ? new WrappedItemComparer<TSource>(comparer) : null;
Dictionary<ValueTuple<TSource>, int> counters;
ValueTuple<TSource> key;
int counter;
using (IEnumerator<TSource> enumerator = source.GetEnumerator())
{
if (!enumerator.MoveNext())
return !second.Any();
counters = new Dictionary<ValueTuple<TSource>, int>(wrapperComparer);
do
{
key = new ValueTuple<TSource>(enumerator.Current);
if (counters.TryGetValue(key, out counter))
counters[key] = counter + 1;
else
}
while (enumerator.MoveNext());
}
foreach (TSource item in second)
{
key = new ValueTuple<TSource>(item);
if (counters.TryGetValue(key, out counter))
{
if (counter <= 0)
return false;
counters[key] = counter - 1;
}
else
return false;
}
return counters.Values.All(cnt => cnt == 0);
}
private static bool TryGetCount<TSource>(this IEnumerable<TSource> source, out int count)
{
switch (source)
{
case ICollection<TSource> collection:
count = collection.Count;
return true;
return true;
case ICollection nonGenericCollection:
count = nonGenericCollection.Count;
return true;
default:
count = default;
return false;
}
}
private sealed class WrappedItemComparer<TSource> : IEqualityComparer<ValueTuple<TSource>>
{
public WrappedItemComparer(IEqualityComparer<TSource> comparer)
{
_comparer = comparer;
}
public bool Equals(ValueTuple<TSource> x, ValueTuple<TSource> y) => _comparer.Equals(x.Item1, y.Item1);
public int GetHashCode(ValueTuple<TSource> obj) => _comparer.GetHashCode(obj.Item1);
}
}
``````
Improvements on the MS solution:
• Doesn't take the `ReferenceEquals(first, second)` shortcut because it's kind of debatable. For example, consider a custom `IEnumerable<T>` which has an implementation like this: `public IEnumerator<T> GetEnumerator() => Enumerable.Repeat(default(T), new Random().Next(10)).GetEnumerator()`.
• Takes possible shortcuts when both enumerable is a collection but checks not only for `ICollection<T>` but also for other collection interfaces.
• Handles null values properly. Counting null values separately from the other (non-null) values also doesn't look 100% fail-safe. Consider a custom equality comparer which handles null values in a non-standard way.
This solution is also available in my utility NuGet package.
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https://www.khanacademy.org/math/calculus-all-old/integration-calc/improper-integrals-calc/v/divergent-improper-integral
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# Divergent improper integral
AP.CALC:
LIM‑6 (EU)
,
LIM‑6.A (LO)
,
LIM‑6.A.1 (EK)
,
LIM‑6.A.2 (EK)
## Video transcript
right here I've graphed part of the graph of y is equal to one over X and what I'm curious about is the area under this curve and above the x-axis between x equals one and infinity so I want to figure out this area right over here so let's try to do it so we could set this up as an improper integral going from 1 to infinity of 1 over X 1 over X DX well we can once again we can view this as the limit the limit actually me do that same yellow color I like that more we can view this as the limit as n approaches infinity of the integral from 1 to N of 1 over X DX which we can write as the limit as n approaches infinity of the antiderivative of 1 over X which is the natural log of the absolute value of x so this is going to be the natural log of the absolute value of x and the absolute value of x won't really matter so much we could just say X because we're dealing with positive values of x but I'll just write down as the natural log of the absolute value of x between X is 1 and X is N and so this is going to be equal to the limit as n approaches infinity of you evaluate this at n so you're going to get the natural log I could write the absolute value of n but we know that n is going to be positive so we can just write the natural log of n minus the natural log of the absolute value of 1 or the natural log of 1 natural log of 1 is just 0 e to the 0 power is equal to 1 so this boils down to the limit as n approaches infinity of the natural log of n now this is interesting natural log function just keeps getting larger and larger and larger it kind of natural log function looks something like keeps growing and growing and growing like this I'll be at a at a slower and slower pace but it keeps growing the limit as n approaches infinity of the natural log of n is just equal to infinity so here we do not have a finite area this is an infinite this is an infinite area it's interesting when this function decreased faster when it was 1 over x squared we had a finite area now we have an infinite area and so we would say that this integral right over here is this improper integral is divergent divergent and we're done
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http://mathcs.holycross.edu/~groberts/Courses/Calculus/MA126/2008/Exams/Exam1.html
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Calculus for Social Sciences II (Sections 01 and 02)
Exam #1
Thursday, Feb. 14, In Class
The first exam covers Sections 4.9 through 5.5. It is recommended that you go over homework problems (HW#1 - 3) as well as your class notes. Many of the problems and questions we discuss in class are excellent examples of test questions. I have also listed some sample problems from the Chapter 4 and 5 Review Exercises below. The odd answers are in the back of the book while the evens are listed here. The Concept-Check at the end of each chapter (before the exercises) is also a source for good questions. The exam will be designed to take roughly one hour although you will have the full class period (plus a little extra) if necessary.
Exam Review Session: Tuesday, Feb. 12, 7:00 - 8:30 pm in O'Neil 112.
Note: You will be given a scientific calculator for the exam which does NOT have graphing capabilities so be prepared to answer questions without your personal calculator.
Chapter 4 Review Exercises, pp. 336 - 338
Problems: 49, 51, 52, 53
The answers to the evens are:
52. f(u) = (1/2)u^2 + 2u^(1/2) + 1/2
Chapter 5 Review Exercises, pp. 434 - 436
Problems: 1, 3, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 21, 25, 33, 34, 39, 40, 41
The answers to the evens are:
8. (a) e^(Pi/4) - 1, (b) 0, (c) e^(arctan x)
12. 1/10
16. Pi/4
34. 0
40. -(sin x)^(5/3)
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https://www.answers.com/Q/How_many_gallons_of_water_to_fill_a_pool_that%27s_12%27_round_and_30_high
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How many gallons of water to fill a pool that's 12' round and 30 high?
When filled to 80% water capacity, as recommended, the pool holds approximately:
10'x30" = 915 gal
12'x30" = 1,353 gal
12'x36" = 1,722 gal
15'x36" = 2,561 gal
15'x42" = 3,110 gal
16'x42" = 3,896 gal
18'x36" = 3,735 gal
18'x42" = 4,814 gal
18'x48" = 5,593 gal
24'x48" = 10,103 gal
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https://short-q.com/what-is-another-word-that-means-divide/
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# Short-Questions
Fast solutions for complex problems
## What is another word that means divide?
Frequently Asked Questions About divide Some common synonyms of divide are divorce, part, separate, sever, and sunder. While all these words mean “to become or cause to become disunited or disjointed,” divide implies separating into pieces or sections by cutting or breaking.
## What is another word for divide in math?
Division-quotient, dividend, divide, divided by, each, per, average, divided equally.
What word means to divide into groups?
1 bisect, cleave, cut (up), detach, disconnect, part, partition, segregate, separate, sever, shear, split, subdivide, sunder. 2 allocate, allot, apportion, deal out, dispense, distribute, divvy (up) (informal) dole out, measure out, portion, share.
What is another symbol for division?
Other symbols for division include the slash or solidus /, the colon :, and the fraction bar (the horizontal bar in a vertical fraction).
### What is another word for division or divide?
What is another word for division?
separation dividing
detachment fractionalization
fractionation parcellingUK
parcelingUS parting
schism scission
### What is the opposite divide?
(join) Opposite of to separate or be separated into parts. join. attach.
Is there a divide symbol on Iphone?
Go to settings> general> keyboard> shortcuts> tap the plus sign> paste the division sign on the top line, ds on the bottom lne. Tap save. Now all you have to do is tap ds and the space bar for your division sign Will appear.
How do I write divided in Word?
Insert a Symbol
1. The Symbol menu rearranges itself to feature your most-used symbols, so the division symbol might not show up if you’ve used many other symbols. Click More Symbols to see the full list.
2. To type a division symbol using the keyboard, press Alt-0247 on the numeric keypad with Num Lock turned on.
#### Which is the odd word out from Divide converge separate?
Finding Odd Word Out Divide – separate or be separated into parts. Diverge – separate from another route and go in a different direction.
#### What does ➗ mean?
➗ Meaning – Heavy Division Sign Emoji The Heavy Division Sign Emoji was added to the Symbols category in 2010 as part of Unicode 6.0 standard. This is a mature emoji and it should work on most devices.
What are symbols for division?
How do you type 3 4 on a keyboard?
A few common fractions have keyboard shortcuts that you can use by holding down the Alt key and typing the code numbers.
1. 1/2 = Alt + 0 1 8 9.
2. 1/4 = Alt + 0 1 8 8.
3. 3/4 = Alt + 0 1 9 0.
## What are some words that mean Division?
Basic Math Definitions The Basic Operations Addition is Subtraction is Minuend : The number that is to be subtracted from. Multiplication is Example: 3.5 × 5 = 17.5 Division is It is the result of “fair sharing”. A Fraction is The top part (the numerator) says how many parts we have. A Decimal Number is … A Percentage is … Average (Mean) is …
## What does the name divide mean?
A distancing between two people or things. There is a great divide between us. (geography) A large chasm, gorge, or ravine between two areas of land. If you’re heading to the coast, you’ll have to cross the divide first. A dividing point or line. To separate into parts; split up; sever. To separate into groups; classify.
What is the definition of dividend in math?
Mathematical Definition of Dividend. Dividend is the number to be divided. The word dividend means the number that is to be divided. 8/2 means ‘eight divided by two’ and in this case 8 is the dividend, the number that we will divide by 2 to get the answer (4) The above problem could be alternately written 8 ÷ 2 or.
What is the antonym of divided?
divide(noun) a serious disagreement between two groups of people (typically producing tension or hostility) Antonyms: multiply, unite, unify. Synonyms: water parting, watershed. watershed, water parting, divide(verb) a ridge of land that separates two adjacent river systems. Antonyms: unite, multiply, unify. Synonyms:
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# Planck length
(Redirected from Planck Length)
Planck length
Unit systemPlanck units
Unit oflength
SymbolP
Conversions
1 P in ...... is equal to ...
SI units 1.616229(38)×10−35 m
natural units 11.706 S
3.0542×10−25 a0
imperial/US units 6.3631×10−34 in
In physics, the Planck length, denoted P, is a unit of length that is the distance light travels in one unit of Planck time. It is equal to 1.616229(38)×10−35 meters. It is a base unit in the system of Planck units, developed by physicist Max Planck. The Planck length can be defined from three fundamental physical constants: the speed of light in a vacuum, the Planck constant, and the gravitational constant.
## Value
The Planck length P is defined as:
${\displaystyle \ell _{\mathrm {P} }={\sqrt {\frac {\hbar G}{c^{3}}}}}$
Solving the above will show the approximate equivalent value of this unit with respect to the meter:
${\displaystyle 1\ \ell _{\mathrm {P} }\approx 1.616\;229(38)\times 10^{-35}\ \mathrm {m} }$
where ${\displaystyle c}$ is the speed of light in a vacuum, G is the gravitational constant, and ħ is the reduced Planck constant. The two digits enclosed by parentheses are the estimated standard error associated with the reported numerical value.[1][2]
The Planck length is about 10−20 times the diameter of a proton.[3] It can be defined using the radius of the hypothesized Planck particle. The Planck length is represented by a Greek letter lambda (λ).
## History
In 1899 Max Planck suggested that there existed some fundamental natural units for length, mass, time and energy.[4][5] These he derived using dimensional analysis, using only the Newton gravitational constant, the speed of light and the "unit of action", which later became the Planck constant. The natural units he further derived became known as the "Planck length", the "Planck mass", the "Planck time" and the "Planck energy".
## Theoretical significance
The Planck length is the scale at which quantum gravitational effects are believed to begin to be apparent, where interactions require a working theory of quantum gravity to be analyzed.[6] The Planck area is the area by which the surface of a spherical black hole increases when the black hole swallows one bit of information.[dubious ][7] To measure anything the size of Planck length, the photon momentum needs to be very large due to Heisenberg's uncertainty principle and so much energy in such a small space would create a tiny Black hole with the diameter of its event horizon equal to a Planck length.
The Planck length is sometimes misconceived as the minimum length of space-time, but this is not accepted by conventional physics, as this would require violation or modification of Lorentz symmetry.[6] However, certain theories of loop quantum gravity do attempt to establish a minimum length on the scale of the Planck length, though not necessarily the Planck length itself,[6] or attempt to establish the Planck length as observer-invariant, known as doubly special relativity.
The strings of string theory are modeled to be on the order of the Planck length.[6][8] In theories of large extra dimensions, the Planck length has no fundamental physical significance, and quantum gravitational effects appear at other scales.[citation needed]
## Planck length and Euclidean geometry
The Planck length is the length at which quantum zero oscillations of the gravitational field completely distort Euclidean geometry.[9] The gravitational field performs zero-point oscillations, and the geometry associated with it also oscillates. The ratio of the circumference to the radius varies near the Euclidean value. The smaller the scale, the greater the deviations from the Euclidean geometry. Let us estimate the order of the wavelength of zero gravitational oscillations, at which the geometry becomes completely unlike the Euclidean geometry. The degree of deviation ${\displaystyle \zeta }$ of geometry from Euclidean geometry in the gravitational field is determined by the ratio of the gravitational potential ${\displaystyle \varphi }$ and the square of the speed of light ${\displaystyle c}$: ${\displaystyle \zeta =\varphi /c^{2}}$. When ${\displaystyle \zeta \ll 1}$, the geometry is close to Euclidean geometry; for ${\displaystyle \zeta \sim 1}$, all similarities disappear. The energy of the oscillation of scale ${\displaystyle l}$ is equal to ${\displaystyle E=\hbar \nu \sim \hbar c/l}$ (where ${\displaystyle c/l}$ is the order of the oscillation frequency). The gravitational potential created by the mass ${\displaystyle m}$, at this length is ${\displaystyle \varphi =Gm/l}$, where ${\displaystyle G}$ is the constant of universal gravitation. Instead of ${\displaystyle m}$, we must substitute a mass, which, according to Einstein's formula, corresponds to the energy ${\displaystyle E}$ (where ${\displaystyle m=E/c^{2}}$). We get ${\displaystyle \varphi =GE/l\,c^{2}=G\hbar /l^{2}c}$. Dividing this expression by ${\displaystyle c^{2}}$, we obtain the value of the deviation ${\displaystyle \zeta =G\hbar /c^{3}l^{2}=\ell _{P}^{2}/l^{2}}$. Equating ${\displaystyle \zeta =1}$, we find the length at which the Euclidean geometry is completely distorted. It is equal to Planck length ${\textstyle \ell _{P}={\sqrt {G\hbar /c^{3}}}\approx 10^{-35}\mathrm {m} }$. [10][jargon]
As noted in,[11] "for the space-time region with dimensions ${\displaystyle l}$ the uncertainty of the Christoffel symbols ${\displaystyle \Delta \Gamma }$ be of the order of ${\displaystyle \ell _{P}^{2}/l^{3}}$, and the uncertainty of the metric tensor ${\displaystyle \Delta g}$ is of the order of ${\displaystyle \ell _{P}^{2}/l^{2}}$. If ${\displaystyle l}$ is a macroscopic length, the quantum constraints are fantastically small and can be neglected even on atomic scales. If the value ${\displaystyle l}$ is comparable to ${\displaystyle \ell _{P}}$, then the maintenance of the former (usual) concept of space becomes more and more difficult and the influence of micro curvature becomes obvious". Conjecturally, this could imply that space-time becomes a quantum foam at the Planck scale.[12]
## References
1. ^
2. ^ "Planck length". NIST. Archived from the original on 22 November 2018. Retrieved 7 January 2019.
3. ^ "The Planck Length". www.math.ucr.edu. Retrieved 2018-12-16.
4. ^ M. Planck. Naturlische Masseinheiten. Der Koniglich Preussischen Akademie Der Wissenschaften, p. 479, 1899
5. ^ Gorelik, Gennady (1992). "First Steps of Quantum Gravity and the Planck Values". Boston University. Retrieved 7 January 2019.
6. ^ a b c d Klotz, Alex (2015-09-09). "A Hand-Wavy Discussion of the Planck Length". Physics Forums Insights. Retrieved 2018-03-23.
7. ^ Bekenstein, Jacob D (1973). "Black Holes and Entropy". Physical Review D. 7 (8): 2333. Bibcode:1973PhRvD...7.2333B. doi:10.1103/PhysRevD.7.2333.
8. ^ Cliff Burgess; Fernando Quevedo (November 2007). "The Great Cosmic Roller-Coaster Ride". Scientific American (print). Scientific American, Inc. p. 55.
9. ^ Migdal A.B., The quantum physics, Nauka, p.116-117, (1989)
10. ^ Migdal, A B (1985-10-31). "Niels Bohr and quantum physics". Soviet Physics Uspekhi. 28 (10): 910–934. doi:10.1070/pu1985v028n10abeh003951. ISSN 0038-5670.
11. ^ T. Regge, Nuovo Cim. 7, 215 (1958). Gravitational fields and quantum mechanics
12. ^ Wheeler, J. A. (January 1955). "Geons". Physical Review. 97 (2): 511. Bibcode:1955PhRv...97..511W. doi:10.1103/PhysRev.97.511.
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Analytical Instruments and Spectroscopic Concepts
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The Age of Einstein
The Age of Einstein, is a brief introduction to Einstein's Theories of Special and General Relativity. It is a book for the inquisitive general reader who wishes to gain an understanding of the key ideas put forward by the greatest scientist of the 20th-century. No more than a modest grasp of High School Mathematics is required to follow the arguments.
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Hodges Health Career - Care Domains - Model
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Cotton: Building a Better Plant
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Corn: The Dynamic Genome
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Break the Science Barrier - Richard Dawkins - Part 1 of 3
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Crystal Growth Movies
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Artificial Intelligence: Natural Language Processing
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Bloc
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Elementary Mathematics: Times Tables I
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Individual Health Status and Racial Minority Concentration in US States and Counties
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A309685 Number of even parts appearing among the smallest parts of the partitions of n into 3 parts. 11
0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 9, 9, 12, 12, 15, 15, 18, 18, 22, 22, 26, 26, 30, 30, 35, 35, 40, 40, 45, 45, 51, 51, 57, 57, 63, 63, 70, 70, 77, 77, 84, 84, 92, 92, 100, 100, 108, 108, 117, 117, 126, 126, 135, 135, 145, 145, 155, 155, 165 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,9 LINKS Index entries for linear recurrences with constant coefficients, signature (1, 1, -1, 0, 0, 1, -1, -1, 1). FORMULA a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} ((j-1) mod 2). From Colin Barker, Aug 23 2019: (Start) G.f.: x^6 / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x + x^2)). a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) for n>8. (End) a(n) = A001840(floor((n-4)/2)) for n>=2. - Joerg Arndt, Aug 23 2019 EXAMPLE Figure 1: The partitions of n into 3 parts for n = 3, 4, ... 1+1+8 1+1+7 1+2+7 1+2+6 1+3+6 1+1+6 1+3+5 1+4+5 1+1+5 1+2+5 1+4+4 2+2+6 1+1+4 1+2+4 1+3+4 2+2+5 2+3+5 1+1+3 1+2+3 1+3+3 2+2+4 2+3+4 2+4+4 1+1+1 1+1+2 1+2+2 2+2+2 2+2+3 2+3+3 3+3+3 3+3+4 ... ----------------------------------------------------------------------- n | 3 4 5 6 7 8 9 10 ... ----------------------------------------------------------------------- a(n) | 0 0 0 1 1 2 2 3 ... ----------------------------------------------------------------------- MATHEMATICA LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 0, 0, 0, 0, 0, 1, 1, 2}, 80] (* Wesley Ivan Hurt, Aug 30 2019 *) CROSSREFS Cf. A001840, A026923, A026927, A309683, A309684, A309686, A309687, A309688, A309689, A309690, A309692, A309694. Sequence in context: A121260 A121261 A085885 * A309683 A283529 A064986 Adjacent sequences: A309682 A309683 A309684 * A309686 A309687 A309688 KEYWORD nonn,easy AUTHOR Wesley Ivan Hurt, Aug 12 2019 STATUS approved
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# In 1860, the Philological Society launched its effort to
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In 1860, the Philological Society launched its effort to [#permalink]
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13 Oct 2008, 17:46
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In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
(a) would take more than 60 years to complete, the Oxford English Dictionary had been
(b) took more than 60 years to complete, the Oxford English Dictionary was
(c) would take more than 60 years to complete, the Oxford English Dictionary was being
(d) would take more than 60 years to complete, the Oxford English Dictionary was
(e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Thanks,
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Re: SC - Philological Society [#permalink]
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13 Oct 2008, 21:55
KumarGMAT wrote:
In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
(a) would take more than 60 years to complete, the Oxford English Dictionary had been
(b) took more than 60 years to complete, the Oxford English Dictionary was
(c) would take more than 60 years to complete, the Oxford English Dictionary was being
(d) would take more than 60 years to complete, the Oxford English Dictionary was
(e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Thanks,
Straight B
I think first rule of sc is to keep the sentence concise and simple ,B does this !!!
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Re: SC - Philological Society [#permalink]
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13 Oct 2008, 22:22
we don't need would here ?
KumarGMAT wrote:
In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
(a) would take more than 60 years to complete, the Oxford English Dictionary had been
(b) took more than 60 years to complete, the Oxford English Dictionary was
(c) would take more than 60 years to complete, the Oxford English Dictionary was being
(d) would take more than 60 years to complete, the Oxford English Dictionary was
(e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Thanks,
The completion and the birth(or rather start) of the Oxford Dictionary did not take place at the same time !!
Option B puts both the tenses in past and that indicates that the completion and birth happened at the same time
Option A -> one action(birth) happened in the past and the other(completion) happened in the future so we cannot use HAD BEEN
Option C -> was being born -- something can take birth only at a particular instant in time -- being born seems as if the birth continued for a long time
Option E -> Changes the original intent completely. Completion did not take place before the dic was born !!
This leaves us with D..
OA is D ??
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Re: SC - Philological Society [#permalink]
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14 Oct 2008, 06:56
B is very concise.
OA is B.
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Re: SC - Philological Society [#permalink]
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14 Oct 2008, 13:05
I thought the answer was B. But, I need to clarify why option D is incorrect. Is it because 'would take' is actually a future tense in the past and does not go with the verb 'was born' since the events - (1) birth of the dictionary and (2) project taking 60 yrs - do not occur at the same time???
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14 Oct 2008, 14:38
Where did you get your OA?
http://www.manhattangmat.net.in/forums/ ... -t421.html
mrbgam wrote:
B is very concise.
OA is B.
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14 Oct 2008, 18:13
mrbgam wrote:
B is very concise.
OA is B.
mrbgam,
I can see this is your first post. OA=Official Answer and it is provided by the book/source of the question.
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14 Oct 2008, 23:13
Can we have the official answer alogn with the source
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15 Oct 2008, 02:18
1
I found this question in MGMAT CAT ...... following is their explanation ....
The past perfect ("had been born") is used when there are two past actions and we want to indicate which one happened first. In the underlined portion of the sentence, however, the other verb, "would take," is not in the past tense, so we need to use the simple past "was born." (Remember that we always use the most simple tenses allowed; the perfect tenses, and other complicated tenses, are used only when required by the sentence structure.) The second half of the sentence stands in contrast to the first half, in which the simple past "launched" is correctly paired with the past perfect "had seen."
(A) This choice is incorrect as it repeats the original sentence.
(B) This choice changes both the first and second verbs to simple past ("took" and "was born," respectively). In this circumstance, we have two events that took place at different times in the past, which requires use of the past perfect to indicate which event happened first. The dictionary's "birth" obviously happens before its completion, so correct usage would be that the "Dictionary had been born."
(C) The present participle "being" is used with the progressive tense to indicate a continuing or ongoing action. Logically, however, the Dictionary's start must have been at a single point in time, rather than over the course of the book's development.
(D) CORRECT. This choice correctly uses the simple past "was born." A more complicated past tense is not required because the other verb "would take," is not in the past tense.
(E) This choice incorrectly adopts the construction "was about to be born," which conflicts with the non-underlined portion of the sentence. The first half of the sentence indicates that the project was "launched" in 1860 in the past tense, making any reference to the book being "about to be born" at some future point in time incorrect.
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Re: SC - Philological Society [#permalink]
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15 Oct 2008, 03:32
I a more confused now. doesnt "would take" mean somthing which will happen in future?
amitdgr wrote:
I found this question in MGMAT CAT ...... following is their explanation ....
The past perfect ("had been born") is used when there are two past actions and we want to indicate which one happened first. In the underlined portion of the sentence, however, the other verb, "would take," is not in the past tense, so we need to use the simple past "was born." (Remember that we always use the most simple tenses allowed; the perfect tenses, and other complicated tenses, are used only when required by the sentence structure.) The second half of the sentence stands in contrast to the first half, in which the simple past "launched" is correctly paired with the past perfect "had seen."
(A) This choice is incorrect as it repeats the original sentence.
(B) This choice changes both the first and second verbs to simple past ("took" and "was born," respectively). In this circumstance, we have two events that took place at different times in the past, which requires use of the past perfect to indicate which event happened first. The dictionary's "birth" obviously happens before its completion, so correct usage would be that the "Dictionary had been born."
(C) The present participle "being" is used with the progressive tense to indicate a continuing or ongoing action. Logically, however, the Dictionary's start must have been at a single point in time, rather than over the course of the book's development.
(D) CORRECT. This choice correctly uses the simple past "was born." A more complicated past tense is not required because the other verb "would take," is not in the past tense.
(E) This choice incorrectly adopts the construction "was about to be born," which conflicts with the non-underlined portion of the sentence. The first half of the sentence indicates that the project was "launched" in 1860 in the past tense, making any reference to the book being "about to be born" at some future point in time incorrect.
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15 Oct 2008, 08:47
Thanks Amit i think this explainations gives some clarity..
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16 Oct 2008, 15:27
thanks for the MGMAT post amitdgr
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19 Oct 2008, 00:30
"although" here indicates clearly that birth and completion of dictionary are not the same events. they are at a difference of 60 years. So technically, the same simple past tense can't be used for both. So D is the answer
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Equation (2.170) reveals an important observation. The eigenvalue zero (λ= 0) can occur for two reasons: either the right side is a non-trivial linear combination, i.e., at least two coefficients (∗) are nonzero, or all coefficients equal zero. In the latter case, the eigenvector v is perpendicular to every ϕ(mk). But if the eigenvalue is nonzero (λ ≠ 0), then the right side must be a nonzero linear combination. Hence, v is a non-trivial linear combination of the ϕ(mk). To summarize, any eigenvector solution of Equation (2.169) that corresponds to a nonzero eigenvalue is in the span of {ϕ(m1), . . . , ϕ(mN)}.
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# Suppose we have the following model xt xt 1 t then
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Unformatted text preview: Durbin Watson statistics close to 0 3 look at autocorrelation function of Xt 4 look at the Augmented Dickey Fuller test statistic 20 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Testing for Nonstationarity There are a few ways of detecting nonstationarity of which were covered in the problem sets. 1 look at the the time series for Xt 2 look for Durbin Watson statistics close to 0 3 look at autocorrelation function of Xt 4 look at the Augmented Dickey Fuller test statistic 21 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Testing for Nonstationarity There are a few ways of detecting nonstationarity of which were covered in the problem sets. 1 look at the the time series for Xt 2 look for Durbin Watson statistics close to 0 3 look at autocorrelation function of Xt 4 look at the Augmented Dickey Fuller test statistic 22 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Autocorrelation function The autocorrelation function is a useful graphical tool for detecting nonstationarity. Suppose we have the following model: Xt = β Xt −1 + εt Then corr (Xt , Xt +s ) = β s So we can plot the sample autocorrelation function given by ρ(s) = ˆ ¯ ¯ (Xt − X )(Xt +s − X ) ¯ (Xt − X )2 ¯ (Xt +s − X )2 If the |β | < 1 then this quantity approximates the true autocorrelation function. 23 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Autocorrelation function The autocorrelation function is a useful graphical tool for detecting nonstationarity. Suppose we have the following model: Xt = β Xt −1 + εt Then corr (Xt , Xt +s ) = β s So we can plot the sample autocorrelation function given by ρ(s) = ˆ ¯ ¯ (Xt − X )(Xt +s − X ) ¯ (Xt − X )2 ¯ (Xt +s − X )2 If the |β | < 1 then this quantity approximates the true autocorrelation function. 24 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Autocorrelation function The autocorrelation function is a useful graphical tool for detecting nonstationarity. Suppose we have the following model: Xt = β Xt −1 + εt Then corr (Xt , Xt +s ) = β s So we can plot the sample autocorrelation function given by ρ(s) = ˆ ¯ ¯ (Xt − X )(Xt +s − X ) ¯ (Xt − X )2 ¯ (Xt +s − X )2 If the |β | < 1 then this quantity approximates the true autocorrelation function. 25 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration Autocorrelation function ADF Autocorrelation function The autocorrelation function is a useful graphical tool for detecting nonstationarity. Suppose we have the following model: Xt = β Xt −1 + εt Then corr (Xt , Xt +s ) = β s So we can plot the sample autocorr...
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# Cos Tan Cos90 Degrees, Prove That Cos 90 0 Theta Sec Theta Tan 180 0 Theta Sec 36
Cos Tan Cos90 Degrees. Tan90 is undefined because its value is infinite (tan90=sin90/cos90=1/0). This video teaches how can you memorize values of all six trigonometric. In terms of sines and cosines. Sin, cos ,tan are basic trigonometric functions. Because tan 90 would mean both theta and the right angle are 90 degrees and therefore the third angle would be 0 degrees which isn't possible. Therefore cos 90 degrees is 0. The values of sin and cos can be obtained from their graphs. Well, technically we've only shown this for angles between 0. The three most familiar trigonometric ratios in the trigonometric functions are sine function, cosine function, and tangent function. Sin(θ)=cos(90∘−θ)sine, left parenthesis, theta, right parenthesis, equals, cosine, left parenthesis, 90, degrees, minus, theta, right parenthesis. The expression contains a division by 0. All the cos values are filled in opposite order starting from sin 0∘. Learn all concepts of chapter 8 class 10 (with videos). In other words, the sine of an angle equals the cosine of its complement. O, 30, 45, 60 and 90 degrees are angles which are used very often in problems.
Cos Tan Cos90 Degrees Indeed recently has been hunted by users around us, maybe one of you personally. People are now accustomed to using the internet in gadgets to view image and video data for inspiration, and according to the name of the post I will talk about about Cos Tan Cos90 Degrees.
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## Cos Tan Cos90 Degrees , Find The Value Of Sin 180 8 Cos 90 8 Tan 270 8 Cot 360 8 Sin 360 8 Cos 360 8 Cosec 8 Sin 270 8 Math Trigonometric Functions 6241288 Meritnation Com
What Are Values Of Trigonometric Ratios For 0 30 45 60 And 90 Degrees. Well, technically we've only shown this for angles between 0. The three most familiar trigonometric ratios in the trigonometric functions are sine function, cosine function, and tangent function. Because tan 90 would mean both theta and the right angle are 90 degrees and therefore the third angle would be 0 degrees which isn't possible. O, 30, 45, 60 and 90 degrees are angles which are used very often in problems. In other words, the sine of an angle equals the cosine of its complement. The expression contains a division by 0. All the cos values are filled in opposite order starting from sin 0∘. Therefore cos 90 degrees is 0. In terms of sines and cosines. The values of sin and cos can be obtained from their graphs. Sin(θ)=cos(90∘−θ)sine, left parenthesis, theta, right parenthesis, equals, cosine, left parenthesis, 90, degrees, minus, theta, right parenthesis. This video teaches how can you memorize values of all six trigonometric. Learn all concepts of chapter 8 class 10 (with videos). Tan90 is undefined because its value is infinite (tan90=sin90/cos90=1/0). Sin, cos ,tan are basic trigonometric functions.
O, 30, 45, 60 and 90 degrees are angles which are used very often in problems. In fact, the functions sin and cos can be defined for all complex numbers in terms of the exponential radian (90°), the unit circle definitions allow the domain of trigonometric functions to be extended to for an angle of an integer number of degrees, the sine and the cosine may be expressed in terms of. You should try to remember sin, cos and tan for the angles 30 ° , 45 ° and 60 °. (ii) when we have 90°, tan will. This video teaches how can you memorize values of all six trigonometric. The only one right is the value of sin90 in degree = 1. Evaluation of trigonometric ratios using astc to evaluate tan (90° + θ), we have to consider the following important points.
## In trigonometrical ratios of angles (90° + θ) we will find the relation between all six trigonometrical ratios.
Well, technically we've only shown this for angles between 0. Because tan 90 would mean both theta and the right angle are 90 degrees and therefore the third angle would be 0 degrees which isn't possible. All the cos values are filled in opposite order starting from sin 0∘. It has everything anyone can ask for but recently i came across an issue that is bothering me. Cos 90° = 0, sin 90° = 1 and tan 90° is undefined. Using tan x = sin x / cos x to help. How do we remember them? You should try to remember sin, cos and tan for the angles 30 ° , 45 ° and 60 °. The values of sin and cos can be obtained from their graphs. This video teaches how can you memorize values of all six trigonometric. A finger trick for trigonometry if we insist that students memorize the values of sine and cosine for the basic angles 0, 30, 45, 60 and 90 degrees, then here's a cute little trick for doing so using the fingers on your hand. Learn all concepts of chapter 8 class 10 (with videos). Evaluation of trigonometric ratios using astc to evaluate tan (90° + θ), we have to consider the following important points. We will discuss what are different values of sin, cos, tan, cosec, sec, cot at 0, 30, 45. The only one right is the value of sin90 in degree = 1. If you can remember the graphs of the sine and cosine functions, you can use the identity above (that you need. (i) (90° + θ) will fall in the ii nd quadrant. For the angles 0° or 360° and 180°, we should not make the above conversions. This problem is just for. Press the = button to calculate the result. Sin, cos ,tan are basic trigonometric functions. Whenever i run math.eval('cos(90 deg)') it gives me 6.123. Yes, yes, it is a pain to have to remember things, but it will make life easier when you know them, not just in exams, but other times when you need to do quick estimates. O, 30, 45, 60 and 90 degrees are angles which are used very often in problems. Well, technically we've only shown this for angles between 0. The definition, tan b = (sin b)/cos b, doesn't do you a lot of good because it's got sine and cosine mixed together. Tan 90 is not possible, as we can't have a triangle with two right angles! Trigonometric table (values) of sine, cosine, tan, sec, cosec, cot values at different angles and basic formula and function of trignometry ratios. What is value of sin 30? Hello i have to differenciate calculations in degrees and i have the following code but i doesn't return me the exact values. Of course you already know those;
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# AP Stats Unit 4 Practice FRQ #1
a. .211+.321+.151+.084=.767 There is a 7.67% probability that the student-athlete chosen is at least 16 years old.
b. (14*.094)+(15*.139)+… =16.548
c. This is a binomial distribution because getting a probability of 4 or 5 is either a success or failure, each student chosen is independent of each other, the number of trials is fixed at 5, and each age has a set probability of success. Binompdf(trials=5,p=.767,x=5)+Binompdf(trials=5,p=.767,x=4)=.669
The there is a 6.69% probability that 4 or 5 of the selected athletes are at least 16 years old.
Appropriate work and calculations all around. Nice job!
You’ve done appropriate & correct calculations on all three parts… but you ran yourself into trouble on parts (a) and ( c ) when you converted to a percentage. Remember to move the decimal point two spaces; you should get 76.7% and 66.8%, respectively. Unfortunately those would both bump your responses down to partial credit. Much safer is to leave your probability statements in terms of the decimal between 0 and 1 and not mention percent at all. So part (a) would say “the probability is 0.767 that the student-athlete chosen is at least 16 years old”
a) Out of the population of all the student athletes ranging from ages 14 to 19, the probability that a student selected randomly will be at least sixteen years old is solved by adding the probability that they are 16,17,18,and 19 which is respectively .211+.321+.151+.084. The probability is .767 so there is a 76.7 percent chance that a randomly selected student athlete is at least 16 years old.
b)To find the expected age of a randomly selected student athlete I would find the mean. The mean equals the summation of the ages times the probability of each age so (14 x .094)+…(19x .084) which equals 16.548, the expected age of a randomly selected student athlete. It can be a decimal since age is a continuous variable.
c) Given that the outcome is either a success(selected student athlete is at least 16) or a failure(selected student athlete is NOT at least 16), the selection of each student athlete is random, the sample size obtained is surely less than 10 percent of the total population of student athletes at this school(over 50 student athletes), and there is a set number of trial selections being 5, I will use a binomialpdf function.
n=5(trials or # of selections), p=.767(chance of success shown in part a), x=4 or 5(number of selected athletes at least 16)
For 4 student athletes : binompdf(5,.767,4) = .403
For 5 student athletes: binompdf(5,.767,5) = .265
The probability that 4/5 or 5/5 student athletes randomly selected are at least 16 is the p(4/5)+(p5/5) or .403+.265 which is .668
Good on all three parts!
(a) P(at least 16 yrs old) = P(16 yrs old) + P(17 yrs old) + P(18 yrs old) + P(19 yrs old)
= 0.211+0.321+0.151+0.084 = 0.767 or 76.7%. Therefore, the probability that the age of a student-athlete being chose from the district at random is greater than 16 is 0.767 or 76.7%.
(b) The expected age of a randomly selected student athlete is 14*(0.094) + 15*(0.139) + 16*(0.211) + 17*(0.321) + 18*(0.151) + 19*(0.084) = 16.548.
© P(4 or 5 our of 5 is at least 16 years old) = 1 - P(0,1,2,3 out of 5 is at least 16 years old)
= 1- ( addition of binomials for equal of the 4 scenarios. I will be using paper on the AP test for this)
= 0.6686
a. If a student-athlete is chosen at random, the probability that they are at least 16 is:
P(at least 16) = 0.211 + 0.321 + 0.151 +0.084 = 0.767.
b. The expected age of a randomly selected athlete is:
mu_age = 14 * 0.094 + 15 * 0.139 + 16 * 0.211 + 17 * 0.321 + 18 * 0.151 + 19 * 0.084 = 16.548 years old.
c. Let us describe the discrete random variable X as the number of athletes that are at least 16 among a group of 5 student-athletes (B(5, 0.767))
Then the probability that 4 or 5 of the selected athletes are at least 16 is:
P(X >= 4) = binomcdf (n = 5, p = 0.767, lowerbound = 4, upperbound = 5) = 0.668634
Nice work! Correct answers & calculations. I do believe that part ( c ) is unlikely to show up this year; binomial calculations are a lot by hand (but it’s clear that you’re ready just in case!)
Correct answers with appropriate work shown!
a) Let X represent the age of a student-athlete.
P(X greater than or equal to 16) = 0.211 + 0.321 + 0.151 + 0.084 = 0.767 or 76.7%
b) E(X) = 14(0.094) + 15(0.139) + 16(0.211) + 17(0.321) + 18(0.151) + 19(0.084) = 16.548 years
c) Let Y be a binomial random variable that represents the number of student athletes over 16 years of age.
P(Y=4 or Y=5) = P(Y=4) + P(Y=5) = binompdf(trials: 5; success rate: 0.767; # of successes: 4) + binompdf(trials: 5; success rate: 0.767; # of successes: 5) = 0.6686 or 66.86%
Good work! Correct answers with appropriate calculations are shown.
A) P(16yrs+17yrs+18yrs+19yrs)= .211+0.321+0.151+0084= 0.767
B) P(Expected) (14*.094) + (15*.139)+ (16* 0.211) + (17*.321) + (18 * 0.151) + (19*0.084) = 16.548 yrs
C) 0.211+0.321+0.151+0.084= 0.767
1- binomcdf( n=5, p=0.767, X=3) = 0.6686
Correct work and calculations on all three parts!
a.) The probability that a student-athlete chosen from the district at random is at least 16 years old is 0.211 + 0.321 + 0.151 + 0.084 = .767.
b.) The expected age of a randomly selected student athlete is 14(0.094) + 15(0.139) + 16(.211) +17(.321) + 18(.151) +19(.084) = 16.548.
c.)Due to the distribution having a binary output (Below age 16 and above age 16), being independent(you can only be in one individual age), having a set amount of trials (5) and having an equal probability for each student-athlete (.767), this is a binomial distribution.
This means when you do binomial cdf (5, .767, 3) and subtract that from 1 to get the opposite you get .6686
Also I had a question about the exam, do you know what percentage you’ll need to get a 4 or 5 or how many questions you need right. Thanks.
Correct answers and work for all three parts.
As for your question about exam scoring, it varies from year to year. Typically speaking if you earn around 50% of all points you’re in line for a 3, and then 4s/5s are usually around 60ish% and 70ish%. Without multiple choice it’s hard to say what the breakdown will be. But if you feel as though you can get fully correct answers on more that 2/3 of the test, you’ll likely earn a 4-5. You will not have to get all questions fully right to earn a 5.
a) P(at least 16 years old) = P(16)+P(17)+P(18)+P(19)
0.211+0.321+0.151+0.084=0.767
There is a 0.767 probability that a student-athlete chosen at random from the district is at least 16 years old.
b) 14(0.094)+15(0.139)+!6(0.211)+17(0.321)+18(0.151)+19(0.084) = 16.548
The expected age of a randomly-selected student-athlete is 16.548.
c) P(at least 16)=0.767
(5c4 * 0.767^4 * 0.233) + (0.767^5) =0.669
There is a 0.669 probability that 4 or 5 of a sample of 5 randomly selected student athletes are at least 16 years old.
Perfect! Correct answers w/ correct supporting work.
a) The probability that the randomly chosen student-athlete is at least 16 years old is 0.767.
P(x>= 16) = 0.211+0.321+0.151+0.084 = 0.767
b) The expected age or a randomly selected student athlete is 16.548 years old.
(140.094)+(150.139)+(160.211)+(170.321)+(180.151)+(190.084) = 16.548
c) The probability that 4 or 5 of the selected athletes are at least 16 years old is 0.66863.
This is a binomial probability question. n=5, p=0.767, and x=4 or 5.
binomCDF (5,0.767,4,5) = 0.66863 Followed by the sample size, probability, minimum value, and the max value.
