url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.teachoo.com/7274/843/Euclid-s-postulates/category/Postulates/ | 1,680,336,240,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00235.warc.gz | 1,108,270,406 | 32,251 | Postulates
Chapter 5 Class 9 Introduction to Euclid's Geometry
Concept wise
## Postulate 1:
A straight line may be drawn from any one point to any other point.
The postulate says that a line passes through two point.
But, it does not say that only one line passes through 2 distinct points .
So, we make an axiom of it - Axiom 5.1
## Postulate 2:
Terminated line means line segment can be extended from either
side to form a lines.
Here, Line AB can be extended from both sides
## Postulate 3:
A circle can be drawn with any center and any radius
Eg: Circle of radius 1.5 cm & centre (0, 0)
Circle of radius 2 cm & centre (0, 0)
Circle of radius 1 cm & centre (1, 1)
## Postulate 4:
All right angles are equal to one another .
Here, ∠ B = ∠ Q = 90°
## Postulate 5:
If a straight line falling on two straight lines makes the interior
angles on the same side of it taken together less than two right angles, then the
two straight lines, if produced indefinitely, meet on that side on which the sum of
angles is less than two right angles
We discuss more about it in
Get live Maths 1-on-1 Classs - Class 6 to 12 | 304 | 1,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-14 | longest | en | 0.861824 |
https://www.varsitytutors.com/high_school_physics-help/kirchoff-s-laws | 1,669,911,445,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00076.warc.gz | 1,109,352,968 | 64,191 | # High School Physics : Kirchoff's Laws
## Example Questions
### Example Question #1 : Kirchoff's Laws
Calculate the current through the 10 ohm resistor.
Possible Answers:
Correct answer:
Explanation:
To begin, let us start with the resistor and the resistor that are in parallel. In parallel we can add resistors through the equation
This new resistor is now in series with the two resistors. In series we can just add these resistors up.
This new resistor is now in parallel with the resistor. In parallel we can add resistors through the equation
This new resistor is now in series with the and resistor. In series we can just add these resistors up.
Now that we have the equivalent resistance of the circuit. We can now determine the current flowing out of the battery.
Rearrange to solve for current.
One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.
We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series. So we can put this information into our chart.
Using Ohm’s Law we can determine the voltage for each of these two resistors.
We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.
Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors. Kirchoff’s laws state that the sum of the current flowing in and out of junction must equal 0.
We have flowing in from the resistor and flowing out to go through the resistor.
So the current going through the 4 Ohm resistor is 3 amps. This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them. We can then use this to determine the voltage drop across each of the resistors.
Now let’s add this to our chart.
We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor. Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.
We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.
### Example Question #2 : Kirchoff's Laws
Calculate the current in the 15 Ohm Resistor.
Possible Answers:
Correct answer:
Explanation:
To begin, let us start with the resistor and the resistor that are in parallel. In parallel we can add resistors through the equation
This new resistor is now in series with the two resistors. In series we can just add these resistors up.
This new resistor is now in parallel with the resistor. In parallel we can add resistors through the equation
This new resistor is now in series with the and resistor. In series we can just add these resistors up.
Now that we have the equivalent resistance of the circuit. We can now determine the current flowing out of the battery.
Rearrange to solve for current.
One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.
We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series. So we can put this information into our chart.
Using Ohm’s Law we can determine the voltage for each of these two resistors.
We can now use Kirchoff’s loop law through the loop of the 5, 3, and 14 Ohm resistor to determine the voltage that is traveling through the 14 Ohm resistor.
We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.
We can now add this information to our chart.
Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors. Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.
We have 4A flowing in from the 5 Ohm resistor and 1A flowing out to go through the 14 Ohm resistor.
So the current going through the 4 Ohm resistor is 3 amps. This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them. We can then use this to determine the voltage drop across each of the resistors.
Now let’s add this to our chart.
We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor. Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.
We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.
We can now add this information to our chart.
We can now analyze the junction between the 4, 10 and 15 Ohm resistor.
Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.
We have 2A flowing in from the 4 Ohm resistor and 1.2A flowing out to go through the 10 Ohm resistor.
### Example Question #3 : Kirchoff's Laws
Calculate the current through the 6 ohm resistor.
Possible Answers:
Correct answer:
Explanation:
To begin we need to simplify the circuit to get the equivalent resistance. Let’s start with the 3 and 6 Ohm resistors in parallel.
Now we can add this resistor to the 4 Ohm resistor as they are in series.
We can now determine the current coming out of the battery using Ohm’s Law.
Rearrange to solve for current.
The current coming out of the battery will be the same current that moves through the 4 Ohm resistor. So we can determine the voltage drop across the 4 ohm resistor.
We can then use Kirchoff’s loop law to determine the voltage drop across the 6 Ohm resistor. Let’s analyze the loop that goes from the battery to the 4 ohm resistor and through the 6 ohm resistor.
### Example Question #4 : Kirchoff's Laws
Calculate the voltage drop across the 4 ohm resistor.
Possible Answers:
Correct answer:
Explanation:
To begin we need to simplify the circuit to get the equivalent resistance. Let’s start with the 3 and 6 Ohm resistors in parallel.
Now we can add this resistor to the 4 Ohm resistor as they are in series.
We can now determine the current coming out of the battery using Ohm’s Law.
Rearrange to solve for current.
The current coming out of the battery will be the same current that moves through the 4 Ohm resistor. So we can determine the voltage drop across the 4 ohm resistor.
### Example Question #1 : Kirchoff's Laws
Kirchoff’s junction rule is an example of
Possible Answers:
None of the givens answers
Conservation of charge
Conservation of energy
Conservation of momentum
Correct answer:
Conservation of charge
Explanation:
Kirchoff’s loop rules states the sum of the current going into and out of the junction must equal 0. In other words, the current going in must equal the current going on. Current is a measure of the flow of charge. Therefore, this law is conservation of charge as the number of electrons going into a junction must equal the number of electrons flowing out.
### Example Question #6 : Kirchoff's Laws
Calculate the voltage drop across the 14 ohm resistor.
Possible Answers:
Correct answer:
Explanation:
To begin, let us start with the resistor and the resistor that are in parallel. In parallel we can add resistors through the equation
This new resistor is now in series with the two 4Ω resistors. In series we can just add these resistors up.
This new resistor is now in parallel with the resistor. In parallel we can add resistors through the equation
This new resistor is now in series with the and resistor. In series we can just add these resistors up.
Now that we have the equivalent resistance of the circuit. We can now determine the current flowing out of the battery.
Rearrange to solve for current.
One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.
We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series. So we can put this information into our chart.
Using Ohm’s Law we can determine the voltage for each of these two resistors.
We can now use Kirchoff’s loop law through the loop of the 5, 3, and 14 Ohm resistor to determine the voltage that is traveling through the 14 Ohm resistor.
### Example Question #1 : Kirchoff's Laws
Kirchoff’s loop rule is an example of
Possible Answers:
Conservation of charge
None of the givens answers
Conservation of momentum
Conservation of energy
Correct answer:
Conservation of energy
Explanation:
Kirchoff’s loop law states that the sum of the voltage around a loop must equal zero. In other words, the voltage that is being provided by the batteries in the circuit must equal the voltage being used by the objects in the circuit. Voltage is a measure of the potential difference, or energy within the circuit. In other words, the battery does a certain amount of work and provides energy to the circuit which is then used by all the parts of the circuit. Therefore this is an example of conservation of energy.
### Example Question #8 : Kirchoff's Laws
Calculate the voltage drop from point to point .
Possible Answers:
Correct answer:
Explanation:
To begin we need to simplify the circuit to get the equivalent resistance. Let’s start with the 3 and 6 Ohm resistors in parallel.
Now we can add this resistor to the 4 Ohm resistor as they are in series.
We can now determine the current coming out of the battery using Ohm’s Law.
Rearrange to solve for current.
The current coming out of the battery will be the same current that moves through the 4 Ohm resistor. So we can determine the voltage drop across the 4 ohm resistor.
We can then use Kirchoff’s loop law to determine the voltage drop from point to point .
Let’s analyze the loop that goes from the battery to the 4 ohm resistor and through the 3 ohm resistor.
### Example Question #9 : Kirchoff's Laws
Which of the equations here is valid for the circuit shown?
Possible Answers:
Correct answer:
Explanation:
To answer this question we must consider Kirchoff’s Loop Law. This law states that the voltage around any loop must equal 0. In this case there are two different loops at play. To begin, let’s start on the left with the 2 Volt battery.
As we start with the 2 Volt battery, we then move into the 1 Ohm resistor with going through it. Ohm’s law states that the voltage is equal to the current times the resistance. Therefore the voltage through this circuit is Since this resistor is using the voltage this will be a negative voltage when we sum around the loop.
We will continue our loop through the middle of the circuit into the 4 Volt battery. This battery is facing the opposite direction from our 2 Volt battery and therefore will be a negative when it comes to our equation.
Next is the 2 Ohm resistor with going through it. According to Ohm’s law the voltage being used by this resistor is equal to
When summarizing all of these parts we get an equation that looks like which simplifies down to This is one of the equations available to us and therefore there is no need to analyze any other loops. | 2,617 | 11,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-49 | longest | en | 0.915274 |
https://eklausmeier.goip.de/blog/2021/02-09-poisson-log-normal-distributed-random-numbers | 1,722,910,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00121.warc.gz | 174,872,074 | 14,547 | # Poisson Log-Normal Distributed Random Numbers
Task at hand: Generate random numbers which follow a lognormal distribution, but this drawing is governed by a Poisson distribution. I.e., the Poisson distribution governs how many lognormal random values are drawn. Input to the program are $\lambda$ of the Poisson distribution, modal value and either 95% or 99% percentile of the lognormal distribution.
From Wikipedia's entry on Log-normal distribution we find the formula for the quantile $q$ for the $p$-percentage of the percentile $(0<p<1)$, given mean $\mu$ and standard deviation $\sigma$:
$$q = \exp\left( \mu + \sqrt{2}\,\sigma\, \hbox{erf}^{-1}(2p-1)\right)$$
and the modal value $m$ as
$$m = \exp\left( \mu - \sigma^2 \right).$$
So if $q$ and $m$ are given, we can compute $\mu$ and $\sigma$:
$$\mu = \log m + \sigma^2,$$
and $\sigma$ is the solution of the quadratic equation:
$$\log q = \log m + \sigma^2 + \sqrt{2}\,\sigma\, \hbox{erf}^{-1}(2p-1),$$
hence
$$\sigma_{1/2} = -{\sqrt{2}\over2}\, \hbox{erf}^{-1}(2p-1) \pm\sqrt{ {1\over2}\left(\hbox{erf}^{-1}(2p-1)\right)^2 - \log(m/q) },$$
or more simple
$$\sigma_{1/2} = -R/2 \pm \sqrt{R^2/4 - \log(m/q) },$$
with
$$R = \sqrt{2}\,\hbox{erf}^{-1}(2p-1).$$
For quantiles 95% and 99% one gets $R$ as 1.64485362695147 and 2.32634787404084 respectively. For computing the inverse error function I used erfinv.c from lakshayg.
Actual generation of random numbers according Poisson- and lognormal-distribution is done using GSL. My program is here: gslSoris.c.
Poisson distribution looks like this (from GSL documentation):
Lognormal distribution looks like this (from GSL): | 534 | 1,649 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.728379 |
http://www.pairlist.net/pipermail/retros/2021-January/004901.html | 1,674,865,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00788.warc.gz | 82,834,577 | 1,902 | # [Retros] Retro problems from the Stuttgarter Zeitung/solutions
keilhack at aol.com keilhack at aol.com
Tue Jan 19 01:30:52 EST 2021
```1) Werner Keym, shortest mate? (why?)
Q7/ppk5/2P1P3/b1KpP3/8/2P5/8/8 w - - 0 1Solution: Not 1.e5:d6#?, as Kc8/d8:Bc7 Bb8-c7+ were possible last moves, but 1.Q:b7+ Kd8 2.Qd7#.
2) Werner Keym, shortest mate? (why?)
7R/6kp/4R1P1/4PpKp/7n/8/6P1/8 w - - 0 1
Solution: Not 1.e5:f6 e.p.+? K:h8 2.Re8#, as Kg8-g7 g7:h8R+ Kf8-g8 are possible last moves, so 1.R:h7+ Kf8 2.Rf7+ Kg8 3.Re8#.
3) Werner Keym/Bernd Schwarzkopf2k5/4K3/8/8/8/8/8/8 w - - 0 1Add 4 chessman. In the resulting position there shall be as much as possible ways to reach the same position after one white and one black move. Which is the position, and how many different moves by White are possible?Add wPg7 and black rooks on f8, g8, h8. g7:f8Q/R/B/K+ R:f8 and g7:h8Q/R/B/K R:h8, so eight different way to reach the same position!
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``` | 440 | 1,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-06 | latest | en | 0.738827 |
https://oeis.org/A126062 | 1,621,175,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991224.58/warc/CC-MAIN-20210516140441-20210516170441-00079.warc.gz | 454,221,495 | 4,244 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A126062 Array read by antidiagonals: see A128195 for details. 3
1, 1, 1, 1, 4, 1, 1, 9, 15, 1, 1, 16, 65, 64, 1, 1, 25, 175, 511, 325, 1, 1, 36, 369, 2020, 4743, 1956, 1, 1, 49, 671, 5629, 27313, 52525, 13699, 1, 1, 64, 1105, 12736, 100045, 440896, 683657, 109600, 1, 1, 81, 1695, 25099, 280581, 2122449, 8390875, 10256775 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,5 LINKS P. Luschny, Variants of Variations. EXAMPLE Array begins: [0]..1,..1,...1,.....1,......1,.......1,.........1,..........1,............1 [1]..1,..4,..15,....64,....325,....1956,.....13699,.....109600,.......986409 [2]..1,..9,..65,...511,...4743,...52525,....683657,...10256775,....174369527 [3]..1,.16,.175,..2020,..27313,..440896,...8390875,..184647364,...4616348125 [4]..1,.25,.369,..5629,.100045,.2122449,..53163625,.1542220261,..50895431301 [5]..1,.36,.671,.12736,.280581,.7376356,.229151411,.8252263296,.338358810761 MAPLE A126062 := proc(k, n) if n = 0 then 1 ; else (n*k+1)*(A126062(k, n-1)+k^n) ; fi ; end: for diag from 0 to 10 do for k from diag to 0 by -1 do n := diag-k ; printf("%d, ", A126062(k, n)) ; od ; od ; # R. J. Mathar, May 18 2007 MATHEMATICA a[_, 0] = 1; a[k_, n_] := a[k, n] = (n*k+1)*(a[k, n-1]+k^n); Table[a[k-n, n], {k, 0, 10}, {n, 0, k}] // Flatten (* Jean-François Alcover, Jan 08 2014 *) CROSSREFS Sequence in context: A039756 A126065 A299427 * A243608 A219207 A157108 Adjacent sequences: A126059 A126060 A126061 * A126063 A126064 A126065 KEYWORD nonn,tabl,easy AUTHOR N. J. A. Sloane, Feb 28 2007 EXTENSIONS More terms from R. J. Mathar, May 18 2007 STATUS approved
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Last modified May 16 10:30 EDT 2021. Contains 343941 sequences. (Running on oeis4.) | 848 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.545873 |
https://www.vedantu.com/question-answer/a-hollow-steel-ball-and-a-solid-steel-ball-of-class-11-physics-cbse-60a275193abc385357c43a1d | 1,702,195,905,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00843.warc.gz | 1,121,042,207 | 28,394 | Courses
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# When a hollow steel ball and a solid steel ball of same size are submerged in water, do the two experience the same upthrust.
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Hint:The upthrust or the buoyant force is the upward force experienced by the object, when the object is completely or partially immersed. The Archimedes’ principle states that when a body is immersed fully or completely. The body will experience an upthrust which is equal to the weight of the liquid displaced by it.
When an object of volume $V$ and density $\rho$ is immersed in a liquid of density ${\rho _L}$.If $g$ is the acceleration due to gravity. The weight of the body is given by $V\rho g$.The upthrust or buoyant force on the body is given by $V{\rho _L}g$. From this equation it is clear that upthrust or buoyant force depends on following factors: | 241 | 950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-50 | longest | en | 0.914103 |
https://perastudio.pl/Oct/density_10766.html | 1,642,827,813,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00166.warc.gz | 505,978,661 | 5,879 | ## Density of crusher run per cubic metre
Crusher run density kg m3 hat is the density of crusher run per cubic metre 876 kgm 3 is the densityowever, you might want it in different units, such as grams per cubic centimeter, for examplee would then get 876,000gmillion cm 3 or 6 g cm 3 , which is a bit less dense than wateret price. Density Of Crusher Run Malaysia
Get Price
• density of crusher run kg m3
Answers,14/05/2014 Density of normal concrete is in the order of about 2400 kg per cubic metre The densityof lightweight concrete will be less than about density 1900 kg per cubic metreThe high density concrete What is the density of crusher run in metric?
• compacted density of crusher run kg m3 - BINQ Mining
Apr 19, 2013 What is the standard density of topsoil expressed in tonnes per cubic … How do you convert density in grammes per cubic centremetre to tonnes per cubic metre? they are equivalent, if is 1g/cm 3 then it has a density of 1 tonne/m 2 (also 1 kg/litre). What is the … 1 m3 Crusher Run = Tonnes (without wastage ) … More detailed
• How many ton crusher run per cubic meter? - Answers
Jan 13, 2011 1 m3 Crusher Run = 2.45 Tonnes (without wastage) Solid marble weighs 160 lbs. per cubic ft 160lbs / 2000 lbs/ton = 0.08 tons 1 cubic ft = 0.028316846592 cubic meter 0.08 tons per cubic ft
• Density Of Granite Crusher Run In 1Cubic Meter
Density Of Granite Crusher Run In Cubic Meter. As 1000kg of pure water 1 cubic metre those materials under 1000kgcum will float more dense will sink ie those materials with a Granite broken 103 1650 Iron ore crushed see metals table 131181 21002900
• density of crusher run stone
Density Of Crusher Run Stone&New Projects: stone, crushed weighs 1.602 gram per cubic centimeter or 602 kilogram per cubic meter, i.e. density of stone, crushed is equal to 602 kgm In imperial or US customary measurement system, the density is equal to 100.0096 pound per cubic foot lbft or 0.92601 ounce per cubic inch ozinch
• Crusher Run Tonnage Per Meter Cube - Industris Mining
Convert 1 cubic yard of crusher run to metric ton. convert cubic yards of crusher run to tons beltconveyers.net we use an estimate of 2900 lbs per cubic yard of dry crusher run BUT 3250 lbs crusher cubic.Crushed gravel weight per cubic meter crusher 100 mesh
• bulk density of oil stone crusher
Crushed li ne bulk density. Stone crushers, Jaw, Cone, Impact, Vsi crusher for Crushed li ne bulk density As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any sizereduction requirements including, Crushed li ne bulk density, quarry, aggregate, and different kinds of mineralsbulk density for 19mm crusher run nicosoccorsoit
• Stone, crushed volume to weight conversion
About Stone, crushed; 1 cubic meter of Stone, crushed weighs 1 602 kilograms [kg] 1 cubic foot of Stone, crushed weighs 100.00959 pounds [lbs] Stone, crushed weighs 1.602 gram per cubic centimeter or 1 602 kilogram per cubic meter, i.e. density of stone, crushed is equal to 1 602 kg/m .In Imperial or US customary measurement system, the density is equal to 100.0096 pound per cubic
• Density of asphalt premix = 2.3ton per m3 Density of
Density of asphalt premix = 2.3ton per m3 Density of crusher run = 1.8ton per m3 Density of concrete = 2.4ton per m3
• weight of 1m3 of crusher run
What is the density of crusher run in metric. How many ton crusher run per cubic metre? 1 m3 Crusher Run = 2.45 Tonnes (without wastage).One crusher run per cubic meter is typically the weight of two tons. However, this conversion may vary in specific instances. Read more
• Density Of Crusher Run Kg M
Weight Of Crusher Run Per Cubic Meter. Crusher run 1 m3 per kg - m-e-geu crusher run density kg m3 mobile crushers all over the what is the weight of 1 yard of crush and run how much does crusher run gravel weigh per cubic yard what is the weight of crusher stone sand for cubic meter wha
• Crusher Run Gravel Weight
Weight Of 2 5 Crusher Run Per M. Weight of cubic meter of crusher run stone. What Does 1 Cubic Meter Of Crusher Run WeightHow much does Crusher Run Gravel weigh per Cubic yard The How much Standard Weights for Crushed Rock Per Meter asphalt aggregate blends all weigh in at 1.07 tons
• convert m3 to tonnes for crusher run
Dec 11 2012 convert 4 cubic meters crush and run to tons – Crusher South 1 ton of fire wood equals to how many cubic meter 25 cubic meters convert tonne m3 crusher run How many tons does a cubic meter of crusher run weigh More detailed. More Detail; What is
• What is the density of crusher run in metric? - Answers
May 08, 2012 What is the density of crusher run per cubic metre? One crusher run per cubic meter is typically the weight of two tons. However, this conversion may vary in specific instances
• compacted density of crusher run kg m3 - BINQ
Apr 19, 2013 crusher run density kg m3 – What is the density of crusher run – The Q&A wiki. What is the density of crusher run in metric? 2082 kg/m to 2243 … More detailed
• Crusher run density kg m3 in malaysia
What is the density of crusher run per cubic metre - Answers.com. The density of copper is 8940 kilograms per cubic meter. 8 people ... 1 m3 Crusher Run = 2.45 Tonnes (without wastage) ... 1 cubic meter aluminum = 270 kg. Asphalt and Composites Kuala Lumpur Malaysia - Kuala | 1,366 | 5,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.73018 |
http://www.bccvietnam.com/how-to-nlxxdvr/phase-of-oscillation-formula-b42274 | 1,620,588,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00632.warc.gz | 64,292,939 | 13,790 | ## phase of oscillation formula
• Describe the advantages of buffered phase shift oscillators. However, in this case the time constant (inverse of the oscillation frequency) is identical to the first formula as given by you (involving SQRT[3]). Then apply Barkhausen's criteria for phase shift: the sum of the phase shifts from the two transmittance must be equal to zero for an oscillation to exist. Let's start by the Limit Cycle . \$\endgroup\$ – LvW Apr 22 '14 at 14:34 add a comment | Fig. Its circuit is shown in Fig. This is due to the presence of an extra capacitor. • Figure illustrates an oscillator with a small amount of damping. As an example, … First find the transmittance of the 3 RC blocks and the one of your amplifier. For example, in a transverse wave traveling along a string, each point in the string oscillates back and forth in the transverse direc-tion (not along the direction of the string). A phase-shift oscillator is a linear electronic oscillator circuit that produces a sine wave output. At least his equations give discontinuity. This formula is only applicable if the phase shift network uses same Resistance and capacitance value, that means R1 = R2 and C1 = C2 = C3. Moreover, the feedback factor gets affected in Colpitts oscillator. • Know relevant formulae for the frequency of oscillation. Oscillation occurs at the frequency where the total phase shift through the 3 RC circuits is 180°. Find the phase difference between a point 0.3m from the peak of a wave and another point 0.7m further along from the same peak. When solving questions for both these topics always keep in mind that your concepts should be clear. The frequency and period of the oscillation are both determined by the constant , which appears in the simple harmonic oscillator equation, whereas the amplitude, , and phase angle, , are determined by the initial conditions. Acceleration – we can calculate the acceleration of the object at any point in it’s oscillation using the equation below. Answer W2. Similarity between electrical and mechanical oscillations - definition The LC oscillation is similar to the mechanical oscillation of a blocka ttached to a spring. The Phase shift oscillator can be made as variable phase shift oscillator which can produce a wide range of frequencies depending on the pre-set value determined. • The mechanical energy of a damped oscillator decreases continuously. However, we still have little understanding of how some cells can exhibit calcium oscillations with a period of less than a second, whereas other cells have oscillations with a period of hundreds of seconds. One has its maximum excursion at a different time than the other for example. Under forced oscillation, the phase of harmonic motion of the particle differs from the phase of the driving force. By applying phase analysis methods, we showed that PV cells in the PFC exhibited robust phase-locked firing to high-frequency oscillations (100–250 Hz) and delta rhythms (1–4 Hz), but poor coupling to gamma (30–80 Hz) or theta (4–8 Hz) oscillations. The Van der Pol oscillator can be represented by the following differential equations: \begin{aligned} \dot{x}&=y \\ \dot{y}&=\mu(1-x^2)y-x\end{aligned} where \mu is a scalar parameter indicating the damping strength. The negative gain of the amplifier stage (-K) will add another 180° phase-shift. The significance of using a Clapp oscillator over a colpitt oscillator is that the frequency stability of the Clapp oscillator is more. When the preexisting jet is located more northward (southward), the induced dipole can have a low-over-high (high-over-low) structure and thus can make the center of the stationary wave anomaly shift southward (northward), which can be regarded as an initial state or embryo of a positive (negative) phase North Atlantic Oscillation (NAO). General Equation of sine wave - Phase Difference, Wave speed, How to prepare Oscillations & Waves. It would physically mean the mass-spring system is oscillating BEFORE driving force. Example W3 The equation of a transverse sinusoidal wave is given by: . The proper way to derive the oscillation frequency from this oscillator is to go back to Barkhausen's oscillation criteria. This formula is called the Thomson formula in honor of British physicist William Thomson $$\left(1824-1907\right)$$, who derived it theoretically in $$1853.$$ Damped Oscillations in Series $$RLC$$-Circuit. Phase of oscillation March 4, 2014 The concept of the phase is a way of comparing two oscillations which are occuring at the same time. While the phase shift network of the clapp oscillator consists of three capacitor and one inductor. The image below shows a typical RC phase shift oscillator circuit with a BJT: RC phase shift oscillator with a transistor. Thus, the phase of theta band oscillations may be critical for the coordination of neural activity [22,27]. Using the phase sensitivity functions, collective oscillation of the network under weak perturbation can be described approximately by a one-dimensional phase equation. If you look at the above equations for phase shift and output frequency, it should be obvious that there is a complex nonlinear relationship between these two values. The frequency (f) of an oscillation is measure in hertz (Hz) it is the number of oscillations per second. So, if I just left this as cosine, that would say this thing's gonna get as big as one at some point in time and that's a lie. The key contributions are: (1) to predict the phase noise correctly using the large signal time domain calculations (Bessel functions) and nonlinear CAD simulators and derive a set of algebraic equations for the noise calculations (many The frequency of oscillation is given by and the phase shift is 180 o. An important question is the asymptotic (for $\epsilon \rightarrow 0$) calculation of the phase trajectory of the relaxation oscillation of the system (1), and the establishment of asymptotic formulas for the characteristics of this oscillation — its period, amplitude, etc. The second order differential equation describing the damped oscillations in a series $$RLC$$-circuit we got above can be written as But they do not identically track each other. 3.1.3 Loading Effect . Oscillations. For a block of mass m oscillating with frequency ω 0 , the equation is: d t 2 d 2 x + ω 0 2 x = 0 Here, ω 0 = m k , and k is the spring constant. So, x corresponds to q. Time period for spring oscillator, Time period for simple pendulum, Waves. Formulas for Oscillations & Waves. Mechanical oscillations - Phase of harmonic oscillations: φ - phase , ν - frequency , t - time 4.1 Phase-shift Oscillator using Op-Amp: The op-amp is used in the inverting mode; therefore, any signal that appears at the inverting terminal is shifted by 180° at the output. 3.1.2 Three Cascaded High Pass Filters Fig. How on earth could be phase angle not only discontinuous but also negative. • The oscillator excess open-loop gain (which is necessary for initial oscillator build-up) should be minimized in order to prevent amplitude fluctuations from being converted into significant frequency fluctuations. So, let's say our amplitude for a particular simple harmonic oscillator happened to be .2 meters, that would mean that this here, I can represent this here with .2 meters, this doesn't even make it to one. Two new quadrature oscillator circuits using operational amplifiers are presented. Oscillations David Morin, morin@physics.harvard.edu A wave is a correlated collection of oscillations. The time for one oscillation is called the period (T) it is measured in seconds. It is related to the period of oscillation $$T$$ by the formula The variable $$\omega$$ is called the circular or cyclic frequency of oscillation. This thing only gets as big as .2, so it's easy though. 16.10. A set of coupled adjoint equations for phase sensitivity functions, which characterize the phase response of the collective oscillation to small perturbations applied to individual elements, is derived. This is due to the fact that, for a regenerative effect, the signal must undergo n*360 degrees phase shift: 180 from the amplifier and another 180 from the feedback network. The frequency of the oscillation (in hertz) is , and the period is . • The decrease in amplitude is called damping and the motion is called damped oscillation. Both oscillations waggle back and forth and we will assume that both do it at the same frequency. 3.1.1 High Pass Fig. Oscillations in the concentration of free cytosolic calcium are an important control mechanism in many cell types. For each pair-wise relation between ROIs for each subject, a PLV was calculated for each frequency of interest (1–49 Hz in 1-Hz intervals). [See Equations -.] Damped oscillations • Real-world systems have some dissipative forces that decrease the amplitude. One of the conditions for oscillation is that the (regenerative) feedback loop must provide a 180 degree phase shift. In these formulas, $$A$$ means the amplitude of oscillation, $${\omega t + {\varphi _0}}$$ is the phase of oscillation, $${{\varphi _0}}$$ is the initial phase at time $$t = 0.$$ Figure 2. Figure 16.10: Operational amplifier phase-shift oscillator The feedback portion of the oscillator can be derived by applying Kirchhoff’s current law at node a and node b respectively. Outputs of two sinusoidal signals with 90° phase difference are available in each circuit configuration. But derviations according to Vibration and Waves by A.P. The operational amplifier phase-shift oscillator is another oscillator type that meets the principles of oscillator design. Resulting in a total phase-shift of 360° or 0° which is the required condition for oscillation. French give discontiunity as i showed u on the grpahs. grounded base oscillator rather than the Colpitts oscillator. In this tutorial, we will learn how to draw the phase portrait of Van-Der-Pol oscillator in LaTeX using TikZ and Pgfplots. 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To draw the phase portrait of Van-Der-Pol oscillator in LaTeX using TikZ and Pgfplots the mass-spring system is oscillating driving. A sine wave - phase difference between a point 0.3m from the peak of damped. A typical RC phase shift oscillator circuit with a small amount of damping is measure in hertz Hz! A total phase-shift of 360° or 0° which is the number of oscillations per second feedback loop must a! Quadrature oscillator circuits using operational amplifiers are presented band oscillations may be critical for the frequency where the total shift. The amplitude through the 3 RC circuits is 180° that meets the principles of oscillator design using Clapp. Systems have some dissipative forces that decrease the amplitude both proposed quadrature oscillators are based on characteristic... Physically mean the mass-spring system is oscillating BEFORE driving force significance of using Clapp... A one-dimensional phase equation • Real-world systems have some dissipative forces that the... Oscillator decreases continuously of your amplifier by A.P your amplifier a different time than the for. Oscillator is more of theta band oscillations may be critical for the frequency f... Forces that decrease the amplitude have some dissipative forces that decrease the.. Occurs at the same peak the advantages of buffered phase shift is 180 o phase of oscillation formula oscillator more! May be critical for the coordination of neural activity [ 22,27 ] the RC! Amplitude is called damping and the motion is called damping and the one your. Energy of a damped oscillator decreases continuously oscillation is given by and the phase shift oscillators operational amplifier phase-shift is... Wave speed, How to draw the phase portrait of Van-Der-Pol oscillator in LaTeX using TikZ and Pgfplots 180.. Free cytosolic calcium are an important control mechanism in many cell types concepts be! Collective oscillation of the conditions for oscillation forth and we will learn How to prepare oscillations & Waves we assume... Physically mean the mass-spring system is oscillating BEFORE driving force physically mean the mass-spring system is oscillating BEFORE force. Point 0.3m from the same peak the object at any point in it ’ s oscillation using equation. Of free cytosolic calcium are an important control mechanism in many cell types french give discontiunity as i showed on! Topics always keep in mind that your concepts should be clear ( f ) of extra! Period of oscillation is called damped oscillation theta band oscillations may be critical for the coordination of activity. Of theta band oscillations may be critical for the coordination of neural activity [ 22,27 ] the period oscillation! | 4,282 | 20,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-21 | longest | en | 0.894291 |
https://www.scienceblogs.com/builtonfacts/2009/03/29/sunday-function-26 | 1,696,033,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00383.warc.gz | 1,052,482,598 | 13,690 | # Sunday Function
The quadratic formula. With the exception of the Pythagorean Theorem, it's probably the single most common mathematical formula people carry from high school. It's not a function as such, it's something that solves a function. Let me give an example:
Pick a number x, square it, add twice x to that, subtract 3 from that. You might want to figure out what x makes the answer equal to zero. It's the kind of thing we have to do in physics all the time. In this case, the answers (and you can find them by several methods) happen to be -3 and 1. Plug them in and you'll see that you end up with 0 as the answer.
In general if you have a quadratic equation like this:
The particular x that solves this equation (and there will be two of them - though they might be complex numbers, and they might be the same number, long story...) will be given by the quadratic formula:
Not so bad. But what if we're trying to solve a cubic equation, like this?
Well, just like the quadratic equation had two solutions, the cubic equation will have three. Here is the first:
Can you even read that? It's pretty tough for me. There's two more of course, for the other two solutions.
What about the quartic equation? It looks like this:
We can do it too. I'll spare you the horror, but it's about as much uglier than the cubic equation than the cubic equation is uglier than the quadratic equation. And there's four solutions, not just three.
What about the quintic equation? The one with an x^5 term?
Surprisingly - or perhaps mercifully - there isn't a solution. Some mathematicians much smarter than me proved it in 1824. At least there isn't a solution in anything with just fractions, powers, and roots like the ones we've seen here. But sometimes you really do need the actual numbers that solve quintic or even higher equations. Fortunately there's tricks we can do. These days computers automate most of them, but even before computers it was possible to get a decimal approximation to the solutions of these equations without needing an exact algebraic solution. And to be totally honest, that's pretty much all we ever do for any polynomial worse than the quadratic.
It's important to remember that there really are real live numbers that solve these polynomials of 5th and higher order, just not numbers we can express in terms of the standard arithmetic operations. But we're physicists. Experiments pretty much always give us decimal approximations of reality anyway, so we're used to it.
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The cubic formula is quite unwieldy; it's something I can't carry in my head. I recall it's often easier to present if you declare 2 variables (often called p and q) and a determinant.
Dude, you really ought to put all these Sunday Function posts together in a book.
It's called "Handbook of Mathematical Functions", Edited by Abramowitz and Stegun", U.S. Department of Commerce (National Bureau of Standards).
To be a bit pedantic: there are algebraic solutions for some versions of the quintic equation, but no general algebraic solution that works on all quintics.
If people would like a book, I can recommend Mario Livio's book "The equation that couldn't be solved". It's light on equations, but covers the history of attempts to tackle the quintic equations, including the remarkable stories of Abel and Galois.
You may not know that solving that quadratic equation was one of the main problems that drove the creation of algebra itself by Arab mathematicians. Learned that from one of my math colleagues who teaches "math for the masses".
Since trig tables were being constructed at about the same time, I suspect they had to formulate the algorithm for completing the square so there was a means to compute the root needed to evaluate the half angle identities used to work your way from special angles to other ones.
By CCPhysicist (not verified) on 29 Mar 2009 #permalink
There are schemes for computing exactly with the roots of any polynomial, one of them is implemented in Mathematica. After a long exact calculation you can convert the answer to decimal form with arbitrary precision. Problem is, it may take a lot of speed and memory to push the calculation through to the end.
"It's important to remember that there really are real live numbers that solve these polynomials of 5th and higher order, just not numbers we can express in terms of the standard arithmetic operations."
Well the solutions don't have to be real. Some polynomials of even degree do not have any real roots.
You may not know that solving that quadratic equation was one of the main problems that drove the creation of algebra itself by Arab mathematicians. Learned that from one of my math colleagues who teaches "math for the masses"
Iâm sorry to have to tell you this, CCPhysicist but your professor for âmath for the massesâ is wrong! The general solution to the quadratic equation was know to the Babylonians in the so called Old Babylonian period possibly as early as 1700 BCE. It was also known to the ancient Greeks and can be found in its geometrical form in Euclidâs âElementsâ from about 400 BCE; both sources were known to the Islamic algebraists. It is Euclidâs geometric proof that can be found in the book al-KitÄb al-muḫtaá¹£ar fÄ« ḥisÄb al-Äabr wa-l-muqÄbala by Ä al-KhwÄrizmÄ«. The Islamic algebraists further developed algebra in that they developed methods for finding individual solutions to specific forms of the cubic, bi-quadratic and quintic equations and in some cases polynomials of higher degrees, those susceptible to substitution.
Another book that has a great take on the solving of polynomial equations is Marcus du Sautoy's Symmetry. He's does a wonderful job of relating the history of finding solutions and placing the general problem in the context of symmetry. | 1,267 | 5,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | latest | en | 0.970128 |
https://infinitylearn.com/question-answer/there-are-two-identical-small-holes-of-area-of-cro-63a0306c15d12f0e636f27f8 | 1,701,958,702,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00616.warc.gz | 347,279,714 | 12,852 | Questions
# There are two identical small holes of area of cross section a on the opposite sides of a tank containing a liquid of density $\rho$ . The difference in height between the holes is h. The tank is resting on a smooth horizontal surface. The horizontal force which will have to be applied on the tank to kept it in equilibrium is
Moderate
A
B
$\frac{2\mathrm{gh}}{\rho \mathrm{a}}$
C
D
$\frac{\rho gh}{a}$ | 109 | 422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.862833 |
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hw9_solutions
hw9_solutions - Due November 6 2008 CS 257(Luke Olson...
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Due: November 6, 2008 CS 257 (Luke Olson): Homework #9 Problem 1 Problem 1 We are coding for a particle physics team and asked to compute the mass of a region with density ρ = ln ( x ) x - 1 . That is, compute m = Z 1 0 ρ ( x ) dx. We convince the team use n -point Gauss Quadrature, praising its ability to achieve spectral convergence as n increases. Is this a good idea in this situation? Why not Newton-Cotes? Use int gauss.m and hand in a plot (with code) of n versus error along with comments on the outcome . ( hint: the integral should be π 2 / 6 . See dilog and polylogarithm in Wikipedia for more information ). Solution We test the accuracy of Gauss Quadrature with the following script (a reference case of f ( x ) = cos ( πx ) is commented out): 1 clear ; 2 f=inline( ’log(x)./(x-1)’ ); 3 exact= pi ˆ2 / 6; 4 5 %f=inline(’cos(pi * x/2)’); 6 %exact=2/pi; 7 8 nlist=2:100; 9 err= length (nlist); 10 err2= length (nlist); 11 12 for k=1: length (nlist) 13 n=nlist(k); 14 err(k) = abs (exact - int_gauss(f,0,1,1,n)); 15 end 16 17 figure ; 18 semilogy (nlist,err); xlabel ( ’points in quadrature’ ); ylabel (
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{[ snackBarMessage ]} | 459 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-17 | latest | en | 0.74815 |
https://www.travelmath.com/distance/from/Chesterfield,+MO/to/Chicago,+IL | 1,576,241,022,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00348.warc.gz | 900,786,387 | 14,154 | # The distance from Chesterfield, Missouri to Chicago, Illinois is:
### 318 miles / 512 km driving271 miles / 436 km flying
Get: distance driving distance stopping points halfway point From: To:
## Map of distance from Chesterfield, MO to Chicago, IL
For a quick answer, you can use DistanceCalc.com to get the distance from Chesterfield to Chicago.
## Distance from Chesterfield, MO to Chicago, IL
The total driving distance from Chesterfield, MO to Chicago, IL is 318 miles or 512 kilometers.
The total straight line flight distance from Chesterfield, MO to Chicago, IL is 271 miles.
This is equivalent to 436 kilometers or 236 nautical miles.
Your trip begins in Chesterfield, Missouri. It ends in Chicago, Illinois.
Your flight direction from Chesterfield, MO to Chicago, IL is Northeast (34 degrees from North).
The distance calculator helps you figure out how far it is to get from Chesterfield, MO to Chicago, IL. It does this by computing the straight line flying distance ("as the crow flies") and the driving distance if the route is drivable. It uses all this data to compute the total travel mileage.
## Chesterfield, Missouri
City: Chesterfield
State: Missouri
Country: United States
Category: cities
## Chicago, Illinois
City: Chicago
State: Illinois
Country: United States
Category: cities
## Distance calculator
Travelmath helps you find distances based on actual road trip directions, or the straight line flight distance. You can get the distance between cities, airports, states, countries, or zip codes to figure out the best route to travel to your destination. Compare the results to the straight line distance to determine whether it's better to drive or fly. The database uses the latitude and longitude of each location to calculate distance using the great circle distance formula. The calculation is done using the Vincenty algorithm and the WGS84 ellipsoid model of the Earth, which is the same one used by most GPS receivers. This gives you the flying distance "as the crow flies." Find your flight distances quickly to estimate the number of frequent flyer miles you'll accumulate. Or ask how far is it between cities to solve your homework problems. You can lookup U.S. cities, or expand your search to get the world distance for international trips. You can also print out pages with a travel map. | 491 | 2,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-51 | longest | en | 0.857853 |
https://www.numbersaplenty.com/10271 | 1,643,056,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00505.warc.gz | 965,766,232 | 3,289 | Search a number
10271 is a prime number
BaseRepresentation
bin10100000011111
3112002102
42200133
5312041
6115315
741642
oct24037
915072
1010271
117798
125b3b
1348a1
143a59
15309b
hex281f
10271 has 2 divisors, whose sum is σ = 10272. Its totient is φ = 10270.
The previous prime is 10267. The next prime is 10273. The reversal of 10271 is 17201.
10271 is digitally balanced in base 2 and base 3, because in such bases it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 10271 - 22 = 10267 is a prime.
It is a Sophie Germain prime.
Together with 10273, it forms a pair of twin primes.
It is a Chen prime.
It is an Ulam number.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (10273) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5135 + 5136.
It is an arithmetic number, because the mean of its divisors is an integer number (5136).
210271 is an apocalyptic number.
10271 is a deficient number, since it is larger than the sum of its proper divisors (1).
10271 is an equidigital number, since it uses as much as digits as its factorization.
10271 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 14, while the sum is 11.
The square root of 10271 is about 101.3459421980. The cubic root of 10271 is about 21.7372321241.
Adding to 10271 its reverse (17201), we get a palindrome (27472).
The spelling of 10271 in words is "ten thousand, two hundred seventy-one", and thus it is an iban number. | 507 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.898959 |
https://nrich.maths.org/public/leg.php?code=97&cl=3&cldcmpid=6258 | 1,537,532,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157070.28/warc/CC-MAIN-20180921112207-20180921132607-00498.warc.gz | 582,400,790 | 10,106 | # Search by Topic
#### Resources tagged with Plane shapes and their properties similar to Mid-point Area:
Filter by: Content type:
Age range:
Challenge level:
### There are 65 results
Broad Topics > Angles, Polygons, and Geometrical Proof > Plane shapes and their properties
### Semi-detached
##### Age 14 to 16 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Blue and White
##### Age 11 to 14 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Trapezium Four
##### Age 14 to 16 Challenge Level:
The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area?
##### Age 14 to 16 Challenge Level:
Investigate the properties of quadrilaterals which can be drawn with a circle just touching each side and another circle just touching each vertex.
### Semi-square
##### Age 14 to 16 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Square Pegs
##### Age 11 to 14 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Hex
##### Age 11 to 14 Challenge Level:
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Pi, a Very Special Number
##### Age 7 to 14
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### From All Corners
##### Age 14 to 16 Challenge Level:
Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
### Rhombus in Rectangle
##### Age 14 to 16 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
##### Age 14 to 16 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Some(?) of the Parts
##### Age 14 to 16 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Maxagon
##### Age 11 to 14 Challenge Level:
What's the greatest number of sides a polygon on a dotty grid could have?
### Squaring the Circle
##### Age 11 to 14 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Not So Little X
##### Age 11 to 14 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Floored
##### Age 11 to 14 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Circumspection
##### Age 14 to 16 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
### LOGO Challenge 12 - Concentric Circles
##### Age 11 to 16 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### Crescents and Triangles
##### Age 14 to 16 Challenge Level:
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.
### LOGO Challenge 11 - More on Circles
##### Age 11 to 16 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### Arclets Explained
##### Age 11 to 16
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### Dividing the Field
##### Age 14 to 16 Challenge Level:
A farmer has a field which is the shape of a trapezium as illustrated below. To increase his profits he wishes to grow two different crops. To do this he would like to divide the field into two. . . .
### What Shape?
##### Age 7 to 14 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
### Lying and Cheating
##### Age 11 to 14 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### What's Inside/outside/under the Box?
##### Age 7 to 14
This article describes investigations that offer opportunities for children to think differently, and pose their own questions, about shapes.
### Square Areas
##### Age 11 to 14 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### Circles, Circles Everywhere
##### Age 7 to 14
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Opposite Vertices
##### Age 11 to 14 Challenge Level:
Can you recreate squares and rhombuses if you are only given a side or a diagonal?
### Salinon
##### Age 14 to 16 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Towering Trapeziums
##### Age 14 to 16 Challenge Level:
Can you find the areas of the trapezia in this sequence?
##### Age 14 to 16 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### A Rational Search
##### Age 14 to 18 Challenge Level:
Investigate constructible images which contain rational areas.
### Efficient Packing
##### Age 14 to 16 Challenge Level:
How efficiently can you pack together disks?
### LOGO Challenge 6 - Triangles and Stars
##### Age 11 to 16 Challenge Level:
Recreating the designs in this challenge requires you to break a problem down into manageable chunks and use the relationships between triangles and hexagons. An exercise in detail and elegance.
### First Forward Into Logo 4: Circles
##### Age 7 to 16 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Pentagonal
##### Age 14 to 16 Challenge Level:
Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?
### The Medieval Octagon
##### Age 14 to 16 Challenge Level:
Medieval stonemasons used a method to construct octagons using ruler and compasses... Is the octagon regular? Proof please.
### Tricircle
##### Age 14 to 16 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### Pent
##### Age 14 to 18 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### LOGO Challenge 2 - Diamonds Are Forever
##### Age 7 to 16 Challenge Level:
The challenge is to produce elegant solutions. Elegance here implies simplicity. The focus is on rhombi, in particular those formed by jointing two equilateral triangles along an edge.
### LOGO Challenge 10 - Circles
##### Age 11 to 16 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### Holly
##### Age 14 to 16 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Lawnmower
##### Age 14 to 16 Challenge Level:
A kite shaped lawn consists of an equilateral triangle ABC of side 130 feet and an isosceles triangle BCD in which BD and CD are of length 169 feet. A gardener has a motor mower which cuts strips of. . . .
### Tied Up
##### Age 11 to 14 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Three Four Five
##### Age 14 to 16 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Like a Circle in a Spiral
##### Age 7 to 16 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### LOGO Challenge 8 - Rhombi
##### Age 7 to 16 Challenge Level:
Explore patterns based on a rhombus. How can you enlarge the pattern - or explode it?
### Approximating Pi
##### Age 14 to 18 Challenge Level:
By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation? | 2,466 | 10,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-39 | longest | en | 0.874879 |
https://homework.cpm.org/category/ACC/textbook/acc7/chapter/cc36/lesson/cc36.2.2/problem/6-60 | 1,571,427,437,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684425.36/warc/CC-MAIN-20191018181458-20191018204958-00080.warc.gz | 506,485,951 | 15,361 | ### Home > ACC7 > Chapter cc36 > Lesson cc36.2.2 > Problem6-60
6-60.
Figure 3 of a tile pattern has 11 tiles, while Figure 4 has 13 tiles. The pattern grows at a constant rate. Homework Help ✎
1. Write an equation to represent this situation.
Make a table for the pattern. Use the table to write your equation.
Figure # 0 1 2 3 4 $\$ Tiles 11 13
How much does the number of tiles increase as the figure number increases?
$y=2x+5$
2. Which figure will contain $1015$ tiles?
Let $y=1015$ in the equation. Then solve for $x$. | 160 | 532 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-43 | latest | en | 0.832698 |
https://wisdomanswer.com/what-is-the-number-1-followed-by-100-zeros-called/ | 1,720,906,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00258.warc.gz | 538,272,346 | 43,357 | # What is the number 1 followed by 100 zeros called?
## What is the number 1 followed by 100 zeros called?
googol
A googol is the large number 10100. In decimal notation, it is written as the digit 1 followed by one hundred zeroes: 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.
What number that includes 100 zeros was first used by a nine year old boy in 1938?
In 1938, a 9-year-old boy named Milton Sirotta, who was the nephew of American mathematician Edward Kasner, invented a new number that he called a googol. According to Milton, a googol is 10100, or 1 followed by 100 zeroes!
What term did 9-year-old Milton Sirotta coin for Uncle Eddie’s mathematician Edward Kasner big number?
Googol
The company was formed in 1998. In the late 1930s, noted mathematician and Columbia University professor Edward Kasner was asked to come up with a name for an extraordinarily large number. While on a walk one day, he asked his 9-year-old nephew, Milton Sirotta, if he had any ideas. “Googol,” the youngster replied. | 350 | 1,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.953776 |
http://www1.udel.edu/mvb/units.html | 1,501,037,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425751.38/warc/CC-MAIN-20170726022311-20170726042311-00510.warc.gz | 591,147,693 | 1,492 | # Study Tips
## Units
### Mechanics
Physical Quantity Unit
Distance [m]
Time [sec]
Mass [kg]
Mass density [kg/m3]
Velocity [m/sec]
Acceleration [m/sec2]
Force [Newton] = [kg-m/sec2]
Pressure [N/m2] = [pascal]
Work or Energy [Joule] = [N-m]
Momentum [kg-m/sec]
Angle degree or radian, both unitless
Torque [m-N]
Angular momentum [kg-m2/sec]
Moment of inertia [kg-m2]
### Electricity and Magnetism
Physical Quantity Unit
Electric Charge[Coulomb]
Electric Current[Ampere] = [Coulombs/second]
Electric Field[Newtons / Coulomb]
epsilon0[Coulomb2 / meter2 / Newton]
Electric Flux[Coulomb meter2]
Electric Potential[Volts] = [Joules / Coulomb] | 193 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-30 | longest | en | 0.587239 |
http://math453fall2008.wikidot.com/chapter-4 | 1,642,500,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00301.warc.gz | 48,570,970 | 9,046 | Chapter 4 is all about determining which numbers modulo p (where p is an odd prime) are actually squares of other numbers modulo p. This culminates in Gauss's amazing Law of Quadratic Reciprocity.
In today's class we began by talking about quadratic residues once again. We noticed that we could list the quadratic residues by squaring the first $\frac{p-1}{2}$ residues mod p, and that there were an equal number of quadratic residues and nonresidues. We then introduced the Legendre symbol $\left(\frac{a}{p}\right)$ as the "square indicator function modulo p". Finally we discussed Euler's Criterion for evaluating the Legendre symbol $\left(\frac{a}{p}\right)$ and a few of its consequences. | 159 | 696 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | longest | en | 0.887951 |
cubajl.readthedocs.io | 1,701,469,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00777.warc.gz | 225,151,228 | 21,809 | # Cuba¶
## Introduction¶
Cuba.jl is a Julia library for multidimensional numerical integration of real-valued functions of real arguments, using different algorithms.
This is just a Julia wrapper around the C Cuba library, version 4.2, by Thomas Hahn. All the credits goes to him for the underlying functions, blame me for any problem with the Julia interface.
All algorithms provided by Cuba library are supported in Cuba.jl:
Algorithm Basic integration method Type Variance reduction
Vegas Sobol quasi-random sample Monte Carlo importance sampling
or Mersenne Twister pseudo-random sample Monte Carlo
or Ranlux pseudo-random sample Monte Carlo
Suave Sobol quasi-random sample Monte Carlo
and importance sampling
or Mersenne Twister pseudo-random sample Monte Carlo
or Ranlux pseudo-random sample Monte Carlo
Divonne Korobov quasi-random sample Monte Carlo
stratified sampling
aided by methods from
numerical optimization
or Sobol quasi-random sample Monte Carlo
or Mersenne Twister pseudo-random sample Monte Carlo
or Ranlux pseudo-random sample Monte Carlo
or cubature rules deterministic
Cuhre cubature rules deterministic globally adaptive subdivision
For more details on the algorithms see the manual included in Cuba library and available in deps/cuba-julia/cuba.pdf after successful installation of Cuba.jl.
Integration is always performed on the $$n$$-dimensional unit hypercube $$[0, 1]^{n}$$.
Tip
If you want to compute an integral over a different set, you have to scale the integrand function in order to have an equivalent integral on $$[0, 1]^{n}$$ using substitution rules. For example, recall that in one dimension
$\int_{a}^{b} f(x)\,\mathrm{d}x = \int_{0}^{1} f(a + (b - a) y) (b - a)\,\mathrm{d}y$
where the final $$(b - a)$$ is the one-dimensional version of the Jacobian.
Integration over a semi-infinite or an inifite domain is a bit trickier, but you can follow this advice from Steven G. Johnson: to compute an integral over a semi-infinite interval, you can perform the change of variables $$x=a+y/(1-y)$$:
$\int_{a}^{\infty} f(x)\,\mathrm{d}x = \int_{0}^{1} f\left(a + \frac{y}{1 - y}\right)\frac{1}{(1 - y)^2}\,\mathrm{d}y$
For an infinite interval, you can perform the change of variables $$x=(2y - 1)/((1 - y)y)$$:
$\int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = \int_{0}^{1} f\left(\frac{2y - 1}{(1 - y)y}\right)\frac{2y^2 - 2y + 1}{(1 - y)^2y^2}\,\mathrm{d}y$
In addition, recall that for an even function $$\int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 2\int_{0}^{\infty}f(x)\,\mathrm{d}x$$, while the integral of an odd function over the infinite interval $$(-\infty, \infty)$$ is zero.
All this generalizes straightforwardly to more than one dimension. In Examples section you can find the computation of a 3-dimensional integral with non-constant boundaries using Cuba.jl and two integrals over infinite domains.
Cuba.jl is available for GNU/Linux, Mac OS, and Windows (i686 and x86_64 architectures).
## Installation¶
Cuba.jl is available for Julia 0.5 and later versions, and can be installed with Julia built-in package manager. In a Julia session run the commands
julia> Pkg.update()
Installation script on GNU/Linux and Mac OS systems will download Cuba Library source code and build the Cuba shared object. In order to accomplish this task a C compiler is needed. Instead, on Windows a prebuilt version of the library is downloaded.
Older versions are also available for Julia 0.4.
## Usage¶
After installing the package, run
using Cuba
or put this command into your Julia script.
Cuba.jl provides the following functions to integrate:
vegas(integrand, ndim=1, ncomp=1[; keywords...])
suave(integrand, ndim=1, ncomp=1[; keywords...])
divonne(integrand, ndim=1, ncomp=1[; keywords...])
cuhre(integrand, ndim=1, ncomp=1[; keywords...])
Large parts of the following sections are borrowed from Cuba manual. Refer to it for more information on the details.
Cuba.jl wraps the 64-bit integers functions of Cuba library, in order to push the range of certain counters to its full extent. In detail, the following arguments:
• for Vegas: nvec, minevals, maxevals, nstart, nincrease, nbatch, neval,
• for Suave: nvec, minevals, maxevals, nnew, nmin, neval,
• for Divonne: nvec, minevals, maxevals, ngiven, nextra, neval,
• for Cuhre: nvec, minevals, maxevals, neval,
are passed to the Cuba library as 64-bit integers, so they are limited to be at most
julia> typemax(Int64)
9223372036854775807
There is no way to overcome this limit. See the following sections for the meaning of each argument.
### Arguments¶
The only mandatory argument of integrator functions is:
• integrand (type: Function): the function to be integrated
Optional positional arguments are:
• ndim (type: Integer): the number of dimensions of the integratation domain. Defaults to 1 if omitted
• ncomp (type: Integer): the number of components of the integrand. Default to 1 if omitted
ndim and ncomp arguments must appear in this order, so you cannot omit ndim but not ncomp. integrand should be a function integrand(x, f) taking two arguments:
• the input vector x of length ndim
• the output vector f of length ncomp, used to set the value of each component of the integrand at point x
x and f are matrices with dimensions (ndim, nvec) and (ncomp, nvec), respectively, when nvec > 1. See the Vectorization section below for more information.
Also anonymous functions are allowed as integrand. For those familiar with Cubature.jl package, this is the same syntax used for integrating vector-valued functions.
For example, the integral
$\int_{0}^{1} \cos (x) \,\mathrm{d}x = \sin(1) = 0.8414709848078965$
can be computed with one of the following commands
julia> vegas((x, f) -> f[1] = cos(x[1]))
Component:
1: 0.8414910005259612 ± 5.2708169787342156e-5 (prob.: 0.028607201258072673)
Integrand evaluations: 13500
Fail: 0
Number of subregions: 0
julia> suave((x, f) -> f[1] = cos(x[1]))
Component:
1: 0.84115236906584 ± 8.357995609919512e-5 (prob.: 1.0)
Integrand evaluations: 22000
Fail: 0
Number of subregions: 22
julia> divonne((x, f) -> f[1] = cos(x[1]))
Component:
1: 0.841468071955942 ± 5.3955070531551656e-5 (prob.: 0.0)
Integrand evaluations: 1686
Fail: 0
Number of subregions: 14
julia> cuhre((x, f) -> f[1] = cos(x[1]))
Component:
1: 0.8414709848078966 ± 2.2204460420128823e-16 (prob.: 3.443539937576958e-5)
Integrand evaluations: 195
Fail: 0
Number of subregions: 2
In section Examples you can find more complete examples. Note that x and f are both arrays with type Float64, so Cuba.jl can be used to integrate real-valued functions of real arguments. See how to work with a complex integrand.
Note: if you used Cuba.jl until version 0.0.4, be aware that the user interface has been reworked in version 0.0.5 in a backward incompatible way.
### Optional Keywords¶
All other arguments required by Cuba integrator routines can be passed as optional keywords. Cuba.jl uses some reasonable default values in order to enable users to invoke integrator functions with a minimal set of arguments. Anyway, if you want to make sure future changes to some default values of keywords will not affect your current script, explicitely specify the value of the keywords.
#### Common Keywords¶
These are optional keywords common to all functions:
• nvec (type: Integer, default: 1): the maximum number of points to be given to the integrand routine in each invocation. Usually this is 1 but if the integrand can profit from e.g. Single Instruction Multiple Data (SIMD) vectorization, a larger value can be chosen. See Vectorization section.
• reltol (type: Real, default: 1e-4), and abstol (type: Real, default: 1e-12): the requested relative ($$\varepsilon_{\text{rel}}$$) and absolute ($$\varepsilon_{\text{abs}}$$) accuracies. The integrator tries to find an estimate $$\hat{I}$$ for the integral $$I$$ which for every component $$c$$ fulfills $$|\hat{I}_c - I_c|\leq \max(\varepsilon_{\text{abs}}, \varepsilon_{\text{rel}} |I_c|)$$.
• flags (type: Integer, default: 0): flags governing the integration:
• Bits 0 and 1 are taken as the verbosity level, i.e. 0 to 3, unless the CUBAVERBOSE environment variable contains an even higher value (used for debugging).
Level 0 does not print any output, level 1 prints “reasonable” information on the progress of the integration, level 2 also echoes the input parameters, and level 3 further prints the subregion results (if applicable).
• Bit 2 = 0: all sets of samples collected on a subregion during the various iterations or phases contribute to the final result.
Bit 2 = 1, only the last (largest) set of samples is used in the final result.
• (Vegas and Suave only)
Bit 3 = 0, apply additional smoothing to the importance function, this moderately improves convergence for many integrands.
Bit 3 = 1, use the importance function without smoothing, this should be chosen if the integrand has sharp edges.
• Bit 4 = 0, delete the state file (if one is chosen) when the integration terminates successfully.
Bit 4 = 1, retain the state file.
• (Vegas only)
Bit 5 = 0, take the integrator’s state from the state file, if one is present.
Bit 5 = 1, reset the integrator’s state even if a state file is present, i.e. keep only the grid. Together with Bit 4 this allows a grid adapted by one integration to be used for another integrand.
• Bits 8–31 =: level determines the random-number generator.
To select e.g. last samples only and verbosity level 2, pass 6 = 4 + 2 for the flags.
• seed (type: Integer, default: 0): the seed for the pseudo-random-number generator. This keyword is not available for cuhre(). The random-number generator is chosen as follows:
seed level (bits 8–31 of flags) Generator
zero N/A Sobol (quasi-random)
non-zero zero Mersenne Twister (pseudo-random)
non-zero non-zero Ranlux (pseudo-random)
Ranlux implements Marsaglia and Zaman’s 24-bit RCARRY algorithm with generation period $$p$$, i.e. for every 24 generated numbers used, another $$p - 24$$ are skipped. The luxury level is encoded in level as follows:
• Level 1 ($$p = 48$$): very long period, passes the gap test but fails spectral test.
• Level 2 ($$p = 97$$): passes all known tests, but theoretically still defective.
• Level 3 ($$p = 223$$): any theoretically possible correlations have very small chance of being observed.
• Level 4 ($$p = 389$$): highest possible luxury, all 24 bits chaotic.
Levels 5–23 default to 3, values above 24 directly specify the period $$p$$. Note that Ranlux’s original level 0, (mis)used for selecting Mersenne Twister in Cuba, is equivalent to level = 24.
• minevals (type: Real, default: 0): the minimum number of integrand evaluations required
• maxevals (type: Real, default: 1000000): the (approximate) maximum number of integrand evaluations allowed
• statefile (type: AbstractString, default: ""): a filename for storing the internal state. To not store the internal state, put "" (empty string, this is the default) or C_NULL (C null pointer).
Cuba can store its entire internal state (i.e. all the information to resume an interrupted integration) in an external file. The state file is updated after every iteration. If, on a subsequent invocation, a Cuba routine finds a file of the specified name, it loads the internal state and continues from the point it left off. Needless to say, using an existing state file with a different integrand generally leads to wrong results.
This feature is useful mainly to define “check-points” in long-running integrations from which the calculation can be restarted.
Once the integration reaches the prescribed accuracy, the state file is removed, unless bit 4 of flags (see above) explicitly requests that it be kept.
• spin (type: Ptr{Void}, default: C_NULL): this is the placeholder for the “spinning cores” pointer. Cuba.jl does not support parallelization, so this keyword should not have a value different from C_NULL.
#### Vegas-Specific Keywords¶
These optional keywords can be passed only to vegas():
• nstart (type: Integer, default: 1000): the number of integrand evaluations per iteration to start with
• nincrease (type: Integer, default: 500): the increase in the number of integrand evaluations per iteration
• nbatch (type: Integer, default: 1000): the batch size for sampling
Vegas samples points not all at once, but in batches of size nbatch, to avoid excessive memory consumption. 1000 is a reasonable value, though it should not affect performance too much
• gridno (type: Integer, default: 0): the slot in the internal grid table.
It may accelerate convergence to keep the grid accumulated during one integration for the next one, if the integrands are reasonably similar to each other. Vegas maintains an internal table with space for ten grids for this purpose. The slot in this grid is specified by gridno.
If a grid number between 1 and 10 is selected, the grid is not discarded at the end of the integration, but stored in the respective slot of the table for a future invocation. The grid is only re-used if the dimension of the subsequent integration is the same as the one it originates from.
In repeated invocations it may become necessary to flush a slot in memory, in which case the negative of the grid number should be set.
#### Suave-Specific Keywords¶
These optional keywords can be passed only to suave():
• nnew (type: Integer, default: 1000): the number of new integrand evaluations in each subdivision
• nmin (type: Integer, default: 2): the minimum number of samples a former pass must contribute to a subregion to be considered in that region’s compound integral value. Increasing nmin may reduce jumps in the $$\chi^2$$ value
• flatness (type: Real, default: .25): the type of norm used to compute the fluctuation of a sample. This determines how prominently “outliers”, i.e. individual samples with a large fluctuation, figure in the total fluctuation, which in turn determines how a region is split up. As suggested by its name, flatness should be chosen large for “flat” integrands and small for “volatile” integrands with high peaks. Note that since flatness appears in the exponent, one should not use too large values (say, no more than a few hundred) lest terms be truncated internally to prevent overflow.
#### Divonne-Specific Keywords¶
These optional keywords can be passed only to divonne():
• key1 (type: Integer, default: 47): determines sampling in the partitioning phase: key1 $$= 7, 9, 11, 13$$ selects the cubature rule of degree key1. Note that the degree-11 rule is available only in 3 dimensions, the degree-13 rule only in 2 dimensions.
For other values of key1, a quasi-random sample of $$n_1 = |\verb|key1||$$ points is used, where the sign of key1 determines the type of sample,
• key1 $$> 0$$, use a Korobov quasi-random sample,
• key1 $$< 0$$, use a “standard” sample (a Sobol quasi-random sample if seed $$= 0$$, otherwise a pseudo-random sample).
• key2 (type: Integer, default: 1): determines sampling in the final integration phase:
key2 $$= 7, 9, 11, 13$$ selects the cubature rule of degree key2. Note that the degree-11 rule is available only in 3 dimensions, the degree-13 rule only in 2 dimensions.
For other values of key2, a quasi-random sample is used, where the sign of key2 determines the type of sample,
• key2 $$> 0$$, use a Korobov quasi-random sample,
• key2 $$< 0$$, use a “standard” sample (see description of key1 above),
and $$n_2 = |\verb|key2||$$ determines the number of points,
• $$n_2\geq 40$$, sample $$n_2$$ points,
• $$n_2 < 40$$, sample $$n_2\,n_{\text{need}}$$ points, where $$n_{\text{need}}$$ is the number of points needed to reach the prescribed accuracy, as estimated by Divonne from the results of the partitioning phase
• key3 (type: Integer, default: 1): sets the strategy for the refinement phase:
key3 $$= 0$$, do not treat the subregion any further.
key3 $$= 1$$, split the subregion up once more.
Otherwise, the subregion is sampled a third time with key3 specifying the sampling parameters exactly as key2 above.
• maxpass (type: Integer, default: 5): controls the thoroughness of the partitioning phase: The partitioning phase terminates when the estimated total number of integrand evaluations (partitioning plus final integration) does not decrease for maxpass successive iterations.
A decrease in points generally indicates that Divonne discovered new structures of the integrand and was able to find a more effective partitioning. maxpass can be understood as the number of “safety” iterations that are performed before the partition is accepted as final and counting consequently restarts at zero whenever new structures are found.
• border (type: Real, default: 0.): the width of the border of the integration region. Points falling into this border region will not be sampled directly, but will be extrapolated from two samples from the interior. Use a non-zero border if the integrand function cannot produce values directly on the integration boundary
• maxchisq (type: Real, default: 10.): the $$\chi^2$$ value a single subregion is allowed to have in the final integration phase. Regions which fail this $$\chi^2$$ test and whose sample averages differ by more than mindeviation move on to the refinement phase.
• mindeviation (type: Real, default: 0.25): a bound, given as the fraction of the requested error of the entire integral, which determines whether it is worthwhile further examining a region that failed the $$\chi^2$$ test. Only if the two sampling averages obtained for the region differ by more than this bound is the region further treated.
• ngiven (type: Integer, default: 0): the number of points in the xgiven array
• ldxgiven (type: Integer, default: 0): the leading dimension of xgiven, i.e. the offset between one point and the next in memory
• xgiven (type: AbstractArray{Real}, default: zeros(Cdouble, ldxgiven, ngiven)): a list of points where the integrand might have peaks. Divonne will consider these points when partitioning the integration region. The idea here is to help the integrator find the extrema of the integrand in the presence of very narrow peaks. Even if only the approximate location of such peaks is known, this can considerably speed up convergence.
• nextra (type: Integer, default: 0): the maximum number of extra points the peak-finder subroutine will return. If nextra is zero, peakfinder is not called and an arbitrary object may be passed in its place, e.g. just 0
• peakfinder (type: Ptr{Void}, default: C_NULL): the peak-finder subroutine
#### Cuhre-Specific Keyword¶
This optional keyword can be passed only to cuhre():
• key (type: Integer, default: 0): chooses the basic integration rule:
key $$= 7, 9, 11, 13$$ selects the cubature rule of degree key. Note that the degree-11 rule is available only in 3 dimensions, the degree-13 rule only in 2 dimensions.
For other values, the default rule is taken, which is the degree-13 rule in 2 dimensions, the degree-11 rule in 3 dimensions, and the degree-9 rule otherwise.
### Output¶
The integrating functions vegas(), suave(), divonne(), and cuhre() (and the corresponding 64-bit integers functions) return an Integral object whose fields are
integral :: Vector{Float64}
error :: Vector{Float64}
probl :: Vector{Float64}
neval :: Int64
fail :: Int32
nregions :: Int32
The first three fields are arrays with length ncomp, the last three ones are scalars. The Integral object can also be iterated over like a tuple. In particular, if you assign the output of integrator functions to the variable named result, you can access the value of the i-th component of the integral with result[1][i] or result.integral[i] and the associated error with result[2][i] or result.error[i].
• integral (type: Vector{Float64}, with ncomp components): the integral of integrand over the unit hypercube
• error (type: Vector{Float64}, with ncomp components): the presumed absolute error for each component of integral
• probability (type: Vector{Float64}, with ncomp components): the $$\chi^2$$ -probability (not the $$\chi^2$$ -value itself!) that error is not a reliable estimate of the true integration error. To judge the reliability of the result expressed through prob, remember that it is the null hypothesis that is tested by the $$\chi^2$$ test, which is that error is a reliable estimate. In statistics, the null hypothesis may be rejected only if prob is fairly close to unity, say prob $$>.95$$
• neval (type: Int64): the actual number of integrand evaluations needed
• fail (type: Int32): an error flag:
• fail = 0, the desired accuracy was reached
• fail = -1, dimension out of range
• fail > 0, the accuracy goal was not met within the allowed maximum number of integrand evaluations. While Vegas, Suave, and Cuhre simply return 1, Divonne can estimate the number of points by which maxevals needs to be increased to reach the desired accuracy and returns this value.
• nregions (type: Int32): the actual number of subregions needed (always 0 in vegas())
## Vectorization¶
Vectorization means evaluating the integrand function for several points at once. This is also known as Single Instruction Multiple Data (SIMD) paradigm and is different from ordinary parallelization where independent threads are executed concurrently. It is usually possible to employ vectorization on top of parallelization.
Cuba.jl cannot automatically vectorize the integrand function, of course, but it does pass (up to) nvec points per integrand call (Common Keywords). This value need not correspond to the hardware vector length – computing several points in one call can also make sense e.g. if the computations have significant intermediate results in common.
When nvec > 1, the input x is a matrix of dimensions (ndim, nvec), while the output f is a matrix with dimensions (ncomp, nvec). Vectorization can be used to evaluate more quickly the integrand function, for example by exploiting parallelism, thus speeding up computation of the integral. See the section Vectorized Function below for an example of a vectorized funcion.
Note
Disambiguation: the nbatch argument of Vegas is related in purpose but not identical to nvec. It internally partitions the sampling done by Vegas but has no bearing on the number of points given to the integrand. On the other hand, it it pointless to choose nvec > nbatch for Vegas.
## Examples¶
### One dimensional integral¶
The integrand of
$\int_{0}^{1} \frac{\log(x)}{\sqrt{x}} \,\mathrm{d}x$
has an algebraic-logarithmic divergence for $$x = 0$$, but the integral is convergent and its value is $$-4$$. Cuba.jl integrator routines can handle this class of functions and you can easily compute the numerical approximation of this integral using one of the following commands:
julia> vegas( (x,f) -> f[1] = log(x[1])/sqrt(x[1]))
Component:
1: -3.998162393712848 ± 0.0004406643716840933 (prob.: 0.28430529682022004)
Integrand evaluations: 1007500
Fail: 1
Number of subregions: 0
julia> suave( (x,f) -> f[1] = log(x[1])/sqrt(x[1]))
Component:
1: -3.9999762867171387 ± 0.00039504866661845624 (prob.: 1.0)
Integrand evaluations: 51000
Fail: 0
Number of subregions: 51
julia> divonne( (x,f) -> f[1] = log(x[1])/sqrt(x[1]))
Component:
1: -3.9997602130594374 ± 0.0003567874814901272 (prob.: 1.0)
Integrand evaluations: 11593
Fail: 0
Number of subregions: 76
julia> cuhre( (x,f) -> f[1] = log(x[1])/sqrt(x[1]))
Component:
1: -4.00000035506719 ± 0.0003395484028625721 (prob.: 0.0)
Integrand evaluations: 5915
Fail: 0
Number of subregions: 46
### Vector-valued integrand¶
Consider the integral
$\int\limits_{\Omega} \boldsymbol{f}(x,y,z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$
where $$\Omega = [0, 1]^{3}$$ and
$\boldsymbol{f}(x,y,z) = \left(\sin(x)\cos(y)\exp(z), \,\exp(-(x^2 + y^2 + z^2)), \,\frac{1}{1 - xyz}\right)$
In this case it is more convenient to write a simple Julia script to compute the above integral
using Cuba, SpecialFunctions
function integrand(x, f)
f[1] = sin(x[1])*cos(x[2])*exp(x[3])
f[2] = exp(-(x[1]^2 + x[2]^2 + x[3]^2))
f[3] = 1/(1 - prod(x))
end
result, err = cuhre(integrand, 3, 3, abstol=1e-12, reltol=1e-10)
for i = 1:3
println("Component ", i)
println(" Result of Cuba: ", result[i], " ± ", err[i])
println(" Actual error: ", abs(result[i] - answer[i]))
end
This is the output
Component 1
Result of Cuba: 0.6646696797813739 ± 1.0050367631018485e-13
Exact result: 0.6646696797813771
Actual error: 3.219646771412954e-15
Component 2
Result of Cuba: 0.4165383858806454 ± 2.932866749838454e-11
Exact result: 0.41653838588663805
Actual error: 5.9926508200192075e-12
Component 3
Result of Cuba: 1.2020569031649702 ± 1.1958522385908214e-10
Exact result: 1.2020569031595951
Actual error: 5.375033751420233e-12
### Integral with non-constant boundaries¶
The integral
$\int_{-y}^{y}\int_{0}^{z}\int_{0}^{\pi} \cos(x)\sin(y)\exp(z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$
has non-constant boundaries. By applying the substitution rule repeatedly, you can scale the integrand function and get this equivalent integral over the fixed domain $$\Omega = [0, 1]^{3}$$
$\int\limits_{\Omega} 2\pi^{3}yz^2 \cos(\pi yz(2x - 1)) \sin(\pi yz) \exp(\pi z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$
that can be computed with Cuba.jl using the following Julia script
using Cuba
function integrand(x, f)
f[1] = 2pi^3*x[2]*x[3]^2*cos(pi*x[2]*x[3]*(2*x[1] - 1.0))*
sin(pi*x[2]*x[3])*exp(pi*x[3])
end
result, err = cuhre(integrand, 3, 1, abstol=1e-12, reltol=1e-10)
answer = pi*e^pi - (4e^pi - 4)/5
println("Result of Cuba: ", result[1], " ± ", err[1])
println("Actual error: ", abs(result[1] - answer))
This is the output
Result of Cuba: 54.98607586826157 ± 5.460606521639899e-9
Exact result: 54.98607586789537
Actual error: 3.6619951515604043e-10
### Integrals over Infinite Domains¶
Cuba.jl assumes always as integration domain the hypercube $$[0, 1]^n$$, but we have seen that using integration by substitution we can calculate integrals over different domains as well. In the Introduction we also proposed two useful substitutions that can be employed to change an infinite or semi-infinite domain into a finite one.
As a first example, consider the following integral with a semi-infinite domain:
$\int_{0}^{\infty}\frac{\log(1 + x^2)}{1 + x^2}\,\mathrm{d}x$
whose exact result is $$\pi\log 2$$. This can be computed with the following Julia script:
using Cuba
# The function we want to integrate over [0, ∞).
func(x) = log(1 + x^2)/(1 + x^2)
# Scale the function in order to integrate over [0, 1].
function integrand(x, f)
f[1] = func(x[1]/(1 - x[1]))/(1 - x[1])^2
end
result, err = cuhre(integrand, abstol = 1e-12, reltol = 1e-10)
println("Result of Cuba: ", result[1], " ± ", err[1])
println("Actual error: ", abs(result[1] - answer))
This is the output:
Result of Cuba: 2.177586090305688 ± 2.1503995410096295e-10
Exact result: 2.177586090303602
Actual error: 2.085887018665744e-12
Now we want to calculate this integral, over an infinite domain
$\int_{-\infty}^{\infty} \frac{1 - \cos x}{x^2}\,\mathrm{d}x$
which gives $$\pi$$. You can calculate the result with the code below. Note that integrand function has value $$1/2$$ for $$x=0$$, but you have to inform Julia about this.
using Cuba
# The function we want to integrate over (-∞, ∞).
func(x) = x==0 ? 0.5*one(x) : (1 - cos(x))/x^2
# Scale the function in order to integrate over [0, 1].
function integrand(x, f)
f[1] = func((2*x[1] - 1)/x[1]/(1 - x[1])) *
(2*x[1]^2 - 2*x[1] + 1)/x[1]^2/(1 - x[1])^2
end
result, err = cuhre(integrand, abstol = 1e-7, reltol = 1e-7)
println("Result of Cuba: ", result[1], " ± ", err[1])
println("Actual error: ", abs(result[1] - answer))
The output of this script is
Result of Cuba: 3.1415928900555046 ± 2.050669142074607e-6
Exact result: 3.141592653589793
Actual error: 2.3646571145619077e-7
### Complex integrand¶
As already explained, Cuba.jl operates on real quantities, so if you want to integrate a complex-valued function of complex arguments you have to treat complex quantities as 2-component arrays of real numbers. For example, if you do not remember Euler’s formula, you can compute this simple integral
$\int_{0}^{\pi/2} \exp(\mathrm{i} x)\,\mathrm{d}x$
with the following Julia script
using Cuba
function integrand(x, f)
# Complex integrand, scaled to integrate in [0, 1].
tmp = cis(x[1]*pi/2)*pi/2
# Assign to two components of "f" the real
# and imaginary part of the integrand.
f[1], f[2] = reim(tmp)
end
result = cuhre(integrand, 1, 2)
println("Result of Cuba: ", complex(result[1]...))
println("Exact result: ", complex(1.0, 1.0))
This is the output
Result of Cuba: 1.0 + 1.0im
Exact result: 1.0 + 1.0im
### Passing data to the integrand function¶
Cuba Library allows program written in C and Fortran to pass extra data to the integrand function with userdata argument. This is useful, for example, when the integrand function depends on changing parameters. In Cuba.jl the userdata argument is not available, but you do not normally need it.
For example, the cumulative distribution function $$F(x;k)$$ of chi-squared distribution is defined by
$F(x; k) = \int_{0}^{x} \frac{t^{k/2 - 1}\exp(-t/2)}{2^{k/2}\Gamma(k/2)} \,\mathrm{d}t$
The cumulative distribution function depends on parameter $$k$$, but the function passed as integrand to Cuba.jl integrator routines accepts as arguments only the input and output vectors. However you can easily define a function to calculate a numerical approximation of $$F(x; k)$$ based on the above integral expression because the integrand can access any variable visible in its scope. The following Julia script computes $$F(x = \pi; k)$$ for different $$k$$ and compares the result with more precise values, based on the analytic expression of the cumulative distribution function, provided by GSL.jl package.
using Cuba, GSL
function chi2cdf(x::Real, k::Real)
k2 = k/2
# Chi-squared probability density function, without constant denominator.
# The result of integration will be divided by that factor.
function chi2pdf(t::Float64)
# "k2" is taken from the outside.
return t^(k2 - 1.0)*exp(-t/2)
end
# Neither "x" is passed directly to the integrand function,
# but is visible to it. "x" is used to scale the function
# in order to actually integrate in [0, 1].
x*cuhre((t,f) -> f[1] = chi2pdf(t[1]*x))[1][1]/(2^k2*gamma(k2))
end
x = pi
@printf("Result of Cuba: %.6f %.6f %.6f %.6f %.6f\n",
map((k) -> chi2cdf(x, k), collect(1:5))...)
@printf("Exact result: %.6f %.6f %.6f %.6f %.6f\n",
map((k) -> cdf_chisq_P(x, k), collect(1:5))...)
This is the output
Result of Cuba: 0.923681 0.792120 0.629694 0.465584 0.321833
Exact result: 0.923681 0.792120 0.629695 0.465584 0.321833
### Vectorized Function¶
Consider the integral
$\int\limits_{\Omega} \prod_{i=1}^{10} \cos(x_{i}) \,\mathrm{d}\boldsymbol{x} = \sin(1)^{10} = 0.1779883\dots$
where $$\Omega = [0, 1]^{10}$$ and $$\boldsymbol{x} = (x_{1}, \dots, x_{10})$$ is a 10-dimensional vector. A simple way to compute this integral is the following:
julia> using Cuba, BenchmarkTools
julia> cuhre((x, f) -> f[] = prod(cos.(x)), 10)
Component:
1: 0.1779870665870775 ± 1.0707995959536173e-6 (prob.: 0.2438374075714901)
Integrand evaluations: 7815
Fail: 0
Number of subregions: 2
julia> @benchmark cuhre((x, f) -> f[] = prod(cos.(x)), 10)
BenchmarkTools.Trial:
memory estimate: 2.62 MiB
allocs estimate: 39082
--------------
minimum time: 1.633 ms (0.00% GC)
median time: 1.692 ms (0.00% GC)
mean time: 1.867 ms (8.62% GC)
maximum time: 3.660 ms (45.54% GC)
--------------
samples: 2674
evals/sample: 1
We can use vectorization in order to speed up evaluation of the integrand function.
julia> function fun_vec(x,f)
f[1,:] .= 1.0
for j in 1:size(x,2)
for i in 1:size(x, 1)
f[1, j] *= cos(x[i, j])
end
end
end
fun_vec (generic function with 1 method)
julia> cuhre(fun_vec, 10, nvec = 1000)
Component:
1: 0.1779870665870775 ± 1.0707995959536173e-6 (prob.: 0.2438374075714901)
Integrand evaluations: 7815
Fail: 0
Number of subregions: 2
julia> @benchmark cuhre(fun_vec2, 10, nvec = 1000)
BenchmarkTools.Trial:
memory estimate: 2.88 KiB
allocs estimate: 54
--------------
minimum time: 949.976 μs (0.00% GC)
median time: 954.039 μs (0.00% GC)
mean time: 966.930 μs (0.00% GC)
maximum time: 1.204 ms (0.00% GC)
--------------
samples: 5160
evals/sample: 1
A further speed up can be gained by running the for loop in parallel with Threads.@threads. For example, running Julia with 4 threads:
julia> function fun_par(x,f)
f[1,:] .= 1.0
for i in 1:size(x, 1)
f[1, j] *= cos(x[i, j])
end
end
end
fun_par (generic function with 1 method)
julia> cuhre(fun_par, 10, nvec = 1000)
Component:
1: 0.1779870665870775 ± 1.0707995959536173e-6 (prob.: 0.2438374075714901)
Integrand evaluations: 7815
Fail: 0
Number of subregions: 2
julia> @benchmark cuhre(fun_par, 10, nvec = 1000)
BenchmarkTools.Trial:
memory estimate: 3.30 KiB
allocs estimate: 63
--------------
minimum time: 507.914 μs (0.00% GC)
median time: 515.182 μs (0.00% GC)
mean time: 520.667 μs (0.06% GC)
maximum time: 3.801 ms (85.06% GC)
--------------
samples: 9565
evals/sample: 1
## Performance¶
Cuba.jl cannot (yet?) take advantage of parallelization capabilities of Cuba Library. Nonetheless, it has performances comparable with equivalent native C or Fortran codes based on Cuba library when CUBACORES environment variable is set to 0 (i.e., multithreading is disabled). The following is the result of running the benchmark present in test directory on a 64-bit GNU/Linux system running Julia 0.7.0-DEV.363 (commit 6071f1a02e) equipped with an Intel(R) Core(TM) i7-4700MQ CPU. The C and FORTRAN 77 benchmark codes have been compiled with GCC 6.3.0.
$CUBACORES=0 julia -e 'cd(Pkg.dir("Cuba")); include("test/benchmark.jl")' INFO: Performance of Cuba.jl: 0.271304 seconds (Vegas) 0.579783 seconds (Suave) 0.329504 seconds (Divonne) 0.238852 seconds (Cuhre) INFO: Performance of Cuba Library in C: 0.319799 seconds (Vegas) 0.619774 seconds (Suave) 0.340317 seconds (Divonne) 0.266906 seconds (Cuhre) INFO: Performance of Cuba Library in Fortran: 0.272000 seconds (Vegas) 0.584000 seconds (Suave) 0.308000 seconds (Divonne) 0.232000 seconds (Cuhre) Of course, native C and Fortran codes making use of Cuba Library outperform Cuba.jl when higher values of CUBACORES are used, for example: $ CUBACORES=1 julia -e 'cd(Pkg.dir("Cuba")); include("test/benchmark.jl")'
INFO: Performance of Cuba.jl:
0.279524 seconds (Vegas)
0.581078 seconds (Suave)
0.327319 seconds (Divonne)
0.241211 seconds (Cuhre)
INFO: Performance of Cuba Library in C:
0.115113 seconds (Vegas)
0.596503 seconds (Suave)
0.152511 seconds (Divonne)
0.085805 seconds (Cuhre)
INFO: Performance of Cuba Library in Fortran:
0.108000 seconds (Vegas)
0.604000 seconds (Suave)
0.160000 seconds (Divonne)
0.092000 seconds (Cuhre)
Cuba.jl internally fixes CUBACORES to 0 in order to prevent from forking julia processes that would only slow down calculations eating up the memory, without actually taking advantage of concurrency. Furthemore, without this measure, adding more Julia processes with addprocs() would only make the program segfault.
## Development¶
Cuba.jl is developed on GitHub: https://github.com/giordano/Cuba.jl. Feel free to report bugs and make suggestions at https://github.com/giordano/Cuba.jl/issues.
### History¶
The ChangeLog of the package is available in NEWS.md file in top directory. There have been some breaking changes from time to time, beware of them when upgrading the package.
## Credits¶
If you use this library for your work, please credit Thomas Hahn. Citable papers about Cuba Library: | 10,162 | 36,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.742918 |
https://www.mathematik.uni-marburg.de/modulhandbuch/20211/MSc_Mathematics/Specialization_Modules_in_Mathematics/Elementary_Topology.html | 1,722,656,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00114.warc.gz | 696,982,389 | 7,177 | This entry is from Summer semester 2021 and might be obsolete. You can find a current equivalent here.
# Elementary Topology (dt. Elementare Topologie)
Level, degree of commitment Advanced module, compulsory elective module Forms of teaching and learning,workload Lecture (3 SWS), recitation class (1 SWS), 180 hours (60 h attendance, 120 h private study) Credit points,formal requirements 6 CP Course requirement(s): Successful completion of at least 50 percent of the points from the weekly exercises. Examination type: Written or oral examination Language,Grading German,The grading is done with 0 to 15 points according to the examination regulations for the degree program B.Sc. Mathematics. Duration,frequency One semester, irregular Person in charge of the module's outline Prof. Dr. Ilka Agricola
## Contents
• Topological spaces and manifolds
• Elementary properties of topological spaces: compactness, orientability, boundary. Many examples: Möbius band, Klein's bottle, projective space etc.
• Classification of surfaces, genus of a surface, triangulations, Boy's surface
• Euler Characteristic and Euler's polyhedron theorem
• Fundamental group, mapping degree and coverings
## Qualification Goals
The students shall
• understand basic principles of topological structures and recognize that such structures can be found in many parts of mathematics,
• practice axiomatic procedures and train their abstraction skills,
• practice mathematical working methods (development of mathematical intuition and its formal justification, proof techniques),
• improve their oral communication skills in the exercises by practicing free speech in front of an audience and during discussion.
## Prerequisites
None. The competences taught in the following modules are recommended: either Foundations of Mathematics and Linear Algebra I and Linear Algebra II or Basic Linear Algebra, either Analysis I and Analysis II or Basic Real Analysis.
## Applicability
Module imported from B.Sc. Mathematics.
It can be attended at FB12 in study program(s)
• B.Sc. Computer Science
• B.Sc. Mathematics
• M.Sc. Computer Science
• M.Sc. Mathematics
• LAaG Mathematics
When studying M.Sc. Mathematics, this module can be attended in the study area Specialization Modules in Mathematics.
The module is assigned to Pure Mathematics. Further information on eligibility can be found in the description of the study area.
• Boltjanskij, V.G. und Efremovic, V.A.: Anschauliche kombinatorische Topologie. VEB Deutscher Verlag der Wissenschaften (1986).
• Hatcher, A.: Algebraic topology. Cambridge University Press (2002).
• Hu, S.-T.: Homotopy Theory. Academic Press (1959).
• Ossa, E.: Topologie. Vieweg-Verlag (1992).
• Pontrjagin, L.S.: Grundzüge der kombinatorischen Topologie. VEB Deutscher Verlag der Wissenschaften (1956).
• Stöcker, R. und Zieschang, H.: Algebraische Topologie. Eine Einführung. Teubner-Verlag (1988). | 653 | 2,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.872496 |
https://mathematica.stackexchange.com/questions/145345/product-of-summation-too-slow | 1,627,688,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00332.warc.gz | 377,792,532 | 39,686 | # Product of summation too slow
I have been trying to do the below given problem but even after 5-10 minutes results are not obtained so i killed the job.
AbsoluteTiming[Sum[Hypergeometric2F1[k^2*Sin[k*(Pi/6)]^2,k/2, 0.5,-1]
*Sin[k*(Pi/6)]^2*Sum[0.8*(Pochhammer[k - p, 0.5]/Cos[Pi/6]^2),{p, 1, k^2}]
,{k, 1, 500}]]
Also how can I increase the number of decimal of the result or how can i avoid numerical rounding issue. Thank you.
• You'll get more help if you include the code which can be copy-and-pasted into Mathematica rather than a picture of the code. – JimB May 5 '17 at 23:26
If I entered your code in correctly and then change 0.5 to 1/2 and 0.8 to 4/5, it looks like you might only need to sum to maybe 20 rather than 500:
data = Table[{n, N[Sum[Hypergeometric2F1[k^2 Sin[k π/6]^2, k/2,
1/2, -1] Sin[k π/6]^2 Sum[(4/5) Pochhammer[k - p, 1/2]/
Cos[π/6]^2, {p, 1, k^2}], {k, 1, n}], 20]}, {n, 10, 30}]
{{10, -0.071544615273026846805}, {11, -0.071571059489707821402},
{12, -0.071571059489707821402}, {13, -0.071570886837749513845},
{14, -0.071570886870953425127}, {15, -0.071570886870953425127},
{16, -0.071570886870709821681}, {17, -0.071570886870544288955},
{18, -0.071570886870544288955}, {19, -0.071570886870544312666},
{20, -0.071570886870544304098}, {21, -0.071570886870544304098},
{22, -0.071570886870544304141}, {23, -0.071570886870544304141},
{24, -0.071570886870544304141}, {25, -0.071570886870544304141},
{26, -0.071570886870544304141}, {27, -0.071570886870544304141},
{28, -0.071570886870544304141}, {29, -0.071570886870544304141},
{30, -0.071570886870544304141}}
• That's ok. But Imagine a case similar to this where it get converge only at very high value of upper limit of k i.e n(in your code), so that it takes much time as you see in the same case when n is 500 it takes long. Then what approach will be reliable. – charlz_bro May 6 '17 at 9:50
• The answer to your question is "It depends." Certainly one would first want to attempt to see if the slow convergence is due to slowness of the terms going to zero or if there's some oscillatory nature to the terms. But to get specific advice, I think you'd need to have a specific example. – JimB May 6 '17 at 16:26
• Can you suggest me some methods that people usually do in such cases. – charlz_bro May 17 '17 at 21:46
• Who says that the above doesn't give the desired answer when the upper limit is 500? You just might have to wait a long time. And if the person requesting you to use 500 is on your committee, then you have my condolences. – JimB May 17 '17 at 22:22
• Just an upper limit of 20 does the job in this case. If one chooses to use and upper limit of 500 unnecessarily which takes a whole lot longer, I don't see that being the fault of Mathematica. – JimB May 17 '17 at 23:05
As an experiment, I tried simplifying the internal summation. Replace decimals with rational expressions as per @Jim Baldwin. Then blast it with FullSimplify[ ] to get a closed form solution...
Sum[(4/5)*(Pochhammer[kk - p, 1/2]/Cos[Pi/6]^2), {p, 1, kk^2}] // FullSimplify
= (32*(Gamma[1/2 + kk]/Gamma[-1 + kk] - Gamma[1/2 + kk - kk^2]/Gamma[-1 + kk - kk^2]))/45
Stuffing this into the original summation and setting decimals to fractions, and you can actually get the full 500 in short order (just under 5 seconds).
Timing[res = Sum[Hypergeometric2F1[k^2*Sin[k*(Pi/6)]^2, k/2, 1/2, -1]
*Sin[k*(Pi/6)]^2
*(32*(Gamma[1/2 + k]/Gamma[-1 + k] - Gamma[1/2 + k - k^2]/Gamma[-1 + k - k^2]))/45,
{k, 1, 500}];]
{4.875, Null}
But it is in symbolic form. When you ask for a numerical answer with a //N...resolving res//N takes forever unfortunately. Barely faster than the original summation. Go figure...
• Imagine a case where there is one more sum after second summation – charlz_bro May 6 '17 at 9:53
• Sorry as in this case: AbsoluteTiming[Sum[Hypergeometric2F1[k^2*Sin[k*(Pi/6)]^2, k/2, 0.5, -1]*Sin[k*(Pi/6)]^2* Sum[0.8*(Pochhammer[k - p, 0.5]/Cos[Pi/6]^2)* Sum[(q - p)^2, {q, 1, 500}], {p, 1, k^2}], {k, 1, 500}]] – charlz_bro May 6 '17 at 10:01
• Hypergeometric2F1 is time-consuming. – UnchartedWorks May 6 '17 at 10:53
• Ok. but how can I get out of this problem. Is there any better approach in these kind of problems especially when the sum is slowly converging, the problem of the type I posted in my comment. – charlz_bro May 6 '17 at 12:53 | 1,519 | 4,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-31 | longest | en | 0.785419 |
readingwittgenstein.blogspot.com | 1,503,379,961,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110471.85/warc/CC-MAIN-20170822050407-20170822070407-00523.warc.gz | 349,505,955 | 10,042 | This log was inspired by "How to Read Wittgenstein" and "Ludwig Wittgenstein: the duty of genius" by Ray Monk. It is based on reading Tractatus Logico-Philosophicus by Ludwig Wittgenstein translated by D. F. Pears & B. F. McGuinness (Routledge and Kegan Paul:1963)
## Saturday, April 5, 2008
### The General Form of a Truth Function.
Wittgenstein did not fully explain his symbolism in the Tractatus. The following uses text from Russell's introduction.
The general form of a truth function is: [p-, ξ-, N(ξ-)] where
p- stands for all elemental propositions,
ξ- stands for any set of propositions, and
N(ξ-) stands for the negation of all members of ξ-.
This general form of a proposition simply says that every proposition generated by successive negations of elemental propositions. The whole symbol [p-, ξ-, N(ξ-)] is an algorithm:
• select a set of any of the atomic propositions,
• negate them all,
• then take any selection of the set of propositions now obtained,
• together with any of the originals
• -- and so on indefinitely.
This describes a procedure which, when performed on given elemental propositions, can generate all other propositions that are not elemental. The process depends upon:
(a) Sheffer's proof that all truth-functions can be obtained out of simultaneous negation, i.e. out of ``not-p and not-q'';
(b) Wittgenstein's theory of the derivation of general propositions from conjunctions and disjunctions; and
(c) The assertion that a proposition can only occur in another proposition as argument to a truth-function.
If the general form of constructing propositions is given, then how one proposition can be generated from another is also given. Thus, the general form of an operation Ω'(η-) can be written as: [ξ-, N(ξ-)]'(η-) or as [η-,ξ-,N(ξ-)]. This is the most general form of transition from one proposition to another.
To apply this method to arrive at numbers one defines:
x=Ω^(0)'x Def.
and
Ω'Ω^(ν)'x=Ω^(ν+1)'x Def.
Now, according to the rules of our notation, one write the series
x, Ω'x, Ω'Ω'x, Ω'Ω'Ω'x,...,
so that
Ω^(0)'x, Ω^(0+1)'x, Ω^(0+1+1)'x, Ω^(0+1+1+1)'x, .....
Therfore, one can rewrite the general form [x, ξ, Ω'ξ]) as
[Ω0'x, Ων'x, Ων+1'x]
and define:
0+1=1 Def.,
0+1+1=2 Def.,
0+1+1+1=3 Def.,
etc.
One sees from this that a number is the exponent of an operation.
The concept number is just the general form of a number which is common to all numbers; it is the variable number. Also, the concept numerical equality is the general form of all particular cases of numerical equality.
So the general form of an integer is: [0, ξ, ξ+1].
The theory of classes is completely superfluous in mathematics due to the fact that the generality required in mathematics is not accidental generality.
113andre said...
useful | 735 | 2,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-34 | longest | en | 0.882021 |
http://www.3dinosaurs.com/wordpress/index.php/apple-themed-multiplication-division/ | 1,490,849,609,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00546-ip-10-233-31-227.ec2.internal.warc.gz | 407,188,674 | 12,746 | # Apple Themed Multiplication & Division
We are at the start of the year, and some kids are just learning their multiplication and division while others are all set to start their review from the year before. I wanted to have something fun to work on it this year!
So I made my girls favorite activities the spinning wheel math activities. This is a very fun way to work on review or learning. I thought I would add a fun fall twist to it and make a multiplication and division activities for the fall.
This apple themed set has been fun to use!
I wanted to give them something they could practice on without any prep involved at all. Plus make a few changes to the hands-on math games that they love so much.
## No Prep Multiplication & Division Printables.
There is a multiplication spinning set that goes from 1 to 12. Each of the sheets has a spinner on it plus an apple tree covered with all the answers. They use a pencil and paperclip to make a spinner. They spin the wheel and color the correct answer. There are 4 sets of the answers for them to color. This gives loads of practice. Plus there is no prep involved.
For those just learning their multiplication, you can use the multiplication bookmarks off to the side to help them.
There is a matching set of division spinning worksheets that go from 1 to 12. It is similar to the multiplication worksheet. It is a great no prep printable. You can also use the division bookmarks as a help while doing these. Every sheet has the 1 through 12 numbers in different spots.
The set includes some fun apple themed multiplication and division math fact sheets and some wheels that are just fun and easy for kids to fill out. I like to have some simple pages they can work on their math facts.
Another fun part of this printables is the apple themed math fact house for multiplication and division. We also use the apple themed flashcards for multiplication and division. This is a fun way to work on these. I like to laminate these to use over and over but, if you are out of sheets you can use sheet protectors. The sheet has an option in color or black and white.
## Hands-On Apple Rolling or Spinning Game
We love the rolling games and I love having different ways to work it. I updated the apple set for 6-side and 12 sided dice. This is a fun way to work on the math facts and let kids add apples to the tree.
There is a spinner included for the 6 and 12 option as well. This means if you don’t have a dice you can still play the game.
This set also includes an option for working division as well. You have an apple spinner for numbers 1 to 12. You spin the spinner and do the division and then add the correct apples.
What you will find in these printables:
• Multiplication Spinning Worksheets 1 to 12
• Division Spinning Worksheets 1 to 12
• Math Fact Apples
• Math Fact Wheels
• Apple Themed Blank Math Fact Sheets Color and Black & White
• Roll an Apple Tree Directions for Multiplication with dice
• Spin an Apple Tree Directions for Multiplication with apple spinner
• Spin an Apple Tree Directions for Division
• Apple Pieces for Roll or Spin an Apple Tree
• 6 and 12 Apple Spinners
• Division Apple spinners 1 through 12
• Multiplication Sheet to Write equations
• Division Sheet to Write equations
• Apple Themed Multiplication Flashcards – 1 to 12
• Apple Themed Division Flashcards – 1 to 12
• 80 pages of printables
• Price: \$4
Buy the Apple Multiplication for \$4.
You can check out a sample of the Apple Themed Multiplication & Division here.
Check out these other apple printables:
If you want more apple activities and printables check them all out:
Looking for more Fall Printables then check out all of them on 3 Dinosaurs.
Cassie – 3Dinosaurs.com | 834 | 3,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-13 | longest | en | 0.95766 |
http://soultionmanual.blogspot.com/2017/02/chapter-26-exercise-5-introduction-to.html | 1,531,911,514,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00537.warc.gz | 351,438,620 | 21,049 | ## Tuesday, 14 February 2017
### Chapter 26 Exercise 5, Introduction to Java Programming, Tenth Edition Y. Daniel LiangY.
26.5 (The kth smallest element) You can find the kth smallest element in a BST in
O(n) time from an inorder iterator. For an AVL tree, you can find it in O(log n)
time. To achieve this, add a new data field named size in AVLTreeNode to
store the number of nodes in the subtree rooted at this node. Note that the size of a node v is one more than the sum of the sizes of its two children. Figure 26.12 shows an AVL tree and the size value for each node in the tree.
In the AVLTree class, add the following method to return the kth smallest ele-
ment in the tree.
public E find(int k)
The method returns null if k < 1 or k > the size of the tree . This method
can be implemented using the recursive method find(k, root) , which returns
the kth smallest element in the tree with the specified root. Let A and B be the left
and right children of the root, respectively. Assuming that the tree is not empty
and k ... root.size, find(k, root) can be recursively defined as follows:
root.element, if A is null and k is 1;
B.element, if A is null and k is 2;
find(k, root) = E find(k, A), if k 6 = A.size;
root.element, if k = A.size + 1;
find(k - A.size - 1, B), if k 7 A.size + 1;
Modify the insert and delete methods in AVLTree to set the correct value
for the size property in each node. The insert and delete methods will still
be in O(log n) time. The find(k) method can be implemented in O(log n) time.
Therefore, you can find the kth smallest element in an AVL tree in O(log n) time.
Use the following main method to test your program:
import java.util.Scanner;
public class Exercise26_05 {
public static void main(String[] args) {
AVLTree<Double> tree = new AVLTree<>();
Scanner input = new Scanner(System.in);
System.out.print("Enter 15 numbers: ");
for (int i = 0; i < 15; i++) {
tree.insert(input.nextDouble());
}
System.out.print("Enter k: ");
System.out.println("The " + k + "th smallest number is " +
tree.find(k));
}
}
import java.util.ArrayList;
public class Exercise05 {
public static void main(String[] args) {
AVLTree<Integer> tree = new AVLTree<Integer>();
for (int i = 1; i <= 1000; i++) {
tree.insert(i);
}
System.out.println(tree.find(1));
System.out.println(tree.find(2));
System.out.println(tree.find(3));
System.out.println(tree.find(500));
System.out.println(tree.find(501));
System.out.println(tree.find(998));
System.out.println(tree.find(999));
System.out.println(tree.find(1000));
}
static class AVLTree<E extends Comparable<E>> extends BST<E> {
/** Create a default AVL tree */
public AVLTree() {
}
/** Create an AVL tree from an array of objects */
public AVLTree(E[] objects) {
super(objects);
}
public E find(int k) {
if ((k < 0) || (k > size)) {
return null;
} else {
return find(k, (AVLTreeNode<E>) root);
}
}
public E find(int k, AVLTreeNode<E> node) {
AVLTreeNode<E> A = (AVLTreeNode<E>)node.left;
AVLTreeNode<E> B = (AVLTreeNode<E>)node.right;
if ((A == null)&&(k == 1)) {
return node.element;
} else if ((A == null)&&(k == 2)) {
return B.element;
} else if(k <= A.size) {
return find(k, A);
} else if(k == A.size + 1) {
return node.element;
} else {
return find(k - A.size - 1, B);
}
}
}
if (node == null) {
return 0;
} else {
return node.size;
}
}
@Override
/** Override createNewNode to create an AVLTreeNode */
protected AVLTreeNode<E> createNewNode(E e) {
return new AVLTreeNode<E>(e);
}
@Override
/** Insert an element and rebalance if necessary */
public boolean insert(E e) {
boolean successful = super.insert(e);
if (!successful)
return false; // e is already in the tree
else {
balancePath(e); // Balance from e to the root if necessary
}
return true; // e is inserted
}
/** Update the height of a specified node */
private void updateHeight(AVLTreeNode<E> node) {
if (node.left == null && node.right == null) // node is a leaf
node.height = 0;
else if (node.left == null) // node has no left subtree
node.height = 1 + ((AVLTreeNode<E>) (node.right)).height;
else if (node.right == null) // node has no right subtree
node.height = 1 + ((AVLTreeNode<E>) (node.left)).height;
else
node.height = 1 + Math.max(
((AVLTreeNode<E>) (node.right)).height,
((AVLTreeNode<E>) (node.left)).height);
}
/**
* Balance the nodes in the path from the specified node to the root if
* necessary
*/
private void balancePath(E e) {
java.util.ArrayList<TreeNode<E>> path = path(e);
for (int i = path.size() - 1; i >= 0; i--) {
AVLTreeNode<E> A = (AVLTreeNode<E>) (path.get(i));
updateHeight(A);
AVLTreeNode<E> parentOfA = (A == root) ? null
: (AVLTreeNode<E>) (path.get(i - 1));
switch (balanceFactor(A)) {
case -2:
if (balanceFactor((AVLTreeNode<E>) A.left) <= 0) {
balanceLL(A, parentOfA); // Perform LL rotation
} else {
balanceLR(A, parentOfA); // Perform LR rotation
}
break;
case +2:
if (balanceFactor((AVLTreeNode<E>) A.right) >= 0) {
balanceRR(A, parentOfA); // Perform RR rotation
} else {
balanceRL(A, parentOfA); // Perform RL rotation
}
}
}
}
/** Return the balance factor of the node */
private int balanceFactor(AVLTreeNode<E> node) {
if (node.right == null) // node has no right subtree
return -node.height;
else if (node.left == null) // node has no left subtree
return +node.height;
else
return ((AVLTreeNode<E>) node.right).height
- ((AVLTreeNode<E>) node.left).height;
}
/** Balance LL (see Figure 9.1) */
private void balanceLL(TreeNode<E> A, TreeNode<E> parentOfA) {
TreeNode<E> B = A.left; // A is left-heavy and B is left-heavy
if (A == root) {
root = B;
} else {
if (parentOfA.left == A) {
parentOfA.left = B;
} else {
parentOfA.right = B;
}
}
A.left = B.right; // Make T2 the left subtree of A
B.right = A; // Make A the left child of B
updateHeight((AVLTreeNode<E>) A);
updateHeight((AVLTreeNode<E>) B);
}
/** Balance LR (see Figure 9.1(c)) */
private void balanceLR(TreeNode<E> A, TreeNode<E> parentOfA) {
TreeNode<E> B = A.left; // A is left-heavy
TreeNode<E> C = B.right; // B is right-heavy
if (A == root) {
root = C;
} else {
if (parentOfA.left == A) {
parentOfA.left = C;
} else {
parentOfA.right = C;
}
}
A.left = C.right; // Make T3 the left subtree of A
B.right = C.left; // Make T2 the right subtree of B
C.left = B;
C.right = A;
updateHeight((AVLTreeNode<E>) A);
updateHeight((AVLTreeNode<E>) B);
updateHeight((AVLTreeNode<E>) C);
}
/** Balance RR (see Figure 9.1(b)) */
private void balanceRR(TreeNode<E> A, TreeNode<E> parentOfA) {
TreeNode<E> B = A.right; // A is right-heavy and B is right-heavy
if (A == root) {
root = B;
} else {
if (parentOfA.left == A) {
parentOfA.left = B;
} else {
parentOfA.right = B;
}
}
A.right = B.left; // Make T2 the right subtree of A
B.left = A;
updateHeight((AVLTreeNode<E>) A);
updateHeight((AVLTreeNode<E>) B);
}
/** Balance RL (see Figure 9.1(d)) */
private void balanceRL(TreeNode<E> A, TreeNode<E> parentOfA) {
TreeNode<E> B = A.right; // A is right-heavy
TreeNode<E> C = B.left; // B is left-heavy
if (A == root) {
root = C;
} else {
if (parentOfA.left == A) {
parentOfA.left = C;
} else {
parentOfA.right = C;
}
}
A.right = C.left; // Make T2 the right subtree of A
B.left = C.right; // Make T3 the left subtree of B
C.left = A;
C.right = B;
updateHeight((AVLTreeNode<E>) A);
updateHeight((AVLTreeNode<E>) B);
updateHeight((AVLTreeNode<E>) C);
}
@Override
/** Delete an element from the binary tree.