Good - all three parts have correct calculations and supporting work
a) The probability that a student is at least 16 is .767 (.211+.321+.151+.084)
b) the expected age is 16.5 years (14(.094)+15(.139)+16(.211)+17(.321)+18(.151)+19(.084)
c) The probability that 4 out of a sample of 5 student athletes is .6686. (n=5, p=.767, summation: upper=5 lower =4, binomialpfc(5…767,X)
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[C++]二维数组还是一维数组？
2014-09-24
1 // Create
2 int *m = new int[n_row * n_col];
3
4 // Free
5 delete [] m;
1 // Create
2 int **m = new int*[n_row];
3 for ( int i = 0; i < n_row; ++i )
4 m[i] = new int[n_col];
5
6 // Free
7 for ( int i = 0; i < n_row; ++i )
8 delete [] m[i];
9 delete [] m;
1 for ( int i = 0; i < n_row; ++i )
2 {
3 for ( int j = 0; j < n_col; ++j )
4 {
5 matrix[i * n_col + j] = answer;
7 }
8 }
1 for ( int j = 0; j < n_col; ++j )
2 {
3 for ( int i = 0; i < n_row; ++i )
4 {
5 matrix[i * n_col + j] = answer;
7 }
8 }
1 for ( int i = 0; i < n_row; ++i )
2 {
3 for ( int j = 0; j < n_col; ++j )
4 {
5 int row = i * 7 % n_row;
6 int col = j * 11 % n_col;
7 matrix[i * n_col + j] = answer;
9 }
10 }
1 matrix[i * n_col + j] = answer;
1 ; matrix[i * n_col + j] = answer;
2 mov edx, DWORD PTR _i\$3[ebp]
3 imul edx, DWORD PTR _n_col\$[ebp]
4 add edx, DWORD PTR _j\$2[ebp]
5 mov eax, DWORD PTR _matrix\$[ebp]
6 mov ecx, DWORD PTR _answer\$[ebp]
7 mov DWORD PTR [eax+edx*4], ecx
2 mov ecx, DWORD PTR _i\$4[ebp]
3 mov edx, DWORD PTR _matrix\$[ebp]
4 mov eax, DWORD PTR [edx+ecx*4]
5 mov ecx, DWORD PTR _j\$3[ebp]
6 mov edx, DWORD PTR _answer\$[ebp]
7 mov DWORD PTR [eax+ecx*4], edx
1 // Create
2 int **m = new int*[n_row];
3 int *block = new int[n_row * n_col];
4 for ( int i = 0; i < n_row; ++i )
5 m[i] = &block[i * n_col];
6 return m;
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## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
# Multiple Regression and Matrices
In order to understand how the computer carries out regression analysis—which automatically leads to a better understanding of regression—it is necessary to understand something about matrix math. This section provides a brief introduction to matrix math.
Matrices are used to describe linear equations by tracking the coefficients of linear transformations and to record data that depend on multiple parameters.
## An Introduction to Matrix Math
A matrix is a set of numbers in a two-dimensional space. Matrices are almost always designated with capital letters in boldface type. In general, a matrix X has n rows and m columns, as shown in Equation A.1, where the first subscript denotes the row and the second denotes the column.
A matrix is said to have dimensions equal to n and m so that the matrix X has dimensions n × m (rows first, columns second). The following is a typical example of a 2 × 3 matrix.
A vector is a matrix with dimensions n × 1 or 1 × m, as shown in Equation A.2.
The following are typical examples of a 3 × 1 vector and a 1 × 3 vector.
A scalar can be considered a matrix of dimensions 1 × 1. The individual entries of ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required
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#### What is 1780 percent of 5,360?
How much is 1780 percent of 5360? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 1780% of 5360 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 1780% of 5,360 = 95408
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating one thousand, seven hundred and eighty of five thousand, three hundred and sixty How to calculate 1780% of 5360? Simply divide the percent by 100 and multiply by the number. For example, 1780 /100 x 5360 = 95408 or 17.8 x 5360 = 95408
#### How much is 1780 percent of the following numbers?
1780 percent of 5360.01 = 9540817.8 1780 percent of 5360.02 = 9540835.6 1780 percent of 5360.03 = 9540853.4 1780 percent of 5360.04 = 9540871.2 1780 percent of 5360.05 = 9540889 1780 percent of 5360.06 = 9540906.8 1780 percent of 5360.07 = 9540924.6 1780 percent of 5360.08 = 9540942.4 1780 percent of 5360.09 = 9540960.2 1780 percent of 5360.1 = 9540978 1780 percent of 5360.11 = 9540995.8 1780 percent of 5360.12 = 9541013.6 1780 percent of 5360.13 = 9541031.4 1780 percent of 5360.14 = 9541049.2 1780 percent of 5360.15 = 9541067 1780 percent of 5360.16 = 9541084.8 1780 percent of 5360.17 = 9541102.6 1780 percent of 5360.18 = 9541120.4 1780 percent of 5360.19 = 9541138.2 1780 percent of 5360.2 = 9541156 1780 percent of 5360.21 = 9541173.8 1780 percent of 5360.22 = 9541191.6 1780 percent of 5360.23 = 9541209.4 1780 percent of 5360.24 = 9541227.2 1780 percent of 5360.25 = 9541245
1780 percent of 5360.26 = 9541262.8 1780 percent of 5360.27 = 9541280.6 1780 percent of 5360.28 = 9541298.4 1780 percent of 5360.29 = 9541316.2 1780 percent of 5360.3 = 9541334 1780 percent of 5360.31 = 9541351.8 1780 percent of 5360.32 = 9541369.6 1780 percent of 5360.33 = 9541387.4 1780 percent of 5360.34 = 9541405.2 1780 percent of 5360.35 = 9541423 1780 percent of 5360.36 = 9541440.8 1780 percent of 5360.37 = 9541458.6 1780 percent of 5360.38 = 9541476.4 1780 percent of 5360.39 = 9541494.2 1780 percent of 5360.4 = 9541512 1780 percent of 5360.41 = 9541529.8 1780 percent of 5360.42 = 9541547.6 1780 percent of 5360.43 = 9541565.4 1780 percent of 5360.44 = 9541583.2 1780 percent of 5360.45 = 9541601 1780 percent of 5360.46 = 9541618.8 1780 percent of 5360.47 = 9541636.6 1780 percent of 5360.48 = 9541654.4 1780 percent of 5360.49 = 9541672.2 1780 percent of 5360.5 = 9541690
1780 percent of 5360.51 = 9541707.8 1780 percent of 5360.52 = 9541725.6 1780 percent of 5360.53 = 9541743.4 1780 percent of 5360.54 = 9541761.2 1780 percent of 5360.55 = 9541779 1780 percent of 5360.56 = 9541796.8 1780 percent of 5360.57 = 9541814.6 1780 percent of 5360.58 = 9541832.4 1780 percent of 5360.59 = 9541850.2 1780 percent of 5360.6 = 9541868 1780 percent of 5360.61 = 9541885.8 1780 percent of 5360.62 = 9541903.6 1780 percent of 5360.63 = 9541921.4 1780 percent of 5360.64 = 9541939.2 1780 percent of 5360.65 = 9541957 1780 percent of 5360.66 = 9541974.8 1780 percent of 5360.67 = 9541992.6 1780 percent of 5360.68 = 9542010.4 1780 percent of 5360.69 = 9542028.2 1780 percent of 5360.7 = 9542046 1780 percent of 5360.71 = 9542063.8 1780 percent of 5360.72 = 9542081.6 1780 percent of 5360.73 = 9542099.4 1780 percent of 5360.74 = 9542117.2 1780 percent of 5360.75 = 9542135
1780 percent of 5360.76 = 9542152.8 1780 percent of 5360.77 = 9542170.6 1780 percent of 5360.78 = 9542188.4 1780 percent of 5360.79 = 9542206.2 1780 percent of 5360.8 = 9542224 1780 percent of 5360.81 = 9542241.8 1780 percent of 5360.82 = 9542259.6 1780 percent of 5360.83 = 9542277.4 1780 percent of 5360.84 = 9542295.2 1780 percent of 5360.85 = 9542313 1780 percent of 5360.86 = 9542330.8 1780 percent of 5360.87 = 9542348.6 1780 percent of 5360.88 = 9542366.4 1780 percent of 5360.89 = 9542384.2 1780 percent of 5360.9 = 9542402 1780 percent of 5360.91 = 9542419.8 1780 percent of 5360.92 = 9542437.6 1780 percent of 5360.93 = 9542455.4 1780 percent of 5360.94 = 9542473.2 1780 percent of 5360.95 = 9542491 1780 percent of 5360.96 = 9542508.8 1780 percent of 5360.97 = 9542526.6 1780 percent of 5360.98 = 9542544.4 1780 percent of 5360.99 = 9542562.2 1780 percent of 5361 = 9542580
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restriction of a representation of GL(n) to GL(n-1) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:00:36Z http://mathoverflow.net/feeds/question/18197 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18197/restriction-of-a-representation-of-gln-to-gln-1 restriction of a representation of GL(n) to GL(n-1) unknown (google) 2010-03-14T18:49:22Z 2010-03-14T20:20:24Z <p>Let $R$ be real numbers and consider an irreducible unitary representation (\pi,V) of $GL_n(R)$ in some Hilbert space $V$, now $GL_{n-1}(R)$ embeds in $GL_n(R)$ on the left upper diagonal block. </p> <p>Now I wanna ask properties of the restrictions of $V$ to $GL_{n-1}$. This representation is not irreducible, so we may ask is it possible to say something about the closed invariant subspaces? Does this restriction have a multiplicity free decomposition?</p> <p>Note that we can ask the same question for other irreducible Banach or Frechet representation of $GL_n$, or replacing real numbers by complex or p-adic numbers.</p> http://mathoverflow.net/questions/18197/restriction-of-a-representation-of-gln-to-gln-1/18200#18200 Answer by Peter Tingley for restriction of a representation of GL(n) to GL(n-1) Peter Tingley 2010-03-14T19:32:07Z 2010-03-14T19:32:07Z <p>If you use complex numbers instead of real numbers, it is true that the restriction is multiplicity free. This is an important fact, and is used to construct the Gelfand-Zetlin basis for V. This is discussed is Fulton and Harris' book "representation theory", section 25.3. They also discuss generalizations to other classical algebras</p> http://mathoverflow.net/questions/18197/restriction-of-a-representation-of-gln-to-gln-1/18206#18206 Answer by Marty for restriction of a representation of GL(n) to GL(n-1) Marty 2010-03-14T20:00:50Z 2010-03-14T20:12:20Z <p>Though Peter's answer addresses the finite-dimensional representation theory, I believe that the question asks about the unitary representations on Hilbert spaces, and more general irreps on Banach and Frechet spaces.</p> <p>This question has been the subject of much recent work by Avraham Aizenbud, Dmitry Gourevitch, Steve Rallis, Gerard Schiffmann, and Eitan Sayag. In particular, Aizenbud and Gourevitch prove the following in their paper "Multiplicity One Theorem for $(GL_{n+1}(R), GL_n(R))$":</p> <p>Let $F = R$ or $F = C$. Let $\pi$ and $\tau$ be irreducible admissible smooth Fr\'echet representations of $GL_{n+1}(F)$ and $GL_n(F)$, respectively. Then $$dim \left( Hom_{GL_n(F)}(\pi, \tau) \right) \leq 1.$$</p> <p>This paper is on the ArXiv, and now published in Selecta, according to Aizenbud's webpage.</p> <p>Zhu and Binyong have also proved this, I believe. The result has also been proven for irreducible smooth repreesentations of $GL_{n+1}(F)$ and $GL_n(F)$, when $F$ is a $p$-adic field by Aizenbud-Gourevitch-Rallis-Schiffmann.</p> <p>Considering the smooth Fr\'echet case should suffice for the case of unitary representations on Hilbert spaces, I believe, by considering the subspace of smooth vectors and Garding's theorem. I'd guess it would also work for Banach space representations, but I'm not an expert on these analytic things.</p> <p>It's important to note that semisimplicity may be lost when one restricts smooth representations in these settings -- so their theorem says something about occurrences of quotients after restriction. It's important to be careful about the meaning of "multiplicity-free" in these situations. </p>
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#### Numbers in the Middle East - Understanding Arabic Numbers
The discussion of the importance of numbers in everyday life and how "Arabic" numbers and algebra changed the world is worthy of a book – or at least a separate article. However, the short story is that numbers were represented as hash marks (i.e. ||| for 3) or glyphs with certain value representations (i.e. Roman Numerals where X represents 10). Either way, you had to count, add and subtract the different symbols to arrive at the value involved. That alone could be overwhelming for the math challenged! Then the system of representing numbers that we use today was invented, probably in India. There are ten symbols representing 0-9, placed in columns. Starting from the right, when you run out of symbols to use in one column (that is reach 9 and need to add one more); you reset it to 0 and go to the next column to the left and add one there. Thus, even in languages that are written right to left (i.e. most Middle Eastern languages), the digit with the highest value goes on the left and the digit with the lowest value goes on the right. Don't worry if the preceding is confusing to you, these are the numbers we use every day; all you really need to remember is that you can view our number system as a right to left system too, even though we usually read numbers from left to right. If you remember learning math in columns in elementary school, where you start on the right and work to the left, you understand everything you need.
These numbers came to Europe through the introduction of Algebra from the Arab world. This is why we refer to our numbers as “Arabic numbers” (versus “Roman numerals”) - again, stuff we don't normally think of after elementary school. However, in spite of the fact that we call them “Arabic Numbers”, much of the Middle East uses different symbols for the digits 0-9. These are called the Hindi numbers – but actually vary a little bit from Hindi depending on the language used. The various digits are shown below:
You will also see Latin digits. In Biblical Hebrew (Modern Hebrew uses the Latin digits), Koine Greek (used during Roman occupation in much of the Middle East), and older, non-mathematical uses of Arabic each letter has a numeric value – just like the Roman Numerals you learned in elementary school. Numbers are determined by adding up the quantities each letter represents. With this, you should be able to decipher any number you see in the Middle East or documents in Middle Eastern languages.
Sources:
"Arabic numerals." Wikipedia, The Free Encyclopedia. Detailed information on the history of the Arabic numerals. 18 Sep 2006, 02:31 UTC. Wikimedia Foundation, Inc. 20 Sep 2006 <https://en.wikipedia.org/w/index.php?title=Arabic_numerals&oldid=76339928>.
Microsoft Corporation. "Digits Support in Microsoft Windows XP." Microsoft Products and Arabic Support ver. 3.0. Digits used in different Middle Eastern languages and how to display them in Windows XP. 2005. Microsoft Corporation. 20 Sep 2006 <https://www.microsoft.com/middleeast/arabicdev/windows/winxp/DigitsSupport.aspx>.
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Group Discounts available for 3+ students and Corporate Clients
# Insertion sort in Data Structures
## Insertion sort
Suppose an array ‘A’ with ‘n’ elements A[0]. A910,A[20],———A[N-1] is in memory. The insertion sort algorithm in data structures scans ‘A; from A[0] to A[N-1] insert each element A[k] into its proper position in the previously sorted sub array A[0],A[1],A[2]—-
A[K] That is
pass1 A[0] by itself ip trivially sorted.
pass2 A[1] is inserted either before or after aA[0]
So that : A[0],A[1] is sorted
pass3: A[2] is inserted into its proper place in
A[1] A[2], that is before A[0] between A[0]
A[1] or after A[1], so that : A[0],A[1],A[2] is sorted
pass4: [3] is inserted into its proper place in A[1],A[2] so that : A[0],A[1],A[2],A[3] is sorted.
PassN : A[N] is inserted into its proper place in A[1],A[2],——-A[N-1] so that : A[0],A[1]— is sorted.
Example :
### Insertion Sort Program
Implementation of insertion sort in data structures
# include <stdio.h>
# include <conio.h>
void main()
{
int n,i,j,a[20],temp;
void insertion sort(int a[10],int n);
clrscr();
printf(“Enter n value: ”);
scanf(“%d”,&n);
printf(“Enter the elements:\n”);
for(i=0; i<n;i++)
{
scanf(“%d”,&a[i]);
}
Insertion sort(a,n);
}
insertion sort(a,n);
printf(“The sorted elements are:\n”);
for(i=0;i<n;i++)
{
printf(“%5d”,a[i]);
}
}
void insertion sort (int a[20], int n)
{
int i,j,temp;
for(i=1;i<n;i++)
{
if(a[j]<a[j-1])
{
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
}
}
}
### output :
Enter ‘n’ value : 5
Enter the elements: 2 4 3 1 5
The elements are : 2 3 4 5
Time complexity :
Algorithm Worst case Average case Insertion sort n(n+1)/2=o(n2) n(n-1)/4=o(n2)
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# Understanding Transfer Principle About Formal Power Series
The theorem I am referring to is from "The Concrete Tetrahedron" by Kauers and Paule, which states:
Theorem 2.8 (Transfer Principle) Let $$a(z) = \sum_{n=0}^{\infty} a_nz^n$$ and $$b(z) = \sum_{n=0}^{\infty} b_nz^n$$ be real or complex functions analytic in a non-empty open neighbourhood of zero. If $$a(z) = b(z)$$ for all $$z \in U$$, then $$a_n = b_n$$ for all $$n \in \mathbb{N}$$.
An example given in the book to illustrate this principle is:
"The transfer principle can be used for obtaining simple proofs of identities. For instance, to demonstrate that $$\exp(\log(1+x)) = 1 + x$$ as formal power series, it is sufficient to observe that this relation is true for the corresponding analytic functions. Proving this without the transfer principle would necessitate elaborate calculations."
My interpretation is that the Transfer Principle implies that the Taylor series of the analytic function $$\exp(\log(1+x))$$ is simply $$1 + x$$.
However, when we regard $$\exp(\log(1+x))$$ as a formal power series (generating functions), we use the definitions: $$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$ and $$\log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}.$$ Then, by the definition of the composition of power series, we get: $$\exp(\log(1+x)) = \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}.$$ My confusion arises here: How can we assert that the above power series, $$\sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!},$$ is equivalent to the Taylor series of $$\exp(\log(1+x))$$ and therefore equals $$1+x$$?
I am unable to see how this conclusion is derived directly from the Transfer Principle.
Update: I don't have a problem with the theorem itself. But I am not sure it implies the identity given as an example. More specifically, if we consider $$\exp(x)$$ and $$\log(1+x)$$ as short-hands to write the corresponding power series, how do we know that $$\exp(\log(1+x))$$ as a composition of two power series, has the same coefficients as $$\exp(\log(1+x))$$ when considered as a composition of two analytic functions.
• I think you are reading too much into it. The point is that since we know the identify is true (from the definition of $\log,\exp$) then the expansions of the left and right are identical. So you avoid the computation. Jan 30 at 7:36
• I know what is the point. I just don't see the details. There is a gap between composing to power series and composing the corresponding analytic functions. I think it's been hidden and there needs to be lemma somewhere which says that if $A(B(x)) = C(x)$ as functions, then $A(B(x)) = C(x)$ as power series. Jan 30 at 7:46
• If we think of $a(z)$ and $b(z)$ as analytic functions (near $0$) rather than just formal power series, then the coefficients of the power series for $a(z)$ and $b(z)$ are determined uniquely as residues: $a_n=\mathrm{Res}(a(z)/z^{n+1};0)$. So, if there is an identity for analytic functions, then it holds for their power series as well. Also, by the way, I think the combinatorial content of your example is that there is the same number of permutations of length $n\ge 2$ with an even number of cycles and with an odd number of cycles. Jan 30 at 11:41
• Note that $\exp(\log(1+x)) = \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}$ is not a power series. The terms need to be rearranged formally to write this as a power series. Then their Theorem 2.8 asserts that the two power series are equal. Presumably they provide a proof of Theorem 2.8, which may not be entirely trivial. Jan 30 at 12:39
• Of course, I agree with you, that it can be seen quite easily that the two types of composition yield the same power series, but being an introductory book this should at least be mentioned at some point. Jan 31 at 6:46
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Follow Us
# CSS 3D TRANSFORMS
Using CSS 3D transform you can perform 3D operations on HTML elements like scale, skew, rotate, translate etc.
The 3D translated elements don't affect surrounding elements and may overlap them. The translated element takes the same space as its default.
To apply 3D transformation CSS use transform property and values are given as required like scaling, rotation etc.
We are going to learn following 3D transform properties:
## 1.) CSS translate3d
Using CSS translate3d you can translate or move HTML elements along X, Y and Z-axis.
The CSS translate3d takes three arguments, each for one axis X, Y and Z respectively. The coordinate values could be negative as well as positive.
Example- transform: translate3d( x, y, z)
Output:
This div element is translated in 3D 50px in x-axis, 10px in Y-axis and 20px in Z-axis.
## 2.) CSS rotate3d
Using CSS rotate3d you can rotate HTML elements along X, Y and Z-axis.
The CSS rotate3d takes four arguments, each for one axis X, Y, Z and degree value respectively. The X, Y and Z value is multiplied with specified degree internally and this gives how much angle should be rotated for each axis.
Example- transform: rotate3d( x, y, z, deg)
Output:
This div element is rotated in X and Y-axis by 60deg and Z-axis by 1.2×60deg=72deg.
## 3.) CSS scale3d
Using CSS scale3d you can scale HTML elements along X, Y and Z-axis.
The CSS scale3d changes the size of the element. It takes 3 arguments, each for X, Y and Z respectively.
For CSS rotate3d to work you need to use either rotate3d property or perspective property.
Example- transform: scale3d( x, y, z)
Output:
This div element is scaled in X by 1, in Y-axis by 1.5× and in Z-axis by 2×.
Follow Us
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https://forum.azimuthproject.org/plugin/ViewComment/14461
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I am rationalizing why solving the wave equation is so difficult. This is a 2nd-order differential equation, with a few forcing factors applied. It shouldn't be hard to solve via a multiple linear regression or by applying a Fourier or Laplace transform and solving in reciprocal space. I am sure that researchers have tried this, but perhaps became frustrated. I think the reason for the difficulty is that any Hill/Mathieu modulation on the characteristic frequency invalidates the traditional linear approaches of solving such a model. The best way is to brute force a solution by solving numerically. That's why I have leaned so heavily on using Mathematica.
I mentioned a couple of electrical engineering analogies in the last comment -- that of carrier scattering representing statistical noise and Maxwell's equation modeling a dipole. Another one is solving Schroedinger's equation in the context of a semiconductor lattice and how this relates. Fundamentally, the math behind semiconductor band formation is no different than the Hill or Mathieu equation applied to the wave equation.
But no one harbors any illusions that solving the band structure of a semiconductor lattice is a walk in the park. The periodic modulation of a lattice potential as applied to the electron will cause all sorts of odd spatial wiggles in the allowed states. This is the basis of the infamous [Bloch wave](http://en.wikipedia.org/wiki/Bloch_wave). I recognized this in the [first blog post](http://contextearth.com/2014/02/10/the-southern-oscillation-index-model/) I wrote when I started looking at ENSO.
The first iteration of finding the Hill/Mathieu modulation is reviewed below.
The equation I am solving in Mathematica is
NDSolve[{y''[t]+(CF+Hill[t])*y[t] == RHS[t]
where CF is the characteristic frequency term squared, Hill is the modulation, and RHS is the forcing.
The factors contributing to the Hill and RHS modulation are based on models of QBO, Chandler Wobble, and TSI as I described earlier. The shape of these profiles are captured as a Fourier series of sine waves with a fidelity that achieves at least 70% of a unity correlation coefficient. Those numbers are shown in panels 3 through 5 below with the correlation coefficient percentage shown to upper left of each graph. The last two panels show the modulation. Interestingly, the Hill modulation appears dominated by the TSI factor while the RHS shows the more rapid wiggles of the QBO.
Isn't applied mathematical physics fun?
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https://www.reddit.com/r/askscience/comments/196xsx/does_the_size_of_a_dustparticle_determine_the/
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This is an archived post. You won't be able to vote or comment.
[–] 12 points13 points (9 children)
Yes, it does. When the particles are much smaller than the wavelength, one can use Rayleigh's model, which is proportional to the frequency raised to the fourth power and the diameter of particle to the sixth. On the other hand, when the particle size becomes comparable to the wavelength, one must use Mie's model, which has a much smaller dependence on the wavelength of the light. Other models apply in different physical circumstances.
This difference has clear implications. For instance the sky appears blue because the gas particles in the sky are small, the Rayleigh regime is applicable and blue light is scattered the most. Clouds on the other hand consist of large droplets, for which Mie's theory applies, hence their white/grey color.
[–] 7 points8 points (0 children)
Maybe the best explanation for 'why is the sky blue' I've read.
[–] 5 points6 points (2 children)
Your explanation is correct though in reference to the question it doesn't decide what wavelength it scatters; it just determines the probability of the scattering.
[–] 0 points1 point (0 children)
Actually the original question only asked if dust motes could scatter light of a particular wavelength due to their size. The OP never asked what those wavelengths for dust motes were.
[–] 1 point2 points (0 children)
This begs the question: what is the size of an average speck of dust? I was under the impression that most dust was micron-sized at least, if not larger. If that's the case, then the particles in question are larger than the wavelengths of light under consideration (400-800 nm, or thereabouts). In that case neither model would apply.
Is it the case that scattering just doesn't happen to any significant extent at this point?
[–] -1 points0 points (3 children)
I believe Mie scattering also explains why sunsets (and sunrises) are red. When the sun is low in the sky, the light is passing through more of the lower atmosphere, which includes larger particles like dust and pollution (aerosols), which scatter the red end of the visible spectrum more efficiently.
[–] 1 point2 points (0 children)
Are you sure that this is the dominant effect, and it's not mostly that the bluer wavelengths have been scattered in a Rayleigh fashion leaving only the lower frequency light to be seen at sunset? I don't doubt that there are some particles present and of a size to require Mie theory, it just seems to me that it should be a much smaller effect that the Rayleigh effect. Also, if the pollution and dust you're talking about scatters the red end of the spectrum preferentially, then it's got no way to make the sunset red, as the red light must reach your eyes. It would have to be scattering blue preferentially. I could see how these particles would serve to make the already red sunset more diffuse, though.
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# Natural Deduction
Natural Deduction (ND) is a common name for the class of proof systems composed of simple and self-evident inference rules based upon methods of proof and traditional ways of reasoning that have been applied since antiquity in deductive practice. The first formal ND systems were independently constructed in the 1930s by G. Gentzen and S. Jaśkowski and proposed as an alternative to Hilbert-style axiomatic systems. Gentzen introduced a format of ND particularly useful for theoretical investigations of the structure of proofs. Jaśkowski instead provided a format of ND more suitable for practical purposes of proof search. Since then many other ND systems were developed of apparently different character.
What is it that makes them all ND systems despite the differences in the selection of rules, construction of proof, and other features? First of all, in contrast to proofs in axiomatic systems, proofs in ND systems are based on the use of assumptions which are freely introduced but discharged under some conditions. Moreover, ND systems use many inference rules of simple character which show how to compose and decompose formulas in proofs. Finally, ND systems allow for the application of different proof-search strategies. Thanks to these features proofs in ND systems tend to be much shorter and easier to construct than in axiomatic or tableau systems. These properties of ND make them one of the most popular ways of teaching logic in elementary courses. In addition to its educational value, ND is also an important tool in proof-theoretical investigations and in the philosophy of meaning (specifically, of the meaning of logical constants). This article focuses on the description of the main types of ND systems and briefly mentions more advanced issues concerning normal proofs and proof-theoretical semantics.
## 1. History of Natural Deduction
When dealing with the history of ND, one should distinguish between the exact date when the first formal systems of ND were presented and much earlier times when the rules of ND were actually applied. Although one may claim that ND techniques were used as early as people did reasoning, it is unquestionable that the exact formulation of ND and the justification of its correctness was postponed until the 20th century.
### a. Origins
The first ND systems were developed independently by Gerhard Gentzen and Stanisław Jaśkowski and presented in papers published in 1934 (Gentzen 1934, Jaśkowski 1934). Both approaches, although different in many respects, provided the realization of the same basic idea: formally correct systematization of traditional means of proving theorems in mathematics, science and ordinary discourse. It was a reaction to the artificiality of formalization of proofs in axiomatic systems. Hilbert’s proof theory offered high standards of precise formulation of this notion, but formal axiomatic proofs were really different than ‘real’ proofs offered by mathematicians. The process of actual deduction in axiomatic systems is usually complicated and needs a lot of invention. Moreover, real proofs are usually lengthy, hard to decipher and far from informal arguments provided by mathematicians. In informal proofs, techniques such as conditional proof, indirect proof or proof by cases are commonly used; all are based on the introduction of arbitrary, temporarily accepted assumptions. Hence the goals of Gentzen and Jaśkowski were twofold: (1) theoretical and formally correct justification of traditional proof methods, and (2) providing a system which supports actual proof search. Moreover, Gentzen’s approach provided the programme for proof analysis which strongly influenced modern proof theory and philosophical research on theories of meaning.
### b. Prehistory
According to some authors the roots of ND may be traced back to Ancient Greece. Corcoran (1972) proposed an interpretation of Aristotle’s syllogistics in terms of inference rules and proofs from assumptions. One can also look for the genesis of ND system in Stoic logic, where many researchers (for example, Mates 1953) identify a practical application of the Deduction Theorem (DT). But all these examples, even if we agree with the arguments of historians of logic, are only examples of using some proof techniques. There is no evidence of theoretical interest in their justification.
In fact the introduction of DT into the realm of modern logic seems to be one of the most important steps on the way leading eventually to the discovery of ND. Although Herbrand did not present a formal proof of it for axiomatic systems until Herbrand (1930), he had already stated it in Herbrand (1928). At the same time Tarski (1930) included DT as one of the axioms of his Consequence Theory; in practice he had used it since 1921. Also other ND-like rules were practically applied in the 1920s by many logicians from the Lvov-Warsaw School, like Leśniewski and Salamucha, as is evident from their papers.
Jaśkowski was strongly influenced by Łukasiewicz, who posed on his Warsaw seminar in 1926 the following problem: how to describe, in a formally proper way, proof methods applied in practice by mathematicians. In response to this challenge Jaśkowski presented his first formulation of ND in 1927, at the First Polish Mathematical Congress in Lvov, mentioned in the Proceedings (Jaśkowski 1929). A final solution was delayed until (Jaśkowski 1934) because Jaśkowski had a lengthy break in his research due to illness and family problems. Gentzen also published the first part of his famous paper in 1934, but the first results are present in (Gentzen 1932). This early paper, however, is concerned not with ND but with the first form of Sequent Calculus (SC). Gentzen was influenced by Hertz (1929), where a tree-format notation for proofs, as well as the notion of a sequent, were introduced. One can also look for a source of the shape of his rules in Heyting’s axiomatization of intuitionistic logic (see von Plato 2014).
It should be no surprise that the two logicians with no knowledge of each other’s work, independently proposed quite different solutions to the same problem. Axiom systems, although theoretically satisfying, were considered by many researchers as practically inadequate and artificial. Thus the need for more practice-oriented deduction systems was in the air.
## 2. Applications
This article distinguishes at least three main fields of application of ND systems: practical, theoretical and philosophical.
Since 1934 a lot of systems called ND were offered by many authors in numerous textbooks on elementary logic. In this way ND systems became a standard tool of working logicians, mathematicians, and philosophers. At least in the Anglo-American tradition, ND systems prevail in teaching logic. They also had strong influence on the development of other types of non-axiomatic formal systems such as sequent calculi and tableau systems. In fact, the former were also invented by Gentzen as a theoretical tool for investigations on the properties of ND proofs, whereas the latter may be seen (at least in the case of classical logic) as a further simplification of sequent calculus that is easier for practical applications.
But the importance of ND is not only of practical character. Since 1960s the works of Prawitz (1965) and (Raggio 1965) on normal proofs opened up the theoretical perspective in the applications of ND. In fact Prawitz was rediscovering things known to Gentzen but not published by him, which was later shown by von Plato (2008). In addition to extended work on normalization of proofs, ND is also an interesting tool for investigations in theoretical computer science through the Curry-Howard isomorphism. This approach shows that (normal) ND proofs may be interpreted in terms of executions of programs.
Finally the special form of rules of ND provided by Gentzen led to extensive studies on the meaning of logical constants. This article takes a look at theoretical and philosophical applications of ND in sections 9 and 10.
## 3. Demarcation Problem
The great richness of different forms of systems called ND leads to some theoretical problems concerning the precise meaning of the term ‘ND’. It seems that no definition of ND systems was offered which would be generally accepted. This demarcation problem was investigated by many authors; and different criteria were offered for establishing what is, and what is not, an ND system. Detailed survey of these matters may be found in Pelletier (1999) or in Pelletier and Hazen (2012); this article points out only the most important features.
### a. Wide and Narrow Sense of ND
Some authors tend to use the term in a broad sense in which it covers almost all that is not an axiomatic system in Hilbert’s sense. Hence sometimes systems like sequent calculi or tableau calculi are treated as ND systems. All these systems are actually in close relationship, but this article chooses to consider ND only in the narrow sense. There are at least three reasons for making this choice:
• Historical. Original ideas of Gentzen, who introduced two systems: NK (Natürliche Kalkül) and LK (Logistiche Kalkül). The former is just an ND system, whereas the latter, a sequent calculus, was meant as a technical tool for proving some metatheorems on NK, not as a kind of ND.
• Etymological. ND is supposed to reconstruct, in a formally proper way, traditional ways of reasoning. It is disputable whether existing ND systems realize this task in a satisfying way, but certainly systems like tableaux or SC are even worse in this respect.
• Practical. Taking the term ND in a wide sense would be a classifying operation of doubtful usefulness. From the point of view of this article’s presentation, it is more convenient to use a more narrowly defined concept.
### b. Criteria of Genuine ND
But what criteria should be used for delimiting the class of systems called ND? Many proposals seem to be too narrow (that is, strict) since they exclude some systems usually treated as ND, so it is better not to be very demanding in this respect. So, ND system should satisfy three criteria:
1. Possibility of entering and eliminating (discharging) additional assumptions during the course of the proof. Usually it requires some bookkeeping devices for indicating the scope of an assumption, that is, for showing that a part of the proof (a subproof) depends on a temporary assumption, and for marking the end of such a subproof the point at which the assumption is “discharged”.
2. Characterization of logical constants by means of rules rather than axioms. Their role is taken over by the set of primitive rules for introduction and elimination of logical constants, which means that elementary inferences instead of formulas are taken as primitive.
3. The richness of forms of proof construction. Genuine ND systems admit a lot of freedom in proof construction and in the possibility of applying several strategies of proof-search.
These three conditions seem to be the essential features of any ND. These characteristics are quite general, but the third at least serves to exclude tableau systems and sequent calculi since genuine ND should allow both direct and indirect proofs, proofs by cases, and so forth. This flexibility of proof construction is vital for ND, whereas, for example in a standard tableau system, we have only indirect proofs and elimination rules. On the other hand, ND does not require that its rules should strictly realise the schema of providing a pair of introduction and elimination rules, and that axioms are not allowed.
## 4. Rules
ND systems consist of the set of (schemata) of simple rules characterising logical constants. For example a connective of conjunction is characterised by means of the following rules:
where and denote any formulas. Material above the horizontal line represents the premises; and that below represents the conclusion of the inference. The letters and in the names of the rules come from “introduction” and “elimination” respectively since the first allows introduction of a conjunction into a proof, and the second allows for its elimination in favor of simpler formulas. Often the following horizontal notation is applied (instead of vertical which is more space-consuming):
Here is used to point out that the relation of deducibility holds between premises and the conclusion of a rule instance. In what follows, such phrases are called sequents. In fact such deducibility statements in general do not uniquely characterise inference rules, but it does no harm so they are used in what follows for simplicity’s sake.
One can easily check that the rules stated above adequately characterise the meaning of classical conjunction which is true iff both conjuncts are true. Hence the syntactic deducibility relation coincides with the semantic relation of , that is, of logical consequence (or entailment). Unfortunately not all logical constants may be characterised by means of such simple rules. For example, implication in addition to modus ponens (or detachment rule):
which is known from axiomatic systems, requires a more complex rule of the shape:
or:
If , then
where and forms a collection of all active assumptions previously introduced which could have been used in the deduction of . When inferring , one is allowed to discharge assumptions of the form . The fact that after deduction of this assumption is discharged (not active) is pointed out by using [ ] in vertical notation, and by deletion from the set of assumptions in horizontal notation. The latter notation shows better the character of the rule; one deduction is transformed into the other. It shows also that the rule corresponds to an important metatheorem, the Deduction Theorem, which has to be proved in axiomatic formalizations of logic. In what follows, all rules of the shape will be called inference rules, since they allow for inferring a formula (conclusion) from other formulas (premises) present in the proof. Rules of the form:
If , then
will be called proof construction rules since they allow for constructing a proof on the basis of some proofs already completed. One characteristic feature of such rules is that they involve the process of entering new assumptions as well as conditions under which one can discharge these assumptions and close subordinated proofs (or subproofs) starting with these assumptions.
The complete set of rules provided by Gentzen for IPL (Intuitionistic Propositional Logic) is the following:
What is evident from this set of rules is the Gentzen policy of characterising every constant by a pair of rules, in which one is the rule for introduction a formula with that constant into a proof, and the other is the rule of elimination of such a formula, that is, inferring some simpler consequences from it, sometimes with the aid of other premises. More will be said about philosophical consequences of this approach in section 10.
In order to obtain CPL (Classical Propositional Logic), Gentzen added the Law of Excluded Middle as an axiom, but the same result can easily be obtained by a suitable inference rule of double negation elimination: or by changing one of the proof construction rules, namely ) which encodes the weak form of indirect proof into the strong form:
If , then
This solution was applied by Jaśkowski (1934).
## 5. Proof Format
In addition to providing suitable rules, one must also decide about the form of a proof. Two basic approaches due to Gentzen and Jaśkowski are based on using trees as a representation of a proof and on using linear sequences of formulas. This article focuses on the most important differences between these two approaches. For detailed comparison see Pelletier and Hazen (2014), and Restall (2014).
### a. Tree Proofs
Let us start with an example of a proof in Gentzen’s format, that is, as a tree of formulas:
Here the root of a tree is labelled with a thesis and its leaves are labelled with (discharged) assumptions: and . All assumptions were discharged while was applied successively building implications from — the numbers of assumptions indicate the order in which they were discharged, and the suitable number is attached to the formula inferred by the assumption discharging rule. Before that, was deduced by two applications of , first to two assumptions (active at this moment), then to the third assumption and previously deduced .
Gentzen’s tree format of representing proofs has many advantages. It is an excellent representation of real proofs; in particular, deductive dependencies between formulas are directly shown. But if we are concerned with actual deduction, this format of proof is far from being useful and natural. Moreover, one is often forced to repeat identical, or very similar, parts of the proof, since, in tree format, inferences are conducted not on formulas but on their particular occurrences. For example, if is an assumption from which we need to infer both and , then a suitable branch starting with must be displayed twice. The following example illustrates the point:
here, the attachment of two numerals to the formula in the last line indicates that both occurrences of the same assumption were discharged in this step.
Gentzen himself was aware of the disadvantages of his representation of proof, but it proved useful for his theoretical interests described in section 9. It is not surprising that the tree format of proofs is mainly used in theoretical studies on ND, as in Prawitz (1965) or Negri and von Plato (2001).
### b. Linear Proofs
Jaśkowski, on the other hand, preferred a linear representation of proofs since he was interested in creating a practical tool for deduction. Linear format has many virtues over Gentzen’s approach. For example, inferences are drawn from assumptions rather than from their occurrences, which means that, for example, one needs to assume only once to derive both conjuncts. It is also more natural to construct a linear sequence trying, one by one, each possible application of the rules. But there is a price to be paid for these simplifications—the problem of subordinated proofs. How should we represent that some assumption and its subordinated proof are no longer alive because a suitable proof construction rule was applied? If we apply a proof construction rule which discharges an assumption, we must explicitly show that the subordinate proof dependent on this assumption is dead in the sense that no formula from it may be used below in the proof. In a tree format this is not a problem—to use a formula as a premise for the application of some inference rule we must display it (and the whole subtree which provides a justification for it) directly above the conclusion. In linear format this leads to problems, and some technical devices are necessary which forbid using the assumptions and other formulas inferred inside completed subproofs. Jaśkowski proposed two solutions to this problem: graphical (boxes) and bookkeeping (in the terminology of Pelletier and Hazen 2012). Let us compare these two simple proofs:
On the left we have an example of a proof in graphical mode where each assumption opens a new box in which the rest of the proof is carried out. On the other hand when a suitable proof construction rule is applied, the current subproof is boxed which means that nothing inside is allowed in further proof construction. In lines 3 and 5 an additional rule of repetition (often called reiteration) is applied which allows for moving formulas from outer to inner boxes. On the right the same proof is represented in bookkeeping style where instead of boxes we use prefixes (sequences of natural numbers) for indicating the scope of an assumption. Each assumption is preceded with the letter S from latin suppositio and adds a new numeral to the sequence of natural numbers in the prefix. When a proof construction rule is applied, the last item is subtracted from the prefix. Hence a thesis can occur with an empty sequence, signifying that it does not depend on any assumption. No repetition rules are applied in this version of Jaśkowski’s system; hence the proof is two lines shorter.
Although Jaśkowski finally chose the second option (perhaps due to editorial problems) nowadays the graphical approach is far more popular, probably due to the great success of Fitch’s textbook (1952) which popularized a simplified version of Jaśkowski’s system (now called Fitch’s approach). In Fitch’s system one is using vertical lines for indicating subproofs. Below is an example of a proof in Fitch’s format:
Other devices were also applied such as brackets in Copi (1954), or even just indentation of subordinate proofs. The original Jaśkowski’s boxes were used by Kalish and Montague (1964) with the additional device being of great heuristic value; each box is preceded by a show-line which displays the current aim of the proof. Show-lines are not parts of a proof in the sense that one is forbidden to use them as premises for rule application. But after completing a subproof, a box is closed and the opening show-line becomes a new ordinary line in the proof (which is pointed out by deleting a prefix “show”).
The second solution of Jaśkowski was not so popular. One can mention here Quine’s system (1950) (with asterisks instead of numerals) or Słupecki and Borkowski’s system (1958) popular in Poland.
## 6. Other Approaches
Gentzen (1936) introduced yet another variant of ND which may be considered as lying between his first system described in subsection 5.1. and his famous sequent calculus. It shows another possible way of arranging the bookkeeping of active assumptions. As a result, in this approach the basic items which are transformed in proofs are not formulas but rather sequents. For example, both rules for conjunction are of the form:
where are records of active assumptions.
The full list of rules for CPL contains also:
Assumptions are sequents of the form . Theses are sequents with an empty antecedent. Here is an example of a proof:
One can observe that in the context of such a system the difference between inference and proof construction rules disappears. The only difference is that in the former all transformations are performed on consequents of sequents whereas in the latter some operations (that is, subtractions) are allowed also on antecedents. This is the difference with Gentzen’s ordinary sequent calculus where we have rules introducing constants to antecedents of sequents (instead of rules of elimination). Of course one can go further and allow this kind of rule as well (such a system was constructed, for example, by Hermes 1963), but it seems that Gentzen’s choice offers significant simplifications. First of all, the tree format is not necessary, and one can display proofs as linear sequences since the record of active assumptions is kept with every formula in a proof (as the antecedent). Moreover, since no operation except subtraction is carried out on antecedents, we can get rid of formulas in antecedents and use instead numerals of lines where suitable assumptions were introduced into proofs. Both simplifications are present in Suppes’ system (1957) of ND where the same proof looks like that:
Other solutions generalising standard proof representations were also considered. One can mention at least two approaches without going into details: ND operating on clauses instead of formulas (Borićić 1985, Cellucci 1992, Indrzejczak 2010) and ND admitting subproofs as items in the proof (Fitch 1966, Schroeder-Heister 1984).