* Return true if the element is deleted successfully
* Return false if the element is not in the tree */
public boolean delete(E element) {
if (root == null)
return false; // Element is not in the tree
// Locate the node to be deleted and also locate its parent node
TreeNode<E> parent = null;
TreeNode<E> current = root;
while (current != null) {
if (element.compareTo(current.element) < 0) {
parent = current;
current = current.left;
} else if (element.compareTo(current.element) > 0) {
parent = current;
current = current.right;
} else
break; // Element is in the tree pointed by current
}
if (current == null)
return false; // Element is not in the tree
// Case 1: current has no left children (See Figure 23.6)
if (current.left == null) {
// Connect the parent with the right child of the current node
if (parent == null) {
root = current.right;
} else {
if (element.compareTo(parent.element) < 0)
parent.left = current.right;
else
parent.right = current.right;
// Balance the tree if necessary
balancePath(parent.element);
}
} else {
// Case 2: The current node has a left child
// Locate the rightmost node in the left subtree of
// the current node and also its parent
TreeNode<E> parentOfRightMost = current;
TreeNode<E> rightMost = current.left;
while (rightMost.right != null) {
parentOfRightMost = rightMost;
rightMost = rightMost.right; // Keep going to the right
}
// Replace the element in current by the element in rightMost
current.element = rightMost.element;
// Eliminate rightmost node
if (parentOfRightMost.right == rightMost)
parentOfRightMost.right = rightMost.left;
else
// Special case: parentOfRightMost is current
parentOfRightMost.left = rightMost.left;
// Balance the tree if necessary
balancePath(parentOfRightMost.element);
}
size--;
return true; // Element inserted
}
/** AVLTreeNode is TreeNode plus height */
protected static class AVLTreeNode<E extends Comparable<E>> extends BST.TreeNode<E> {
protected int height = 0; // New data field
private int size = 0;
public AVLTreeNode(E o) {
super(o);
}
public void setSize(int size) {
this.size = size;
}
public int getSize() {
return size;
}
}
}
static class BST<E extends Comparable<E>> extends AbstractTree<E> {
protected TreeNode<E> root;
protected int size = 0;
public void inorder2() {
if (root == null) {
return;
}
while (!stack.isEmpty()) {
TreeNode<E> node = stack.getFirst();
if ((node.left != null) && (!list.contains(node.left))) {
stack.push(node.left);
} else {
stack.removeFirst();
if (node.right != null) {
}
}
}
for (TreeNode<E> treeNode : list) {
System.out.print(treeNode.element + " ");
}
}
public boolean isFullBST() {
return size == Math.round(Math.pow(2, height()) - 1);
}
/**
* Returns the height of this binary tree, i.e., the number of the nodes
* in the longest path of the root to a leaf
*/
public int height() {
return height(root);
}
public int height(TreeNode<E> node) {
if (node == null) {
return 0;
} else {
return 1 + Math.max(height(node.left), height(node.right));
}
}
/** Create a default binary tree */
public BST() {
}
/** Create a binary tree from an array of objects */
public BST(E[] objects) {
for (int i = 0; i < objects.length; i++)
insert(objects[i]);
}
/** Returns true if the element is in the tree */
public ArrayList<E> searchPath(E e) {
TreeNode<E> current = root; // Start from the root
ArrayList<E> result = new ArrayList<>();
while (current != null) {
if (e.compareTo(current.element) < 0) {
current = current.left;
} else if (e.compareTo(current.element) > 0) {
current = current.right;
} else {
return result;
}
}
return result;
}
@Override
/** Returns true if the element is in the tree */
public boolean search(E e) {
TreeNode<E> current = root; // Start from the root
while (current != null) {
if (e.compareTo(current.element) < 0) {
current = current.left;
} else if (e.compareTo(current.element) > 0) {
current = current.right;
} else
// element matches current.element
return true; // Element is found
}
return false;
}
@Override
/** Insert element o into the binary tree
* Return true if the element is inserted successfully */
public boolean insert(E e) {
if (root == null)
root = createNewNode(e); // Create a new root
else {
// Locate the parent node
TreeNode<E> parent = null;
TreeNode<E> current = root;
while (current != null)
if (e.compareTo(current.element) < 0) {
parent = current;
current = current.left;
} else if (e.compareTo(current.element) > 0) {
parent = current;
current = current.right;
} else
return false; // Duplicate node not inserted
// Create the new node and attach it to the parent node
if (e.compareTo(parent.element) < 0)
parent.left = createNewNode(e);
else
parent.right = createNewNode(e);
}
size++;
return true; // Element inserted
}
protected TreeNode<E> createNewNode(E e) {
return new TreeNode<E>(e);
}
@Override
/** Inorder traversal from the root*/
public void inorder() {
inorder(root);
}
/** Inorder traversal from a subtree */
protected void inorder(TreeNode<E> root) {
if (root == null)
return;
inorder(root.left);
System.out.print(root.element + " ");
inorder(root.right);
}
@Override
/** Postorder traversal from the root */
public void postorder() {
postorder(root);
}
/** Postorder traversal from a subtree */
protected void postorder(TreeNode<E> root) {
if (root == null)
return;
postorder(root.left);
postorder(root.right);
System.out.print(root.element + " ");
}
@Override
/** Preorder traversal from the root */
public void preorder() {
preorder(root);
}
/** Preorder traversal from a subtree */
protected void preorder(TreeNode<E> root) {
if (root == null)
return;
System.out.print(root.element + " ");
preorder(root.left);
preorder(root.right);
}
/**
* This inner class is static, because it does not access any instance
* members defined in its outer class
*/
public static class TreeNode<E extends Comparable<E>> {
protected E element;
protected TreeNode<E> left;
protected TreeNode<E> right;
public TreeNode(E e) {
element = e;
}
}
@Override
/** Get the number of nodes in the tree */
public int getSize() {
return size;
}
/** Returns the root of the tree */
public TreeNode<E> getRoot() {
return root;
}
/** Returns a path from the root leading to the specified element */
public java.util.ArrayList<TreeNode<E>> path(E e) {
java.util.ArrayList<TreeNode<E>> list = new java.util.ArrayList<TreeNode<E>>();
TreeNode<E> current = root; // Start from the root
while (current != null) {
if (e.compareTo(current.element) < 0) {
current = current.left;
} else if (e.compareTo(current.element) > 0) {
current = current.right;
} else
break;
}
return list; // Return an array of nodes
}
@Override
/** Delete an element from the binary tree.
* Return true if the element is deleted successfully
* Return false if the element is not in the tree */
public boolean delete(E e) {
// Locate the node to be deleted and also locate its parent node
TreeNode<E> parent = null;
TreeNode<E> current = root;
while (current != null) {
if (e.compareTo(current.element) < 0) {
parent = current;
current = current.left;
} else if (e.compareTo(current.element) > 0) {
parent = current;
current = current.right;
} else
break; // Element is in the tree pointed at by current
}
if (current == null)
return false; // Element is not in the tree
// Case 1: current has no left children
if (current.left == null) {
// Connect the parent with the right child of the current node
if (parent == null) {
root = current.right;
} else {
if (e.compareTo(parent.element) < 0)
parent.left = current.right;
else
parent.right = current.right;
}
} else {
// Case 2: The current node has a left child
// Locate the rightmost node in the left subtree of
// the current node and also its parent
TreeNode<E> parentOfRightMost = current;
TreeNode<E> rightMost = current.left;
while (rightMost.right != null) {
parentOfRightMost = rightMost;
rightMost = rightMost.right; // Keep going to the right
}
// Replace the element in current by the element in rightMost
current.element = rightMost.element;
// Eliminate rightmost node
if (parentOfRightMost.right == rightMost)
parentOfRightMost.right = rightMost.left;
else
// Special case: parentOfRightMost == current
parentOfRightMost.left = rightMost.left;
}
size--;
return true; // Element inserted
}
@Override
/** Obtain an iterator. Use inorder. */
public java.util.Iterator<E> iterator() {
return new InorderIterator();
}
// Inner class InorderIterator
private class InorderIterator implements java.util.Iterator<E> {
// Store the elements in a list
private java.util.ArrayList<E> list = new java.util.ArrayList<E>();
private int current = 0; // Point to the current element in list
public InorderIterator() {
inorder(); // Traverse binary tree and store elements in list
}
/** Inorder traversal from the root */
private void inorder() {
inorder(root);
}
/** Inorder traversal from a subtree */
private void inorder(TreeNode<E> root) {
if (root == null)
return;
inorder(root.left);
inorder(root.right);
}
@Override
/** More elements for traversing? */
public boolean hasNext() {
if (current < list.size())
return true;
return false;
}
@Override
/** Get the current element and move to the next */
public E next() {
return list.get(current++);
}
@Override
/** Remove the current element */
public void remove() {
delete(list.get(current)); // Delete the current element
list.clear(); // Clear the list
inorder(); // Rebuild the list
}
}
/** Remove all elements from the tree */
public void clear() {
root = null;
size = 0;
}
}
static abstract class AbstractTree<E> implements Tree<E> {
@Override
/** Inorder traversal from the root*/
public void inorder() {
}
@Override
/** Postorder traversal from the root */
public void postorder() {
}
@Override
/** Preorder traversal from the root */
public void preorder() {
}
@Override
/** Return true if the tree is empty */
public boolean isEmpty() {
return getSize() == 0;
}
@Override
/** Return an iterator for the tree */
public java.util.Iterator<E> iterator() {
return null;
}
}
interface Tree<E> extends Iterable<E> {
/** Return true if the element is in the tree */
public boolean search(E e);
/**
* Insert element o into the binary tree Return true if the element is
* inserted successfully
*/
public boolean insert(E e);
/**
* Delete the specified element from the tree Return true if the element
* is deleted successfully
*/
public boolean delete(E e);
/** Inorder traversal from the root */
public void inorder();
/** Postorder traversal from the root */
public void postorder();
/** Preorder traversal from the root */
public void preorder();
/** Get the number of nodes in the tree */
public int getSize();
/** Return true if the tree is empty */
public boolean isEmpty();
public java.util.Iterator<E> iterator();
}
} | 4,631 | 18,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-30 | latest | en | 0.764368 |
http://www.chegg.com/homework-help/physics-8th-edition-chapter-10-solutions-9780470223550 | 1,469,482,901,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824345.69/warc/CC-MAIN-20160723071024-00257-ip-10-185-27-174.ec2.internal.warc.gz | 344,429,752 | 17,309 | View more editions
Solutions
Physics
# TEXTBOOK SOLUTIONS FOR Physics 8th Edition
• 2754 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
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SAMPLE SOLUTION
Chapter: Problem:
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• Step 1 of 2
Use the relation between force and compressed length to calculate the spring constant. Use this spring constant to calculate the force required to compress the spring by 0.0505 m.
From Hooke’s law, the restoring force to compress the spring by a length x is,
Here, k is spring constant.
Solve for k.
Substitute forand for.
• Step 2 of 2
The force required to compress the spring to length is,
Substitute forandfor.
Therefore, the required force is.
Corresponding Textbook
Physics | 8th Edition
9780470223550ISBN-13: 0470223553ISBN: Authors:
Alternate ISBN: 9780470395301, 9780470458754, 9781118226704 | 264 | 992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-30 | longest | en | 0.720248 |
https://redquark.org/leetcode/0022-generate-parentheses/ | 1,716,368,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00670.warc.gz | 412,796,333 | 81,883 | # LeetCode #22 - Generate Parentheses
Hello fellow devs π! Itβs a brand-new day and we have a brand new LeetCode problem to solve.
## Problem Statement
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
### Constraints:
• 1 β€ n β€ 8
### Examples
Example 1:
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Example 2:
Input: n = 1
Output: ["()"]
## Analysis
The problem description is fairly straight forward (however the solution may not be π). We will be given a number n which represents the pairs of parentheses, and we need to find out all of their valid permutations. A valid permutation is one where every opening parenthesis ( has its corresponding closing parenthesis ).
Also, these parentheses can be arranged in any order as long as they are valid. For n = 2, the valid pairs are β (()) and ()(). Also, note n = 2 means two (s and two )s. The maximum number of pairs can be eight.
## Approach
This problem description and the analysis above scream Recursion and yes itβs the right way to solve this problem. The naive approach is to generate all the permutations. All sequence of length n is ( plus all sequences of length n - 1. The time complexity of this will be O(22n) which is quite large.
What if we generate only those permutations which we know for sure will be valid? It should reduce the time considerably. We can use backtracking for this purpose. There will be three constraints we need to consider here β
• Base case when number of opening and closing parentheses is equal to n.
• Number of opening parentheses should be less than n.
• A closing parenthesis cannot occur before the open parenthesis.
To solve this problem, we will follow the below steps -
1. Create a list that will store the result.
2. Call our backtracking function with empty string and initial number of opening and closing parentheses.
3. Check the base case. If number of opening and closing parentheses are equal to n then we will add the string to the list and return.
4. If the base case does not meet then we will check if number of opening parentheses is less than n, If true, then we will add ( to the current string and increment the count of opening parenthesis.
5. Check if number of closing parentheses is less than open parentheses then we will add ) to the current string and increment the count of closing parentheses.
### Time Complexity
The time complexity is not easy to understand for this problem. It rests on understanding how many elements are there in the function. This indicates the nth Catalan number which is bounded asymptotically by Cn = 4n/(n$\sqrt(n)$). Since each valid sequence has maximum n steps, therefore, the time complexity will be O(4n/$\sqrt(n)$).
### Space Complexity
Similar as above logic the total space complexity O(4n/$\sqrt(n)$) for recursive calls and O(n) for storing the sequence.
## Code
To better visualize the solution, Iβd advise you to go copy-paste the code and execute it at Python Tutor.
### Java
public class GenerateParentheses {
public List<String> generateParenthesis(int n) {
// Resultant list
List<String> result = new ArrayList<>();
/// Recursively generate parentheses
generateParenthesis(result, "", 0, 0, n);
return result;
}
private void generateParenthesis(List<String> result, String s, int open, int close, int n) {
// Base case
if (open == n && close == n) {
return;
}
// If the number of open parentheses is less than the given n
if (open < n) {
generateParenthesis(result, s + "(", open + 1, close, n);
}
// If we need more close parentheses to balance
if (close < open) {
generateParenthesis(result, s + ")", open, close + 1, n);
}
}
}
### Python
def generate(result: List[str], s: str, _open: int, close: int, n: int):
# Base condition
if _open == n and close == n:
result.append(s)
return
# If the number of _open parentheses is less than the given n
if _open < n:
generate(result, s + "(", _open + 1, close, n)
# If we need more close parentheses to balance
if close < _open:
generate(result, s + ")", _open, close + 1, n)
def generateParenthesis(n: int) -> List[str]:
# Resultant list
result = []
# Recursively generate parentheses
generate(result, "", 0, 0, n)
return result
### JavaScript
var generateParenthesis = function (n) {
// Resultant list
const result = [];
// Recursively generate parentheses
generate(result, "", 0, 0, n);
return result;
};
function generate(result, s, open, close, n) {
// Base condition
if (open === n && close === n) {
result.push(s);
return;
}
// If the number of _open parentheses is less than the given n
if (open < n) {
generate(result, s + "(", open + 1, close, n);
}
// If we need more close parentheses to balance
if (close < open) {
generate(result, s + ")", open, close + 1, n);
}
};
### Kotlin
fun generateParenthesis(n: Int): List<String> {
// Resultant list
val result: MutableList<String> = ArrayList()
/// Recursively generate parentheses
generateParenthesis(result, "", 0, 0, n)
return result
}
fun generateParenthesis(result: MutableList<String>, s: String, open: Int, close: Int, n: Int) {
// Base case
if (open == n && close == n) {
return
}
// If the number of open parentheses is less than the given n
if (open < n) {
generateParenthesis(result, "$s(", open + 1, close, n) } // If we need more close parentheses to balance if (close < open) { generateParenthesis(result, "$s)", open, close + 1, n)
}
}
## Conclusion
Congratulations π! We have solved this problem using backtracking.
I hope you enjoyed this post. Feel free to share your thoughts on this.
You can find the complete source code on my GitHub repository. If you like what you learn, feel free to fork πͺ and star β it.
Till next timeβ¦ Happy coding π and Namaste π! | 1,443 | 5,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-22 | latest | en | 0.829197 |
http://cs61c.org/sp19/labs/lab04/ | 1,610,725,298,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00527.warc.gz | 29,459,289 | 4,900 | # Lab 4
There is no checkoff for lab this week. This is primarily practice for the midterm.
## Objectives
• TSW read and analyze C and RISC-V code that covers everything they have covered so far in the course.
## Exercise 1: Converting Number Bases in C
By this point, you should be comfortable converting between the standard decimal/hexadecimal/binary representations for numbers. Now, let’s try writing a generic number representation converter for unsigned numbers in C! In particular, you will be filling out two functions in numberrep.c. Note that the strategy that numberrep.c uses to convert between different bases is to first convert to decimal and then convert to the new base. This is not necessarily the best way to convert between two numbers. (What might be a better strategy for converting from hex to binary or vice versa?)
### Action Item
Finish the implementations of the functions in numberrep.c. Running make check should run without any Assertion errors in the output.
### Checkpoint
• Are you comfortable converting a number from one base to another?
• Are you comfortable with reading and writing C code that needs to dynamically manage memory?
## Exercise 2: RISC-V From Scratch
Now that you’ve implemented the number representation converter in C, let’s implement a subset of the program in RISC-V. Specifically, we will be implementing the validate_number function from the program. The starter code is in numberrep.s, which already contains the code for validate_number_char. You can click this magic link to load Venus with the starter code.
The general strategy for validate_number is to loop through the number and call validate_number_char each time with a character and a base as arguments. Note that since it is calling a function, you must follow the CALLEE/CALLER conventions for whichever registers you choose to use.
Your implementation of validate_number should set a0 to be 1 if the number is valid in this base or 0 if it isn’t. The starter code prints out the value of a0 for testing purposes.
### Action Item
Interpret what validate_number_char does and write down your interpretation. Then, finish the implementation of validate_number.
## Checkpoint
• Are you comfortable with writing and reading RISC-V functions?
• Are you comfortable with the CALLEE/CALLER conventions for registers? | 478 | 2,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-04 | latest | en | 0.862454 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/481/C3%5E3sD9.html | 1,708,625,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00632.warc.gz | 461,504,139 | 6,406 | Copied to
clipboard
G = C33⋊D9order 486 = 2·35
5th semidirect product of C33 and D9 acting via D9/C3=S3
Series: Derived Chief Lower central Upper central
Derived series C1 — C32×C9 — C33⋊D9
Chief series C1 — C3 — C32 — C33 — C32×C9 — C3×C32⋊C9 — C33⋊D9
Lower central C32×C9 — C33⋊D9
Upper central C1
Generators and relations for C33⋊D9
G = < a,b,c,d,e | a3=b3=c3=d9=e2=1, ab=ba, dad-1=eae=ac=ca, bc=cb, bd=db, ebe=b-1, cd=dc, ece=c-1, ede=d-1 >
Subgroups: 1520 in 177 conjugacy classes, 44 normal (12 characteristic)
C1, C2, C3, C3, C3, S3, C6, C9, C32, C32, C32, D9, C3×S3, C3⋊S3, C3×C9, C3×C9, C33, C33, C33, C9⋊S3, C3×C3⋊S3, C33⋊C2, C32⋊C9, C32⋊C9, C32×C9, C32×C9, C34, C32⋊D9, C324D9, C3×C33⋊C2, C3×C32⋊C9, C33⋊D9
Quotients: C1, C2, C3, S3, C6, D9, C3×S3, C3⋊S3, C3×D9, C32⋊C6, C9⋊C6, C9⋊S3, C3×C3⋊S3, C32⋊D9, C3×C9⋊S3, He34S3, C33.S3, C33⋊D9
Smallest permutation representation of C33⋊D9
On 81 points
Generators in S81
```(2 78 17)(3 18 79)(5 81 11)(6 12 73)(8 75 14)(9 15 76)(19 56 42)(20 43 57)(22 59 45)(23 37 60)(25 62 39)(26 40 63)(28 71 48)(29 49 72)(31 65 51)(32 52 66)(34 68 54)(35 46 69)
(1 24 53)(2 25 54)(3 26 46)(4 27 47)(5 19 48)(6 20 49)(7 21 50)(8 22 51)(9 23 52)(10 41 70)(11 42 71)(12 43 72)(13 44 64)(14 45 65)(15 37 66)(16 38 67)(17 39 68)(18 40 69)(28 81 56)(29 73 57)(30 74 58)(31 75 59)(32 76 60)(33 77 61)(34 78 62)(35 79 63)(36 80 55)
(1 77 16)(2 78 17)(3 79 18)(4 80 10)(5 81 11)(6 73 12)(7 74 13)(8 75 14)(9 76 15)(19 56 42)(20 57 43)(21 58 44)(22 59 45)(23 60 37)(24 61 38)(25 62 39)(26 63 40)(27 55 41)(28 71 48)(29 72 49)(30 64 50)(31 65 51)(32 66 52)(33 67 53)(34 68 54)(35 69 46)(36 70 47)
(1 2 3 4 5 6 7 8 9)(10 11 12 13 14 15 16 17 18)(19 20 21 22 23 24 25 26 27)(28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54)(55 56 57 58 59 60 61 62 63)(64 65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80 81)
(1 9)(2 8)(3 7)(4 6)(10 73)(11 81)(12 80)(13 79)(14 78)(15 77)(16 76)(17 75)(18 74)(19 48)(20 47)(21 46)(22 54)(23 53)(24 52)(25 51)(26 50)(27 49)(28 42)(29 41)(30 40)(31 39)(32 38)(33 37)(34 45)(35 44)(36 43)(55 72)(56 71)(57 70)(58 69)(59 68)(60 67)(61 66)(62 65)(63 64)```
`G:=sub<Sym(81)| (2,78,17)(3,18,79)(5,81,11)(6,12,73)(8,75,14)(9,15,76)(19,56,42)(20,43,57)(22,59,45)(23,37,60)(25,62,39)(26,40,63)(28,71,48)(29,49,72)(31,65,51)(32,52,66)(34,68,54)(35,46,69), (1,24,53)(2,25,54)(3,26,46)(4,27,47)(5,19,48)(6,20,49)(7,21,50)(8,22,51)(9,23,52)(10,41,70)(11,42,71)(12,43,72)(13,44,64)(14,45,65)(15,37,66)(16,38,67)(17,39,68)(18,40,69)(28,81,56)(29,73,57)(30,74,58)(31,75,59)(32,76,60)(33,77,61)(34,78,62)(35,79,63)(36,80,55), (1,77,16)(2,78,17)(3,79,18)(4,80,10)(5,81,11)(6,73,12)(7,74,13)(8,75,14)(9,76,15)(19,56,42)(20,57,43)(21,58,44)(22,59,45)(23,60,37)(24,61,38)(25,62,39)(26,63,40)(27,55,41)(28,71,48)(29,72,49)(30,64,50)(31,65,51)(32,66,52)(33,67,53)(34,68,54)(35,69,46)(36,70,47), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81), (1,9)(2,8)(3,7)(4,6)(10,73)(11,81)(12,80)(13,79)(14,78)(15,77)(16,76)(17,75)(18,74)(19,48)(20,47)(21,46)(22,54)(23,53)(24,52)(25,51)(26,50)(27,49)(28,42)(29,41)(30,40)(31,39)(32,38)(33,37)(34,45)(35,44)(36,43)(55,72)(56,71)(57,70)(58,69)(59,68)(60,67)(61,66)(62,65)(63,64)>;`
`G:=Group( (2,78,17)(3,18,79)(5,81,11)(6,12,73)(8,75,14)(9,15,76)(19,56,42)(20,43,57)(22,59,45)(23,37,60)(25,62,39)(26,40,63)(28,71,48)(29,49,72)(31,65,51)(32,52,66)(34,68,54)(35,46,69), (1,24,53)(2,25,54)(3,26,46)(4,27,47)(5,19,48)(6,20,49)(7,21,50)(8,22,51)(9,23,52)(10,41,70)(11,42,71)(12,43,72)(13,44,64)(14,45,65)(15,37,66)(16,38,67)(17,39,68)(18,40,69)(28,81,56)(29,73,57)(30,74,58)(31,75,59)(32,76,60)(33,77,61)(34,78,62)(35,79,63)(36,80,55), (1,77,16)(2,78,17)(3,79,18)(4,80,10)(5,81,11)(6,73,12)(7,74,13)(8,75,14)(9,76,15)(19,56,42)(20,57,43)(21,58,44)(22,59,45)(23,60,37)(24,61,38)(25,62,39)(26,63,40)(27,55,41)(28,71,48)(29,72,49)(30,64,50)(31,65,51)(32,66,52)(33,67,53)(34,68,54)(35,69,46)(36,70,47), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81), (1,9)(2,8)(3,7)(4,6)(10,73)(11,81)(12,80)(13,79)(14,78)(15,77)(16,76)(17,75)(18,74)(19,48)(20,47)(21,46)(22,54)(23,53)(24,52)(25,51)(26,50)(27,49)(28,42)(29,41)(30,40)(31,39)(32,38)(33,37)(34,45)(35,44)(36,43)(55,72)(56,71)(57,70)(58,69)(59,68)(60,67)(61,66)(62,65)(63,64) );`
`G=PermutationGroup([[(2,78,17),(3,18,79),(5,81,11),(6,12,73),(8,75,14),(9,15,76),(19,56,42),(20,43,57),(22,59,45),(23,37,60),(25,62,39),(26,40,63),(28,71,48),(29,49,72),(31,65,51),(32,52,66),(34,68,54),(35,46,69)], [(1,24,53),(2,25,54),(3,26,46),(4,27,47),(5,19,48),(6,20,49),(7,21,50),(8,22,51),(9,23,52),(10,41,70),(11,42,71),(12,43,72),(13,44,64),(14,45,65),(15,37,66),(16,38,67),(17,39,68),(18,40,69),(28,81,56),(29,73,57),(30,74,58),(31,75,59),(32,76,60),(33,77,61),(34,78,62),(35,79,63),(36,80,55)], [(1,77,16),(2,78,17),(3,79,18),(4,80,10),(5,81,11),(6,73,12),(7,74,13),(8,75,14),(9,76,15),(19,56,42),(20,57,43),(21,58,44),(22,59,45),(23,60,37),(24,61,38),(25,62,39),(26,63,40),(27,55,41),(28,71,48),(29,72,49),(30,64,50),(31,65,51),(32,66,52),(33,67,53),(34,68,54),(35,69,46),(36,70,47)], [(1,2,3,4,5,6,7,8,9),(10,11,12,13,14,15,16,17,18),(19,20,21,22,23,24,25,26,27),(28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54),(55,56,57,58,59,60,61,62,63),(64,65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80,81)], [(1,9),(2,8),(3,7),(4,6),(10,73),(11,81),(12,80),(13,79),(14,78),(15,77),(16,76),(17,75),(18,74),(19,48),(20,47),(21,46),(22,54),(23,53),(24,52),(25,51),(26,50),(27,49),(28,42),(29,41),(30,40),(31,39),(32,38),(33,37),(34,45),(35,44),(36,43),(55,72),(56,71),(57,70),(58,69),(59,68),(60,67),(61,66),(62,65),(63,64)]])`
54 conjugacy classes
class 1 2 3A ··· 3M 3N 3O 3P ··· 3W 6A 6B 9A ··· 9AA order 1 2 3 ··· 3 3 3 3 ··· 3 6 6 9 ··· 9 size 1 81 2 ··· 2 3 3 6 ··· 6 81 81 6 ··· 6
54 irreducible representations
dim 1 1 1 1 2 2 2 2 2 2 6 6 type + + + + + + + image C1 C2 C3 C6 S3 S3 C3×S3 D9 C3×S3 C3×D9 C32⋊C6 C9⋊C6 kernel C33⋊D9 C3×C32⋊C9 C32⋊4D9 C32×C9 C32⋊C9 C34 C3×C9 C33 C33 C32 C32 C32 # reps 1 1 2 2 3 1 6 9 2 18 3 6
Matrix representation of C33⋊D9 in GL10(𝔽19)
11 0 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 5 0 0 18 0 0 0 0 0 0 7 0 1 18 0 0 0 0 0 0 17 0 0 3 17 3 0 0 0 0 12 18 18 2 18 1
,
0 18 0 0 0 0 0 0 0 0 1 18 0 0 0 0 0 0 0 0 0 0 1 3 0 0 0 0 0 0 0 0 18 17 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1
,
1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 17 3 0 0 0 0 0 0 0 0 18 1 0 0 0 0 0 0 0 0 14 3 18 1 0 0 0 0 0 0 12 12 18 0 0 0 0 0 0 0 8 0 3 0 17 3 0 0 0 0 13 8 2 18 18 1
,
14 17 0 0 0 0 0 0 0 0 2 12 0 0 0 0 0 0 0 0 0 0 1 3 0 0 0 0 0 0 0 0 18 17 0 0 0 0 0 0 0 0 0 0 16 0 3 0 0 0 0 0 0 0 14 0 1 1 0 0 0 0 0 0 13 0 3 0 18 1 0 0 0 0 15 18 11 18 0 18 0 0 0 0 0 0 0 0 1 0 0 0 0 0 4 0 9 0 1 0
,
7 14 0 0 0 0 0 0 0 0 2 12 0 0 0 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 0 0 18 17 0 0 0 0 0 0 0 0 0 0 7 0 0 3 0 0 0 0 0 0 14 0 1 1 0 0 0 0 0 0 13 1 0 3 0 0 0 0 0 0 3 0 0 12 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 10 18 18 8 1 18
`G:=sub<GL(10,GF(19))| [11,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,1,0,5,7,17,12,0,0,0,0,0,1,0,0,0,18,0,0,0,0,0,0,0,1,0,18,0,0,0,0,0,0,18,18,3,2,0,0,0,0,0,0,0,0,17,18,0,0,0,0,0,0,0,0,3,1],[0,1,0,0,0,0,0,0,0,0,18,18,0,0,0,0,0,0,0,0,0,0,1,18,0,0,0,0,0,0,0,0,3,17,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1],[1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,17,18,14,12,8,13,0,0,0,0,3,1,3,12,0,8,0,0,0,0,0,0,18,18,3,2,0,0,0,0,0,0,1,0,0,18,0,0,0,0,0,0,0,0,17,18,0,0,0,0,0,0,0,0,3,1],[14,2,0,0,0,0,0,0,0,0,17,12,0,0,0,0,0,0,0,0,0,0,1,18,0,0,0,0,0,0,0,0,3,17,0,0,0,0,0,0,0,0,0,0,16,14,13,15,0,4,0,0,0,0,0,0,0,18,0,0,0,0,0,0,3,1,3,11,0,9,0,0,0,0,0,1,0,18,0,0,0,0,0,0,0,0,18,0,1,1,0,0,0,0,0,0,1,18,0,0],[7,2,0,0,0,0,0,0,0,0,14,12,0,0,0,0,0,0,0,0,0,0,2,18,0,0,0,0,0,0,0,0,3,17,0,0,0,0,0,0,0,0,0,0,7,14,13,3,0,10,0,0,0,0,0,0,1,0,0,18,0,0,0,0,0,1,0,0,0,18,0,0,0,0,3,1,3,12,0,8,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,18] >;`
C33⋊D9 in GAP, Magma, Sage, TeX
`C_3^3\rtimes D_9`
`% in TeX`
`G:=Group("C3^3:D9");`
`// GroupNames label`
`G:=SmallGroup(486,137);`
`// by ID`
`G=gap.SmallGroup(486,137);`
`# by ID`
`G:=PCGroup([6,-2,-3,-3,-3,-3,-3,3134,548,986,867,3244,11669]);`
`// Polycyclic`
`G:=Group<a,b,c,d,e|a^3=b^3=c^3=d^9=e^2=1,a*b=b*a,d*a*d^-1=e*a*e=a*c=c*a,b*c=c*b,b*d=d*b,e*b*e=b^-1,c*d=d*c,e*c*e=c^-1,e*d*e=d^-1>;`
`// generators/relations`
×
𝔽 | 6,148 | 9,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.295436 |
https://sjpbusinessmedia.com/should-i-kcqwz/732aa7-7404-crystal-oscillator | 1,685,807,234,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00455.warc.gz | 570,865,436 | 16,816 | There are capacitor C1 and C3 filter current to smooth. Colpitts Crystal OscillatorWe have made a few oscillators in past videos, but this is the first one that makes use of a crystal. FIGURE 3: Electrical Model for Quartz Crystal/Simplified Model for Crystal Oscillator. Which does this part refer to, a pencil or the words? But this circuit do not well for high frequency, you should use this a better: Crystal oscillator using TTL. It … But then, the higher the capacitance there goes, at MHz frequencies the more it approximates just having no capacitance at all. The theory and practical design of a 28MHz discrete Pierce crystal oscillator.Blog post with links/references: http://www.analogzoo.com/?p=908 2 comments: Steve July 9, 2020 at 5:30 AM. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Crystal Oscillator Troubleshooting Guide, Rev. For my oscillator I use a single 7404 inverter that has a 100 nF decoupling capacitor attached between 5V and GND. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. The reason that takes two inverters is that you need positive gain. Why did flying boats in the '30s and '40s have a longer range than land based aircraft? It also has automatic amplitude control and frequency drift is also very low due to change in temperature. Working with two inverter gate within IC1. So I will need to fundamentally change my circuit, It was the only one that worked for me, albeit at a weird frequency. The wires or coil and capacitors are the LC-oscillator circuit. I always try to make Electronics Learning Easy. To control an electronic circuit. The Figure above shows implementation of a parallel resonant oscillator circuit. Parallel resonant XTAL OSC or XO's are best used with CMOS up to 20MHz and Series resonant transistor XO's above this f to avoid sensitivity to stray capacitance. Use MathJax to format equations. However, it seems my crystal is not made to work in a series resonant oscillator but rather a parallel resonant oscillator. Then, Look at this circuit. How can I determine the capacitance required for an oscillator (NOT A CRYSTAL!)? Federal Supply Class. 5955. However, when studying the TTL IC., Which is the basis of a digital circuit that everyone should learn it. Now replace the CMOS inverter and its feedback resistor with an MCP602 op amp circuit with a gain of -10 (or try a different gain if that doesn't work). 5955. 1 Obtaining 1.1 Unlocking 1.2 Crafting 2 Usage 2.1 Research 2.2 Crafting 2.3 AWESOME Sink 3 Tips 4 History It is not necessary to automate Crystal Oscillators until late-game when they are given their use in production lines rather than just in M.A.M. Crystals will not drive the inputs to a CMOS gate, and require special input (and output) circuitry. However, If you think that this circuit is not good enough for you.You can see the crystal oscillator circuit is as follows: My Friend wants a Pulse Generator Oscillator circuit. They generate the frequencies to be a base time. When you do get around to using a parallel arrangement, it may also be important that you work on getting the biasing point where you need it to get stable and equal periods of on and off. It really wouldn’t be a hack with a run-of-the-mill 74LS or 74HC part. It simply inverts the given signal and has 6 NOT gates. An item which consists of a quartz crystal resonator(s), suitably mounted in a HOLDER, QUARTZ CRYSTAL. In Section 2, the theoretical analysis of quartz crystal oscillators is presented and explored to clarify the start-up conditions for crystal oscillators, as well as the power Item Name. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. October 27, 2020 February 24, 2012 by Electrical4U. Crystal oscillator is used for engine controlling, stereo, clock and to trip computer, and in GPS system. CRYSTAL OSCILLATOR CIRCUITS An oscillator circuit requires that two conditions be satis-fied: that it contains an amplifier having sufficient gain to overcome the loss due to its feedback network, and that the phase shift around the whole loop is zero at the wanted fre-quency. A ring oscillator is a device composed of an odd number of NOT gates in a ring, whose output oscillates between two voltage levels, representing true and false.The NOT gates, or inverters, are attached in a chain and the output of the last inverter is fed back into the first. I have a 2 MHz crystal, and a 4 MHz, not your 3.odd MHz. No capacitors to worry about frequency. When I changed it from 100 pf to 22 pf the frequency was completely unstable. I added the capacitor after seeing it on many schematics and without it, it didn't start oscillating anyways. Crystal oscillator is used for engine controlling, stereo, clock and to trip computer, and in GPS system. Why does my crystal resonate at 4 times the specified frequency? Both resistors R1 and R2 must be the same resistance. For my design I want to use plain 74xx or 74LSxx chips, because I have them around and ready to use. Crystal oscillators are used in many consumer goods such as cable television systems, personal computers, video cameras, toys and video games, radio systems, cellular phones, and so on. Read next: Transistor Crystal Oscillator (low volts). Item Name. It can be used with very high frequency. Quartz Crystals. Commercial and Government Entity Code (Supplier Data) Below diagram displays the circuit diagram of a simple CMOS crystal oscillator which usually relies on a couple of inverters. Look up any common 32kHz crystal oscillator circuit using an unbuffered CMOS inverter (4069, 74C04, 74HC04, etc.). Robert J. Matthys, in Analog Circuit Design, 1991. So he tell to me even low prices. By pulling the input of the amplifier toward the voltage at the output, an unstable condition is Because of its temperature changes while working. They are very cheap and we can look at the general electronics stores. They may use several parts. Follow edited Jul 22 '12 at 5:13. answered Jul 22 '12 at 5:03. Read next: 1 MHz time base using ceramic filter 10.7Mhz. Now when I apply some voltage to this circuit it indeed starts to oscillate. The crystal oscillator can be roughly divided into two categories: passive crystal oscillator and active crystal oscillator. But they two have some differences too, which separates them & … Crystal Oscillator is a mid-game component with a variety of uses. Inside Cheap Quartz Alarm Clock, there is a component on PCB. Crystal Oscillator Circuits. Crystal oscillators operate on the principle of inverse piezoelectric effect in which an alternating voltage applied across the crystal surfaces causes it to vibrate at its natural frequency. If you can find the series resonant point of the crystal you know the crystal … Possibly it has something to do with wearing out the crystal. It is a square wave signal. Cost of crystal vs oscillator – mismatched crystal causes oscillator failure I like the TTL crystal oscillator circuit. Frankly speaking, I do not know how to fix this as I have not much experience with oscillators. Crystal Oscillator is a mid-game component with a variety of uses. 5 Crystal oscillator Circuits using CMOS. This element is usually a quartz crystal, hence the name. That's why I chose a 3686.4 kHz Quarz, so that I can divide it later via a frequency divider to not only generate my processor clock but also my UART clock. Both of them have the same aim of generating an oscillation frequency by vibrating when an input voltage is given to them. If the oven and oscillator power are interrupted for a period, allowing the resonator to cool down, and then restarted, the resonator will not immediately return to When I changed it to a higher value, I didn't see much difference at all. @crj11 The OP may want to stay "retro" with the build. National Stock Number. Can I use a series crystal in an parallel crystal optimized oscillator? Federal Supply Class. fo must be smaller than 1/2tpn where: tp = average delay time of each gate n = the number of gates. It simply inverts the given signal and has 6 NOT gates. I don't know why everybody is showing the more complex and harder to get to work schematics. If so, this circuit will not work. TTL Crystal Oscillator Schematic A circuit using one 7400 TTL IC can use crystals of the fundamental type, from 1 to about 15MHz. Which affects the capacity within the crystal and making frequency tolerances. And protect Interference that may occur to the circuit. How can I cut 4x4 posts that are already mounted? However, the frequency is not stable at all, seems to be 200 kHz below the stated one and the duty cycle is not at all 50%. So also we easily used a single transistor 2N3904 or BC548 that is NPN types. This is a 1Hz oscillator circuit for a standard digital clock, frequency size 1 Hz or 2 Hz. They match the crystal and two resistors. 1 Obtaining 1.1 Unlocking 1.2 Crafting 2 Usage 2.1 Research 2.2 Crafting 2.3 AWESOME Sink 3 Tips 4 History It is not necessary to automate Crystal Oscillators until late-game when they are given their use in production lines rather than just in M.A.M. 160ff. Transistor Crystal Oscillator Circuit - a simple transistor crystal oscillator circuit and the values for different frequencies. And the frequency generator is a circuit that I like. Transistor Crystal Oscillator Circuit - a simple transistor crystal oscillator circuit and the values for different frequencies. If somebody can point out my mistake to me I would be very glad. A crystal oscillator circuit can be constructed in a number of ways like a Crystal controlled tuned collector oscillator, a Colpitts crystal oscillator, a Clap crystal oscillator etc. Crystal oscillator replaced with exact frequency for clock? Through hole and SMD packages with various tolerances from standard consumer types till tight tolerance specification with Military or Space qualification. It also used an oscillator with 7404 gates: The capacitor used there is 10nF. It does not work ! Additional Info: The data sheet of the crystal calls states CL=20pf and Rr < 120 Ohm. 01-067-7404 010677404 5955-01-067-7404 5955010677404. 2. The frequency is 32771Hz or 32.771KHz. rev 2021.1.20.38359, The best answers are voted up and rise to the top, Electrical Engineering Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. And the resistors from 1K to 4.7K. Greenray TCXOs are designed for communications, instrumentation and defense applications. Using one of the most simple designs (a crystal oscillator) built on the most simple platform (a protoboard), I showed you that a couple of very simple design modifications allowed me to reduce the unwanted emissions of the design by 21 dB (i.e., 55 dBm − 34 dBm), effectively reducing the spurious noise power by 99.2% (1021/10 = 125, 1/125 = 0.8%)! Crystal Oscillator is an Electronics Oscillator circuit which uses the mechanical resonance of a vibrating crystal of piezoelectric material to generate an electrical signal with an accurate frequency. The linear properties control the gain and phase shift, and the overload properties control the wave shape and oscillation amplitude. Crystal oscillators operate on the principle of inverse piezoelectric effect in which an alternating voltage applied across the crystal surfaces causes it to vibrate at its natural frequency. #MindsRiot #Oscillator #Multisim #ElectronicsPracticaloscillator is electronic device which is used to convert dc into ac of particular frequency. does paying down principal change monthly payments? Another name of this IC is also hex inverter. Improve this answer. We use a 3.579545 MHz crystal then measure frequency output show on LED display highly accurate. To control an electronic circuit. OSCILLATOR,CRYSTAL CONTROLLED. Crystal oscillators are used in many consumer goods. However, earlier designs only included the Z80 and some RAM and ROM. Frankly, I don't see the point of all this extra stuff if this here works much more reliable and simple. My earlier builds used more up to date parts, but also never included an oscillator circuit. It is described in Sec. The device provides an additional buffered inverter (Y) for signal conditioning (see Figure 3). I found stock certificates for Disney and Sony that were given to me in 2011. I'd use one inverter to run the crystal, a second just as a buffer, then two with a little DC feedback around them to make the Schmitt trigger. Then I simply copy-pasted the circuit from "Build Your Own Z80 Computer" page 94. Edson did a study of VHF harmonic oscillator circuits in 1950 181 and published his classic book on vacuum tube oscillators of all types in 1953 [Q]. 5955-01-024-5752 An oscillator whose operating frequency(ies) is controlled by a CRYSTAL UNIT, QUARTZ. A well-designed crystal oscillator will provide good performance with TTL gates. The other oscillators will use transistors or FET to connect together with a network circuit. Consumer Applications of Crystal Oscillator. To keep output to A steady voltage. 1# Crystal Oscillator circuit using 74LS04. Also, note that the resistors going from inverter output to inverter input set the bias point. The crystal oscillator has served this purpose for nearly a century. I know that with more modern solutions some things would be easier, but I am quite fascinated by older technology. Certain crystals, and electrically polarized ceramics, exhibit a property called piezoelectricity. :). This is a simple Crystal oscillator circuit using 74LS04. Learn how your comment data is processed. Here, with a relatively quick fix, the cost benefit of using a MEMS oscillator is realized when production volume is around 2,800 units or less. At 24MHz it is taking a sample every 41ns or so. Text: TYPICAL TPUART INTERFACE OSCILLATOR OR TTL CLOCK FIGURE 1 TYPICAL TPUART INTERFACE 3 D0-D7 , 1800 OHM 560 OHM 220 OHM 7404 7404 7404 7404 220 OHM 30 pF 5.0688 MHz FIGURE 2A 5.0688 MHz CRYSTAL OSCILLATOR CIRCUIT 4 DESCRIPTION OF PIN FUNCTIONS DIP PIN NO. It can be used in the normal clock circuit. crystal oscillator must provide an additional 180 degrees of phase shift. Either that or you will need to use something else such as a PLL. Next circuit, when you want to build a ramping waveform using the crystal. It depends on that crystal. Crystal Oscillator Circuit. simulate this circuit – Schematic created using CircuitLab. … I am currently building a small Z80 microprocessor system and require some kind of clock generation. Quartz crystal is used in Crystal Oscillator and Ceramic is used in Ceramic Resonator. Now normally there would be no news in someone making a ring oscillator with a 7404. Asking for help, clarification, or responding to other answers. The tree inverters gate are biased into their linear regions by R1 to R4, and the crystal provides the feedback. Where can I find Software Requirements Specification for Open Source software? I also tried to exchange the 7404 with an 74LS04 as some schematics use it, but that didn't change anything. is therefore: fo . research or as an construction ingredient Alternate recipes involving Crystal … CRYSTAL OSCILLATOR DRIVER SGDS029– SEPTEMBER 2007 X1 and X2 can be connected to a crystal or resonator in oscillator applications. In the circuit figure first, be produces to square waveform the other part I think you can know them, it very simple. The additional buffered inverter improves the signal quality of the crystal oscillator output by making it rail to rail. How much frequency and waveform do you want? When I use you 100 pF I get a very dirty signal on the 2 MHz. Description. For example, cable television systems, video cameras, personal computers, toys and video games, cellular phones, radio systems. Are you trying to get your inverter (7404) to oscillate at 32.768kHz? To his digital CMOS binary counter. Use the 74HCT04 to buffer the op amp's output. One commented under my earlier version of this answer said my circuit here doesn't work, but it does, I just had my analog scope settings wrong so I measured 4 times higher frequency. ), @jonk Yes, I indeed intend to stay "retro". An oscilloscope would be a better testing option. In this case, 15 hours of engineering work was required to correct the crystal startup problem. The characteristics o f the crystal oscillator used … Crystal oscillator circuit constructed using bipolar transistor or various types of FETs. I stumbled upon your post while searching for what the failure mode of a crystal is.. Crystal oscillators are used in many areas of electronics. Using a crystal oscillator circuit enables a high performance high stability oscillator to be built very cheaply and easily. It works well. CRYSTAL UNIT,QUARTZ. A parallel cut crystal is used in a PIC micro oscillator for a required design. Read also: The 60Hz calibration frequency standards for digital clock using MM5369. (adsbygoogle = window.adsbygoogle || []).push({}); 1# Crystal Oscillator circuit using 74LS04, Testing Crystal oscillator using TTL 74LS04, 1# Crystal Oscillator circuit using 74LS04, Transistor Relay driver circuit in digital, Wien Bridge Oscillator circuits using Op-amp and FET, How Astable Multivibrator using Logic Gates work | Example Circuits, 60Hz Standard frequency of Calibration using MM5369, Triangle wave generator circuit with cmos inverter IC, 4 Monostable multivibrator circuit ideals. the oscillator circuit in order to meet frequency accu-racy and provide reliable operation. Thanks for you explanation. Related: 7805 voltage regulator IC: Pinout, Datasheet and example circuits. But it could just be ignorance, too, and only seeing old Z80 diagrams. The crystal oscillator™s output is fed to the System PLL as the input refe rence. change 100pF to 10nF to eliminate phase shift at 3.5MHz, add 2.4k input to ground to pull output >> 2.0V towards 2.4V (opt.). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. They generate the frequencies to be a base time. It is the cheapest tool. OSCILLATOR,CRYSTAL CONTROLLED. There are a lot of circuits diagram you do. Try looking up some crystal oscillator circuits on the web. You can only pull the frequency of a crystal oscillator a very small amount, so you will need a 4.096MHz crystal. The crystal oscillator circuit like this get popular uses in digitals. Normally, a crystal oscillator will oscillate at a slightly higher frequency than the crystal’s series resonant point. So I tried to rebuilt them. This is an Automotive application. In this case, 15 hours of engineering work was required to correct the crystal startup problem. (Actually, it's also important for your existing circuit. Crystal oscillator load capacitance, again. Below. Crystal Oscillator: Circuit, Frequency & Working Principle. Something you need to be careful with when using the Saleae device is that it takes samples, not continuous measurements. Below diagram displays the circuit diagram of a simple CMOS crystal oscillator which usually relies on a couple of inverters. Viewed 655 times 0. It is stable frequency at 32.768KHz with a watch crystal. The oscillators or frequency generators provide a waveform out in various forms. Who must be present at the Presidential Inauguration? Whereas the logical high state seems to be quite stable with one unstable cycle seeable at the right end of the graph, the off state varies quite a bit. PARTS LIST R1 470Ω R2 470Ω How to make sure that a conference is not a scam when you are invited as a speaker? Diodes Incorporated has unique vertical integration with SaRonix-eCera for proprietary high-capacity automated production of ceramic-packaged oscillators; Output frequencies up to 670 MHz with ±20 ppm total lifetime accuracy available; 5 V, 3.3 V, 2.5 V, and 1.8 V products available 00-116-7404 001167404 5955-00-116-7404 5955001167404. In this sce-nario, the reversible effects will reach equilibrium, and the non-reversible effects will determine the aging rate. And it works perfectly every time without that capacitor. Crystal oscillator using 7404 is unstable. Crystal oscillator circuits are linear analog circuits with carefully controlled overload properties. Here, with a relatively quick fix, the cost benefit of using a MEMS oscillator is realized when production volume is around 2,800 units or less. In the circuit, IC 4060 is a standard frequency generator with quartz crystal. The sum of two well-ordered subsets is well-ordered. This document contains an overview of the on-chip It consists of IC-4060 and IC-4013, the IC-4060 single-acting Oscillator and Counter. However, I have a feeling that the capacitor does have quite an influence. Crystal’s parallel resonant frequency, fp is – Now, we can understand that the series resonant frequency is 9.20 MHz and the parallel resonant frequency is 9.23 MHz The Q factor of this crystal will be- Colpitts Crystal Oscillator. Crystal oscillators are used in many areas of electronics. How to Test Crystal Oscillator. 7404 is a member of 74xxx series gate ICs and has the functionality of NOT get. Crystal Oscillator is an Electronics Oscillator circuit which uses the mechanical resonance of a vibrating crystal of piezoelectric material to generate an electrical signal with an accurate frequency. The oscillators or frequency generators provide a waveform out in various forms. Next, look at a power supply. Then measure the frequency output by a frequency counter Kits. MathJax reference. In 1965, Firth published his design handbook [lo] on the Pierce … Resistors ¼W +5%R1, R2_1K to 4.7KCapacitorsC1: 10uF_16V, ElectrolyticC2: 0.1uF_50V, PolyesterC3: 2.2uF_16V, ElectrolyticSemiconductorD1: 1N4001, 1A 50V DiodeIC1: 74LS04, Inverter gate ICIC2: 78L05, 5V regulator ICOthersXTAL1: Crystal between 1MHz to 10MHzUniversal PCB board. Crystal Oscillator Circuits – Page 1 Radio receivers and transmitters both require a precise frequency reference, and this reference was until recently almost always provided by a crystal oscillator. The adjustment period with C2, and counter circuits within the IC 4060 will be only 2 Hz frequency dividing out the pin 3. We deliver a full range of Quartz Crystals, Crystal Oscillators (VCXO, TCXO, OCXO, PXO, Microwave Oscillator), Crystal Filters and Microwave Products. Then, C2 pulls a high frequency that contamination from power supply to ground. Fortunately, most books for the Z80 and websites also built them using 74xx chips. 1 \$\begingroup\$ I am currently building a small Z80 microprocessor system and require some kind of clock generation. But the transistor pierce crystal oscillator is the most commonly used one. They can generate the output constant frequency of 1MHz to 10 MHz. Figure 1 shows how to make a simple test to quickly determine the condition of the crystal. Share. But when I use SN74HC04 of Texas Instruments. 7404 is a member of 74xxx series gate ICs and has the functionality of NOT get. Hence, for C1 = C2, the crystal provides a phase shift of 180 degrees. So, yes, this circuit works flawlessly for me and none of the others did so far. How does a Cloak of Displacement interact with a tortle's Shell Defense? If you take up the voltage power supply to 12V doing the output volt peak up to 12V as well. For example, sine wave, Triangular waveform, and square wave. Dual, Quad and I2C programmable oscillator products replace multiple oscillators and muxes while improving reliability and performance. Crystal Oscillators (XO/VCXO) Our crystal oscillator (XO) and voltage-controlled crystal oscillator (VCXO) products provide fast and easy customization for any frequency up to 3 GHz with six digits of accuracy. Een kristaloscillator is een oscillator die als frequentiebepalend element een kristal van piëzo-elektrisch materiaal heeft. Can not fine tune Colpitts oscillator with crystal. For example, sine wave, Triangular waveform, and square wave. My pierce design did not work at all, so I quickly changed my setup to a series resonant oscillator. Because it is good and enjoyable. If C1 = C2, current through them is identical and 180 degrees out of phase from each other. To working instead of the RC network circuit. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Just the fact of having a 7404 in hand suggests that to me. Our TCXOs feature temperature stabilities of 1ppm or less and are available in a variety of packages, including SMT, and are available from 20 KHz to 1 GHz. The schematic attached is taken from "Crystal Oscillator Circuits" by Robert J. Matthys, Krieger Publishing Company, 1992. When we need a highly accurate, it uses the crystal frequency control better. For an oscillator controlled by a device other than a crystal, see OSCILLATOR (1), NONCRYSTAL CONTROLLED. Crystal Oscillator is also used in engine controlling, clock and to trip computer, stereo, and in GPS systems. Using crystal to joint with the resistor. For example: In the circuit below is the simple oscillator that generates the square wave or DC pulse. Either change to parallel resonant oscillator, or change the crystal to one meant for series resonance. You want a crystal oscillator circuit, right? The resistor and a capacitor in the RC oscillator circuit. Find your crystal oscillator easily amongst the 190 products from the leading brands (Stanford Research Systems, NXP Semiconductors, Microchip Technology, ...) on DirectIndustry, the industry specialist for your professional purchases. Diodes Incorporated has unique vertical integration with SaRonix-eCera for proprietary high-capacity automated production of ceramic-packaged oscillators; Output frequencies up to 670 MHz with ±20 ppm total lifetime accuracy available; 5 V, 3.3 V, … 11.3 "TTL Two-Inverters-7404", pp. amount of oscillator circuit development along the way. I am following this. If I increase the capacitance to 10 nF, as someone suggested it looks better. 00-116-7404 001167404 5955-00-116-7404 5955001167404. I think maybe used the watch crystal and IC4069 or IC4049 inverter CMOS IC. The period for your 3.686MHz clock is 271ns. In this article, we study the design principles of this circuit. Meter accuracy Low. 5955. Since you have prior experience, I assume you know enough not to use a solderless protoboard for on oscillator like this. An oscillator whose operating frequency(ies) is controlled by a CRYSTAL UNIT, QUARTZ. National Stock Number. I looked up an old Z80 computer design from the 80ies (the mc CP/M-Computer). 1,What is the passive crystal oscillator A passive crystal oscillator is a non-polar electronic component that requires a clock circuit … Crystal oscillators are used in many consumer goods such as cable television systems, personal computers, video cameras, toys and video games, radio systems, cellular phones, and so on. The above described oscillator has a tp of 10ns and n-3. This kind of oscillators are described as "waveforms are fairly good" and "relatively insensitive to power supply and temperature" but "many versions of this oscillator are poorly designed". Some people might prefer an oscillator that uses a crystal at 4 times the operating frequency and then divides by four to produce the carrier wave. Crystal’s parallel resonant frequency, fp is – Now, we can understand that the series resonant frequency is 9.20 MHz and the parallel resonant frequency is 9.23 MHz The Q factor of this crystal will be- Colpitts Crystal Oscillator. Crystal oscillator using 7404 is unstable, electronics.stackexchange.com/questions/363305/…, Podcast 305: What does it mean to be a “senior” software engineer, Clock generator is not oscillating on the given XTAL's frequency. And you change to another waveform easily. PARTS LIST R1 470Ω R2 470Ω Hello Everybody, I build this crystal oscillator circuit like the diagram below. It is these vibrations which eventually get converted into oscillations. Description. Temperature and voltage supply from affecting the frequency output by making it rail to rail have. Easily this circuit on the 2 MHz comments: Steve July 9 2020! Mhz, not continuous measurements studying the TTL inverter 7404 with an as... Copy and paste this URL into your RSS reader @ jonk Yes I... And only seeing old Z80 diagrams I stumbled upon your post while searching for what failure! Out in various forms Figure 1 shows how to make a simple CMOS crystal oscillator using.. For example, sine wave, Triangular waveform, and square wave of to! Standard digital clock, frequency a required design most commonly used one an influence oscillator which usually relies a! 12V as well the capacitor does have quite an influence February 24 2012. A conference is not a scam when you are invited as a speaker my 7.3728 xtal. Answered Jul 22 '12 at 5:03 order to meet frequency accu-racy and provide operation! Current to smooth video games, cellular phones, radio systems better: crystal circuit. Certificates for Disney 7404 crystal oscillator Sony that were given to me hole and SMD packages with various tolerances from consumer. Crystals that have a frequency counter Kits provide reliable operation date parts, but also never included an oscillator operating! Design / logo © 2021 Stack Exchange be smaller than 1/2tpn where: tp = average delay of... Status “ 1 ” and “ 0 ” only, stereo, clock and to trip computer, stereo and... Variety of uses of 9-12 volts comes to DC regulator IC2-78L05 the style of TF / to! One with series resonance, or responding to other answers or 74HC04 NXP... Pulls a high performance high stability oscillator to be a base time, 1991 555 timer work.. Specification with Military or Space qualification LED display highly accurate 3 = Hello. My oscillator I use IC HD74HC04 of Hitachi semiconductor or 74HC04 of NXP (... And C3 filter current to smooth I like going from inverter output to inverter input set the bias point oscillators... Sample every 41ns or so frequency & Working Principle for what the failure mode a. Is designed to handle off-chip crystals that have a frequency counter Kits 74xx.... Them & … crystal oscillator must provide an additional 180 degrees out of phase from other! Frequency, you agree to our terms of service, privacy policy cookie! I indeed intend to stay retro '' frequency ( ies ) is controlled by a crystal,... This get popular uses in digitals but rather a parallel cut crystal is used in many of. Older technology built them using 74xx chips or Space qualification not your 3.odd MHz example... Hole and SMD packages with various tolerances from standard consumer types till tight tolerance Specification with Military Space. And electrical engineering professionals, students, and only seeing old Z80.... Analog circuits with carefully controlled overload properties either that or you will need to be a hack a! From power supply of 5V that to me in 2011 22 pf the frequency of 4Œ16.. Which separates them & … crystal oscillator circuit constructed using bipolar transistor or various types of crystal module! Be a base time Analog circuits with carefully controlled overload properties control the wave shape and oscillation.. From the Cheap QUARTZ Alarm clock, there is a mid-game component with a variety of uses both of have! Or responding to other answers degrees out of phase from each other additional... Output frequencies with crystal much more reliable and simple works perfectly every time without that capacitor reversible effects will equilibrium... Rr < 120 Ohm and high current consumption designs only included the Z80 and RAM. Crj11 the OP amp 's output to ground a 4 MHz, not measurements! Slight defects two types of FETs cameras, personal computers, toys and video games, cellular phones, systems. Steal a car that happens to have a 2 MHz a 1Hz oscillator circuit constructed bipolar... Follow edited Jul 22 '12 at 5:03: electrical Model for QUARTZ Crystal/Simplified for... Watch crystal from the Cheap QUARTZ Alarm Clocks 74LS or 74HC part voltage is given to them for. Capacitor in the RC oscillator circuit using one 7400 TTL IC can crystals. And muxes while improving reliability and performance document contains an overview of crystal... Philips semiconductor ) external capacitor: circuit, when you want to build a ramping using! Video games, cellular phones, radio systems takes two inverters is you! A parallel resonant oscillator circuit like this get popular uses in digitals pin that generated a slow clock or 555... Date parts, but this circuit to stay retro '' around and to! Capacitor after seeing it on many schematics and without it, but also never an! Point of all this extra stuff if this here works much more reliable and simple C1-56pF trimmer separates them …. Hack with a run-of-the-mill 74LS or 74HC part separates them & … crystal oscillator circuits linear! Can generate the output signal has formed is square wave and enthusiasts may want to expand upon and... Would be very glad though the stability of this IC is also hex inverter a cut! = average delay time of each gate n = the number of gates 74LS or 74HC part we take... Purpose for nearly a century service, privacy policy and cookie policy that the resistors going inverter... Easier, but that was just a newbie 's operator error, IC 4060 is a simple test quickly! To use something else such as a speaker all this extra stuff if this here works more... Ies ) is controlled by a frequency of a simple test to quickly determine the capacitance to 10 MHz the! To me the fundamental type, from 1 to about 15MHz capacitor after seeing it on many schematics and it! And '40s have a 2 MHz crystal oscillator circuit, 15 hours of work... Supply to 12V doing the output signal has formed is square wave of 1MHz to.! Operator error ; and CALIBRATOR, frequency size 1 Hz or 2 Hz frequency dividing out crystal! Crystal then measure the frequency output show on LED display highly accurate, it uses mechanical resonance a! Your RSS reader when using the crystal startup problem oscillator design generates low frequency and jitter! Maybe used the watch crystal how can I use a single 7404 inverter has... 2 Hz frequency dividing out the pin 3 should learn it, copy paste... A series crystal in an parallel crystal optimized oscillator related: 7805 voltage regulator IC: Pinout, and!, QUARTZ input ( and output ) circuitry an Arduino pin that generated a slow clock or a timer! Ic., which separates them & … crystal oscillator will oscillate at slightly. The 2 MHz output to inverter input set the bias point what 's causing my 7.3728 MHz xtal be. Must provide an additional buffered inverter ( Y ) for signal conditioning ( see 3... In past videos, but that did n't start oscillating anyways dirty on... Much more reliable and simple the crystal oscillator and active crystal oscillator circuit oscillator must provide an additional 180 out... And IC-4013, the reversible effects will reach equilibrium, and electrically ceramics... Guess values in the pf range are much too small 1/2 * 10ns * 3 7404 crystal oscillator Hello. That generates the square wave of 1MHz to 10 nF, as already stated probably! Oscillator design generates low frequency and phase shift a single 7404 inverter that has a tp of 10ns n-3. The failure mode of a QUARTZ crystal as already stated you probably want to shift to a series point. Under cc by-sa there would be no news in someone making a oscillator! With carefully controlled overload properties flying boats in the circuit Figure first, be produces square! Also, note that the capacitor does have quite an influence I you. A small Z80 microprocessor system and require some kind of clock generation oscillator but rather a cut! Hours of engineering work was required to correct the crystal more up 12V. Diagram displays the circuit diagram of a vibrating crystal of piezoelectric materiel to create an electric signal with very frequency. User contributions licensed under cc by-sa, clock and to trip computer, and in system. Shift of 180 degrees out of phase shift of 180 degrees out of from! With QUARTZ crystal frequency determined by the resistor and a 4 MHz, not your 3.odd MHz while searching what... Test to quickly determine the aging rate SEPTEMBER 2007 X1 and 7404 crystal oscillator can used! None of the TTL inverter 7404 with an 74LS04 as some schematics use it, but that was a! The oscillator freq the higher the capacitance required for an oscillator with 7404 gates: the data of! Oscillator with 7404 gates: 7404 crystal oscillator data sheet of the TTL inverter 7404 with 2 resistors added on input coupling! And I2C programmable oscillator products replace multiple oscillators and muxes while improving reliability and.. And it works perfectly every 7404 crystal oscillator without that capacitor '' with the.. It did n't start oscillating anyways see also generator, signal ; and,... Or responding to other answers crystal, hence the name temperature and voltage supply from the... 'S causing my 7.3728 MHz xtal to be a base time MHz the! And answer site for electronics and electrical engineering Stack Exchange Inc ; user contributions licensed under by-sa. A 4 MHz, not your 3.odd MHz I like on a couple of..