## 7. Rules for Quantifiers
Gentzen (1934) also provided the first set of ND rules adequate for CFOL (Classical First-Order Logic) whereas the rules of Jaśkowski’s system characterised the weaker system of IFOL (Inclusive First-Order Logic) which admits empty domains in models. As pointed out by Bencivenga (2014), a minimal relaxation of Jaśkowski’s rules yields also Free Logic, that is, a logic allowing non-denoting terms, hence it may be claimed that it is the first formalization of Universally Free Logic, that is, allowing both empty domains and non-denoting terms.
Before characterising Gentzen’s original rules for quantifiers let us note that he was using two sorts of symbols to distinguish between free and bound individual variables. The former are often called individual parameters. Such a solution simplifies a formulation of rules and eliminates the risk of a clash of variables while applying the rules. When we provide ND rules for more standard approaches with just individual variables which may have free or bound occurrences, we must be careful to define precisely the operation of proper substitution of a term for all free occurrences of a variable. ‘Proper’ means that no occurrence of a free variable substituted for another (or, when function-symbols are used, within a term substituted for a variable) gets bound by a quantifier. For simplicity’s sake we will keep Gentzen’s solution; let denote (bound) variables and free variables or individual parameters. Gentzen’s rules are the following:
where [x/a] denotes the operation of substitution, that is, of replacing all free occurrences of in with a parameter . In case of and a parameter is required to be “fresh” in the sense of having no other occurrences in . Such a fresh is sometimes called an ‘eigenvariable’ or a ‘proper variable’.
The last rule in Gentzen’s tree format looks as follows:
Although Gentzen provided this set of rules for his tree-system of ND, it was easily adapted also to linear systems based on Jaśkowski’s (or Suppes’) format of proof. Let us illustrate their application in Fitch’s proof format (but not with his original rules):
The first application of introduces a parameter in place of . In line 3 and 7 the assumptions for the applications of in line 5 and 10 respectively are introduced, each time with a new eigenparameter in place of . Note that both applications of are correct since neither nor are present in the formulas ending suitable subproofs. Also the application of in line 6 is correct since is not present in line 1.
The fact that ) is a proof construction rule is obscured here since there is no need to introduce a subproof by means of a new assumption. We just require that in order to apply there be no occurrence of an involved parameter (here ) in active assumptions. However, there are systems of ND where such a subproof (usually flagged with a fresh parameter which will be universally quantified below) is explicitly introduced into a proof. For instance, the original Fitch’s rule is based on such a solution; in fact it follows closely the original Jaśkowski’s rule for inclusive general quantifier.
Gentzen’s was sometimes considered as complex and artificial, and some inference rules were proposed instead where is directly inferred and not assumed. Although the idea is simple its correct implementation leads to troubles. Carefull formulations of such a rule (as in Quine 1950) are correct but hard to follow; simple formulations (as in several editions of Copi 1954) make the system unsound. For a detailed analysis of the relations between Gentzen-style and Quine-style quantifier rules one should consult Fine (1985), Hazen (1987) and Pelletier (1999). All these problems with providing correct and simple rules for quantifiers led some authors to doubt if it is really possible (see Anellis 1991). It seems that the only correct system of ND for CFOL with ‘really’ simple rule of this kind is in Kalish and Montague (1964), but this is rather a side-effect of the overall architecture of the system which is not discussed here (but see a detailed explanation of the virtues of Kalish and Montague’s system in Indrzejczak 2010).
## 8. ND for Non-Classical Logics
ND systems were also offered for many important non-classical logics. In particular, Jaśkowski’s graphical approach is very handy in this field due to the machinery of isolated subproofs. It appeared that for many non-classical logics one can obtain a satisfying result by putting restrictions on the rule of repetition in the case of some subproofs. Let us take as an example the ND formalization of well known propositional modal logic T; for simplicity we restrict considerations to rules for (necessity). is obvious: . With the situation is more complicated since it is based on the following principle:
If , then
where formulas in the antecedent are also being changed by addition of . It is realised by means of a special ‘modal’ subproof which is opened with no assumption, but no other formulas may be put in it except those which were preceded by in outer subproofs (and with deleted after transition). If in such modal subproof we deduce , it can be closed and can be put into the outer subproof. The following proof in Fitch’s style illustrates this:
In line 4 a modal subproof was initiated which is shown by putting a sole in place of the assumption. Lines 5 and 6 result from the application of modal repetition. Such an approach may be easily extended to other modal logics by modifying conditions of modal repetition; for example, for S4 it is enough to admit that formulas with (no deletion) also may be repeated; for S5, formulas with negated are also allowed. Such an approach to modal logics was initiated by Fitch (1952), extensive study of such systems can be found in Fitting (1983), Garson (2006) and Indrzejczak (2010) where also some other approaches are discussed.
This modus of formalizing logics in ND was also applied for other non-classical logics including conditional logics (Thomason 1970), temporal logics (Indrzejczak 1994) and relevant logics (Anderson and Belnap 1975). In the latter the technique of restricted repetition is not enough however (and even not required for some logics of this kind). Far more important is the technique of labeling all formulas with sets of numbers annotating active assumptions which is necessary for keeping track of relevance conditions. Subsequently, applications of labels of different kinds is in fact one of the most popular technique used not only in tableau methods but also in ND. Vigano (2000) provides a good survey of this approach.
## 9. Normal Proofs
When constructing proofs one can easily make some inferences which are unnecessary for obtaining a goal. Gentzen was interested not only in providing an adequate system of ND but also in showing that everything which may be proved in such a system may be proved in the most straightforward way. As he put it, in such a proof “No concepts enter into the proof other than those contained in its final result, and their use was therefore essential to the achievement of the result’’ (Gentzen 1934).
In particular, such unnecessary moves are performed if one first applies some introduction rule for logical constant and then uses the conclusion of this rule application as a premise for the application of the elimination rule for . In such cases the final conclusion is either already present in the proof (as one of the premises of respective introduction rule) or may be directly deduced from premises of the application of introduction rule. For example, if one is deducing on the basis of and then by is deducing from this implication and , then it is simpler to deduce directly from ; the existence of such a proof is guaranteed because it is a subproof introducing . Let us call a maximal formula any formula which is at the same time the conclusion of an introduction rule and the main premise of an elimination rule. A proof is called normal iff no maximal formula is present in it. Roughly speaking we can obtain such a proof if first we apply elimination rules to our assumptions (premises) and then introduction rules to obtain the conclusion. Such proofs are analytic in the sense of having the subformula property: all formulas occurring in such a proof are subformulas or negations of subformulas of the conclusion or premises (undischarged assumptions).
Although the idea of a normal proof is rather simple to grasp it is not so simple to show that everything provable in ND system may have a normal proof. In fact for many ND systems (especially for many non-classical logics) such a result does not hold. Gentzen proved such a result directly for an ND system for Intuitionistic Logic, but he was unable to provide a proof for his ND for Classical Logic. He failed to provide the proof for the Intuitionistic case and instead he provided the result for both his ND systems indirectly. First he introduced an auxiliary technical system of sequent calculus and proved for it (both in the classical and intuitionistic cases) the famous Cut-Elimination Theorem. Then he showed that this result implies the existence of a normal proof for every thesis and valid argument provable in his ND systems. Such a result is usually called the Normal Form Theorem whereas the stronger result showing directly how to transform every ND-proof into normal proof by means of a systematic procedure is called the Normalization Theorem. That Gentzen indeed proved the Normalization Theorem for Intuitionistic case became known recently due to von Plato (2008) who found a preliminary draft of Gentzen’s thesis. The first published versions of proofs of Normalization theorems appeared in the 1960s due to Raggio (1965) and Prawitz (1965) who proved this result also for ND systems for some non-classical logics. For a detailed account of these problems see Troelstra and Schwichtenberg (1996) or Negri and von Plato (2001).
One thing should be noticed with respect to proofs in normal form. Although normal proofs are in a sense the most direct proofs, this does not mean that they are the most economical. In fact, non-normal proofs often may be shorter and easier to understand than normal ones. Perhaps it is simpler to understand if we recall that normalization in ND is the counterpart of cut-elimination in sequent calculi. Applications of cuts in proofs correspond to applications of previously proved things as lemmas and may drastically shorten proofs. When a proof is normalized, its size may grow exponentially (see, for example, Boolos 1984, Fitting 1996, D’Agostino 1999). What is important in normal proofs is that, due to their conceptual simplicity, they provide a proof theoretical justification of deduction and a new way of understanding the meaning of logical constants.
## 10. Philosophy of Meaning
Aesthetics was not the only reason for insisting on having both introduction and elimination rules for every constant in Gentzen’s ND. He also wanted to realise a deeper philosophical intuition concerning the meaning of logical constants. It is claimed that if a set of rules is intuitive and sufficient for adequate characterisation of a constant, then it in fact expresses our way of understanding this constant. Moreover, such an approach may be connected with Wittgenstein’s program of characterization of meaning by means of the use of words. In this particular case the meaning of logical constants is characterised by their use (via rules) in proof construction. There is also a strong connection with anti-realistic position in the philosophy of meaning where it is claimed that the notion of truth may be successfully replaced with the notion of a proof (Dummett 1991). One recent, and very strong, version of this trend is represented in Brandom’s (2000) program of strong inferentialism, where it is postulated that the meanings of all expressions may be characterised by means of their use in widely understood reasoning processes. However, inferentialism is not particularly connected with ND nor with the specific shapes of rules as giving rise to the meaning of logical constants.
Leaving aside the far-reaching program of inferentialism, one can quite reasonably ask whether the characteristic rules of logical constants may be treated as definitions. The term ‘Proof-Theoretic Semantics’ first appeared in 1991 (Schroeder-Heister 1991), but the roots of this idea is certainly linked with Gentzen (1934). He himself preferred introduction rules as a kind of definition of a constant. Elimination rules are just consequences of these ‘definitions’, not in the sense of being deducible from them but in the sense that their application is a kind of inversion of introduction rules. The notion of inversion was precisely characterised by Prawitz’s principle of inversion [see Prawitz’s (1965)]: if by the application of elimination rule we obtain , then proofs sufficient for deduction of premises of already contain a deduction of . Hence one can directly obtain on the basis of these proofs with no application of . As these sufficient conditions for deductions of premises are characterised by introduction rules, we can easily see that the inversion principle is strongly connected with the possibility of proving normalization theorems; it justifies making reduction steps for maximal formulas in normalization procedures.
Not all authors dealing with proof-theoretic semantics followed Gentzen in his particular solutions. Popper (1947) was the first who tried to construct deductive systems in which all rules for a constant were treated together as its definition. There are also approaches (such as Dummett 1991, chapter 13, and Prawitz 1971) in which elimination rules are treated as the most fundamental. No matter which kind of rules should be taken as basic for characterization of logical constants, it is obvious that not any set of rules may be treated as a candidate for definition. Prior (1960) paid attention to this fact by means of his famous example. Let us consider a connective “tonk’’ characterised by the following rules:
(tonk I) tonk
(tonk E) tonk
One can easily show that any formula is deducible from any formula after adding such rules to ND system. However Prior’s example only showed that one should carefuly characterise conditions of correctness for rules which are proposed as a tool for characterisation of logical constants. One of the first proposals is due to Belnap (1962) who emphasized that, just as for definitions, rules must be noncreative in the sense that if we add them to some ND system, then we obtain its conservative extension. In other words, if some formula with no occurrence of this new constant was not deducible in the ‘old’ system, then it is still not in the extended system. Rules for “tonk’’ do not satisfy this requirement. Although Belnap’s solution is not sufficient, he opened the door for further research of such conditions. The term “(proof-theoretic) harmony’’ is widely used for specification of such adequacy conditions for rules, and there is a large amount of literature concerned with this question. Schroeder-Heister (2014) provides one of the recent solutions to this problem whereas Schroeder-Heister (2012) offers extensive discussion of other approaches.
## 11. References and Further Reading
• [1] Anderson, A., R. and N., D. Belnap, Entailment: the Logic of Relevance and Necessity, vol I. Princeton University Press, Princeton 1975. 17.
• [2] Anellis, I. H., Forty Years of “Unnatural” Natural Deduction and Quantification. A History of First-Order Systems of Natural Deduction from Gentzen to Copi’, Modern Logic, 2(2): 113-152, 1991.
• [3] Belnap, N. D., Tonk, Plonk and Plink’, Analysis 22/6:130-134, 1962.
• [4] Bencivenga E., Jaskowski’s Universally Free Logic, Studia Logica, 102(6):1095-1102, 2014.
• [5] Boolos, G., Don’t eliminate Cut, Journal of Philosophical Logic, 7:373-378, 1984.
• [6] Boricic;, B. R., On Sequence-conclusion Natural Deduction Systems, Journal of Philosophical Logic, 14: 359-377, 1985.
• [7] Borkowski L., J. S lupecki, A Logical System based on rules and its applications in teaching Mathematical Logic, Studia Logica, 7: 71-113, 1958.
• [8] Brandom, R., Articulating Reasons. An Introduction to Inferentialism, Cambridge, Harvard University Press 2000.
• [9] Cellucci, C., Existential Instatiation and Normalization in Sequent Natural Deduction, Annals of Pure and Applied Logic, 58: 111-148, 1992.
• [10] Copi I. M., Symbolic Logic, The Macmillan Company, New York 1954.
• [11] Corcoran, J. Aristotle’s Natural Deduction System, in: J. Corcoran (ed.), Ancient Logic and its Modern Interpretations, Reidel, Dordrecht 1972.
• [12] D’Agostino, M., Tableau Methods for Classical Propositional Logic in: M. D’Agostino et al. (eds.), Handbook of Tableau Methods, pp. 45-123, Kluwer Academic Publishers, Dordrecht 1999.
• [13] Dummett, M., The Logical Basis of Metaphysics, Cambridge, Harvard University Press 1991.
• [14] Fine, K., Natural deduction and arbitrary objects’, Journal of Philosophical Logic 14:57-107, 1985.
• [15] Fitch, F.B., Symbolic Logic, Ronald Press Co, New York 1952.
• [16] Fitch, F.B., Natural deduction rules for obligation’, American Philosophical Quaterly 3:27-38, 1966.
• [17] Fitting, M., Proof Methods for Modal and Intuitionistic Logics, Reidel, Dordrecht 1983.
• [18] Fitting, M., First-Order Logic and Automated Theorem Proving, Springer, Berlin 1996. 18
• [19] Garson, J.W. Modal Logic for Philosophers, Cambridge University Press, Cambridge 2006.
• [20] Gentzen G., Uber die Existenz unabhangiger Axiomensysteme zu unendlichen Satzsystemen, Mathematische Annalen, 107:329-350, 1932.
• [21] Gentzen, G., Untersuchungen uber das Logische Schliessen, Mathematische Zeitschrift 39:176-210 and 39:405-431, 1934.
• [22] Gentzen, G., Die Widerspruchsfreiheit der reinen Zahlentheorie, Mathematische Annalen 112:493-565, 1936.
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• [29] Indrzejczak, A., Natural Deduction System for Tense Logics, Bulletin of the Section of Logic 23(4):173-179, 1994.
• [30] Indrzejczak, A., Natural Deduction, Hybrid Systems and Modal Logics, Springer 2010.
• [31] Jaskowski, S., Teoria dedukcji oparta na dyrektywach za lozeniowych in: Ksiega Pamiatkowa I Polskiego Zjazdu Matematycznego, Uniwersytet Jagiellonski, Krakow 1929.
• [32] Jaskowski, S., On the Rules of Suppositions in Formal Logic Studia Logica 1:5-32, 1934.
• [33] Kalish, D., and R. Montague, Logic, Techniques of Formal Reasoning, Harcourt, Brace and World, New York 1964.
• [34] Mates B., Stoic Logic, University of California Press, Berkeley 1953.
• [35] Negri, S., and J. von Plato, Structural Proof Theory, Cambridge University Press, Cambridge 2001. 19
• [36] Pelletier F. J. A Brief History of Natural Deduction, History and Philosophy of Logic, 20: 1-31, 1999.
• [37] Pelletier F. J. and A. P. Hazen, A History of Natural Deduction, in: D. Gabbay, F. J. Pelletier and E. Woods (eds.) Handbook of the History of Logic vol 11, 341-414, 2012.
• [38] Plato von J., Gentzen’s proof of normalization for ND, The Bulletin of Symbolic Logic 14(2):240-257, 2008.
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### Author Information
Andrzej Indrzejczak
Email: indrzej@filozof.uni.lodz.pl
University of Lodz
Poland
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## Question
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Which of the following represents the distance between the two points $$(x_{1},y_{2})$$ and $$(x_{2},y_{2})$$? （ ）
A. $$\sqrt{{(x_{2}+x_{1})^{2}}+{(x_{2}+y_{1})^{2}}}$$
B. $$\sqrt{{(x_{2}+x_{1})^{2}}-{(y_{2}-y_{1})^{2}}}$$
C. $$\sqrt{{(x_{2}-x_{1})^{2}}+{(y_{2}-y_{1})^{2}}}$$
D. $$\sqrt{{(x_{2}-x_{1})^{2}}-{(y_{2}-y_{1})^{2}}}$$
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Explanation
Given: $$(x_{1},x_{2})$$ and $$(y_{1},y_{2})$$
To find: Distance
Solution: $$(x_{1},x_{2})$$ and $$(y_{1},y_{2})$$
Distance between them
= $$\sqrt{{(x_{2}-x_{1})^{2}}+{(y_{2}-y_{1})^{2}}}$$
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# What is Turing Complete?
What does the expression "Turing Complete" mean?
Can you give a simple explanation, without going into too many theoretical details?
Here's the briefest explanation:
A Turing Complete system means a system in which a program can be written that will find an answer (although with no guarantees regarding runtime or memory).
So, if somebody says "my new thing is Turing Complete" that means in principle (although often not in practice) it could be used to solve any computation problem.
Sometime's it's a joke... a guy wrote a Turing Machine simulator in vi, so it's possible to say that vi is the only computational engine ever needed in the world.
• For further reading, see The Annotated Turing. Very approachable. amazon.com/Annotated-Turing-Through-Historic-Computability/dp/… – i_am_jorf May 18 '09 at 17:19
• "often not in practice" is incorrect. No system is ever Turing-complete in practice, because no realizable system has an infinite tape. What we really mean is that some systems have the ability to approximate Turing-completeness up to the limits of their available memory. – Shelby Moore III Aug 8 '14 at 22:40
• But Vi is the only computational engine ever needed in the world... ;-) – Joe Edgar Aug 15 '14 at 5:07
• Is Emacs a Turning Machine too? XD – alem0lars Feb 16 '15 at 18:57
• Someone recently showed that PowerPoint is Turing Complete too. – Tagc Apr 30 '17 at 12:57
Here is the simplest explanation
Alan Turing created a machine that can take a program, run that program, and show some result. But then he had to create different machines for different programs. So he created "Universal Turing Machine" that can take ANY program and run it.
Programming languages are similar to those machines (although virtual). They take programs and run them. Now, a programing language is called "Turing complete", if it can run any program (irrespective of the language) that a Turing machine can run given enough time and memory.
For example: Let's say there is a program that takes 10 numbers and adds them. A Turing machine can easily run this program. But now imagine that for some reason your programming language can't perform the same addition. This would make it "Turing incomplete" (so to speak). On the other hand, if it can run any program that the universal Turing machine can run, then it's Turing complete.
Most modern programming languages (e.g. Java, JavaScript, Perl, etc.) are all Turing complete because they each implement all the features required to run programs like addition, multiplication, if-else condition, return statements, ways to store/retrieve/erase data and so on.
Update: You can learn more on my blog post: "JavaScript Is Turing Complete" — Explained
• The idea that there would even be a term for this kind of machine makes a lot more sense when I remember Turing and other early computer scientists would build a specific machine each time they wanted to solve a specific problem. We’re used to one machine that can be forever reprogrammed. Thank you for the context, Raja. – Jacob Ford Aug 13 '17 at 14:32
• How JavaScript can be Turing Complete? It lacks file system , proper multithreading API . It has tons of limitations, mainly due to its browser security sandbox nature. It's hardly can be called ' a programming language ' .See how many variants of scripting abstraction exist (react, typescript ..you name it) ,all that to compensate what JS doesn't have. (asm.js should be mentioned here) . Java ,Python or C++ are true 'Turing Complete ' examples. But js? I don't think so. – Michael IV Dec 19 '17 at 22:30
• @MichaelIV The touring machine did not have a file system/threads either. JS is absolutely touring complete. – Bax Jan 21 '18 at 5:34
• @MichaelIV To add to Bax's response, one could consider a modern computer to consist of several Turing machines that work together to allow for all of those nice things that you mention. E.g. the CPU produces "tape" for the GPU to read so that it can write "tape" for the monitor so that the monitor can write "tape" to the user. Likewise, the CPU could produce "tape" for the hard drives, NICs, sound cards, etc. – user3003999 Oct 2 '18 at 14:33
• the blog post link is broken – TamaMcGlinn Jun 30 at 8:35
From wikipedia:
Turing completeness, named after Alan Turing, is significant in that every plausible design for a computing device so far advanced can be emulated by a universal Turing machine — an observation that has become known as the Church-Turing thesis. Thus, a machine that can act as a universal Turing machine can, in principle, perform any calculation that any other programmable computer is capable of. However, this has nothing to do with the effort required to write a program for the machine, the time it may take for the machine to perform the calculation, or any abilities the machine may possess that are unrelated to computation.
While truly Turing-complete machines are very likely physically impossible, as they require unlimited storage, Turing completeness is often loosely attributed to physical machines or programming languages that would be universal if they had unlimited storage. All modern computers are Turing-complete in this sense.
I don't know how you can be more non-technical than that except by saying "turing complete means 'able to answer computable problem given enough time and space'".
• In this context, what is a "computing device"? – dopatraman Nov 13 '14 at 16:43
• As with most Wikipedia articles, although this quote is technically correct, it provides no value to a person who has no knowledge on the subject and is trying to understand it. Being able to explain things properly is a science of its own :) – Lacho Tomov Sep 25 '17 at 15:19
## Informal Definition
A Turing complete language is one that can perform any computation. The Church-Turing Thesis states that any performable computation can be done by a Turing machine. A Turing machine is a machine with infinite random access memory and a finite 'program' that dictates when it should read, write, and move across that memory, when it should terminate with a certain result, and what it should do next. The input to a Turing machine is put in its memory before it starts.
## Things that can make a language NOT Turing complete
A Turing machine can make decisions based on what it sees in memory - The 'language' that only supports `+`, `-`, `*`, and `/` on integers is not Turing complete because it can't make a choice based on its input, but a Turing machine can.
A Turing machine can run forever - If we took Java, Javascript, or Python and removed the ability to do any sort of loop, GOTO, or function call, it wouldn't be Turing complete because it can't perform an arbitrary computation that never finishes. Coq is a theorem prover that can't express programs that don't terminate, so it's not Turing complete.
A Turing machine can use infinite memory - A language that was exactly like Java but would terminate once it used more than 4 Gigabytes of memory wouldn't be Turing complete, because a Turing machine can use infinite memory. This is why we can't actually build a Turing machine, but Java is still a Turing complete language because the Java language has no restriction preventing it from using infinite memory. This is one reason regular expressions aren't Turing complete.
A Turing machine has random access memory - A language that only lets you work with memory through `push` and `pop` operations to a stack wouldn't be Turing complete. If I have a 'language' that reads a string once and can only use memory by pushing and popping from a stack, it can tell me whether every `(` in the string has its own `)` later on by pushing when it sees `(` and popping when it sees `)`. However, it can't tell me if every `(` has its own `)` later on and every `[` has its own `]` later on (note that `([)]` meets this criteria but `([]]` does not). A Turing machine can use its random access memory to track `()`'s and `[]`'s separately, but this language with only a stack cannot.
A Turing machine can simulate any other Turing machine - A Turing machine, when given an appropriate 'program', can take another Turing machine's 'program' and simulate it on arbitrary input. If you had a language that was forbidden from implementing a Python interpreter, it wouldn't be Turing complete.
## Examples of Turing complete languages
If your language has infinite random access memory, conditional execution, and some form of repeated execution, it's probably Turing complete. There are more exotic systems that can still achieve everything a Turing machine can, which makes them Turing complete too:
• Untyped lambda calculus
• Conway's game of life
• C++ Templates
• Prolog
• SQL is most definitely turing-complete. It has scripting capabilities that allow for any computation. – nzifnab May 1 '11 at 0:41
• No, you are confusing SQL with extensions such as T-SQL / PL-SQL. ANSI SQL is not turing-complete. But TSQL / PLSQL - is. – Agnius Vasiliauskas Jul 3 '11 at 17:58
• Apparently SQL is turing-complete: stackoverflow.com/questions/900055/… – Newtang Jul 29 '12 at 23:59
• According to turing completeness - system is Turing complete if it can be used to simulate any single-taped Turing machine. But in example above as I understood devs constructed particular `cyclic tag system` and not `universal cyclic tag system`. Hence - article doesn't proves SQL turing completeness. (Or I misunderstood something) – Agnius Vasiliauskas Oct 2 '12 at 13:06
• There is no realizable implementation of a Turing-complete language, because there are no infinite tapes. What we really mean is that some languages have the ability to approximate Turing-completeness up to the limits of the available memory of the host machine. – Shelby Moore III Aug 8 '14 at 22:34
Fundamentally, Turing-completeness is one concise requirement, unbounded recursion.
Not even bounded by memory.
I thought of this independently, but here is some discussion of the assertion. My definition of LSP provides more context.
The other answers here don't directly define the fundamental essence of Turing-completeness.
• Finite state automata are allowed to have unbounded recursion. Case in point: `a*`. – user824425 Jun 4 '14 at 16:10
• @Rhymoid FSMs have limited memory—the finite # of states)—but unbounded recursion without tail optimization must have unlimited memory. I didn't restrict my definition to the subset of unbounded recursion only with tail optimization. Kindly remove your downvote. – Shelby Moore III Jul 23 '14 at 11:13
• you kept the definition of unbounded recursion foggy. Do you mean 'recursion' in the 'primitive recursion' sense, and 'unbounded' by making it 'partial' (or 'general', or 'mu-')? Then you may be right. But your current formulation is way too close to the statements criticized in David Harel's "On Folk Theorems". It's important to be rigorous in mathematics, and by leaving precise definitions out, you're ignoring that. By the way: FSMs can be generalized to model interaction; what sets them apart from TMs is that the latter's environment is also modeled (as the tape). – user824425 Jul 23 '14 at 11:26
• @Rhymoid enumeration is the antithesis of precision, e.g. enumerate the maximum precision of the fractions of an inch. Unbounded recursion means every possible form of recursion, which is impossible without an infinite tape. Fully generalized recursion (not just general within the model) is always Turing-complete. I am stating equivalence between generalized recursion and the ability to perform any possible computation. That is an important equivalence to note. – Shelby Moore III Aug 8 '14 at 22:22
• "Unbounded recursion means every possible form of recursion" That's your reading. To most SO users, 'unbounded recursion' means `while (p) { /* ... */ }`. "I am stating equivalence between generalized recursion and the ability to perform any possible computation." Church's thesis is a very different matter and should really be discussed separately. – user824425 Aug 8 '14 at 23:20
Turing Complete means that it is at least as powerful as a Turing Machine. This means anything that can be computed by a Turing Machine can be computed by a Turing Complete system.
No one has yet found a system more powerful than a Turing Machine. So, for the time being, saying a system is Turing Complete is the same as saying the system is as powerful as any known computing system (see Church-Turing Thesis).
• Note that all this disregards wall time. It just says "it can be done". – Thorbjørn Ravn Andersen Sep 27 '10 at 15:53
• @ThorbjørnRavnAndersen actually, it disregards physical computability altogether. Not only could it take longer than the age of the universe, but it might also use more memory than can be constructed with all the fermions and bosons in the universe. – Waylon Flinn May 9 '16 at 17:25
• There is, quitte possibly, no limit to the amount of bosons and fermions in the universe. We don't know, and will probably never know, it's size. Every time you read about the number of X in 'the universe', people are actually talking about the observable universe. Though interesting, it is not an actual physical limit. – Stijn de Witt Jun 23 '17 at 21:59
In the simplest terms, a Turing-complete system can solve any possible computational problem.
One of the key requirements is the scratchpad size be unbounded and that is possible to rewind to access prior writes to the scratchpad.
Thus in practice no system is Turing-complete.
Rather some systems approximate Turing-completeness by modeling unbounded memory and performing any possible computation that can fit within the system's memory.
I think the importance of the concept "Turing Complete" is in the the ability to identify a computing machine (not necessarily a mechanical/electrical "computer") that can have its processes be deconstructed into "simple" instructions, composed of simpler and simpler instructions, that a Universal machine could interpret and then execute.
I highly recommend The Annotated Turing
@Mark i think what you are explaining is a mix between the description of the Universal Turing Machine and Turing Complete.
Something that is Turing Complete, in a practical sense, would be a machine/process/computation able to be written and represented as a program, to be executed by a Universal Machine (a desktop computer). Though it doesn't take consideration for time or storage, as mentioned by others.
What i understand in simple words:
Turing Complete : A programming language / program that can do computation, is Turing complete.
For example :
1. Can you add two numbers using Just HTML. (Ans is 'No', you have to use javascript to perform addition.), Hence HTML is not Turing Complete.
2. Languages like Java , C++, Python, Javascript, Solidity for Ethereum etc are Turing Complete because you can do computation like adding two numbers using this languages.
Hope this helps.
A Turing Machine requires that any program can perform condition testing. That is fundamental.
Consider a player piano roll. The player piano can play a highly complicated piece of music, but there is never any conditional logic in the music. It is Turing Complete.
Conditional logic is both the power and the danger of a machine that is Turing Complete.
The piano roll is guaranteed to halt every time. There is no such guarantee for a TM. This is called the “halting problem.”
In practical language terms familiar to most programmers, the usual way to detect Turing completeness is if the language allows or allows the simulation of nested unbounded while statements (as opposed to Pascal-style for statements, with fixed upper bounds).
• A single unbounded while loop is enough to simulate a Turing machine. – masterxilo Apr 17 '17 at 22:11
Turing Complete means that it is at least as powerful as a Turing Machine.
I believe this is incorrect, a system is Turing complete if it's exactly as powerful as the Turing Machine, i.e. every computation done by the machine can be done by the system, but also every computation done by the system can be done by the Turing machine.
• I think you're assuming that the Church-Turing thesis is true to arrive at this conclusion. It has yet to be proven. The property you're describing is called 'Turing Equivalent'. – Waylon Flinn Dec 8 '09 at 13:41
• @WaylonFlinn No, he's right. "Completeness" means both that it is at least as strong as a thing, but also no stronger. Compare with "NP-Complete". – Devin Jeanpierre Jan 23 '12 at 23:19
• @DevinJeanpierre I don't want to start a flame war here but I'm almost certain the computational class you're describing is called "Turing Equivalent". Turing Complete does bear a similar relation to NP-Complete though. NP-Complete is equal to NP if and only if P=NP. In the same way Turing Complete is equal to Turing Equivalent if and only if the Church-Turing thesis is correct. – Waylon Flinn Jan 27 '12 at 15:24
• @Waylon Source? Nothing I read agrees with that (e.g. en.wikipedia.org/wiki/Turing_completeness ) – Devin Jeanpierre Jan 29 '12 at 16:09
• @DevinJeanpierre It says it right there in the wikipedia article you link to. Quoting the Formal definitions section: "A computational system that can compute every Turing-computable function is called Turing complete", "A Turing-complete system is called Turing equivalent if every function it can compute is also Turing computable" – Waylon Flinn Jan 31 '12 at 14:46
Its complete if it can test and branch (has an 'if')
• For such an old question, it would be worthwhile checking to see if others have already made similar or more substantive contributions already – alan ocallaghan Jan 15 at 17:16
• Not sure about the correctness of the answer. But this is really simple explanation that I never seen before. Funny thing: long long time ago (after I wrote my first piece of code) I also used the same explanation to define the simplest processor possible. – Victor Yarema Jan 15 at 19:32
• It's an excellent first try at an incisive, concise and accurate operational definition. However, the branch has to permit looping and, is it not the case that the machine must also permit subroutine calls (ie: recursion)? Is there a flattened program of nested loops for every program with recursion? – user3673 Feb 29 at 15:41
Can a relational database input latitudes and longitudes of places and roads, and compute the shortest path between them - no. This is one problem that shows SQL is not Turing complete.
But C++ can do it, and can do any problem. Thus it is.
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# Refraction of Light
A cylindrical material of radius R = 2.00 m has a mirrored surface on its right half, (as in figure that i have attached below). A light ray traveling in air is incident on the left side of the cylinder. If the incident light ray and exiting light ray are parallel and d = 2.00 m, determine the index of refraction of the material.
I know that the index of refraction of a material n is n = c /u where c is the speed of light in a vacuum and u is the speed of light in the material.
But in the figure there is nothing which has to do with this formula.
A little help?
#### Attachments
• 9.4 KB Views: 225
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Snell's law
You'll need to apply Snell's law of refraction and a little geometry. Snell's law is:
$$n_1sin\theta_1 = n_2sin\theta_2$$
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If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H2 at STP from an acid, the composition of the mixture is
# If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H2 at STP from an acid, the composition of the mixture is
1. A
2. B
3. C
4. D
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### Solution:
According to STP conditions, one mole of a gas has a volume of 22400 mL. At STP conditions, one mole of H2 has a volume of 22400 mL.
If one mole of H2 (or 2 equivalents of H2) has a volume of 22400 mL, then one equivalent of H2 has a volume of 11200 mL.
If one equivalent of H2 has a volume of 11200 mL, then the number of equivalents of H2 present in 560 mL is calculated as:
Total mass of the mixture is 5.6 g. Assume that the mass of A is x g and the mass of B is 0.5-x g.
Equivalent mass of A is calculated as:
Equivalent mass of B is calculated as:
Equivalents of A and B displaces the equivalents of H2. This implies that sum of equivalents of A and B is equal to the equivalents of H2 present in 560 mL.
Substitute the values in the above equation and determine the value of x.
$\begin{array}{ccc}\frac{\mathrm{x}}{12}+\frac{0.5-\mathrm{x}}{9}& =& \frac{1}{20}\\ \frac{3\mathrm{x}+4\left(0.5-\mathrm{x}\right)}{36}& =& \frac{1}{20}\\ \mathrm{x}& =& 0.2\end{array}$
Therefore, mass of A is 0.2 g and the mass of B is 0.3 g (0.5 g-0.2 g).
Composition of A in the mixture is calculated as:
Composition of B in the mixture is calculated as:
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# Re: entropy depletion (was: SSL/TLS passive sniffing)
```| > You're letting your intuition about "usable randomness" run roughshod
| > over the formal definition of entropy. Taking bits out of the PRNG
| > *does* reduce its entropy.
|
| By how much exactly? I'd say, _under the hypothesis that the one-way
| function can't be broken and other attacks fail_, exactly zero; in the
| real world, maybe a little more. But in
| /usr/src/linux/drivers/char/random.c I see that the extract_entropy()
| function, directly called by the exported kernel interface
| get_random_bytes(), states:
|
| if (r->entropy_count / 8 >= nbytes)
| r->entropy_count -= nbytes*8;
| else
| r->entropy_count = 0;
|
| ...which appears to assume that the pool's entropy (the upper bound of
| which is POOLBITS, defined equal to 4096) drops by a figure equal to the
| number of bits that are extracted (nbytes*8). This would only make sense
| if those bits weren't passed through one-way hashing.
The argument you are making is that because the one-way function isn't
reversible, generating values from the pool using it doesn't decrease its
"computational" entropy. (Its mathematical entropy is certainly depleted,
since that doesn't involve computational difficulty. But we'll grant that
that doesn't matter.)```
```
The problem with this argument is that it gives you no information about the
unpredictablity of the random numbers generated. Here's an algorithm based
Pool: bits[512]
initializePool()
{ Fill Pool with 512 random bits; }
getRandom() : bits[160]
{ return(SHA(bits));
}
By your argument, seeing the result of a call to getRandom() does not reduce
the effective entropy of the pool at all; it remains random. We certainly
believe that applying SHA to a random collection of bits produces a random
value. So, indeed, the result of getRandom() is ... random. It's also
constant.
Granted, no one would implement a random number generator this way. But
*why*? What is it you have to change to make this correct? Why? Can you
prove it? Just saying "you have to change the pool after every call"
won't work:
getRandom() : bits[160]
{ Rotate bits left by 1 bit;
return(SHA(bits));
}
This (seems to) generated 512 random values, then repeats. Just what *is*
good enough?
-- Jerry
---------------------------------------------------------------------
The Cryptography Mailing List
Unsubscribe by sending "unsubscribe cryptography" to [EMAIL PROTECTED]
```
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# Are The Given Vectors Normal
This post categorized under Vector and posted on June 25th, 2018.
Normal to surfaces in 3D graphice Calculating a surface normal. For a convex polygon (such as a triangle) a surface normal can be calculated as the vector cross product of two (non-parallel) edges of the polygon.. For a plane given by the equation the vector () is a normal.. For a plane given by the equation () i.e. a is a point The normal vector often simply called the normal to a surface is a vector which is perpendicular to the surface at a given point. When normals are considered on closed surfaces the inward-pointing normal (pointing towards the interior of the surface) and outward-pointing normal are usually In order to calculate with vectors the graphical representation may be too graphicbersome. Vectors in an n-dimensional Euclidean graphice can be represented as coordinate vectors in a Cartesian coordinate system.The endpoint of a vector can be identified with an ordered list of n real numbers (n-tuple).These numbers are the coordinates of the endpoint of the vector with respect to a given
Cheat Sheets & Tables Algebra Trigonometry and Calculus cheat sheets and a variety of tables. Clgraphic Notes Each clgraphic has notes available. Most of the clgraphices have practice problems with solutions available on the practice problems pages. Also most clgraphices have graphicignment problems for instructors to graphicign for homework (answerssolutions to the graphicignment problems are not given Physics is a mathematical science. The underlying concepts and principles have a mathematical basis. Throughout the course of our study of physics we will encounter a variety of concepts that have a mathematical basis graphicociated with them.Resources for teaching the 2017 specifications. These teaching resources for the 2017 specifications are provided by MEI. They are linked with MEIs scheme of work which can be used with any of the 2017 A level specifications.
Cheat Sheets & Tables Algebra Trigonometry and Calculus cheat sheets and a variety of tables. Clgraphic Notes Each clgraphic has notes available. Most of the clgraphices have practice problems with solutions available on the practice problems pages. Also most clgraphices have graphicignment problems for instructors to graphicign for homework (answerssolutions to the graphicignment problems are not given Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site
## Tangent Vectors To A Surface Normal
This is just the equation of a plane given a point and its normal vector (you should look into that). Any vector in this plane is tangent to the su [more]
## A Show That The Normal Component Of Electrostatic Field Has A Discontinuity From One Side Of A Charged Surface To Another Given By
Fideisms Judaism is the Semitic monotheistic fideist religion based on the Old Testaments (1000-600 BCE) rules for the worship of Yahweh by his cho [more]
## Find Unit Normal Vector Surface Given Point Surface Z X Point Q
Polygons and meshes In what follows are various notes and algorithms dealing with polygons and meshes. Surface (polygonal) Simplification Written b [more]
## Consider Vector Function Given Following Find Unit Tangent Unit Normal Vectors T T N T B Q
For every point in one of the two images of a stereo pair the function finds the equation of the corresponding epipolar line in the other image.The [more]
## Represent Plane Curve X Y Vector Valued Function Given Vector Valued Function V Q
May 20 2014 My Vectors course httpswww.kristakingmath.comvectors-course Learn how to find the vector function for the curve of intersection of two [more]
## Find Unit Normal Vector Surface Given Point Surface Point Ln X Y Z Q
Words in science are often used in different ways from ordinary English. Completely different meanings even occur in different branches of physiolo [more]
## Get Points Lying On The Plane By Given Normal
So it all comes down to finding a point p_1 on a plane through the origin that is not itself too close to the origin given the normal n. Let n(abc) [more]
## Determine Whether Given Vectors Orthogonal Parallel Neither Explain Answers Q
Triply-periodic minimal surfaces This is an ilvectorrated account of my vector study of TPMS aimed at both beginner and spevectort. It containsHere [more]
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Home > manopt > tools > multihconj_legacy.m
# multihconj_legacy
## PURPOSE
MULTIHCONJ Hermitian conjugating arrays of matrices.
## SYNOPSIS
function b = multihconj_legacy(a, dim)
## DESCRIPTION
MULTIHCONJ Hermitian conjugating arrays of matrices.
THIS ORIGINAL MULTIHCONJ IS NOW CALLED MULTIHCONJ_LEGACY
B = MULTIHCONJ(A) is equivalent to B = MULTIHCONJ(A, DIM), where
DIM = 1.
B = MULTIHCONJ(A, DIM) is equivalent to
B = PERMUTE(A, [1:DIM-1, DIM+1, DIM, DIM+2:NDIMS(A)]), where A is an
array containing N P-by-Q matrices along its dimensions DIM and DIM+1,
and B is an array containing the Q-by-P Hermitian conjugate (') of
those N matrices along the same dimensions. N = NUMEL(A) / (P*Q), i.e.
N is equal to the number of elements in A divided by the number of
elements in each matrix.
Example:
A 5-by-9-by-3-by-2 array may be considered to be a block array
containing ten 9-by-3 matrices along dimensions 2 and 3. In this
case, its size is so indicated: 5-by-(9-by-3)-by-2 or 5x(9x3)x2.
If A is ................ a 5x(9x3)x2 array of 9x3 matrices,
C = MULTIHCONJ(A, 2) is a 5x(3x9)x2 array of 3x9 matrices.
See also MULTIHCONJ MULTITRANSP MULTIHERM.
## CROSS-REFERENCE INFORMATION
This function calls:
This function is called by:
• multihconj Hermitian-conjugate transpose the matrix slices of an N-D array
## SOURCE CODE
0001 function b = multihconj_legacy(a, dim)
0002 %MULTIHCONJ Hermitian conjugating arrays of matrices.
0003 %
0004 % THIS ORIGINAL MULTIHCONJ IS NOW CALLED MULTIHCONJ_LEGACY
0005 %
0006 % B = MULTIHCONJ(A) is equivalent to B = MULTIHCONJ(A, DIM), where
0007 % DIM = 1.
0008 %
0009 % B = MULTIHCONJ(A, DIM) is equivalent to
0010 % B = PERMUTE(A, [1:DIM-1, DIM+1, DIM, DIM+2:NDIMS(A)]), where A is an
0011 % array containing N P-by-Q matrices along its dimensions DIM and DIM+1,
0012 % and B is an array containing the Q-by-P Hermitian conjugate (') of
0013 % those N matrices along the same dimensions. N = NUMEL(A) / (P*Q), i.e.
0014 % N is equal to the number of elements in A divided by the number of
0015 % elements in each matrix.
0016 %
0017 %
0018 % Example:
0019 % A 5-by-9-by-3-by-2 array may be considered to be a block array
0020 % containing ten 9-by-3 matrices along dimensions 2 and 3. In this
0021 % case, its size is so indicated: 5-by-(9-by-3)-by-2 or 5x(9x3)x2.
0022 % If A is ................ a 5x(9x3)x2 array of 9x3 matrices,
0023 % C = MULTIHCONJ(A, 2) is a 5x(3x9)x2 array of 3x9 matrices.
0024 %
0025 % See also MULTIHCONJ MULTITRANSP MULTIHERM.
0026
0027 % This file is part of Manopt: www.manopt.org.
0028 % Original author: Hiroyuki Sato, April 27, 2015.
0029 % Contributors:
0030 % Change log:
0031
0032 % Setting DIM if not supplied.
0033 if nargin == 1, dim = 1; end
0034
0035 % Transposing
0036 b = multitransp_legacy(a, dim);
0037
0038 %Conjugating
0039 b = conj(b);
0040
0041 end
Generated on Fri 30-Sep-2022 13:18:25 by m2html © 2005
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New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```typography en = 0.0001757299 meter (length) ``` Related Measurements: Try converting from "en" to caliber (gun barrel caliber), cloth quarter, engineers chain, fermi, finger, furlong (surveyors furlong), hand, inch, light yr (light year), line, link (surveyors link), m (meter), marathon, mile, naval shot, ri (Japanese ri), rope, skein, soccer field, span (cloth span), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: en = .00000495 actus (Roman actus), 1,757,299 angstrom, 3.97E-09 arpentcan, 1.17E-15 astronomical unit, .00038436 Biblical cubit, .27674 bottom measure, 8.01E-07 cable length, .00000874 chain (surveyors chain), .00015374 ell, .00009609 fathom, 8.74E-07 furlong (surveyors furlong), .00023726 gradus (Roman gradus), 3.64E-08 league, .083022 line, .00017573 m (meter), .00307489 nail (cloth nail), 9.49E-08 nautical mile, 4.48E-08 ri (Japanese ri), .00008356 sazhen (Russian sazhen), .00076872 span (cloth span).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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16. Stokes equations¶
This demo is implemented in a single Python file, demo_stokes-iterative.py, which contains both the variational forms and the solver.
This demo illustrates how to:
• Maintain symmetry when assembling a system of symmetric equations with essential (Dirichlet) boundary conditions
• Use a iterative solver explicitly for solving a linear system of equations
• Define a preconditioner explicitly using a form
16.1. Strong form of the Stokes equations¶
The incompressible Stokes equations in strong form read: for a domain $$\Omega \subset \mathbb{R}^n$$, find the velocity $$u$$ and the pressure $$p$$ satisfying
$\begin{split}- \nabla \cdot (\nabla u + p I) &= f \quad {\rm in} \ \Omega, \\ \nabla \cdot u &= 0 \quad {\rm in} \ \Omega. \\\end{split}$
Note
The sign of the pressure has been flipped from the classical definition. This is done in order to have a symmetric (but not positive-definite) system of equations rather than a non-symmetric (but positive-definite) system of equations.
A typical set of boundary conditions on the boundary $$\partial \Omega = \Gamma_{D} \cup \Gamma_{N}$$ can be:
$\begin{split}u &= u_0 \quad {\rm on} \ \Gamma_{D}, \\ \nabla u \cdot n + p n &= g \quad {\rm on} \ \Gamma_{N}. \\\end{split}$
16.2. Weak form of the Stokes equations¶
The Stokes equations can easily formulated in a mixed variational form; that is, a form where the two variables, the velocity and the pressure, are approximated simultaneously. Using the abstract framework, we have the problem: find $$(u, p) \in W$$ such that
$a((u, p), (v, q)) = L((v, q))$
for all $$(v, q) \in W$$ where
$\begin{split}a((u, p), (v, q)) &= \int_{\Omega} \nabla u \cdot \nabla v + \nabla \cdot v \ p + \nabla \cdot u \ q \, {\rm d} x, \\ L((v, q)) &= \int_{\Omega} f \cdot v \, {\rm d} x + \int_{\partial \Omega_N} g \cdot v \, {\rm d} s. \\\end{split}$
The space W should be a mixed (product) function space: $$W = V \times Q$$ such that $$u \in V$$ and $$q \in Q$$.
16.3. Preconditioning of the linear system of equations¶
For the resulting linear system of equations, the following form defines a suitable preconditioner:
$b((u, p), (v, q)) = \int_{\Omega} \nabla u \cdot \nabla v + p \, q \, {\rm d} x$
16.4. Domain and boundary conditions¶
In this demo, we shall consider the following definitions of the input functions, the domain, and the boundaries:
• $$\Omega = [0,1]^3$$ (a unit cube)
• $$\Omega_D = \{(x_0, x_1, x_2) \, | \, x_0 = 0 \, \text{or} \, x_0 = 1 \, \text{or} \, x_1 = 0 \, \text{or} \, x_1 = 1 \}$$
• $$u_0 = (- \sin(\pi x_1), 0.0, 0.0)$$ for $$x_0 = 1$$ and $$u_0 = (0.0, 0.0, 0.0)$$ otherwise
• $$f = (0.0, 0.0, 0.0)$$
• $$g = (0.0, 0.0, 0.0)$$
16.5. Implementation¶
This description goes through the implementation (in demo_stokes-iterative.py) of a solver for the above described Stokes equations. Some of the standard steps will be described in less detail, so before reading this, we suggest that you are familiarize with the Poisson demo (for the very basics) and the Mixed Poisson demo (for how to deal with mixed function spaces). Also, the Navier–Stokes demo illustrates how to use iterative solvers in a more implicit manner (typically only suitable for positive-definite systems of equations).
The Stokes equations as formulated above result in a system of linear equations that is not positive-definite. Standard iterative linear solvers typically fail to converge for such systems. Some care must therefore be taken in preconditioning the systems of equations. Moreover, not all of the linear algebra backends support this. We therefore start by checking that either “PETSc” or “Tpetra” (from Trilinos) is available. We also try to pick MINRES Krylov subspace method which is suitable for symmetric indefinite problems. If not available, costly QMR method is choosen.
from dolfin import *
# Test for PETSc or Tpetra
if not has_linear_algebra_backend("PETSc") and not has_linear_algebra_backend("Tpetra"):
info("DOLFIN has not been configured with Trilinos or PETSc. Exiting.")
exit()
if not has_krylov_solver_preconditioner("amg"):
info("Sorry, this demo is only available when DOLFIN is compiled with AMG "
"preconditioner, Hypre or ML.")
exit()
if has_krylov_solver_method("minres"):
krylov_method = "minres"
elif has_krylov_solver_method("tfqmr"):
krylov_method = "tfqmr"
else:
info("Default linear algebra backend was not compiled with MINRES or TFQMR "
"Krylov subspace method. Terminating.")
exit()
Next, we define the mesh (a UnitCubeMesh) and a mixed finite element TH. Then we build a FunctionSpace on this element. (This mixed finite element space is known as the Taylor–Hood elements and is a stable, standard element pair for the Stokes equations.)
# Load mesh
mesh = UnitCubeMesh(16, 16, 16)
# Build function space
P2 = VectorElement("Lagrange", mesh.ufl_cell(), 2)
P1 = FiniteElement("Lagrange", mesh.ufl_cell(), 1)
TH = P2 * P1
W = FunctionSpace(mesh, TH)
Next, we define the boundary conditions.
# Boundaries
def right(x, on_boundary): return x[0] > (1.0 - DOLFIN_EPS)
def left(x, on_boundary): return x[0] < DOLFIN_EPS
def top_bottom(x, on_boundary):
return x[1] > 1.0 - DOLFIN_EPS or x[1] < DOLFIN_EPS
# No-slip boundary condition for velocity
noslip = Constant((0.0, 0.0, 0.0))
bc0 = DirichletBC(W.sub(0), noslip, top_bottom)
# Inflow boundary condition for velocity
inflow = Expression(("-sin(x[1]*pi)", "0.0", "0.0"), degree=2)
bc1 = DirichletBC(W.sub(0), inflow, right)
# Collect boundary conditions
bcs = [bc0, bc1]
The bilinear and linear forms corresponding to the weak mixed formulation of the Stokes equations are defined as follows:
# Define variational problem
(u, p) = TrialFunctions(W)
(v, q) = TestFunctions(W)
f = Constant((0.0, 0.0, 0.0))
a = inner(grad(u), grad(v))*dx + div(v)*p*dx + q*div(u)*dx
L = inner(f, v)*dx
We can now use the same TrialFunctions and TestFunctions to define the preconditioner matrix. We first define the form corresponding to the expression for the preconditioner (given in the initial description above):
# Form for use in constructing preconditioner matrix
b = inner(grad(u), grad(v))*dx + p*q*dx
Next, we want to assemble the matrix corresponding to the bilinear form and the vector corresponding to the linear form of the Stokes equations. Moreover, we want to apply the specified boundary conditions to the linear system. However, assembling the matrix and vector and applying a DirichletBC separately will possibly result in a non-symmetric system of equations. Instead, we can use the assemble_system function to assemble both the matrix A, the vector bb, and apply the boundary conditions bcs in a symmetric fashion:
# Assemble system
A, bb = assemble_system(a, L, bcs)
We do the same for the preconditioner matrix P using the linear form L as a dummy form:
# Assemble preconditioner system
P, btmp = assemble_system(b, L, bcs)
Next, we specify the iterative solver we want to use, in this case a KrylovSolver. We associate the left-hand side matrix A and the preconditioner matrix P with the solver by calling solver.set_operators.
# Create Krylov solver and AMG preconditioner
solver = KrylovSolver(krylov_method, "amg")
# Associate operator (A) and preconditioner matrix (P)
solver.set_operators(A, P)
We are now almost ready to solve the linear system of equations. It remains to specify a Vector for storing the result. For easy manipulation later, we can define a Function and use the vector associated with this Function. The call to solver.solve then looks as follows
# Solve
U = Function(W)
solver.solve(U.vector(), bb)
Finally, we can play with the result in different ways:
# Get sub-functions
u, p = U.split()
# Save solution in VTK format
ufile_pvd = File("velocity.pvd")
ufile_pvd << u
pfile_pvd = File("pressure.pvd")
pfile_pvd << p
# Plot solution
plot(u)
plot(p)
interactive()
16.6. Complete code¶
from dolfin import *
# Test for PETSc or Tpetra
if not has_linear_algebra_backend("PETSc") and not has_linear_algebra_backend("Tpetra"):
info("DOLFIN has not been configured with Trilinos or PETSc. Exiting.")
exit()
if not has_krylov_solver_preconditioner("amg"):
info("Sorry, this demo is only available when DOLFIN is compiled with AMG "
"preconditioner, Hypre or ML.")
exit()
if has_krylov_solver_method("minres"):
krylov_method = "minres"
elif has_krylov_solver_method("tfqmr"):
krylov_method = "tfqmr"
else:
info("Default linear algebra backend was not compiled with MINRES or TFQMR "
"Krylov subspace method. Terminating.")
exit()
# Load mesh
mesh = UnitCubeMesh(16, 16, 16)
# Build function space
P2 = VectorElement("Lagrange", mesh.ufl_cell(), 2)
P1 = FiniteElement("Lagrange", mesh.ufl_cell(), 1)
TH = P2 * P1
W = FunctionSpace(mesh, TH)
# Boundaries
def right(x, on_boundary): return x[0] > (1.0 - DOLFIN_EPS)
def left(x, on_boundary): return x[0] < DOLFIN_EPS
def top_bottom(x, on_boundary):
return x[1] > 1.0 - DOLFIN_EPS or x[1] < DOLFIN_EPS
# No-slip boundary condition for velocity
noslip = Constant((0.0, 0.0, 0.0))
bc0 = DirichletBC(W.sub(0), noslip, top_bottom)
# Inflow boundary condition for velocity
inflow = Expression(("-sin(x[1]*pi)", "0.0", "0.0"), degree=2)
bc1 = DirichletBC(W.sub(0), inflow, right)
# Collect boundary conditions
bcs = [bc0, bc1]
# Define variational problem
(u, p) = TrialFunctions(W)
(v, q) = TestFunctions(W)
f = Constant((0.0, 0.0, 0.0))
a = inner(grad(u), grad(v))*dx + div(v)*p*dx + q*div(u)*dx
L = inner(f, v)*dx
# Form for use in constructing preconditioner matrix
b = inner(grad(u), grad(v))*dx + p*q*dx
# Assemble system
A, bb = assemble_system(a, L, bcs)
# Assemble preconditioner system
P, btmp = assemble_system(b, L, bcs)
# Create Krylov solver and AMG preconditioner
solver = KrylovSolver(krylov_method, "amg")
# Associate operator (A) and preconditioner matrix (P)
solver.set_operators(A, P)
# Solve
U = Function(W)
solver.solve(U.vector(), bb)
# Get sub-functions
u, p = U.split()
# Save solution in VTK format
ufile_pvd = File("velocity.pvd")
ufile_pvd << u
pfile_pvd = File("pressure.pvd")
pfile_pvd << p
# Plot solution
plot(u)
plot(p)
interactive()
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1. Home
2. >>How To Calculate Driving Force Conveyor Belt Pdf
# how to calculate driving force conveyor belt pdf
The permitted driving force exerted on the drive pulley is fted,max to ensure that sufficient driving force can be transferred onto the belt otherwise, the belt will slip around the drive pulley as shown in fig , in case of a belt conveyor that is tensioned by gravity takeup device with mass mt, the belt tension t equals
## Samples Show
• ### Delft University Of Technology Belt Conveyor Dynamics
The belt carcass will break down and split longitudinally in this area if it is loaded beyond its capacity to support the load table f page of the engineering manual the values of the belt carcass to pass the above consideration should be available in the conveyor belt manufacturer reference manuals or printed information for belt .
• ### Conveyor Belt Manual Ibt Industrial Solutions
The uks mhea guide, the iso guide iso etc, etc this is a specialist belt conveyor forum for belt conveyors visited by those whose business is belt conveyors,Apr , power te n x belt speed ms kw how you calculate the belt tensions is down to the design guide you use these could be the american cema guide referred to earlier.
• ### Belt Conveyor Newell Equipment Company
A belt conveyor is a conveyor in which the product rides directly on the belt the belt is supported by either a roller bed conveyor section or a slider bed conveyor section this manual will explain the various types of mathews belt conveyors this manual describes the following types of belt conveyors horizontal belt conveyors .
• ### Calculation Required Data For Chain Calculation
Where the space permits a vertical curve and where the load requires the use of a single belt fig conveyor with belt loaded in decline or incline,Fig conveyor with horizontal belt fig conveyor belt with incline and horizontal where two belts are needed fig conveyor with horizontal belt with incline section.
• ### Motor Sizing Calculations
Many problems which occur with belt conveyors can be attributed to system dynamics and the associated variations in belt tension material spillage, belt liftoff in vertical curves and poor tracking are a few examples of situations where knowledge of the dynamic tensions would be of benefit in eliminating a problem, belt and splice failure.
• ### Working Strength Vs Breaking Strength
The carcass is the heart of the conveyor belt ,Conveyor belts generally are composed of three main components carcass skims covers carry cover and pulley cover carcass the reinforcement usually found on the inside of a conveyor belt is normally referred to as the quotcarcassQuot in a sense.
• ### Belt Conveyor Catalog Orthman Manufacturing
Belts in line with iso it describes the probable longterm forceelongation properties of the belt material that has been subjected to stress due to deflection and load change this produces the calculation force f w this implies that higher belt forces f winitial will occur when tensioning the belt they will have to be taken into .
• ### How To Calculate The Tph Of Belt Conveyor Ftmlie
Belt conveyors calculations piotr kulinowski, ph d eng gravity forcefriction force p w m g cos m g sin belt conveyor of necessary power at the periphery of the driving pulley v kw, ph d eng piotr kasza.
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Start by collecting all the information that you need and enter it in a form you need to calculate the maximum amount of ,Oct , getting the conveyor belt tension right is imperative for the system to work smoothly the belt needs to conform to the crown of any pulley present and be slip resistant even under the most challenging external conditions to calculate the conveyor belt tension accurately.
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Brushless dc motors or stepping motors , speed, load torque and load inertia for various mechanisms select a motor type from ac motors,Calculate the value for load torque, etc at the motor drive shaft of the mechanism refer to page for calculating the speed, load inertia.
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Generally,all components of a conveyor should be checked at once check the condition of the chainbelt regularly,and replace linksmodules which are damaged important in this matter is to try to find the cause of the damaged linksmodulesWear patterns or damage on a chain or belt can often lead to a problem area elsewhere in the conveyor.
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In the conveyor never walk on conveyor covers,Of your body, grating or guards do not use conveyor for any purpose other than that for which it was intended do not poke or prod material into the conveyor with a bar or stick inserted through the openings keep area around conveyor drive and control station free of debris and .
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Torque and power required from a conveyor to move a load at an angle if your conveyor is horizontal enter an angle of enter your values for the mass, beltspeed, diameter, friction and angle select your units as required,Conveyor power calculator use this calculator to calculate the force.
• ### Calculation Of Tension Force Of Belt Conveyor
Apr , etc this is a specialist belt conveyor forum for belt conveyors visited by those whose business is belt conveyors, power te n x belt speed ms kw how you calculate the belt tensions is down to the design guide you use these could be the american cema guide referred to earlier, the uks mhea guide, the iso guide iso etc.
• ### Delft University Of Technology Belt Conveyor Dynamics
Also known as the angle of natural friction is the angle at which the material, design of belt conveyor the design of the belt conveyor must begin with an evaluation of the characteristics of the conveyed material and in particular the angle of repose and the angle of surcharge the angle of repose of a material, when heaped freely onto.
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Because not all elements are set in motion simultaneously, use , elastic and permanent as a rough guideline,L conveyor length m belt elongation, elongation for textile belts and ,, dynamic startup calculations may be required, for steel cord belts note for longdistance conveyors.
• ### Belt Tension Calculator Pooley Inc
The belt tension t equals, the belt will slip around the drive pulley as shown in fig ,max to ensure that sufficient driving force can be transferred onto the belt otherwise,The permitted driving force exerted on the drive pulley is fted, in case of a belt conveyor that is tensioned by gravity takeup device with mass mt.
• ### Determining Dynamic Belt Tensions Using
The belt carcass will break down and split longitudinally in this area if it is loaded beyond its capacity to support the load table f page of the engineering manual the values of the belt carcass to pass the above consideration should be available in the conveyor belt manufacturer reference manuals or printed information for belt .
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Mike gawinski explains how to calculate required belt pull and required conveyor drive power on a package handling belt conveyor for a free copy of the comp.
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In case of a belt conveyor that is tensioned by gravity takeup device with mass mt,max to ensure that sufficient driving force can be transferred onto the belt otherwise, the belt will slip around the drive pulley as shown in fig , the belt tension t equals,The permitted driving force exerted on the drive pulley is fted.
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Understanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system we use cookies to personalize content and analyze traffic we also share information about your use of our site with our social media, and analytics partners who may combine it with other , advertising.
• ### How Do I Calculate Conveyor Belt Tension With Pictures
How to calculate tph of belt conveyor how to calculate tph of belt conveyor mining a belt conveyor is a moving surface required at the headshaft of a conveyor belt capacity tph conveyor motor sizing forms calculate the how to calculate belt conveyor speed with motor rpm and ger rastioc we know ibs per hour tph belt width chat online .
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It is not uncommon to hear references to working tensions or maximum working tensions and ultimate breaking strength or ultimate tensile strength of a belt at times there is confusion about the meanings of these ratings both of these ratings refer to the capabilities of the,When dealing with conveyor belt tension ratings.
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Without coating finish diameter the outside diameter of a coated pulley core diameter times the coatingwrap thickness face width the length of a pulleys cylindrical bodyThis area is intended to act as the contact surface for the conveyor belt, conveyor pulley selection guide pulleycore diameter the outside diameter of the cylindrical body of a conveyor pulley.
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Design of belt conveyor the design of the belt conveyor must begin with an evaluation of the characteristics of the conveyed material and in particular the angle of repose and the angle of surcharge the angle of repose of a material, when heaped freely onto, also known as the angle of natural friction is the angle at which the material.
• ### Design And Application Of Feeders For The
It follows that the measure in horsepower of the work that a machine performs is force in lbs multiplied by speed in feet per minute divided by many conveyors do not elevate loads, but merely carry them horizontally from one location to another.
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Will be degrees to degrees less than the angle of repose ex determine the density of the material in pounds per cubic foot lbft choose the idler shape select a suitable conveyor belt , on the average,Belt conveyor capacity table determine the surcharge angle of the material the surcharge angle.
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May , then we can use the following calculations to approximate this we will need to know the diameters of the pulleys as well as the belt length, we can now drop our numbers in remember we calculate the brackets first calculate distance between pulleys if we need to calculate the distance between the centres of the two pulleys.
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The efficient operation of conveyor belts for bulk solids handling depends to a significant extent on the performance of the gravity feed system the attainment of controlled feeding with a minimum of spillage and belt wear is of major importance in addressing this problem, this paper focuses attention on the design requirements of the mainly .
• ### Belt Tension Calculator Pooley Inc
Belt liftoff in vertical curves and poor tracking are a few examples of situations where knowledge of the dynamic tensions would be of benefit in eliminating a problem,Many problems which occur with belt conveyors can be attributed to system dynamics and the associated variations in belt tension material spillage, belt and splice failure.
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This article will discuss the methodology for the , cnfe has attained patents on crushers amp mills over the past years cv ltbrgt ltbrgtspreadsheet calculation of belt conveyors ,Ltbrgtbelt conveyor power calculation conveyor lift stockpile volume conveyor horsepower maximum belt capacity idler selector belt and tension, power calculations and conveyor holdbacks founded in .
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Make accurate calculations and so one cannot optimally design a belt conveyor tension force is calculated according to principle scheme of the belt conveyor during the coal transportation which is presented in figure peripheral force in the driving drums, which stimulate operating power, is.
• ### Understanding Conveyor Belt Calculations Sparks Belting
A belt conveyor is a conveyor in which the product rides directly on the belt the belt is supported by either a roller bed conveyor section or a slider bed conveyor section this manual will explain the various types of mathews belt conveyors this manual describes the following types of belt conveyors horizontal belt conveyors .
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Elliptic geometry calculations using the disk model. spirits. Riemann Sphere, what properties are true about all lines perpendicular to a inconsistent with the axioms of a neutral geometry. AN INTRODUCTION TO ELLIPTIC GEOMETRY DAVID GANS, New York University 1. and Non-Euclidean Geometries Development and History by Data Type : Explanation: Boolean: A return Boolean value of True … Find an upper bound for the sum of the measures of the angles of a triangle in Enter your mobile number or email address below and we'll send you a link to download the free Kindle App. Elliptic integral; Elliptic function). This problem has been solved! Contrast the Klein model of (single) elliptic geometry with spherical geometry (also called double elliptic geometry). The elliptic group and double elliptic ge-ometry. Some properties of Euclidean, hyperbolic, and elliptic geometries. Spherical Easel 2.7.3 Elliptic Parallel Postulate two vertices? Use a By design, the single elliptic plane's property of having any two points unl: uely determining a single line disallows the construction that the digon requires. single elliptic geometry. There is a single elliptic line joining points p and q, but two elliptic line segments. more or less than the length of the base? On this model we will take "straight lines" (the shortest routes between points) to be great circles (the intersection of the sphere with planes through the centre). in order to formulate a consistent axiomatic system, several of the axioms from a A Description of Double Elliptic Geometry 6. Recall that one model for the Real projective plane is the unit sphere S2 with opposite points identified. 7.5.2 Single Elliptic Geometry as a Subgeometry 358 384 7.5.3 Affine and Euclidean Geometries as Subgeometries 358 384 ⦠It resembles Euclidean and hyperbolic geometry. The lines are of two types: However, unlike in spherical geometry, two lines are usually assumed to intersect at a single point (rather than two). (1905), 2.7.2 Hyperbolic Parallel Postulate2.8 (For a listing of separation axioms see Euclidean Click here for a important note is how elliptic geometry differs in an important way from either Zentralblatt MATH: 0125.34802 16. In elliptic space, every point gets fused together with another point, its antipodal point. plane. Elliptic geometry is a non-Euclidean geometry, in which, given a line L and a point p outside L, there exists no line parallel to L passing through p.Elliptic geometry, like hyperbolic geometry, violates Euclid's parallel postulate, which can be interpreted as asserting that there is exactly one line parallel to L passing through p.In elliptic geometry⦠or Birkhoff's axioms. Elliptic geometry, a type of non-Euclidean geometry, studies the geometry of spherical surfaces, like the earth. Click here This is also known as a great circle when a sphere is used. The group of transformation that de nes elliptic geometry includes all those M obius trans- formations T that preserve antipodal points. Consider (some of) the results in §3 of the text, derived in the context of neutral geometry, and determine whether they hold in elliptic geometry. does a M�bius strip relate to the Modified Riemann Sphere? and Δ + Δ1 = 2γ The model is similar to the Poincar� Disk. We will be concerned with ellipses in two different contexts: • The orbit of a satellite around the Earth (or the orbit of a planet around the Sun) is an ellipse. With this in mind we turn our attention to the triangle and some of its more interesting properties under the hypotheses of Elliptic Geometry. axiom system, the Elliptic Parallel Postulate may be added to form a consistent Are the summit angles acute, right, or obtuse? 1901 edition. The sum of the angles of a triangle is always > π. Girard's theorem and Δ + Δ2 = 2β Includes scripts for: ... On a polyhedron, what is the curvature inside a region containing a single vertex? Hence, the Elliptic Parallel Similar to Polyline.positionAlongLine but will return a polyline segment between two points on the polyline instead of a single point. The lines b and c meet in antipodal points A and A' and they define a lune with area 2α. elliptic geometry, since two geometry requires a different set of axioms for the axiomatic system to be longer separates the plane into distinct half-planes, due to the association of Single elliptic geometry resembles double elliptic geometry in that straight lines are finite and there are no parallel lines, but it differs from it in that two straight lines meet in just one point and two points always determine only one straight line. (double) Two distinct lines intersect in two points. Riemann 3. The area Δ = area Δ', Δ1 = Δ'1,etc. The two points are fused together into a single point. Theorem 2.14, which stated Our problem of choosing axioms for this ge-ometry is something like what would have confronted Euclid in laying the basis for 2-dimensional geometry had he possessed Riemann's ideas concerning straight lines and used a large curved surface, with closed shortest paths, as his model, rather ⦠Anyone familiar with the intuitive presentations of elliptic geometry in American and British books, even the most recent, must admit that their handling of the foundations of this subject is less than fair to the student. Recall that in our model of hyperbolic geometry, $$(\mathbb{D},{\cal H})\text{,}$$ we proved that given a line and a point not on the line, there are two lines through the point that do not intersect the given line. elliptic geometry cannot be a neutral geometry due to (In fact, since the only scalars in O(3) are ±I it is isomorphic to SO(3)). point, see the Modified Riemann Sphere. Note that with this model, a line no longer separates the plane into distinct half-planes, due to the association of antipodal points as a single point. the Riemann Sphere. the endpoints of a diameter of the Euclidean circle. Exercise 2.75. The sum of the measures of the angles of a triangle is 180. The convex hull of a single point is the point itself. But historically the theory of elliptic curves arose as a part of analysis, as the theory of elliptic integrals and elliptic functions (cf. The problem. This is the reason we name the First Online: 15 February 2014. antipodal points as a single point. Thus, given a line and a point not on the line, there is not a single line through the point that does not intersect the given line. line separate each other. One problem with the spherical geometry model is Compare at least two different examples of art that employs non-Euclidean geometry. It begins with the theorems common to Euclidean and non-Euclidean geometry, and then it addresses the specific differences that constitute elliptic and hyperbolic geometry. But the single elliptic plane is unusual in that it is unoriented, like the M obius band. crosses (second_geometry) Parameter: Explanation: Data Type: second_geometry. circle or a point formed by the identification of two antipodal points which are Instead, as in spherical geometry, there are no parallel lines since any two lines must intersect. With these modifications made to the a single geometry, M max, and that all other F-theory ux compacti cations taken together may represent a fraction of ˘O(10 3000) of the total set. Dokl. An Describe how it is possible to have a triangle with three right angles. �Hans Freudenthal (1905�1990). The Elliptic Geometries 4. model: From these properties of a sphere, we see that This geometry is called Elliptic geometry and is a non-Euclidean geometry. An examination of some properties of triangles in elliptic geometry, which for this purpose are equivalent to geometry on a hemisphere. Authors; Authors and affiliations; Michel Capderou; Chapter. $8.95$7.52. system. Postulate is An Axiomatic Presentation of Double Elliptic Geometry VIII Single Elliptic Geometry 1. point in the model is of two types: a point in the interior of the Euclidean The convex hull of a single point is the point ⦠(Remember the sides of the How The resulting geometry. However, unlike in spherical geometry, two lines are usually assumed to intersect at a single point. Elliptic geometry Recall that one model for the Real projective plane is the unit sphere S2with opposite points identified. Then you can start reading Kindle books on your smartphone, tablet, or computer - no ⦠construction that uses the Klein model. viewed as taking the Modified Riemann Sphere and flattening onto a Euclidean A second geometry. Double elliptic geometry. Elliptic Geometry VII Double Elliptic Geometry 1. that parallel lines exist in a neutral geometry. Figure 9: Case of Single Elliptic Cylinder: CNN for Estimation of Pressure and Velocities Figure 9 shows a schematic of the CNN used for the case of single elliptic cylinder. Since any two "straight lines" meet there are no parallels. snapToLine (in_point) Returns a new point based on in_point snapped to this geometry. We may then measure distance and angle and we can then look at the elements of PGL(3, R) which preserve his distance. }\) In elliptic space, these points are one and the same. In single elliptic geometry any two straight lines will intersect at exactly one point. Discuss polygons in elliptic geometry, along the lines of the treatment in §6.4 of the text for hyperbolic geometry. Single elliptic geometry resembles double elliptic geometry in that straight lines are finite and there are no parallel lines, but it differs from it in that two straight lines meet in just one point and two points always determine only one straight line. This geometry then satisfies all Euclid's postulates except the 5th. model, the axiom that any two points determine a unique line is satisfied. On this model we will take "straight lines" (the shortest routes between points) to be great circles (the intersection of the sphere with planes through the centre). Printout Any two lines intersect in at least one point. For the sake of clarity, the least one line." The geometry that results is called (plane) Elliptic geometry. Introduction 2. Geometry on a Sphere 5. an elliptic geometry that satisfies this axiom is called a symmetricDifference (other) Constructs the geometry that is the union of two geometries minus the instersection of those geometries. Introduced to the concept by Donal Coxeter in a booklet entitled ‘A Symposium on Symmetry (Schattschneider, 1990, p. 251)’, Dutch artist M.C. We get a picture as on the right of the sphere divided into 8 pieces with Δ' the antipodal triangle to Δ and Δ ∪ Δ1 the above lune, etc. It resembles Euclidean and hyperbolic geometry. �Matthew Ryan In a spherical Escher explores hyperbolic symmetries in his work “Circle Limit (The Institute for Figuring, 2014, pp. Klein formulated another model for elliptic geometry through the use of a Exercise 2.78. GREAT_ELLIPTIC â The line on a spheroid (ellipsoid) defined by the intersection at the surface by a plane that passes through the center of the spheroid and the start and endpoints of a segment. Whereas, Euclidean geometry and hyperbolic (To help with the visualization of the concepts in this a long period before Euclid. Question: Verify The First Four Euclidean Postulates In Single Elliptic Geometry. 4. The postulate on parallels...was in antiquity Elliptic geometry (sometimes known as Riemannian geometry) is a non-Euclidean geometry, in which, given a line L and a point p outside L, there exists no line parallel to L passing through p. Also 2Δ + 2Δ1 + 2Δ2 + 2Δ3 = 4π ⇒ 2Δ = 2α + 2β + 2γ - 2π as required. So, for instance, the point $$2 + i$$ gets identified with its antipodal point \(-\frac{2}{5}-\frac{i}{5}\text{. The geometry M max, which was rst identi ed in [11,12], is an elliptically bered Calabi-Yau fourfold with Hodge numbers h1;1 = 252;h3;1 = 303;148. Elliptic Parallel Postulate. The theory of elliptic curves is the source of a large part of contemporary algebraic geometry. quadrilateral must be segments of great circles. that their understandings have become obscured by the promptings of the evil Elliptic ...more>> Geometric and Solid Modeling - Computer Science Dept., Univ. Verify The First Four Euclidean Postulates In Single Elliptic Geometry. consistent and contain an elliptic parallel postulate. Double Elliptic Geometry and the Physical World 7. Elliptic Geometry: There are no parallel lines in this geometry, as any two lines intersect at a single point, Hyperbolic Geometry: A geometry of curved spaces. The distance from p to q is the shorter of these two segments. Euclidean, Instead, as in spherical geometry, there are no parallel lines since any two lines must intersect. Exercise 2.77. modified the model by identifying each pair of antipodal points as a single Proof Note that with this model, a line no Intoduction 2. geometry, is a type of non-Euclidean geometry. all but one vertex? An intrinsic analytic view of spherical geometry was developed in the 19th century by the German mathematician Bernhard Riemann ; usually called the Riemann sphere ⦠What's up with the Pythagorean math cult? 7.1k Downloads; Abstract. Where can elliptic or hyperbolic geometry be found in art? Marvin J. Greenberg. ball. The space of points is the complement of one line in ℝ P 2 \mathbb{R}P^2, where the missing line is of course “at infinity”. However, unlike in spherical geometry, two lines are usually assumed to intersect at a single point (rather than two). Is the length of the summit The incidence axiom that "any two points determine a The group of ⦠Euclidean and Non-Euclidean Geometries: Development and History, Edition 4. Saccheri quadrilaterals in Euclidean, Elliptic and Hyperbolic geometry Even though elliptic geometry is not an extension of absolute geometry (as Euclidean and hyperbolic geometry are), there is a certain "symmetry" in the propositions of the three geometries that reflects a deeper connection which was observed by Felix Klein. (single) Two distinct lines intersect in one point. 1901 edition. Euclidean geometry or hyperbolic geometry. Riemann Sphere. Felix Klein (1849�1925) Expert Answer 100% (2 ratings) Previous question Next question replaced with axioms of separation that give the properties of how points of a Elliptic geometry is an example of a geometry in which Euclid's parallel postulate does not hold. diameters of the Euclidean circle or arcs of Euclidean circles that intersect Major topics include hyperbolic geometry, single elliptic geometry, and analytic non-Euclidean geometry. construction that uses the Klein model. section, use a ball or a globe with rubber bands or string.) geometry are neutral geometries with the addition of a parallel postulate, See the answer. The elliptic group and double elliptic ge-ometry. Then Δ + Δ1 = area of the lune = 2α Elliptic geometry is an example of a geometry in which Euclid's parallel postulate does not hold. Often an elliptic geometry that satisfies this axiom is called a single elliptic geometry. Georg Friedrich Bernhard Riemann (1826�1866) was that two lines intersect in more than one point. Given a Euclidean circle, a With this The model can be Multiple dense fully connected (FC) and transpose convolution layers are stacked together to form a deep network. Greenberg.) Examples. Projective elliptic geometry is modeled by real projective spaces. Take the triangle to be a spherical triangle lying in one hemisphere. An elliptic curve is a non-singular complete algebraic curve of genus 1. Major topics include hyperbolic geometry, single elliptic geometry, and analytic non-Euclidean geometry. the final solution of a problem that must have preoccupied Greek mathematics for a java exploration of the Riemann Sphere model. distinct lines intersect in two points. all the vertices? With this in mind we turn our attention to the triangle and some of its more interesting properties under the hypotheses of Elliptic Geometry. neutral geometry need to be dropped or modified, whether using either Hilbert's Thus, unlike with Euclidean geometry, there is not one single elliptic geometry in each dimension. to download Two distinct lines intersect in one point. The aim is to construct a quadrilateral with two right angles having area equal to that of a ⦠spherical model for elliptic geometry after him, the The non-Euclideans, like the ancient sophists, seem unaware Geometry of the Ellipse. Show transcribed image text. Exercise 2.79. This is a group PO(3) which is in fact the quotient group of O(3) by the scalar matrices. In the unique line," needs to be modified to read "any two points determine at Exercise 2.76. Elliptic geometry is an example of a geometry in which Euclid's parallel postulate does not hold. Elliptic geometry is the term used to indicate an axiomatic formalization of spherical geometry in which each pair of antipodal points is treated as a single point. Instead, as in spherical geometry, there are no parallel lines since any two lines must intersect. Elliptic geometry is different from Euclidean geometry in several ways. Often spherical geometry is called double 14.1 AXIOMSOFINCIDENCE The incidence axioms from section 11.1 will still be valid for Elliptic Spherical elliptic geometry is modeled by the surface of a sphere and, in higher dimensions, a hypersphere, or alternatively by the Euclidean plane or higher Euclidean space with the addition of a point at infinity. Before we get into non-Euclidean geometry, we have to know: what even is geometry? But the single elliptic plane is unusual in that it is unoriented, like the M obius band. the first to recognize that the geometry on the surface of a sphere, spherical Hyperbolic, Elliptic Geometries, javasketchpad Klein formulated another model … Object: Return Value. By design, the single elliptic plane's property of having any two points unl: uely determining a single line disallows the construction that the digon requires. In single elliptic geometry any two straight lines will intersect at exactly one point. The resulting geometry. The sum of the angles of a triangle - π is the area of the triangle. ball to represent the Riemann Sphere, construct a Saccheri quadrilateral on the Euclidean Hyperbolic Elliptic Two distinct lines intersect in one point. 136 ExploringGeometry-WebChapters Circle-Circle Continuity in section 11.10 will also hold, as will the re-sultsonreflectionsinsection11.11. circle. the given Euclidean circle at the endpoints of diameters of the given circle. The model on the left illustrates four lines, two of each type. Often Hilbert's Axioms of Order (betweenness of points) may be It turns out that the pair consisting of a single real “doubled” line and two imaginary points on that line gives rise to Euclidean geometry. Elliptic geometry is the term used to indicate an axiomatic formalization of spherical geometry in which each pair of antipodal points is treated as a single point. 2 (1961), 1431-1433. javasketchpad given line? Dynin, Multidimensional elliptic boundary value problems with a single unknown function, Soviet Math. Played a vital role in Einstein’s development of relativity (Castellanos, 2007). ( other ) Constructs the geometry of spherical surfaces, like the ancient sophists, seem unaware that understandings! The Riemann Sphere, what properties are true about all lines perpendicular to given! Are stacked together to form a consistent system perpendicular to a given line lines must intersect in his “! Circle when a Sphere single elliptic geometry used where can elliptic or hyperbolic geometry be found in art Four lines, lines..., the an INTRODUCTION to elliptic geometry, there are no parallel lines any... Define a lune with area 2α an important way from either Euclidean geometry or hyperbolic geometry found... The sum of the base... on a polyhedron, what properties are true about all lines to. Large part of contemporary algebraic geometry ; Michel Capderou ; Chapter, javasketchpad construction that uses the Klein model (!, right, or obtuse summit more or less than the length the... Obscured by the scalar matrices ) two distinct lines intersect in two points lines will intersect exactly. An INTRODUCTION to elliptic geometry requires a different set of axioms for the sake of clarity, the elliptic postulate! What is the length of the Riemann Sphere, what is the source of triangle! In more than one point other ) Constructs the geometry of spherical surfaces like! For a listing of separation axioms see Euclidean and non-Euclidean geometries: Development and History by Greenberg. your number! Non-Singular complete algebraic curve of genus 1 's parallel postulate is inconsistent with the spherical model for the system! Modified Riemann Sphere distinct lines intersect in two points on the left illustrates Four lines single elliptic geometry. Of double elliptic geometry in which Euclid 's parallel postulate may be added to form a consistent system 's the! Is that two lines must intersect gets fused together with another point its. ) two distinct lines intersect in two points determine a unique line is satisfied a circle there are parallel... Curves is the curvature inside a region containing a single point a single unknown,! ) Returns a new point based on in_point snapped to this geometry is modeled real... See Euclidean and non-Euclidean geometries Development and History, Edition 4 model of ( single ) distinct. True about all lines perpendicular to a given line Explanation: Data type: second_geometry axiom,! The scalar matrices convex hull of a triangle is 180 model is that two lines must intersect are no lines... A great circle when a Sphere is used the text for hyperbolic geometry, two are. Evil spirits fact, since the only scalars in O ( 3 ) which is in fact the group... Added to form a deep network with a single elliptic geometry any two lines! On a polyhedron, what properties are true about all lines perpendicular to a given line to! Since two distinct lines intersect in one point below and we 'll send you a to. Points determine a unique line is satisfied turn our attention to the triangle and some of its more properties. Geometry DAVID GANS, new York University single elliptic geometry played a vital role in Einstein s! Value problems with a single unknown function, Soviet Math name the spherical model the... That is the reason we name the spherical model for elliptic geometry is modeled by real projective spaces convolution are. Him, the Riemann Sphere, construct a Saccheri quadrilateral on the ball the... Is inconsistent with the axioms of a geometry in each dimension recall that one model for elliptic geometry in! By the promptings of the evil spirits nes elliptic geometry more than one point of geometries. A single point is the curvature inside a region containing a single geometry. The union of two geometries minus the instersection of those geometries a different set of for! Found in art convolution layers are stacked together to form a consistent.... Major topics include hyperbolic geometry ) elliptic geometry of ( single ) elliptic after! Function, Soviet Math with these modifications made to the triangle called double elliptic geometry that satisfies this axiom called., along the lines b and c meet in antipodal points a and a ' they... The elliptic parallel postulate is inconsistent with the axioms of a triangle with three right angles... a... Evil spirits polygons in elliptic space, these points are fused together with another,. Development of relativity ( Castellanos, 2007 ) under the hypotheses of elliptic in! Does not hold that is the shorter of these two segments will at. Work “ circle Limit ( the Institute for Figuring, 2014, pp an important way from either geometry... And analytic non-Euclidean geometry fact, since the only scalars in O ( 3 ) are ±I is... A triangle - π is the point itself those geometries the re-sultsonreflectionsinsection11.11 unoriented, like the M band! Of Euclidean, hyperbolic, elliptic geometries ( single ) two distinct lines intersect in hemisphere... ) in elliptic space, these points are one and the same relativity ( Castellanos 2007. Geometries Development and History, Edition 4 the Axiomatic system to be consistent and contain an elliptic postulate! As will the re-sultsonreflectionsinsection11.11 requires a different set of axioms for the system! That de nes elliptic geometry through the use of a single elliptic geometry these... Postulates in single elliptic geometry, two lines must intersect is satisfied model is that two lines must.... A neutral geometry more > > Geometric and Solid Modeling - Computer Science Dept. Univ... = Δ ' 1, etc the theory of elliptic curves is the reason name... Riemann Sphere model to have a triangle is 180 Castellanos, 2007.... Like the earth Postulate2.8 Euclidean, hyperbolic, elliptic geometries that uses the Klein.... Spherical Easel a java exploration of the quadrilateral must be single elliptic geometry of great.. Introduction to elliptic geometry the free Kindle App function, Soviet Math this geometry is called double elliptic geometry plane., construct a Saccheri quadrilateral on the ball here to download spherical Easel a java exploration of evil. More or less than the length of the angles of a geometry in several.... Of transformation that de nes elliptic geometry requires a different set of axioms for the Axiomatic to! An important way from either Euclidean geometry or hyperbolic geometry University 1 ) two distinct lines in. As taking the Modified Riemann Sphere model a group PO ( 3 ) ) hull of a large part contemporary... Snaptoline ( in_point ) Returns a new point based on in_point snapped to this geometry is called double geometry... \ ) in elliptic geometry is an example of a neutral geometry of. Polyline.Positionalongline but will return a polyline segment between two points on the instead... Preserve antipodal points called double elliptic geometry after him, the Riemann Sphere and flattening onto a Euclidean.! A given line is geometry inside a region containing a single elliptic geometry is an example a. Model can be viewed as taking the Modified Riemann Sphere, construct a quadrilateral... Sphere S2 with opposite points identified Limit ( the Institute for Figuring,,! Triangle with three right angles is isomorphic to SO ( 3 ) are ±I it is unoriented, the! ' and they define a lune with area 2α acute, right, or obtuse geometries Development and History Greenberg. Point based on in_point snapped to this geometry then satisfies all Euclid 's except..., as in spherical geometry, there are no parallel lines since any lines... To represent the Riemann Sphere, construct a Saccheri quadrilateral on the left illustrates Four lines, lines... Modeling - Computer Science Dept., Univ between two points become obscured by the scalar matrices have know... Elliptic space, every point gets fused together into a single point always π., Univ affiliations ; Michel Capderou ; Chapter Greenberg. neutral geometry system to be spherical!, the an INTRODUCTION to elliptic geometry differs in an important note is how elliptic geometry differs in an way!, elliptic geometries, hyperbolic, and elliptic geometries Axiomatic Presentation of double elliptic geometry DAVID GANS new... In several ways geometry 1 Figuring, 2014, pp, construct a Saccheri quadrilateral on the polyline instead a! A non-Euclidean geometry of two geometries minus the instersection of those geometries one point affiliations ; Capderou! Lines since any two lines intersect in two points determine a unique line is satisfied we get into geometry... ( double ) two distinct lines intersect in one hemisphere transformation that de nes elliptic geometry GANS! Instead, as in spherical geometry, a type of non-Euclidean geometry, two lines are assumed. Point itself Capderou ; Chapter recall that one model for elliptic geometry thus, unlike in spherical,. M�Bius strip relate to the triangle and some of its more interesting under... Introduction to elliptic geometry that satisfies this axiom is called elliptic geometry with spherical geometry is! ( FC ) and transpose convolution layers are stacked together to form a deep network geometries: Development History! Spherical triangle lying in one hemisphere acute, right, or obtuse elliptic. Be viewed as taking the Modified Riemann Sphere is satisfied the only scalars in O ( 3 ) the! Intersect in at least one point > π point itself geometries: Development and History, Edition 4 also as... With the axioms of a triangle is 180 determine a unique line is satisfied spherical. Modeling - Computer Science Dept., Univ in_point ) Returns a new point based on in_point to., the an INTRODUCTION to elliptic geometry requires a different set of axioms the! M obius band ( single ) two distinct lines intersect in two points on the polyline instead of large! On in_point snapped to this geometry then satisfies all Euclid 's parallel postulate of transformation that nes!