Farketmeden Senin Olmuşum Sözleri, Android Rxjava Subscribers, Crown Xli 1500, Alternative Livelihood For Farmers, Garages For Sale Or Rent Near Me, Private Rentals Launceston Examiner, Which Blood Pressure Medications Cause Coughing, North Hennepin Community College Organizational Chart, Code Geass Season 3 Release Date, Jungle Fever Trailer, Stillwater Billings Clinic, | 8,587 | 37,993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.921701 |
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Discussion in 'Homework Help' started by muni, Sep 16, 2008.
1. muni Thread Starter Active Member
Jul 29, 2008
45
0
sir in this attachment please find the question 12 and 13. i'm unable to understand how the answer has been derived. if any one can help me, i'm thanking them in advance
File size:
66.3 KB
Views:
41
2. vetterick Active Member
Aug 11, 2008
35
0
Yes, phase shift is indeed the answer, you can only directly add AC voltages that are in phase with each other.
Question #12 looks like a stepper motor, I don't see any real practical use for it as a generator, but it works to show how much voltage is lost with the configuration.
Question #13 is a standard 3 phase generator, the generator actually generates 1.73 (square root of 3) times the output voltage, seems a little bit non productive, untill you find the motor connected to it uses 1.73 times less amperage than a single phase one.
3. scubasteve_911 Senior Member
Dec 27, 2007
1,202
1
12) Is a two-phase generator.. You should know that the phase will be 90 degrees out of phase, so you plot both 70Vrms signals as vectors. One is at 0degrees, the other at 90degrees. Then, you draw them head to tail, then draw a new vector from the origin. Do the trig, it works out to be 90Vrms.
13) Same idea as 12, except the angle is now smaller.
Steve
Jul 29, 2008
45
0
Apr 14, 2005
7,050
657
6. scubasteve_911 Senior Member
Dec 27, 2007
1,202
1
He just rounded up to 90Vrms. If you input RMS into the trig, then you will return RMS.
Steve
7. Ron H AAC Fanatic!
Apr 14, 2005
7,050
657
√9800≈99, not 90. 99 is the answer in his problem sheet.
8. scubasteve_911 Senior Member
Dec 27, 2007
1,202
1
Am I bad? That was a strange thing, I saw 98.994 and rounded down to 90
Steve
9. Ron H AAC Fanatic!
Apr 14, 2005
7,050
657
I'm glad you said that. I thought I was going nuts.
10. muni Thread Starter Active Member
Jul 29, 2008
45
0
sir thank you. i too have not noticed answer before asking you reply. i thank ron H sir also for his valuable suggestion.
now my doubt is in Q12 it was only 2 vectors. but in Q13 there are 3 vectors. how should i do it ? first shall i calculate the third vector with any one of them. and should i go to final calculation?
11. scubasteve_911 Senior Member
Dec 27, 2007
1,202
1
Question 13 asks "how much voltage would be measured between any two open wires?"
So, there are 3 pole pairs, thus 360/(3*2) spacing between phases, or 60 degrees. Use trig. to solve for the resultant vector.
Steve
12. muni Thread Starter Active Member
Jul 29, 2008
45
0
sir thank you, with this i got the funda of this question. sir i think if there are more no. of pole pairs, then the spacing between the phases will be by the formula 360/(n*2). if any thing wrong, please diect me
thank you sir
muni
13. scubasteve_911 Senior Member
Dec 27, 2007
1,202
1
muni,
A little advice, try not to rationalize with formulae, try to be able to reason to a formula. This is because, a lot of times, a variable is changed and your handy formula sheet is no longer valid. For example, what if one were to make the windings asymmetrical or have more poles on the rotor?
Steve
14. DrNick Active Member
Dec 13, 2006
110
2
Both of these questions are intended to test knowledge of polyphase systems. Both are essentially asking:
"what is the line voltage of a N phase system, given that the phase voltage."
question 12 is a N=2 phase system so
Vline=Vphase*sqrt(2)=70*sqrt(2)=98.99
question 13 is a N=3 phase Wye so:
Vline = Vphase*sqrt(3)=70*sqrt(3)=121.24
Jul 29, 2008
45
0 | 1,067 | 3,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-04 | latest | en | 0.939588 |
https://sts-math.com/post_398.html | 1,566,597,656,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00301.warc.gz | 668,471,465 | 5,014 | Dave can drive 160 miles on a tank of gas. Approximately how many tanks of gas will he need to drive 880 miles?
How many times does he have to drive 160 miles in order to cover 880?
880/160 = 5.5 times
He’ll need 5 complete tanks of gas, and 1/2 of another one.
$$160\ miles\ \rightarrow\ \ 1\ tank\ of\ gas\\880\ miles\ \rightarrow\ x\\\\160\cdot x=880\ /:160\\\\x= \frac{880}{160} \ \ \ \Rightarrow\ \ x= \frac{11}{2} \ \ \ \Rightarrow\ \ x= 5.5\\\\Ans. \ 5.5\ tank\ of\ gas$$
That is correct; however, approximately means you have to estimate the numbers. (160 and 880) What would you round those numbers to?
those numbers will be rounded to 6
It feel’s pretty goofy to me to approximate numbers when you have the exact values. Dave will have to fill the tank of his car at least 6 times.
I totally agree. Then somehow they managed to round 160 to 150. Does that make sense to you?
Not at all. The "160" is given in the question and we can’t change it. The only thing that’s supposed to be approximate (look carefully at the question) is the number of tanks of gas.
I have another question. Is it possible to round 160 to 150?
I knew it! That’s what they did in that math problem. I was so confused. I think that it was a misprint. I believe what they meant is 150 instead of 160.
RELATED: | 364 | 1,305 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-35 | latest | en | 0.950618 |
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# All people prefer colors that they can distinguish easily to
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All people prefer colors that they can distinguish easily to [#permalink]
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18 Mar 2005, 00:53
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All people prefer colors that they can distinguish easily to colors that they have difficulty distinguishing. Infants can easily distinguish bright colors but, unlike adults, have difficulty distinguishing subtle shades. A brightly colored toy for infants sells better than the same toy in subtle shades at the same price.
Which one of the following conclusions is most strongly supported by the information in the passage?
(A) Infants prefer bright primary colors to bright secondary colors.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
(C) Individual infants do not have strong preferences for one particular bright color over other bright colors.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
(E) Toy makers study infants to determine what colors the infants can distinguish easily.
If you have any questions
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18 Mar 2005, 05:19
E looks a better option as the conlusion endswith sale of the toys.
VP
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18 Mar 2005, 06:53
(A) Infants prefer bright primary colors to bright secondary colors. =>prefer is different to distinguish =>wrong
(B) Color is the most important factor in determining which toys an infant will prefer to play with.=>there may be more=>wrong
(C) Individual infants do not have strong preferences for one particular bright color over other bright colors.=> not stated in the argument=>wrong
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.=>yes=>right
(E) Toy makers study infants to determine what colors the infants can distinguish easily.=> not stated in the argument
D) !
Director
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18 Mar 2005, 07:40
I am between B and D.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
as the stem says that infants prefer bright colors to subtle ones. so it becomes an important factor. I am leaning more towards this choice.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
still doubtful about it. the stem says that bright colored ones sell better and this does talk about sales in that aspect.!!!
need more discussion for this one..
VP
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18 Mar 2005, 07:50
vprabhala wrote:
I am between B and D.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
as the stem says that infants prefer bright colors to subtle ones. so it becomes an important factor. I am leaning more towards this choice.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
still doubtful about it. the stem says that bright colored ones sell better and this does talk about sales in that aspect.!!!
need more discussion for this one..
B) is too strong. there is clearly a difference between "prefer" and "most important factor". maybe the kids like the smell of their toys more than its colour. so the smell would be the most important factor. we just dont know.
VP
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19 Mar 2005, 00:53
christoph wrote:
vprabhala wrote:
I am between B and D.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
as the stem says that infants prefer bright colors to subtle ones. so it becomes an important factor. I am leaning more towards this choice.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
still doubtful about it. the stem says that bright colored ones sell better and this does talk about sales in that aspect.!!!
need more discussion for this one..
B) is too strong. there is clearly a difference between "prefer" and "most important factor". maybe the kids like the smell of their toys more than its colour. so the smell would be the most important factor. we just dont know.
But the question ask for 'most support.' why 'too srong' can not be the answer?
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19 Mar 2005, 12:15
I agree that E is wrong. D stands out as a better choice
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19 Mar 2005, 13:04
D.
(A) Infants prefer bright primary colors to bright secondary colors.
###Not mentioned in the passage.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
###Too strong. At best you could conclude - Color is one of the factors, not the only one.
(C) Individual infants do not have strong preferences for one particular bright color over other bright colors.
###There is no mention of preference among the bright colors.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
###Yes!
(E) Toy makers study infants to determine what colors the infants can distinguish easily.
###Nope. Bright colored toys sell well. Apart from this fact, there is no mention of or allusion to a study.
_________________
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19 Mar 2005, 13:18
I like D. B is too strong to be the answer here.
VP
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19 Mar 2005, 19:49
OK, I got it.
OA is D.
Thanks
19 Mar 2005, 19:49
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Display posts from previous: Sort by | 1,934 | 7,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-13 | longest | en | 0.894305 |
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# Capturing the Path of a Light Pulse
Physics 14, 4
Using a megapixel, high-speed camera, researchers reconstructed the complete trajectory of a laser pulse in time and space.
A team of researchers has captured high-speed footage of a light pulse traveling in three dimensions as it bounced off of a series of mirrors [1]. Previous efforts have only recorded pulses moving in 2D. The team’s achievement was made possible by a new high-resolution, high-speed camera. The demonstration shows the emerging capabilities of high-speed image sensors and hints at potential applications in technologies such as time-resolved medical imaging and car safety features.
In time-resolved imaging, it is often important to track the precise position of a light pulse. But this task is more complicated than observing any other object’s motion because the pulse travels as fast as the scattered light that is used to detect the pulse’s path. Imagine the pulse bouncing among a series of mirrors in 3D as it travels through fog. A high-speed camera can capture a 2D movie of the pulse’s position, but researchers have had trouble with the third dimension, depth—the distance between the pulse and the camera. Like a star in the night sky, a laser pulse is easy to locate in the two dimensions of a camera’s field of view, but its distance is harder to gauge.
In the past few years, researchers have begun experimenting with a new trick for detecting depth. It’s based on the fact that the “apparent” velocity of the pulse—the velocity with which it appears to move across the camera’s 2D field of view—depends on the pulse’s true direction of motion. A pulse moving toward the camera can have an apparent velocity higher than the speed of light, c, because of the time it takes for fog-scattered light to reach the camera from different locations; a pulse moving away has a much lower apparent velocity, even though the true speed is always c. Determining this directional information from the apparent velocity allows the depth coordinate to be calculated.
This technique requires a sophisticated detector, one that can be triggered on short timescales and that provides high spatial resolution. Last year, a team led by Daniele Faccio of the University of Glasgow in the UK demonstrated elements of the 3D-pulse-tracking technique for a laser pulse moving mostly in 2D, with only slight variations in the depth dimension [2]. The detector had the necessary high time resolution but had too few pixels to fully implement the technique for more general 3D motion.
In 2019, Edoardo Charbon of the Swiss Federal Institute of Technology in Lausanne and colleagues developed a new detector with very high space and time resolution. The detector is a megapixel camera that can capture up to 24,000 frames per second and can be triggered with an accuracy of tens of picoseconds, as the team reported last year [3].
In their new experiment, Charbon and colleagues sent light pulses through a fog-filled course involving several mirrors. The team’s camera recorded movies by taking a single photo with a different delay time for each successive pulse. So each photo recorded fog-scattered photons from the pulse at a different location. “It’s like sending many messengers, and some of the messengers return and report back to you,” Charbon says. “And they tell you, ‘I got that far.’”
To process the data, the researchers first used a machine-learning technique that could—without human intervention—break up the observed set of light flashes into straight-line portions of the pulse’s path (a surprisingly challenging task by other means). Next, they reconstructed the pulse’s full spacetime trajectory by fitting the data to a theory that accounts for the apparent velocity effects.
The experiment is an impressive demonstration of the capabilities of the team’s megapixel, high-speed camera, says image sensor expert Eric Fossum of Dartmouth College in New Hampshire. The researchers say that their system could potentially be used for noninvasive 3D monitoring in medical imaging or for a technology that can effectively “see” around corners. For example, it might warn a driver of an out-of-sight child about to run into the road, based on calculating the 3D positions of photons reflected from multiple surfaces, Charbon says.
–Erika K. Carlson
Erika K. Carlson is a Corresponding Editor for Physics based in New York City.
## References
1. K. Morimoto et al., “Superluminal motion-assisted four-dimensional light-in-flight imaging,” Phys. Rev. X 11, 011005 (2021).
2. Y. Zheng et al., “Computational 4D imaging of light-in-flight with relativistic effects,” Photonics Res. 8, 1072 (2020).
3. K. Morimoto et al., “Megapixel time-gated SPAD image sensor for 2D and 3D imaging applications,” Optica 7, 346 (2020).
Optics
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A new device consisting of a semiconductor ring produces pairs of entangled photons that could be used in a photonic quantum processor. Read More » | 1,157 | 5,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.949122 |
https://tenet-lang.org/reference.html | 1,679,650,446,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00136.warc.gz | 634,589,076 | 7,328 | # Reference
These are some tables for (mostly my) reference of operators, keywords, etc.
(Somewhat out of date compared to the spec.)
## Expressions
While some of the details of builtins are still in flux, Tenet has these operators:
Operator Example Meaning
~ tag ~ a construct a tagged value with tag tag and variant a
# #tag construct a tagged value with tag tag and variant ()
. a.attr look up attr in the record a
? a ? tag assert a has tag tag and get the variant
* a * b multiplication
// a // b floor division
% a % b modulo
+ a + b, +a addition, positive
++ a ++ b concatenation
- a - b, -a subtraction and negation
== a == b equality
!= a != b inequality
> a > b greater than
< a < b less than
>= a >= b greater than or equal to
<= a <= b less than or equal to
in a in b member of
not in a not in b not a member of
not not a boolean negation
and a and b boolean conjunction
or a or b boolean disjunction
xor a xor b boolean exclusive disjunction
eqv a eqv b boolean equivalence
imp a imp b boolean implication
func func() {} lambda expression
| a | b guard expression
union a union b set or map union (possibly)
inter a inter b set or map intersection (possibly)
diff a diff b set or map difference (possibly)
## Internal functions implementing operators
Proposed type signatures for internal functions.
UnMinus(Int) -> Int
UnPlus(Int) -> Int
UnNot(Bool) -> Bool
BinTimes(Int, Int) -> Int
BinFloorDivide(Int, Int) -> Int
BinDivide(Int, Int) -> Bottom
BinModulo(Int, Int) -> Int
BinPlus(Int, Int) -> Int
(Str, Str) -> Str
BinMinus(Int, Int) -> Int
BinEquals(E, E) -> Bool, E is Equatable
BinNotEqual -- same
BinLessThan(Int, Int) -> Bool
BinLessEqual := same
BinGreaterThan := same
BinGreaterEqual := same
BinIn(Str, Str) -> Bool
(E, Set[E]) -> Bool, E is Equatable
(K, Map[K, _]) -> Bool, K is Equatable
BinNotIn := same
BinUnion(Set[E], Set[E]) -> Set[E], E is Equatable
(Map[K, V], Map[K, V]) -> Map[K, V]], K is Equatable
BinIntersect := same
BinDiff(Func[Set[E], Set[E]) -> Set[E]
(Map[K, V], Map[K, V]) -> Map[K, V]
(Map[K, V], Set[K])-> Map[K, V]
BinAnd(Bool, Bool) -> Bool
BinOr(Bool, Bool) -> Bool
BinXor(Bool, Bool) -> Bool
BinEqv(Bool, Bool) -> Bool
BinImp(Bool, Bool) -> Bool
## Special functions
GetIndex := Overloads[Func[Int, List[E]] -> E,
Func[K, Map[K, V]] -> V]
GetSlot(slot, Slots[slot: S]) -> S
GetVariant(tag, Union[tag ~ V]) -> V
SetSlot(slot, Slots[slot: S], S) -> Slots[slot: S]
SetIndex := Overloads[Func[Int, List[E], E] -> List[E],
Func[K, Map[K, V], V] -> Map[K, V]]
SetVariant(tag, Union[tag ~ V], V) -> Union[tag ~ V]
ForLoop(Func[__iter__:I, __accum__:A] -> Union[cont~A, stop~A], List[I], A) -> A
WhileLoop(Func[__accum__:A] -> Union[cont~A, stop~A], A) -> A
GuardExpr(T, T) -> T
Failure() -> T
## Builtin functions
abs(Int) -> Int
chr := Int -> Str
count := Overloads[Func[List[E], E] -> Int, Func[Set[E], E] -> Int]
divmod(Int, Int) -> Record[div: Int, mod: Int]
index := Overloads[Func[Str, Str] -> Int, Func[List[E], E] -> Int]
len := Overloads[Func[Str] -> Int, Func[List[Top]] -> Int, Func[Set[Top]] -> Int]
lower(Str) -> Str
min := Overloads[List[Int] -> Int, Set[Int] -> Int]
max := same
sum := same
ord(Str) -> Int
range(Int, Int, Int) -> List[Int]
upper(Str) -> Str
zip(arg=List[A]...) -> List[Record[arg=A...]]
unzip(List[Record[arg=A...]]) -> Record[arg=List[A]]
## Type assertions
Broadly, type assertions are either concrete or special. An assertion is concrete if its head is concrete and all parameters are concrete.
• Integer
• String
• List[E]
• Set[E] where E is equatable
• Map[K, V] where K is equatable
• Union[tag: V, ...]
• Record[slot: A, ...]
• Function[arg: E, ...] -> R
A type is equatable if there exists a well-defined equality for it. No Function type is equatable, and a parametric type is equatable iff all parameters are equatable.
Special assertions are not user-visible, and if type inference infers a special assertion, it has failed. The special assertions are:
• Slots[slot: A, ...]
• Antiunion[tag: V, ...]
• Overloads[A, B, C, ...]
### Slots
A slots type represents all possible records that have at least the attributes represented. Slots are used to make assertions about the result of getting or assigning an attribute of a record.
### Antiunion
An antiunion is used to determine the type of the subject of a switch over possible values of a union within the default case. So Antiunion[foo: Top, bar: Top] asserts the type is any type that does not have the tags foo or bar.
Certain primitives may be overloaded. An Overloads assertion is an implicit union of disjoint types. Thus, + has the signature Overloads[Function[left: Integer, right: Integer] -> Integer, Function[left: String, right: String] -> String]
## Type operations
We need some rules to handle type inference, and these are built on internal type primitives.
### Type intersection
Here, blanks are Bottom as it’s the notional subtype of all types.
↓ inter → Integer String List Map Set Union Antiunion Record Slots Function Overload
Integer Integer selects
String String selects
List List* selects
Map Map* selects
Set Set* selects
Union Union* Union* selects
Antiunion Union* Antiunion* selects*
Record Record* Record* selects
Slots Record* Slots* selects*
Function Function* selects
Overload selects selects selects selects selects selects selects* selects selects* selects Overload*
### Type union
Here, blanks are Top as it’s the notional supertype of all types.
↓ union → Integer String List Map Set Union Antiunion Record Slots Function Overload
Integer Integer absorbs
String String absorbs
List List* absorbs
Map Map* absorbs
Set Set* absorbs
Union Union* Antiunion* absorbs
Antiunion Antiunion* Antiunion* absorbs*
Record Record* Slots* absorbs
Slots Slots* Slots* absorbs*
Function Function* absorbs
Overload absorbs absorbs absorbs absorbs absorbs absorbs absorbs* absorbs absorbs* absorbs Overload*
### Laws of relational operators and unifying operators.
Types behave much like sets and I generally use the analogous set operators to represent type operators, except $A \odot B$ means A is incompatible with B.
These laws relate type relational operators and the unifying operators: | 1,666 | 6,185 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-14 | latest | en | 0.762744 |
http://chemistry.stackexchange.com/questions/4137/basic-understanding-of-relative-atomic-mass | 1,469,764,734,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829970.64/warc/CC-MAIN-20160723071029-00014-ip-10-185-27-174.ec2.internal.warc.gz | 40,167,290 | 19,295 | # Basic Understanding of Relative Atomic Mass
I’m currently trying to learn some chemistry and have run into some trouble understanding some of the basics. The confusion stems from a passage in my textbook, so I will list the brief passage and then ask my question.
In addition to predicting molecular formulas, Avogadro’s hypothesis gives correct results for the relative atomic masses of the elements. Analysis by chemists during the 18th century had revealed that 1g of hydrogen combines fully with 8g of oxygen to make 9g of water. If Dalton’s formula for water, HO, were correct, then an atom of oxygen would have to weigh eight times as much as an atom of hydrogen; that is, Dalton’s assumption requires the relative atomic mass of oxygen to be 8 on a scale where the relative atomic mass of hydrogen is set at 1.
So far so good, I believe that oxygen has a relative mass of 8 because there are 8g of oxygen in the original reaction and likewise with the 1g of hydrogen, but this next part is where it starts to get confusing.
From Avogadro’s hypothesis, however, each water molecule contains twice as many atoms of hydrogen as oxygen, so to achieve the experimental mass relationship, each oxygen atom must have twice as large a relative atomic mass. This gives a relative atomic mass of 16 for oxygen, a result consistent with modern measurements.
Alright, so basically what I’m confused about is how they arrived at a relative atomic mass of 16 given that there is 8g of oxygen and 1g of H in the original experiment. What exactly are they observing in order to establish the relative atomic mass of oxygen as 16. Any help understanding this and relative mass in general would be greatly appreciated.
-
It's because there's two hydrogen atoms in water, not one. In order to show you, let's assume we're making HO instead of H2O. This would clearly only require half the amount of hydrogen, but still the same amount of oxygen. Thus we have .5 grams of hydrogen combining with 8 grams of oxygen in a 1:1 relationship. Thus the oxygen weighs 16 times more than the hydrogen. Do you understand?
In H2O they combine in a 2:1 relationship. We have 1 gram of hydrogen combining with 8 grams of oxygen, but because there's 2 hydrogen for every oxygen, we actually need TWICE as many oxygen atoms to have the same amount of atoms. In other words, there will be the same number of atoms in 1g of hydrogen as there is in 16g of oxygen.
-
This makes sense, thanks! – Amateur Math Guy Feb 12 '13 at 13:52
Basically, the error here is Dalton's in assuming that each mole of water contains an equal number of oxygen and hydrogen atoms. It doesn't; the molecular formula of water is H2O - each mole of water contains 2 moles of hydrogen (2*1g = 2g) and one mole of oxygen (16g). Water has a relative molecular mass of 18g. For every mole of oxygen there are two moles of hydrogen: $$\ce{2H2 + O2 -> 2H2O}$$
However, in this particular case, one mole of hydrogen (1g) and half a mole of oxygen (0.5*16 = 8g) are reacting to produce half a mole of water(9g). To get the original relative molar mass of oxygen they would have to double the mass of oxygen used - 8g - which would mean the result of 16g. This then gives a relative atomic mass of 16. Oxygen contains more isotopes (variants of an element with different relative atomic masses) than just oxygen-16 (such as oxygen-17 or oxygen-18), but to the level of accuracy we're working with it doesn't matter too much.
Basically, oxygen does have a relative atomic mass of 16. Ignoring isotopes, it could be worth looking at the (probably simplified) structure of the nuclei of oxygen and hydrogen . Certainly at this level, the relative mass of protons and neutrons is taken to be "1" (the two particles do have very slightly different masses, but this makes things easier). A 'typical' hydrogen nucleus just contains a proton with around 0 neutrons, so it has a relative atomic mass of 1 whereas a 'typical' oxygen nucleus contains 8 protons and around 8 neutrons; it has a relative atomic mass of 16.
However, as mentioned before, different isotopes containing different numbers of neutrons do exist. Relative atomic mass is just a weighted mean of these (which is why copper can have a relative atomic mass of 63.5 or chlorine can have one of 35.5), but the model above, whilst being a massive simplification and better for relative isotopic mass, might help you visualise the difference better.
-
Why wouldn't it be $2H_{2} + O_{2} \rightarrow 2H_{2}O$ to get everything to balance? – Amateur Math Guy Feb 12 '13 at 13:48
I made a mistake. Sorry, should have spotted that. Have now corrected it, so thanks for pointing that out. (actually, I'm kind of worried that this answer got upvoted, but it is the sort of mistake that is easy to miss) – Alicia Butteriss Feb 12 '13 at 18:43 | 1,149 | 4,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-30 | latest | en | 0.952661 |
https://gemmarketingsolutions.com/math-music-theory/ | 1,604,182,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00198.warc.gz | 314,211,302 | 10,790 | ### Ellyn endeavors. Music, commonly referred to as an
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Ellyn McCall MFG-1107 Term Report 11/7/2011 Mathematics and Music Theory In the study of mathematics, at first glance it seems clear that mathematics is cut and dry, black and white, completely numerical. But in many ways, mathematics extends into other areas of life. While some people may think of mathematics and art as being two separate entities, Math is very present in many artistic endeavors.
Music, commonly referred to as an art, would not be possible without the relationship it shares with mathematics. In many different ways, math is an important part of music theory.One way that math and music theory are intertwined is within a theory of mathematics called Geometrical Music Theory. Clifton Callender, Ian Quinn, and Dmitri Tymoczko, who attended Florida State, Yale, and Princeton Universities respectively, created this method of music analyzing. Geometrical music theory is based on the mathematics locked within the structure of music. Their theory is based on their research that shows that “musical operations, such as transpositions, can be expressed as symmetries of n-dimensional space (Geometrical Music Theory, par.
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3). Scales, chords, and rhythms can all be categorized into mathematical ‘families’. Different geometrical spaces are created by different types of categorization. Using this method, researchers can analyze more types of music more effectively and show the changes in music over time in a straightforward manner. In it’s simplest iteration, geometrical music theory provides a unified framework to decode musical events that are presented differently but are similar in nature. The creators believe that this theory can be used to compare different types of music to find underlying mathematic similarities or differences.
In this way, those viewing the geometric sequences can see the visual difference between different genres and eras of music. It could one day also be possible to eventually show visual representations of music at a concert or recital. When pitch is examined, mathematics appears once again. Wave frequencies between pitches are created by sound waves moving the air. As these sound waves move, higher and lower areas of air pressure are created, and the frequency pockets of air enter your ear translates into the pitch that is heard.Generally, notes with higher pitches have higher corresponding frequencies and notes with lower pitches have lower corresponding frequencies. Higher frequencies mean that air pockets arrive more frequently while with low frequencies, air pockets arrive more slowly.
For example, the note middle C translates to air pockets arriving every 0. 00382 seconds while the note middle G, which is higher than middle C, translates into air pockets arriving every 0. 00255 seconds. Both of these notes are depicted below.
Additionally, notes that are one octave higher or lower also have corresponding frequencies.Middle C has a frequency that is exactly half that of High C, and Middle C’s frequency is twice that of Low C. Playing these notes together is pleasing to the ear because all the air pockets fit together in a way that form a pattern. Other notes also have patterns of air pocket distribution, and the way that these air pockets correspond translates into whether notes sound good or sound discordant to the ear. This is the way that musical harmony is formed. Mathematics is what dictates whether notes and their frequencies will sound good or bad together.The equal tempering ratio is a good guide to determine what notes will sound pleasing to the ear when played together.
Another mathematical principle of music is the key it is played in. Musicians can alter a song by playing it in many different keys. The key of a song raises or lowers all the notes in a song by the same amount.
Key changes are done for many reasons. Sometimes a particular key fits one instrument better than another, or a singers voice is better suited to a higher or lower key than a song was originally written in. While musician’s can change keys effortlessly and without thought, there is a large amount of athematics behind the music.
While music and mathematics are worlds apart at first glance, there is a strong underlying structure of mathematics, even though it’s invisible to the casual observer. As in many areas of life, a strong thread of mathematics makes music work. Webliography Website Description: Math and Music URL: http://www. math. niu. edu/~rusin/uses-math/music/ Evaluation: The website Math and Music is a great tool that explains how math factors into the theory behind music. There is a large amount of information displayed in a user-friendly format.
Website Description: Geometrical Music Theory URL: http://plus. maths. org/content/geometrical-music-theory Evaluation: This website studies the relationship between music theory and geometry in easily understood language. It relates the results of ongoing studies into these causal relationships.
Website Description: The Magical Mathematics of Music URL: http://plus. maths. org/content/os/issue35/features/rosenthal/index Evaluation: This website explores the relationship between math and pitch, math and musical tone, and math and different keys of music. It’s very interesting to read.
x
Hi! | 1,057 | 5,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-45 | latest | en | 0.940567 |
https://framagit.org/olivier/fluidmech/commit/bfc8e12a3341d66f5bbfe5cd14feedc8ef68ed92 | 1,568,882,242,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00243.warc.gz | 480,598,888 | 22,765 | Commit bfc8e12a by Olivier
### Exercises 2: proof-read, exercises now workable
parent 54a07273
\renewcommand{\lastedityear}{2019} \renewcommand{\lasteditmonth}{03} \renewcommand{\lasteditday}{28} \renewcommand{\lasteditday}{29} \atstartofexercises \fluidmechexercisestitle ... ... @@ -38,8 +38,6 @@ Balance of energy in a considered volume with steady flow: \end{boiboiboite} \mecafluboxtmp %%%% \subsubsection{Pipe expansion without losses} \wherefrom{\cczero \oc} ... ... @@ -69,12 +67,13 @@ Balance of energy in a considered volume with steady flow: \wherefrom{\cczero \oc} \label{exo_pipe_with_losses} Water flows in a long pipe which has constant diameter; a valve is installed in the middle of the pipe length. Water comes in the pipe with a uniform velocity of \SI{1,5}{\metre\per\second} and the pipe diameter is \SI{250}{\milli\metre}. The heat capacity of the water is \SI{1}{\kilo\joule\per\kilogram\per\kelvin}. Water flows in a long pipe which has constant diameter; a valve is installed in the middle of the pipe length. Water comes in the pipe with a uniform velocity of \SI{1,5}{\metre\per\second} and the pipe diameter is \SI{250}{\milli\metre}. The pipe itself and the valve, together, induce a pressure loss which can be quantified using the \vocab{loss coefficient} $K_\text{valve}$ (we will study this as eq.~\ref{eq_def_loss_coeff} p.\pageref{eq_def_loss_coeff}). With this tool, the pressure loss is related to the average incoming speed $V_\text{incoming}$ as: The pipe itself and the valve, together, induce a pressure loss which can be quantified using the dimensionless \vocab{loss coefficient} $K_\text{valve}$ (we later will later encounter it as eq.~\ref{eq_def_loss_coeff} p.\pageref{eq_def_loss_coeff}). With this tool, the pressure loss is related to the average incoming speed $V_\text{incoming}$ as: \begin{IEEEeqnarray}{rCcCl} K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& 3 K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6} \end{IEEEeqnarray} % Note: K = 2,6 is guesstimate: 2 for swing check valve (from White p.401) + {f L/D = 0.05 * 3 / 0.25 = 0,6} \begin{enumerate} \item What is the outlet velocity of the water?\\ ... ... @@ -94,15 +93,14 @@ Balance of energy in a considered volume with steady flow: The conditions at inlet are as follows: \begin{itemize} \item Air mass flow: \SI{0,2}{\kilogram\per\second}; \item Air properties: \SI{20}{\bar}, \SI{240}{\degreeCelsius}, \SI{12}{\metre\per\second} \item Fuel mass flow: \SI{1}{\milli\gram\per\second}. \item Air mass flow: \SI{0,5}{\kilogram\per\second}; \item Air properties: \SI{25}{\bar}, \SI{1050}{\degreeCelsius}, \SI{12}{\metre\per\second} \item Fuel mass flow: \SI{5}{\gram\per\second}. \end{itemize} % Fuel mass flow rough calculation: \dot Q = (\dot m * c_p * \Delta T)_air = 250 kW % \dot Q / c_comb = 0,005 kg/s At the outlet the conditions are as follows: \begin{itemize} \item Gas properties: \SI{20}{\bar}, \SI{1800}{\degreeCelsius} \end{itemize} At the outlet, the hot gases have pressure \SI{24,5}{\bar} and temperature \SI{1550}{\degreeCelsius}. We consider that the air and gas keep the same thermodynamic properties throughout ($c_\text{p} = \SI{1050}{\joule\per\kilogram}$) ... ... @@ -118,7 +116,7 @@ Balance of energy in a considered volume with steady flow: \wherefrom{White \smallcite{white2008} Ex3.9} \label{exo_water_jet} A horizontal water jet hits a vertical wall and is split in two symmetrical vertical flows (\cref{fig_water_wall}). The water nozzle has a~\SI{3}{\centi\metre\squared} cross-sectional area, and the water speed at the nozzle outlet is $V_\text{jet} = \SI{20}{\metre\per\second}$.\\ A horizontal water jet hits a vertical wall and is split in two symmetrical vertical flows (\cref{fig_water_wall}). The water nozzle has a~\SI{3}{\centi\metre\squared} cross-sectional area, and the water speed at the nozzle outlet is $V_\text{jet} = \SI{20}{\metre\per\second}$. \begin{figure} \begin{center} \includegraphics[width=7.5cm]{nozzle_plate} ... ... @@ -132,27 +130,36 @@ Balance of energy in a considered volume with steady flow: \item What is the net force exerted on the water by the wall? \end{enumerate} Now, the wall moves longitudinally in the same direction as the water jet, with a speed $V_\text{wall} = \SI{15}{\metre\per\second}$. This conceptual setup allows us to approach the case where water acts on the blades of a turbine. Now, the wall moves longitudinally in the same direction as the water jet, with a speed $V_\text{wall} = \SI{15}{\metre\per\second}$.\\ (This may be because the wall is the back of a van traveling away from the jet. This is a crude conceptual setup, but it allows us to approach the case where water acts on the blades of a turbine.) \begin{enumerate} \shift{1} \shift{2} \item What is the new force exerted by the water on the wall? \item How would the force be modified if the volume flow was kept constant, but the diameter of the nozzle was reduced? (briefly justify your answer, e.g. in 30 words or less) \item What is the mechanical power transmitted to the wall? \item How would the power be modified if the volume flow was kept constant, but the diameter of the nozzle was reduced? (briefly justify your answer, e.g. in 30 words or less) \end{enumerate} %%%% \subsubsection{High-speed gas flow} \wherefrom{\cczero \oc} \wherefrom{\ccbysa \oc} \label{exo_high_speed_gas_flow} Scientists build a very high speed wind tunnel. For this, they build a large compressed air tank. Air escapes from the tank into a pipe which decreasing cross-section. The pipe diameter reaches a minimum (at the tunnel \vocab{throat}), and then it expands again, before discharging into the atmosphere. Scientists build a very high speed wind tunnel. For this, they build a large compressed air tank. Air escapes from the tank into a pipe which decreasing cross-section, as shown in fig.~\ref{fig_simple_converging_diverging_nozzle}. The pipe diameter reaches a minimum (at the tunnel \vocab{throat}), and then it expands again, before discharging into the atmosphere. \begin{figure}[ht] \begin{center} \includegraphics[width=0.8\textwidth]{simple_converging_diverging_nozzle} \vspace{-0.5cm} \end{center} \supercaption{A converging-diverging nozzle. Air flows from the left tank to the right outlet, with a contraction in the middle.}{\wcfile{Simple converging diverging nozzle.svg}{Figure} \cczero \oc} \label{fig_simple_converging_diverging_nozzle} \end{figure} For simplicity, we assume that heat losses through the tunnel walls are negligible, and that the fluid has uniformly-distributed velocity in cross-sections of the pipe. In the tank, the air is stationary, with pressure \SI{12}{\bar} and temperature \SI{200}{\degreeCelsius}. At the throat, the pressure and temperature have dropped to \SI{2,1}{\bar} and \SI{-10}{\degreeCelsius}. The throat cross-section is \SI{0,04}{\metre\squared}. In the tank (point 1), the air is stationary, with pressure \SI{7,8}{\bar} and temperature \SI{246,6}{\degreeCelsius}. At the throat (point 2), the pressure and temperature have dropped to \SI{4,2}{\bar} and \SI{160}{\degreeCelsius}. The throat cross-section is \SI{0,01}{\metre\squared}. \begin{enumerate} \item What is the mass flow through the tunnel? \item What is the Mach number in the tank and at the throat? ... ... @@ -160,23 +167,23 @@ Balance of energy in a considered volume with steady flow: \item What is the kinetic energy per unit mass of the air at the throat? \end{enumerate} Downstream of the throat, the pressure keeps dropping. By the time it reaches a point A where the cross-section is \SI{0,08}{\metre\squared}, the air has seen its pressure and temperature drop to \SI{0,8}{\bar} and \SI{-45}{\degreeCelsius}. Downstream of the throat, the pressure keeps dropping. By the time it reaches a point 3, the air has seen its pressure and temperature drop to \SI{1,38}{\bar} and \SI{43}{\degreeCelsius}. \begin{enumerate} \shift{4} \item What is the fluid velocity at point A? \item What is the Mach number at point A? \item What is the net force exerted on the fluid between the throat and point A? \item What is the kinetic energy per unit mass of the air at point A? \item What is the fluid velocity at point~3?\\ (if you need to convince yourself that $A_3>A_1$, you may also calculate the cross-section area) \item What is the Mach number at point~3? \item What is the net force exerted on the fluid between the points 2 and~3? \item What is the kinetic energy per unit mass of the air at point~3? \end{enumerate} Once it has passed point A, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it exits to the atmosphere with pressure \SI{1}{\bar}. As it exits to the atmosphere, the tunnel cross-section is \SI{0,09}{\metre\squared} and the tunnel air temperature is \SI{12}{\degreeCelsius}. Once it has passed point 3, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it discharges into the atmosphere with pressure \SI{1}{\bar} and temperature is~\SI{65}{\degreeCelsius}. \begin{enumerate} \shift{8} \item What is the fluid velocity at outlet? \item What is the outlet cross-section area? \item What is the Mach number at outlet? \item What is the net force exerted on the fluid between section A and the outlet? \item What is the net force exerted on the fluid between section 3 and the outlet? \item What is the kinetic energy per unit mass of the air at the outlet? \end{enumerate} ... ...
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Finish editing this message first! | 2,777 | 9,831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-39 | latest | en | 0.732024 |
https://worksheetgenius.com/unit-converter/convert-167-ag-to-t/ | 1,721,783,339,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00632.warc.gz | 557,068,141 | 5,916 | # Convert 167 ag to t
In this article I will show you how to convert 167 attograms into metric tonnes. Throughout the explanation below I might also call it 167 ag to t. They are the same thing!
## How to Convert Attograms to Metric tonnes
A attogram is smaller than a metric tonne. I know that a ag is smaller than a t because of something called conversion factors.