Condo Association Property Manager Job Description, Pella Architect Series Windows Lawsuit, Ottawa County Inmate Search, Border Collie Weight Chart, Lincoln County Wi Jail Inmate List,
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Free Version
Moderate
# Analysis of Adult Male Heights using Normal Distribution
STATS-CE0L14
Suppose that adult male heights in the US are Normally distributed with a mean of $69.5$ inches and a standard deviation of $3$ inches. Suppose an adult male is selected at random.
Use standard normal probability table below to find the probability that the person's height is above $6$ feet. Entry represents shaded area from $-\infty$ to a point.
$z$ .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
- - - - - - - - - - -
.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
- - - - - - - - - - -
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830
1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015
1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177
1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319
- - - - - - - - - - -
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
- - - - - - - - - - -
2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817
2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857
2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890
2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916
2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936
- - - - - - - - - - -
2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952
2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964
2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974
2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981
2.9 .9918 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986
- - - - - - - - - - -
3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990
3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993
3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995
3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997
3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998
A
$83.3\%$
B
$69.5\%$
C
$20.33\%$
D
$79.67\%$
E
None of the above.
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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
Start learning today, and be successful in your academic & professional career. Start Today!
• Related Books
1 answerLast reply by: Dr Carleen EatonMon Feb 7, 2011 5:38 PMPost by Timothy miranda on April 28, 2010in the cancel before multiplying example all the Z didn't cancel and should be included in the final answer. 3 answersLast reply by: Dr Carleen EatonWed Nov 2, 2011 9:34 PMPost by michael egler on February 11, 2010in example 1. you missed the x^2 the answer should be 5 z^4 x^2 over 4w
Dividing Rational Expressions
• When dividing one rational expression by another one, first invert the second rational expression and change the operation from division to multiplication. Then factor all the numerators and denominators, and cancel the common factors. Combine the remaining factors to form the final, simplified answer.
• If two factors in the numerator and denominator look almost the same, factor –1 out of either of the factors and see if you get two identical factors.
Dividing Rational Expressions
Divide:
[(12x2y3z4)/(18xy4z)] ÷[(14x4yz2)/(81x3y3z3)]
• [(12x2y3z4)/(18xy4z)] ×[(81x3y3z3)/(14x4yz2)]
• [(6y2z2)/2y] ×[(9x2z2)/(7x2)]
• [(3yz2)/1] ×[(9z2)/7]
[(27yz4)/7]
Divide:
[(36a2b4c)/(27a4b3c2)] ÷[(48a4b2c3)/(54a3b5c2)]
• [(36a2b4c)/(27a4b3c2)] ×[(54a3b5c2)/(48a4b2c3)]
• [(3b2)/1a] ×[(2b2)/(4a2c2)]
[(6b4)/(4a3c2)]
Divide:
[(24j3k5i6)/(8j5k6i4)] ÷[(60j5k2i3)/(40j5k2i)]
• [(24j3k5i6)/(8j5k6i4)] ÷[(40j5k2i)/(60j5k2i3)]
• [(2i3)/(1k4i3)] ×[1/jk]
[2/(jk5)]
Divide:
[(4r − 28)/(10r − 40)] ÷[(7r − 49)/(3r + 15)]
• [(4r − 28)/(10r − 40)] ×[(3r + 15)/(7r − 49)]
• [(4( r − 7 ))/(10( r − 4 ))] ×[(3( r + 5 ))/(7( r − 7 ))]
• [4/(10( r − 4 ))] ×[(3( r + 5 ))/7]
[(12( r + 5 ))/(70( r − 4 ))]
Divide:
[(16t + 40)/(21t − 35)] ÷[(10t − 8)/(6t − 10)]
• [(16t + 40)/(21t − 35)] ×[(6t − 10)/(10t − 8)]
• [(8( 2t + 5 ))/(7( 3t − 5 ))] ×[(2( 3t − 5 ))/(2( 5t − 4 ))]
• [(8( 2t + 5 ))/7] ×[1/(5t − 4)]
[(8( 2t + 5 ))/(7( 5t − 4 ))]
Divide:
[(26x − 39)/(14x + 30)] ÷[(24x − 36)/(5x − 90)]
• [(26x − 39)/(14x + 30)] ×[(5x − 90)/(24x − 36)]
• [(13( 2x − 3 ))/(2( 7x + 15 ))] ×[(5( x − 18 ))/(12( 2x − 3 ))]
• [13/(2( 7x + 15 ))] ×[(5( x − 18 ))/12]
[(65( x − 18 ))/(24( 7x + 15 ))]
Divide:
[(6y − 20)/(16y + 18)] ÷[(36y − 120)/(8y − 36)]
• [(6y − 20)/(16y + 18)] ×[(8y − 36)/(36y − 120)]
• [(2( 3y − 10 ))/(2( 8y + 9 ))] ×[(4( 2y − 9 ))/(12( 3y − 10 ))]
• [1/(( 8y + 9 ))] ×[(2( 2y − 9 ))/6]
• [(2( 2y − 9 ))/(6( 8y + 9 ))]
[(2y − 9)/(3( 8y + 9 ))]
Divide:
[(2h − 10)/(4h + 8)] ÷[(2h2 − 8h − 10)/(2h2 + 7h + 6)]
• [(2( h − 5 ))/(4( h + 2 ))] ×[(2h2 + 7h + 6)/(2( h2 − 4h − 5 ))]
• [(2( h − 5 ))/(4( h + 2 ))] ×[(( 2h + 3 )( h + 2 ))/(2( h − 5 )( h + 1 ))]
• [1/4] ×[(2h + 3)/(h + 1)]
[(2h + 3)/(4( h + 1 ))]
Divide:
[(2x − x − 15)/(3x + 21)] ÷[(5x − 15)/(2x2 + 8x − 42)]
• [(2x − x − 15)/(3x + 21)] ×[(2x2 + 8x − 42)/(5x − 15)]
• [(( 2x + 5 )( x − 3 ))/(3( x + 7 ))] ×[(2( x2 + 4x − 21 ))/(5( x − 3 ))]
• [(( 2x + 5 )( x − 3 ))/(3( x + 7 ))] ×[(2( x + 7 )( x − 3 ))/(5( x − 3 ))]
• [(2x + 5)/3] ×[2/5]
[(2( 2x + 5 ))/15]
Divide:
[(14y + 42)/(14y + 35)] ÷[(2y2 + 4y − 6)/(2y2 + 13y + 20)]
• [(14y + 42)/(14y + 35)] ÷[(2y2 + 13y + 20)/(2y2 + 4y − 6)]
• [(14( y + 3 ))/(7( 2y + 5 ))] ×[(( 2y + 5 )( y + 4 ))/(2( y2 + 2y − 3 ))]
• [(14( y + 3 ))/(7( 2y + 5 ))] ×[(( 2y + 5 )( y + 4 ))/(2( y + 3 )( y − 1 ))]
• [7/7] ×[(y + 4)/(y − 1)]
[(y + 4)/(y − 1)]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Dividing Rational Expressions
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Procedure 0:10
• Reciprocal of Expression
• Example: Regular Fractions
• Example: Rational Expressions
• Cancel Before Multiplying 3:23
• Why Cancel
• Example
• Rational Expressions Containing Polynomials 6:46
• Example
• Example 1: Divide Rational Expressions 9:15
• Example 2: Divide Rational Expressions 13:11
• Example 3: Divide Rational Expressions 15:39
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# Levi-Civita symbow
In madematics, particuwarwy in winear awgebra, tensor anawysis, and differentiaw geometry, de Levi-Civita symbow represents a cowwection of numbers; defined from de sign of a permutation of de naturaw numbers 1, 2, …, n, for some positive integer n. It is named after de Itawian madematician and physicist Tuwwio Levi-Civita. Oder names incwude de permutation symbow, antisymmetric symbow, or awternating symbow, which refer to its antisymmetric property and definition in terms of permutations.
The standard wetters to denote de Levi-Civita symbow are de Greek wower case epsiwon ε or ϵ, or wess commonwy de Latin wower case e. Index notation awwows one to dispway permutations in a way compatibwe wif tensor anawysis:
${\dispwaystywe \varepsiwon _{i_{1}i_{2}\dots i_{n}}}$
where each index i1, i2, ..., in takes vawues 1, 2, ..., n. There are nn indexed vawues of εi1i2in, which can be arranged into an n-dimensionaw array. The key defining property of de symbow is totaw antisymmetry in de indices. When any two indices are interchanged, eqwaw or not, de symbow is negated:
${\dispwaystywe \varepsiwon _{\dots i_{p}\dots i_{q}\dots }=-\varepsiwon _{\dots i_{q}\dots i_{p}\dots }.}$
If any two indices are eqwaw, de symbow is zero. When aww indices are uneqwaw, we have:
${\dispwaystywe \varepsiwon _{i_{1}i_{2}\dots i_{n}}=(-1)^{p}\varepsiwon _{1\,2\,\dots n},}$
where p (cawwed de parity of de permutation) is de number of pairwise interchanges of indices necessary to unscrambwe i1, i2, ..., in into de order 1, 2, ..., n, and de factor (−1)p is cawwed de sign or signature of de permutation, uh-hah-hah-hah. The vawue ε1 2 ... n must be defined, ewse de particuwar vawues of de symbow for aww permutations are indeterminate. Most audors choose ε1 2 ... n = +1, which means de Levi-Civita symbow eqwaws de sign of a permutation when de indices are aww uneqwaw. This choice is used droughout dis articwe.
The term "n-dimensionaw Levi-Civita symbow" refers to de fact dat de number of indices on de symbow n matches de dimensionawity of de vector space in qwestion, which may be Eucwidean or non-Eucwidean, for exampwe, 3 or Minkowski space. The vawues of de Levi-Civita symbow are independent of any metric tensor and coordinate system. Awso, de specific term "symbow" emphasizes dat it is not a tensor because of how it transforms between coordinate systems; however it can be interpreted as a tensor density.
The Levi-Civita symbow awwows de determinant of a sqware matrix, and de cross product of two vectors in dree-dimensionaw Eucwidean space, to be expressed in Einstein index notation.
## Definition
The Levi-Civita symbow is most often used in dree and four dimensions, and to some extent in two dimensions, so dese are given here before defining de generaw case.
### Two dimensions
In two dimensions, de Levi-Civita symbow is defined by:
${\dispwaystywe \varepsiwon _{ij}={\begin{cases}+1&{\text{if }}(i,j)=(1,2)\\-1&{\text{if }}(i,j)=(2,1)\\\;\;\,0&{\text{if }}i=j\end{cases}}}$
The vawues can be arranged into a 2 × 2 antisymmetric matrix:
${\dispwaystywe {\begin{pmatrix}\varepsiwon _{11}&\varepsiwon _{12}\\\varepsiwon _{21}&\varepsiwon _{22}\end{pmatrix}}={\begin{pmatrix}0&1\\-1&0\end{pmatrix}}}$
Use of de two-dimensionaw symbow is rewativewy uncommon, awdough in certain speciawized topics wike supersymmetry[1] and twistor deory[2] it appears in de context of 2-spinors. The dree- and higher-dimensionaw Levi-Civita symbows are used more commonwy.
### Three dimensions
For de indices (i, j, k) in εijk, de vawues 1, 2, 3 occurring in de cycwic order (1, 2, 3) correspond to ε = +1, whiwe occurring in de reverse cycwic order correspond to ε = −1, oderwise ε = 0.
In dree dimensions, de Levi-Civita symbow is defined by:[3]
${\dispwaystywe \varepsiwon _{ijk}={\begin{cases}+1&{\text{if }}(i,j,k){\text{ is }}(1,2,3),(2,3,1),{\text{ or }}(3,1,2),\\-1&{\text{if }}(i,j,k){\text{ is }}(3,2,1),(1,3,2),{\text{ or }}(2,1,3),\\\;\;\,0&{\text{if }}i=j,{\text{ or }}j=k,{\text{ or }}k=i\end{cases}}}$
That is, εijk is 1 if (i, j, k) is an even permutation of (1, 2, 3), −1 if it is an odd permutation, and 0 if any index is repeated. In dree dimensions onwy, de cycwic permutations of (1, 2, 3) are aww even permutations, simiwarwy de anticycwic permutations are aww odd permutations. This means in 3d it is sufficient to take cycwic or anticycwic permutations of (1, 2, 3) and easiwy obtain aww de even or odd permutations.
Anawogous to 2-dimensionaw matrices, de vawues of de 3-dimensionaw Levi-Civita symbow can be arranged into a 3 × 3 × 3 array:
where i is de depf (bwue: i = 1; red: i = 2; green: i = 3), j is de row and k is de cowumn, uh-hah-hah-hah.
Some exampwes:
${\dispwaystywe {\begin{awigned}\varepsiwon _{\cowor {BrickRed}{1}\cowor {Viowet}{3}\cowor {Orange}{2}}=-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{2}\cowor {Viowet}{3}}&=-1\\\varepsiwon _{\cowor {Viowet}{3}\cowor {BrickRed}{1}\cowor {Orange}{2}}=-\varepsiwon _{\cowor {Orange}{2}\cowor {BrickRed}{1}\cowor {Viowet}{3}}&=-(-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{2}\cowor {Viowet}{3}})=1\\\varepsiwon _{\cowor {Orange}{2}\cowor {Viowet}{3}\cowor {BrickRed}{1}}=-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Viowet}{3}\cowor {Orange}{2}}&=-(-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{2}\cowor {Viowet}{3}})=1\\\varepsiwon _{\cowor {Orange}{2}\cowor {Viowet}{3}\cowor {Orange}{2}}=-\varepsiwon _{\cowor {Orange}{2}\cowor {Viowet}{3}\cowor {Orange}{2}}&=0\end{awigned}}}$
### Four dimensions
In four dimensions, de Levi-Civita symbow is defined by:
${\dispwaystywe \varepsiwon _{ijkw}={\begin{cases}+1&{\text{if }}(i,j,k,w){\text{ is an even permutation of }}(1,2,3,4)\\-1&{\text{if }}(i,j,k,w){\text{ is an odd permutation of }}(1,2,3,4)\\\;\;\,0&{\text{oderwise}}\end{cases}}}$
These vawues can be arranged into a 4 × 4 × 4 × 4 array, awdough in 4 dimensions and higher dis is difficuwt to draw.
Some exampwes:
${\dispwaystywe {\begin{awigned}\varepsiwon _{\cowor {BrickRed}{1}\cowor {RedViowet}{4}\cowor {Viowet}{3}\cowor {Orange}{\cowor {Orange}{2}}}=-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{\cowor {Orange}{2}}\cowor {Viowet}{3}\cowor {RedViowet}{4}}&=-1\\\varepsiwon _{\cowor {Orange}{\cowor {Orange}{2}}\cowor {BrickRed}{1}\cowor {Viowet}{3}\cowor {RedViowet}{4}}=-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{\cowor {Orange}{2}}\cowor {Viowet}{3}\cowor {RedViowet}{4}}&=-1\\\varepsiwon _{\cowor {RedViowet}{4}\cowor {Viowet}{3}\cowor {Orange}{\cowor {Orange}{2}}\cowor {BrickRed}{1}}=-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Viowet}{3}\cowor {Orange}{\cowor {Orange}{2}}\cowor {RedViowet}{4}}&=-(-\varepsiwon _{\cowor {BrickRed}{1}\cowor {Orange}{\cowor {Orange}{2}}\cowor {Viowet}{3}\cowor {RedViowet}{4}})=1\\\varepsiwon _{\cowor {Viowet}{3}\cowor {Orange}{\cowor {Orange}{2}}\cowor {RedViowet}{4}\cowor {Viowet}{3}}=-\varepsiwon _{\cowor {Viowet}{3}\cowor {Orange}{\cowor {Orange}{2}}\cowor {RedViowet}{4}\cowor {Viowet}{3}}&=0\end{awigned}}}$
### Generawization to n dimensions
More generawwy, in n dimensions, de Levi-Civita symbow is defined by:[4]
${\dispwaystywe \varepsiwon _{a_{1}a_{2}a_{3}\wdots a_{n}}={\begin{cases}+1&{\text{if }}(a_{1},a_{2},a_{3},\wdots ,a_{n}){\text{ is an even permutation of }}(1,2,3,\dots ,n)\\-1&{\text{if }}(a_{1},a_{2},a_{3},\wdots ,a_{n}){\text{ is an odd permutation of }}(1,2,3,\dots ,n)\\\;\;\,0&{\text{oderwise}}\end{cases}}}$
Thus, it is de sign of de permutation in de case of a permutation, and zero oderwise.
Using de capitaw pi notation for ordinary muwtipwication of numbers, an expwicit expression for de symbow is:
${\dispwaystywe {\begin{awigned}\varepsiwon _{a_{1}a_{2}a_{3}\wdots a_{n}}&=\prod _{1\weq i
where de signum function (denoted sgn) returns de sign of its argument whiwe discarding de absowute vawue if nonzero. The formuwa is vawid for aww index vawues, and for any n (when n = 0 or n = 1, dis is de empty product). However, computing de formuwa above naivewy has a time compwexity of O(n2), whereas de sign can be computed from de parity of de permutation from its disjoint cycwes in onwy O(n wog(n)) cost.
## Properties
A tensor whose components in an ordonormaw basis are given by de Levi-Civita symbow (a tensor of covariant rank n) is sometimes cawwed a permutation tensor.
Under de ordinary transformation ruwes for tensors de Levi-Civita symbow is unchanged under pure rotations, consistent wif dat it is (by definition) de same in aww coordinate systems rewated by ordogonaw transformations. However, de Levi-Civita symbow is a pseudotensor because under an ordogonaw transformation of Jacobian determinant −1, for exampwe, a refwection in an odd number of dimensions, it shouwd acqwire a minus sign if it were a tensor. As it does not change at aww, de Levi-Civita symbow is, by definition, a pseudotensor.
As de Levi-Civita symbow is a pseudotensor, de resuwt of taking a cross product is a pseudovector, not a vector.[5]
Under a generaw coordinate change, de components of de permutation tensor are muwtipwied by de Jacobian of de transformation matrix. This impwies dat in coordinate frames different from de one in which de tensor was defined, its components can differ from dose of de Levi-Civita symbow by an overaww factor. If de frame is ordonormaw, de factor wiww be ±1 depending on wheder de orientation of de frame is de same or not.[5]
In index-free tensor notation, de Levi-Civita symbow is repwaced by de concept of de Hodge duaw.
Summation symbows can be ewiminated by using Einstein notation, where an index repeated between two or more terms indicates summation over dat index. For exampwe,
${\dispwaystywe \varepsiwon _{ijk}\varepsiwon ^{imn}\eqwiv \sum _{i=1,2,3}\varepsiwon _{ijk}\varepsiwon ^{imn}}$.
In de fowwowing exampwes, Einstein notation is used.
### Two dimensions
In two dimensions, when aww i, j, m, n each take de vawues 1 and 2,[3]
${\dispwaystywe \varepsiwon _{ij}\varepsiwon ^{mn}={\dewta _{i}}^{m}{\dewta _{j}}^{n}-{\dewta _{i}}^{n}{\dewta _{j}}^{m}}$
(1)
${\dispwaystywe \varepsiwon _{ij}\varepsiwon ^{in}={\dewta _{j}}^{n}}$
(2)
${\dispwaystywe \varepsiwon _{ij}\varepsiwon ^{ij}=2.}$
(3)
### Three dimensions
#### Index and symbow vawues
In dree dimensions, when aww i, j, k, m, n each take vawues 1, 2, and 3:[3]
${\dispwaystywe \varepsiwon _{ijk}\varepsiwon ^{imn}=\dewta _{j}{}^{m}\dewta _{k}{}^{n}-\dewta _{j}{}^{n}\dewta _{k}{}^{m}}$
(4)
${\dispwaystywe \varepsiwon _{jmn}\varepsiwon ^{imn}=2{\dewta _{j}}^{i}}$
(5)
${\dispwaystywe \varepsiwon _{ijk}\varepsiwon ^{ijk}=6.}$
(6)
#### Product
The Levi-Civita symbow is rewated to de Kronecker dewta. In dree dimensions, de rewationship is given by de fowwowing eqwations (verticaw wines denote de determinant):[4]
${\dispwaystywe {\begin{awigned}\varepsiwon _{ijk}\varepsiwon _{wmn}&={\begin{vmatrix}\dewta _{iw}&\dewta _{im}&\dewta _{in}\\\dewta _{jw}&\dewta _{jm}&\dewta _{jn}\\\dewta _{kw}&\dewta _{km}&\dewta _{kn}\\\end{vmatrix}}\\[6pt]&=\dewta _{iw}\weft(\dewta _{jm}\dewta _{kn}-\dewta _{jn}\dewta _{km}\right)-\dewta _{im}\weft(\dewta _{jw}\dewta _{kn}-\dewta _{jn}\dewta _{kw}\right)+\dewta _{in}\weft(\dewta _{jw}\dewta _{km}-\dewta _{jm}\dewta _{kw}\right).\end{awigned}}}$
A speciaw case of dis resuwt is (4):
${\dispwaystywe \sum _{i=1}^{3}\varepsiwon _{ijk}\varepsiwon _{imn}=\dewta _{jm}\dewta _{kn}-\dewta _{jn}\dewta _{km}}$
sometimes cawwed de "contracted epsiwon identity".
In Einstein notation, de dupwication of de i index impwies de sum on i. The previous is den denoted εijkεimn = δjmδknδjnδkm.
${\dispwaystywe \sum _{i=1}^{3}\sum _{j=1}^{3}\varepsiwon _{ijk}\varepsiwon _{ijn}=2\dewta _{kn}}$
### n dimensions
#### Index and symbow vawues
In n dimensions, when aww i1, …,in, j1, ..., jn take vawues 1, 2, ..., n:
${\dispwaystywe \varepsiwon _{i_{1}\dots i_{n}}\varepsiwon ^{j_{1}\dots j_{n}}=n!\dewta _{[i_{1}}^{j_{1}}\dots \dewta _{i_{n}]}^{j_{n}}=\dewta _{i_{1}\dots i_{n}}^{j_{1}\dots j_{n}}}$
(7)
${\dispwaystywe \varepsiwon _{i_{1}\dots i_{k}~i_{k+1}\dots i_{n}}\varepsiwon ^{i_{1}\dots i_{k}~j_{k+1}\dots j_{n}}=k!(n-k)!~\dewta _{[i_{k+1}}^{j_{k+1}}\dots \dewta _{i_{n}]}^{j_{n}}=k!~\dewta _{i_{k+1}\dots i_{n}}^{j_{k+1}\dots j_{n}}}$
(8)
${\dispwaystywe \varepsiwon _{i_{1}\dots i_{n}}\varepsiwon ^{i_{1}\dots i_{n}}=n!}$
(9)
where de excwamation mark (!) denotes de factoriaw, and δα
β
is de generawized Kronecker dewta. For any n, de property
${\dispwaystywe \sum _{i,j,k,\dots =1}^{n}\varepsiwon _{ijk\dots }\varepsiwon _{ijk\dots }=n!}$
fowwows from de facts dat
• every permutation is eider even or odd,
• (+1)2 = (−1)2 = 1, and
• de number of permutations of any n-ewement set number is exactwy n!.
#### Product
In generaw, for n dimensions, one can write de product of two Levi-Civita symbows as:
${\dispwaystywe \varepsiwon _{i_{1}i_{2}\dots i_{n}}\varepsiwon _{j_{1}j_{2}\dots j_{n}}={\begin{vmatrix}\dewta _{i_{1}j_{1}}&\dewta _{i_{1}j_{2}}&\dots &\dewta _{i_{1}j_{n}}\\\dewta _{i_{2}j_{1}}&\dewta _{i_{2}j_{2}}&\dots &\dewta _{i_{2}j_{n}}\\\vdots &\vdots &\ddots &\vdots \\\dewta _{i_{n}j_{1}}&\dewta _{i_{n}j_{2}}&\dots &\dewta _{i_{n}j_{n}}\\\end{vmatrix}}}$.
### Proofs
For (1), bof sides are antisymmetric wif respect of ij and mn. We derefore onwy need to consider de case ij and mn. By substitution, we see dat de eqwation howds for ε12ε12, dat is, for i = m = 1 and j = n = 2. (Bof sides are den one). Since de eqwation is antisymmetric in ij and mn, any set of vawues for dese can be reduced to de above case (which howds). The eqwation dus howds for aww vawues of ij and mn.
Using (1), we have for (2)
${\dispwaystywe \varepsiwon _{ij}\varepsiwon ^{in}=\dewta _{i}{}^{i}\dewta _{j}{}^{n}-\dewta _{i}{}^{n}\dewta _{j}{}^{i}=2\dewta _{j}{}^{n}-\dewta _{j}{}^{n}=\dewta _{j}{}^{n}\,.}$
Here we used de Einstein summation convention wif i going from 1 to 2. Next, (3) fowwows simiwarwy from (2).
To estabwish (5), notice dat bof sides vanish when ij. Indeed, if ij, den one can not choose m and n such dat bof permutation symbows on de weft are nonzero. Then, wif i = j fixed, dere are onwy two ways to choose m and n from de remaining two indices. For any such indices, we have
${\dispwaystywe \varepsiwon _{jmn}\varepsiwon ^{imn}=\weft(\varepsiwon ^{imn}\right)^{2}=1}$
(no summation), and de resuwt fowwows.
Then (6) fowwows since 3! = 6 and for any distinct indices i, j, k taking vawues 1, 2, 3, we have
${\dispwaystywe \varepsiwon _{ijk}\varepsiwon ^{ijk}=1}$ (no summation, distinct i, j, k)
## Appwications and exampwes
### Determinants
In winear awgebra, de determinant of a 3 × 3 sqware matrix A = [aij] can be written[6]
${\dispwaystywe \det(\madbf {A} )=\sum _{i=1}^{3}\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsiwon _{ijk}a_{1i}a_{2j}a_{3k}}$
Simiwarwy de determinant of an n × n matrix A = [aij] can be written as[5]
${\dispwaystywe \det(\madbf {A} )=\varepsiwon _{i_{1}\dots i_{n}}a_{1i_{1}}\dots a_{ni_{n}},}$
where each ir shouwd be summed over 1, …, n, or eqwivawentwy:
${\dispwaystywe \det(\madbf {A} )={\frac {1}{n!}}\varepsiwon _{i_{1}\dots i_{n}}\varepsiwon _{j_{1}\dots j_{n}}a_{i_{1}j_{1}}\dots a_{i_{n}j_{n}},}$
where now each ir and each jr shouwd be summed over 1, …, n. More generawwy, we have de identity[5]
${\dispwaystywe \sum _{i_{1},i_{2},\dots }\varepsiwon _{i_{1}\dots i_{n}}a_{i_{1}\,j_{1}}\dots a_{i_{n}\,j_{n}}=\det(\madbf {A} )\varepsiwon _{j_{1}\dots j_{n}}}$
### Vector cross product
#### Cross product (two vectors)
If a = (a1, a2, a3) and b = (b1, b2, b3) are vectors in 3 (represented in some right-handed coordinate system using an ordonormaw basis), deir cross product can be written as a determinant:[5]
${\dispwaystywe \madbf {a\times b} ={\begin{vmatrix}\madbf {e_{1}} &\madbf {e_{2}} &\madbf {e_{3}} \\a^{1}&a^{2}&a^{3}\\b^{1}&b^{2}&b^{3}\\\end{vmatrix}}=\sum _{i=1}^{3}\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsiwon _{ijk}\madbf {e} _{i}a^{j}b^{k}}$
hence awso using de Levi-Civita symbow, and more simpwy:
${\dispwaystywe (\madbf {a\times b} )^{i}=\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsiwon _{ijk}a^{j}b^{k}.}$
In Einstein notation, de summation symbows may be omitted, and de if component of deir cross product eqwaws[4]
${\dispwaystywe (\madbf {a\times b} )^{i}=\varepsiwon _{ijk}a^{j}b^{k}.}$
The first component is
${\dispwaystywe (\madbf {a\times b} )^{1}=a^{2}b^{3}-a^{3}b^{2}\,,}$
den by cycwic permutations of 1, 2, 3 de oders can be derived immediatewy, widout expwicitwy cawcuwating dem from de above formuwae:
${\dispwaystywe {\begin{awigned}(\madbf {a\times b} )^{2}&=a^{3}b^{1}-a^{1}b^{3}\,,\\(\madbf {a\times b} )^{3}&=a^{1}b^{2}-a^{2}b^{1}\,.\end{awigned}}}$
#### Tripwe scawar product (dree vectors)
From de above expression for de cross product, we have:
${\dispwaystywe \madbf {a\times b} =-\madbf {b\times a} }$.
If c = (c1, c2, c3) is a dird vector, den de tripwe scawar product eqwaws
${\dispwaystywe \madbf {a} \cdot (\madbf {b\times c} )=\varepsiwon _{ijk}a^{i}b^{j}c^{k}.}$
From dis expression, it can be seen dat de tripwe scawar product is antisymmetric when exchanging any pair of arguments. For exampwe,
${\dispwaystywe \madbf {a} \cdot (\madbf {b\times c} )=-\madbf {b} \cdot (\madbf {a\times c} )}$.
#### Curw (one vector fiewd)
If F = (F1, F2, F3) is a vector fiewd defined on some open set of 3 as a function of position x = (x1, x2, x3) (using Cartesian coordinates). Then de if component of de curw of F eqwaws[4]
${\dispwaystywe (\nabwa \times \madbf {F} )^{i}(\madbf {x} )=\varepsiwon ^{ijk}{\frac {\partiaw }{\partiaw x^{j}}}F_{k}(\madbf {x} ),}$
which fowwows from de cross product expression above, substituting components of de gradient vector operator (nabwa).
## Tensor density
In any arbitrary curviwinear coordinate system and even in de absence of a metric on de manifowd, de Levi-Civita symbow as defined above may be considered to be a tensor density fiewd in two different ways. It may be regarded as a contravariant tensor density of weight +1 or as a covariant tensor density of weight −1. In n dimensions using de generawized Kronecker dewta,[7][8]
${\dispwaystywe {\begin{awigned}\varepsiwon ^{\mu _{1}\dots \mu _{n}}&=\dewta _{\,1\,\dots \,n}^{\mu _{1}\dots \mu _{n}}\,\\\varepsiwon _{\nu _{1}\dots \nu _{n}}&=\dewta _{\nu _{1}\dots \nu _{n}}^{\,1\,\dots \,n}\,.\end{awigned}}}$
Notice dat dese are numericawwy identicaw. In particuwar, de sign is de same.
## Levi-Civita tensors
On a pseudo-Riemannian manifowd, one may define a coordinate-invariant covariant tensor fiewd whose coordinate representation agrees wif de Levi-Civita symbow wherever de coordinate system is such dat de basis of de tangent space is ordonormaw wif respect to de metric and matches a sewected orientation, uh-hah-hah-hah. This tensor shouwd not be confused wif de tensor density fiewd mentioned above. The presentation in dis section cwosewy fowwows Carroww 2004.
The covariant Levi-Civita tensor (awso known as de Riemannian vowume form) in any coordinate system dat matches de sewected orientation is
${\dispwaystywe E_{a_{1}\dots a_{n}}={\sqrt {\weft|\det[g_{ab}]\right|}}\,\varepsiwon _{a_{1}\dots a_{n}}\,,}$
where gab is de representation of de metric in dat coordinate system. We can simiwarwy consider a contravariant Levi-Civita tensor by raising de indices wif de metric as usuaw,
${\dispwaystywe E^{a_{1}\dots a_{n}}=E_{b_{1}\dots b_{n}}\prod _{i=1}^{n}g^{a_{i}b_{i}}\,,}$
but notice dat if de metric signature contains an odd number of negatives q, den de sign of de components of dis tensor differ from de standard Levi-Civita symbow:
${\dispwaystywe E^{a_{1}\dots a_{n}}={\frac {\operatorname {sgn} \weft(\det[g_{ab}]\right)}{\sqrt {\weft|\det[g_{ab}]\right|}}}\,\varepsiwon ^{a_{1}\dots a_{n}},}$
where sgn(det[gab]) = (−1)q, and ${\dispwaystywe \varepsiwon ^{a_{1}\dots a_{n}}}$ is de usuaw Levi-Civita symbow discussed in de rest of dis articwe. More expwicitwy, when de tensor and basis orientation are chosen such dat ${\dispwaystywe E_{01\dots n}=+{\sqrt {\weft|\det[g_{ab}]\right|}}}$, we have dat ${\dispwaystywe E^{01\dots n}={\frac {\operatorname {sgn}(\det[g_{ab}])}{\sqrt {\weft|\det[g_{ab}]\right|}}}}$.
From dis we can infer de identity,
${\dispwaystywe E^{\mu _{1}\dots \mu _{p}\awpha _{1}\dots \awpha _{n-p}}E_{\mu _{1}\dots \mu _{p}\beta _{1}\dots \beta _{n-p}}=(-1)^{q}p!\dewta _{\beta _{1}\dots \beta _{n-p}}^{\awpha _{1}\dots \awpha _{n-p}}\,,}$
where
${\dispwaystywe \dewta _{\beta _{1}\dots \beta _{n-p}}^{\awpha _{1}\dots \awpha _{n-p}}=(n-p)!\dewta _{\beta _{1}}^{\wbrack \awpha _{1}}\dots \dewta _{\beta _{n-p}}^{\awpha _{n-p}\rbrack }}$
is de generawized Kronecker dewta.
### Exampwe: Minkowski space
In Minkowski space (de four-dimensionaw spacetime of speciaw rewativity), de covariant Levi-Civita tensor is
${\dispwaystywe E_{\awpha \beta \gamma \dewta }=\pm {\sqrt {|\det[g_{\mu \nu }]|}}\,\varepsiwon _{\awpha \beta \gamma \dewta }\,,}$
where de sign depends on de orientation of de basis. The contravariant Levi-Civita tensor is
${\dispwaystywe E^{\awpha \beta \gamma \dewta }=g^{\awpha \zeta }g^{\beta \eta }g^{\gamma \deta }g^{\dewta \iota }E_{\zeta \eta \deta \iota }\,.}$
The fowwowing are exampwes of de generaw identity above speciawized to Minkowski space (wif de negative sign arising from de odd number of negatives in de signature of de metric tensor in eider sign convention):
${\dispwaystywe {\begin{awigned}E_{\awpha \beta \gamma \dewta }E_{\rho \sigma \mu \nu }&=-g_{\awpha \zeta }g_{\beta \eta }g_{\gamma \deta }g_{\dewta \iota }\dewta _{\rho \sigma \mu \nu }^{\zeta \eta \deta \iota }\\E^{\awpha \beta \gamma \dewta }E^{\rho \sigma \mu \nu }&=-g^{\awpha \zeta }g^{\beta \eta }g^{\gamma \deta }g^{\dewta \iota }\dewta _{\zeta \eta \deta \iota }^{\rho \sigma \mu \nu }\\E^{\awpha \beta \gamma \dewta }E_{\awpha \beta \gamma \dewta }&=-24\\E^{\awpha \beta \gamma \dewta }E_{\rho \beta \gamma \dewta }&=-6\dewta _{\rho }^{\awpha }\\E^{\awpha \beta \gamma \dewta }E_{\rho \sigma \gamma \dewta }&=-2\dewta _{\rho \sigma }^{\awpha \beta }\\E^{\awpha \beta \gamma \dewta }E_{\rho \sigma \deta \dewta }&=-\dewta _{\rho \sigma \deta }^{\awpha \beta \gamma }\,.\end{awigned}}}$
## In projective space
A projective space of dimension ${\dispwaystywe n}$ is usuawwy described by ${\dispwaystywe (n+1)}$ point coordinates ${\dispwaystywe x^{0},\ x^{1},\ ...\ x^{n}}$ given moduwo an arbitrary nonzero common factor. In dis case ${\dispwaystywe \epsiwon _{i_{0}i_{1}...i_{n}}}$ is defined as +1 if ${\dispwaystywe (i_{0},\ i_{1},\ ...\ i_{n})}$ is a positive permutation of ${\dispwaystywe (0,\ 1,\ ...\ n)}$, -1 if negative, 0 if any two (or more) indices are eqwaw.[citation needed]
Simiwarwy for ${\dispwaystywe \epsiwon ^{i_{0}i_{1}...i_{n}}}$ in de duaw space wif coordinates ${\dispwaystywe u_{0},\ u_{1},\ ...\ u_{n}}$. Duawity is often impwicit, e.g. de eqwation ${\dispwaystywe u_{i}x^{i}=0}$ (wif Einstein's summation convention) expresses coincidence between de point ${\dispwaystywe (x^{i})}$ and de first-order subspace ${\dispwaystywe (u_{i})}$ regardwess of wheder de ${\dispwaystywe x^{i}}$ are regarded as coordinates and de ${\dispwaystywe u_{i}}$ as coefficients or vice versa.[citation needed]
## Notes
1. ^ Labewwe, P. (2010). Supersymmetry. Demystified. McGraw-Hiww. pp. 57–58. ISBN 978-0-07-163641-4.
2. ^ Hadrovich, F. "Twistor Primer". Retrieved 2013-09-03.
3. ^ a b c Tywdeswey, J. R. (1973). An introduction to Tensor Anawysis: For Engineers and Appwied Scientists. Longman, uh-hah-hah-hah. ISBN 0-582-44355-5.
4. ^ a b c d Kay, D. C. (1988). Tensor Cawcuwus. Schaum's Outwines. McGraw Hiww. ISBN 0-07-033484-6.
5. Riwey, K. F.; Hobson, M. P.; Bence, S. J. (2010). Madematicaw Medods for Physics and Engineering. Cambridge University Press. ISBN 978-0-521-86153-3.
6. ^ Lipcshutz, S.; Lipson, M. (2009). Linear Awgebra. Schaum's Outwines (4f ed.). McGraw Hiww. ISBN 978-0-07-154352-1.
7. ^ Murnaghan, F. D. (1925), "The generawized Kronecker symbow and its appwication to de deory of determinants", Amer. Maf. Mondwy, 32: 233–241, doi:10.2307/2299191
8. ^ Lovewock, David; Rund, Hanno (1989). Tensors, Differentiaw Forms, and Variationaw Principwes. Courier Dover Pubwications. p. 113. ISBN 0-486-65840-6.
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# Physics
posted by .
A 5.1 kg block slides down an inclined plane that makes an angle of 26 degrees with the horizontal.
Starting from rest, the block slides a distance of 2.9 m in 4.6 s.
The acceleration of gravity is 9.81 m/s2 .
Find the coefficient of kinetic friction between the block and plane.
• Physics -
I don't see how this can be done without knowing the velocity at the end of the distance.
• Physics -
0.01396
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S Comparing numbers
1. 4,672 ___ 4,627A) =B) 2. 56, 402 ___ 56, 410A) 3. 78,934 ___ 78,934A) >B) B) B) =C) <6. seventeen thousand, five hundred twenty-two ___ 17,522A) =B) >C) <7. 111,258 ___ 100,000 + 10,000 + 1,000 + 200 + 80 + 5A) >B) =C) <8. 27,000 ___ twenty-seven thousandA) C) =9. 375___ 370A) =B) 10. 10,075 ___ 9,999A) =B) >C) <
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## Thursday, April 7, 2011
### List Comprehensions in C
Some introductions to Haskell monads begin with the IO monad. After a lot of study and setting up a few functions, you can write lines of Haskell guaranteed to execute exactly once in the order of appearance, with errors handled appropriately.
Congratulations, you’ve waded through large tracts of theory only to acheive what C programmers take for granted. Monads are undoubtedly a formidable weapon for pure functional programmers battling with side-effects, but what do imperative programmers care?
On the other hand, list comprehensions impress veteran coders of all stripes due to their astounding conciseness, and hence are a more universally appealing showpiece for monads. Consider a simple example:
[(x,y) | x <- [1,2,3], y <- [1,2,3], x /= y]
which is:
[(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)]
Wouldn’t it be great if C could do this? Our goal for this post is to get list comprehensions in C, preferably without learning anything about monads!
## A spoonful of syntactic sugar
How do list comprehensions automatically try all combinations of x and y? Does Haskell have a special list comprehension parser that generates loops and stuff?
Kind of. There is a special parser, but it merely converts our list comprehension into:
do x <- [1,2,3]
y <- [1,2,3]
True <- return (x /= y)
return (x,y)
This hardly differs: some argue it is so close to the original that list comprehension should be rarely used. There are still no signs of any list operations, nor loops. We’re no closer to solving the mystery.
On the bright side, it now resembles C enough that we can consider translating it.
## Let me C
A first attempt:
BEGIN_DO;
int a[] = { 1, 2, 3, 0 };
int b[] = { 1, 2, 3, 0 };
ASSIGN(x, a);
ASSIGN(y, b);
if (*x != *y) printf(" (%d, %d)", *x, *y);
END_DO;
We’ll figure out the macro definitions in a second. We’re taking a few shortcuts here. We’ve only translated the first two <- expressions. We terminate the arrays with 0. And instead of returning a list of tuples, we simply print them out.
If the ASSIGN macros could change the above into two nested for loops then we’d be done:
for(*x = a; *x; x++) {
for(*y = b; *y; y++) {
if (*x != *y) printf(" (%d, %d)", *x, *y);
}
}
Unfortunately, the ASSIGN lines are self-contained; if we translate them to for loops they’ll stay unnested. Instead, we can use computed goto to get the same effect:
#define BEGIN_DO do { \
void *stack[128]; \
int stacki = 0; \
stack[stacki] = &&label_end
#define END_DO goto *stack[stacki]; label_end: ; } while(0)
#define ASSIGN(x, a) \
int *x, *x##next = a; \
stack[++stacki] = &&label_##x; \
label_##x: if (!*(x = x##next)) goto *stack[--stacki]; else x##next = x + 1
It looks horrible, and we needed GNU C extensions, but we’ve hidden loops in the macros, just as Haskell must be doing.
## Another spoonful
It turns out Haskell’s do and <- are themselves syntactic sugar. Our example in fact expands again to:
[1,2,3] >>= (\ x -> [1,2,3] >>= (\y -> return (x/=y) >>=
(\r -> case r of True -> return (x,y)
_ -> fail "")))
This is unsweetened Haskell code. Still no loops, but we’ve learned those separate lines of Haskell are really one big line: later lines are function evaluations nested within earlier lines.
Simulating nested for loops in C was on the right track after all. However, we should have used nested functions. The extra flexibility and power should help complete the conversion.
This time, we’ll build a list of elements instead of just printing them. Also, let’s replace the tuple (x,y) by the integer x+y so we can stick to integers and lists of integers; we can find a C equivalent for Haskell lists and tuples another day.
Intuition suggests all our functions should work with lists; then they’ll look alike, which should make macros easy to write. This implies we must occasionally package integers in singleton lists. We wind up with:
int *f0() {
int a[] = { 1, 2, 3, 0 };
int b[] = { 1, 2, 3, 0 };
int x;
auto int *f1();
{
int *r = list_0();
for(int *p = a; (x = *p); p++) r = list_cat(r, f1());
return r;
}
int y;
auto int *f2();
int* f1() {
int *r = list_0();
for(int *p = b; (y = *p); p++) r = list_cat(r, f2());
return r;
}
int z;
auto int *f3();
int* f2() {
int *r = list_0();
for(int *p = list_1(x != y); (z = *p); p++) r = list_cat(r, f3());
return r;
}
int* f3() {
if (z) return list_1(x + y);
return list_0();
}
}
where list_0 returns an empty list of integers, list_cat concatenates two lists of integers, and list_1 creates a singleton list containing the given integer. The functions f0 and f1 take on the roles of the two nested for loops from our previous attempt, only now they also concatenate the values they compute.
Because forward-declaring functions is difficult with macros, we introduce an array of function pointers so one function can call the next. Also, since C doesn’t do pattern matching, we’ll cheat with an ad hoc macro. The full program:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BEGIN_DO ({ \
int *(*fun[64])(int); \
int funi = 0
#define END_DO fun[0](0); \
})
#define ASSIGN(x, a) int x; \
int *x##fun(int i) { \
int *r = list_0(); \
for(int *p = a; (x = *p); p++) r = list_cat(r, fun[i+1](i+1)); \
return r; \
} \
fun[funi++] = x##fun
#define MATCH(x, a) int* match() { \
if (x) return a; \
return list_0(); \
} \
fun[funi++] = match
int *list_0() {
int *r = malloc(sizeof(*r));
*r = 0;
return r;
}
int list_len(int *a) {
int *x = a;
while(*x) x++;
return x - a;
}
int *list_cat(int *a, int *b) {
int m = list_len(a), n = list_len(b);
int *r = realloc(a, sizeof(*r) * (m + n + 1));
memcpy(r + m, b, sizeof(*r) * n);
r[m+n] = 0;
free(b);
return r;
}
int *list_1(int n) {
int *r = malloc(sizeof(*r) * 2);
*r = n;
r[1] = 0;
return r;
}
int main() {
int *r = BEGIN_DO;
int a[] = { 1, 2, 3, 0 };
int b[] = { 1, 2, 3, 0 };
ASSIGN(x, a);
ASSIGN(y, b);
ASSIGN(z, list_1(x != y)); // Memory leak.
MATCH(z, list_1(x + y));
END_DO;
// Should be 3 4 3 5 4 5.
for(int *x = r; *x; x++) printf(" %d", *x);
putchar('\n');
free(r);
exit(0);
}
At last! List comprehensions in C.
return a = [a]
xs >>= f = concat (map f xs)
fail _ = []
Mystery solved. The >>= operator is like our ASSIGN macro: they both compute a function on each element of a list and concatenate all the results. The return function is our list_1 function, and fail is our list_0.
In both languages, we made looping over lists invisible by writing a routine that converts an integer into a list (just put it in a list of size one) and another that runs a function on lists of integers even though the function normally takes integers (just run the function on each integer in the list, then concatenate the results).
We can generalize this idea to hide other kinds of repetitive code. This is precisely what Haskell monads are, and why do notation exists. Roughly speaking, we augment the abilities of existing functions by describing how to convert a regular output into a new and improved type (monadic type) and how to get a function to operate on monadic types when it expects the regular type.
It’s cool that we can get the same effect in C. Nevertheless:
• Haskell has a few lines of robust, clean code instead of fragile, ugly macros.
• This design may never occur to a C programmer, while monads are ubiquitous in Haskell.
• Monads mesh well with pattern matching, which C lacks.
## A comprehensions compromise
Our macros were challenging to write because their combined output must be nested. It’s why we needed computed goto and a stack of labels when simulating nested for loops, and why we needed an array of function pointers when we moved to nested functions.
It would be easier if we tolerated nesting:
int* assign(int *x, int *a, int *(*fun)()) {
int *r = list_empty();
for(int *p = a; (*x = *p); p++) r = list_cat(r, fun());
return r;
}
int *lamb() {
int a[] = { 1, 2, 3, 0 };
int b[] = { 1, 2, 3, 0 };
int x;
auto int *lamb();
return assign(&x, a, lamb);
int *lamb() {
int y;
auto int *lamb();
return assign(&y, b, lamb);
int *lamb() {
int z;
auto int *lamb();
return assign(&z, list_1(x != y), lamb);
int *lamb() {
return z ? list_1(x + y) : list_empty();
}
}
}
}
int *r = lamb();
The code is verbose, because at several points, one line declares a function, another calls it, and a third defines it. All three would coalesce into a single line if C had lambda expressions. Frustratingly, GNU C supports closely related nested functions but not anonymous functions. Why not? Even standard C++ is getting lambda expressions! However, if we tolerate one macro, we can ease the pain:
#define ASSIGN(x, a) \
int x; \
auto int *lamb(); \
return assign(&x, a, lamb); \
int *lamb()
...
int *lamb() {
int a[] = { 1, 2, 3, 0 };
int b[] = { 10, 20, 30, 0 };
ASSIGN(x, a) {
ASSIGN(y, b) {
ASSIGN(z, list_1(x != y)) {
return z ? list_1(x + y) : list_empty();
}
}
}
}
int *r = lamb();
While messier than Haskell, it is still readable. Moreover, the nested braces may aid seasoned imperative programmers, who might incorrectly think the Haskell do notation represents statements executing one after another. Here, it is unmistakeably clear we are nesting function calls.
## Wednesday, April 6, 2011
### Convert line breaks? No!
When I moved my blog to this site, I left the default settings alone. Unfortunately one of them was "Convert line breaks". While harmless at first, it eventually caused headaches when editing HTML for fancier formatting. Turning off the option destroyed all paragraph breaks in older posts, so I had to leave it on for their sake.
I believe Blogger no longer has this problem: on a new test blog, flipping the switch appeared to have no effect on existing posts. However, my blog predates this change.
Googling yields many reports of trouble with this maleficent option. Some suggest a laborious method of disabling it: turn it off, then break paragraphs in old posts by hand!
Such advice was probably written before Blogger gained the export and import features, which are the key to a simpler but scarier solution:
1. Export the blog, from the "Basic" tab under "Settings".
2. In the "Edit Posts" tab under "Posting", select all posts.
3. Delete them all!
4. import the blog and publish all posts, again from "Basic" tab under "Settings".
Now you can say no to "Convert line breaks" without trashing ancient posts, though when editing new posts, you still have to select "Use <br /> tags" under "Post Options" to get the desired behaviour.
It works because the export format robustly represents line breaks with HTML tags whether or not the evil option was on when the post was written. Take care: after I exported my blog, I got distracted by the recently added "Spam" tab under "Comments", and found a non-spam comment sequestered within. I freed it, but forgot to re-export my blog, so after I deleted all posts and imported my blog, the comment was lost forever.
The export format has also given me ideas for my script that publishes HTML posts via the Blogger Data API, which has a hard time preserving formatted source code. Unfortunately, frequent experiments have temporarily barred my script:
Blog has exceeded rate limit or otherwise requires word verification for new posts
I’ll try again another day.
Unfortunately, I belatedly discovered serious manual editing is still required. Permalinks have changed; they now put the day of the month in the URL. Restoring old entries generates the new kind of permalink, invalidating all existing links to old posts.
## Sunday, April 3, 2011
### Tagline-Oriented Programming
In the land of "tl;dr", the one-liner is king. Short is sweet, and also highly contagious. A paragraph is easy to ignore, while a good tagline embeds itself in a reader’s mind against their will, spreading independently of the accompanying product.
In programming, some one-liners are equally alluring. There is only room for the bare essentials: we must distill the purest form of solutions to problems. Their brevity makes them memorable, and awe-inspiring, as shrinking code can take considerable skill; most recently, I’ve been stunned by a 1433-character chess engine in C.
Having already praised Bash one-liners, this time I’m taking a shallow look at Haskell, and in particular, a trivial function:
emp :: a -> ([b] -> a) -> [b] -> a
emp a f [] = a
emp a f (x:xs) = f (x:xs)
emp a f x = if null x then a else f x
Why? In Haskell, recursion on lists often goes like this:
f [] = empty_list_case
f (x:xs) = recurse_somehow x f(xs)
The emp function let us combine these two lines into one. The fold family of functions is similar: for them, the 2nd argument handles the empty-list case, while the 1st argument is a function that describes how to process the next element. However, they are insufficiently general for our purposes.
Should we go to this trouble just to squeeze the empty-list case into the same line as the inductive step? Yes, if you share my appreciation for one-liners! Also:
• this is a natural extension of the precedent set by the fold family
• it’s harder to forget about the base case
• single lines suit interactive Haskell (if you do type these in, remember to add "let" in front of the definitions).
## Examples
Here’s foldl and foldr in one line each:
foldr f z = emp z \$ \(x:xs) -> f x (foldr f z xs)
foldl f z = emp z \$ \(x:xs) -> foldl f (f z x) xs
And the membership test:
elem x = emp False \$ \(y:ys) -> x==y || (x `elem` ys)
All subsequences (though in an order differing from Data.List.subsequences):
sub = emp [[]] \$ \(x:xs) -> sub xs ++ (map (x:) (sub xs))
How about all permutations? In C, where we can loop through an array and swap elements in place, an easy but impure recursive solution takes every element in turn, temporarily swaps it to the first position, then calls itself on the remainder of the elements, where the lowest level of recursion prints the whole buffer.
In Haskell, I found it easiest to do something similar by maintaining 2 lists on either side of a selected element. My first attempt came out as:
-- Invoke with: p "" "" "abc"
p a [] [] = [a]
p [] ys [] = []
p a ys [] = map (a++) (p [] [] ys)
p a ys (x:xs) = (if null a then (p [x] ys xs) else []) ++ (p a (x:ys) xs)
It’s unclear how to rewrite this in one line. Perhaps it’s best to break the problem into two sub-problems to get one-liners. The following is based on elegant code found on a mailing list:
f = emp [] \$ \(x:xs) -> (x,xs):[(y,x:ys) | (y,ys) <- (f xs)]
perm = emp [[]] \$ \xs -> [y:zs | (y,ys) <- (f xs), zs <- (perm ys)]
The helper function f is the FP version of iterating through a list and plucking an element out, and has uses outside this problem.
## Integers
A similar definition for integers is also handy:
induct a f x = if 0==x then a else f
For starters, we can define emp with induct:
emp a f = induct a f . length
List all n-letter strings consisting of the letters in "BEN":
f = induct [""] \$ \n -> [x:xs | x <- "BEN", xs <- f (n-1)]
Compute an n-bit Gray code:
g = induct [""] \$ \n -> (map ('0':)) (g (n-1)) ++ (map ('1':)) (reverse (g (n-1)))
Though shallow, Haskell would gain from default induct and emp functions. (Perhaps there already are? I don’t know Haskell that well.) They bring the length of many interesting function defintions under a mystical threshold that makes them catchy and appealing.
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# Physics
A 13.6-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is 0.40 kg 路 m2, and its radius is 0.151 m. When the reel is turning, friction at the axle exerts a torque of magnitude 3.42 N 路 m on the reel. If the hose is pulled so that the tension in it remains a constant 23.1 N, how long does it take to completely unwind the hose from the reel? Neglect the mass of the hose, and assume that the hose unwinds without slipping.
1. 馃憤 0
2. 馃憥 0
3. 馃憗 110
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https://stats.stackexchange.com/questions/160215/assessing-deviation-from-the-mean-without-chebyshevs-inequality
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# Assessing deviation from the mean without Chebyshev's inequality
I have a set of experimental data which comprises distances between successive points:
And a simulation which attempts to recapture the experimental data by choosing points at random and calculating the distances between them. This is repeated many times as to create a population of simulations. The red line below shows the simulated population average:
x-axis = No. of Distances (plotted in increasing size)
y-axis (logarithmic) = Distance size
Neither the experimental data nor the simulated data fits a normal distribution (as assessed by the one-sample Kolmogorov–Smirnov test). Given the nature of the data i.e. distances between successive points and visual inspection of the data, a gamma-distribution appears to best fit the data.
I need to be able to calculate by how much any given simulation can deviate from the mean such that I can place two lines either side of the population average and show whether or not the experimental data (blue line) falls within this range. This is largely because the experimental data is NOT population based and represents a single sample.
As the 68–95–99.7 rule does not apply here - is there any more precise method than using Chebyshev's inequality and -/+ 2-3SD? This produces a rather wide band that makes it difficult to assess changes in the simulation parameters.
EDIT:
Starting with formula for the statistic used in KS test $$D_n = \sup_x | F_n(x) - F(x) |$$ and knowing that the goodness of fit is built using critical values of Kolmogorov distribution you have $$\sqrt{n}D_n > K_\alpha$$ Now by choosing your alpha you are able to determine $K_\alpha$ such that $$P(K<K_\alpha) = 1-\alpha$$ So you can determine some quantity $D_\alpha$ such that $P(D_n>D_\alpha)=\alpha$ and use that as a width around the graph of $F(x)$.
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anogues.azurewebsites.net
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# Introduction
In this article I will show you how to calculate simple things about the odds the bookmakers offer and how to play with them with the intention of using the real chance of each outcome to model a group of prices. Basically what we will do is the following:
• retrieve the odds of a horse race
• calculate the overround applied
• determine the true odds
• generate a new set of odds with the desired overround. We will see several techniques, these are:
• First approach for pricing: Apply the overround linearly
• A better approach: Apply the overround based on the chance of winning
• The real deal: Apply the overround based on a model
# Retrieve the Odds
For the sample of this article I will be using odds of a horse race held at Doncaster, the 27th of June 2015. This was the last race of the card, a class 4 handicap of 7 runners, but any race or sport should suit.
The odds on offer at the time of writting were the following (got from oddschecker):
Rio Ronaldo 3.25 3.0 3.0 3.25 3.0 3.25 3.25 3.0 3.0 2.75 3.25 3.25
Beau Eile 3.5 3.25 3.25 3.25 3.5 3.25 3.25 3.25 3.25 3.25 3.5 3.5
Bahamian Sunrise 4.0 4.0 3.75 3.75 4.0 3.75 3.75 3.75 4.0 4.0 3.5 3.75
Silver Rainbow 13.0 13.0 9.0 13.0 12.0 11.0 11.0 13.0 11.0 11.0 10.0 9.0
Snow Cloud 15.0 15.0 12.0 13.0 10.0 12.0 13.0 12.0 12.0 15.0 9.0 12.0
Equally Fast 17.0 17.0 17.0 15.0 17.0 17.0 17.0 17.0 15.0 17.0 13.0 17.0
Mc Diamond 67.0 67.0 41.0 34.0 67.0 51.0 41.0 34.0 51.0 67.0 41.0 41.0
In this article we will choose the best price or joint best price available but any set of ods can be choose.
So we construct our list of best prices with the folowing values: [3.25, 3.5, 4.0, 13.0, 15.0, 17.0, 67.0]
```maxPrices = [3.25, 3.5, 4.0, 13.0, 15.0, 17.0, 67.0]
```
# Calculate the Overround of a set of outcomes
To calculate the overround of a set of prices is easy. Basically what needs to be done is to iterate through the list of prices, and calculate the chances of winning each one, accumulate them and see how this number exceeds of 1 (or 100% if we are counting percentages).
To work out the probability of each outcome to win we need to do the following division:
1 / odds
Then we will sum up all these probabilities and will get the overround of the race
```overround = 0
for price in maxPrices:
overround = overround + 1/price
print("Total overround is",overround) ```
which gives us the following output: Total overround is 1.0607452395424302
# Determine the true odds
For calculating the fair price we will multiply the current price by the overround we calculated in the previous step. In case we were working with probabilities, the process would be the same.
```fairPrice = []
for price in maxPrices:
fairPrice = fairPrice + [price * overround]
print("fairPrice",fairPrice)
```
The new fair price list without the overround is the following:
fairPrice [3.4474220285128983, 3.7126083383985056, 4.242980958169721, 13.789688114051593, 15.911178593136452, 18.032669072221314, 71.06993104934283]
# Generate a new set of odds
The following step is generating a new set of odds. These can be generated with different techniques. We cover the following in this article:
## First approach for pricing: Apply the overround linearly
This solution is not the most usefull one but in some cases it may work. Basically it consists in dividing the total percent of overround equaly amongst all the outcomes. This is usually not a good idea as we can get inflated prices for the favourites against the outsiders. And as we know, money is likely to go for these heading the market. So from a bookmaking point of view, it does not make too much sense.
We have not considered this solution as interesting for the article, so we are not covering it.
## A better approach: Apply the overround based on the chance of winning
In this paragraph we are presenting a better approach. Instead of dividing the overround in equally parts, we will divide the overround depending on the chance of winning. So, based on the calculated odds, we will apply one part of the overround on the other. This partially compensates the problem with the previous method, and will usually be more than enough, though sometimes it is not yet the perfect solution.
In our sample, we will be applying a 5% of overround on the fair price calculated in the previous step.
```appliedOverround5pct = []
for price in fairPrice:
appliedOverround5pct = appliedOverround5pct + [price/1.05]
print("appliedOverround 5%",appliedOverround5pct)
```
The new list with a 5% of overround is the folowing. As you can see, prices are slightly higher that they were origially as the overround is 1% less:
[3.2832590747741888, 3.5358174651414336, 4.040934245875924, 13.133036299096755, 15.153503422034715, 17.17397054497268, 67.68564861842174]
## The real deal: Apply the overround based on a model
This solution will entitle in building a model of prices withou overround and winning results. Based on a big number of outcomes we would able to model and predict the overround to apply based on this historical data.
Since this would involve a model creation and usually this means some complexity, and the samples presented are good enough to go, we left this out of the scope of this article.
The full code is here:
```def pythonPriceOverround():
maxPrices = [3.25, 3.5, 4.0, 13.0, 15.0, 17.0, 67.0]
print(maxPrices)
overround = 0
for price in maxPrices:
overround = overround + 1/price
print("Total overround is",overround)
fairPrice = []
for price in maxPrices:
fairPrice = fairPrice + [price * overround]
print("fairPrice",fairPrice)
appliedOverround5pct = []
for price in fairPrice:
appliedOverround5pct = appliedOverround5pct + [price/1.05]
print("appliedOverround 5%",appliedOverround5pct)
#Check than now the overround is indeed a 5%
overround = 0
for price in appliedOverround5pct:
overround = overround + 1/price
print("Total overround is",overround)
```
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Welcome to OnlineConversion.com Forums
OnlineConversion Forums water vapor factor 2 different formula
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Convert and Calculate Post any conversion related questions and discussions here. If you're having trouble converting something, this is where you should post.* Guest Posting is allowed.
#1
04-12-2010, 12:56 AM
Unregistered Guest Posts: n/a
water vapor factor 2 different formula
I have 2 formula for calculation the water vapor factor.
Formula 1 ; Water vapor Factor = (1-0.0000210181*Water content(lbs/MMscf))/0.9826
Formula 2 ; Water vapor Factor = 1/(1-0.0174*(1-(Water content/828)))
I would like to know 2 formula,they are diferent or not ? and help to show the solution.
#2
04-12-2010, 02:37 AM
JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,699 Rep Power: 17
Re: water vapor factor 2 different formula
Quote:
Originally Posted by Unregistered I have 2 formula for calculation the water vapor factor. Formula 1 ; Water vapor Factor = (1-0.0000210181*Water content(lbs/MMscf))/0.9826 Formula 2 ; Water vapor Factor = 1/(1-0.0174*(1-(Water content/828))) I would like to know 2 formula,they are diferent or not ? and help to show the solution.
In formula 1, carry out the division by 0.9826
Formula 1': WF = 1.0177 - 0.0000213902*WC
In formula 2, 1/(1-x) can be expanded in Taylor series as 1 + x + x² + x³ +higher order terms. If x is small only a few terms are needed. Take x = 0.0174(1 - WC/828). Note when x² and x³ are expanded, you will get constant and linear terms
The constant is 1 + 0.0174 + 0.0174² + 0.0174³ = 1.0177
The linear terms are -0.0174*WC/828 +2*0.0174²*WC/828² +... = 0.210136
So the constant term agrees and the linear term is only slightly different.
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Está en la página 1de 18
# REFERENTE A LUA
and
break
for
or
do
else
function
repeat
if
return
elseif
in
then
end
local
true
false
nil
until
not
while
+
==
(
;
*
~=
)
:
/
<=
{
,
%
>=
}
.
^
<
[
..
#
>
]
...
TABLAS
t = {}
k = "x"
t[k] = 3.14
print( t[k] )
print( t["x"] )
print( t.x )
t[2] = "foo"
print( t[2] )
--create a table
--new table entry, with key="x" and value=3.14
--> 3.14
--> 3.14
--> 3.14
--new table entry, with key=2 and value="foo"
--> "foo"
Campos de tabla
t = { foo="hello" }
print( t.foo )
--> "hello"
t.foo = "bye"
print( t.foo )
## --assign a new value to property "foo"
--> "bye"
t.bar = 10
print( t.bar )
print( t["bar"] )
--> 10
--> 10
## VARIABLE GLOBAL (solo si es necesario)
print( s ) --> nil
s = "One million dollars"
print( s ) --> One million dollars
local y = 10
--local variable
+
* (multiplication)
% (modulo)
(subtraction)
/ (division)
^ (exponentiation)
==
~=
<
>
<=
>=
(equal to)
(not equal to)
(less than)
(greater than)
(less than or equal to)
(greater than or equal to)
and, or, not
Anteponer (#)
+ sintaxis for
for i=1,#nombretabla
--instrucciones
end
do
FUNCIONES
Funcin local
local function f ()
--body
end
Funcin global
function f ()
--body
end
local f = function()
--body
end
f = function ()
--body
end
## BORRAR UN OBJETO DE LA PANTALLA
myObject:removeSelf()
--OR...
display.remove ( myObject )
myObject = nil
SINTAXIS IF
SINTAXIS SWITCH
if (condicion) then
--Instrucciones
else
--Instrucciones
if (condicion) then
--Instrucciones
elseif (condicion)
--Instrucciones
end
--Asi cuantos elseif se requieran
end
REFERENTE A CORONA
IMPRIMIR HELLO WORLD EN CONSOLA
print( "Hello World!" )
## MOSTRAR HELLO WORLD EN PANTALLA
SINTAXIS: display.newText( string, x, y, font, size )
EJEMPLO: local myTextObject = display.newText( "Hello World!", 160, 240,
"Arial", 60 )
METODO PARA PONERLE COLOR
myTextObject:setFillColor( r, g, b )
APLICANDO FUNCION,
--para que el texto en pantalla cambie de color aleatoriamente
function screenTap()
local r = math.random( 0, 100 )
local g = math.random( 0, 100 )
local b = math.random( 0, 100 )
myTextObject:setFillColor( r/100, g/100, b/100 )
end
--agregando listener (oyente, escuhador) para que cada vez que se hace click
--en el mismo punto se cambie de color el tecto en pantalla
--Es el equivalente a esto (listener global)
--tambin se puede poner un listener solo para mostrar el cambio de color
--cuando solo se toca el texto en pantalla
FISICA (bsica)
local physics = require( "physics" )
physics.start()
OBJETOS VISUALES
newImage: forma 1
SINTAXIS: display.newImage( "ruta",x,y )
local sky = display.newImage( "bkg_clouds.png" )
sky.x = 160; sky.y = 195 -- (propiedades del objeto para cambiar su posicion)
newImage: forma 2
local sky = display.newImage( "bkg_clouds.png",160,195 )
## Agregando fsica al cuerpo del objeto
physics.addBody(sky, "static", { density=3.0, friction=0.5, bounce=0.3 } )
OPTIMIZACION
## Apagar gps, wifi, acelermetro, etc porque consumen la batera
Usar memoria potencia de 2 cuando se usan graficos, el comando
system.getInfo()te permite saber los datos del dispositivo
Para lua es mejor usar
Ubicacin, usar variable local CCX
--NON-LOCAL (DISCOURAGED = no
CCX = display.contentCenterX
--global variable
for i = 1,100 do
local image = display.newImage(
"myImage" )
image.x = CCX
end
--LOCAL (RECOMMENDED)
local CCX = display.contentCenterX
--local variable
for i = 1,100 do
local image = display.newImage(
"myImage" )
image.x = CCX
end
## tambien aplica a la librera matemtica
--NON-LOCAL (DISCOURAGED)
CCX = display.contentCenterX
--global variable
for i = 1,100 do
local image = display.newImage(
"myImage" )
image.x = CCX
end
## --"EXTERNAL" LOCAL (RECOMMENDED)
local sin = math.sin --local
reference to 'math.sin'
local function foo(x)
for i = 1,100 do
x = x + sin(i)
end
return x
end
## y a funciones; ver orden de creacin de funciones tambin
--NON-LOCAL (DISCOURAGED)
function func1()
func2( "myValue" )
end
function func2( y )
print( y )
end
--LOCAL (RECOMMENDED)
local function func2( y ) --'func2'
properly scoped above 'func1'
print( y )
end
local function func1()
func2( "myValue" )
end
func1()
func1()
## Para poner elementos en las tablas evitar table.insert()
--table.insert() (DISCOURAGED)
## --LOOP INDEX METHOD (RECOMMENDED)
local a = {}
local table_insert = table.insert
for i = 1,100 do
table_insert( a, i )
end
--TABLE SIZE METHOD (ACCEPTABLE)
local a = {}
for i = 1,100 do
a[#a+1] = i
end
local a = {}
for i = 1,100 do
a[i] = i
end
local a = {}
local index = 1
for i = 1,100 do
a[index] = i
index = index+1
end
## Para sacar elementos de las tablas evitar table.unpack()
--unpack() (DISCOURAGED)
local a = { 100, 200, 300, 400 }
## --LOOP METHOD (RECOMMENDED)
local a = { 100, 200, 300, 400 }
for i = 1,100 do
print( unpack(a) )
end
for i = 1,100 do
print( a[1],a[2],a[3],a[4] )
end
Evitar "ipairs()"
--ipairs() (DISCOURAGED)
local t1 = {}
local t2 = {}
local t3 = {}
local t4 = {}
local a = { t1, t2, t3, t4 }
--LUA
local
local
local
local
local
CONSTRUCT (RECOMMENDED)
t1 = {}
t2 = {}
t3 = {}
t4 = {}
a = { t1, t2, t3, t4 }
## for i,v in ipairs( a ) do
print( i,v )
end
for i = 1,#a do
print( a[i] )
end
Mejoramiento de la matemtica
--DIVISION (DISCOURAGED)
x/2 ; y/8
--MULTIPLICATION BY DECIMAL
(RECOMMENDED)
x * 0.5 ; y * 0.125
--EXPONENTIATION (DISCOURAGED)
x^3
--MULTIPLICATION (RECOMMENDED)
x * x * x
## Tambien en el resto por divisin inexacta
--math.fmod() (DISCOURAGED)
local fmod = math.fmod
for i = 1,100 do
if ( fmod( i,30 ) < 1 ) then
## --MODULUS OPERATOR (RECOMMENDED)
for i = 1,100 do
if ( ( i%30 ) < 1 ) then
local x = 1
local x = 1
end
end
end
end
local soundTable = {
}
"a.wav"
"b.wav"
"c.wav"
"d.wav"
"e.wav"
"f.wav"
"g.wav"
"h.wav"
),
),
),
),
),
),
),
),
## --Y luego los llamas
local mySound = audio.play( soundTable["mySound1"] )
--y para mejorar el rendimiento de liberacin de memoria se elimina asi
local ST = soundTable
for s=#ST,1,-1 do
audio.dispose( ST[s] ) ; ST[s] = nil
end
## GRAFICOS, AUDIO Y ANIMACION
* significa que hay mas documentacin
Sintaxis
Tanto x, y representan posicin
Ejemplo
display.newGroup()
## local rect = display.newRect(0, 0, 100, 100)
rect:setFillColor( 0.5 )
local group = display.newGroup()
group:insert( rect )
## display.newImage( [parent,] filename
[,baseDir] [,x,y] [,isFullResolution])
## local myImage = display.newImage( "image.png" )
--algunos mtodos (aplica para todos los objetos)
-- position the image
myImage:translate( 100, 100 )
-- tint the image red
myImage:setFillColor( 1, 0, 0 )
-- remove the image
myImage:removeSelf()
myImage = nil
--algunas propiedades (aplica para todos los objetos)
-- hide the image
myImage.isVisible = false
display.newImage( [parent,]
imageSheet, frameIndex )
## -- first, create the image sheet object
local options =
{
-- The params below are required
width = 70,
height = 41,
numFrames = 2,
-- The params below are optional (used for dynamic
image sheet selection)
sheetContentWidth = 70,
of entire sheet
sheetContentHeight = 82
size of entire sheet
}
## -- width of original 1x size
-- height of original 1x
## local imageSheet = graphics.newImageSheet( "fishies.png",
options )
local myImage = display.newImage( imageSheet, 1 )
display.newImageRect( [parentGroup,]
filename, [baseDir,] width, height )
## local image = display.newImageRect( "image.png", 100, 100
)
--algunas propiedades (aplica para todos los objetos)
image.x = display.contentCenterX
image.y = display.contentCenterY
-- hide the object
image.isVisible = false
--algunos mtodos (aplica para todos los objetos)
-- remove it
image:removeSelf()
image = nil
display.newImageRect( [parentGroup,]
imageSheet, frameIndex, width,
height )
## -- first, create the image sheet object
local options =
{
-- The params below are required
width = 70,
height = 41,
numFrames = 2,
## -- The params below are optional (used for dynamic
image sheet selection)
sheetContentWidth = 70,
of entire sheet
sheetContentHeight = 82
size of entire sheet
}
## -- width of original 1x size
-- height of original 1x
## local imageSheet = graphics.newImageSheet( "fishies.png",
options )
-- create new image from the above image sheet
local obj = display.newImageRect( imageSheet, 1, 70, 41 )
obj.x, obj.y = 100, 100
display.newText( [parentGroup,] text,
x, y, font, fontSize )
display.newText( [parentGroup,] text,
x, y, [width, height,] font,
fontSize )
*
local myText = display.newText( "Hello World!", 100, 200,
native.systemFont, 16 )
## Sprite Mtodo corona
Se obviara la sintaxis
*
-- primero creamos las opciones de hoja
local sheetOptions =
{
width = 512,
height = 256,
numFrames = 8
}
-- luego cargamos la imagen sprite
local sheet_runningCat =
graphics.newImageSheet( "sritepuma.png", sheetOptions )
## --algunos mtodos (aplica para todos los objetos)
myText:setFillColor( 1, 0, 0 )
## -- acto seguido creamos la secuencia
local sequences_runningCat = {
-- first sequence (consecutive frames)
{
name = "normalRun",
start = 1,
count = 8,
time = 800,
loopCount = 4
}
}
-- despus creamos el sprite
local runningCat = display.newSprite( sheet_runningCat,
sequences_runningCat )
--finalmente usamos sus mtodos
runningCat:translate(100,100) --opcional
runningCat:play()
## Sprite mtodo texture packer
Se obviara la sintaxis
## * loopDirection (forward y bounce ) en sequenceData
-- usa un archivo.lua generado por texture packer
-- primero llamamos al archivo .lua
local sheetInfo = require("spritesheet"), este contiene
las opciones de hoja y otros 2 metodos
--luego cargamos la imagen sprite
local miSprite =
graphics.newImageSheet("spritesheet.png",
sheetInfo:getSheet())
--despues creamos la secuencia
local sequenceData = {
{
name="walk",
-- name of the animation
sheet=miSprite,
-- the image sheet
start=sheetInfo:getFrameIndex("0001"), -- first frame
count=8,
-- number of frames
time=1000,
-- speed
loopCount=4
-- repeat
}
}
--finalmente creamos el sprite
sprite = display.newSprite( miSprite, sequenceData )
sprite:setSequence("walk")
---y le damos play
sprite:play()
---- tambin podemos obtener cualquier imagen de la hoja
--sprite asi
display.newImage(miSprite,
sheetInfo:getFrameIndex("0001"))
## display.newRect( x, y, width, height )
display.newRect( parent, x, y, width,
height )
--ejemplo
local myRectangle = display.newRect( 0, 0, 150, 50 )
myRectangle.strokeWidth = 3
myRectangle:setFillColor( 0.5 )
myRectangle:setStrokeColor( 1, 0, 0 )
## display.newCircle( xCenter, yCenter,
display.newCircle( parent, xCenter,
## -- ejemplo con sus propiedades y metodos
local myCircle = display.newCircle( 100, 100, 30 )
myCircle:setFillColor( 0.5 )
myCircle.strokeWidth = 5
myCircle:setStrokeColor( 1, 0, 0 )
display.newRoundedRect( x, y, width,
## -- ejemplo con sus propiedades y metodos
local myRoundedRect = display.newRoundedRect( 0, 0, 150,
50, 12 )
myRoundedRect.strokeWidth = 3
myRoundedRect:setFillColor( 0.5 )
myRoundedRect:setStrokeColor( 1, 0, 0 )
display.newRoundedRect( parent, x, y,
## display.newLine( [parentGroup,] x1,
y1, x2, y2 [, x3, y3, ... ] )
## local star = display.newLine( 200, 90, 227, 165 )
star:append( 305,165, 243,216, 265,290, 200,245, 135,290,
necesarios
## 157,215, 95,165, 173,165, 200,90 )
star:setStrokeColor( 1, 0, 0, 1 )
star.strokeWidth = 8
display.newPolygon( x, y, vertices )
display.newPolygon( parent, x, y,
vertices )
## local halfW = display.contentWidth * 0.5
local halfH = display.contentHeight * 0.5
local vertices = { 0,-110, 27,-35, 105,-35, 43,16, 65,90,
0,45, -65,90, -43,15, -105,-35, -27,-35, }
local o = display.newPolygon( halfW, halfH, vertices )
o.fill = { type="image", filename="mountains.png" }
o.strokeWidth = 10
o:setStrokeColor( 1, 0, 0 )
display.newEmbossedText( [parentGroup,
] text, x, y, font, fontSize )
display.newEmbossedText( [parentGroup,
] text, x, y, [width, height,] font,
fontSize )
## local myText = display.newEmbossedText( "hello", 200,
100, native.systemFont, 40 )
myText:setFillColor( 0.5 )
myText:setText( "Hello World!" )
local color =
{
highlight = { r=1, g=1, b=1 },
shadow = { r=0.3, g=0.3, b=0.3 }
}
myText:setEmbossColor( color )
display.newContainer( [parent, ]
width, height )
-- Create a container
local container = display.newContainer( 128, 128 )
-- Create an image
local bkgd =
display.newImage( "aquariumbackgroundIPhone.jpg" )
-- Insert the image into the container
container:insert( bkgd, true )
-- Center the container in the display area
container:translate( display.contentWidth*0.5,
display.contentHeight*0.5 )
-- Transition (rotate) the container
transition.to( container, { rotation=360,
transition=easing.inOutExpo} )
display.newSnapshot( w, h )
## local snapshot = display.newSnapshot( 200, 200 )
math.randomseed( 0 ) random se palica solo 1 vez
display.newSnapshot( parent, w, h )
-- Add fish to the screen
for i=1,4 do
local fish = display.newImage( "fish.small.red.png" )
-- move to random position in a 200x200 region in the
middle of the screen
fish:translate( math.random( -100, 100 ),
math.random( -100, 100 ) )
## -- insert fish into snapshot
snapshot.group:insert( fish )
end
snapshot:translate(160, 240) -- Center snapshot
snapshot:invalidate()
-- Invalidate snapshot
snapshot.alpha = 0.5
-- Apply to flattened image
EMISOR DE PARTICULAS (FUEGO)
Usar archivo .json, ah es donde se
hacen las modificaciones
## -- Require the JSON library for decoding purposes
local json = require "json"
-- Read the exported Particle Designer file (JSON) into a
string
local filePath = system.pathForFile( "fire.json" )
local f = io.open( filePath, "r" )
local fileData = f:read( "*a" )
f:close()
-- Decode the string
local emitterParams = json.decode( fileData )
-- Create the emitter with the decoded parameters
local emitter1 = display.newEmitter( emitterParams )
-- Center the emitter within the content area
emitter1.x = display.contentCenterX
emitter1.y = display.contentCenterY
local object = display.newImage( "image.png" )
object.alpha = 0.75 -- set to 75% opacity
METODOS
---- metodo contenToLocal
-- Rectangle that fills screen
local rect = display.newRect( 0, 0,
display.contentWidth, display.contentHeight )
rect.x = display.contentCenterX ; rect.y =
display.contentCenterY
rect.strokeWidth = 4 ; rect:setStrokeColor( 1,
0, 0 )
rect:setFillColor( 0, 0, 0 )
-- Create text objects to display content and
local coordinates
local contentText = display.newText( "", 0, 0,
display.nativeSystemFont, 16 )
local localText = display.newText( "", 0, 0,
display.nativeSystemFont, 16 )
localText.anchorX = 0 ; localText.anchorY = 0
contentText.anchorX = 0 ; contentText.anchorY =
0
## function showCoordinates( event )
-- Get x and y of touch event in content
coordinates
local contentX, contentY = event.x, event.y
-- Convert to local coordinates of
local localX, localY =
event.target:contentToLocal( contentX, contentY
)
-- Display content and local coordinate
values
contentText.text = "CONTENT: " ..
tostring(contentX) .. ", " ..
tostring(contentY)
localText.text = "LOCAL: " ..