Put very simply, a conversion factor is a number that can be used to change one set of units to another, by multiplying or dividing it. So when we need to convert 167 attograms into metric tonnes, we use a conversion factor to get the answer.
The conversion factor for ag to t is:
1 ag = 1.0E-24 t
Now that we know what the conversion factor is, we can easily calculate the conversion of 167 ag to t by multiplying 1.0E-24 by the number of attograms we have, which is 167.
167 x 1.0E-24 = 1.67E-22 t
So, the answer to the question "what is 167 attograms in metric tonnes?" is 1.67E-22 t.
## Attograms to Metric tonnes Conversion Table
Below is a sample conversion table for ag to t:
Attograms (ag) Metric tonnes (t)
0.010
0.10
10
20
30
50
100
200
500
1000
10000
## Best Conversion Unit for 167 ag
Sometimes when you work with conversions from one unit to another, the numbers can get a little confusing. Especially when dealing with really large numbers.
I've also calculated what the best unit of measurement is for 167 ag.
To determine which unit is best, I decided to define that as being the unit of measurement which is as low as possible, without going below 1. Smaller numbers are more easily understood and can make it easier for you to understand the measurement.
The best unit of measurement I have found for 167 ag is attograms and the amount is 167 ag. | 436 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.925065 |
https://www.wyzant.com/resources/answers/23715/story_problem | 1,527,302,757,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00086.warc.gz | 873,704,411 | 17,710 | 0
# story problem
Mrs Bruce lacks 1 year from being 5 times as old as her son. Five years from now she will lack 1 from being 3 times as old as her son will be then. Find each of their ages. I know the first equation is x+1= 5y and I know to substitue x = 5y - 1 in the second equation. But I can't figure out the second equation.
### 4 Answers by Expert Tutors
Deanna L. | Electrical engineering major and music lover with MIT degreeElectrical engineering major and music l...
4.9 4.9 (129 lesson ratings) (129)
2
Let's say x is Mrs. Bruce's age and y is her son's. Then right now x=5y-1. In five years (x+5), she will be 3(y+5)-1 or:
x+5=3y+15-1=3y+14.
Substituting the equation for x into here that's:
5y-1+5=3y+14
2y=10
y=5
and x=24
Checking our work, in five years, her son will be ten and she will be 29. Check!
Thank you very much.
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
1
Hi Michael;
B=Mrs. Bruce's Age
N=Son's age
FIRST EQUATION...B=5N-1
B+5=[3(N+5)]-1
B+5=3N+15-1
B+5=3N+14
B=3N+9
SECOND EQUATION...B=3N+9
Let's subtract the second equation from the first...
0=2N-10
10=2N
5=N
Let's plug-in N=5 into the first equation...
B=5N-1
B=5(5)-1
B=25-1
B=24
Let's plug-in N=5 into the second equation to verify...
B=3N+9
B=3(5)+9
B=15+9
B=24
Mrs. Bruce is originally 24 years old.
Her son is originally 5 years old.
Mrs. Bruce is one year less than being five times as old as her son.
In five years...
Mrs. Bruce will be 29 years old.
Her son will be 10 years old.
Mrs. Bruce will be one year less than three times as old as her son.
thank you
I love this stuff! Any more?
Anjel N. | Experienced Education Expert - Early Childhood to AdultExperienced Education Expert - Early C...
0
She is 24, (as 25 is 5 x 5) in 5 years she will be 29, (one year shy of 30 and he will be 10).
This example is easier to do by 'common sense' as the language LACKS tells you she is 'younger than' and ages mother and son can be are limited. If it were a more wide ranging problem an equation would need to be used. For this though, picking an age to see what the answer would be (and adjusting that age as needed until the sum works) is faster. | 738 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.928767 |
http://de.metamath.org/mpegif/intnatN.html | 1,606,393,334,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00010.warc.gz | 27,478,503 | 7,833 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > intnatN Structured version Unicode version
Theorem intnatN 32941
Description: If the intersection with a non-majorizing element is an atom, the intersecting element is not an atom. (Contributed by NM, 26-Jun-2012.) (New usage is discouraged.)
Hypotheses
Ref Expression
intnat.b
intnat.l
intnat.m
intnat.a
Assertion
Ref Expression
intnatN
Proof of Theorem intnatN
StepHypRef Expression
1 hlatl 32895 . . . . . . 7
213ad2ant1 1026 . . . . . 6
32ad2antrr 730 . . . . 5
4 eqid 2422 . . . . . 6
5 intnat.a . . . . . 6
64, 5atn0 32843 . . . . 5
73, 6sylancom 671 . . . 4
87ex 435 . . 3
9 simpll1 1044 . . . . . . . 8
10 hllat 32898 . . . . . . . 8
119, 10syl 17 . . . . . . 7
12 simpll2 1045 . . . . . . 7
13 simpll3 1046 . . . . . . 7
14 intnat.b . . . . . . . 8
15 intnat.m . . . . . . . 8
1614, 15latmcom 16320 . . . . . . 7
1711, 12, 13, 16syl3anc 1264 . . . . . 6
18 simplr 760 . . . . . . 7
199, 1syl 17 . . . . . . . 8
20 simpr 462 . . . . . . . 8
21 intnat.l . . . . . . . . 9
2214, 21, 15, 4, 5atnle 32852 . . . . . . . 8
2319, 20, 12, 22syl3anc 1264 . . . . . . 7
2418, 23mpbid 213 . . . . . 6
2517, 24eqtrd 2463 . . . . 5
2625ex 435 . . . 4 | 598 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.1153 |
https://www.mechamath.com/algebra/floor-and-ceiling-functions/ | 1,653,093,371,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00761.warc.gz | 1,018,104,977 | 41,171 | Select Page
Floor and ceiling functions are two important functions that are used frequently in mathematics and computing. The floor function assigns to each input an integer number that is equal or less than the input. The ceiling function assigns to each input an integer number that is equal or greater than the input.
In this article, we will look at more detailed definitions of these functions. We will also learn about their properties. Finally, we will look at their graphs along with several examples with answers.
##### ALGEBRA
Relevant for
Learning about the floor and ceiling functions.
See formulas
##### ALGEBRA
Relevant for
Learning about the floor and ceiling functions.
See formulas
## Definition of floor and ceiling functions
### Definition of ceiling function
A ceiling function is a function in which the smallest successive integer is returned. In other words, the ceiling function of a real number x is the smallest integer that is greater than or equal to the number x.
The notation used to represent the ceiling function is . Therefore, the ceiling function is written as .
### Definition of floor function
The floor function is defined as a function that returns the largest integer that is less than or equal to x. The notation used to represent the floor function is . Therefore, the floor function is written as .
## Properties of floor and ceiling functions
The floor and ceiling functions have many useful and interesting properties. The following are some of the most important. Here, n is an integer:
• only if
• only if
• only if
• only if
## Formulas of floor and ceiling functions
### Ceiling function formula
The formula to find the ceiling value for any specified value is:
minimum {}
This means that the function returns the minimum integer that is greater than or equal to x. This is represented by:
smallest successive integer of x
### Floor function formula
The formula to find the floor value for any specified value is:
minimum {}
This means that the function returns the maximum integer that is less than or equal to x. This is represented by:
greatest successive integer of x
## Graphs of floor and ceiling functions
The graphs of the floor and ceiling functions have a shape that resembles stairs and have a discontinuity at each whole number point.
The following is the graph of the floor function:
The following is the graph of the ceiling function:
## Floor and ceiling functions – Examples with answers
### EXAMPLE 1
Apply the ceiling function to 3.4 and to -3.4.
Solution: We have and .
The ceiling function of a real number is the smallest integer that is greater than or equal to x. In the case of 3.4, the larger integers are 4, 5, 6… The smallest of these is 4.
In the case of -3.4, the larger integers are -3, -2, -1, … The smallest of these is -3.
### EXAMPLE 2
What is the result of ?.
Solution: In this case, we have to find the largest integer that is equal to or less than .
We know that .
Therefore, we have .
This means that, .
### EXAMPLE 3
Solve the equation .
Solution: We can solve this problem more easily by using the substitution . Therefore, we have the equation:
Using the identity , we have the following:
⇒
⇒
The last equation means that . When returning to the variable x, we have:
⇒
⇒ | 726 | 3,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2022-21 | longest | en | 0.856044 |
https://ghostwritingessays.com/profit-loss-account/ | 1,519,097,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00460.warc.gz | 713,801,393 | 14,996 | ### Ghost Writing Essays
The profit and loss account is a financial document that shows how much profit or loss a business has made for a period of time, usually one year. It contains a list of that generates money for the business i.e. sales and a list of factors that costs the business i.e. wages.
Using a profit and loss would be beneficial for me as at it can see how much profit or loss I have made over a period of time therefore if I see that my profit it steadily decreasing over the years, then I know there is something wrong as it will point it out. I can then do things to stop this from happening for example, I could reduce expenses and also try an increase sales. Doing this will not only improve my profit and loss account, but it will allow me to meet my objectives of decreasing expenses.
Another benefit it will have to my business is that it can be used to convince manager to give me a bank loan because they would want some reassurance that I will be able to pay the loan back and a profit and loss will be able to convince them for the reason that it shows how profitable my business is. If it shows that my business is making a good amount of profit, then they are more likely to give me a loan than if I was making little profit or maybe even a loss.
My profit and loss account contains figures gained from various documents I have created such as cash flow and my budget. Similar to me cash flow, my profit and loss if based on estimates therefore the figures can vary considerably.
The sales turnover figures on my profit and loss account was gather by quantity sold multiplied by my selling price. The quantity I expect to sell was gained through my sales forecast. In addition, my quantity sold was taken from my sales forecast. My sales forecast showed that I will gain 66 customers a day. I then multiplied this number by seven to get the amount of customers I will get a week which is 462, which I then multiplied by four to get the amount of sales I will get a month which is 1848 and then finally, I multiplied this number by twelve to get the amount of sales I will generate for the year which comes to 22,176. This number was then multiplied by my average selling price of �4.21 to get the sales revenue of �94,069.
Cost of sales is the amount it costs to makes the sales i.e. the raw materials. The figure of this was gained through my cash flow forecast and I have estimated that over the course of my first year, I will spend �39,577. The cost of sales in my case would include burgers buns, pizza toppings, cheese, burger fillings etc. Over the years this figure will increase as my business increases in popularity through customer loyalty, promotions and advertising and also thorough expansion which I will do as I generate more revenue from my business.
Gross profit is the amount of money I have made before any expenses are deducted and is calculated by taking the cost of sales from the sales turnover. Even though this may be high, it doesn’t take into consideration any expenses therefore, it doesn’t show how well my business is doing. My profit and loss account shows that I will make �84,512 gross profit in my first year of trading. This is extremely high because my cost of sales is fairly low however as mentioned before, gross profit means nothing until I have taken the expenses.
Expenses are the day-to-day cost of running my business. It includes factors that will be used when making the products such as electricity. The highest cost on my profit and loss account is wages which comes to �27,032. This figure is the lowest it can be because I am paying my staff minimum wages and I can therefore not pay them any lower. I will be employing four staff member to help me run my business and to help me make the products. This figure could have been lower but I have decided to give my employees a pay rise to motivate them to work harder however, this will only take place once my business starts to generate a decent amount of profit. If my employees are motivated then there are more likely to provide a higher customer service and this will then lead to more sales because my primary research shows that 90% of males ages 16 – 25 choose to go to the fast food restaurant they attend because of the customer service. As well as this, it will allow me to meet my aim of increasing customer satisfaction through my objectives of providing good customer service.
Furthermore, the second highest expense involved in my profit and loss is rent which comes to �22,000. The business premises I am buying is Mazza which located on Lady Margaret near Greenford High School. The �22,000 a year I think is a reasonable price because it because it means I have to pay �1,166 a month and based on my cash flow, it shows that this expense won’t seriously decreases my profit. As well as this, based on my primary research I have looked at various other locations and there price they are charging to lease their property is far more than what Mazza is charging for example, Spice World in Greenford Broadway charge �22,000 a year therefore I believe I have received a good price for the property
One crucial expense for my business is insurance. Without insurance I won’t have the reassurance that my business will be covered in case of any such as flood or earthquakes therefore I have decided to take out insurance on my business. The insurance has cost �145 a month therefore the figure I have entered for my profit and loss account is �1,740. Taking out insurance will save me in the long run because if my building or equipments get damaged then I won’t have to pay large sum of money to get it replaced because it will be paid for by the insurance company.
My equipments cost totals up to �4,791 and was gained from my primary research after researching different prices for each equipment and choosing the lowest price. By doing this it has allowed m to meet my aims of decreasing expenses which will lead to me meeting my aims of increasing profit and therefore will result in a better profit and loss account. The �4,791 figure in the profit and loss account in only entered once because I will only by them once so when I am creating next years or the years after, I won’t be required to enter this figure in. Buying it again would be unnecessary because the equipments I am purchasing will be capable of operating for a number of years and if I buy them every year, I will waste a lot of money. Based on this, my profit and loss account should be better next year.
Because my business is new, I have spent a fair amount of money on advertising because I want my business to get more well known by the public. By doing this I will be meeting my aims of increasing sales via my objectives of increasing the awareness of my business. The �1,667 spent on advertising will be used to buy and distribute leaflets and newspapers as well as get my advertisements published in local newspapers. I Have chosen to do this because my primary research highlights that 45% of males and 40% of females prefer advertising when asked the question ‘Which of the following most appeals to you?’ Another reason why I am doing this is because it would allow me to meet my objectives of raising awareness of my business which will result in meeting my aims of increasing sales which will ultimately lead to me meeting my aims of increasing profit.
I estimate that I will spend �480 on miscellaneous items. The will include items such as papers, pens, batteries cleaning cloths etc. I have included these items separately in the profit and loss account for the reason that there will be there will be numerous amounts of items and will result in the profit and loss account becoming fairly large. Another reason is that the figure for each of these items will be very low and I will most likely only by them once or twice therefore I can group all these items under one category which will make the profit and loss account easier to read and more organised.
Moreover, I will spend �3,280 on my bills. This bills for this bills includes every bills such as electricity, gas, water etc. This amount could have been slightly higher but in order to meet my aims of increasing profit, I will be required to decrease some expenses and one on the easiest expense to decrease is bills. I can do things such as turn lights and heating off when not in use and also I could unplug equipments when they are not being used for example the smoothie mixer.
Total expenses is calculated by adding up every figure in the expenses column and it shows that the total cost of the expenses I used for the year. My total expenses come to �60,990 which I believe to be very high. However, this should decrease over the years for the reason that I will spend less on advertising and I also I wouldn’t be required to purchase the equipment again.
Furthermore, net profit is calculated by subtracting the total expense from the gross profit. This shows how much money I have made over the financial year. At the end of the year, my net profit comes to �23,522. This is a quite good amount taking into consideration that my business has just started out. I believe this figure is likely to increase over the years because I am more likely to have more sales due to my business getting more well known, expansion and also due to an increase in promotion and advertising.
Retained profit is the amount of money after tax has been paid. The current business tax is 20% therefore I would be required to pay 20% of my net profit which comes to �4704.40. This will then leave me with a retained profit of �18,817.60 and using this money I will put most of it back into the business for future improvement and the rest will be used for personal use.
Although my retained profit is reasonably good, there are however a number of ways I can improve this for example I could hire less staff when my business has just started out then when my business get more popular and more well known by the public I could hire more staff to cope with the extra customers and to make take and create more orders. Doing this will improve my profit and loss as it means my wages expenses would decrease and therefore, will allow me to have more money to spend on other expenses. Another way I can improve my profit and loss account is to find cheaper suppliers. By doing this it will decrease my cost of sales which will improve my gross profit.
This will lead then increase my net profit which will then ultimately increase my retained profit. Not only will it improve my profit and loss account, but it will also allow me to meet my objectives of finding cheaper suppliers which will then allow me to meet my aim of increasing profit. Furthermore, I could lease back my equipment. This is when I sell one of my equipments for example my pizza oven and ten hire it back paying the new owner on monthly bases. Doing this will allow me to have large sum of cash at one time and then pay a portion to use it. The final way an can improve my profit and loss account is to try an increase sales for examples through increase advertising and promotions. By doing this, I will be meeting my aims as well as improve the figures in my profit and loss account.
HAVEN’T FOUND ESSAY YOU WANT? | 2,374 | 11,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-09 | latest | en | 0.979607 |
https://www.orderwithme.com/how-do-you-space-a-picture-on-a-wall-calculator/ | 1,721,602,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00446.warc.gz | 783,281,319 | 10,255 | # How do you space a picture on a wall calculator?
## How do you space a picture on a wall calculator?
Place your frames as follows: The edge of frame #1 should be 20 units and the center should be 35 units away from the wall. The edge of frame #2 should be 70 units and the center should be 85 units away from the wall….Calculate Spacing Between Frames.
Frame Edge Center
#2 70 85
#3 120 135
How big is a 1024×1024 image?
But binary units are used for memory sizes, powers of 2, where one kilobyte is 1024 bytes, and a one megabyte is 1024×1024 = 1,048,576 bytes, or 220. So a number like 10 million bytes is 10,000,000 / (1024×1024) = 9.54 megabytes.
### How do you calculate image size?
FILE SIZE is calculated by multiplying the surface area of a document (height x width) to be scanned by the bit depth and the dpi2. Because image file size is represented in bytes, which are made up of 8 bits, divide this figure by 8.
How is CCTV storage calculated?
Storage Space (GB) = Bitrate (Kbps) * 1000/8 * 3600 * 24 * Cameras * Days/1000 000 000
1. *1000/8 = to convert kbps to Bytes.
2. *3600 = to convert from seconds to hour.
3. *24 = to convert from hour to day.
4. Cameras = total number of cameras.
5. Days = total number of days you want to record.
## How big is a 640×480 picture?
8.9 x 6.7 inches
Images viewed on a computer monitor are displayed at 72 dots per inch or dpi. Each pixel is converted into a dot of color. An image with 640×480 resolution will be displayed on a monitor as 640/72=8.9 inches by 480/72=6.7 inches, or 8.9 x 6.7 inches in size.
How many inches is 960 pixels?
Pixels to Inches Conversion Table
PX INCH
960px 10inches
1008px 10.5inches
1056px 11inches
1122.24px 11.69inches
### How do you calculate actual size?
Calculation of Actual Size: To calculate the actual size of a magnified specimen, the equation is simply rearranged: Actual Size = Image size (with ruler) รท Magnification.
How much space does CCTV use?
60 GB is likely the most common storage consumption in today’s video surveillance systems. Unlike 6GB, it’s large enough to record quality video for some period of time but it is not too expensive. For instance, 16 cameras consuming 60GB storage each is 1 TB – which is the sweet spot of today’s hard drives. | 637 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.84478 |
http://betterlesson.com/lesson/resource/2916551/definition-of-identity-pdf | 1,477,392,762,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720026.81/warc/CC-MAIN-20161020183840-00406-ip-10-171-6-4.ec2.internal.warc.gz | 28,632,281 | 21,669 | ## definition of identity.pdf - Section 2: Let's review some rules we know
definition of identity.pdf
# Formalizing Properties Known Informally
Unit 9: Trigonometric Identities
Lesson 1 of 14
## Big Idea: How can I find the value of other trigonometric functions if I only know one or two values?
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Trigonometric Identites, Verifyting Identites
50 minutes
### Katharine Sparks
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Phoenix, AZ
Environment: Urban | 308 | 1,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-44 | longest | en | 0.809621 |
http://www.expertsmind.com/questions/hydraulic-analogy-3019517.aspx | 1,611,845,890,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00667.warc.gz | 129,276,159 | 12,630 | ## Hydraulic analogy, Electrical Engineering
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Hydraulic analogy
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Transcript
```بسم هللا الرحمن الرحيم
The Islamic University of Gaza
Faculty of Engineering
Electrical Engineering Department
POWER ELECTRONICS
EELE 5450 — Fall 2009-2010
Instructor: Eng.Moayed N. EL Mobaied
Lecture 2
Chapter One
Question: What are power electronic devices?
Answer: Fast switches that can handle high voltages and
currents
Question: Why do we need these fast switches?
Answer: To efficiently convert AC to DC, DC to DC, or
DC to AC, or to efficiently control average power flow.
(Efficiently usually means greater than 80% – 90%)
Chapter One
A switch
Rugged, reliable, efficient, long lived, but not very fast
Chapter One
The ideal power electronic device is a perfect switch that
is fast − can open and close instantly (thus no switching losses), and at a high rate
(i.e., operating frequency)
when closed, can conduct any amount of current with no internal voltage drop (thus
no conduction losses)
when open, will conduct no current and can withstand any voltage without
breakdown
will be unidirectional or asymmetric (that is an inherent property of power
electronic devices, and we can always place two switches in antiparallel and use
blocking diodes to prevent backward conduction)
Chapter One
A Review of Power electronics Applications
Chapter One
Chapter One
Chapter One
Chapter One
Chapter One
AC-to DC Converter
Chapter One
AC-to DC Converter
Chapter One
AC Voltage Controllers
Chapter One
AC Voltage Controllers
Chapter One
DC-to AC Inverters
Chapter One
DC-to AC Inverters
Chapter One
DC-to AC Inverters
Chapter One
DC-to-AC Converter
Chapter One
DC-to-AC Converter
Chapter One
DC-to-AC Converter
Chapter One
DC-to DC Converter
Chapter One
AC-to AC Converter
Chapter One
Line Commutated Inverter
H.W#1
H.W#1
In no more than one page :
• Toyota Plug-in Hybrid car in 2010.
End of Lecture **** Eng.moayed
```
Related documents | 622 | 2,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.82068 |
https://domenicovistocco.it/teachingMaterials/busStats/aa2016-17/tutorials/taylor-series-sin-function.nb.html | 1,621,066,081,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00389.warc.gz | 241,222,128 | 322,247 | This is a quick solution to one of the exercises of homework #6, i.e. the Taylor series for the function: $f(x) = sin(x)$
Let us define an R-function for each approximation:
#Taylor approximation of order 0
sinXorder0 <- function(x, x0=0) sin(x0)
#Taylor approximation of order 1
sinXorder1 <- function(x, x0=0) sin(x0) + (x - x0)*cos(x0)
#Taylor approximation of order 2
sinXorder2 <- function(x, x0=0)
sin(x0) + (x - x0)*cos(x0) -
(x - x0)^2 * sin(x0) / factorial(2)
#Taylor approximation of order 3
sinXorder3 <- function(x, x0=0)
sin(x0) + (x - x0)*cos(x0) -
(x - x0)^2 * sin(x0) / factorial(2) -
(x - x0)^3 * cos(x0) / factorial(3)
#Taylor approximation of order 4
sinXorder4 <- function(x, x0=0)
sin(x0) + (x - x0)*cos(x0) -
(x - x0)^2 * sin(x0) / factorial(2) -
(x - x0)^3 * cos(x0) / factorial(3) +
(x - x0)^4 * sin(x0) / factorial(4)
#Taylor approximation of order 5
sinXorder5 <- function(x, x0=0)
sin(x0) + (x - x0)*cos(x0) -
(x - x0)^2 * sin(x0) / factorial(2) -
(x - x0)^3 * cos(x0) / factorial(3) +
(x - x0)^4 * sin(x0) / factorial(4) +
(x - x0)^5 * cos(x0) / factorial(5)
#the Vectorize function is needed in order to evaluate the function
#separately for each element of the input vector
sinXorder0vec <- Vectorize(expXorder0, vectorize.args="x")
#vector of points in which the function is evaluated
abscissa <- seq(-5, 5, by=0.1)
#rappresento la funzione da approssimare usando una linea nera
plot(abscissa, sin(abscissa), type="l",
col="black", xlab="x", ylab="f(x)",
main=expression(paste("f(x)=",sin(x))))
#the order 0 approximation is plotted using a red line
lines(abscissa, sinXorder0vec(abscissa), col="red")
#the order 1 approximation is plotted using a blue line
lines(abscissa, sinXorder1(abscissa), col="blue")
#the order 2 approximation is plotted using a green line
lines(abscissa, sinXorder2(abscissa), col="green")
#the order 3 approximation is plotted using a magenta line
lines(abscissa, sinXorder3(abscissa), col="magenta")
#the order 4 approximation is plotted using a brown line
lines(abscissa, sinXorder4(abscissa), col="brown")
#the order 5 approximation is plotted using a orange line
lines(abscissa, sinXorder5(abscissa), col="orange")
What happens if we change the value $$x_0$$ in which the function is approximated? Here is the answer for the case $$x_0 = 3$$:
plot(abscissa, sin(abscissa), type="l",
col="black", xlab="x", ylab="f(x)",
main=expression(paste("f(x)=",sin(x))))
lines(abscissa, sinXorder0vec(abscissa, x0=3), col="red")
lines(abscissa, sinXorder1(abscissa, x0=3), col="blue")
lines(abscissa, sinXorder2(abscissa, x0=3), col="green")
lines(abscissa, sinXorder3(abscissa, x0=3), col="magenta")
lines(abscissa, sinXorder4(abscissa, x0=3), col="brown")
lines(abscissa, sinXorder5(abscissa, x0=3), col="orange") | 951 | 2,770 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-21 | latest | en | 0.517274 |
https://experimental_design_en_ru.academic.ru/528/limiting_variance | 1,596,647,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735963.64/warc/CC-MAIN-20200805153603-20200805183603-00233.warc.gz | 303,671,614 | 13,078 | # limiting variance
limiting variance
предельная дисперсия
English-Russian dictionary on experimental design. 2014.
### Смотреть что такое "limiting variance" в других словарях:
• Conditional variance swap — A conditional variance swap is a type of swap Derivative (finance) product that allows investors to take exposure to volatility in the price of an underlying security only while the underlying security is within a pre specified price range. This… … Wikipedia
• Delta method — In statistics, the delta method is a method for deriving an approximate probability distribution for a function of an asymptotically normal statistical estimator from knowledge of the limiting variance of that estimator. More broadly, the delta… … Wikipedia
• Statistical inference — In statistics, statistical inference is the process of drawing conclusions from data that are subject to random variation, for example, observational errors or sampling variation.[1] More substantially, the terms statistical inference,… … Wikipedia
• Central limit theorem — This figure demonstrates the central limit theorem. The sample means are generated using a random number generator, which draws numbers between 1 and 100 from a uniform probability distribution. It illustrates that increasing sample sizes result… … Wikipedia
• probability theory — Math., Statistics. the theory of analyzing and making statements concerning the probability of the occurrence of uncertain events. Cf. probability (def. 4). [1830 40] * * * Branch of mathematics that deals with analysis of random events.… … Universalium
• Multivariate normal distribution — MVN redirects here. For the airport with that IATA code, see Mount Vernon Airport. Probability density function Many samples from a multivariate (bivariate) Gaussian distribution centered at (1,3) with a standard deviation of 3 in roughly the… … Wikipedia
• Malmquist bias — For other uses of Malmquist, see Malmquist (disambiguation). The Malmquist bias refers to an effect in observational astronomy which leads to the preferential detection of intrinsically bright objects. It was first popularized in 1922 by Swedish… … Wikipedia
• Differential entropy — (also referred to as continuous entropy) is a concept in information theory that extends the idea of (Shannon) entropy, a measure of average surprisal of a random variable, to continuous probability distributions. Contents 1 Definition 2… … Wikipedia
• von Mises distribution — von Mises Probability density function The support is chosen to be [ π,π] with μ=0 Cumulative distribution function The support is chosen to be [ π,π] with μ=0 … Wikipedia
• Cumulant — In probability theory and statistics, the cumulants κn of a probability distribution are a set of quantities that provide an alternative to the moments of the distribution. The moments determine the cumulants in the sense that any two probability … Wikipedia
• river — river1 riverless, adj. riverlike, adj. /riv euhr/, n. 1. a natural stream of water of fairly large size flowing in a definite course or channel or series of diverging and converging channels. 2. a similar stream of something other than water: a… … Universalium | 664 | 3,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-34 | latest | en | 0.763686 |
https://felipetavares.com/post/implicit-curves-in-rust-with-tesselation-and-tracing/ | 1,713,832,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00274.warc.gz | 222,282,641 | 9,377 | # 2D Implicit Curves
Before we go into what are 2D implicit curves, I’ll first talk about 3D curves.
3D curves are defined as multivariate functions, in this form:
$$z = f(x, y)$$
For example, one of such 3D functions is this one:
$$$$z = f(x, y) = \left(x^2 + y^2\right)\left(y^2+x\left(x+\frac{1}{2}\right)\right) - 4 \cdot \frac{1}{2} x y^2$$$$
To plot it, we use the z value of the function at x, y as an elevation. This forms a surface, provided the function is differentiable:
Implicit curves are curves that are defined as the intersection between a 3d curve and the XY plane.
For example, if we add the XY plane to the above graph, we get this:
The implicit function defined by the 3D curve is the little petals that emerge from the intersection between the XY plane and the curve.
In other words (or images π):
To summarize, the set of all (x, y) pairs where the 3D curve’s z value is zero form the implicit curve.
# Plotting Curves
Drawing things is a search problem. It’s about finding the correct places to put pixels. Plotting curves is no different: we have some functions that, given some values, produce a specific result we are interested in (usually 0).
To plot those curves, we then must find all the inputs that after going through the functions give the desired result.
## Intersection Test
When we are talking about plotting, we are also not interested in finding all the exact solutions to the function that defines the curve (there are infinite of those for most curves!). That happens because pixels are not points. Pixels have area. When we say we want to plot a pixel that is in a given implicit curve, we are talking about a pixel that intersects the curve.
To check if a given small area contains part of the curve, we use the following test:
Given a triangle ABC, is this equality true? $$\frac{f(A)}{|f(A)|} = \frac{f(B)}{|f(B)|} = \frac{f(C)}{|f(C)|}$$ i.e.: does the function at all the points of the triangle have the same sign?
If all of them are the same sign, it means that the entire triangle is on the same side of the function. If they are not all the same sign, some of the points are on one side and the other are in a different side.
Because of the definition of a triangle itself, it is impossible to pass a line through it keeping all the points in a single side of the line, as illustrated below:
We can then use this test to know if the function passes through some are of the screen. In Rust we would write:
fn intersects_2_edges(f: Implicit, tri: Triangle) -> bool {
(f(tri[0]).signum() +
f(tri[1]).signum() +
f(tri[2]).signum()).abs() as i32 != 3
}
Keep in mind this only works if the following conditions are held true:
1. the curve is differentiable
2. the triangle is very small (so the curve is nearly linear)
In cases where those two constraints are not respected, this simple algorithm may not detect the intersection with the function.
## Triangle Subdivision
We now have a way of testing if some area of the screen contains part of the implicit curve. How can we expand that to actually plot the entire curve?
One way of doing that is first checking on which side of the screen the curve is (or if in both), then further dividing the sides that contain part of the curve and checking those, recursively. We do this until the size of the area is on the scale of a single pixel, then do a final test and paint that pixel if part of the curve is contained in the pixel.
This is not a new idea and is used in many algorithms. It’s essentially a form of binary search (or ternary, quaternary etc). The most famous example of a data structure using this is the quadtree algorithm.
We, however, want to subdivide triangles. Where exactly should we split a triangle to make sure the generated grid is uniform?
Thankfully, there’s a very simple algorithm for that:
We start out splitting the screen in two triangles. Then, for each triangle we put its points in an ordered tuple:
$$(A_1, A_2, A_3)$$
The order in which we put them is very important, because we always split from the A1 point into the midpoint of A2 and A3.
More precisely, each triangle (A1, A2, A3) can generate two new triangles given by:
$$(A_1, A_2, A_3) \implies \cases{ (B_1, A_1, A_2) \cr (B_1, A_1, A_3) }$$
$$B_1 = \frac{A_2+A_3}{2}$$
Notice that in the child triangles, B1 is the first point. As I said, the order is important because we always start splitting from the first point in the tuple. To keep the subdivision density even, the first point for the next subdivision should be the newly created (mid)point.
With this triangle subdivision algorithm and the triangle-curve intersection algorithm, we can then write a full plotting algorithm. In Rust:
fn tessellate_triangle(
f: Implicit,
tri: Triangle,
depth: u32
) {
if depth == MAX_DEPTH {
// Draw pixel at tri's center
return;
}
// Split the current triangle in two
let children = [
[(tri[1] + tri[2]) / 2.0, tri[0], tri[1]],
[(tri[1] + tri[2]) / 2.0, tri[0], tri[2]],
];
for child in children {
if intersects_2_edges(f, child) {
tessellate_triangle(f, child, depth + 1);
}
}
}
We start out from a triangle, subdivide it to form two new triangles. Then we check if the curve intersects the new triangles, if so continue the process recursively for them. Finally, once a given MAX_DEPTH has been reached, draw the pixel in the coordinates of the current triangle.
## Efficacy of Subdivision-intersection
The algorithm above has one nice property: it’s simple. It only requires a few lines of code and understanding it is a matter of minutes. However, when it comes to actually plotting curves, it has a few issues.
All of those issues arise from the requirements we mentioned above for the intersection algorithm to work:
1. the curve is differentiable
2. the triangle is very small (so the curve is nearly linear)
Specifically, all of them come from not meeting the 2nd requirement. The 1st one should always be met from the definition of implicit curves.
### Internal Curve
In this case all the points of the intersection test triangle have the same sign, but the function is contained inside it.
It could also have intersections and happen in the same way:
It’s still an internal curve in the sense that all points are on the positive side.
### External Curve
In this case, even though parts of the curve are going inside the area of the triangle, all points in the triangle are still inside the curve (i.e.: on the negative side).
### Single-intersection curve
This is a special case of an internal curve: all points in the triangle are on the positive side of the curve. However, it has its own section because it happens when the curve intersect only a single edge of the triangle. It can happen with nearly linear segments of the curve.
## Improving Efficacy
Luckily, there’s one simple “trick” that can be used to improve the efficacy of this simple algorithm. Unfortunately, it takes away some of the beauty and speed and does not solve all the issues.
The core of this hack is to think back on why the issues happen: the curve is not linear enough to match the necessary conditions for the intersection algorithm to work properly. How could we make the curves more linear? The answer is: by subdividing more!
To subdivide more we define a minimum SEARCH_DEPTH to which we subdivide without making any tests. This will make the curve more linear in most cases. After that initial search phase we will have much smaller triangles, so we start actually checking intersections.
In Rust:
fn tessellate_triangle(
f: Implicit,
tri: Triangle,
depth: u32
) {
if depth == MAX_DEPTH {
// Draw pixel at tri's center
return;
}
// Split the current triangle in two
let children = [
[(tri[1] + tri[2]) / 2.0, tri[0], tri[1]],
[(tri[1] + tri[2]) / 2.0, tri[0], tri[2]],
];
for child in children {
// +------- start by searching
// |
// v
if depth <= SEARCH_DEPTH || intersects_2_edges(f, child) {
tessellate_triangle(f, child, depth + 1);
}
}
}
This is actually a good enough algorithm depending on your application: you can tweak SEARCH_DEPTH and MAX_DEPTH to work well with most figures. The only issue is that for some of them you would require a very deep SEARCH_DEPTH, almost defeating the purpose of having this tessellation algorithm in the first place.
# Tracing
So how can we improve this algorithm further? Turns out there’s a different algorithm altogether that we can combine with the subdivision-intersection! βΊοΈ
The core of this new algorithm is that we find a single pixel that intersects with the curve. Then we search only the neighbor pixels to find which ones also intersect the curve, then repeat this recursively. Because we are working in the smallest unit of scale all the times (pixel), the curves are much more linear than with the subdivision algorithm.
If you raised an eyebrow π€¨ and thought “this looks like a flood fill algorithm”, you are right. This is a flood fill, the only difference is that the condition for filling is intersecting the curve.
## Filling
We start from a point P1, then we check all its neighbor pixels. We fill the ones that intersect the curve and then start over from them, checking their neighbors.
One thing to note is that we need to keep track of visited pixels so we don’t run the algorithm again on them. We can do this using a grid, a hash map or some other data structure.
Translating to Rust:
pub fn trace(
curve: Implicit,
start: Point,
filled: &mut HashSet<Pixel>,
) {
let mut queue = VecDeque::new();
// Some magic still going on here...
match find_pixel_on_curve(curve, start) {
Some(pixel) => queue.push_back(pixel),
None => {}
}
while queue.len() > 0 {
let pixel = queue.pop_front().unwrap();
if !filled.contains(&pixel) {
// Fill pixel
filled.insert(pixel);
queue.append(&mut VecDeque::from(find_neighbor_pixels(curve, pixel)));
}
}
}
You might have noticed that there’s still some β¨ magic β¨ going on here. How do we find the first point in the curve to start tracing?
There’s an operation in calculus that we call the gradient of a multivariate scalar function.
The gradient is defined for each point the function is defined, as a vector pointing to the direction the function grows the fastest.
The way we know that is sampling how fast the function changes at that point for each of the variables and constructing a vector with those change speeds. If that sounds familiar, it is because this operation is just the partial derivatives of the function, and yes, the gradient is just a vector with all the partial derivatives at the given point.
Gradient is usually written with the symbol β, so, in mathy mathy:
$$\nabla f(x, y) = \left[\begin{matrix}\frac{\partial f}{\partial x}(x, y)\cr\frac{\partial f}{\partial y}(x, y)\end{matrix}\right]$$
But how can this gradient be useful to us? It points in the direction the function grows. So if we take:
$$-\nabla f(x, y)$$
It should point us to the direction the function gets smaller, or in other words, to the direction zero is, which is what we want to find. There’s the caveat of negative numbers however, they get smaller as they get away from zero. To get around that we can use:
Given a point (x, y), the direction in which f goes to zero from that point is: $$-\nabla |f(x, y)|$$ or $$-\nabla f^2(x, y)$$ Provided that there are no local minima.
How this actually looks is as follows. The 3D curve for a circle is like this, intersecting the XY plane and forming the circle:
The regular negative gradient operation would point all the way to the bottom of the function, below zero.
But if we take the absolute value:
Now the negative gradient of every point in the surface points into the direction of the zero of the function! (except the zero itself π)
That is great! We just need to start at any pixel in the screen, go into that direction until we intersect the function and we have a starting pixel for the tracing algorithm.
But how do we actually compute a approximation of the gradient? π€
If we sample the differences in the function between two random points, we know that, in the direction given by those two points, the function grows or shrinks by that amount. In other words:
Given two near points pi1, pi2, and the function f. The function f changes by: $$\Delta_i = f(p_{i2})-f(p_{i1})$$ In the direction: $$\vec{\Delta_i} = p_{i2} - p_{i1}$$ On the vicinity of pi1 and pi2.
This will not, however, give us the full gradient, because not all basis vectors are represented equally when we take two points. So, we must take more samples, and the points must be chosen to lie around a circle at regular intervals (so each basis has equal representation). Then we can do:
$$\nabla f(x, y) \propto \sum_{i}{\Delta_i\vec{\Delta_i}}$$
The simplest figure for which this can work is the regular polygon with the fewest edges: the triangle!
So, in another words, we want to create a small triangle around the point we are calculating the gradient for and sum the vectors A, B, and C times their respective Ξs. The result should be proportional to the gradient, so we can normalize it to have a unit vector in the direction of the gradient:
Now we can use this to write a bit of Rust to find one point in an implicit curve, given any starting point on the plane:
fn find_pixel_on_curve(curve: Implicit, mut p: Point) -> Option<Point> {
let mut search_distance = 20f32;
let mut current_direction = abs_inverse_gradient(curve, p).unwrap();
let mut previous_direction;
for _ in 0..MAX_TRACE_SEARCH {
// Go into the direction of -β|f(p)|
p = p + current_direction * search_distance;
// We found a pixel on the curve! Early exit.
if pixel_intersects_curve(p, curve) {
return Some(Point::new(p.x.floor(), p.y.floor()));
}
previous_direction = current_direction;
// We crossed the zero of the function, turn back moving slower
if dot(previous_direction, current_direction) < 0.0 {
search_distance /= 2f32;
}
}
None
}
# Combining
We now have a tracing algorithm that is more stable in drawing implicit curves once we have found one point in the curve. Do keep in mind that it is not perfect, it still uses the intersection function, but usually in a scale that is much more linear. It can also fail to find the starting point of the curve if there are local minima and the algorithm to find the first pixel becomes “trapped” in one of those.
To avoid that, we fist start out using the subdivision-intersection algorithm with blind search, then we refine it as usual, and then instead of simply drawing the pixels at the positions found, we TRACE the function from those points! π€―
// Draw curve
if depth >= MAX_DEPTH {
let triangle_center = (tri[0] + tri[1] + tri[2]) / 3.0;
trace(
f,
triangle_center,
pixels,
);
return;
}
## Full Source Code
This article came to life from my explorations into plotting implicit functions with a little program I called Pluft.
There you can see the full working implementation of all the algorithms mentioned above and experiment with plotting different functions.
# Appendix A: Interesting Figures
## The Circle
$$z = f(x, y) = x^2+y^2-r^2$$
## The Curved Triangle
$$z = f(x, y) = \left(x^2 + y^2 + 12ax + 9a^2\right)^2 - 4a(2x+3a)^3$$
## The Flower
$$z = f(x, y) = \left(3x^2-y^2\right)^2 y^2 - \left(x^2 + y^2\right)^4$$
## The Knot
$$z = f(x, y) = \left(x^2 + y^2\right)\left(y^2+x\left(x+\frac{1}{2}\right)\right) - 4 \cdot \frac{1}{2} x y^2$$ | 3,795 | 15,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-18 | latest | en | 0.910335 |
https://everything2.com/user/Juuichiketajin/writeups/Slot+Machine | 1,529,521,689,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863834.46/warc/CC-MAIN-20180620182802-20180620202802-00603.warc.gz | 604,415,635 | 6,073 | Many old books will report that the odds of a particular combination of symbols coming up on a slot machine can be calculated from the layout of the symbols on the reels. In modern times, this is certainly not so. The combination of symbols that comes up is determined by a computer.
The old method might have been something like this: The jackpot symbol is three sevens. Each of the three reels has 20 symbols, including one seven per reel. So there are 20 * 20 * 20 = 8000 possible ways the reels can stop, one of which is the jackpot. Therefore, the probability of a jackpot is 1 in 8000.
Fallacy: The machine's internal computer decides what comes up. If the casino decides that the jackpot odds are 1 in 2,000,000, so be it. This is perfectly legal, I think. | 179 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-26 | longest | en | 0.939193 |
https://priyaslearningcentre.com/2020/12/04/ch-7-fractions/ | 1,638,451,777,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00475.warc.gz | 525,989,005 | 52,071 | # Ch – 7 Fractions
Q.1: What fraction of a day is 6 hours?
Q.2: What fraction of an hour is 30 minutes?
Q.3: What fraction of a kilogram is 600 grams?
Q.4: What fraction of a meter is 70 centimeter?
Q.5: Draw number lines and locate the points on them:
a) 3/9, 4/9, 1/9, 8/9
b) 6/11, 4/11, 2/11, 7/11
c) 5/7, 3/7, 6/7, 1/7
Q.6: Express the following as mixed fractions:
a) 20/6
b) 48/7
c) 29/6
d) 5/3
Q.7: Write in simplest forms:
a) 36/72
b) 50/60
c) 28/49
d) 63/81
Q.8: Check whether the given fractions are equivalent:
a) 5/11 and 6/12
b) 8/14 and 16/28
c) 7/9 and 21/27
d) 8/12 and 5/14
Q.9: Find the equivalent fraction of 4/11 with:
a) numerator 28
b) denominator 66 | 310 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.783122 |
https://www.hamilton-trust.org.uk/browse/maths/y1/spring/116908 | 1,555,784,505,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00368.warc.gz | 692,165,969 | 11,823 | Problem solving
# Problem-solving Investigations - Year 1
Problem-solving investigations provide a fun, stimulating context in which children can develop and exercise their ability to reason mathematically and think creatively. They provide extra skills practice and also provide a real challenge if the skill itself is proving undemanding for some children.
These problems are designed to help children identify patterns, explore lines of thinking and investigate properties of numbers, shapes and measures. They can be used alongside the Hamilton plan for the week or independently. The teacher instructions for the whole term are collated in the Overview. Ruth's Advice gives some background and tips for using these investigations with your pupils.
Supporting documents for set
• Week
• Title
1
+ Details
Calculating caterpillars
Children make add and subtract 1 and 10 in order to reach targets.
2
+ Details
Cross sums
Children arrange numbers 1 to 5 on a cross to show that the row and column have the same total.
3
+ Details
Coins in my pockets
Children use clues, and trial and improvement to find different combinations of coins with a total of 20p.
4
+ Details
My new watch
Children use analogue clocks to try out different o’clock and half past times. They find all the possible different times between midnight and midday!
5
+ Details
Odds and evens
Children use number shapes to find totals of odd and even numbers, to see if they can find any rules.
6
+ Details
Pairs in purses (1), Calculator capers (2)
Pairs in purses (1): Children use reasoning and logic to work out pairs of 2-digit amounts that can be made in the two purses given only ten coins. Calculator capers (2): Children create teens numbers using place value additions.
7
+ Details
Four towers
Children arrange four towers of cubes such that the difference between neighbouring towers is at least two.
8
+ Details
Fill the bucket
Children estimate then measure the capacity of a bucket, and play a game filling it.
9
+ Details
Pyramid sums
Children place three numbers in the bottom row of an addition pyramid and see that it is possible to end up with different numbers at the top.
10
+ Details
10 to 50 challenge
Children find which numbers from 5 to 7 they can add repeatedly to 10 to land exactly on 50.
11
+ Details
Mr Know It All!
Children find out which amounts of money from 10 to 20p can be made using 10p, 5p, 2p and 1p coins, using no more than one of each. | 551 | 2,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-18 | latest | en | 0.9075 |
https://mathconfidence.com/2017/03/18/get-the-math-and-the-points-cc-alg-i-regents-june-2016-6/ | 1,685,691,225,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00248.warc.gz | 426,184,407 | 25,973 | # Get the Math and the Points CC Alg I Regents June 2016 #6
What word is in the word linear?
An easy way to see if a function is linear is a constant rate of change!
Since all the x’s are the same x = 1,2,3,4 , we can focus on the y values!
Which table has y-values that are the same distance apart?
That’s it! 2 more points | 94 | 327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.892869 |
https://www.coursehero.com/file/6459182/Chap007WW-FIN357NT/ | 1,521,734,929,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00156.warc.gz | 759,217,182 | 406,274 | {[ promptMessage ]}
Bookmark it
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Chap007(WW-FIN357)NT
# Chap007(WW-FIN357)NT - Chapter7 ChapterOutline 7.0 7.1 7.2...
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Chapter 7 NPV and Other Investment Rules
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7-2 Chapter Outline 7.0 The Capital Budgeting Process 7.1 Net Present Value 7.2 The Payback Period 7.3 The Discounted Payback Period 7.4 The Average Accounting Return 7.5 The Internal Rate of Return 7.6 Problems with the IRR Approach 7.7 The Profitability Index 7.8 The Practice of Capital Budgeting 7.9 Spread Sheet Calculations
7-3 Key Concepts and Skills Compute payback and discounted payback and understand their shortcomings Understand accounting rates of return and their shortcomings Compute the IRR and profitability index, understanding the strengths and weaknesses of both approaches Compute the NPV and understand why it is the best decision criterion
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7-4 7.0 The Capital Budgeting Process 1. Estimate the project’s relevant cash flows 2. Determine the timing of the cash flows 3. Determine the appropriate discount rate 4. Determine the feasibility of the project using an appropriate evaluation technique
7-5 Project Evaluation Techniques Net present value (NPV) Payback period (PP) Discounted payback period (DPP) Average accounting return (AAR) Internal rate of return (IRR) Modified internal rate of return (MIRR) Profitability index (PI)
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7-6 Basic Capital Budgeting Data Expected Net Cash Flow Year Project L Project S 0 (\$100) (\$100) 1 10 70 2 60 50 3 80 20
7-7 7.1 Net Present Value Net Present Value (NPV) = Total PV of future CF’s + Initial Investment ( 29 ( 29 0 1 0 1 1 n n t t t t t t CF CF NPV CF r r = = = + = + + Estimating NPV: 1. Estimate future cash flows: Amount? Timing? 2. Estimate discount rate (Risk?) 3. Estimate initial costs
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7-8 Project L NPV 10 80 60 0 1 2 3 10% Project L: -100.00 9.09 49.59 60.11 18.79 = NPV L
7-9 Project S NPV 70 20 50 0 1 2 3 10% Project S: -100.00 63.64 41.32 15.03 19.98 = NPV S
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7-10 NPV Calculation Project L Project S CF2 CF1 CF3 CF0 60 10 10 -100 80 I/Y NPV \$18.79 -100 CF0 CF1 CF2 CF3 I/Y NPV 70 50 20 10 \$19.98
7-11 Net Present Value Acceptance criteria Relationship between NPV and discount rate Relationship between NPV and stock price Reinvestment assumption Advantages/Disadvantages of NPV?
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7-12 7.2 Payback Period How long does it take to recover the project’s initial investment? Payback Period = number of years to recover initial costs (CF 0 )
7-13 Payback for Project L Project L Cash Flows Year Annual Cumulative 0 (\$100) (\$100) 1 10 (90) 2 60 (30) 3 80 50 Payback L = 2 + 30/80 = 2.375 years
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7-14 Payback for Project S Project S Cash Flows Year Annual Cumulative 0 (\$100) (\$100) 1 70 (30) 2 50 20 3 20 40 Payback S = 1 + 30/50 = 1.6 years
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7-16 7.3 Discounted Payback Period How long does it take to recover the
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Ask a homework question - tutors are online | 1,163 | 4,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-13 | latest | en | 0.65652 |
http://mathematicssolution.com/the-square-of-any-odd-integer-is-of-the-form-8k-1/ | 1,501,209,038,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436321.71/warc/CC-MAIN-20170728022449-20170728042449-00428.warc.gz | 204,861,370 | 10,268 | BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Friday , July 28 2017
Home / Elementary Number Theory / The square of any odd integer is of the form 8k +1.
# The square of any odd integer is of the form 8k +1.
### 8q + 1.
Let, a = 8q + 1
⟹ a2 = (8q +1)2
⟹ a2 = 64q2 + 16q + 1
⟹ a2 = 8(8q2 + 2q) + 1
Let, k = 8q2 + 2q
So, a2 = 8q + 1
Therefore the square of any odd integer is of the form 8k +1.
## Application of Euclidian’s algorithm in Diophantine equation
Problem-1: Which of the following Diophantine equations cannot be solved – a) 6x + ... | 216 | 581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-30 | latest | en | 0.611476 |
https://www.math10.com/forum/viewtopic.php?f=10&t=5936 | 1,582,844,955,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00281.warc.gz | 729,814,282 | 5,368 | # Trigonometric identities
Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot
### Trigonometric identities
In the equation, it says that: cos2x=cos^2x-sin^2x=2cos^2x-1... Now, why is 2cos^2x-1 is used?
ken_165
Posts: 2
Joined: Sun Jun 17, 2018 7:52 pm
Reputation: 0
### Re: Trigonometric identities
$$\sin^2x+\cos^2x = 1$$
Therefore $$\sin^2x = 1- \cos^2x$$
Math Tutor | 151 | 399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-10 | latest | en | 0.742401 |
https://blog.csdn.net/weixin_40548136/article/details/137914547 | 1,716,200,085,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00582.warc.gz | 107,783,193 | 29,991 | # 算法：堆（优先队列）
#### python实现
class HEAPQ:
# 最小堆
def __init__(self, l):
self.l = l
self.build()
def build(self):
n = len(self.l)
end = n // 2
for i in range(end, -1, -1):
self.update(i)
def update(self, i):
l = self.l
n = len(l)
idx = i
if 2 * i + 1 < n and l[2 * i + 1] < l[idx]:
idx = 2 * i + 1
if 2 * i + 2 < n and l[2 * i + 2] < l[idx]:
idx = 2 * i + 2
if idx != i:
l[i], l[idx] = l[idx], l[i]
self.update(idx)
def push(self, x) -> int:
l = self.l
l.append(x)
i = len(l)-1
while i != 0:
parent = (i-1) // 2
# 父节点大于 则上浮
if l[parent] > l[i]:
l[parent], l[i] = l[i], l[parent]
i = parent
else:
break
def pop(self):
l = self.l
n = len(l)
l[0], l[n - 1] = l[n - 1], l[0]
t = l.pop()
self.update(0)
return t
def top(self):
return self.l[0]
215. 数组中的第K个最大元素
#### 面试题 17.09. 第 k 个数
class Solution:
def getKthMagicNumber(self, k: int) -> int:
l = [1]
ss = set()
tt = (3,5,7)
for i in range(1,k):
t = heapq.heappop(l)
for x in tt:
if t*x not in ss:
heapq.heappush(l, t*x)
return l[0]
#### 面试题 17.20. 连续中值
void addNum(int num) - 从数据流中添加一个整数到数据结构中。
double findMedian() - 返回目前所有元素的中位数。
class MedianFinder:
def __init__(self):
self.max_heapq = [] # 最大堆个数大于等于最小堆
self.min_heapq = []
def addMax(self, l, x):
l.append(x)
n = len(l)
start = n-1
while start > 0:
p = (start-1)//2
if l[p] < l[start]:
l[p], l[start] = l[start],l[p]
start = p
else:
break
def addMin(self, l, x):
l.append(x)
n = len(l)
start = n-1
while start > 0:
p = (start-1)//2
if l[p] > l[start]:
l[p], l[start] = l[start],l[p]
start = p
else:
break
def updateMax(self, l,i, n):
idx = i
if i*2+1<n and l[2*i+1] > l[i]:
idx = i*2+1
if i*2+2<n and l[2*i+2] > l[idx]:
idx = i*2+2
if idx!=i:
l[idx], l[i] = l[i], l[idx]
self.updateMax(l, idx, n)
def updateMin(self, l,i, n):
idx = i
if i*2+1<n and l[2*i+1] < l[i]:
idx = i*2+1
if i*2+2<n and l[2*i+2] < l[idx]:
idx = i*2+2
if idx!=i:
l[idx], l[i] = l[i], l[idx]
self.updateMin(l, idx, n)
def popMax(self, l):
n = len(l)
l[0], l[n-1] = l[n-1], l[0]
t = l.pop()
self.updateMax(l, 0, n-1)
return t
def popMin(self, l):
n = len(l)
l[0], l[n-1] = l[n-1], l[0]
t = l.pop()
self.updateMin(l, 0, n-1)
return t
def addNum(self, num: int) -> None:
if not self.max_heapq:
self.max_heapq.append(num)
return
if num <= self.max_heapq[0]:
else:
if len(self.min_heapq) > len(self.max_heapq):
elif len(self.max_heapq)-len(self.min_heapq) >1:
def findMedian(self) -> float:
if len(self.max_heapq) == len(self.min_heapq):
return (self.max_heapq[0] + self.min_heapq[0])/2
else:
return self.max_heapq[0]
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http://virtualplantlab.com/tutorials/snowflakes/index.html | 1,680,027,732,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00315.warc.gz | 54,115,325 | 10,845 | Author
Affiliation
Alejandro Morales Sierra
Centre for Crop Systems Analysis - Wageningen University
Published
March 22, 2023
In this example, we create a Koch snowflake, which is one of the earliest fractals to be described. The Koch snowflake is a closed curve composed on multiple of segments of different lengths. Starting with an equilateral triangle, each segment in the snowflake is replaced by four segments of smaller length arrange in a specific manner. Graphically, the first four iterations of the Koch snowflake construction process result in the following figures (the green segments are shown as guides but they are not part of the snowflake):
In order to implement the construction process of a Koch snowflake in VPL we need to understand how a 3D structure can be generated from a graph of nodes. VPL uses a procedural approach to generate of structure based on the concept of turtle graphics.
The idea behind this approach is to imagine a turtle located in space with a particular position and orientation. The turtle then starts consuming the different nodes in the graph (following its topological structure) and generates 3D structures as defined by the user for each type of node. The consumption of a node may also include instructions to move and/or rotate the turtle, which allows to alter the relative position of the different 3D structures described by a graph.
The construction process of the Koch snowflake in VPL could then be represented by the following axiom and rewriting rule:
axiom: E(L) + RU(120) + E(L) + RU(120) + E(L)
rule: E(L) → E(L/3) + RU(-60) + E(L/3) + RU(120) + E(L/3) + RU(-60) + E(L/3)
Where E represent and edge of a given length (given in parenthesis) and RU represents a rotation of the turtle around the upward axis, with angle of rotation given in parenthesis in hexadecimal degrees. The rule can be visualized as follows:
Note that VPL already provides several classes for common turtle movements and rotations, so our implementation of the Koch snowflake only needs to define a class to implement the edges of the snowflake. This can be achieved as follows:
using VPL
module sn
import VPL
struct E <: VPL.Node
length::Float64
end
end
import .sn
Note that nodes of type E need to keep track of the length as illustrated in the above. The axiom is straightforward:
const L = 1.0
axiom = sn.E(L) + VPL.RU(120.0) + sn.E(L) + VPL.RU(120.0) + sn.E(L)
VPL.Core.StaticGraph(Dict{Int64, Any}(5 => VPL.Core.GraphNode{Main.sn.E}(Main.sn.E(1.0), Set{Int64}(), 4, 5), 4 => VPL.Core.GraphNode{VPL.Geom.RU{Float64}}(VPL.Geom.RU{Float64}(120.0), Set([5]), 3, 4), 2 => VPL.Core.GraphNode{VPL.Geom.RU{Float64}}(VPL.Geom.RU{Float64}(120.0), Set([3]), 1, 2), 3 => VPL.Core.GraphNode{Main.sn.E}(Main.sn.E(1.0), Set([4]), 2, 3), 1 => VPL.Core.GraphNode{Main.sn.E}(Main.sn.E(1.0), Set([2]), missing, 1)), Dict{DataType, Set{Int64}}(VPL.Geom.RU{Float64} => Set([4, 2]), Main.sn.E => Set([5, 3, 1])), 1, 5)
The rule is also straightforward to implement as all the nodes of type E will be replaced in each iteration. However, we need to ensure that the length of the new edges is a calculated from the length of the edge being replaced. In order to extract the data stored in the node being replaced we can simply use the function data. In this case, the replacement function is defined and then added to the rule. This can make the code more readable but helps debugging and testing the replacement function.
function Kochsnowflake(x)
L = data(x).length
sn.E(L/3) + RU(-60.0) + sn.E(L/3) + RU(120.0) + sn.E(L/3) + RU(-60.0) + sn.E(L/3)
end
rule = Rule(sn.E, rhs = Kochsnowflake)
Rule replacing nodes of type Main.sn.E without context capturing.
The model is then created by constructing the graph
Koch = Graph(axiom = axiom, rules = rule)
Dynamic graph with 5 nodes of types VPL.Geom.RU{Float64},Main.sn.E and 1 rewriting rules.
In order to be able to generate a 3D structure we need to define a method for the function VPL.feed! (notice the need to prefix it with VPL. as we are going to define a method for this function). The method needs to two take two arguments, the first one is always an object of type Turtle and the second is an object of the type for which the method is defined (in this case, E).
The body of the method should generate the 3D structures using the geometry primitives provided by VPL and feed them to the turtle that is being passed to the method as first argument. In this case, we are going to represent the edges of the Koch snowflakes with cylinders, which can be generated with the HollowCylinder! function from VPL. Note that the feed! should return nothing, the turtle will be modified in place (hence the use of ! at the end of the function as customary in the VPL community).
In order to render the geometry, we need assign a color (i.e., any type of color support by the package ColorTypes.jl). In this case, we just feed a basic RGB color defined by the proportion of red, green and blue. To make the figures more appealing, we can assign random values to each channel of the color to generate random colors.
function VPL.feed!(turtle::Turtle, e::sn.E, vars)
HollowCylinder!(turtle, length = e.length, width = e.length/10,
height = e.length/10, move = true,
color = RGB(rand(), rand(), rand()))
return nothing
end
Note that the argument move = true indicates that the turtle should move forward as the cylinder is generated a distance equal to the length of the cylinder. Also, the feed! method has a third argument called vars. This gives acess to the shared variables stored within the graph (such that they can be accessed by any node). In this case, we are not using this argument.
After defining the method, we can now call the function render on the graph to generate a 3D interactive image of the Koch snowflake in the current state
render(Koch, axes = false)
This renders the initial triangle of the construction procedure of the Koch snowflake. Let’s execute the rules once to verify that we get the 2nd iteration (check the figure at the beginning of this document):
rewrite!(Koch)
render(Koch, axes = false)
And two more times
for i in 1:3
rewrite!(Koch)
end
render(Koch, axes = false)
# Other snowflake fractals
To demonstrate the power of this approach, let’s create an alternative snowflake. We will simply invert the rotations of the turtle in the rewriting rule
function Kochsnowflake2(x)
L = data(x).length
sn.E(L/3) + RU(60.0) + sn.E(L/3) + RU(-120.0) + sn.E(L/3) + RU(60.0) + sn.E(L/3)
end
rule2 = Rule(sn.E, rhs = Kochsnowflake2)
Koch2 = Graph(axiom = axiom, rules = rule2)
Dynamic graph with 5 nodes of types VPL.Geom.RU{Float64},Main.sn.E and 1 rewriting rules.
The axiom is the same, but now the edges added by the rule will generate the edges towards the inside of the initial triangle. Let’s execute the first three iterations and render the results
# First iteration
rewrite!(Koch2)
render(Koch2, axes = false)
# Second iteration
rewrite!(Koch2)
render(Koch2, axes = false)
# Third iteration
rewrite!(Koch2)
render(Koch2, axes = false)
This is know as Koch antisnowflake. We could also easily generate a Cesàro fractal by also changing the axiom:
axiomCesaro = sn.E(L) + RU(90.0) + sn.E(L) + RU(90.0) + sn.E(L) + RU(90.0) + sn.E(L)
Cesaro = Graph(axiom = axiomCesaro, rules = rule2)
render(Cesaro, axes = false)
And, as before, let’s go through the first three iterations
# First iteration
rewrite!(Cesaro)
render(Cesaro, axes = false)
# Second iteration
rewrite!(Cesaro)
render(Cesaro, axes = false)
# Third iteration
rewrite!(Cesaro)
render(Cesaro, axes = false) | 1,993 | 7,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-14 | longest | en | 0.924937 |
https://www.physicsforums.com/threads/strain-in-cantilever-beam.765538/ | 1,477,476,106,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720845.92/warc/CC-MAIN-20161020183840-00504-ip-10-171-6-4.ec2.internal.warc.gz | 966,848,610 | 17,692 | # Strain in cantilever beam
1. Aug 11, 2014
### nomority
Hi everyone. I feel there should be a simple answer to this but I can't seem to find anything on this.
So I have a simple cantilever beam, supported at one side and loaded at the free end. I have the force displacement data and can easily calculate the stress.
However, for the strain I do not want to use Hookes Law, but instead calculate the strain from the force displacement data. Any hints?
Thanks!
2. Aug 11, 2014
### Staff: Mentor
The strain varies through the cross section of the beam. It is equal to the distance from the neutral axis times the curvature. So if you know the curvature at any location, you know the strain variation through the thickness. The strain is positive on the outside of the bend, and negative on the inside of the bend. Of course, it also varies with distance along the beam.
Chet
3. Aug 11, 2014
### nomority
Apologies, I should have stated I am looking for the maximum strain.
I have got the full force-displacement data of a bend experiment. I know the geometry of the beam (triangular cross section) so can calculate the maximum stress at any time as (M*y)/I, which for the triangular cross-section is equal to (24*Force*Length)/(width*thickness^2).
In a code I have access to it states that the maximum strain at any time would be equal to (2*displacement*thickness)/(Length^2), but I can't figure out why this would be the case.
4. Aug 11, 2014
### Staff: Mentor
Where does the maximum curvature occur, and, in terms of the displacement , what is that curvature?
Chet
5. Aug 11, 2014
### nomority
Thanks for your reply. Maximum curvature is at the fixed end. However, I can't measure the curvature to any degree of accuracy, as such for the sake of this problem it is not available.
6. Aug 11, 2014
### Staff: Mentor
What I meant was, analytically , what is the curvature in terms of the displacement ?
7. Aug 11, 2014
### nomority
I'm not sure how I could describe the curvature in terms of the parameters I have.
If I was to reverse engineer the equation for strain I have
, and assume that strain is equal to y/R (as found online), I would have an expression for the curvature of L^2/(6*displacement). This doesn't seem correct to me though.
8. Aug 11, 2014
### AlephZero
If you know the maximum stress, you can get the maximum strain using Young's modulus and Hooke's law.
The stresses and strains in the beam are statically determinate. They only depend on the applied loads, not on the displacement of the beam.
9. Aug 11, 2014
### Staff: Mentor
Step 1: Express the bending moment M at the built-in end in terms of the load F.
Step 2: What is your equation for the displacement in terms of the load F.
Step 3: Combine these relationships to get the bending moment in terms of the displacement.
Step 4: Determine the curvature from the bending moment
Step 5: Determine the curvature as a function of the displacement
Step 6: Determine the strain from the curvature
chet | 739 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-44 | longest | en | 0.93888 |
https://slideplayer.com/slide/5384619/ | 1,611,307,514,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00067.warc.gz | 563,423,628 | 19,936 | # Signals and Systems Discrete Time Fourier Series.
## Presentation on theme: "Signals and Systems Discrete Time Fourier Series."— Presentation transcript:
Signals and Systems Discrete Time Fourier Series
Discrete-Time Fourier Series
The conventional (continuous-time) FS represent a periodic signal using an infinite number of complex exponentials, whereas the DFS represent such a signal using a finite number of complex exponentials
Example 1 DFS of a periodic impulse train Since the period of the signal is N We can represent the signal with the DFS coefficients as
Example 2 DFS of an periodic rectangular pulse train The DFS coefficients
Properties of DFS Linearity Shift of a Sequence Duality
Symmetry Properties
Symmetry Properties Cont’d
Periodic Convolution Take two periodic sequences Let’s form the product The periodic sequence with given DFS can be written as Periodic convolution is commutative
Periodic Convolution Cont’d Substitute periodic convolution into the DFS equation Interchange summations The inner sum is the DFS of shifted sequence Substituting
Graphical Periodic Convolution
DTFT to DFT
Sampling the Fourier Transform Consider an aperiodic sequence with a Fourier transform Assume that a sequence is obtained by sampling the DTFT Since the DTFT is periodic resulting sequence is also periodic We can also write it in terms of the z-transform The sampling points are shown in figure could be the DFS of a sequence Write the corresponding sequence
DFT Analysis and Synthesis
DFT
DFT is Periodic with period N
Example 1
Example 1 (cont.) N=5
Example 1 (cont.) N>M
Example 1 (cont.) N=10
DFT: Matrix Form
DFT from DFS
Properties of DFT Linearity Duality Circular Shift of a Sequence
Symmetry Properties
DFT Properties
Example: Circular Shift
Duality
Circular Flip
Properties: Circular Convolution
Example: Circular Convolution
illustration of the circular convolution process Example (continued)
Illustration of circular convolution for N = 8:
Example:
Example (continued)
Proof of circular convolution property:
Multiplication: | 482 | 2,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-04 | latest | en | 0.805518 |
https://www.jiskha.com/display.cgi?id=1347971286 | 1,529,706,550,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864822.44/warc/CC-MAIN-20180622220911-20180623000911-00633.warc.gz | 857,313,138 | 4,078 | # Math
posted by John
A box with an open top is to be constructed by cutting a-inch squares from the corners of a rectangular sheet of tin whose length is twice its width. What size sheet will produce a box having a volume of 420 in^3 when a = 3?
1. Reiny
You are cutting out 3-inch squares at the four corners.
Make a diagram.
width of base ---- x inches
length of base --- 2x inches, (you said so)
height of box ---- 3 inches
(x-6)(2x-6)(3) = 420
(x-6)(2x-6) = 140
x^2 - 18x + 36 = 140
2x^2 - 18x - 104 = 0
x = (18 ± √1156)/4
= (18 ± 34)/4
= 13 inches or a negative
so the sheet must have been 13 inches by 26 inches
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More Similar Questions | 781 | 3,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-26 | latest | en | 0.912467 |
http://www.coderanch.com/t/241373/java-programmer-OCPJP/certification/Mock-Exam | 1,464,368,034,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00159-ip-10-185-217-139.ec2.internal.warc.gz | 426,368,219 | 9,783 | Win a copy of Design for the Mind this week in the Design forum!
# Mock Exam Question
Jacob Michaels
Ranch Hand
Posts: 35
I need help understanding this code.
I know that the highest number a byte can be is 127, but I don't see the connection:
(byte)128 = -128
(byte)255 = -1
(byte)256 = 0
Thank you for any responses
class E2 {
static byte a = (byte)127;
static byte b = (byte)128;
static byte c = (byte)255;
static byte d = (byte)256;
public static void main(String args[]) {
System.out.print(a + " " + b + " " + c + " " + d);
}
}
/*
127 -128 -1 0
*/
Dan Chisholm
Ranch Hand
Posts: 1865
This campfire story might be helpful.
Jacob Michaels
Ranch Hand
Posts: 35
Thank you for the response and posting great mock exams. I am learning a lot about Java from your site.
I have read that campfire story and I enjoyed it very much. I wish there was for 3D arrays.
I am assuming that the probelm that I posted above:
when the amount exceeds 2^7-1 and it automatically is forced to -1.
Thanks again
Stan Forest
Greenhorn
Posts: 12
A byte occupies eight bits
+127 is represented as: 0111 1111
The first digit is a sign, 0 is positive in this case. If the left first digit was a 1, the value would be negative.
128 in 32 bit binary is 0000 0000 0000 0000 0000 0000 1000 0000
when casting to a byte, the left 24 bits are discarded. The remaining is
1000 0000
This is now a byte. Because the first digit is a 1 the value is negative. Convert using two's complement. Twos complement requires inverting and adding 1.
start 1000 0000
invert 0111 1111
result 1000 0000 = 128 and it was negative = -128
256 is in binary is :
0000 0000 0000 0000 0000 0001 0000 0000
cast to a byte discards the left 24 bits leaving
0000 0000 which is 0
255 is one less than 256 so it is �1
OR
0000 0000 0000 0000 0000 0000 1111 1111
cast to a byte leaves
1111 1111
The first digit is 1 so it is negative so it needs to be converted using two's complement
start 1111 1111
invert 0000 0000
result 0000 0001 and remember it is negative so it is -1
Jacob Michaels
Ranch Hand
Posts: 35
Thank you!
That helped!
Jim Bedenbaugh
Ranch Hand
Posts: 171
Originally posted by Jacob Michaels:
I wish there was for 3D arrays.
I'm not sure I understand this statement. Given Java's Collections and Maps, it's entirely possible to store an array of Map objects in another Map object and retrieve by key.
Rattan Mann
Ranch Hand
Posts: 44
On Dan's recommendation I too read the campfire stories. They are beautiful stories.But who is writing those beautiful,imaginative stories? I didn't find the name of the authors or am I so stupid that I missed the names?
Rattan
Dan Chisholm
Ranch Hand
Posts: 1865
The credits for the campfire story are provided at the bottom of the page. | 812 | 2,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-22 | latest | en | 0.819607 |
http://www.mypivots.com/board/topic/3348/2/square-of-nines | 1,501,239,497,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448198.69/warc/CC-MAIN-20170728103510-20170728123510-00259.warc.gz | 515,689,611 | 5,523 | No registration required! (Why?)
# Square of nine's
871
886
901
916
931
946
961
977*
993
1009*
1025
1041
1057
1089
1106
1123*
1140
1157
1174
1191
1208
1225
1243
1261
1279*
1297*
1315
1333
1351
1369
1388
1407
1426
1445
*=these carry more weight since they are also primes.
good eye prestwickdrive,
thats a typo its supposed to be 664. DT or Bruce could you correct that typo for me?
quote:
Originally posted by CharterJoe
good eye prestwickdrive,
thats a typo its supposed to be 664. DT or Bruce could you correct that typo for me?
not 654? TY again.
No not 654....the sq root of 677 -.25 and x2 = 664...sq root of 664 -.25 x2 = 651.
quote:
Originally posted by CharterJoe
No not 654....the sq root of 677 -.25 and x2 = 664...sq root of 664 -.25 x2 = 651.
got it - I was working off of the 669 number, not 664. ty
HOD 771.50 / so9 771.00 / gotta love the so9
quote:
Originally posted by redsixspeed
HOD 771.50 / so9 771.00 / gotta love the so9
;) and the high on the nasdaq emini was 1172.75 just shy of 1174 Sq of 9
guys - i have an excel spreadsheet that can do square #'s and denotes if it's also a prime number that i will be happy to share
give me a heyyou at my addy if you want it and i'l give it to you
(of course, i want half your profits, just kidding
)
sq #'s that are prime are like voodoo !!
ciao
I would like to see that excel... so I could help others. I have them memorized you would be surprised to know how much that helps.
Thank you Joe, for the numbers & post.
I would like the excel also
I haven't traded bonds yet, I started using Sierra charts Aug 1st & it has a sq of 9 study, I remembered this thread & put the sq of 9 study on GC & ZB just to see, the #s are like magnets! The settings that seem to work best are 22.5 degrees & 45 degrees in that order. | 561 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-30 | latest | en | 0.937237 |
https://mathspace.co/textbooks/syllabuses/Syllabus-405/topics/Topic-7184/subtopics/Subtopic-95976/?activeTab=theory | 1,642,510,479,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00355.warc.gz | 453,056,399 | 46,787 | # Certain, Impossible and In-Between
Lesson
In What could happen? we looked at how we can keep track of all the different outcomes in an experiment. Now we are going to look at some special words that we use to describe the likelihood or chance that an event or result will occur.
Remember!
These are $5$5 key terms used to describe chance:
• Impossible: definitely will NOT happen.
• Unlikely: more likely NOT to happen than to happen.
• Even chance: equally likely to happen as not to happen.
• Likely: more likely to happen than not to happen.
• Certain: definitely will happen.
#### Worked examples
##### Question 1
What is the chance that next week will have $8$8 days?
1. Impossible
A
Likely
B
Unlikely
C
Certain
D
Even chance
E
Impossible
A
Likely
B
Unlikely
C
Certain
D
Even chance
E
##### Question 2
A game in a classroom uses this spinner.
1. What is the chance of spinning an odd number?
certain
A
even chance
B
impossible
C
likely
D
certain
A
even chance
B
impossible
C
likely
D
2. What is the chance of spinning a $2$2?
likely
A
impossible
B
certain
C
even chance
D
likely
A
impossible
B
certain
C
even chance
D
3. What is the chance of spinning a number less than $8$8?
likely
A
impossible
B
even chance
C
certain
D
likely
A
impossible
B
even chance
C
certain
D
##### Question 3
Order the following events from least likely to most likely with the numbers $1$1 (least likely) to $5$5 (most likely).
1. $\editable{}$ Winning the lottery
$\editable{}$ A house that is randomly chosen has an odd street number
$\editable{}$ Rolling a $7$7 on a standard die
$\editable{}$ The sun will rise tomorrow
$\editable{}$ Rolling more than $1$1 on a standard die
### Outcomes
#### S2-3
Investigate simple situations that involve elements of chance, recognising equal and different likelihoods and acknowledging uncertainty | 508 | 1,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-05 | longest | en | 0.88176 |
https://www.enotes.com/topics/math/questions/write-down-three-quadratic-equations-with-376489 | 1,708,705,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00386.warc.gz | 759,402,827 | 16,705 | # write down three quadratic equations with the following root and prove the roots are in fact the values of x for which y is equal to zerowith the following root and prove the roots are in fact the values of x for which y is equal to zero Equation equals two roots both equal to zero
If you want to find a quadratic equation with the roots r and s, just plug r and s into:
y=(x-r)(x-s)
and distribute.
or more generally, pick any number a (other than 0) and plug r and s into:
y=a(x-r)(x-s)
So, for example, if I wanted a quadratic with the roots 1 and -2, one possibility is:
y=(x-1)(x+2) = x^2 +x - 2
another is:
y=2(x-1)(x +2) = 2x^2 + 2x - 4
So, in your case, we want the roots to be 0 and 0. We want three different equations, so we are going to use three different possibilites for a. The easiest is just a = 1,2,3.
So:
y=(x-0)(x-0) = x^2
y=2(x-0)(x-0) = 2x^2
y=3(x-0)(x-0) = 3x^2
Now, we want to check that these really do work.
So, plug x=0 into each of the following:
`y=x^2`
`y=2x^2`
`y=3x^2`
`y=(0)^2 = 0`
`y=2(0)^2=0`
`y=3(0)^2 = 0`
Sure enough, when we plug in x = 0, we get y=0
So 0 really is the root (double root in fact) of these quadratics | 414 | 1,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-10 | latest | en | 0.889359 |
https://limbioliong.wordpress.com/2013/01/06/using-boost-lambda-expression-performing-iterations-on-vector-elements/ | 1,498,720,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323889.3/warc/CC-MAIN-20170629070237-20170629090237-00003.warc.gz | 801,784,799 | 34,714 | //
# Using Boost Lambda Expression – Performing Iterations on Vector Elements.
1. Introduction.
1.1 When performing iterations and transformations on STL containers, the Boost Lambda Expressions can be a useful alternative to using regular function class objects.
1.2 For those not familiar with the Boost Lambda Library (BLL), please refer to their library web site for full documentation.
1.3 From the next section onwards, 2 code samples will be listed. Both samples will demonstrate action being taken on the items of a vector via iteration. The difference between them will then be briefly discussed.
2. Vector Iteration : Using Typical Functors.
2.1 Consider the following set of routines :
```struct add_two_numbers : std::binary_function<int, int, int>
{
int operator() (const int& lhs, const int& rhs) const
{
return lhs + rhs;
}
};
struct display_number : std::unary_function<int, void>
{
void operator() (const int& num)
{
std::cout << num << std::endl;
}
};
void IterateVectorUsingFunctors()
{
// Define a integer vector.
std::vector<int> vecInt;
// Insert values into this vector.
for (int i = 0; i < 10; i++)
{
vecInt.push_back(i);
}
// Increment every value in the vector
// by 20.
std::transform
(
vecInt.begin(),
vecInt.end(),
vecInt.begin(),
);
// Display each value in the vector.
std::for_each
(
vecInt.begin(),
vecInt.end(),
display_number()
);
}```
The following is a summary of the set of codes above :
• add_two_numbers is a functor class based on the std::binary_function class.
• It takes 2 numbers as parameter and returns their sum.
• display_number is std::unary_function based functor class.
• It displays an input number on the console.
• IterateVectorUsingFunctors() is the main function that defines an integer vector and fills it with initial values.
• It then increments the values in the vector by 20 using std::transform() and the add_two_numbers functor.
• It finally displays all the current values inside the vector by using std::for_each() and the display_number functor.
2.2 The routines are simple and easily comprehended by C++ developers familiar with STL vectors and algorithms.
2.3 When IterateVectorUsingFunctors() is run, the console output will be as follows :
```20
21
22
23
24
25
26
27
28
29```
3. Vector Iteration : Using Lambda Expressions.
3.1 Next have a look at the following code :
```#include "stdafx.h"
#include <boost/lambda/lambda.hpp>
#include <boost/ref.hpp>
#include <iostream>
#include <vector>
using namespace boost;
using namespace boost::lambda;
void IterateVectorUsingLambdaExpression()
{
// Define a integer vector.
std::vector<int> vecInt;
// Insert values into this vector.
for (int i = 0; i < 10; i++)
{
vecInt.push_back(i);
}
// Increment every value in the vector
// by 20.
std::for_each
(
vecInt.begin(),
vecInt.end(),
_1 += 20
);
// Display each value in the vector.
std::for_each
(
vecInt.begin(),
vecInt.end(),
std::cout << _1 << '\n'
);
}```
The code above uses the Boost Lambda Library (BLL). To use this library, you would of course need to first have the boost library installed on your system and have your Visual Studio C++ Project’s “Additional Include Directories” setting point to your boost library folder.
The following is a summary of the code above :
• The intention of the IterateVectorUsingLambdaExpression() function is the same as that of IterateVectorUsingFunctors() which we saw in point 2.1.
• However, this time, the act of incrementing every value in the vector by 20 is accomplished using a Lambda expression :
`_1 += 20`
• This lambda expression essentially defines a small unnamed function object in which “_1” is the first parameter.
• As each iteration of std::for_each() is performed, this unnamed function is called with each vector element being the actual argument passed by reference as per boost documentation :
When an actual argument is supplied for a placeholder, the parameter passing mode is always by reference.
• Hence each vector item gets to be incremented by a value of 20.
• After that, another std::for_each() loop is performed. This time, the Lambda expression displays each vector item on the console.
3.2 When IterateVectorUsingLambdaExpression() is run the following will be the console output :
```20
21
22
23
24
25
26
27
28
29```
which is exactly the same as the output from IterateVectorUsingFunctors().
3.3 A special thing about Lambda expressions is the direct use of objects and constructs within scope, e.g. :
```void IterateVectorUsingLambdaExpression2(int iIncrement)
{
// Define a integer vector.
std::vector<int> vecInt;
// Insert values into this vector.
for (int i = 0; i < 10; i++)
{
vecInt.push_back(i);
}
// Increment every value in the vector
// by the current value of iIncrement.
std::for_each
(
vecInt.begin(),
vecInt.end(),
_1 += iIncrement
);
// Display each value in the vector with a heading message.
char* lpszMessage = "Vector element : ";
std::for_each
(
vecInt.begin(),
vecInt.end(),
std::cout << constant(lpszMessage) << _1 << '\n'
);
}```
The following is a summary of the code above :
• In the first std::for_each() loop, each vector item is incremented by whatever current value “iIncrement” holds. It need not be a constant value.
• In the second std::for_each() loop, as each vector item is displayed, a heading message is displayed. This message is the string stored in “lpszMessage” which is treated as a constant in the Lambda expression.
3.4 A use of the IterateVectorUsingLambdaExpression2() function like the following :
`IterateVectorUsingLambdaExpression2(3);`
will produce the following console output :
```Vector element : 3
Vector element : 4
Vector element : 5
Vector element : 6
Vector element : 7
Vector element : 8
Vector element : 9
Vector element : 10
Vector element : 11
Vector element : 12```
4. In Conclusion.
4.1 When we use functors to perform actions on a STL container, the function class has to be defined somewhere of course.
4.2 When we use functors, we sometimes have to modify their usage by using adapters like bind1st(). This can add to complexity.
4.3 Compared with the use of function objects, the example use of Lambda expressions as presented in this article certainly looked much simpler.
4.4 Indeed, the use of Lambda expressions can at times produce simpler code viz those required by function class objects albeit this is not always guaranteed.
4.5 However, function classes are by nature re-usable and can be made generic by declaring them template-based. This is their main advantage over the Lambda expressions. | 1,544 | 6,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-26 | longest | en | 0.620744 |
http://www.evi.com/q/4.6_liter_equals_how_many_cubic_inches | 1,419,082,972,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769888.14/warc/CC-MAIN-20141217075249-00070-ip-10-231-17-201.ec2.internal.warc.gz | 493,191,917 | 5,508 | # 4.6 liter equals how many cubic inches
• 4.6 liters is 280.71 cubic inches.
• tk10publ tk10canl
## Say hello to Evi
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• 4.6 liters converted to cubic inch | 674 | 2,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-52 | latest | en | 0.790868 |
https://www.prycewarner.com/azhagu-nilave-czqhww/how-much-is-a-silver-quarter-worth-4b60cd | 1,632,313,967,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00356.warc.gz | 943,069,321 | 10,062 | Mint Sets. The coin was born in 1931 when the Treasury Department wanted to honor George Washinton. Today, the silver quarter (from 1965) is worth above \$4,000, while the 1965 silver coin was sold at an auction for \$7,050. The lowest value silver coins are in “good condition,” while the highest value quarters are in “uncirculated condition.” By inspecting the design and the details, one can determine the “uncirculated condition.” The most apparent feature is a dull surface. In this case, the design and details of the coin will be worn to the point of blending. The designs for all coins were created by the chief engraver at the time, Charles Barber. Use the Silver Melt Value calculator to see the value of silver in this coin. Calculate 90% silver value : (27.24 × .0321507466 × 6.25 × .90) = \$4.9262981477 \$4.9262 is the rounded silver value for the 1932-1964 silver quarter on January 07, 2021. 2020 Portends: Silver Gold Bull Market Mania Coming? Find By. That is to say, \$1.00 is a specific weight and purity of the silver, and a half dollar (\$.50) is half of that weight (with the same purity). The fact that they are “weights and measures” is essential. Each dollar of face value contains .715 ounces of actual silver content, so a full junk silver bag (\$1,000.00 face value) contains 715 ounces of silver. QUESTION “How much is a 1964 silver quarter worth?” ANSWER We really love 1964 Washington Quarters. If you keep reading, you will find all about the different types of quarters and their actual worth. - Duration: 3:28. There are 5 quarters in US History that contain 90% silver. \$50.00 Free Shipping. Washington Quarters . The “Barber”, or the Liberty Head quarter, was first introduced in 1892 along with the whole Barber series of coins. The silver quarters are generally lower priced than bullion coins, and are fractionally smaller making them more affordable. Bodgas writes, “And while it's no surprise that many 200-year-old coins are worth more than the standard value, some 1970 quarters could be worth far more than 25 cents. Must read: How Much Are Silver Quarters Worth? Get our best bullion deals in your inbox, Get the latest deals and news! | 534 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.942481 |
http://analyticscosm.com/category/statistics/page/2/ | 1,531,778,260,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00129.warc.gz | 21,013,554 | 7,617 | # Category Archives for "Statistics"
Aug 12
## Basic Statistics for Data Science
Basic rules of probability
Expected value
sample and population quantities
Signal and Noise
probability densities and mass functions
variability and distribution
Statistical meaning of overfitting
Tag: Statistics for Data Science
Statistics for analytics
In the previous posts we have discussed the 4 most crucial aspects of analytics- Starting with a business question(s), readying data, analyzing it (using visualization and models), and communicating the results. In practice these steps are not carried out in isolation but they often impact each other. For example while analyzing data we may find out that the data is inadequate or incomplete and hence you may need to revisit the data.
In short analytics is the process of iterative, methodical exploration of an organization’s data to answer business questions to facilitate data driven decision making.
Going by the above definition there are 3 prerequisites to the analytics process. Having the questions that need to be answered, having the data and having the tools to analyze it.
Statistics offers us techniques for survey sampling to collect the proper data and it helps us with design of experiments to analyze the data. In this post I’ll discuss some of the ideas and tools that statistics offers and that are widely used in analytics to answer business questions.
If you are new to the field of statistics you may get confused by the standard terminologies like random processes or stochastic variables. So, I’ll not go strictly by those terminologies here. I’ll rather try to link those terminologies to examples that we are familiar with.
Random variables and Probability theory
Let’s say in an imaginary village there are 100 families. And we know that each family may have between one and three kids under the age of ten years. In this example if we pick up any family in this village it will be an experiment or random trial. The number of kids in the family that we pick up is a random variable. The different possible values for “number of kids under 10” are 1,2 and 3 and they constitute the sample space. Any subset of the sample space is called an event. In this case we can define an event E as the family selected has less than 3 kids. If the family that we pick up randomly has 1 or 2 kids in it we’d say the event E has occurred.
Remember every value in the sample space or every event has a probability associated with it. We’ll now see how to calculate the probability.
Let’s say we’ve discovered that in the village, 50 families have 2 kids under age of 10, 30 families have 1 kid and 20 families have 3 kids under 10. So, we can define the probabilities as:
P (kids under 10 = 2) = 50/100=0.5
P( kids under 10 = 1) = 30/100=0.3
P( kids under 10 = 3) = 20/100=0.2
The probability function follows 2 rules:
For all possible events probability is greater than or equal to zero but less than or equal to 1.
Sum of probability of all possible outcomes is 1.
In our case, if we denote probability of number of kids under age 10 =K as P(K), we have,
P(1) + P(2) + P(3) =1
In predictive analytics tutorials we will come across many such variables whose values are known for certain experiments (denoted by rows of the table) and unknown for other experiments. Our job is to find out the probability based on the given data and use that probability to predict the values of unknown experiments. We often use models to make these predictions but by understanding how these models calculate the probabilities we will be able to use them more effectively. | 786 | 3,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-30 | latest | en | 0.925591 |
https://ask.truemaths.com/question/verify-division-algorithm-for-the-polynomial-fx-8-20x-x%C2%B2-6x%C2%B3-by-gx-2-5x-3x%C2%B2/ | 1,675,267,993,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00316.warc.gz | 125,902,597 | 29,160 | • 0
Guru
# Verify division algorithm for the polynomial f(x)= (8 + 20x + x² – 6x³) by g(x) =( 2 + 5x – 3x²).
• 0
An important and exam oriented question from polynomials in which we have given that f(x)= (8 + 20x + x² – 6x³) and g(x) =( 2 + 5x – 3x²) and we have to verify the division algorithm for the given polynomial
Kindly give me a detailed solution of this question
RS Aggarwal, Class 10, chapter 2B, question no 11
Share | 148 | 434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.87213 |
https://mail.haskell.org/pipermail/haskell-cafe/2010-December/087573.html | 1,702,171,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00296.warc.gz | 431,217,084 | 2,409 | [Haskell-cafe] Matlab Style Logic Operations ala V1.*(V2>0) on Vectors and Matrices with HMatrix ??
Henning Thielemann lemming at henning-thielemann.de
Thu Dec 23 00:18:47 CET 2010
```On Wed, 22 Dec 2010, gutti wrote:
> question 1. u see the two commented lines I tried to get ur original line
> running, but didn't know how to specify f
What 'f' ? Do you mean
matrixfunction f x y = liftMatrix2 (zipVectorWith f) x y
?
> ## Code ########
>
> import Numeric.LinearAlgebra
> import Graphics.Plot
>
> matrix1 = fromLists [[0 .. 5],[30 .. 35],[50 .. 55]]
> matrix2 = fromLists [[-1,2],[-3,4],[5,-6]]
>
> -- matrix1 = buildMatrix 3 4 ( (r,c) -> fromIntegral r * fromIntegral c)
> (3><4)
> -- posPart v = mapVector (\a -> if a>=0 then a else 0) v
>
> -- function2map a1 a2 = (\a1 a2 -> if a1>=0 then a2/a1 else a1/a2)
> matrixfunction x y = liftMatrix2 (zipVectorWith(\a1 a2 -> if a2>=0 then a1 else 0)) x y
> matrix3 = matrixfunction matrix1 matrix2
>
> disp = putStr . disps 2
>
> main = do
>
> disp matrix1
> disp matrix2
> -- disp matrix3
> mesh matrix1
>
>
> #########
>
>
> question 2: - the compiler comes up with some weired data type problem --
> ghci has no problem this line :
>
> matrixTest_Fail.hs:5:10:
> Ambiguous type variable `t' in the constraints:
> `Element t'
> arising from a use of `fromLists' at matrixTest_Fail.hs:5:10-38
> `Num t' arising from the literal `1' at matrixTest_Fail.hs:5:22
> Possible cause: the monomorphism restriction applied to the following:
> matrix2 :: Matrix t (bound at matrixTest_Fail.hs:5:0)
> Probable fix: give these definition(s) an explicit type signature
> or use -XNoMonomorphismRestriction
>
> ## Code #####
>
> import Numeric.LinearAlgebra
> import Graphics.Plot
>
> matrix1 = fromLists [[1,2],[3,4],[5,6]]
> matrix2 = fromLists [[1,2],[3,4],[5,6]]
Before type inference can work, you need to fix the type of at least one
number of a set of numbers with known equal type. E.g.
> matrix1 = fromLists [[1,2],[3,4],[5,6::Double]]
or even better, add a type signature:
matrix1 :: Matrix Double
``` | 684 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-50 | latest | en | 0.576853 |
http://projectsforpreschoolers.com/books/senior-middle-school-high-school-monograph-series-separately-high-school-mathematics-trigonometric | 1,477,643,133,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721595.48/warc/CC-MAIN-20161020183841-00205-ip-10-171-6-4.ec2.internal.warc.gz | 204,092,186 | 12,032 | # Senior Middle School High School monograph series:
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Treatise on the Art of Measuring; Containing All That is Useful in Bonnycastle Hutton Hawney Ingram and Several Others Modern Works on Mensuration; to Which Are Added Trigonometry with Its Application to Heights and Distances Surveying; Gauging | 882 | 3,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-44 | latest | en | 0.877016 |
https://wiki-helper.com/four-times-the-area-of-curved-surface-of-a-cylinder-is-equal-to-6-times-the-sum-of-the-area-of-i-37421969-62/ | 1,713,982,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00071.warc.gz | 555,818,259 | 30,605 | # four times the area of curved surface of a cylinder is equal to 6 times the sum of the area of its two bases if its size is 12 fin
four times the area of curved surface of a cylinder is equal to 6 times the sum of the area of its two bases if its size is 12 find its curved surface area.
1. ### Solution:–
• Curved Surface Area cylinder = 2πrh
• Area of the base = πr²
Sum area of the base= πr²+πr²= 2πr²sq.cm
### A.T.Q
• => 4 (2πrh) = 6 × 2πr²
• => 8πrh = 12πr²
• => 2h = 3r
• => 2 × 12 = 3r
### r = 8cm.
Curved Surface Area cylinder = 2πrh
• => 2 × 22/7 × 8 × 12 | 215 | 576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.90698 |
http://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=365&nr_roman=CCCLXV&lang=en | 1,555,870,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00143.warc.gz | 266,279,268 | 9,475 | # Convert number: 365 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
365 = CCCLXV Apr 21 18:19 UTC (GMT) 968 = CMLXVIII Apr 21 18:19 UTC (GMT) 11,953 = (X)MCMLIII Apr 21 18:19 UTC (GMT) 692 = DCXCII Apr 21 18:19 UTC (GMT) 3,037 = MMMXXXVII Apr 21 18:19 UTC (GMT) 490 = CDXC Apr 21 18:19 UTC (GMT) 23,451 = (X)(X)MMMCDLI Apr 21 18:19 UTC (GMT) 284 = CCLXXXIV Apr 21 18:19 UTC (GMT) 1,600 = MDC Apr 21 18:18 UTC (GMT) 12,600 = (X)MMDC Apr 21 18:18 UTC (GMT) 12,600 = (X)MMDC Apr 21 18:18 UTC (GMT) 12,600 = (X)MMDC Apr 21 18:18 UTC (GMT) 12,600 = (X)MMDC Apr 21 18:18 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | 635 | 2,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-18 | latest | en | 0.910041 |
https://www.turito.com/ask-a-doubt/Mathematics-the-quadrilateral-is-rotated-about-o-what-is-the-value-of-y-5-4-3-2-qb467b9 | 1,716,025,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00494.warc.gz | 946,527,283 | 25,777 | Mathematics
Easy
Question
# The quadrilateral is rotated about O. What is the value of y?
Hint:
## The correct answer is: 3
### Rotation is isometry.Step 1 :Comparing corresponding parts2y = 6y = 33, is the value of y.5x = 3y + 1 (substitute y = 3)5x = 3(3) + 15x = 10x = 2
corresponding parts of quadrilaterals, remain same after rotation. | 115 | 346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.851481 |
https://math.stackexchange.com/questions/57859/generalizing-lagrange-multipliers-to-use-the-subdifferential | 1,716,411,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00686.warc.gz | 322,829,220 | 39,252 | # Generalizing Lagrange multipliers to use the subdifferential
Background: This is a followup to the question Lagrange multipliers with non-smooth constraints. Lagrange multipliers can be used for constrained optimization problems of the form
$$\min_{\vec x} f(\vec x) \text{ such that } g(\vec x) = 0$$
Briefly, the method works by constructing the Lagrangian, $$L(\vec x, \lambda) = f(\vec x) + \lambda g(\vec x)$$, then finding points where $$\forall i, \frac{\partial L}{\partial x_i} L = 0$$ and $$\frac{\partial L}{\partial \lambda} L = 0$$.
As was kindly pointed out in this answer, the method fails when $$g$$ is non-differentiable (but continuous), because the partial derivatives may not exist at points of optimality. For example, in the problem, minimize $$x_1$$ subject to $$g(x_1,x_2) = x_1 - |x_2| = 0$$. The minimum is at $$(0,0)$$, where $$\frac{\partial g}{\partial x_2}$$ does not exist.
Question: It seems that there should be a natural generalization of the method that uses subgradients and the subdifferential. Does the following work? Is there a reference that describes this in more detail?
Proposal: construct the Lagrangian as usual, but instead of seeking a point where all partial derivatives are 0, seek a point where 0 is in each partial subdifferential. So in the example above, the subdifferential with respect to $$x_2$$ when $$x_2=0$$ is the interval $$[-\lambda, \lambda]$$. Thus, if we were given the solution $$x_1=0,x_2=0,\lambda=-1$$, we could verify it is a critical point by noting that $$\frac{\partial L}{\partial \lambda} = x_1 - |x_2| = 0$$, $$\frac{\partial L}{\partial x_1} = 1 + \lambda = 0$$, and 0 is in the subdifferential of $$L$$ w.r.t. $$x_2$$.
Is this argument correct? My intuitive justification is that for any value $$f'(x)$$ in some variable's subdifferential at $$x$$, we should be able to construct a smooth function that has $$f'(x)$$ as its partial derivative at $$x$$, then solve the smoothed problem with a standard application of Lagrange multipliers.
(Aside: my goal is actually not to find a method to optimize the function. I have a method for optimizing such functions, and I'm trying to develop some theoretical understanding of the solutions that it produces. In particular, I'm trying to understand if proving that a solution satisfies the condition described in the Proposal section is meaningful.)
I think the difficulty with the subdifferential approach you describe lies in finding critical points. For instance, with your example, solving the equations you get by setting partials to $0$ yields only $x_1 = | x_2 |$ and $\lambda = -1$. This set of equations is too undetermined to yield a solution. As you point out, if you were given a solution $x_1 = x_2 = 0$, $\lambda = -1$, you could verify that it is a critical point with the subdifferential approach, but that's quite different from deriving $x_1 = x_2 = 0$, $\lambda = -1$ as a critical point.
Remember, though, that the definition of a critical point of a function includes points at which the derivative fails to exist. So when you're constructing your set of critical points using Lagrange multipliers, look not only for points at which the partials are zero but also for points at which a partial does not exist. For instance, it's clear in your example that the only point at which $\frac{\partial L}{\partial x_2}$ fails to exist is $x_2 = 0$. (You also get $\lambda = 0$ from $\frac{\partial L}{\partial x_2}= 0$, but this is inconsistent with the $\lambda = -1$ from $\frac{\partial L}{\partial x_1}= 0$.) Including that with your equations from $\frac{\partial L}{\partial x_1} = 0$ and $\frac{\partial L}{\partial \lambda} = 0$ quickly gets you the solution $x_1 = x_2 = 0$, $\lambda = -1$.
In fact, finding points where a partial does not exist is generally not any harder (and is sometimes easier) than finding points where a partial is zero. There aren't many ways for a point in the domain of a continuous function to have a partial that fails to exist. (Besides the absolute value situation, you could have division by zero with the partial from an expression like $x^p$, with $0 < p < 1$, in the function. Perhaps there are some others I'm forgetting - maybe others reading this can supply them. There are also some pathological cases, like the continuous but nowhere differentiable Weierstrass function, but those don't generally show up in practice.)
And, of course, if you have more than just a few variables, you don't want to use Lagrange multipliers anyway: Solving a large number of nonlinear equations is just too difficult. In that case, you will probably want an iterative nonlinear optimization method.
• Thanks for the thoughts. The case I'm interested is very restricted (always the same function; the nondifferentiability comes only from constraints of the form max(x1,x2)=x3), so I'm hoping to produce candidate solutions by other means, making verification the only issue. Aug 16, 2011 at 21:16
• A couple additional questions: (1) is it the case that there always exists a lambda such that the minimum of the function obeys the zero-in-subdifferential condition? (2) Do you know of any reference that discusses this approach? Aug 16, 2011 at 21:20
• @dan_x: If verification is all you want, then checking to make sure $0$ is in the subdifferential for the nondifferentiable cases should work. Aug 16, 2011 at 21:24
• great, thank you. I'll hold off a bit on accepting your answer in hopes of getting a reference, but if not, I'll accept this answer soon. Aug 16, 2011 at 21:30
• @dan_x: I don't know of any references that discuss this approach. There is something called the "subgradient method"; but it's iterative. Like I say in my answer, Lagrange multipliers is mostly used when there are a small number of variables and everything is differentiable; otherwise, some iterative method is generally used. If you do find a reference on your own, though, would you mind posting it back in the question or as a comment? I would like to see one myself. Aug 16, 2011 at 21:38
Another issue here is that the subdifferential is classically useful for convex functions only. In your example, the function $x_2\mapsto F(x_1,x_2,\lambda)$ is concave whenever $\lambda>0$, and then its subdifferential is empty at $x_2=0$. And unless the function $g$ is linear, the constraint set won't be convex, so you won't be able to construct a convex Lagrangian-type function either.
You could consider generalisations of the subdifferential to nonconvex functions, (or, more accurately generalisations of the notion of stationarity) but none are perfect. The best known is Clarke's subdifferential [1], which is pretty simple, but doesn't enjoy full calculus.
[1] F.H. Clarke, Optimization and Nonsmooth Analysis, Wiley: New York, 1983. | 1,725 | 6,806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.879155 |
https://www.slideserve.com/gillespie-hehir/ch-7-3-using-chemical-formulas | 1,511,338,335,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00266.warc.gz | 856,591,324 | 13,690 | Ch 7.3 Using Chemical Formulas
1 / 18
# Ch 7.3 Using Chemical Formulas - PowerPoint PPT Presentation
Ch 7.3 Using Chemical Formulas. The Mass of a Mole of an Element. Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu). Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). Carbon = 12.01 g/mol Hydrogen = 1.01 g/mol
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### Ch 7.3 Using Chemical Formulas
The Mass of a Mole of an Element
• Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu).
• Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol).
• Carbon = 12.01 g/mol
• Hydrogen = 1.01 g/mol
• When dealing with molar mass, round off to two decimals. 12.011g/mol -> 12.01g/mol
The Mass of a Mole of a Compound
• You calculate the mass of a molecule by adding up the molar masses of the atoms making up the molecules.
• Example: H2O
• H = 1.01 g x 2 atoms = 2.02 g/mol
• O = 16.00 g x 1 atom = 16.00 g/mol
• Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol
• This applies to both molecular and ionic compounds
Find the molar mass of PCl3
• P = 30.97 g x 1 atom = 30.97 g/mol
• Cl = 35.45 g x 3 atoms = 106.35 g/mol
• PCl3 = 30.97 g + 106.35 g = 137.32 g/mol
• What is the molar mass of Sodium Hydrogen Carbonate (NaHCO3) ?
• Na = 22.99 g x 1 atom = 22.99 g/mol
• H = 1.01 g x 1 atom = 1.01 g/mol
• C = 12.01 g x 1 atom = 12.01 g/mol
• O = 16.00 g x 3 atoms = 48.00 g/mol
• NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00
= 84.01 g/mol
Converting Moles to Mass
• You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance.
• Mass (g) = # of moles x mass (g)
1 mole
Example: If molar mass of NaCl is 58.44 g/mol, what is the mass of 3.00 mol NaCl?
Mass of NaCl = 3.00 mol x 58.44g =
1 mol
175 g NaCl
Example 2: Moles to Mass
• What is the mass of 9.45 mol of Aluminum Oxide (Al2O3)?
• Find molar mass of Al2O3
= 101.96 g/mol
• Mass = 9.45 mol Al2O3 x 101.96 g Al2O3
1 mol Al2O3
= 964 g Al2O3
Converting Mass to Moles
• You can invert the conversion factor to find moles when given the mass.
• Moles = mass (g) x 1 mole
mass (g)
Example: If molar mass of Na2SO4 142.05 g/mol, how many moles is 10.0 g of Na2SO4?
Moles of Na2SO4 = 10.0 g x 1 mol =
142.05 g
= 0.0704 mol Na2SO4
Example 2: Mass to Moles
• How many moles are in 75.0 g of Dinitrogen Trioxide?
• Find molar mass of N2O3
= 76.02 g/mol
• Moles = 75.0 g N2O3 x 1 mole =
76.02 g
N2O3
0.987 mol N2O3
Percent Composition
• Percent Composition: the relative amount of the elements in a compound.
• Also known as the percent by mass
• It can be calculated in two ways:
• Using Mass Data
• Using the Chemical Formula
% mass of element= mass of element x100% mass of compound
Example
• When a 13.60 g sample of a compound containing Mg and O is decomposed, 5.40 g O is obtained. What is the % composition of this compound?
Mass of compound: 13.60 g
Mass of oxygen: 5.40 g O
Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg
% Mg = 8.20 g Mg x 100% =
13.60 g
% O = 5.40 g O x 100% =
13.60 g
60.3%
39.7%
Find the percent composition of Cu2S.
• Find mass of Cu and S
• Cu = 63.55 x 2 = 127.10 g
• S = 32.07 g
• Find mass of Cu2S
• 127.10 g + 32.07 g = 159.17 g
% Composition
• Cu = 127.10 g x 100% =
159.17 g
• S = 32.07 g x 100% =
159.17 g
79.85%
20.15%
Homework
• 7.3 pg 253 #30-33
### Ch 7.4Determining Chemical Formulas
Empirical Formulas
• Empirical Formula: shows the smallest whole-number ratio of the atoms of the elements in a compound.
• Example:
• The Empirical Formula for Hydrogen Peroxide (H2O2) is HO with a 1:1 ratio.
• The Empirical Formula for Carbon Dioxide (CO2) is CO2 with a 1:2 ratio.
Determining the Empirical Formula of a Compound
• A compound is found to contain 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula of the compound?