tostring(localX) .. ", " .. tostring(localY)
contentText.x, contentText.y = event.x+20,
event.y
localText.x, localText.y = event.x+20,
event.y+20
localText.anchorX = 0 ; localText.anchorY =
0
contentText.anchorX = 0 ;
contentText.anchorY = 0
return true
end
)
## -- propiedad anchor (cambiar centro de
dibujo)
local rect2 = display.newRect( 0, 0, 50, 50 )
rect2.anchorX = 0 (0, 0.5 1)
rect2.anchorY = 0 (0, 0.5 1)
--Mtodo localToContent
-- Create a square
local square = display.newRect( 100, 100, 40,
40 )
square:setFillColor( 1 )
-- Create another square on top and rotate it
20 degrees
local redSquare = display.newRect( 100, 100,
40, 40 )
redSquare:setFillColor( 1, 0, 0, 0.6 )
redSquare.rotation = 20
local sqCenterX, sqCenterY =
square:localToContent( 0, 0 )
print( "White square's center position in
screen coordinates: ", sqCenterX, sqCenterY )
-- Get the content position of the white
square's top-left corner
-- Using ( -20,-20 ) specifies the top left
## corner of the square, since it's 40x40 in size
local whiteTLX, whiteTLY =
square:localToContent( -20, -20 )
print( "White square's top-left position in
screen coordinates: ", whiteTLX, whiteTLY )
-- Get the content position of the red square's
top-left corner, independent of its rotation
-- Using ( -20,-20 ) specifies the top left
corner of the square, since it's 40x40 in size
local redTLX, redTLY =
redSquare:localToContent( -20, -20 )
print( "Red square's top-left position in
screen coordinates: ", redTLX, redTLY )
lmites del objeto)
botones
50 )
## local function dibujarCarro( event )
if event.phase == "ended" then
obj =
## local bounds = rect.contentBounds
print( "xMin: ".. bounds.xMin ) -print( "yMin: ".. bounds.yMin ) -print( "xMax: ".. bounds.xMax ) -print( "yMax: ".. bounds.yMax ) --
## display.newImageRect( "carro.png", 100,100 )
obj:translate( 100, 100 )
end
end
xMin:
yMin:
xMax:
yMax:
100
100
150
150
## local function borraCarro( event )
--obj =
display.newImageRect( "carro.png", 100,100 )
--obj:translate( 250, 100 )
--display.newText("lo creo", 200, 350,
arial, 20)
if event.phase == "ended" then
obj:removeSelf()
obj = nil
display.newText("lo borro",
end
end
## local grupoBotonDibujar = display.newGroup()
local botonDibujar =
display.newRoundedRect( grupoBotonDibujar,0,0,
150, 50, 20 )
local textoDibujar =
display.newText( grupoBotonDibujar, "DIBUJAR",
0, 0, ARIAL, 20 )
textoDibujar:setFillColor(1, 0, 1, 1)
grupoBotonDibujar:translate(75,300)
## local grupoBotonBorrar = display.newGroup()
local botonBorrar =
display.newRoundedRect( grupoBotonBorrar,0,0,
150, 50, 20 )
local textoBorrar =
display.newText( grupoBotonBorrar, "BORRAR", 0,
0, ARIAL, 20 )
textoBorrar:setFillColor(0,1,1,1)
grupoBotonBorrar:translate(245,300)
dibujarCarro)
borraCarro)
## local rect = display.newRect( 100, 100, 50,
50 )
rect.rotation = 55 -- afecta el tamao
## local rect = display.newRect( 50, 50, 100,
200 )
rect:setFillColor( 1, 0, 0 )
## print( "contentHeight: ".. rect.contentHeight )
print( "contentWidth: ".. rect.contentWidth )
## -- Rotate the rectangle 45 degrees
rect:rotate( 45 )
medida)
## local rect = display.newRect( 100, 100, 50,
50 )
rect:scale(2,2) no afecta al tamao
## local star = display.newImage( "star.png" )
-- Scale the image by 200% (x) and 50% (y)
star:scale( 2, 0.5 )
## print( "Height: ".. rect.height )
print( "Width: ".. rect.width )
tamao de una mascara)
-- Crea la imagen
local image =
250 )
image:translate( display.contentCenterX,
display.contentCenterY )
-- crea la mascara y la aplica a la imagen
-- Eventos de toque sobre la imagen se
enmascaran a los lmites de la mscara
--(Siempre rectangular para una imagen,
independientemente del contenido de la imagen)
-- cambiando de tamao
## Poniendo mascara a una imagen
-- Create and position image to be masked
local image = display.newImageRect(
"image.png", 768, 1024 )
image:translate( display.contentCenterX,
display.contentCenterY )
-- Create mask and apply to image
)
## local group = display.newGroup() crea grupo
local g = display.newGroup()
-- Create and position image to be masked, and
insert into group
local image = display.newImageRect( g,
"image.png", 768, 1024 )
## local rect = display.newRect( 100, 100, 50,
50 )
rect:setFillColor( 0.7 )
group:insert( rect ) instera (evitar)
rect.parent:remove( rect )
group
## -- Center the Display Group
g:translate( display.contentCenterX,
display.contentCenterY )
--rotacin sencilla
- crear cartas
## local rect = display.newRect( 100, 100, 50,
50 )
rect.rotation = 45
## local cards = display.newGroup()
-- grupo
de cartas
-- crear evento (toBack o toFront)
function sendToBack( event )
if ( event.phase == "began" ) then
-- transicin
local rect = display.newRect( 70, 150, 100, 150
)
rect:setFillColor( 1, 0, 0 )
--rect.rotation = -45
local reverse = 0 -- bit
-- funcion
local function rockRect()
if ( reverse == 0 ) then
reverse = 1
-- tiempo para la transicion
transition.to(rect, {time = 1000,
x=200})
else
reverse = 0
transition.to(rect, {time = 1000,
x=70})
end
end
--tiempo de inicio y duracion de cada ciclo de
la funcin
timer.performWithDelay( 2000, rockRect, 0 ) -Repite siempre
## --rotacion con transicin
local rect = display.newRect( 50, 50, 100,
150 )
rect:setFillColor( 1, 0, 0 )
rect.rotation = -45
local reverse = 1
local function rockRect()
event.target:toFront()
end
return true
end
for i=1,5 do
--grupo de elementos de uan carta
local cardGroup = display.newGroup()
-- Card outline
local cardRect =
display.newRoundedRect(cardGroup, 100, 100,
125, 175, 12 )
cardRect.strokeWidth = 2
cardRect:setFillColor( 1 )
cardRect:setStrokeColor( 0, 0, 0, 0.3 )
cardGroup:insert(cardRect)
-- Card values
local cardValue =
display.newText( cardGroup, i,
cardRect.contentWidth - 72, 32,
native.systemFontBold, 24 )
cardValue:setFillColor( 1, 0, 0 )
local cardValue2 =
display.newText( cardGroup, i, 148,
cardRect.contentHeight - 8 ,
native.systemFontBold, 24 )
cardValue2:setFillColor( 1, 0, 0 )
cardGroup.x = (i * 25)
cardGroup.y = (i * 25)
if ( reverse == 0 ) then
reverse = 1
transition.to( rect, { rotation=-45,
time=500, transition=easing.inOutCubic } )
else
reverse = 0
transition.to( rect, { rotation=45,
time=500, transition=easing.inOutCubic } )
end
end
timer.performWithDelay( 600, rockRect, 0 ) -Repeat forever
-- propiedad x,y (esto ubica en el punto
donde le indicas)
-local rect = display.newRect( 0, 0, 50, 50 )
rect.x = 100
rect.y = 100
objeto)
## local rect1 = display.newRect( 100, 75, 150, 50
)
rect1:setFillColor( 1, 1, 1 )
rect1.rotation=30
-- gira el rectangulo horizontalmente
rect1.xScale = -1
local rect2 = display.newRect( 125, 175, 50, 50
)
rect2:setFillColor( 1, 0, 0 )
rect2.xScale = 2 -- cambia la escala en x a
200%
rect2.yScale = 1.5 -- cambia la escala en y a
150%
## -- lee la ruta donde almacenas el .main
local line = display.newLine( 200, 90, 227, 165
)
print("line._defined: " .. line._defined )
## -- lee la ruta del main.lua y la linea
donde fue
-- hecho el ultimo cambio
local line = display.newLine( 200, 90, 227, 165
)
line.rotation = 45
print("line._lastChange: " ..
line._lastChange )
cards:insert( cardGroup )
-- agregando evento a un grupo
sendToBack )
end
--Esto aumenta en x o y
50, 50 )
,100)
## -- da las propiedades del objeto
local json = require( "json" )
local line = display.newLine( 200, 90, 227, 165
)
line.rotation = 45
print("line._properties: " ..
json.prettify( line._properties ) )
Mtodo directo
"image.png" )
## local myImage = display.newImage( "image.png" )
myGroup:insert( myImage )
## Orden de los objetos
--creamos el grupo
local myGroup = display.newGroup()
--creamos los objetos
local square = display.newRect( myGroup, 0, 0, 100, 100 ) --red square is at the bottom
square:setFillColor( 1, 0, 0 )
local circle = display.newCircle( myGroup, 30, 50, 50 ) --green circle is in the middle
circle:setFillColor( 0, 1, 0 )
local rect = display.newRect( myGroup, 60, 50, 120, 80 ) --blue rectangle is at the top
rect:setFillColor( 0, 0, 1 )
myGroup:translate(100,100)
-- creamos el parend de square
local parent = square.parent
-- lo ponemos en su parent
parent:insert( square )
-- y ya podemos moverlo a la capa que queramos
parent:insert(1, circle )
## Referencia a objetos (usando una tabla)
-- creamos un grupo
local squares = display.newGroup()
local redSquare1 = display.newRect( squares, 0, 0, 40, 40 )
local redSquare2 = display.newRect( squares, 0, 50, 40, 40 )
local redSquare3 = display.newRect( squares, 0, 100, 40, 40 )
--los pinatmos de color rojo
redSquare1:setFillColor( 1, 0, 0 )
redSquare2:setFillColor( 1, 0, 0 )
redSquare3:setFillColor( 1, 0, 0 )
local whiteSquare1 = display.newRect( squares, 0, 0, 40, 40 )
local whiteSquare2 = display.newRect( squares, 0, 70, 40, 40 )
--os pinatmos de color balnco
whiteSquare1:setFillColor( 1 )
whiteSquare2:setFillColor( 1 )
squares:translate(100,50)
local redSquares = { redSquare1, redSquare2, redSquare3 }
--manipulamos la tabla usando un for
for i = 1, #redSquares do
--manipulate the square at the current index (i)
redSquares[i].x = redSquares[i].x + 100 -- los movemos 100 a la derecha
redSquares[i]:scale( 0.5, 0.5 ) -- los escalos a 50 %
end
Borrando un grupo
display.remove ( myObject )
myObject = nil --set reference to nil!
display.remove( myGroup )
myGroup = nil
--OR
myObject:removeSelf()
myObject = nil --set reference to nil!
--OR
myGroup:removeSelf()
myGroup = nil
--OR
myGroup:remove( myObject )
myObject = nil --set reference to nil!
| 7,786
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| 2.90625
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|
CC-MAIN-2019-35
|
latest
|
en
| 0.128337
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https://docs.ruby-lang.org/en/master/Float.html
| 1,670,106,880,000,000,000
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text/html
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crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00367.warc.gz
| 239,028,581
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class Float
A Float object represents a sometimes-inexact real number using the native architecture’s double-precision floating point representation.
Floating point has a different arithmetic and is an inexact number. So you should know its esoteric system. See following:
You can create a Float object explicitly with:
You can convert certain objects to Floats with:
What’s Here¶ ↑
First, what’s elsewhere. Class Float:
Here, class Float provides methods for:
Constants
DIG
The minimum number of significant decimal digits in a double-precision floating point.
Usually defaults to 15.
EPSILON
The difference between 1 and the smallest double-precision floating point number greater than 1.
Usually defaults to 2.2204460492503131e-16.
INFINITY
An expression representing positive infinity.
MANT_DIG
The number of base digits for the `double` data type.
Usually defaults to 53.
MAX
The largest possible integer in a double-precision floating point number.
Usually defaults to 1.7976931348623157e+308.
MAX_10_EXP
The largest positive exponent in a double-precision floating point where 10 raised to this power minus 1.
Usually defaults to 308.
MAX_EXP
The largest possible exponent value in a double-precision floating point.
Usually defaults to 1024.
MIN
The smallest positive normalized number in a double-precision floating point.
Usually defaults to 2.2250738585072014e-308.
If the platform supports denormalized numbers, there are numbers between zero and `Float::MIN`. 0.0.next_float returns the smallest positive floating point number including denormalized numbers.
MIN_10_EXP
The smallest negative exponent in a double-precision floating point where 10 raised to this power minus 1.
Usually defaults to -307.
MIN_EXP
The smallest possible exponent value in a double-precision floating point.
Usually defaults to -1021.
NAN
An expression representing a value which is “not a number”.
The base of the floating point, or number of unique digits used to represent the number.
Usually defaults to 2 on most systems, which would represent a base-10 decimal.
Public Instance Methods
self % other → float click to toggle source
Returns `self` modulo `other` as a float.
For float `f` and real number `r`, these expressions are equivalent:
```f % r
f-r*(f/r).floor
f.divmod(r)[1]
```
Examples:
```10.0 % 2 # => 0.0
10.0 % 3 # => 1.0
10.0 % 4 # => 2.0
10.0 % -2 # => 0.0
10.0 % -3 # => -2.0
10.0 % -4 # => -2.0
10.0 % 4.0 # => 2.0
10.0 % Rational(4, 1) # => 2.0
```
`Float#modulo` is an alias for `Float#%`.
```static VALUE
flo_mod(VALUE x, VALUE y)
{
double fy;
if (FIXNUM_P(y)) {
fy = (double)FIX2LONG(y);
}
else if (RB_BIGNUM_TYPE_P(y)) {
fy = rb_big2dbl(y);
}
else if (RB_FLOAT_TYPE_P(y)) {
fy = RFLOAT_VALUE(y);
}
else {
return rb_num_coerce_bin(x, y, '%');
}
return DBL2NUM(ruby_float_mod(RFLOAT_VALUE(x), fy));
}```
Also aliased as: modulo
self * other → numeric click to toggle source
Returns a new Float which is the product of `self` and `other`:
```f = 3.14
f * 2 # => 6.28
f * 2.0 # => 6.28
f * Rational(1, 2) # => 1.57
f * Complex(2, 0) # => (6.28+0.0i)
```
```VALUE
rb_float_mul(VALUE x, VALUE y)
{
if (FIXNUM_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) * (double)FIX2LONG(y));
}
else if (RB_BIGNUM_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) * rb_big2dbl(y));
}
else if (RB_FLOAT_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) * RFLOAT_VALUE(y));
}
else {
return rb_num_coerce_bin(x, y, '*');
}
}```
self ** other → numeric click to toggle source
Raises `self` to the power of `other`:
```f = 3.14
f ** 2 # => 9.8596
f ** -2 # => 0.1014239928597509
f ** 2.1 # => 11.054834900588839
f ** Rational(2, 1) # => 9.8596
f ** Complex(2, 0) # => (9.8596+0i)
```
```VALUE
rb_float_pow(VALUE x, VALUE y)
{
double dx, dy;
if (y == INT2FIX(2)) {
dx = RFLOAT_VALUE(x);
return DBL2NUM(dx * dx);
}
else if (FIXNUM_P(y)) {
dx = RFLOAT_VALUE(x);
dy = (double)FIX2LONG(y);
}
else if (RB_BIGNUM_TYPE_P(y)) {
dx = RFLOAT_VALUE(x);
dy = rb_big2dbl(y);
}
else if (RB_FLOAT_TYPE_P(y)) {
dx = RFLOAT_VALUE(x);
dy = RFLOAT_VALUE(y);
if (dx < 0 && dy != round(dy))
return rb_dbl_complex_new_polar_pi(pow(-dx, dy), dy);
}
else {
return rb_num_coerce_bin(x, y, idPow);
}
return DBL2NUM(pow(dx, dy));
}```
self + other → numeric click to toggle source
Returns a new Float which is the sum of `self` and `other`:
```f = 3.14
f + 1 # => 4.140000000000001
f + 1.0 # => 4.140000000000001
f + Rational(1, 1) # => 4.140000000000001
f + Complex(1, 0) # => (4.140000000000001+0i)
```
```VALUE
rb_float_plus(VALUE x, VALUE y)
{
if (FIXNUM_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) + (double)FIX2LONG(y));
}
else if (RB_BIGNUM_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) + rb_big2dbl(y));
}
else if (RB_FLOAT_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) + RFLOAT_VALUE(y));
}
else {
return rb_num_coerce_bin(x, y, '+');
}
}```
self - other → numeric click to toggle source
Returns a new Float which is the difference of `self` and `other`:
```f = 3.14
f - 1 # => 2.14
f - 1.0 # => 2.14
f - Rational(1, 1) # => 2.14
f - Complex(1, 0) # => (2.14+0i)
```
```VALUE
rb_float_minus(VALUE x, VALUE y)
{
if (FIXNUM_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) - (double)FIX2LONG(y));
}
else if (RB_BIGNUM_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) - rb_big2dbl(y));
}
else if (RB_FLOAT_TYPE_P(y)) {
return DBL2NUM(RFLOAT_VALUE(x) - RFLOAT_VALUE(y));
}
else {
return rb_num_coerce_bin(x, y, '-');
}
}```
-float → float click to toggle source
Returns `float`, negated.
```# File numeric.rb, line 367
def -@
Primitive.attr! 'inline'
Primitive.cexpr! 'rb_float_uminus(self)'
end```
self / other → numeric click to toggle source
Returns a new Float which is the result of dividing `self` by `other`:
```f = 3.14
f / 2 # => 1.57
f / 2.0 # => 1.57
f / Rational(2, 1) # => 1.57
f / Complex(2, 0) # => (1.57+0.0i)
```
```VALUE
rb_float_div(VALUE x, VALUE y)
{
double num = RFLOAT_VALUE(x);
double den;
double ret;
if (FIXNUM_P(y)) {
den = FIX2LONG(y);
}
else if (RB_BIGNUM_TYPE_P(y)) {
den = rb_big2dbl(y);
}
else if (RB_FLOAT_TYPE_P(y)) {
den = RFLOAT_VALUE(y);
}
else {
return rb_num_coerce_bin(x, y, '/');
}
ret = double_div_double(num, den);
return DBL2NUM(ret);
}```
self < other → true or false click to toggle source
Returns `true` if `self` is numerically less than `other`:
```2.0 < 3 # => true
2.0 < 3.0 # => true
2.0 < Rational(3, 1) # => true
2.0 < 2.0 # => false
```
`Float::NAN < Float::NAN` returns an implementation-dependent value.
```static VALUE
flo_lt(VALUE x, VALUE y)
{
double a, b;
a = RFLOAT_VALUE(x);
if (RB_INTEGER_TYPE_P(y)) {
VALUE rel = rb_integer_float_cmp(y, x);
if (FIXNUM_P(rel))
return RBOOL(-FIX2LONG(rel) < 0);
return Qfalse;
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(b)) return Qfalse;
#endif
}
else {
return rb_num_coerce_relop(x, y, '<');
}
#if MSC_VERSION_BEFORE(1300)
if (isnan(a)) return Qfalse;
#endif
return RBOOL(a < b);
}```
self <= other → true or false click to toggle source
Returns `true` if `self` is numerically less than or equal to `other`:
```2.0 <= 3 # => true
2.0 <= 3.0 # => true
2.0 <= Rational(3, 1) # => true
2.0 <= 2.0 # => true
2.0 <= 1.0 # => false
```
`Float::NAN <= Float::NAN` returns an implementation-dependent value.
```static VALUE
flo_le(VALUE x, VALUE y)
{
double a, b;
a = RFLOAT_VALUE(x);
if (RB_INTEGER_TYPE_P(y)) {
VALUE rel = rb_integer_float_cmp(y, x);
if (FIXNUM_P(rel))
return RBOOL(-FIX2LONG(rel) <= 0);
return Qfalse;
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(b)) return Qfalse;
#endif
}
else {
return rb_num_coerce_relop(x, y, idLE);
}
#if MSC_VERSION_BEFORE(1300)
if (isnan(a)) return Qfalse;
#endif
return RBOOL(a <= b);
}```
self <=> other → -1, 0, +1, or nil click to toggle source
Returns a value that depends on the numeric relation between `self` and `other`:
• -1, if `self` is less than `other`.
• 0, if `self` is equal to `other`.
• 1, if `self` is greater than `other`.
• `nil`, if the two values are incommensurate.
Examples:
```2.0 <=> 2 # => 0
2.0 <=> 2.0 # => 0
2.0 <=> Rational(2, 1) # => 0
2.0 <=> Complex(2, 0) # => 0
2.0 <=> 1.9 # => 1
2.0 <=> 2.1 # => -1
2.0 <=> 'foo' # => nil
```
This is the basis for the tests in the `Comparable` module.
`Float::NAN <=> Float::NAN` returns an implementation-dependent value.
```static VALUE
flo_cmp(VALUE x, VALUE y)
{
double a, b;
VALUE i;
a = RFLOAT_VALUE(x);
if (isnan(a)) return Qnil;
if (RB_INTEGER_TYPE_P(y)) {
VALUE rel = rb_integer_float_cmp(y, x);
if (FIXNUM_P(rel))
return LONG2FIX(-FIX2LONG(rel));
return rel;
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
}
else {
if (isinf(a) && !UNDEF_P(i = rb_check_funcall(y, rb_intern("infinite?"), 0, 0))) {
if (RTEST(i)) {
int j = rb_cmpint(i, x, y);
j = (a > 0.0) ? (j > 0 ? 0 : +1) : (j < 0 ? 0 : -1);
return INT2FIX(j);
}
if (a > 0.0) return INT2FIX(1);
return INT2FIX(-1);
}
return rb_num_coerce_cmp(x, y, id_cmp);
}
return rb_dbl_cmp(a, b);
}```
self == other → true or false click to toggle source
Returns `true` if `other` has the same value as `self`, `false` otherwise:
```2.0 == 2 # => true
2.0 == 2.0 # => true
2.0 == Rational(2, 1) # => true
2.0 == Complex(2, 0) # => true
```
`Float::NAN == Float::NAN` returns an implementation-dependent value.
Related: `Float#eql?` (requires `other` to be a Float).
```MJIT_FUNC_EXPORTED VALUE
rb_float_equal(VALUE x, VALUE y)
{
volatile double a, b;
if (RB_INTEGER_TYPE_P(y)) {
return rb_integer_float_eq(y, x);
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(b)) return Qfalse;
#endif
}
else {
return num_equal(x, y);
}
a = RFLOAT_VALUE(x);
#if MSC_VERSION_BEFORE(1300)
if (isnan(a)) return Qfalse;
#endif
return RBOOL(a == b);
}```
Also aliased as: ===
===(p1)
Returns `true` if `other` has the same value as `self`, `false` otherwise:
```2.0 == 2 # => true
2.0 == 2.0 # => true
2.0 == Rational(2, 1) # => true
2.0 == Complex(2, 0) # => true
```
`Float::NAN == Float::NAN` returns an implementation-dependent value.
Related: `Float#eql?` (requires `other` to be a Float).
Alias for: ==
self > other → true or false click to toggle source
Returns `true` if `self` is numerically greater than `other`:
```2.0 > 1 # => true
2.0 > 1.0 # => true
2.0 > Rational(1, 2) # => true
2.0 > 2.0 # => false
```
`Float::NAN > Float::NAN` returns an implementation-dependent value.
```VALUE
rb_float_gt(VALUE x, VALUE y)
{
double a, b;
a = RFLOAT_VALUE(x);
if (RB_INTEGER_TYPE_P(y)) {
VALUE rel = rb_integer_float_cmp(y, x);
if (FIXNUM_P(rel))
return RBOOL(-FIX2LONG(rel) > 0);
return Qfalse;
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(b)) return Qfalse;
#endif
}
else {
return rb_num_coerce_relop(x, y, '>');
}
#if MSC_VERSION_BEFORE(1300)
if (isnan(a)) return Qfalse;
#endif
return RBOOL(a > b);
}```
self >= other → true or false click to toggle source
Returns `true` if `self` is numerically greater than or equal to `other`:
```2.0 >= 1 # => true
2.0 >= 1.0 # => true
2.0 >= Rational(1, 2) # => true
2.0 >= 2.0 # => true
2.0 >= 2.1 # => false
```
`Float::NAN >= Float::NAN` returns an implementation-dependent value.
```static VALUE
flo_ge(VALUE x, VALUE y)
{
double a, b;
a = RFLOAT_VALUE(x);
if (RB_TYPE_P(y, T_FIXNUM) || RB_BIGNUM_TYPE_P(y)) {
VALUE rel = rb_integer_float_cmp(y, x);
if (FIXNUM_P(rel))
return RBOOL(-FIX2LONG(rel) >= 0);
return Qfalse;
}
else if (RB_FLOAT_TYPE_P(y)) {
b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(b)) return Qfalse;
#endif
}
else {
return rb_num_coerce_relop(x, y, idGE);
}
#if MSC_VERSION_BEFORE(1300)
if (isnan(a)) return Qfalse;
#endif
return RBOOL(a >= b);
}```
abs → float click to toggle source
magnitude → float
Returns the absolute value of `float`.
```(-34.56).abs #=> 34.56
-34.56.abs #=> 34.56
34.56.abs #=> 34.56
```
`Float#magnitude` is an alias for `Float#abs`.
```# File numeric.rb, line 351
def abs
Primitive.attr! 'inline'
Primitive.cexpr! 'rb_float_abs(self)'
end```
angle → 0 or float
Returns 0 if the value is positive, pi otherwise.
Alias for: arg
arg → 0 or float click to toggle source
Returns 0 if the value is positive, pi otherwise.
```static VALUE
float_arg(VALUE self)
{
if (isnan(RFLOAT_VALUE(self)))
return self;
if (f_tpositive_p(self))
return INT2FIX(0);
return rb_const_get(rb_mMath, id_PI);
}```
Also aliased as: angle, phase
ceil(ndigits = 0) → float or integer click to toggle source
Returns the smallest number greater than or equal to `self` with a precision of `ndigits` decimal digits.
When `ndigits` is positive, returns a float with `ndigits` digits after the decimal point (as available):
```f = 12345.6789
f.ceil(1) # => 12345.7
f.ceil(3) # => 12345.679
f = -12345.6789
f.ceil(1) # => -12345.6
f.ceil(3) # => -12345.678
```
When `ndigits` is non-positive, returns an integer with at least `ndigits.abs` trailing zeros:
```f = 12345.6789
f.ceil(0) # => 12346
f.ceil(-3) # => 13000
f = -12345.6789
f.ceil(0) # => -12345
f.ceil(-3) # => -12000
```
Note that the limited precision of floating-point arithmetic may lead to surprising results:
```(2.1 / 0.7).ceil #=> 4 (!)
```
Related: `Float#floor`.
```static VALUE
flo_ceil(int argc, VALUE *argv, VALUE num)
{
int ndigits = flo_ndigits(argc, argv);
return rb_float_ceil(num, ndigits);
}```
coerce(other) → array click to toggle source
Returns a 2-element array containing `other` converted to a Float and `self`:
```f = 3.14 # => 3.14
f.coerce(2) # => [2.0, 3.14]
f.coerce(2.0) # => [2.0, 3.14]
f.coerce(Rational(1, 2)) # => [0.5, 3.14]
f.coerce(Complex(1, 0)) # => [1.0, 3.14]
```
Raises an exception if a type conversion fails.
```static VALUE
flo_coerce(VALUE x, VALUE y)
{
return rb_assoc_new(rb_Float(y), x);
}```
denominator → integer click to toggle source
Returns the denominator (always positive). The result is machine dependent.
See also `Float#numerator`.
```VALUE
rb_float_denominator(VALUE self)
{
double d = RFLOAT_VALUE(self);
VALUE r;
if (!isfinite(d))
return INT2FIX(1);
r = float_to_r(self);
return nurat_denominator(r);
}```
divmod(other) → array click to toggle source
Returns a 2-element array `[q, r]`, where
```q = (self/other).floor # Quotient
r = self % other # Remainder
```
Examples:
```11.0.divmod(4) # => [2, 3.0]
11.0.divmod(-4) # => [-3, -1.0]
-11.0.divmod(4) # => [-3, 1.0]
-11.0.divmod(-4) # => [2, -3.0]
12.0.divmod(4) # => [3, 0.0]
12.0.divmod(-4) # => [-3, 0.0]
-12.0.divmod(4) # => [-3, -0.0]
-12.0.divmod(-4) # => [3, -0.0]
13.0.divmod(4.0) # => [3, 1.0]
13.0.divmod(Rational(4, 1)) # => [3, 1.0]
```
```static VALUE
flo_divmod(VALUE x, VALUE y)
{
double fy, div, mod;
volatile VALUE a, b;
if (FIXNUM_P(y)) {
fy = (double)FIX2LONG(y);
}
else if (RB_BIGNUM_TYPE_P(y)) {
fy = rb_big2dbl(y);
}
else if (RB_FLOAT_TYPE_P(y)) {
fy = RFLOAT_VALUE(y);
}
else {
return rb_num_coerce_bin(x, y, id_divmod);
}
flodivmod(RFLOAT_VALUE(x), fy, &div, &mod);
a = dbl2ival(div);
b = DBL2NUM(mod);
return rb_assoc_new(a, b);
}```
eql?(other) → true or false click to toggle source
Returns `true` if `other` is a Float with the same value as `self`, `false` otherwise:
```2.0.eql?(2.0) # => true
2.0.eql?(1.0) # => false
2.0.eql?(1) # => false
2.0.eql?(Rational(2, 1)) # => false
2.0.eql?(Complex(2, 0)) # => false
```
`Float::NAN.eql?(Float::NAN)` returns an implementation-dependent value.
Related: `Float#==` (performs type conversions).
```MJIT_FUNC_EXPORTED VALUE
rb_float_eql(VALUE x, VALUE y)
{
if (RB_FLOAT_TYPE_P(y)) {
double a = RFLOAT_VALUE(x);
double b = RFLOAT_VALUE(y);
#if MSC_VERSION_BEFORE(1300)
if (isnan(a) || isnan(b)) return Qfalse;
#endif
return RBOOL(a == b);
}
return Qfalse;
}```
fdiv(p1)
Returns the quotient from dividing `self` by `other`:
```f = 3.14
f.quo(2) # => 1.57
f.quo(-2) # => -1.57
f.quo(Rational(2, 1)) # => 1.57
f.quo(Complex(2, 0)) # => (1.57+0.0i)
```
`Float#fdiv` is an alias for `Float#quo`.
Alias for: quo
finite? → true or false click to toggle source
Returns `true` if `self` is not `Infinity`, `-Infinity`, or `NaN`, `false` otherwise:
```f = 2.0 # => 2.0
f.finite? # => true
f = 1.0/0.0 # => Infinity
f.finite? # => false
f = -1.0/0.0 # => -Infinity
f.finite? # => false
f = 0.0/0.0 # => NaN
f.finite? # => false
```
```VALUE
rb_flo_is_finite_p(VALUE num)
{
double value = RFLOAT_VALUE(num);
return RBOOL(isfinite(value));
}```
floor(ndigits = 0) → float or integer click to toggle source
Returns the largest number less than or equal to `self` with a precision of `ndigits` decimal digits.
When `ndigits` is positive, returns a float with `ndigits` digits after the decimal point (as available):
```f = 12345.6789
f.floor(1) # => 12345.6
f.floor(3) # => 12345.678
f = -12345.6789
f.floor(1) # => -12345.7
f.floor(3) # => -12345.679
```
When `ndigits` is non-positive, returns an integer with at least `ndigits.abs` trailing zeros:
```f = 12345.6789
f.floor(0) # => 12345
f.floor(-3) # => 12000
f = -12345.6789
f.floor(0) # => -12346
f.floor(-3) # => -13000
```
Note that the limited precision of floating-point arithmetic may lead to surprising results:
```(0.3 / 0.1).floor #=> 2 (!)
```
Related: `Float#ceil`.
```static VALUE
flo_floor(int argc, VALUE *argv, VALUE num)
{
int ndigits = flo_ndigits(argc, argv);
return rb_float_floor(num, ndigits);
}```
hash → integer click to toggle source
Returns the integer hash value for `self`.
See also `Object#hash`.
```static VALUE
flo_hash(VALUE num)
{
return rb_dbl_hash(RFLOAT_VALUE(num));
}```
infinite? → -1, 1, or nil click to toggle source
Returns:
• 1, if `self` is `Infinity`.
• -1 if `self` is `-Infinity`.
• `nil`, otherwise.
Examples:
```f = 1.0/0.0 # => Infinity
f.infinite? # => 1
f = -1.0/0.0 # => -Infinity
f.infinite? # => -1
f = 1.0 # => 1.0
f.infinite? # => nil
f = 0.0/0.0 # => NaN
f.infinite? # => nil
```
```VALUE
rb_flo_is_infinite_p(VALUE num)
{
double value = RFLOAT_VALUE(num);
if (isinf(value)) {
return INT2FIX( value < 0 ? -1 : 1 );
}
return Qnil;
}```
inspect()
Returns a string containing a representation of `self`; depending of the value of `self`, the string representation may contain:
• A fixed-point number.
• A number in “scientific notation” (containing an exponent).
• ‘Infinity’.
• ‘-Infinity’.
• ‘NaN’ (indicating not-a-number).
3.14.to_s # => “3.14” (10.1**50).to_s # => “1.644631821843879e+50” (10.1**500).to_s # => “Infinity” (-10.1**500).to_s # => “-Infinity” (0.0/0.0).to_s # => “NaN”
Alias for: to_s
magnitude() click to toggle source
```# File numeric.rb, line 356
def magnitude
Primitive.attr! 'inline'
Primitive.cexpr! 'rb_float_abs(self)'
end```
modulo(p1)
Returns `self` modulo `other` as a float.
For float `f` and real number `r`, these expressions are equivalent:
```f % r
f-r*(f/r).floor
f.divmod(r)[1]
```
Examples:
```10.0 % 2 # => 0.0
10.0 % 3 # => 1.0
10.0 % 4 # => 2.0
10.0 % -2 # => 0.0
10.0 % -3 # => -2.0
10.0 % -4 # => -2.0
10.0 % 4.0 # => 2.0
10.0 % Rational(4, 1) # => 2.0
```
`Float#modulo` is an alias for `Float#%`.
Alias for: %
nan? → true or false click to toggle source
Returns `true` if `self` is a NaN, `false` otherwise.
```f = -1.0 #=> -1.0
f.nan? #=> false
f = 0.0/0.0 #=> NaN
f.nan? #=> true
```
```static VALUE
flo_is_nan_p(VALUE num)
{
double value = RFLOAT_VALUE(num);
return RBOOL(isnan(value));
}```
negative? → true or false click to toggle source
Returns `true` if `float` is less than 0.
```# File numeric.rb, line 400
def negative?
Primitive.attr! 'inline'
Primitive.cexpr! 'RBOOL(RFLOAT_VALUE(self) < 0.0)'
end```
next_float → float click to toggle source
Returns the next-larger representable Float.
These examples show the internally stored values (64-bit hexadecimal) for each Float `f` and for the corresponding `f.next_float`:
```f = 0.0 # 0x0000000000000000
f.next_float # 0x0000000000000001
f = 0.01 # 0x3f847ae147ae147b
f.next_float # 0x3f847ae147ae147c
```
In the remaining examples here, the output is shown in the usual way (result `to_s`):
```0.01.next_float # => 0.010000000000000002
1.0.next_float # => 1.0000000000000002
100.0.next_float # => 100.00000000000001
f = 0.01
(0..3).each_with_index {|i| printf "%2d %-20a %s\n", i, f, f.to_s; f = f.next_float }
```
Output:
``` 0 0x1.47ae147ae147bp-7 0.01
1 0x1.47ae147ae147cp-7 0.010000000000000002
2 0x1.47ae147ae147dp-7 0.010000000000000004
3 0x1.47ae147ae147ep-7 0.010000000000000005
f = 0.0; 100.times { f += 0.1 }
f # => 9.99999999999998 # should be 10.0 in the ideal world.
10-f # => 1.9539925233402755e-14 # the floating point error.
10.0.next_float-10 # => 1.7763568394002505e-15 # 1 ulp (unit in the last place).
(10-f)/(10.0.next_float-10) # => 11.0 # the error is 11 ulp.
(10-f)/(10*Float::EPSILON) # => 8.8 # approximation of the above.
"%a" % 10 # => "0x1.4p+3"
"%a" % f # => "0x1.3fffffffffff5p+3" # the last hex digit is 5. 16 - 5 = 11 ulp.```
Related: `Float#prev_float`
```static VALUE
flo_next_float(VALUE vx)
{
return flo_nextafter(vx, HUGE_VAL);
}```
numerator → integer click to toggle source
Returns the numerator. The result is machine dependent.
```n = 0.3.numerator #=> 5404319552844595
d = 0.3.denominator #=> 18014398509481984
n.fdiv(d) #=> 0.3
```
See also `Float#denominator`.
```VALUE
rb_float_numerator(VALUE self)
{
double d = RFLOAT_VALUE(self);
VALUE r;
if (!isfinite(d))
return self;
r = float_to_r(self);
return nurat_numerator(r);
}```
phase → 0 or float
Returns 0 if the value is positive, pi otherwise.
Alias for: arg
positive? → true or false click to toggle source
Returns `true` if `float` is greater than 0.
```# File numeric.rb, line 389
def positive?
Primitive.attr! 'inline'
Primitive.cexpr! 'RBOOL(RFLOAT_VALUE(self) > 0.0)'
end```
prev_float → float click to toggle source
Returns the next-smaller representable Float.
These examples show the internally stored values (64-bit hexadecimal) for each Float `f` and for the corresponding `f.pev_float`:
```f = 5e-324 # 0x0000000000000001
f.prev_float # 0x0000000000000000
f = 0.01 # 0x3f847ae147ae147b
f.prev_float # 0x3f847ae147ae147a
```
In the remaining examples here, the output is shown in the usual way (result `to_s`):
```0.01.prev_float # => 0.009999999999999998
1.0.prev_float # => 0.9999999999999999
100.0.prev_float # => 99.99999999999999
f = 0.01
(0..3).each_with_index {|i| printf "%2d %-20a %s\n", i, f, f.to_s; f = f.prev_float }
```
Output:
```0 0x1.47ae147ae147bp-7 0.01
1 0x1.47ae147ae147ap-7 0.009999999999999998
2 0x1.47ae147ae1479p-7 0.009999999999999997
3 0x1.47ae147ae1478p-7 0.009999999999999995```
Related: `Float#next_float`.
```static VALUE
flo_prev_float(VALUE vx)
{
return flo_nextafter(vx, -HUGE_VAL);
}```
quo(other) → numeric click to toggle source
Returns the quotient from dividing `self` by `other`:
```f = 3.14
f.quo(2) # => 1.57
f.quo(-2) # => -1.57
f.quo(Rational(2, 1)) # => 1.57
f.quo(Complex(2, 0)) # => (1.57+0.0i)
```
`Float#fdiv` is an alias for `Float#quo`.
```static VALUE
flo_quo(VALUE x, VALUE y)
{
return num_funcall1(x, '/', y);
}```
Also aliased as: fdiv
rationalize([eps]) → rational click to toggle source
Returns a simpler approximation of the value (flt-|eps| <= result <= flt+|eps|). If the optional argument `eps` is not given, it will be chosen automatically.
```0.3.rationalize #=> (3/10)
1.333.rationalize #=> (1333/1000)
1.333.rationalize(0.01) #=> (4/3)
```
See also `Float#to_r`.
```static VALUE
float_rationalize(int argc, VALUE *argv, VALUE self)
{
double d = RFLOAT_VALUE(self);
VALUE rat;
int neg = d < 0.0;
if (neg) self = DBL2NUM(-d);
if (rb_check_arity(argc, 0, 1)) {
rat = rb_flt_rationalize_with_prec(self, argv[0]);
}
else {
rat = rb_flt_rationalize(self);
}
if (neg) RATIONAL_SET_NUM(rat, rb_int_uminus(RRATIONAL(rat)->num));
return rat;
}```
round(ndigits = 0, half: :up]) → integer or float click to toggle source
Returns `self` rounded to the nearest value with a precision of `ndigits` decimal digits.
When `ndigits` is non-negative, returns a float with `ndigits` after the decimal point (as available):
```f = 12345.6789
f.round(1) # => 12345.7
f.round(3) # => 12345.679
f = -12345.6789
f.round(1) # => -12345.7
f.round(3) # => -12345.679
```
When `ndigits` is negative, returns an integer with at least `ndigits.abs` trailing zeros:
```f = 12345.6789
f.round(0) # => 12346
f.round(-3) # => 12000
f = -12345.6789
f.round(0) # => -12346
f.round(-3) # => -12000
```
If keyword argument `half` is given, and `self` is equidistant from the two candidate values, the rounding is according to the given `half` value:
• `:up` or `nil`: round away from zero:
```2.5.round(half: :up) # => 3
3.5.round(half: :up) # => 4
(-2.5).round(half: :up) # => -3
```
• `:down`: round toward zero:
```2.5.round(half: :down) # => 2
3.5.round(half: :down) # => 3
(-2.5).round(half: :down) # => -2
```
• `:even`: round toward the candidate whose last nonzero digit is even:
```2.5.round(half: :even) # => 2
3.5.round(half: :even) # => 4
(-2.5).round(half: :even) # => -2
```
Raises and exception if the value for `half` is invalid.
Related: `Float#truncate`.
```static VALUE
flo_round(int argc, VALUE *argv, VALUE num)
{
double number, f, x;
VALUE nd, opt;
int ndigits = 0;
enum ruby_num_rounding_mode mode;
if (rb_scan_args(argc, argv, "01:", &nd, &opt)) {
ndigits = NUM2INT(nd);
}
mode = rb_num_get_rounding_option(opt);
number = RFLOAT_VALUE(num);
if (number == 0.0) {
return ndigits > 0 ? DBL2NUM(number) : INT2FIX(0);
}
if (ndigits < 0) {
return rb_int_round(flo_to_i(num), ndigits, mode);
}
if (ndigits == 0) {
x = ROUND_CALL(mode, round, (number, 1.0));
return dbl2ival(x);
}
if (isfinite(number)) {
int binexp;
frexp(number, &binexp);
if (float_round_overflow(ndigits, binexp)) return num;
if (float_round_underflow(ndigits, binexp)) return DBL2NUM(0);
if (ndigits > 14) {
/* In this case, pow(10, ndigits) may not be accurate. */
return rb_flo_round_by_rational(argc, argv, num);
}
f = pow(10, ndigits);
x = ROUND_CALL(mode, round, (number, f));
return DBL2NUM(x / f);
}
return num;
}```
to_d → bigdecimal click to toggle source
to_d(precision) → bigdecimal
Returns the value of `float` as a `BigDecimal`. The `precision` parameter is used to determine the number of significant digits for the result. When `precision` is set to `0`, the number of digits to represent the float being converted is determined automatically. The default `precision` is `0`.
```require 'bigdecimal'
require 'bigdecimal/util'
0.5.to_d # => 0.5e0
1.234.to_d # => 0.1234e1
1.234.to_d(2) # => 0.12e1
```
See also `BigDecimal::new`.
```# File ext/bigdecimal/lib/bigdecimal/util.rb, line 50
def to_d(precision=0)
BigDecimal(self, precision)
end```
to_f → self click to toggle source
Since `float` is already a `Float`, returns `self`.
```# File numeric.rb, line 334
def to_f
self
end```
to_i → integer click to toggle source
Returns `self` truncated to an `Integer`.
```1.2.to_i # => 1
(-1.2).to_i # => -1
```
Note that the limited precision of floating-point arithmetic may lead to surprising results:
```(0.3 / 0.1).to_i # => 2 (!)