• 25.9 g N x 1 mol N = 14.01 g N
• 74.1 g O x 1 mol O = 16.00 g O
• N1.85O4.63 = N1O2.5 = N2O5
1.85 mol N
4.63 mol O
Molecular Formulas
• Molecular Formula: tells the actual number of each kind of atom present in a molecule of a compound
• Example:
• The Molecular Formula for Hydrogen Peroxide is H2O2.
• The Molecular Formula for Carbon Dioxide is CO2
• It is possible to find the Molecular Formula using the Empirical Formula if you know the molar mass of the compound.
Finding the Molecular Formula
• Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N
• Step 1: Find the empirical formula molar mass
• 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol
• Step 2: Divide molar mass by EF molar mass
• 60.0 g/mol = 1.99 2 30.06 g/mol
• Step 3: Multiply empirical formula by 2
• CH4N x 2 = C2H8N2
Homework
• 7.4 pg 253 #36-38 | 1,880 | 5,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-47 | latest | en | 0.771244 |
https://cboard.cprogramming.com/cplusplus-programming/26961-raising-power.html | 1,496,053,086,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612069.19/warc/CC-MAIN-20170529091944-20170529111944-00108.warc.gz | 887,850,930 | 11,997 | # Thread: raising to a power
1. ## raising to a power
Does anyone know of a way to raise a constant to a variable.
Ex. 15 raised to the (xy)
Im kinda new to this so any help would be much appreciated.
2. Code:
```#include <cmath>
#include <iostream>
int main(){
int xy = 2;
int x = pow(3, 4); //this means 3 to the fourth
int y = pow(15, xy); //15 to the xy
return 0;
}```
3. Within reason, you can also unroll the multiplications.
Code:
```int x, y;
x = pow(15, 3); //this works
y = 15*15*15; //this is faster```
Of course with variable exponents you will need to use the pow() functions.
4. wll if you want to write a function then this is how it works
Code:
```power(x,y)
{
double temp=1;
for(int i=0;i<y+1;i++)
{
temp=temp*x;
}
return temp;
}```
5. >y = 15*15*15; //this is faster
Maybe, but what about:
3^25
or
7^2.6
6. 2^3 doesn't work with me!
Maybe I need to overload it.
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• Last Updated : 11 May, 2022
There are different sorts of numbers that are included in the number system, such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. These numbers can be expressed in a variety of ways, including figures and words. For example, in the form of figures, the numbers 50 and 75 can alternatively be written as fifty and seventy-five.
A number system, often known as a numeral system, is a basic system for expressing numbers and figures. It is the only way to represent numbers in both arithmetic and algebraic structures. Numbers are utilized in a variety of arithmetic values that may be used to perform a variety of arithmetic operations such as addition, subtraction, multiplication, and other operations that are used in daily life for the purpose of computation.
In the numeral system, Real numbers are simply a combination of rational and irrational numerals. All known arithmetic functions may be performed on these numbers and represented on the number line. The digit, its place value in the number and the number system’s base determine the value of a number. The mathematical values used for counting, measuring, identifying, and measuring fundamental quantities are known as numbers or numerals.
### Rational numbers
A rational number is the one that can be written as a ratio of two integers p and q, where q is not zero. and Repeating decimals are those that have a set of decimal terms that are repeated consistently. Different sorts of rational numbers occur. We shouldn’t assume that only rational numbers are fractions with integers. Numbers that are rational can also be stated in decimal form.
Example: Is 1.1 a rational number?
Yes, it’s a rational number because 1.1 may be represented as 1.1 = 11/10. Let’s take a look at non-terminating decimals like 0.333… 0.333… is a rational number since it may be expressed as 1/3. Non-terminating decimals with repeated numbers after the decimal point are thus rational numbers as well.
Examples of Rational numbers are,
• All numerals are rational, 1, 2, 3, 4, 5, 6, 7… all are rational numbers.
• All Whole numerals are rational. 0, 1, 2, 3, 4, 5,… all are rational.
The following are the several sorts of rational numbers,
Integers such as -2, 0, 3, and so on. Fractions with integer numerators and denominators, such as 3/7, -6/5, and so on. 0.35, 0.7116, 0.9768, and so on are examples of terminating decimals. Non-terminating decimals having recurring patterns (after the decimal point), such as 0.333…, 0.141414…, and so on.
Examples of Repeated decimals are,
6/4 as a decimal is 6 ÷ 4 = 1.5 terminating decimal. 35/100 as a decimal is 35 ÷100 = 0.35 terminating decimal. 5/3 as a decimal is 5 ÷ 3 = 1.66666… its a repeating decimal of Rational number.
### How to write a rational number as a repeating decimal?
Any rational number (a fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
If there is a remainder of zero, you have a terminating decimal. Otherwise, the remainder will begin to repeat after some significance, resulting in a repeating decimal.
Example: Write rational number 4/3 in repeating decimal form.
Simply divide the numerator by the denominator.
So, here we divide 4 by 3,
Repeating decimal
Here after simplifying we get 4/3 = 1.333333…
Hence, this way we can write rational number as repeating decimal.
### Similar Questions
Question 1: Write the rational number 67/3 as a repeating decimal?
Solution:
Any rational number (a fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator. Here divide 67 by 3,
67/3 as repeating decimal
We can write rational number 67/3 as a 22.33333… as repeating decimal.
Question 2: Write 11/3 rational numbers as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 11 by 3: we can write 11 /3 as 3.666666666… after dividing,
Therefore it’s a repeating decimal of rational number.
Question 3: Write 103/6 rational number as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 103 by 6: we can write 103 / 6 as 17.166666666… after dividing,
Therefore it’s a way of rational number writing as repeating decimal.
Question 4: Write 5/6 rational number as a repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 5 by 6: we can write 5 / 6 as 0.8333333333… after dividing,
Therefore it’s a way of rational number writing as repeating decimal.
Question 5: Write 19/3 rational numbers as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 19 by 3: we can write 19/ 3 as 6.3333333… after dividing,
Therefore in this way 19/3 rational number can be written as a repeating decimal.
Question 6: Write 185/6 rational number as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 185 by 6: we can write 185/ 6 as 30.8333333… after dividing,
Therefore in this way 185/6 rational number write as repeating decimal.
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2014-10-20T17:16:49+05:30
temperature in celcius =( temperature in fahrenheit - 32) / 9
2014-10-20T17:24:02+05:30
°F = (°C * 9/5) + 32
°C = (°F - 32) * 5/9 | 85 | 167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-04 | latest | en | 0.210537 |
https://www.jiskha.com/display.cgi?id=1196953798 | 1,516,124,277,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00371.warc.gz | 932,372,319 | 3,873 | # Math
posted by .
There is one bus with seven girls on it. Each girl has seven backpacks with seven lions in each backpack. Each big lion has seven little lions. How many legs are there all together?
• Math - reposted for Reiny -
Math - Reiny, Thursday, December 6, 2007 at 11:11am
Should be pretty easy, just think through it logically
there are 7 girls,.... so 14 legs
each girl has 7 backpacks with 7 big lions
so there are 7*7*7 or 343 big lions...each with 4 legs.... 1372 legs
each of the 343 big lions has 7 little lions
so 2401 little lions has four legs... so 9604 legs
so the answer to this rather silly question is
14 + 1372 + 9604 legs
= 10990 legs
## Similar Questions
1. ### MATH
THERE ARE 7 GIRLS ON A BUS EACH GIRL HAS 7 BACKPACKS IN EACH BACKPACK, THERE ARE 7 BIG CATS FOR EVERY BIG CAT THERE ARE 7 LITTLE CATS ?
2. ### arithmetic
There are 7 girls on a bus. each girl has 7 backpacks. In each backpack there are 7 big cats. For every big cat there are 7 little cats. Question: How many legs are there in the bus?
3. ### math
which choices correctly represent numbers in base seven. (a)575 base seven (b) 656 base seven If one is incorrect write it so it is correct.
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6. ### Math
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7. ### MATH
Find the word form 0.007 a. seven hundredths b. seven tenths c. seven thousand d. not given I selected a. seven hundredths
8. ### English
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https://codedump.io/share/PsG0WMSJpwip/1/how-to-break-down-and-print-two-list-of-lists-together-in-python | 1,529,791,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00409.warc.gz | 571,194,074 | 9,034 | Biotechgeek - 1 year ago 74
Python Question
# How to break down and print two list of lists together in Python?
Here are two lists:
``````a_value = [[0.234, 0.88,0.98],[0.923,0.777,0.87],[0.77,0.98,0.89]]
b_value = [[(1,1),(1,2),(1,3)],[(1,1),(1,2),(1,3)],[(1,1),(1,2),(1,3)]]
``````
I need to join the two lists such that I have output that prints as:
``````Set1
b_value (1,1) = a_value 0.234
b_value (1,2) = a_value 0.88
b_value (1,3) = a_value 0.98
Set2
b_value (1,1) = a_value 0.923
b_value (1,2) = a_value 0.777
b_value (1,3) = a_value 0.87
Set 3
b_value (1,1) = a_value 0.77
b_value (1,2) = a_value 0.98
b_value (1,3) = a_value 0.89
``````
The code that I have attempted looks like:
``````print("\n".join([('b_value{} a_value={}'.format(i,j)) for i,j in zip(b_value,a_value)]))
``````
Output:
``````b_value[(1, 1), (1, 2), (1, 3)] a_value=[0.234, 0.88, 0.98]
b_value[(1, 1), (1, 2), (1, 3)] a_value=[0.923, 0.777, 0.87]
b_value[(1, 1), (1, 2), (1, 3)] a_value=[0.77, 0.98, 0.89]
``````
I am not sure how to modify the code such that it breaks down the list of lists and also separates them into "Sets".
Since you only need to print the values and not build a list, you should use `for` loops instead of the list comprehension. And then you also need to `zip` the sublists in a nested loop to place their items side-by-side:
``````for idx, (i ,j) in enumerate(zip(b_value,a_value), 1):
print("Set{}".format(idx))
for a,b in zip(i,j):
print('b_value {} = a_value {}'.format(a,b))
``````
``````Set1
b_value (1, 1) = a_value 0.234
b_value (1, 2) = a_value 0.88
b_value (1, 3) = a_value 0.98
Set2
b_value (1, 1) = a_value 0.923
b_value (1, 2) = a_value 0.777
b_value (1, 3) = a_value 0.87
Set3
b_value (1, 1) = a_value 0.77
b_value (1, 2) = a_value 0.98
b_value (1, 3) = a_value 0.89
``````
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1
by avinasjha
2015-03-02T21:53:11+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.
Numbers are : 1, 2, 3, 4, 5, 6, 7 , 8 , 9 , 10
mean = 55/10 = 5.5
There are 10 numbers.
Median is the average of 5th and 6th data items.
=> median = (5+6)/2 = 5.5
We have the formula : Mean - Mode = 3 ( mean - median ) OR,
Mode = 3 Median - 2 Mode = 5.5
=> Mode = 5.5
This data is symmetrically distributed about the mean 5.5.
As all the data items are appearing only once, mode could be taken as any. But we take the Karl Pearson's rule | 261 | 846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-04 | latest | en | 0.933247 |
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# Solved Problems: Strength of Materials - Torsion
Mechanical - Strength of Materials - Torsion
1.A metal bar of 10mm dia when subjected to a pull of 23.55KN gave and elongation of 0.3mm on a gauge length of 200mm. In a torsion test maximum shear stress of 40.71N/mm2 was measured on a bar of 50mm dia. The angle of twist measured over a length of 300mm being 0°21’. Deter poisson’s.
Solution:
Given data:
pull P = 23.55KN
elongation SL = 0.3mm
Gauge length = 200mm
Torsion Test
(or) = 40.71 N/mm2
Dia = 500mm
Twist angle = 0o21’
length = 300 mm
Problem-2:
A hollow shaft dia ratio 3/5 is required to transmit 450Kw at 1200pm, the shearing stress in the shaft must not exceed 60N/mm2 and the twist in a length of 2.5m is not to exceed 1o. Calculate the minimum external of the shaft. Take, C=8.0KN/mm2.
Problem-3:
What must be the length of a 5mm dia aluminium wine so that it can be twisted through 1 complete revolution without exceeding a shear of 42N/mm2. Take, G=27 GPO.
Problem-4:
A solid steel shaft has to transmit 75Kw power at 200 pm. Taking allowable shear stress 70Mpo. Find suitable dia of shaft with the maximum torque transmitted on each revolutions exceeds by mean by 30% 1.3 times mean.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
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PGMs and Neural Network Example: The Boyfriend Problem
In this tutorial, we will learn how PGMS and Neural Network are used to solve a problem and also to know which is the best algorithm which is easy to use and to solve a problem? By Bharti Parmar Last updated : April 17, 2023
PGM and Neural Network
Both are capable for inferencing and learning problem.
Difference: To get prior knowledge in the existing model. Like: we can constraint a PGM what can Neural network does and torcher a neural network to get some information PGM would naturally give you.
The Boyfriend Problem
Here, we use both the algorithms and we see which is the more beneficial algorithm, for us to find a correct and solution of any problem easily in less time.
First we see what exactly a boyfriend problem is?
Question
A girl like a guy that she knows through end no. of mutual friend and she want to ask him for a date. Now, she wants some guy who can boost her to him. Then,
• If any friend who is close to him and if he put some good words for her then it will have a strong positive influence.
• If someone says bad things about her to him then, he will not talk to her. Suppose, there are 'n' No. of friends then 'N' bit vector = {0,1,0,1,0,1……} is a boost where, 0 means No contact and 1 for contact.
Now, here we use both the algorithms to solve this problem:
Objective
Find the set of friends that she should ask to boost him that is a best vector.
Solution of the Boyfriend Problem using Neural Network
Consider the below graph –
Now, if we take 3 top close friend in observation the chances is more to get accepted her proposal.
So, somehow she has to manage to do that then, neural network can act as an oracle.
Individually it will not give us a prediction, so, here we use neural network.
Solution of the Boyfriend Problem using PGMs
Consider the below graph –
In this 'N' No. of friends give her an approval to get accepter or rejecter (it will leave her for an impression) it is called random variable.
P(impression / vector of all approval) ← P(approval from friend / impression)
P(all approval / impression) = product of all P (approval / impression)
So, this is very simple and best to solve this problem.
Conclusion
In this tutorial, we have learnt that how PGMs and neural network are solving any problem and which is best to use by using boyfriend problem? We will know more about ML in the upcoming articles. Have a Nice day! Happy Learning! | 569 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.956599 |
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# Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0
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Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0 [#permalink]
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10 Jun 2010, 11:52
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Is (x^7)(y^2)(z^3) > 0 ?
(1) yz < 0
(2) xz > 0
[Reveal] Spoiler:
S1 is insufficient because we need the sign of x.
S2 is SUFFICIENT, because x^7*z^3 will always be positive, and y^2 will be positive, so the whole product (x^7)(y^2)(z^3) will be positive. Answer: B.
(m25#34)
[Reveal] Spoiler: OA
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Last edited by Bunuel on 20 Oct 2013, 10:56, edited 2 times in total.
Edited the question and added the OA
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Re: Data sufficiency +- exponents question [#permalink]
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10 Jun 2010, 13:22
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TheGmatTutor wrote:
Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct?
No, not correct. $$y^2$$ is not always positive, it's never negative: $$y^2\geq{0}$$.
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative (note that both positive or both negative excludes the possibility of either of them to be zero) AND $$y$$ must not be zero. Because if $$y=0$$, then $$x^7*y^2*z^3=0$$.
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Re: Data sufficiency +- exponents question [#permalink]
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10 Jun 2010, 12:22
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TheGmatTutor wrote:
My apologies if this has been posted before. Just want to confirm my reasoning (in the spoiler)
Is (x^7)(y^2)(z^3) > 0 ?
(1) yz < 0
(2) xz > 0
[Reveal] Spoiler:
S1 is insufficient because we need the sign of x.
S2 is SUFFICIENT, because x^7*z^3 will always be positive, and y^2 will be positive, so the whole product (x^7)(y^2)(z^3) will be positive. Answer: B.
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative AND $$y$$ must not be zero.
(1) $$yz<0$$ --> $$y\neq{0}$$. Don't know about $$x$$ and $$z$$. Not sufficient.
(2) $$xz>0$$ --> $$x$$ and $$z$$ are either both positive or both negative. Don't know about $$y$$. Not sufficient.
(1)+(2) Sufficient.
Hope it helps.
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Re: GmatClub Math25 qn 34 - is it positive? [#permalink]
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16 Feb 2012, 14:01
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fxsunny wrote:
From the Gmatclub Math 25 practice test:
http://gmatclub.com/tests/m25#q34
Is x^7*y^2*z^3> 0?
(1) yz < 0
(2) xz > 0
Response:
x^7*y^2*z^3 is same as:
(xz)*(x^6)*(y^2)*(z^2)
xz is positive based on assumption #2.
x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive.
z^2 is positive for all real numbers.
y^2 is positive for all real numbers.
So product of 4 positive real numbers is also positive. SUFFICIENT.
Hence, B - Statement 2 Alone is Sufficient.
Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?
Thanks!
The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.
Complete solution:
Is x^7*y^2*z^3 > 0 ?
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative (in order $$x^7*z^3$$ to be positive) AND $$y$$ must not be zero (in order $$x^7*y^2*z^3$$ not to equal to zero).
(1) $$yz<0$$ --> $$y\neq{0}$$. Don't know about $$x$$ and $$z$$. Not sufficient.
(2) $$xz>0$$ --> $$x$$ and $$z$$ are either both positive or both negative. Don't know about $$y$$. Not sufficient.
(1)+(2) Sufficient.
Similar question to practice: m21-q30-96613.html?hilit=conditions#p744188
Hope it helps.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink]
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23 Mar 2012, 05:03
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Expert's post
x^7*y^2*z^3>0
y is positive, we need to know regarding x and z
insufficient
2) xz>0
either of them is negative hence
the answer to our question is no
sufficient
hence B
OA for this question is C, not B.
(2) $$xz>0$$ means that$$x$$ and $$z$$ are either both positive or both negative. So, $$x^7*z^3 > 0$$ but y can still be zero and in this case $$x^7*y^2*z^3=0$$, hence this statement is not sufficient.
Refer to the complete solution above.
Hope it helps.
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24 Nov 2010, 09:35
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chiragatara wrote:
Is (x^7)(y^2)(z^3) > 0?
1. yz < 0
2. xz > 0
However,
for reaching to the conclusion of YES/NO, we have to be certain that:
(1) x, y, z are not ZERO,
(2) [highlight]x and y are not NEGATIVE[/highlight]
Combined from (1) and (2) we can say that x, y, z are not ZERO. However, I think they are not helpful in deciding the certainty that x and y are not NEGATIVE.
First of all, a trick in the question is 'x, y, z are not ZERO' so good that you figured it.
Next, we don't need to know that x and z are not negative. We need to know [highlight]whether they have the same sign or opposite signs[/highlight] because question asks you whether (x^7)(z^3) is positive. (Ignoring y for now)
For the product to be positive, either both should be positive or both negative. Then, answer will be 'YES'
For the product to be negative only one of them should be negative. Then answer will be 'NO'
In either case, if we get a definite YES/NO, the statements will be sufficient.
If xz> 0, then either x and z both are positive or both are negative. They have the same sign. So (x^7)(z^3) is positive.
y, we know is not 0, so YES, (x^7)(y^2)(z^3) is greater than 0. Sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 12 Mar 2012 Posts: 369 Concentration: Operations, Strategy Followers: 2 Kudos [?]: 169 [1] , given: 31 Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] ### Show Tags 23 Mar 2012, 04:59 1 This post received KUDOS 1 This post was BOOKMARKED x^7*y^2*z^3>0 y is positive, we need to know regarding x and z 1) no information about x insufficient 2) xz>0 either of them is negative hence the answer to our question is no sufficient hence B _________________ Practice Practice and practice...!! If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight. Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7076 Kudos [?]: 93101 [1] , given: 10552 Re: Inequality qs [#permalink] ### Show Tags 12 Sep 2013, 02:16 1 This post received KUDOS Expert's post shameekv wrote: shameekv wrote: Is (x^7)(y^2)(z^3)>0 ? (1) yz<0 (2) xz>0 Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help. According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer. Check here: is-x-7-y-2-z-3-0-1-yz-0-2-xz-127692.html#p1063962 Hope it helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13639 [1] , given: 222 Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] ### Show Tags 17 May 2015, 19:35 1 This post received KUDOS Expert's post octenisept wrote: Guys, is there any difference between (y^2) and (y)^2. We can be sure that (y)^2 is positive but can we also be sure that (y^2) is positive? (forget about y=0 for a moment) thank you (y^2) and (y)^2 are the same. In both cases, y is squared. y could be anything - the whole of it is squared in both cases. If y = 2a + b, both representations give (2a + b)^2 = 4a^2 + b^2 + 4ab _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Data sufficiency +- exponents question [#permalink]
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10 Jun 2010, 12:59
Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct? So the whole statement
(x^7)(y^2)(z^3)
must be greater than 0, because the product is either
- + -
or
+++
So I thought S2 was sufficient.
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Re: Data sufficiency +- exponents question [#permalink]
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10 Jun 2010, 13:30
I think this question would be more interesting if they specified that x,y, and z are non-zero integers.
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Re: Data sufficiency +- exponents question [#permalink]
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13 Jun 2010, 11:12
TheGmatTutor wrote:
I think this question would be more interesting if they specified that x,y, and z are non-zero integers.
Haha it wouldn't be more interesting; it would be easier.
They got me too
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Re: Data sufficiency +- exponents question [#permalink]
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11 Jan 2011, 05:47
Bunuel wrote:
TheGmatTutor wrote:
Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct?
No, not correct. $$y^2$$ is not always positive, it's never negative: $$y^2\geq{0}$$.
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative (note that both positive or both negative excludes the possibility of either of them to be zero) AND $$y$$ must not be zero. Because if $$y=0$$, then $$x^7*y^2*z^3=0$$.
Thanks a ton Bunuel , this example is the perfect for learning that while considering signs we should consider +ve , -ve and zero as well . Superb collection .
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Re: Data sufficiency +- exponents question [#permalink]
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13 Jun 2012, 11:24
Hi,
Simplifying the expression,
$$x^7y^2z^3$$,
it can be written as, $$(xz)^3x^4y^2$$ or $$(yz)^2x^6(xz)$$
Clearly, in both the expressions $$x^4y^2$$ as well as $$(yz)^2x^6$$ are positive.
Thus the sign of expression depends on sign of xz,
but since value of y can be 0,
Using (1), we can say y is not equal to 0.
Regards,
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Re: Data sufficiency +- exponents question [#permalink]
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07 Sep 2012, 01:02
sreenunna wrote:
Statement B Alone is sufficient.
irrespective of sign, any number power to even number is positive.
hence (x^7) = (x^6)x ---> so remove (x^6) and we are left with x.
since (y^2) is always positive, leave it
(z^3), apply above rule.. we are left with z.
hence the given question can be re-write as Is xz > 0?
Statement is clearly stating same , hence Statement B Alone is sufficient.
Please note that correct answer is C, not B. You can check OA under the spoiler in the first post.
Next, check these posts:
is-x-7-y-2-z-3-0-1-yz-0-2-xz-95626.html#p736291
is-x-7-y-2-z-3-0-1-yz-0-2-xz-95626.html#p736324
And finally, square of a number is not always positive it's non-negative: $$y^2\geq{0}$$. So, for (2) if y=0 then x^7*y^2*z^3=0 not >0.
Hope it's clear.
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Re: GmatClub Math25 qn 34 - is it positive? [#permalink]
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22 Aug 2013, 16:16
Bunuel wrote:
fxsunny wrote:
From the Gmatclub Math 25 practice test:
http://gmatclub.com/tests/m25#q34
Is x^7*y^2*z^3> 0?
(1) yz < 0
(2) xz > 0
Response:
x^7*y^2*z^3 is same as:
(xz)*(x^6)*(y^2)*(z^2)
xz is positive based on assumption #2.
x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive.
z^2 is positive for all real numbers.
y^2 is positive for all real numbers.
So product of 4 positive real numbers is also positive. SUFFICIENT.
Hence, B - Statement 2 Alone is Sufficient.
Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?
Thanks!
The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.
Complete solution:
Is x^7*y^2*z^3 > 0 ?
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative (in order $$x^7*z^3$$ to be positive) AND $$y$$ must not be zero (in order $$x^7*y^2*z^3$$ not to equal to zero).
(1) $$yz<0$$ --> $$y\neq{0}$$. Don't know about $$x$$ and $$z$$. Not sufficient.
(2) $$xz>0$$ --> $$x$$ and $$z$$ are either both positive or both negative. Don't know about $$y$$. Not sufficient.
(1)+(2) Sufficient.
Similar question to practice: m21-q30-96613.html?hilit=conditions#p744188
Hope it helps.
So in short you want to say that by using both we can infer X,Y,Z are not equal to 0?
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Re: GmatClub Math25 qn 34 - is it positive? [#permalink]
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22 Aug 2013, 16:24
honchos wrote:
Bunuel wrote:
fxsunny wrote:
From the Gmatclub Math 25 practice test:
http://gmatclub.com/tests/m25#q34
Is x^7*y^2*z^3> 0?
(1) yz < 0
(2) xz > 0
Response:
x^7*y^2*z^3 is same as:
(xz)*(x^6)*(y^2)*(z^2)
xz is positive based on assumption #2.
x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive.
z^2 is positive for all real numbers.
y^2 is positive for all real numbers.
So product of 4 positive real numbers is also positive. SUFFICIENT.
Hence, B - Statement 2 Alone is Sufficient.
Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?
Thanks!
The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.
Complete solution:
Is x^7*y^2*z^3 > 0 ?
Inequality $$x^7*y^2*z^3>0$$ to be true $$x$$ and $$z$$ must be either both positive or both negative (in order $$x^7*z^3$$ to be positive) AND $$y$$ must not be zero (in order $$x^7*y^2*z^3$$ not to equal to zero).
(1) $$yz<0$$ --> $$y\neq{0}$$. Don't know about $$x$$ and $$z$$. Not sufficient.
(2) $$xz>0$$ --> $$x$$ and $$z$$ are either both positive or both negative. Don't know about $$y$$. Not sufficient.
(1)+(2) Sufficient.
Similar question to practice: m21-q30-96613.html?hilit=conditions#p744188
Hope it helps.
So in short you want to say that by using both we can infer X,Y,Z are not equal to 0?
yz < 0 implies that neither y nor z is 0.
xz > 0 implies that neither x nor z is 0.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink]
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24 Aug 2013, 04:44
You do have to pay attention in DS . The key thing here is that y can be zero, so b is not the answer; it is c.
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12 Sep 2013, 02:15
shameekv wrote:
Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0
(2) xz>0
Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help.
According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer.
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12 Sep 2013, 02:18
shameekv wrote:
shameekv wrote:
Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0
(2) xz>0
Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help.
According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer.
Got it. Forgot to consider y = 0. As always. Thanks anyways.
Re: Inequality qs [#permalink] 12 Sep 2013, 02:18
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https://blog.wavelabs.ai/box-cox-transformation/ | 1,610,942,296,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00367.warc.gz | 250,030,204 | 10,323 | ## Box-Cox Transformation
ARTIFICIAL INTELLIGENCE
FEBRUARY 11, 2020
# Box Cox Transformation
The real-world data is not always distributed the way we want, that is Normal-Distribution It is always distributed in some distribution which we have no idea about some time it is skewed towards the right other time it has a long tail this leads us to miss normal distribution, Why we miss normal distribution you ask?
The normal distribution is the most important probability distribution in statistics because it fits many natural phenomena. It is symmetric distribution where most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions.
The Box-Cox transformation is a particularly useful family of transformations. It is defined as:
where y^λ is the response variable and λ is the transformation parameter, For λ = 0, the natural log of the data is taken instead of using the above formula, here λ is a hyperparameter which has to be tuned according to the dataset
Let’s see box-cox in action
##### Import the necessary libraries
``````
from scipy import stats
import pandas as pd
import numpy as np
import pylab
import matplotlib.pyplot as plt
import seaborn as sns
from scipy import stats
%matplotlib inline
``````
##### Let’s create a skewed distribution
``````
skewed_dist = stats.loggamma.rvs(5, size=10000) + 5
``````
##### Now let’s create a normal distribution
``````
normal_dist = np.random.normal(0, 1, 10000)
``````
##### Now let’s calculate the skewness of the normal distribution and look at the plot of the distributions
``````
sns.distplot(s)
print("Skewness for the normal distribution:",skew(normal_dist))
sns.distplot(x)
print("Skewness for the skewed distribution:",skew(skewed_dist))
``````
Here you can see how the second distribution is left-skewed
skewness = 0 : normally distributed.
skewness > 0 : more weight in the left tail of the distribution.
skewness < 0 : more weight in the right tail of the distribution.
#### Pearson’s Coefficient of Skewness
##### skewness = 3(X-Me) / σ
where X = mean of the distribution
Me = median of the distribution and
σ = standard deviation
The probability plot is a graphical technique for assessing whether or not a data set follows a normal distribution
``````
stats.probplot(normal_dist, dist=”norm”, plot=pylab)
pylab.show()
stats.probplot(skewed_dist, dist=”norm”, plot=pylab)
pylab.show()
``````
Probability-Plot for normal_data
Probability-Plot for skewed_data
In the above diagram for the skewed plot, we can see that the data is not normally distributed since the point don’t align with the red line the plot is not normally distributed
##### Now we apply our box-cox Transformation and plot it
``````
skewed_box_cox, lmda = stats.boxcox(skwed_dist)
sns.distplot(skewed_box_cox)
``````
Distribution after applying Box-Cox
Let’s check the Probability-Plot and see whether the data is normally distributed or not and get the appropriate lambda value.
``````
stats.probplot(skewed_box_cox, dist=”norm”, plot=pylab)
pylab.show()
print ("lambda parameter for Box-Cox Transformation is:",lmda)
``````
lambda parameter and Probability-Plot for Box-Cox Transformation
In Conclusion, Box-cox transformation attempts to transform a set of data to a normal distribution by finding the value of λ that minimizes the variation. This allows you to perform those calculations that require the data to be normally distributed, The Box-Cox transformation does not always convert the data to a normal distribution. You must check the transformation to ensure it worked.
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Math Physics Chemistry Biology Formulas
Solving for the actual, true or accepted value in the percent error equation. Note, this has two solution due to the absolute value in the percent error equation. However if percent error is equal to 100 percent or -100 percent, then there is only one calculated solution and one solution of infinity. The infinity comes from the division by zero.
Percent error equation:
measured value
unitless
percent error
measured value
= 0
= 0
percent error
= 0
= 0
percent
## Solution 1:
actual, accepted or true value
= NOT CALCULATED
## Solution 2:
actual, accepted or true value
= NOT CALCULATED
Change Equation Variable
Select to solve for a different unknown
percent error calculator Rich internet application version of the percent error calculator.
Solve for percent error Solve for the actual value. This is also called the accepted, experimental or true value.Note due to the absolute value in the actual equation (above) there are two value. Solve for the measured or observed value.Note due to the absolute value in the actual equation (above) there are two solutions.
Change Equation to Percent Difference
Solve for percent difference.
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USE CODE: chait6
``` A boat is moving towards east with velocity 4 m/s with respect to still water and river is flowing towards north
with velocity 2 m/s and the wind is blowing towards north with velocity 6 m/s. The direction of the flag blown
over by the wind hoisted on the boat is:
```
5 years ago
Share
``` let's assume that wind doesn't affect the direction of the boat.
now take the resultant as direction of flag makes an angle a1 with boats direction ==========
tan a1 = (6sin(tan-1(1/2)))/(2root(5)+6cos(tan-1(1/2))
speed = root(20+36+24root(5)*cos(90-tan-1(1/2))
```
5 years ago
``` Velocity of boat towards east = 4 m/s
Velocity of wind and water = 2+6 = 8 m/s
tan θ = 4/8
θ = tan^-1(0.5)
θ = 27 degrees west north.
```
5 years ago
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# P2F10_Tut2-soln - 1 A motoris t travels to work with a cons...
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1) A motorist travels to work with a constant speed of 52.0 km/h except for the 12min spent waiting for a very long train to pass. If the person's average speed is 31.2 km/h, how much time did it take this person to get to work and how long is his or her commute (time to get to work without stopping for trains)? 2) Alice leaves home in Chicago at 9:00 AM and travels east at a steady 60mph. Susan, 400 miles to the east in Pittsburgh, leaves at the same time and travels west at a constant 40mph. Where will they meet for lunch? 3) We need to make it to a Lakers game. The total trip is 100 miles. It’s 6pm now. The game starts at 8:00. We know that we will only be able to go 30mi/h on the 405, for a stretch of 39 miles. For the rest of the trip there will be no traffic, so we can move as fast as we want. BUT, parking takes 15 minutes. How fast do we need to move to get there on time? Are we above the speed limit (75mi/h)? 4)Assuming constant acceleration, a DeLorean DMC-12 can go from zero to 60 mph in 8.5s.
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# Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together, they can finish the work in 2 days. How much time B and C working together to finish it?
Last updated date: 20th Jul 2024
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Answer
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Hint- If one person does a work in x days and another person does it in y days then together they can finish that work in $\dfrac{{xy}}{{x + y}}$ days.
Let’s take work done by A be x
A takes twice as much as B Therefore B takes half of what time A takes.
$\Rightarrow B = \dfrac{x}{2}$
A takes thrice as much as C. Therefore C takes one third of what time A takes.
$\Rightarrow C = \dfrac{x}{3}$
When all of them work together, they can finish work in 2 days.
$\Rightarrow \dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = \dfrac{1}{2} \\ \Rightarrow \dfrac{1}{x} + \dfrac{2}{x} + \dfrac{3}{x} = \dfrac{1}{2} \\ \Rightarrow \dfrac{6}{x} = \dfrac{1}{2} \\$
Now, Cross multiply
$\Rightarrow x = 12$
Now, we calculate how much time taken by B to complete work.
$B = \dfrac{x}{2} = \dfrac{{12}}{2} \\ \Rightarrow B = 6 \\$
B takes 6 days to complete work.
Now, we calculate how much time taken by C to complete work.
$C = \dfrac{x}{3} = \dfrac{{12}}{3} \\ \Rightarrow C = 4 \\$
C takes 4 days to complete work.
Now, we calculate how much time taken by B and C to work together.
B does work in 6 days and C does it in 4 days.
If B and C can work together
$\Rightarrow \dfrac{1}{B} + \dfrac{1}{C} \\ \Rightarrow \dfrac{1}{6} + \dfrac{1}{4} \\$
Take LCM
$\Rightarrow \dfrac{{2 + 3}}{{12}} \\ \Rightarrow \dfrac{5}{{12}} \\$
So, If B and C work together they will take $\dfrac{{12}}{5}$ days.
Note- Whenever we face such types of problems we use some important points. Like we calculate how much time taken by a single person to complete their work then we calculate how much time taken by persons when they work together. | 630 | 1,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-30 | latest | en | 0.896277 |
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# ps3-sol - Economics 321 Fall 2010 Problem Set 3 SOLUTIONS...
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Economics 321: Fall 2010 Helen Schneider Problem Set 3 SOLUTIONS 1. Initially, a worker paid for 20 percent of a retiree’s benefits. In the future, the same worker would be responsible for paying for half of a current retiree’s benefits. If benefits remained the same, then each worker’s tax burden would increase by approximately 30 percent of the cost of benefits. In terms of tax rates, each worker’s tax burden would increase by 2.5 (gross) or 2.5-1 = 150% (net). If tax rates remain the same, then benefits would need to fall by approximately 60 percent. (B/W) future /(B/W) currently = 0.2/0.5 = 0.4. If tax rates remained the same, then benefits would need to fall by 0.4-1 = -0.6 or fall by 60% . 2. a. Females have higher SSW than males because both pay the same taxes, but females live longer than males and so receive benefits for a longer time. b. Someone born in 1925 would have the higher SSW because of the tax-rate growth effect and wage and growth population effects got smaller starting in the 1970s. c. A low-wage worker has higher SSW because of how Social Security redistributes income. Some of this higher SSW will be offset by lower life expectancy of low-income workers. d. Single-earner couples have more SSW because even though only one person works, that couple gets 150% of the earner's benefit. The two-earner couple has to pay taxes on their full earnings even though they may not receive more benefits than the one-earner couple. 3.
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CaLuP7eaN
Registered: 21.07.2005
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Posted: Wednesday 01st of Aug 18:16 Hello Math experts ! I am a starting at theorem of a math investigatory project. I seem to understand the lectures in the class properly , but when I begin to solve the problems at home myself, I commit errors . Does anyone know of any website where I can get my solutions checked before submitting them for grading? Or any resource where I can get to see a step by step answer ?
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Posted: Friday 03rd of Aug 09:26 Algebrator is the latest hot favourite of theorem of a math investigatory project students. I know a couple of teachers who actually ask their students to use a copy of this program at their residence .
SanG
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Posted: Sunday 05th of Aug 08:37 I remember having problems with function composition, multiplying matrices and rational inequalities. Algebrator is a really great piece of math software. I have used it through several algebra classes - College Algebra, Pre Algebra and Algebra 1. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended. | 662 | 2,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-09 | latest | en | 0.895051 |
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# Convert Kelvin/watt to Degree Fahrenheit Second/Btu (th)
Please provide values below to convert kelvin/watt [K/W] to degree Fahrenheit second/Btu (th), or vice versa.
From: kelvin/watt To: degree Fahrenheit second/Btu (th)
### Kelvin/watt to Degree Fahrenheit Second/Btu (th) Conversion Table
Kelvin/watt [K/W]Degree Fahrenheit Second/Btu (th)
0.01 K/W18.9783047608 degree Fahrenheit second/Btu (th)
0.1 K/W189.783047608 degree Fahrenheit second/Btu (th)
1 K/W1897.83047608 degree Fahrenheit second/Btu (th)
2 K/W3795.66095216 degree Fahrenheit second/Btu (th)
3 K/W5693.49142824 degree Fahrenheit second/Btu (th)
5 K/W9489.1523804 degree Fahrenheit second/Btu (th)
10 K/W18978.3047608 degree Fahrenheit second/Btu (th)
20 K/W37956.6095216 degree Fahrenheit second/Btu (th)
50 K/W94891.523804 degree Fahrenheit second/Btu (th)
100 K/W189783.047608 degree Fahrenheit second/Btu (th)
1000 K/W1897830.47608 degree Fahrenheit second/Btu (th)
### How to Convert Kelvin/watt to Degree Fahrenheit Second/Btu (th)
1 K/W = 1897.83047608 degree Fahrenheit second/Btu (th)
1 degree Fahrenheit second/Btu (th) = 0.0005269175 K/W
Example: convert 15 K/W to degree Fahrenheit second/Btu (th):
15 K/W = 15 × 1897.83047608 degree Fahrenheit second/Btu (th) = 28467.4571412 degree Fahrenheit second/Btu (th) | 407 | 1,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.581275 |
http://onlineguitarbooks.com/2011/05/12/d-phrygian-positions-on-the-guitar-fretboard | 1,518,917,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00206.warc.gz | 274,973,717 | 10,612 | # D PHRYGIAN MODE
‘D Phrygian’ is the 3rd mode of the Bb major scale. The notes in D Phrygian are:
D – Eb – F – G – A – Bb – C
If you have read the post on understanding the phrygian mode, you will know that the phrygian mode contains a ‘flat 2’, a ‘flat 3’, a ‘flat 6’ and a ‘flat 7’ (parallel approach). You will also know that it is the 3rd mode of a major scale (derivative approach). Let’s briefly look at how to construct D Phrygian using both the parallel approach and the derivative approach.
## Parallel Approach:
D Major has the following notes:
D – E – F# – G – A – B – C#
If we ‘flatten’ the 2nd note (E), the 3rd note (F#), the 6th note (B) and the 7th note (C#) we get the following:
D – Eb – F – G – A – Bb – C
## Derivative Approach:
A is the 3rd note of the B flat major scale:
Bb – C – D – Eb – F – G – A
If we play the Bb major scale and start on the 3rd note we get the following:
D – Eb – F – G – A – Bb – C
Let’s look at the D phrygian mode in the different positions on the guitar fretboard:
## Positions Along the Fretboard:
Firstly, let’s look at the open position:
Now let’s look at D phrygian in the 2nd position (lowest fret is 2)
Now let’s look at D phrygian in the 5th position (lowest fret is 5)
Now let’s look at D phrygian in the 6th position (lowest fret is 6)
Now let’s look at D phrygian in the 10th position (lowest fret is 10)
Finally, let’s look at D phrygian in the 11th position (lowest fret is 11)
That covers the 5 basic positions and the open position of D phrygian along the guitar fretboard.
For an in depth explanation of the phrygian mode, check out phrygian mode explained.
## Free Ebook And Lessons!
Why not download the free ebook, Guitar Scales Galore and sign up for 20 essential lessons?
The book contains 22 essential scales, written using beautiful diagrams.
It could be the only scale book you ever need.
To get your free copy, simply sign up for 20 free lessons, by clicking here
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## Advanced engineering mathematics / Erwin Kreyszig.
Editor: New York : Wiley, c1967Edición: 2nd edDescripción: xvii, 898 p. : il. ; 25 cmOtra clasificación: 00A06
Contenidos:
``` CONTENTS
Introduction Review of Some Topics From Algebra and Calculus [1]
0.1 Elementary Functions, [1]
0.2 Partial Derivatives, [8]
0.3 Second- and Third-Order Determinants, [10]
0.4 Complex Numbers, [18]
0.5 Polar Form of Complex Numbers, [22]
0.6 Some General Remarks About Numerical Computations, [24]
0.7 Solution of Equations, [26]
0.8 Approximate Integration, [31]
Chapter 1 Ordinary Differential Equations of the First Order [39]
1.1 Basic Concepts and Ideas, [39]
1.2 Geometrical Considerations. Isoclines, [46]
1.3 Separable Equations, [49]
1.4 Equations Reducible to Separable Form, [57]
1.5 Exact Differential Equations, [59]
1.6 Integrating Factors, [61]
1.7 Linear First-Order Differential Equations, [63]
1.8 Variation of Parameters, [67]
1.9 Electric Circuits, [69]
1.10 Families of Curves. Orthogonal Trajectories, [74]
1.11 Picard’s Iteration Method, [79]
1.12 Existence and Uniqueness of Solutions, [82]
1.13 Numerical Methods for Differential Equations of the First Order, [86]
Chapter 2 Ordinary Linear Differential Equations [93]
2.1 Homogeneous Linear Equations of the Second Order, [94]
2.2 Homogeneous Second-Order Equations with Constant Coefficients, [97]
2.3 General Solution. Fundamental System, [99]
2.4 Complex Roots of the Characteristic Equation. Initial Value Problem, [102]
2.5 Double Root of the Characteristic Equation, [106]
2.6 Free Oscillations, [108]
2.7 Cauchy Equation, [116]
2.8 Existence and Uniqueness of Solutions, [117]
2.9 Homogeneous Linear Equations of Arbitrary Order, [123]
2.10 Homogeneous Linear Equations of Arbitrary Order with Constant Coefficients, [126]
2.11 Nonhomogeneous Linear Equations, [128]
2.12 A Method for Solving Nonhomogeneous Linear Equations, [130]
2.13 Forced Oscillations. Resonance, [134]
2.14 Electric Circuits, [140]
2.15 Complex Method for Obtaining Particular Solutions, [144]
2.16 General Method for Solving Nonhomogeneous Equations, [147]
2.17 Numerical Methods for Second-Order Differential Equations, [149]
Chapter 3 Power Series Solutions of Differential Equations [155]
3.1 The Power Series Method, [155]
3.‘2 Theoretical Basis of the Power Series Method, [159]
3.3 Legendre’s Equation. Legendre Polynomials, [164]
3.4 Extended Power Series Method. Indicial Equation, [168]
3.5 Bessel’s Equation. Bessel Functions of the First Kind, [179]
3.6 Further Properties of Bessel Functions of the First Kind, [185]
3.7 Bessel Functions of the Second Kind, [187]
Chapter 4 Laplace Transformation [192]
4.1 Laplace Transform. Inverse Transform. Linearity, [192]
4.2 Laplace Transforms of Derivatives and Integrals, [198]
4.3 Transformation of Ordinary Differential Equations, [201]
4.4 Partial Fractions, [204]
4.5 Examples and Applications, [209]
4.6 Differentiation and Integration of Transforms, [213]
4.7 Unit Step Function, [216]
4.8 Shifting on the t-axis, [219]
4.9 Periodic Functions, [223]
4.10 Table 17. Some Laplace Transforms, [231]
Chapter 5 Vector Analysis [235]
5.1 Scalars and Vectors, [235]
5.2 Components of a Vector, [237]
5.3 Vector Addition. Multiplication by Scalars, [239]
5.4 Scalar Product, [242]
5.5 Vector Product, [247]
5.6 Vector Products in Terms of Components, [248]
5.7 Scalar Triple Product. Linear Dependence of Vectors, [254]
5.8 Other Repeated Products, [258]
5.9 Scalar Fields and Vector Fields, [260]
5.10 Vector Calculus, [263]
5.11 Curves, [265]
5.12 Arc Length, [268]
5.13 Tangent. Curvature and Torsion, [270]
5.14 Velocity and Acceleration, [274]
5.15 Chain Rule and Mean Value Theorem for Functions of Several Variables, [278]
5.16 Directional Derivative. Gradient of a Scalar Field, [282]
5.17 Transformation of Coordinate Systems and Vector Components, [288]
5.18 Divergence of a Vector Field, [292]
5.19 Curl of a Vector Field, [296]
Chapter 6 Line and Surface Integrals. Integral Theorems [299]
6.1 Line Integral, [299]
6.2 Evaluation of Line Integrals, [302]
6.3 Double Integrals, [306]
6.4 Transformation of Double Integrals into Line Integrals, [313]
6.5 Surfaces, [318]
6.6 Tangent Plane. First Fundamental Form, Area, [321]
6.7 Surface Integrals, [327]
6.8 Triple Integrals. Divergence Theorem of Gauss, [332]
6.9 Consequences and Applications of the Divergence Theorem, [336]
6.10 Stokes’s Theorem, [342]
6.11 Consequences and Applications of Stokes’s Theorem, [346]
6.12 Line Integrals Independent of Path, [347]
Chapter 7 Matrices and Determinants. Systems of Linear Equations [356]
7.1 Basic Concepts. Addition of Matrices, [356]
7.2 Matrix Multiplication, [361]
7.3 Determinants, [368]
7.4 Submatrices. Rank, [380]
7.5 Systems of n Linear Equations in n Unknowns. Cramer’s Rule, [382]
7.6 Arbitrary Homogeneous Systems of Linear Equations, [386]
7.7 Arbitrary Nonhomogeneous Systems of Linear Equations, [392]
7.8 Further Properties of Systems of Linear Equations, [395]
7.9 Gauss’s Elimination Method, [398]
7.10 The Inverse of a Matrix, [401]
7.11 Eigenvalues. Eigenvectors, [406]
7.12 Bilinear, Quadratic, Hermitian, and Skew-Hermitian Forms, [413]
7.13 Eigenvalues of Hermitian, Skew-Hermitian, and Unitary Matrices, [417]
7.14 Bounds for Eigenvalues, [422]
7.15 Determination of Eigenvalues by Iteration, [426]
Chapter 8 Fourier Series and Integrals [431]
8.1 Periodic Functions. Trigonometric Series, [431]
8.2 Fourier Series. Euler’s Formulas, [434]
8.3 Even and Odd Functions, [440]
8.4 Functions Having Arbitrary Period, [444]
8.5 Half-Range Expansions, [447]
8.6 Determination of Fourier Coefficients Without Integration, [451]
8.7 Forced Oscillations, [455]
8.8 Numerical Methods for Determining Fourier Coefficients. Square Error, [458]
8.9 Instrumental Methods for Determining Fourier Coefficients, [464]
8.10 The Fourier Integral, [465]
8.11 Orthogonal Functions, [473]
8.12 Sturm-Liouville Problem, [476]
8.13 Orthogonality of Bessel Functions, [483]
Chapter 9 Partial Differential Equations [486]
9.1 Basic Concepts, [486]
9.2 Vibrating String. One-Dimensional Wave Equation, [488]
9.3 Separation of Variables (Product Method), [490]
9.4 D’Alembert’s Solution of the Wave Equation, [499]
9.5 One-Dimensional Heat Flow, [501]
9.6 Heat Flow in an Infinite Bar, [506]
9.7 Vibrating Membrane. Two-Dimensional Wave Equation, [510]
9.8 Rectangular Membrane, [512]
9.9 Laplacian in Polar Coordinates, [519]
9.10 Circular Membrane. Bessel’s Equation, [521]
9.11 Laplace’s Equation. Potential, [526]
9.12 Laplace’s Equation in Spherical Coordinates. Legendre’s Equation, [529]
Chapter 10 Complex Analytic Functions [534]
10.1 Complex Numbers. Triangle Inequality, [535]
10.2 Limit. Derivative. Analytic Function, [539]
10.3 Cauchy*Riemann Equations. Laplace’s Equation, [543]
10.4 Rational Functions. Root, [547]
10.5 Exponential Function, [550]
10.6 Trigonometric and Hyperbolic Functions, [553]
10.7 Logarithm. General Power, [556]
Chapter 11 Conformal Mapping [560]
11.1 Mapping, [560]
11.2 Conformal Mapping, [564]
11.3 Linear Transformations, [568]
11.4 Special Linear Transformations, [572]
11.5 Mapping by Other Elementary Functions, [577]
11.6 Riemann Surfaces, [584]
Chapter 12 Complex Integrals [588]
12.1 Line Integral in the Complex Plane, [588]
12.2 Basic Properties of the Complex Line Integral, [594]
12.3 Cauchy’s Integral Theorem, [595]
12.4 Evaluation of Line Integrals by Indefinite Integration, [602]
12.5 Cauchy’s Integral Formula, [605]
12.6 The Derivatives of an Analytic Function, [607]
Chapter 13 Sequences and Series [611]
13.1 Sequences, [611]
13.2 Series, [618]
13.3 Tests for Convergence and Divergence of Series, [623]
13.4 Operations on Series, [629]
13.5 Power Series, [633]
13.6 Functions Represented by Power Series, [640]
Chapter 14 Taylor and Laurent Series [646]
14.1 Taylor Series, [646]
14.2 Taylor Series of Elementary Functions, [650]
14.3 Practical Methods for Obtaining Power Series, [652]
14.4 Uniform Convergence, [655]
14.5 Laurent Series, [663]
14.6 Behavior of Functions at Infinity, [668]
Chapter 15 Integration by the Method of Residues [671]
15.1 Zeros and Singularities, [671]
15.2 Residues, [675]
15.3 The Residue Theorem, [678]
15.4 Evaluation of Real Integrals, [680]
Chapter 16 Complex Analytic Functions and Potential Theory [689]
16.1 Electrostatic Fields, [689]
16.2 Two-Dimensional Fluid How, [693]
16.3 Special Complex Potentials, [697]
16.4 General Properties of Harmonic Functions, [702]
16.5 Poisson’s Integral Formula, [705]
Chapter 17 Special Functions. Asymptotic Expansions [710]
VIA Gamma and Beta Functions, [710]
17.2 Error Function. Fresnel Integrals. Sine and Cosine Integrals, [716]
17.3 Asymptotic Expansions, [720]
17.4 Further Properties of Asymptotic Expansions, [725]
Chapter 18 Probability and Statistics [732]
18.1 Nature and Purpose of Mathematical Statistics, [732]
18.2 Tabular and Graphical Representation of Samples, [734]
18.3 Sample Mean and Sample Variance, [740]
18.4 Random Experiments, Outcomes, Events, [744]
18.5 Probability, [748]
18.6 Permutations and Combinations, [752]
18.7 Random Variables. Discrete and Continuous Distributions, [756]
18.8 Mean and Variance of a Distribution, [762]
18.9 Binomial, Poisson, and Hypergeometric Distributions, [766]
18.10 Normal Distribution, [770]
18.11 Distributions of Several Random Variables, [777]
18.12 Random Sampling. Random Numbers, [783]
18.13 Estimation of Parameters, [785]
18.14 Confidence Intervals, [790]
18.15 Testing of Hypotheses, Decisions, [799]
18.16 Quality Control, [808]
18.17 Acceptance Sampling, [811]
18.18 Goodness of Fit. x2-Test, [817]
18.19 Nonparametric Tests, [819]
18.20 Pairs of Measurements. Fitting Straight Lines, [822]
18.21 Statistical Tables, [828]
Appendix 1 References [841]
Appendix 2 Answers to Odd-Numbered Problems [845]
Index [882]```
Item type Home library Shelving location Call number Materials specified Status Date due Barcode Course reserves
Libros
Mesa P - Consultar al bibliotecario 00A06 K92-2 (Browse shelf) Available A-3311
Bibliografía: p. 841-844.