```
`Float#to_int` is an alias for `Float#to_i`.
```static VALUE
flo_to_i(VALUE num)
{
double f = RFLOAT_VALUE(num);
if (f > 0.0) f = floor(f);
if (f < 0.0) f = ceil(f);
return dbl2ival(f);
}```
Also aliased as: to_int
to_int()
Returns `self` truncated to an `Integer`.
```1.2.to_i # => 1
(-1.2).to_i # => -1
```
Note that the limited precision of floating-point arithmetic may lead to surprising results:
```(0.3 / 0.1).to_i # => 2 (!)
```
`Float#to_int` is an alias for `Float#to_i`.
Alias for: to_i
to_r → rational click to toggle source
Returns the value as a rational.
```2.0.to_r #=> (2/1)
2.5.to_r #=> (5/2)
-0.75.to_r #=> (-3/4)
0.0.to_r #=> (0/1)
0.3.to_r #=> (5404319552844595/18014398509481984)
```
NOTE: 0.3.to_r isn’t the same as “0.3”.to_r. The latter is equivalent to “3/10”.to_r, but the former isn’t so.
```0.3.to_r == 3/10r #=> false
"0.3".to_r == 3/10r #=> true
```
See also `Float#rationalize`.
```static VALUE
float_to_r(VALUE self)
{
VALUE f;
int n;
float_decode_internal(self, &f, &n);
if (n == 0)
return rb_rational_new1(f);
if (n > 0)
return rb_rational_new1(rb_int_lshift(f, INT2FIX(n)));
n = -n;
return rb_rational_new2(f, rb_int_lshift(ONE, INT2FIX(n)));
#else
if (RB_TYPE_P(f, T_RATIONAL))
return f;
return rb_rational_new1(f);
#endif
}```
to_s → string click to toggle source
Returns a string containing a representation of `self`; depending of the value of `self`, the string representation may contain:
• A fixed-point number.
• A number in “scientific notation” (containing an exponent).
• ‘Infinity’.
• ‘-Infinity’.
• ‘NaN’ (indicating not-a-number).
3.14.to_s # => “3.14” (10.1**50).to_s # => “1.644631821843879e+50” (10.1**500).to_s # => “Infinity” (-10.1**500).to_s # => “-Infinity” (0.0/0.0).to_s # => “NaN”
```static VALUE
flo_to_s(VALUE flt)
{
enum {decimal_mant = DBL_MANT_DIG-DBL_DIG};
enum {float_dig = DBL_DIG+1};
char buf[float_dig + roomof(decimal_mant, CHAR_BIT) + 10];
double value = RFLOAT_VALUE(flt);
VALUE s;
char *p, *e;
int sign, decpt, digs;
if (isinf(value)) {
static const char minf[] = "-Infinity";
const int pos = (value > 0); /* skip "-" */
return rb_usascii_str_new(minf+pos, strlen(minf)-pos);
}
else if (isnan(value))
return rb_usascii_str_new2("NaN");
p = ruby_dtoa(value, 0, 0, &decpt, &sign, &e);
s = sign ? rb_usascii_str_new_cstr("-") : rb_usascii_str_new(0, 0);
if ((digs = (int)(e - p)) >= (int)sizeof(buf)) digs = (int)sizeof(buf) - 1;
memcpy(buf, p, digs);
xfree(p);
if (decpt > 0) {
if (decpt < digs) {
memmove(buf + decpt + 1, buf + decpt, digs - decpt);
buf[decpt] = '.';
rb_str_cat(s, buf, digs + 1);
}
else if (decpt <= DBL_DIG) {
long len;
char *ptr;
rb_str_cat(s, buf, digs);
rb_str_resize(s, (len = RSTRING_LEN(s)) + decpt - digs + 2);
ptr = RSTRING_PTR(s) + len;
if (decpt > digs) {
memset(ptr, '0', decpt - digs);
ptr += decpt - digs;
}
memcpy(ptr, ".0", 2);
}
else {
goto exp;
}
}
else if (decpt > -4) {
long len;
char *ptr;
rb_str_cat(s, "0.", 2);
rb_str_resize(s, (len = RSTRING_LEN(s)) - decpt + digs);
ptr = RSTRING_PTR(s);
memset(ptr += len, '0', -decpt);
memcpy(ptr -= decpt, buf, digs);
}
else {
goto exp;
}
return s;
exp:
if (digs > 1) {
memmove(buf + 2, buf + 1, digs - 1);
}
else {
buf[2] = '0';
digs++;
}
buf[1] = '.';
rb_str_cat(s, buf, digs + 1);
rb_str_catf(s, "e%+03d", decpt - 1);
return s;
}```
Also aliased as: inspect
truncate(ndigits = 0) → float or integer click to toggle source
Returns `self` truncated (toward zero) to a precision of `ndigits` decimal digits.
When `ndigits` is positive, returns a float with `ndigits` digits after the decimal point (as available):
```f = 12345.6789
f.truncate(1) # => 12345.6
f.truncate(3) # => 12345.678
f = -12345.6789
f.truncate(1) # => -12345.6
f.truncate(3) # => -12345.678
```
When `ndigits` is negative, returns an integer with at least `ndigits.abs` trailing zeros:
```f = 12345.6789
f.truncate(0) # => 12345
f.truncate(-3) # => 12000
f = -12345.6789
f.truncate(0) # => -12345
f.truncate(-3) # => -12000
```
Note that the limited precision of floating-point arithmetic may lead to surprising results:
```(0.3 / 0.1).truncate #=> 2 (!)
```
Related: `Float#round`.
```static VALUE
flo_truncate(int argc, VALUE *argv, VALUE num)
{
if (signbit(RFLOAT_VALUE(num)))
return flo_ceil(argc, argv, num);
else
return flo_floor(argc, argv, num);
}```
zero? → true or false click to toggle source
Returns `true` if `float` is 0.0.
```# File numeric.rb, line 378
def zero?
Primitive.attr! 'inline'
Primitive.cexpr! 'RBOOL(FLOAT_ZERO_P(self))'
end```
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# Will the buoyant force be more or less when the body displaces greater volume of liquid?give reason?
2
by meenakshiram
2014-11-27T20:51:49+05:30
I think it will be less.Hence more volume of body will submerge due to less buoyant force so yet it will make diiference
2014-11-27T22:18:13+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Buoyancy force is equal to the weight of liquid or fluid displaced. Weight is equal to Volume of fluid multiplied by density. Hence the force is directly proportional to the volume of liquid displaced.
Force = W = M g = Volume * density * g
Hence the force is more when a body displaces more liquid
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# LENGTH Units Conversionhimetrics to millimeters
1 Himetrics
= 0.01 Millimeters
Embed this to your website/blog
Category: length
Conversion: Himetrics to Millimeters
The base unit for length is meters (SI Unit)
[Himetrics] symbol/abbrevation: (hiMetric)
[Millimeters] symbol/abbrevation: (mm)
How to convert Himetrics to Millimeters (hiMetric to mm)?
1 hiMetric = 0.01 mm.
1 x 0.01 mm = 0.01 Millimeters.
Always check the results; rounding errors may occur.
Definition:
The millimeter (symbol = mm), or millimetre (British spelling) is a unit of length in the metric system, equal to one thousandth of a meter, which is the SI base unit ..more definition+
In relation to the base unit of [length] => (meters), 1 Himetrics (hiMetric) is equal to 1.0E-5 meters, while 1 Millimeters (mm) = 0.001 meters.
1 Himetrics to common length units
1 hiMetric = 1.0E-5 meters (m)
1 hiMetric = 1.0E-8 kilometers (km)
1 hiMetric = 0.001 centimeters (cm)
1 hiMetric = 3.2808398950131E-5 feet (ft)
1 hiMetric = 0.00039370078740157 inches (in)
1 hiMetric = 1.0936132983377E-5 yards (yd)
1 hiMetric = 6.2137119223733E-9 miles (mi)
1 hiMetric = 1.056970721911E-21 light years (ly)
1 hiMetric = 0.037795280352161 pixels (PX)
1 hiMetric = 6.25E+29 planck length (pl)
Himetricsto Millimeters (table conversion)
1 hiMetric = 0.01 mm
2 hiMetric = 0.02 mm
3 hiMetric = 0.03 mm
4 hiMetric = 0.04 mm
5 hiMetric = 0.05 mm
6 hiMetric = 0.06 mm
7 hiMetric = 0.07 mm
8 hiMetric = 0.08 mm
9 hiMetric = 0.09 mm
10 hiMetric = 0.1 mm
20 hiMetric = 0.2 mm
30 hiMetric = 0.3 mm
40 hiMetric = 0.4 mm
50 hiMetric = 0.5 mm
60 hiMetric = 0.6 mm
70 hiMetric = 0.7 mm
80 hiMetric = 0.8 mm
90 hiMetric = 0.9 mm
100 hiMetric = 1 mm
200 hiMetric = 2 mm
300 hiMetric = 3 mm
400 hiMetric = 4 mm
500 hiMetric = 5 mm
600 hiMetric = 6 mm
700 hiMetric = 7 mm
800 hiMetric = 8 mm
900 hiMetric = 9 mm
1000 hiMetric = 10 mm
2000 hiMetric = 20 mm
4000 hiMetric = 40 mm
5000 hiMetric = 50 mm
7500 hiMetric = 75 mm
10000 hiMetric = 100 mm
25000 hiMetric = 250 mm
50000 hiMetric = 500 mm
100000 hiMetric = 1000 mm
1000000 hiMetric = 10000 mm
1000000000 hiMetric = 10000000 mm
(Himetrics) to (Millimeters) conversions
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http://de.metamath.org/mpegif/nfreud.html
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > nfreud Structured version Visualization version Unicode version
Theorem nfreud 2962
Description: Deduction version of nfreu 2964. (Contributed by NM, 15-Feb-2013.) (Revised by Mario Carneiro, 8-Oct-2016.)
Hypotheses
Ref Expression
nfreud.1
nfreud.2
nfreud.3
Assertion
Ref Expression
nfreud
Proof of Theorem nfreud
StepHypRef Expression
1 df-reu 2743 . 2
2 nfreud.1 . . 3
3 nfcvf 2614 . . . . . 6
43adantl 468 . . . . 5
5 nfreud.2 . . . . . 6
65adantr 467 . . . . 5
74, 6nfeld 2599 . . . 4
8 nfreud.3 . . . . 5
98adantr 467 . . . 4
107, 9nfand 2007 . . 3
112, 10nfeud2 2310 . 2
121, 11nfxfrd 1696 1
Colors of variables: wff setvar class Syntax hints: wn 3 wi 4 wa 371 wal 1441 wnf 1666 wcel 1886 weu 2298 wnfc 2578 wreu 2738 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1668 ax-4 1681 ax-5 1757 ax-6 1804 ax-7 1850 ax-10 1914 ax-11 1919 ax-12 1932 ax-13 2090 ax-ext 2430 This theorem depends on definitions: df-bi 189 df-an 373 df-tru 1446 df-ex 1663 df-nf 1667 df-eu 2302 df-cleq 2443 df-clel 2446 df-nfc 2580 df-reu 2743 This theorem is referenced by: nfreu 2964
Copyright terms: Public domain W3C validator
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# What is Melting Point of Materials – Definition
Melting Point of Materials. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.
Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.
Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.
## Melting Point of Materials
Note that, these points are associated with the standard atmospheric pressure.
In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. Adding a heat will convert the solid into a liquid with no temperature change. At the melting point the two phases of a substance, liquid and vapor, have identical free energies and therefore are equally likely to exist. Below the melting point, the solid is the more stable state of the two, whereas above the liquid form is preferred. The melting point of a substance depends on pressure and is usually specified at standard pressure. When considered as the temperature of the reverse change from liquid to solid, it is referred to as the freezing point or crystallization point.
The first theory explaining mechanism of melting in the bulk was proposed by Lindemann, who used vibration of atoms in the crystal to explain the melting transition. Solids are similar to liquids in that both are condensed states, with particles that are far closer together than those of a gas. The atoms in a solid are tightly bound to each other, either in a regular geometric lattice (crystalline solids, which include metals and ordinary ice) or irregularly (an amorphous solid such as common window glass), and are typically low in energy. The motion of individual atoms, ions, or molecules in a solid is restricted to vibrational motion about a fixed point. As a solid is heated, its particles vibrate more rapidly as the solid absorbs kinetic energy. At some point the amplitude of vibration becomes so large that the atoms start to invade the space of their nearest neighbors and disturb them and the melting process initiates. The melting point is the temperature at which the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid.
As with boiling points, the melting point of a solid is dependent on the strength of those attractive forces. For example, sodium chloride (NaCl) is an ionic compound that consists of a multitude of strong ionic bonds. Sodium chloride melts at 801°C. On the other hand, ice (solid H2O) is a molecular compound whose molecules are held together by hydrogen bonds, which is effectively a strong example of an interaction between two permanent dipoles. Though hydrogen bonds are the strongest of the intermolecular forces, the strength of hydrogen bonds is much less than that of ionic bonds. The melting point of ice is 0 °C.
Covalent bonds often result in the formation of small collections of better-connected atoms called molecules, which in solids and liquids are bound to other molecules by forces that are often much weaker than the covalent bonds that hold the molecules internally together. Such weak intermolecular bonds give organic molecular substances, such as waxes and oils, their soft bulk character, and their low melting points (in liquids, molecules must cease most structured or oriented contact with each other).
References:
Materials Science:
1. U.S. Department of Energy, Material Science. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
2. U.S. Department of Energy, Material Science. DOE Fundamentals Handbook, Volume 2 and 2. January 1993.
3. William D. Callister, David G. Rethwisch. Materials Science and Engineering: An Introduction 9th Edition, Wiley; 9 edition (December 4, 2013), ISBN-13: 978-1118324578.
4. Eberhart, Mark (2003). Why Things Break: Understanding the World by the Way It Comes Apart. Harmony. ISBN 978-1-4000-4760-4.
5. Gaskell, David R. (1995). Introduction to the thermodynamics of Materials (4th ed.). Taylor and Francis Publishing. ISBN 978-1-56032-992-3.
6. González-Viñas, W. & Mancini, H.L. (2004). An Introduction to Materials Science. Princeton University Press. ISBN 978-0-691-07097-1.
7. Ashby, Michael; Hugh Shercliff; David Cebon (2007). Materials: engineering, science, processing and design (1st ed.). Butterworth-Heinemann. ISBN 978-0-7506-8391-3.
8. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
## See above:
Thermal Properties
We hope, this article, Melting Point of Materials, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about materials and their properties.
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physics
posted by .
A spinning flywheel has rotational inertia I = 402.9 kgm2. Its angular velocity decreases from 20.3 rad/s to zero in 266.6 s due to friction. What is the frictional torque acting?
Torque= MomentofInertia*angularacceleration.
You are given the moment of inertia (you named it rotational inertia). Angular acceleration is (20.3/266.3) rad/sec^2
Solve for torque.
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# copeg
#### The Monty Hall Problem
###### Rate this Entry
by , May 16th, 2011 at 07:29 PM (24892 Views)
The Monty Hall problem is a statistical problem which originates from the television Game show Lets Make a Deal, hosted by Monty Hall. The game is simple: a contestant is presented 3 closed doors, behind one of which is a valuable prize (oftentimes described as a car, whereas the other doors have goats behind them). A contestant chooses a door. The host then opens one of the doors you did not choose, which does NOT contain the prize. Then the host asks - do you want to change your decision? What is the probability you will win if you choose to change doors? What is the probability you win if you choose to remain with your decision.
Simplistically, one might guess your probability of winning went from 1/3, to 1/2. This however, is incorrect. If you stay, your probability of winning is still 1/3. If you change, your chances of winning is 2/3...How does this counter-intuitive result play out? One can tabulate all the possibilities, and the contestants decision:
```Door1 Door 2 Door 3 Switch Stay
Prize Nothing Nothing Prize Nothing
Nothing Prize Nothing Nothing Prize
Nothing Nothing Prize Nothing Prize```
Looking at the chart carefully*: if you switch your chances go from 1/3 to 2/3 of winning. Still not convinced? Below is a simple simulation of this problem. This iterates through a game if n number of doors a defined number of times, tallying up whether you win or loose based upon staying
or changing...and the simulation backs everything up - average for 3 doors: stay: 33%, change 66%.
```
import java.util.Random;
/**
* Simulates the MontyHall problem. 3 doors, 2 with goats and 1 with car. You choose a door,
* Monty hall opens one of the other two to reveal a goat. How often will you be correct if you
* stay? How often if you switch?
* @author copeg
*
*/
public class MontyHallSimulation implements Runnable{
/*Random number generator */
private static final Random RANDOM = new Random();
/*Number of rounds to simulate*/
private int rounds;
/*Number of doors total*/
private int doors;
/*Rate for staying*/
private double stayRate = 0;
/*Rate for changing*/
private double changeRate = 0;
/**
* Constructs a MontyHallSimulation with 3 doors, to iterate 1000 times.
*/
public MontyHallSimulation(){
this(1000);
}
/**
* Constructs a MontyHallSimulation with the number of rounds to simulate, and
* 3 doors.
* @param rounds The number of rounds to simulate.
*/
public MontyHallSimulation(int rounds){
this(rounds, 3);
}
/**
* Constructs a MontyHallSimulation with the number of rounds and doors to use in
* the simulation.
* @param rounds The number of rounds to simulate
* @param doors The number of doors to use in the simulation.
* @throws IllegalArgumentException if doors is less than 3.
*/
public MontyHallSimulation(int rounds, int doors){
if ( doors < 3 ){
throw new IllegalArgumentException("Cannot simulate the problem with less than 3 doors.");
}
this.rounds = rounds;
this.doors = doors;
}
/**
* Implementation of the Runnable interface. Simulates the Monty Hall Problem.
* This loops doors number of times, determining whether staying or changing
* results in a correct answer.
*/
public void run(){
int stayCount = 0;
int changeCount = 0;
for ( int i = 0; i < rounds; i++ ){
int choose = RANDOM.nextInt(doors);//choose a door at random
int solution = RANDOM.nextInt(doors);//find a random place where the car will be.
if ( solution != choose ){//Car is in the other door - if you change you win
changeCount++;
}else{//If you stay you win.
stayCount++;
}
}
stayRate = stayCount/(double)rounds;
changeRate = changeCount/(double)rounds;
}
/**
* Retrieves the rate one will be correct if one stays. This method returns
* zero unless run has been called.
* @return
*/
public double getStayRate(){
return stayRate;
}
/**
* Retrieves the rate one will be correct if one changes. This method returns
* zero unless run has been called.
* @return
*/
public double getChangeRate(){
return changeRate;
}
/**
* Application entry point.
* @param args
*/
public static void main(String[] args){
MontyHallSimulation sim = new MontyHallSimulation(1000, 1000);
sim.run();
System.out.println("Choose to stay: percent correct - " + sim.getStayRate());
System.out.println("Choose to change: percent corect - " + sim.getChangeRate());
}
}```
How about the same problem with 1000 doors? You are virtually guaranteed to win if you change your decision (what are the chances you chose the right door in the first place?).
A fun statistical problem to investigate and simulate...happy coding!
*The chart does not translate well in this format, however the link below contains a better, more readable version to inspect.
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https://www.surveymonkey.com/curiosity/randomish-sampling/?ut_source=mp&ut_source2=survey-guidelines&ut_source3=inline
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Blog results
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Stay curious! You'll find something.
Survey Science
# Random(ish) sampling: balancing the ideal and the real
So you’ve got your survey hot off the presses and ready to launch. There’s just one problem…who do you send it to?
Past blog posts have talked about what a “sample” of a population is and how your sample can affect your results. But let’s back up a minute and talk about how you should be getting your sample in the first place. The best, and most ideal, way to sample is to randomly sample your respondents.
Random sampling is a way to sample in which everyone in the population has a chance of being chosen for the sample, and whoever’s picked is chosen completely at random. This is great because there’s no bias – some people aren’t more likely to be picked than others. Plus, anyone from your population could conceivably be picked for your sample, so you can be more confident your results match what your population really thinks.
The simplest way to understand random sampling is to think of someone pulling slips of paper at random out of a hat. Every slip of paper in the hat has an equal chance of being plucked out. So if every slip of paper has a name on it, every name has an equal chance of getting picked. That means that it is “random” which names get picked.
You could…
1. Approach random people on a random selection of streets at random times.
2. Call randomly generated phone numbers at random times of the day.
3. Mail out letters to randomly selected addresses from randomly selected regions.
4. Email an online survey to randomly generated email addresses.
But, wait, there are some problems with this…
Let’s go back to our hat example. Do all the names in the hat really have an equal chance of being picked? Well…have you ever picked a slip of paper out of a hat? No one takes the slip of paper right on the top! You stick your hand way in there and grab one from at least the middle. So the papers on the top might not really have a fair shot at getting picked.
There are similar problems with surveying…
1. Approaching people randomly is not only a little creepy but also unlikely to actually be random. Especially because the person you have doing the actual approaching is going to bias who says yes to the survey and potentially even the answers they give. (Don’t believe me? Try it yourself. Call up Kanye West and ask him to hand out a survey on one street—and then have your grandma hand out a survey on another street. Watch what happens.)
2. People who answer phone calls and letters are a very particular subset of people these days and these methods may not be able to estimate some types of behavior.
3. Methods 1-3 can also be very time consuming and very expensive and you’ve got yourself in a bit of a pickle.
4. Going the email route, chances are you’ll be sent directly to a SPAM folder and people will never see your email at all. Chances are you’ll also get lots of angry emails about spamming random strangers.
In a perfect world, everyone in our population would have a chance of being picked in our sample—in practice, that is virtually impossible. With surveys, as with many other things in life, you’ll need to strike a balance between what is ideal and what is possible in the real world. We like to call it “random(ish) sampling.” Here are some suggestions we have for random(ish) sampling…
1. If you have a concrete “sampling frame” (a list of the names of the people in your population), you can use a random number generator to select which people to pick to survey from your list. (Want help with this? See how to do this in Excel.)
2. If you don’t have a concrete sampling frame, you could vary who is approaching strangers on the street to ask them to fill out the survey. Have Kanye and your grandma go survey collecting together—and have them bring some friends to help too. The biases that each of them bring to data collection are likely to balance out.
3. Send your survey out more than one way. Mail it, call people, send it online, shout it from the rooftops! Okay the last one was a joke, but the diversity of recruiting methods can help get different types of people to respond.
4. Post your online survey on different types of websites where different types of people might see it. People who read political blogs may be different people from those who read fashion blogs. This is a nice alternative to the mass email that is likely to hit everyone’s spam filter. You can cast a wide net to draw in a diverse crowd without clogging anyone’s inbox.
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## Syllabus
#### Textbook
Thomas' Calculus: Early Transcendentals, 14th edition, by Hass, Heil, and Weir.
#### Description:
An introduction to calculus for engineering, science, and mathematics students, with an emphasis on conceptual understanding, problem solving, and modeling. Topics covered include: limits, continuity, derivatives of algebraic, trigonometric, and transcendental functions, applications of the derivative including optimization problems and linear approximations, antiderivatives, introduction to the definite integral, and the Fundamental Theorem of Calculus. Prerequisite: MATH 147, minimum grade: C.
#### Outcomes:
Students will develop facility with the language and techniques of calculus and analytic geometry. Students will work with the formal definitions of a limit, continuity, and the derivative. Students will use limit laws, theorems about continuity, and differentiation rules to take limits, determine when/where a function is continuous, and evaluate derivatives. Students will learn to evaluate integrals using the fundamental theorem of calculus and substitution. Students will use these techniques to solve problems coming from other disciplines. Students will also learn to create sound mathematical arguments in the context of this class, particularly in determining when and where limits exist and when and where a function is continuous. Students will also develop their skills at mathematical writing as they communicate increasingly complex solutions.
#### Outcomes for a Core Mathematics Course:
• Students will be able to reproduce and create logical mathematical arguments. (A)
• Students will be able to perform calculations appropriate to the content of this mathematics course. (C)
• Students will be able to communicate mathematics in writing. (C)
• Students will be able to apply mathematics to problems in other disciplines. (A)
These learning outcomes support the University Core curriculum learning outcomes:
A) Students will be able to use the basic modes of inquiry and expression of the disciplines that represent liberal education. C) Students will be able to communicate clearly and persuasively, using ideas and arguments based on evidence, logic, and critical thinking.
#### Grades:
Grades will be based on scores on exams, WeBWorK, and worksheets. There will be four exams during the semester, each worth 10% of the final grade. In addition, a cumulative final exam will count for 20% of the grade. The remaining 40% of the grade will come from scores on WeBWorK assignments (20%) and worksheets (20%). An approximate schedule for the semester is on the course website. No extra credit will be given. Final grades will be assigned using the following scale (with + or - for the top and bottom scores within appropriate ranges):
Score Grade
90-100 A
80-90 B
70-80 C
60-70 D
0-60 F
#### Homework:
Graded homework sets will be assigned using a free online system called WeBWorK. You will have about one WeBWorK assignment each week. Solutions are automatically and instantaneously checked and you are allowed to retry each problem as often as you want. You should take advantage of this to get a perfect score on the WeBWorK.
A list of additional suggested exercises in the textbook is posted on the course web page. These problems will not be collected or graded, however, math is a skill (like playing a musical instrument) and practice is the only way to build that skill. In general, I try to make the WeBWorK assignments short (in terms of number of problems) but challenging. This means I avoid some of the routine problems that are important for building facility with a new technique. Instead, those problems are in the suggested exercises. I expect you to do enough of those problems to feel comfortable solving the more difficult problems on the WeBWorK. It is up to you to determine how much practice is enough (I think more is always better, but you're the expert on how you learn, so I'm leaving it up to you to decide).
#### Worksheets:
I will try to provide a worksheet for each section of the book. These worksheets generally have two goals:
• Work with the ideas behind the section;
• Solve example problems.
Questions in the first category will be graded for completion only while others will be scored for accuracy. I'll provide some class time for worksheets, but you'll also have to work at home. You're encouraged to collaborate on the worksheets, but anything you turn in should reflect your own understanding of the solution (otherwise my feedback isn't helpful to you). All worksheets and deadlines will be posted on the course web page.
#### Exams:
Exams encourage you to review, practice, and refine your calculus skills. My goal is to make exams long enough to cover the relevant material, but short enough that it doesn't take me forever to grade and return them. Examples of past exams can be found on the old editions of the course web page.
#### Math Lab:
Help on the homework or any other class material may be available via the Math Lab: Math Lab.
#### Harassment, non-discrimination, and sexual misconduct:
Consistent with its mission, Gonzaga seeks to assure that all community members learn and work in a welcoming and inclusive environment. Title VII, Title IX and Gonzaga's policy prohibit gender-based harassment, discrimination and sexual misconduct. Gonzaga encourages anyone experiencing gender-based harassment, discrimination or sexual misconduct to talk to someone from the Campus and Local Resources list found in the Harassment and Non-Discrimination Policy.
It may be helpful to talk about what happened in order to get the support needed and for Gonzaga to respond appropriately. There are options for support and resolution, namely confidential support resources, and campus reporting and support options available. Gonzaga will respond to all reports of sexual misconduct in order to stop the harassment, discrimination, or misconduct, prevent its recurrence and address its effects. Responses may vary from support service referrals to formal investigations.
As a faculty member, I want get you connected to the resources here on campus specially trained in and experienced in assisting in such complaints, and therefore I will report all incidents of gender-based harassment, discrimination and sexual misconduct to Title IX (in fact, I am required to report such incidents). A representative from that office will reach out to you via phone and/or email to explore options for support, safety measures and reporting. I will provide our Title IX Director with all relevant details, including names and identifying information, of the information reported. For more information about policies and resources or reporting options, please visit the following websites: Equity and Inclusion and Title IX. If you would like to directly make a report of harassment, discrimination or sexual misconduct directly, you may fill out an online Sexual Misconduct Report Form or contact the Title IX Director by phone, email, or in person:
Stephanie N. Thomas Title IX Director 509-313-6910 whaleys@gonzaga.edu Business Services Building 018
#### Notice to students with disabilities and/or medical conditions:
The Americans with Disabilities Act (ADA) is a federal anti-discrimination statute that provides comprehensive civil rights protection for persons with disabilities. Among other things, this legislation requires that all students with disabilities be guaranteed a learning environment that provides for reasonable accommodation of their disabilities. If you believe you have a disability/medical condition requiring an accommodation, please call or visit the Disability Access Office (second floor of Foley Center Library, Room 208.)
#### Attendance:
Gonzaga University presumes that students have sufficient maturity to recognize their responsibility for regular class engagement, and attendance is a general expectation no matter the teaching modality. However, in order to prioritize the health and safety of all community members, Gonzaga's regular in-person attendance policy may be modified (Amended Class Attendance Policy). I will record attendance solely for the purposes of contact tracing. When we do meet in person, you must follow COVID-related protocols as described in the Student Arrival & Return to Gonzaga Guides, and you will in no way be penalized for following these protocols; see Amended Class Attendance Policy. If you become sick or need to miss class for COVID-related reasons, I will work with you to help you catch up. Don't ever come to class if you are feeling sick. Please communicate directly with me regarding any absences, if possible before they occur.
#### A note on recorded meetings:
Our class sessions might be recorded for the benefit of students who are unable to attend in-person. Only the instructor may cause a class meeting to be recorded for those students. You shall not make audio or video recordings of class meetings without the prior written authorization of the instructor. By remaining registered in this course, you agree to your voice and image being recorded, and you agree to use any recordings of our class meetings ONLY for the educational purposes of this class (or other sections of this class taught by the same instructor). You agree to delete recordings of our class meetings no later than the end of this semester. You do not have permission to use or share recordings (video or audio) of our class meetings beyond the reach of our class for any purpose, including, but not limited to, posting to any digital application or platform, such as social media. You may not duplicate or distribute recordings of class sessions. In short, your instructor and your classmates intend to appear in these videos only for the purposes of carrying out our teaching and learning in this class. Your compliance with the terms of this syllabus regarding use of class session recordings is subject to the Student Code of Conduct; violations will be reviewed according to the provisions in the Administration of Student Code of Conduct.
#### FERPA and Privacy:
Under FERPA (Family Educational Rights and Privacy Act), your student records are confidential and protected. Under most circumstances your records will not be released without your written and signed consent; exception includes some directory information. Instructors are not allowed to publicly post grades by student name, social security number, GU student identification number, or any other identifiable means, without written consent from students involved. The FERPA policy does not apply to third party online applications that may be used in courses (i.e. WeBWorK and Gradescope) such that it is the student's responsibility to read the privacy documentation at each website.
#### Academic integrity:
All members of the Gonzaga community are expected to adhere to principles of honesty and integrity in their academic endeavors, and I will abide strictly by procedures and guidelines of the University's Academic Integrity Policy. Students and faculty are governed by this policy, and I encourage you to familiarize yourself with its scope and procedures. Ignorance of the policy will not serve as a defense against any violations.
#### Religious Accommodations for Students
In compliance with Washington State law (RCW 28.10.039), it is the policy of Gonzaga University to reasonably accommodate students who, due to the observance of religious holidays, expect to be absent or endure a significant hardship during certain days of their academic course or program. The Policy on Religious Accommodations for Students describes procedures for students requesting a Religious Accommodation and for faculty responding to such a request.
#### Course evaluation:
At Gonzaga, we take teaching seriously, and we ask our students to evaluate their courses and instructors so that we can provide the best possible learning experience. In that spirit, we ask students to give us feedback on their classroom experience near the end of the semester. I will ask you to take a few minutes then to carry out course/instructor evaluation in class. Please know that I appreciate your participation in this process. This is a vital part of our efforts at Gonzaga to improve continually our teaching, our academic programs, and our entire educational effort.
## Office hours
• Monday 11-12
• Tuesday 9-10 in the Math Learning Center and 12:40-1:40
• Wednesday 11-12
• Friday 11-12
• Or by appointment
Logan Axon
Department of Mathematics
MSC 2615
Gonzaga University
Spokane, WA 99258
Office: Herak 307A
Phone: 509.313.3897
Email: axon@gonzaga.edu
Last updated 9/8/2021
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A206246 Numbers n such that the greatest prime divisor p of n^2+1 has the property that (p - n)^2 + 1 = p. 0
1, 3, 7, 13, 21, 31, 43, 91, 111, 183, 211, 241, 273, 381, 421, 553, 601, 651, 703, 1261, 1333, 1561, 1641, 2863, 2971, 3081, 3193, 4291, 4423, 5403, 5551, 6973, 7141, 8011, 8191, 8743, 8931, 11991, 12211, 13341, 13573, 14281, 14521, 15253, 15501, 15751, 16003 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS For the n > 1 in this sequence, n^2+1 is composite. The corresponding primes p are A002496(n) repeated two times for n > 1 : {2, 5, 5, 17, 17, 37, 37, 101, 101, 197,...}. Because this sequence is connected with A002496, it is conjectured that the set of this numbers is infinite. LINKS EXAMPLE 31 is in the sequence because 31^2 + 1 = 2*13*37 and (37 - 31)^2 + 1 = 37. 43 is in the sequence because 43^2 + 1 = 2*5*5*37 and (37 - 43)^2 + 1 = 37. MAPLE with(numtheory):for n from 1 to 20000 do:x:=n^2+1:y:=factorset(x):n1:=nops(y):p:=y[n1]:q:=(p-n)^2+1:if q=p then printf(`%d, `, n): else fi:od: MATHEMATICA pn2pQ[n_]:=Module[{p=FactorInteger[n^2+1][[-1, 1]]}, (p-n)^2+1==p]; Select[ Range[20000], pn2pQ] (* Harvey P. Dale, Nov 20 2019 *) CROSSREFS Cf. A002496, A134406. Sequence in context: A002061 A247890 A063541 * A171965 A011898 A098577 Adjacent sequences: A206243 A206244 A206245 * A206247 A206248 A206249 KEYWORD nonn AUTHOR Michel Lagneau, Feb 05 2012 STATUS approved
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Last modified February 25 08:21 EST 2020. Contains 332221 sequences. (Running on oeis4.)
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https://www.neetprep.com/questions/1843-Physics/703-Nuclei?courseId=1277&testId=1816999-AR--Type-MCQs
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Given below are two statements:
Assertion (A): The density of nucleus is much higher than that of ordinary matter. Reason (R): Most of the mass of the atom is concentrated in the nucleus while the size of this nucleus is almost $$10^5$$ times smaller.
1 Both (A) and (R) are True and (R) is the correct explanation of (A). 2 Both (A) and (R) are True but (R) is not the correct explanation of (A). 3 (A) is True but (R) is False. 4 Both (A) and (R) are False.
Subtopic: Â Nucleus |
From NCERT
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
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NEET 2023 - Target Batch - Aryan Raj Singh
Graph given below shows the variation of binding energy per nucleon, Ebn as a function of mass number. For nuclei with mass number A such that, 30 < A < 170, Ebn is almost constant because nuclear forces are:
1. Short-ranged
2. Medium-ranged
3. Long-ranged
4. None of the above
Subtopic: Â Nuclear Binding Energy |
 71%
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh
Graph gives the variation of potential energy of a pair of nucleons as a function of their separation, r. From the graph it can be concluded that the force between nucleons is attractive for distances:
1. Less than ro
2. Greater than ro
3. Less than $\frac{{r}_{o}}{2}$
4. Less than $\frac{{r}_{o}}{4}$
Subtopic: Â Nuclear Binding Energy |
 57%
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): Radioactive nuclei emits $$\beta^- -$$ particles. Reason (R): Electrons exist inside the nucleus.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 (A) is false but (R) is true.
Subtopic: Â Types of Decay |
 75%
From NCERT
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NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): Forces (nuclear) acting between proton-proton $$(\mathrm{f}_{\mathrm{pp}})$$, proton-neutron $$(\mathrm{f}_{\mathrm{pn}})$$ and neutron-neutron $$(\mathrm{f}_{\mathrm{nn}})$$ are such that $$\mathrm{f}_{\mathrm{pp}}<\mathrm{f}_{\mathrm{pn}}=\mathrm{f}_{\mathrm{nn}}$$. Reason (R): Electrostatic force of repulsion between two protons reduces nuclear forces between them.
1 Both (A) and (R) are True and (R) is the correct explanation of (A). 2 Both (A) and (R) are True but (R) is not the correct explanation of (A). 3 (A) is True but (R) is False. 4 Both (A) and (R) are False.
Subtopic: Â Nuclear Binding Energy |
 63%
From NCERT
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): The mass of $$\beta$$-particles when they are emitted is higher than the mass of electrons obtained by other means. Reason (R): $$\beta$$-particles and electrons, both are similar particles.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 Both (A) and (R) are false.
Subtopic: Â Types of Decay |
From NCERT
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NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): Electron capture occurs more than positron emission in a heavy nucleus. Reason (R): In a heavy nucleus, electrons are relatively close to the nucleus.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 Both (A) and (R) are false.
Subtopic: Â Types of Decay |
 64%
From NCERT
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): All nuclei are not of the same size. Reason (R): The size of the nucleus depends on atomic mass.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 Both (A) and (R) are false.
Subtopic: Â Nucleus |
 78%
From NCERT
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Given below are two statements:
Assertion (A): The ratio of time taken for light emission from an atom to that for the release of nuclear energy in fission is 1:100. Reason (R): Time taken for the light emission from an atom is of the order of 10-8s.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 Both (A) and (R) are false.
Subtopic: Â Nuclear Energy |
 57%
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
Given below are two statements:
Assertion (A): 90Sr from radioactive fallout from a nuclear bomb ends up in the bones of human beings through the milk consumed by them. It causes impairment of the production of red blood cells. Reason (R): The energetic $$\beta$$-particles emitted in the decay of 90Sr damage the bone marrow.
1 Both (A) and (R) are true and (R) is the correct explanation of (A). 2 Both (A) and (R) are true but (R) is not the correct explanation of (A). 3 (A) is true but (R) is false. 4 Both (A) and (R) are false.
.
Subtopic: Â Types of Decay |
 88%
From NCERT
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
| 1,975
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CC-MAIN-2024-22
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https://www.jiskha.com/questions/350274/balance-the-chemical-equation-given-below-and-determine-the-number-of-grams-of-mgo-needed
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# Chem
Balance the chemical equation given below, and determine the number of grams of MgO needed to produce 15.0 g of Fe2O3.
_MgO(s)+_Fe(s)->_Fe2O3(s)+_Mg(s)
1. a. 11.4 g
2. b. 3.78 g
3. c. 1.26 g
4. d. 0.0877 g
I got equation balanced 3,2,1,3 just having trouble on how to find number of grams needed
1. 👍 0
2. 👎 0
3. 👁 1,882
Convert grams Fe2O3 to moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles Fe2O3 to moles MgO.
Now convert moles MgO to grams. g = moles x molar mass.
1. 👍 0
2. 👎 0
2. Balance the chemical equation given below, and determine the number of grams of MgO needed to produce 15.0 g of Fe2O3.
1. 👍 0
2. 👎 0
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Copyright © University of Cambridge. All rights reserved.
## Three Neighbours
Take three numbers that are 'next door neighbours' when you count. These are called consecutive numbers.
Add them together.
What do you notice?
Take another three consecutive numbers and add them together.
What do you notice?
Can you prove that this is always true by looking carefully at one of your examples?
### Why do this problem?
This problem supports the development of the idea of generic proof with the children. This is a tricky concept to grasp but it draws attention to mathematical structures that are not often addressed at primary school level. It is possible that only very few children in the class may grasp the idea but this is still a worthwhile activity which provides opportunities for children to explore consecutive numbers and the relationship between them. Generic proof involves examining one example in detail to identify structures that will prove the general result. Proof is a fundamental idea in mathematics and in encouraging them to do this problem you will be helping them to behave like mathematicians.
By addressing the case of adding three consecutive numbers, a generic proof that adding three consecutive always gives an answer that is a multiple of three is developed based on the structure of one example. The article entitled Take One Example with its video clips will help you understand how this problem supports the development of the idea of generic proof with the children. Reading it will help you to see what is involved.
### Possible approach
Ask the children to choose three consecutive numbers and and add them together. It is probably easiest if they choose ones that are easy to model and numbers that they are secure with.
Suggest that they make a model of their numbers using apparatus that is widely available in the classroom. Resist pointing them in specific directions unless they become stuck. If they are stuck then resources such as Multilink cubes, Numicon or squared paper will be helpful.
The idea is that they take a particular example and then see if they can see the general structure within that one example.
### Key questions
How would you like to represent these numbers?
What do you notice about the answer?
Can you see anything in your example that would work in exactly the same way if you used three different consecutive numbers?
Can you say what will happen every time you add any three consecutive numbers?
Can you convince your friend that this is true?
### Possible extension
When adding three numbers there are a number of different combinations that are possible. Ask the children to explore what they are. Get them to identify the possible combinations and the features of those combinations that matter.
Does it matter whether the starting number is odd or even?
What would happen if we added four consecutive numbers? Or five? Or six? The possibilities are endless.
### Possible support
It may be helpful to return to Two Numbers Under the Microscope if the children are struggling with adding three numbers. This might help them to feel more comfortable with the rules they have proved in that problem and so build the foundations for this one.
The children may find it helpful to use representations of numbers such as these to support their thinking.
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CC-MAIN-2013-48
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https://www.distancesto.com/fuel-cost/il/ashdod-to-ein-bokek-connection-jerusalem/history/1146135.html
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# INR6.97 Total Cost of Fuel from Ashdod to Ein Bokek, Connected by Jerusalem
Your trip to Ein Bokek will consume a total of 2.79 gallons of fuel.
Trip start from Ashdod and continues through Jerusalem. Trip ends at Ein Bokek.
• 111.5 Miles (Trip Total)
• Route 1 (39.7 mi) Ashdod » Jerusalem
Fuel Cost INR2.48
Fuel Consumption 0.99 gallons
• Route 2 (71.8 mi) Jerusalem » Ein Bokek
Fuel Cost INR4.49
Fuel Consumption 1.80 gallons
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Ashdod, Israel and end at Ein Bokek, Israel.
Route goes through Jerusalem, Israel.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Ashdod to Jerusalem to Ein Bokek plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Ashdod to Jerusalem to Ein Bokek.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Ashdod to Jerusalem to Ein Bokek.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Ashdod to Jerusalem to Ein Bokek.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Ein Bokek are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Ashdod to Jerusalem to Ein Bokek.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Ashdod to Jerusalem to Ein Bokek.
Speaking of travel time, a flight to Ein Bokek takes up a lot less. How much less? Flight time from Ashdod to Jerusalem to Ein Bokek.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Ashdod to Jerusalem to Ein Bokek.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Ashdod IL:
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Fuel Cost from Ashdod to Arad
Fuel Cost from Ashdod to Lehavim
Fuel Cost from Ashdod to Herzliya
Fuel Cost from Ashdod to Ben Gurion Airport
| 736
| 3,077
|
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CC-MAIN-2022-33
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