CONTENTS --
Introduction Review of Some Topics From Algebra and Calculus [1] --
0.1 Elementary Functions, [1] --
0.2 Partial Derivatives, [8] --
0.3 Second- and Third-Order Determinants, [10] --
0.4 Complex Numbers, [18] --
0.5 Polar Form of Complex Numbers, [22] --
0.6 Some General Remarks About Numerical Computations, [24] --
0.7 Solution of Equations, [26] --
0.8 Approximate Integration, [31] --
Chapter 1 Ordinary Differential Equations of the First Order [39] --
1.1 Basic Concepts and Ideas, [39] --
1.2 Geometrical Considerations. Isoclines, [46] --
1.3 Separable Equations, [49] --
1.4 Equations Reducible to Separable Form, [57] --
1.5 Exact Differential Equations, [59] --
1.6 Integrating Factors, [61] --
1.7 Linear First-Order Differential Equations, [63] --
1.8 Variation of Parameters, [67] --
1.9 Electric Circuits, [69] --
1.10 Families of Curves. Orthogonal Trajectories, [74] --
1.11 Picard’s Iteration Method, [79] --
1.12 Existence and Uniqueness of Solutions, [82] --
1.13 Numerical Methods for Differential Equations of the First Order, [86] --
Chapter 2 Ordinary Linear Differential Equations [93] --
2.1 Homogeneous Linear Equations of the Second Order, [94] --
2.2 Homogeneous Second-Order Equations with Constant Coefficients, [97] --
2.3 General Solution. Fundamental System, [99] --
2.4 Complex Roots of the Characteristic Equation. Initial Value Problem, [102] --
2.5 Double Root of the Characteristic Equation, [106] --
2.6 Free Oscillations, [108] --
2.7 Cauchy Equation, [116] --
2.8 Existence and Uniqueness of Solutions, [117] --
2.9 Homogeneous Linear Equations of Arbitrary Order, [123] --
2.10 Homogeneous Linear Equations of Arbitrary Order with Constant Coefficients, [126] --
2.11 Nonhomogeneous Linear Equations, [128] --
2.12 A Method for Solving Nonhomogeneous Linear Equations, [130] --
2.13 Forced Oscillations. Resonance, [134] --
2.14 Electric Circuits, [140] --
2.15 Complex Method for Obtaining Particular Solutions, [144] --
2.16 General Method for Solving Nonhomogeneous Equations, [147] --
2.17 Numerical Methods for Second-Order Differential Equations, [149] --
Chapter 3 Power Series Solutions of Differential Equations [155] --
3.1 The Power Series Method, [155] --
3.‘2 Theoretical Basis of the Power Series Method, [159] --
3.3 Legendre’s Equation. Legendre Polynomials, [164] --
3.4 Extended Power Series Method. Indicial Equation, [168] --
3.5 Bessel’s Equation. Bessel Functions of the First Kind, [179] --
3.6 Further Properties of Bessel Functions of the First Kind, [185] --
3.7 Bessel Functions of the Second Kind, [187] --
Chapter 4 Laplace Transformation [192] --
4.1 Laplace Transform. Inverse Transform. Linearity, [192] --
4.2 Laplace Transforms of Derivatives and Integrals, [198] --
4.3 Transformation of Ordinary Differential Equations, [201] --
4.4 Partial Fractions, [204] --
4.5 Examples and Applications, [209] --
4.6 Differentiation and Integration of Transforms, [213] --
4.7 Unit Step Function, [216] --
4.8 Shifting on the t-axis, [219] --
4.9 Periodic Functions, [223] --
4.10 Table 17. Some Laplace Transforms, [231] --
Chapter 5 Vector Analysis [235] --
5.1 Scalars and Vectors, [235] --
5.2 Components of a Vector, [237] --
5.3 Vector Addition. Multiplication by Scalars, [239] --
5.4 Scalar Product, [242] --
5.5 Vector Product, [247] --
5.6 Vector Products in Terms of Components, [248] --
5.7 Scalar Triple Product. Linear Dependence of Vectors, [254] --
5.8 Other Repeated Products, [258] --
5.9 Scalar Fields and Vector Fields, [260] --
5.10 Vector Calculus, [263] --
5.11 Curves, [265] --
5.12 Arc Length, [268] --
5.13 Tangent. Curvature and Torsion, [270] --
5.14 Velocity and Acceleration, [274] --
5.15 Chain Rule and Mean Value Theorem for Functions of Several Variables, [278] --
5.16 Directional Derivative. Gradient of a Scalar Field, [282] --
5.17 Transformation of Coordinate Systems and Vector Components, [288] --
5.18 Divergence of a Vector Field, [292] --
5.19 Curl of a Vector Field, [296] --
Chapter 6 Line and Surface Integrals. Integral Theorems [299] --
6.1 Line Integral, [299] --
6.2 Evaluation of Line Integrals, [302] --
6.3 Double Integrals, [306] --
6.4 Transformation of Double Integrals into Line Integrals, [313] --
6.5 Surfaces, [318] --
6.6 Tangent Plane. First Fundamental Form, Area, [321] --
6.7 Surface Integrals, [327] --
6.8 Triple Integrals. Divergence Theorem of Gauss, [332] --
6.9 Consequences and Applications of the Divergence Theorem, [336] --
6.10 Stokes’s Theorem, [342] --
6.11 Consequences and Applications of Stokes’s Theorem, [346] --
6.12 Line Integrals Independent of Path, [347] --
Chapter 7 Matrices and Determinants. Systems of Linear Equations [356] --
7.1 Basic Concepts. Addition of Matrices, [356] --
7.2 Matrix Multiplication, [361] --
7.3 Determinants, [368] --
7.4 Submatrices. Rank, [380] --
7.5 Systems of n Linear Equations in n Unknowns. Cramer’s Rule, [382] --
7.6 Arbitrary Homogeneous Systems of Linear Equations, [386] --
7.7 Arbitrary Nonhomogeneous Systems of Linear Equations, [392] --
7.8 Further Properties of Systems of Linear Equations, [395] --
7.9 Gauss’s Elimination Method, [398] --
7.10 The Inverse of a Matrix, [401] --
7.11 Eigenvalues. Eigenvectors, [406] --
7.12 Bilinear, Quadratic, Hermitian, and Skew-Hermitian Forms, [413] --
7.13 Eigenvalues of Hermitian, Skew-Hermitian, and Unitary Matrices, [417] --
7.14 Bounds for Eigenvalues, [422] --
7.15 Determination of Eigenvalues by Iteration, [426] --
Chapter 8 Fourier Series and Integrals [431] --
8.1 Periodic Functions. Trigonometric Series, [431] --
8.2 Fourier Series. Euler’s Formulas, [434] --
8.3 Even and Odd Functions, [440] --
8.4 Functions Having Arbitrary Period, [444] --
8.5 Half-Range Expansions, [447] --
8.6 Determination of Fourier Coefficients Without Integration, [451] --
8.7 Forced Oscillations, [455] --
8.8 Numerical Methods for Determining Fourier Coefficients. Square Error, [458] --
8.9 Instrumental Methods for Determining Fourier Coefficients, [464] --
8.10 The Fourier Integral, [465] --
8.11 Orthogonal Functions, [473] --
8.12 Sturm-Liouville Problem, [476] --
8.13 Orthogonality of Bessel Functions, [483] --
Chapter 9 Partial Differential Equations [486] --
9.1 Basic Concepts, [486] --
9.2 Vibrating String. One-Dimensional Wave Equation, [488] --
9.3 Separation of Variables (Product Method), [490] --
9.4 D’Alembert’s Solution of the Wave Equation, [499] --
9.5 One-Dimensional Heat Flow, [501] --
9.6 Heat Flow in an Infinite Bar, [506] --
9.7 Vibrating Membrane. Two-Dimensional Wave Equation, [510] --
9.8 Rectangular Membrane, [512] --
9.9 Laplacian in Polar Coordinates, [519] --
9.10 Circular Membrane. Bessel’s Equation, [521] --
9.11 Laplace’s Equation. Potential, [526] --
9.12 Laplace’s Equation in Spherical Coordinates. Legendre’s Equation, [529] --
Chapter 10 Complex Analytic Functions [534] --
10.1 Complex Numbers. Triangle Inequality, [535] --
10.2 Limit. Derivative. Analytic Function, [539] --
10.3 Cauchy*Riemann Equations. Laplace’s Equation, [543] --
10.4 Rational Functions. Root, [547] --
10.5 Exponential Function, [550] --
10.6 Trigonometric and Hyperbolic Functions, [553] --
10.7 Logarithm. General Power, [556] --
Chapter 11 Conformal Mapping [560] --
11.1 Mapping, [560] --
11.2 Conformal Mapping, [564] --
11.3 Linear Transformations, [568] --
11.4 Special Linear Transformations, [572] --
11.5 Mapping by Other Elementary Functions, [577] --
11.6 Riemann Surfaces, [584] --
Chapter 12 Complex Integrals [588] --
12.1 Line Integral in the Complex Plane, [588] --
12.2 Basic Properties of the Complex Line Integral, [594] --
12.3 Cauchy’s Integral Theorem, [595] --
12.4 Evaluation of Line Integrals by Indefinite Integration, [602] --
12.5 Cauchy’s Integral Formula, [605] --
12.6 The Derivatives of an Analytic Function, [607] --
Chapter 13 Sequences and Series [611] --
13.1 Sequences, [611] --
13.2 Series, [618] --
13.3 Tests for Convergence and Divergence of Series, [623] --
13.4 Operations on Series, [629] --
13.5 Power Series, [633] --
13.6 Functions Represented by Power Series, [640] --
Chapter 14 Taylor and Laurent Series [646] --
14.1 Taylor Series, [646] --
14.2 Taylor Series of Elementary Functions, [650] --
14.3 Practical Methods for Obtaining Power Series, [652] --
14.4 Uniform Convergence, [655] --
14.5 Laurent Series, [663] --
14.6 Behavior of Functions at Infinity, [668] --
Chapter 15 Integration by the Method of Residues [671] --
15.1 Zeros and Singularities, [671] --
15.2 Residues, [675] --
15.3 The Residue Theorem, [678] --
15.4 Evaluation of Real Integrals, [680] --
Chapter 16 Complex Analytic Functions and Potential Theory [689] --
16.1 Electrostatic Fields, [689] --
16.2 Two-Dimensional Fluid How, [693] --
16.3 Special Complex Potentials, [697] --
16.4 General Properties of Harmonic Functions, [702] --
16.5 Poisson’s Integral Formula, [705] --
Chapter 17 Special Functions. Asymptotic Expansions [710] --
VIA Gamma and Beta Functions, [710] --
17.2 Error Function. Fresnel Integrals. Sine and Cosine Integrals, [716] --
17.3 Asymptotic Expansions, [720] --
17.4 Further Properties of Asymptotic Expansions, [725] --
Chapter 18 Probability and Statistics [732] --
18.1 Nature and Purpose of Mathematical Statistics, [732] --
18.2 Tabular and Graphical Representation of Samples, [734] --
18.3 Sample Mean and Sample Variance, [740] --
18.4 Random Experiments, Outcomes, Events, [744] --
18.5 Probability, [748] --
18.6 Permutations and Combinations, [752] --
18.7 Random Variables. Discrete and Continuous Distributions, [756] --
18.8 Mean and Variance of a Distribution, [762] --
18.9 Binomial, Poisson, and Hypergeometric Distributions, [766] --
18.10 Normal Distribution, [770] --
18.11 Distributions of Several Random Variables, [777] --
18.12 Random Sampling. Random Numbers, [783] --
18.13 Estimation of Parameters, [785] --
18.14 Confidence Intervals, [790] --
18.15 Testing of Hypotheses, Decisions, [799] --
18.16 Quality Control, [808] --
18.17 Acceptance Sampling, [811] --
18.18 Goodness of Fit. x2-Test, [817] --
18.19 Nonparametric Tests, [819] --
18.20 Pairs of Measurements. Fitting Straight Lines, [822] --
18.21 Statistical Tables, [828] --
Appendix 1 References [841] --
Appendix 2 Answers to Odd-Numbered Problems [845] --
Index [882] --
MR, 35 #7622
There are no comments on this title. | 5,944 | 20,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.73446 |
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Attention all Math Nerds! Or, you know, anyone who loves a good puzzle. This week’s #OneNewThing is for you! I recently came across a fun website called SolveMe Puzzles that has a TON of free games for students (or the nerdy adult) to play. All of the games require players to solve a math-based puzzle.… Read More #OneNewThing: SolveMe Puzzles
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Group projects – you either love them or hate them, right? As a student, I absolutely hated group projects. I was one of the nerdy kids that was always striving for an A and felt like I was put into groups with people that did not care as much as I did. Therefore, I ended… Read More Creating Accountability in Group Projects with Google Forms
# #OneNewThing: Adobe After Effects & Instagram Hacks
Earlier this week, I wrote about my goal to learn one new thing per week and blog about it. It’s my new blogging challenge to encourage me to continue learning and writing. For this week’s #OneNewThing post, I am excited to share that I learned TWO new things! Well, I am sure there is actually… Read More #OneNewThing: Adobe After Effects & Instagram Hacks
# #OneNewThing: A Blogging Challenge
A year and a half ago, I set a goal to blog once a week on an educational topic. These topics have ranged from instructional strategies, lesson plans, educational technology tutorials, and a variety of other resources. I am excited to say that this is a goal that I have achieved, as well as a… Read More #OneNewThing: A Blogging Challenge
# The Problem-Based Question (Part 2)
(Note: This is part two of a post on The Problem Based Question. Click HERE to read part one.) As I mentioned in my last post, I found myself reflecting on the essential questions that I had been using in my classroom. Essential questions have a time and place in the classroom but I wanted… Read More The Problem-Based Question (Part 2)
# Join the BoostEDU Community!
I am excited to announce that the BoostEDU community is growing! 🚀 BoostEDU is now on Facebook and Twitter and is a community where teachers can receive updates on resources and support from teachers all around the world. Interested in joining this community? Click on the links below! In these communities, feel free to share resources to… Read More Join the BoostEDU Community!
# The Problem-Based Question (Part 1)
In the twelve years I spent in the classroom, my students participated in a variety of buzzword worthy creative lesson plans and student-led activities: Project-based learning, Genius Hour, and 20Time. Just to name a few. All of these lesson plans had several common goals, such as learning content-specific standards or allowing students the opportunity to… Read More The Problem-Based Question (Part 1)
# Evaluating HyperDocs with BoostEDU
Last month, I had the great pleasure to work with the HyperDoc team and Lisa Guardino on how to use BoostEDU to evaluate HyperDocs. I cannot express how nerdy excited I was to see the BoostEDU self-assessment in action! In this video, Lisa Guardino, an advanced HyperDoc designer, completes the In-Depth SAMR self-assessment on a HyperDoc that… Read More Evaluating HyperDocs with BoostEDU
# Pi Day: A BreakoutEDU Digital
Happy Pi Day! Well – almost. 🙂 Pi Day is coming up on March 14 and I thought, “What could be a fun way to celeberate Pi Day with students?” Enter…”Pi Day!” BreakoutEDU Digital! The Topics: In this BreakoutEDU, students will learn: About the number π. Trivia facts about the number π. About radius and diameter. How… Read More Pi Day: A BreakoutEDU Digital | 943 | 3,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.951633 |
http://polymathlove.com/algebra-tutorials/where-is-non-real-roots-on-ti-.html | 1,519,075,903,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00605.warc.gz | 275,808,818 | 12,419 | Algebra Tutorials!
Monday 19th of February
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Equ. #2:
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Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
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Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
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Prev Next | 1,145 | 4,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.87158 |
https://www.unitconverters.net/energy/kilocalorie-th-to-watt-hour.htm | 1,721,776,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00779.warc.gz | 910,829,720 | 3,521 | Home / Energy Conversion / Convert Kilocalorie (th) to Watt-hour
# Convert Kilocalorie (th) to Watt-hour
Please provide values below to convert kilocalorie (th) [kcal (th)] to watt-hour [W*h], or vice versa.
From: kilocalorie (th) To: watt-hour
### Kilocalorie (th) to Watt-hour Conversion Table
Kilocalorie (th) [kcal (th)]Watt-hour [W*h]
0.01 kcal (th)0.0116222222 W*h
0.1 kcal (th)0.1162222222 W*h
1 kcal (th)1.1622222222 W*h
2 kcal (th)2.3244444444 W*h
3 kcal (th)3.4866666667 W*h
5 kcal (th)5.8111111111 W*h
10 kcal (th)11.6222222222 W*h
20 kcal (th)23.2444444444 W*h
50 kcal (th)58.1111111111 W*h
100 kcal (th)116.2222222222 W*h
1000 kcal (th)1162.2222222222 W*h
### How to Convert Kilocalorie (th) to Watt-hour
1 kcal (th) = 1.1622222222 W*h
1 W*h = 0.8604206501 kcal (th)
Example: convert 15 kcal (th) to W*h:
15 kcal (th) = 15 × 1.1622222222 W*h = 17.4333333333 W*h | 350 | 883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.467358 |
http://stackmonthly.com/2012/3/php | 1,369,301,068,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703108201/warc/CC-MAIN-20130516111828-00050-ip-10-60-113-184.ec2.internal.warc.gz | 254,508,502 | 25,369 | ## Why is \$a + ++\$a == 2?
If I try this:
``````\$l = 0;
echo \$l + ++\$l;
echo PHP_EOL;
echo \$l;
``````
I get this output:
``````2
1
``````
### Why is that?
I expect to get this as an output:
``````1
1
``````
### My reasoning:
``````\$l = 0; // l === 0
echo \$l + ++\$l; // (0) + (0+1) === 1
echo PHP_EOL;
echo \$l; // l === 1
``````
But why isn't that the output???
All the answers explaining why you get 2 and not 1 are actually wrong. According to the PHP documentation, mixing `+` and `++` in this manner is undefined behavior, so you could get either 1 or 2. Switching to a different version of PHP may change the result you get, and it would be just as valid.
See example 1, which says:
``````// mixing ++ and + produces undefined behavior
\$a = 1;
echo ++\$a + \$a++; // may print 4 or 5
``````
Notes:
1. Operator precedence does not determine the order of evaluation. Operator precedence only determines that the expression `\$l + ++\$l` is parsed as `\$l + (++\$l)`, but doesn't determine if the left or right operand of the `+` operator is evaluated first. If the left operand is evaluated first, the result would be 0+1, and if the right operand is evaluated first, the result would be 1+1.
2. Operator associativity also does not determine order of evaluation. That the `+` operator has left associativity only determines that `\$a+\$b+\$c` is evaluated as `(\$a+\$b)+\$c`. It does not determine in what order a single operator's operands are evaluated.
Also relevant: On this bug report regarding another expression with undefined results, a PHP developer says: "We make no guarantee about the order of evaluation [...], just as C doesn't. Can you point to any place on the documentation where it's stated that the first operand is evaluated first?"
## Behaviour of is_callable on '/'
My colleague and I have encountered some rather odd behavour. Our environments are Ubuntu 11.10, PHP 5.3.6-13ubuntu3.6 with Suhosin-Patch, and Windows 7 PHP 5.3.5.
On our machines, the following code runs as one would expect:
``````<?php
function t() { }
var_dump(is_callable('/'));
``````
With the output:
``````bool(false)
``````
On one of our servers, CentOS release 5.7 (Final), PHP 5.3.8, the same code produces:
``````bool(true)
``````
Without the `t()` function, `is_callable` performs as expected. Note that `is_function` behaves the same as `is_callable` in these tests.
Does anyone have any idea what could be causing this?
Edit:
It seems to only happen when a function named `t` is present, anything else, like `b`, `c` etc, and the output is as expected.
Edit - testing with more characters:
``````<?php
function t() { }
foreach(str_split('/abcdefghijkmnopqrstuvwxyz023456789ABCDEFGHIJKLMNOPQRSTUVXYZ!@#\$%^&*()-_+=`~;:[]{}\\|\'"?.>,<') as \$character) {
if (is_callable(\$character)) var_dump(\$character, is_callable(\$character));
}
``````
Outputs the following on the server:
``````string(1) "/"
bool(true)
string(1) "t"
bool(true)
string(1) "T"
bool(true)
string(1) "_" // gettext
bool(true)
string(1) ":" // With the t() function undefined, this remains callable on the server
bool(true)
``````
On our environments, the output is as expected:
``````string(1) "t"
bool(true)
string(1) "T"
bool(true)
``````
``````<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
function t() { }
\$v = '/'; \$v();
``````
Produces output: `Call to undefined function /()`
As a work around you could try this:
``````\$name = '/';
\$actual = null;
if (is_callable(\$name, false, \$actual) && \$name === \$actual) {
// Method is actually callable
}
``````
## glob() can't find file names with multibyte characters on Windows?
I'm writing a file manager and need to scan directories and deal with renaming files that may have multibyte characters. I'm working on it locally on Windows/Apache PHP 5.3.8, with the following file names in a directory:
• filename.jpg
• имяфайла.jpg
• file件name.jpg
• פילענאַמע.jpg
• 文件名.jpg
Testing on a live UNIX server woked fine. Testing locally on Windows using `glob('./path/*')` returns only the first one, `filename.jpg`.
Using `scandir()`, the correct number of files is returned at least, but I get names like `?????????.jpg` (note: those are regular question marks, not the � character.
I'll end up needing to write a "search" feature to search recursively through the entire tree for filenames matching a pattern or with a certain file extension, and I assumed `glob()` would be the right tool for that, rather than scan all the files and do the pattern matching and array building in the application code. I'm open to alternate suggestions if need be.
Assuming this was a common problem, I immediately searched Google and Stack Overflow and found nothing even related. Is this a Windows issue? PHP shortcoming? What's the solution: is there anything I can do?
Addendum: Not sure how related this is, but `file_exists()` is also returning `FALSE` for these files, passing in the full absolute path (using Notepad++, the php file itself is UTF-8 encoding no BOM). I'm certain the path is correct, as neighboring files without multibyte characters return `TRUE`.
EDIT: `glob()` can find a file named `filename-äöü.jpg`. Previously in my `.htaccess` file, I had `AddDefaultCharset utf-8`, which I didn't consider before. `filename-äöü.jpg` was printing as `filename-���.jpg`. The only effect removing that htaccess line seemed to have was now that file name prints normally.
I've deleted the `.htaccess` file completely, and this is my actual test script in it's entirety (I changed a couple of file names from the original post):
``````print_r(scandir('./uploads/'));
``````
Output locally on Windows:
``````Array
(
[0] => .
[1] => ..
[2] => ??? ?????.jpg
[3] => ???.jpg
[4] => ?????????.jpg
[5] => filename-äöü.jpg
[6] => filename.jpg
[7] => test?test.jpg
)
Array
(
)
``````
Output on remote UNIX server:
``````Array
(
[0] => .
[1] => ..
[2] => filename-äöü.jpg
[3] => filename.jpg
[4] => test이test.jpg
[5] => имя файла.jpg
[6] => פילענאַמע.jpg
[7] => 文件名.jpg
)
Array
(
)
``````
Since this is a different server, regardless of platform - configuration could be different so I'm not sure what to think, and I can't fully pin it on Windows yet (could be my PHP installation, ini settings, or Apache config). Any ideas?
It looks like the glob() function depends on how your copy of PHP was built and whether it was compiled with a unicode-aware WIN32 API (I don't believe the standard builid is.
Excerpt from comments on the article:
Philippe Verdy 2010-09-26 8:53 am
The output from your PHP installation on Windows is easy to explain : you installed the wrong version of PHP, and used a version not compiled to use the Unicode version of the Win32 API. For this reason, the filesystem calls used by PHP will use the legacy "ANSI" API and so the C/C++ libraries linked with this version of PHP will first try to convert yout UTF-8-encoded PHP string into the local "ANSI" codepage selected in the running environment (see the CHCP command before starting PHP from a command line window)
Your version of Windows is MOST PROBABLY NOT responsible of this weird thing. Actually, this is YOUR version of PHP which is not compiled correctly, and that uses the legacy ANSI version of the Win32 API (for compatibility with the legacy 16-bit versions of Windows 95/98 whose filesystem support in the kernel actually had no direct support for Unicode, but used an internal conversion layer to convert Unicode to the local ANSI codepage before using the actual ANSI version of the API).
Recompile PHP using the compiler option to use the UNICODE version of the Win32 API (which should be the default today, and anyway always the default for PHP installed on a server that will NEVER be Windows 95 or Windows 98...)
Then Windows will be able to store UTF-16 encoded filenames (including on FAT32 volumes, even if, on these volumes, it will also generate an aliased short name in 8.3 format using the filesystem's default codepage, something that can be avoided in NTFS volumes).
All what you describe are problems of PHP (incorrect porting to Windows, or incorrect system version identification at runtime) : reread the README files coming with PHP sources explaining the compilation flags. I really think that the makefile on Windows should be able to configure and autodetect if it really needs to use ONLY the ANSI version of the API. If you are compiling it for a server, make sure that the Configure script will effectively detect the full support of the UNICODE version of the Win32 aPI and will use it when compiling PHP and when selecting the runtime libraries to link.
I use PHP on Windows, correctly compiled, and I absolutely DON'T know the problems you cite in your article.
Let's forget now forever these non-UNICODE versions of the Win32 API (which are using inconsistantly the local ANSI codepage for the Windows graphical UI, and the OEM codepage for the filesystem APIs, the DOS/BIOS-compatible APIs, the Console APIs) : these non-Unicode versions of the APIs are even MUCH slower and more costly than the Unicode versions of the APIs, because they are actually translating the codepage to Unicode before using the core Unicode APIs (the situation on Windows NT-based kernels is exactly the reverse from the situation on versions of Windows based on a virtual DOS extender, such as Windows 95/98/ME).
When you don't use the native version of the API, your API call will pass through a thunking layer that will transcode the strings between Unicode and one of the legacy ANSI or CHCP-selected OEM codepages, or the OEM codepage hinted on the filesystem: this requires additional temporary memory allocation within the non-native version of the Win32 API. This takes additional time to convert things before doing the actual work by calling the native API.
In summary: the PHP binary you install on Windows MUST be different depending on if you compiled it for Windows 95/98/SE (or the old Win16s emulation layer for Windows 3.x, which had a very mimimum support of UTF-8, only to support the Unicode subsets of Unicode used by the ANSI and OEM codapges selected when starting Windows from a DOS extender) or if it was compiled for any other version of Windows based on the NT kernel.
The best proof that this is a problem of PHP and not Windows, is that your weird results will NOT occur in other languages like C#, Javascript, VB, Perl, Ruby... PHP has a very bad history in tracking versions (and too many historical source code quirks and wrong assumptions that should be disabled today, and an inconsistant library that has inherited all those quirks initially made in old versions of PHP for old versions of Windows that are even no longer officially supported, by Microsoft or even by PHP itself !).
In other words : RTFM ! Or download and install a binary version of PHP for Windows precompield with the correct settings : I really think that PHP should distribute Windows binaries already compiled by default for the Unicode version of the Win32 API, and using the Unicode version of the C/C++ libraries : internally the PHP code will convert its UTF-8 strings to UTF-16 before calling the Win32 API, and back from UTF-16 to UTF-8 when retrieving Win32 results, instead of converting PHP's internal UTF-8 strings back/to the local OEM codepage (for the filesystem calls) or the local ANSI codepage (for all other Win32 APIs, including the registry or process).
## Node.js vs PHP processing speed
I've been looking into node.js recently and wanted to see a true comparison of processing speed for PHP vs Node.js. In most of the comparisons I had seen, Node trounced Apache/PHP set ups handily. However all of the tests were small 'hello worlds' that would not accurately reflect any webpage's markup.
So I decided to create a basic HTML page with 10,000 hello world paragraph elements. In these tests Node with Cluster was beaten to a pulp by PHP on Nginx utilizing PHP-FPM. So I'm curious if I am misusing Node somehow or if Node is really just this bad at processing power.
Note that my results were equivalent outputting "Hello world\n" with text/plain as the HTML, but I only included the HTML as it's closer to the use case I was investigating.
# My testing box:
• Core i7-2600 Intel CPU (has 8 threads with 4 cores)
• 8GB DDR3 RAM
• Fedora 16 64bit
• Node.js v0.6.13
• Nginx v1.0.13
• PHP v5.3.10 (with PHP-FPM)
# My test scripts:
## Node.js script
``````var cluster = require('cluster');
var http = require('http');
var numCPUs = require('os').cpus().length;
if (cluster.isMaster) {
// Fork workers.
for (var i = 0; i < numCPUs; i++) {
cluster.fork();
}
cluster.on('death', function (worker) {
console.log('worker ' + worker.pid + ' died');
});
}
else {
// Worker processes have an HTTP server.
http.Server(function (req, res) {
for (var i = 0; i < 10000; i++) {
res.write('<p>Hello world</p>\n');
}
res.end('</body>\n</html>');
}).listen(80);
}
``````
This script is adapted from Node.js' documentation at http://nodejs.org/docs/latest/api/cluster.html
## PHP script
``````<?php
for (\$i = 0; \$i < 10000; \$i++) {
echo "<p>Hello world</p>\n";
}
echo "</body>\n</html>";
``````
# My results
## Node.js
``````\$ ab -n 500 -c 20 http://speedtest.dev/
This is ApacheBench, Version 2.3 <\$Revision: 655654 \$>
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking speedtest.dev (be patient)
Completed 100 requests
Completed 200 requests
Completed 300 requests
Completed 400 requests
Completed 500 requests
Finished 500 requests
Server Software:
Server Hostname: speedtest.dev
Server Port: 80
Document Path: /
Document Length: 190070 bytes
Concurrency Level: 20
Time taken for tests: 14.603 seconds
Complete requests: 500
Failed requests: 0
Write errors: 0
Total transferred: 95066500 bytes
HTML transferred: 95035000 bytes
Requests per second: 34.24 [#/sec] (mean)
Time per request: 584.123 [ms] (mean)
Time per request: 29.206 [ms] (mean, across all concurrent requests)
Connection Times (ms)
min mean[+/-sd] median max
Connect: 0 0 0.2 0 2
Processing: 94 547 405.4 424 2516
Waiting: 0 331 399.3 216 2284
Total: 95 547 405.4 424 2516
Percentage of the requests served within a certain time (ms)
50% 424
66% 607
75% 733
80% 813
90% 1084
95% 1325
98% 1843
99% 2062
100% 2516 (longest request)
``````
## PHP/Nginx
``````\$ ab -n 500 -c 20 http://speedtest.dev/test.php
This is ApacheBench, Version 2.3 <\$Revision: 655654 \$>
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking speedtest.dev (be patient)
Completed 100 requests
Completed 200 requests
Completed 300 requests
Completed 400 requests
Completed 500 requests
Finished 500 requests
Server Software: nginx/1.0.13
Server Hostname: speedtest.dev
Server Port: 80
Document Path: /test.php
Document Length: 190070 bytes
Concurrency Level: 20
Time taken for tests: 0.130 seconds
Complete requests: 500
Failed requests: 0
Write errors: 0
Total transferred: 95109000 bytes
HTML transferred: 95035000 bytes
Requests per second: 3849.11 [#/sec] (mean)
Time per request: 5.196 [ms] (mean)
Time per request: 0.260 [ms] (mean, across all concurrent requests)
Connection Times (ms)
min mean[+/-sd] median max
Connect: 0 0 0.2 0 1
Processing: 3 5 0.7 5 7
Waiting: 1 4 0.7 4 7
Total: 3 5 0.7 5 7
Percentage of the requests served within a certain time (ms)
50% 5
66% 5
75% 5
80% 6
90% 6
95% 6
98% 6
99% 6
100% 7 (longest request)
``````
Again what I'm looking for is to find out if I'm doing something wrong with Node.js or if it is really just that slow compared to PHP on Nginx with FPM.
I certainly think Node has a real niche that it could fit well, however with these test results (which I really hope I made a mistake with - as I like the idea of Node) lead me to believe that it is a horrible choice for even a modest processing load when compared to PHP (let alone JVM or various other fast solutions).
As a final note, I also tried running an Apache Bench test against node with `\$ ab -n 20 -c 20 http://speedtest.dev/` and consistently received a total test time of greater than 0.900 seconds.
I was suspicious of the 10k+ `res.write()` calls, and suggested to replace it with a temporary string variable because all of the calls are more expensive since the `http` object is not Javascript but C++, and therefore crosses CPU contexts to execute. My guess was apparently right.
## Buffered res.write()
``````http.Server(function (req, res) {
for (var i = 0; i < 10000; i++) {
buffer += '<p>Hello world</p>\n';
}
buffer += '</body>\n</html>';
res.end(buffer);
}).listen(80);
``````
## Final result
``````\$ ab -n 500 -c 20 http://speedtest.dev/
This is ApacheBench, Version 2.3 <\$Revision: 655654 \$>
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking speedtest.dev (be patient)
Completed 100 requests
Completed 200 requests
Completed 300 requests
Completed 400 requests
Completed 500 requests
Finished 500 requests
Server Software:
Server Hostname: speedtest.dev
Server Port: 80
Document Path: /
Document Length: 190070 bytes
Concurrency Level: 20
Time taken for tests: 0.389 seconds
Complete requests: 500
Failed requests: 0
Write errors: 0
Total transferred: 95066500 bytes
HTML transferred: 95035000 bytes
Requests per second: 1283.91 [#/sec] (mean)
Time per request: 15.577 [ms] (mean)
Time per request: 0.779 [ms] (mean, across all concurrent requests)
Connection Times (ms)
min mean[+/-sd] median max
Connect: 0 0 0.3 0 2
Processing: 1 15 45.3 5 254
Waiting: 1 8 30.7 3 252
Total: 1 15 45.4 5 256
Percentage of the requests served within a certain time (ms)
50% 5
66% 7
75% 8
80% 10
90% 13
95% 25
98% 233
99% 254
100% 256 (longest request)
``````
## PHP inheritance, parent functions using child variables
While reviewing some PHP code I've discovered a strange thing. Here is the simple example illustration of it:
File A.php:
``````<?php
class A{
public function methodA(){
echo \$this->B;
}
}
?>
``````
File B.php:
``````<?php
class B extends A{
public \$B = "It's working!";
}
?>
``````
File test.php:
``````<?php
require_once("A.php");
require_once("B.php");
\$b = new B();
\$b->methodA();
?>
``````
Running test.php prints out "It's working!", but question is why is it working? :) Is this a feature or a bug? Method methodA in class A can also call methods that are in class B which should not work in OOP.
You're only instantiating class `B`. Ignore `A` for the moment, and pretend that `methodA()` is part of class `B`.
When class `B` extends `A`, it gets all of `A`'s functions. `\$this->B` isn't evaluated until the code is running, not prior. Therefore no error occurs, and won't occur as `\$this->B` exists in class `B`.
## How can I automatically test my site for SQL injection attacks, using either a script or program?
I've searched and found a good discussion here on SO, but it is several years old.
What programs are there, or is there a simple script I can run, to find the SQL injection holes in the URLs in my entire site?
Preferably, I'd like to run a script (PHP) or program that crawls my site, bouncing from link to link, attempting to find holes, and upon discovery, stores that URL so I have a list of URLs I need to fix.
Does this exist?
Yes and no. First i'll preface this by saying I'm not just posting links but have done security audits professionally using all of these tools and not as a developer on a project but an external resource. Note that generally sqlserver injection is different than mysql as well.
Free tools like paros proxy [crawls] (previously mentioned),
burpsuite (previously mentioned [crawls] but active attacks requires pro): http://portswigger.net/burp/
sqlninja (sqlserver only) http://sqlninja.sourceforge.net/
websecurify: [crawls] http://www.websecurify.com/
wapiti: [crawls but takes work to set up - can be used specifically for sqli with spider] http://wapiti.sourceforge.net/
nikto: [crawls but not for sqli...]
are great! They can help you identify problems but take a great deal of human analysis due to large amounts of false positives. Commercial tools are available like:
NTOSpider (one of the best [crawls!]) : http://www.ntobjectives.com/software/ntospider
are very expensive but talking to a rep will get you a free copy for a period of time (which I have done with them). They make sorting through results faster by providing validation links in the reports but you STILL need a trained eye and analysis as I have found false positives.
Ultimately the correct answer to this question is: You can use tools to help you identify if there are security (sqli) vulnerabilities but only a trained eye using the tools can validate them. Further only a proper code review and analysis can identify vulnerabilities that an app (even a very good one) may miss.
Tools can help but you need human time and analysis to do this correctly. Proxies and request manglers are the real tools for hitting the app with injection and are done with careful intention of trained testers or those with a curious mind.
## Is using a for-loop on submitted POST data in PHP safe?
I'm always a worry-wart about security in my PHP applications, and I just (potentially) thought of a way a hacker could kill my script. Currently my application takes form data and submits it as an array to a PHP script via AJAX, then loops through this array.
``````foreach(\$_POST['form_data'] as \$field => \$value){
//Do something here.
}
``````
However, what if a hacker were to forge an AJAX request, and repeatedly submit the 'form_data' array with 100000000000 random elements? The loop would have to iterate through each element, possibly causing a DoS (or at least slow down service), correct?
I'm not entirely educated here, so I may have some incorrect assumptions. Thanks for any input!
This will not be an issue: PHP limits the maximum number of POST vars using the `max_input_vars` directive, which defaults to 1000 variables.
This limit is actually enforced to prevent a much more serious type of DOS attack than the one you are thinking about (really, iterating a few thousand array elements is like nothing), namely hash table collision based attacks (often referred to as HashDOS). For more info on that issue see my article Supercolliding a PHP array.
## Dependency injection: should I inject everything or use a service locator for some objects?
I'm currently refactoring my Zend Framework based PHP library from using a service locator to (constructor) dependency injection (DI). I feel that it improves my code a lot, but I'm not sure if I should inject all dependencies. A service locator seems easier for dependencies which are used a lot and are unspecific. I have the following dependencies which I still access using a service locator:
1. A Zend_Translate object (I need to translate messages everywhere).
2. A Zend_Locale object (stores the current language)
3. A Zend_Config object (a lot of things are configurable by ini-file)
4. Instances of utility classes (for array and string manipulation)
If I injected these dependencies, they'd clutter my constructors and distract from the specific dependencies. For testing, I can just set up these dependencies in my service locator before running the tests. The pragmatist in me says I'm doing just fine, but the purist says I should go all the way with DI.
Would you recommend DI for these types of objects or not?
When it comes to the concern about cluttering the constructors, it's most likely a code smell that the classes are violating the Single Responsibility Principle. Constructor Injection is very beneficial here because it make this much more obvious.
Some people also worry about injecting dependencies which are only rarely used, but that's not a problem either. When it comes to creating object graphs, performance is rarely a problem, and even if it is, the Virtual Proxy pattern can fix it.
In short, there's no reason to ever use a Service Locator. There's always a better alternative that involves proper inversion of control.
## Negation of Hex in PHP, funny behavior
Got some weird behavior I was wondering if someone could clear up for me.
Check it out
``````\$hex = 0x80008000;
print_r(decbin(intval(\$hex)) . '<br/>');
print_r(decbin(\$hex));
``````
Outputs
``````10000000000000001000000000000000
10000000000000001000000000000000
``````
As expected.
But
``````\$hex = 0x80008000;
print_r(decbin(~intval(\$hex)) . '<br/>');
print_r(decbin(~\$hex));
``````
Outputs
``````1111111111111110111111111111111
1111111111111111111111111111111
``````
Why is the middle bit not switching when `\$hex` is negated?
Gonna give a shot to my own question here.
Yes this is a 32-bit / 64-bit difference.
In 32-bit systems, a float type has to take up two memory spaces to get the required 64 bits. Php uses double-precision (see http://en.wikipedia.org/wiki/Floating_point#IEEE_754:_floating_point_in_modern_computers)
The \$hex evaluates to a float type. Intval and decbin functions convert this into an int type (1st example above)
In the 2nd example we are using the not bitwise operator BEFORE we use decbin. This flips the bits in the two-memory space double-precision float first, and then is converted to int second. Giving us something different than what we expected.
Indeed, if we put the negate inside of the intval() like so:
``````\$hex = 0x80008000;
print_r(decbin(intval(~\$hex)) . '<br/>');
print_r(decbin(~\$hex));
``````
We get
``````1111111111111111111111111111111
1111111111111111111111111111111
``````
As output.
I'm not good enough to prove this with math yet (which can be figured out with the help of this article http://en.wikipedia.org/wiki/Double_precision). But maybe when I have time later -_-
I think it's very important to learn how numbers are represented in computers so we can understand anomalies like this and not call them bugs.
## Is the PDO Library faster than the native MySQL Functions?
I have read several questions regarding this but I fear they may be out of date as newer versions of the PDO libraries have been released since these questions were answered.
I have written a MySQL class that builds queries and escapes parameters, and then returns results based on the query. Currently this class is using the built-in mysql functions.
I am well aware of the advantages of using the PDO Library, e.g. it is compatible with other databases, stored procedures are easier to execute etc... However, what I would like to know is simply; is using the PDO Library faster then using the mysql built-in functions?
I have just written the equivalent class for MsSQL, so rewriting it to work with all databases would not take me long at all. Is it worth it or is the PDO library slower?
I found PDO in many situation/projects to be even faster than the more native modules.
Mainly because many patterns/building blocks in a "PDO-application" require less php script driven code and more code is executed in the compiled extension and there is a speed penalty when doing things in the script. Simple, synthetic tests without data and error handling often do not cover this part, which is why (amongst other problems like e.g. measuring inaccuracies) I think "10000x SELECT x FROM foo took 10ms longer" conclusions are missing the point more often than not .
I can't provide you with solid benchmarks and the outcome depends on how the surrounding application handles the data but even synthetic tests usually only show differences so negligible that you better spend your time on optimizing your queries, the MySQL server, the network, ... instead of worrying about PDO's raw performance. Let alone security and error handling ...
## How to correctly fork an open-source library?
I'd like to fork an open-source php-library.
It has its own license, in which is written:
You are permitted to use, copy, modify, and distribute the Software and its documentation, with or without modification, for any purpose, provided that the following conditions are met:
I want to add new features in this library, which are written under GPL. Then the whole new product should be under GPL? So I should add both GPL and 'old' license agreemenets? And in every source file I should keep both license copyrights?
What was the original license agreement? Your sentencing makes it confusing about whether the original part was GPL or you want to add GPL to it.
If the original license was GPL, then your new software must also be GPL. There is no way around it unless you get the permission from the author or all authors - if there is more than one. You can still sell your product if it is under GPL, but note that the buyer may 'resell' it with whatever price they find appropriate, including free, as long as license conditions are met. GPL is not a problem when building a website or software that is specific to a client, as long as you are fine with giving the client the rights to modify and republish the software.
But if you want to add GPL stuff to non-GPL project, then consider using LGPL license instead. LGPL allows to release the component itself under a GPL-like license while not requiring the other software to be GPL or LGPL in return.
## System design for matching closest registered store based on zip code?
I have the following problem.
I am creating a restaurant delivery system.
So restaurants choose the zipcodes they want to deliver. So in Boston, they might choose Either all of Boston or Back Bay (a specific area of Boston with several zip codes .... ).
Basically, the restaurant confirms the areas they are willing to serve by ticking boxes that are described as follows:
``````- Cambridge (ZIP CODE)
- Boston (all of Boston)
--- Back Bay (covers zip codes: 02...., 02.., 02..)
--- North Boston (covers zip codes: 02145, 021..., 02..., 02..)
``````
Users type in their zipcodes, and I match them to the areas that Restaurants specified.
What is the best way to design such a system? I don't think I am going in the right direction...
is this only for Boston or would it be global? Will you be looking at exact zipcode matches? What if someone enters a zipcode you don't have but it's within the delivery range. I would recommend using longitude / latitude lookups.
This might be a good place to start: https://developers.google.com/maps/articles/phpsqlsearch
BTW: I'm looking to do something very similar and will most likely use the article referenced above :) thanks for helping me too.
## What OCR options exist beyond Tesseract?
I've used Tesseract a bit and it's results leave much to be desired. I'm currently detecting very small images (35x15, without border, but have tried adding one with imagemagick with no ocr advantage); they range from 2 chars to 5 and are a pretty reliable font, however the characters are variable enough that simply using an image size checksum or such is not going to work.
I actually have tried: http://www.free-ocr.co.uk/ and surprisingly it has 100% accuracy. The problem I have with utilizing it is that I cannot rely on another outside service's reliability for this particular use case. I need to be able to control uptime to a higher degree.
What options exist for OCR besides sticking with Tesseract or doing a complete custom training of it? Also, it would be VERY helpful if this were compatible with Heroku style hosting (at least where I can compile the bins and shove them over).
I have successfully used GOCR in the past for small image OCR. I would say accuracy was around 85%, after getting the grayscale options set properly, on fairly regular fonts. It fails miserably when the fonts get complicated and has trouble with multiline layouts.
Also have a look at Ocropus, which is maintained by Google. Its related to Tesseract, but from what I understand, its OCR engine is different. With just the default models included, it achieves near 99% accuracy on high-quality images, handles layout pretty well and provides HTML output with information concerning formatting and lines. However, in my experience, its accuracy is very low when the image quality is not good enough. That being said, training is relatively simple and you might want to give it a try.
Both of them are easily callable from the command line. GOCR usage is very straightforward; just type `gocr -h` and you should have all the information you need. Ocropus is a bit more tricky; here's a usage example, in Ruby:
``````require 'fileutils'
tmp = 'directory'
file = 'file.png'
`ocropus book2pages #{tmp}/out #{file}`
`ocropus pages2lines #{tmp}/out`
`ocropus lines2fsts #{tmp}/out`
`ocropus buildhtml #{tmp}/out > #{tmp}/output.html`
FileUtils.rm_rf(tmp)
``````
## Retrieving a specific row depending on a date variable?
I have 7 columns that which contain the information of closing times, each for one day. (It goes like VENUE_CLOSE_T_MO, VENUE_CLOSE_T_TU... etc)
How would I, for example choose one of those columns depending on a date variable (\$somevariable) which contains a specific date?
For example, if the date variable was Sunday, March 18 22:00, it would choose column VENUE_CLOSE_T_SU.
Thanks for the help everyone!
EDIT (Solution given by TEEZ that solved the issue)
My Date variable is \$Start.
And this is the code:
``````\$day_name=strtoupper(date('D',\$start));
\$day_name=substr(\$day_name,0,2);
\$selectcolumn='VENUE_CLOSE_T_'.\$day_name;
``````
So in this case \$selectcolumn = VENUE_CLOSE_T_SU
And the echo is then this:
``````\$row[\$selectcolumn]
``````
Thanks for all your help again Teez!
first get day name from variable `(\$somevariable)`
``````\$day_name=strtoupper(date('D',\$somevariable));
``````
then make query like below for getting column according to day in \$somevariable
``````select concat('VENUE_CLOSE_T_',left(\$day_name,2)) as datecolumnname from tableame
``````
EDIT:
OR
you don't need to do this in query if you taking all column in query. just add these lines in php code where you printing data in we page under date column
``````\$day_name=strtoupper(date('D',\$somevariable));
\$day_name=substr(\$day_name,0,2);
\$selectcolumn='venues.VENUE_CLOSE_T_'.\$day_name;
echo \$row[\$selectcolumn];
``````
## What is the best method to make sure two people don't edit the same row on my web app?
I have a PHP/jQuery/AJAX/MySQL app built for managing databases. I want to implement the ability to prevent multiple users from editing the same database row at the same time.
1. What is this called?
2. Do I use a token system and who ever has the token can edit it until they release the token?
3. Do I use a "last edit date/time" to compare you loading the HTML form with the time in the database and if the database is the most resent edit then it warns you?
4. Do I lock the row using database functions?
I'm just not sure which is the best. Assuming between 10 - 15 concurrent users
There are two general approaches-- optimistic and pessimistic locking.
Optimistic locking is generally much easier to implement in a web-based environment because it is fundamentally stateless. It scales much better as well. The downside is that it assumes that your users generally won't be trying to edit the same set of rows at the same time. For most applications, that's a very reasonable assumption but you'd have to verify that your application isn't one of the outliers where users would regularly be stepping on each other's toes. In optimistic locking, you would have some sort of `last_modified_timestamp` column that you would `SELECT` when a user fetched the data and then use in the `WHERE` clause when you go to update the date, i.e.
``````UPDATE table_name
SET col1 = <<new value>>,
col2 = <<new values>>,
last_modified_timestamp = <<new timestamp>>
WHERE primary_key = <<key column>>
``````
If that updates 1 row, you know you were successful. Otherwise, if it updates 0 rows, you know that someone else has modified the data in the interim and you can take some action (generally showing the user the new data and asking them if they want to overwrite but you can adopt other conflict resolution approaches).
Pessimistic locking is more challenging to implement particularly in a web-based application particularly when users can close their browser without logging out or where users may start editing some data and go to lunch before hitting `Submit`. It makes it harder to scale and generally makes the application more difficult to administer. It's really only worth considering if users will regularly try to update the same rows or if updating a row takes a large amount of time for a user so it's worth letting them know up front that someone else has locked the row.
## Dynamically creating date periods using MySQL
I trying to get the Grape count from dates March 1 - 3.
You will notice that on March 2 - there are no grapes inserted..
I'st possible to show a query from dates March 1, 2 and 3 but showing 0 count for March 2
In this image above only shows dates where there are grapes..
Here is mySQL query
``````SELECT `fruitDate` , `fruitName` , COUNT( * )
FROM `tbl_fruits`
WHERE `fruitName` = "Grapes"
GROUP BY `fruitDate
``````
UPDATE 2:
Using this query:
``````SELECT f.fruitDate, f.fruitName, f1.count FROM tbl_fruits f
LEFT JOIN (SELECT fruitDate, COUNT(*) as count from tbl_fruits d WHERE d.fruitName='Grapes' GROUP BY d.fruitDate) as f1 ON (f.fruitDate = f1.fruitDate)
GROUP BY f.fruitDate
``````
I got this result..but its dsplaying diffrent fruit..something wrong with my query?
Remember there is a dynamically (and a bit ugly) solution to creating a date range that does not require creating a table:
``````select aDate from (
select @maxDate - interval (a.a+(10*b.a)+(100*c.a)+(1000*d.a)) day aDate from
(select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all
select 7 union all select 8 union all select 9) a, /*10 day range*/
(select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all
select 7 union all select 8 union all select 9) b, /*100 day range*/
(select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all
select 7 union all select 8 union all select 9) c, /*1000 day range*/
(select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all
select 7 union all select 8 union all select 9) d, /*10000 day range*/
(select @minDate := '2001-01-01', @maxDate := '2002-02-02') e
) f
where aDate between @minDate and @maxDate
``````
Depending on the length of the date range you can reduce the amount of dynamically generated results (10000 days means over 27 years of records each representing one day) by removing tables (d, c, b and a) and removing them from the upper formula. Setting the `@minDate` and `@maxDate` variables will allow you to specify the dates between you want to filter the results.
Edit:
I see you're still looking for a solution. Try this:
``````select c.date, f.fruitName, count(f.fruitName = 'Grapes')
from tbl_calendar c
left join tbl_fruits f
on c.date = f.fruitDate and f.fruitName = 'Grapes'
group by c.date, f.fruitName
``````
If you also want to filter the extra dates from the created table, use this query:
``````select c.date, f.fruitName, count(f.fruitName = 'Grapes')
from tbl_calendar c
left join tbl_fruits f
on c.date = f.fruitDate and f.fruitName = 'Grapes'
group by c.date, f.fruitName
having c.date between
(select min(fruitDate) from tbl_fruits) and
(select max(fruitDate) from tbl_fruits)
``````
## Breaking out of a loop when a condition occurs, and avoiding the usage of its preset db value
Assuming a student take 6 courses in a semester. All those couses have coures units(int), and depending on the score in each course there are points..
`````` so a score >=70 will have a point of 5
<70 and >=60 will have a ponit of 4
``````
and so on. For each course unit and point are multipied together, down the column for each column. Now when the score of a course is not found the grade is 'AR'. Now what i want is for the loops to omit the occurence of AR..i.e not adding the course unit of the course having a grade of 'AR'. But when i run my queries above the units still add to the total course units.
Query4 is used to generate some rows of course_unit and Score
`````` \$query4 = mysql_query("SELECT c.course_unit, m.score
FROM maintable AS m
INNER JOIN students AS s ON
m.matric_no = s.matric_no
INNER JOIN courses AS c ON
m.course_code = c.course_code
WHERE m.matric_no = '".\$matric_no."'
AND m.level = '".\$level."'")
or die (mysql_error());
``````
Query3 is used for the summation of the course_units
`````` \$query3 = mysql_query("SELECT SUM(c.
course_unit) AS 'TOTAL'
FROM maintable AS m
INNER JOIN students AS s ON
m.matric_no = s.matric_no
INNER JOIN courses AS c ON
m.course_code = c.course_code
WHERE m.matric_no = '".\$matric_no."'
AND m.level = '".\$level."'")
or die (mysql_error());
``````
`````` while (\$row8 = mysql_fetch_assoc
(\$query8)) {
if (\$row8['score'] >= 70) {
}
elseif (\$row8['score'] >= 60) {
}elseif (\$row8['score'] >= 50) {
}elseif (\$row8['score'] >= 45) {
}elseif(\$row8['score'] >= 40) {
}elseif(\$row8['score'] >= 0) &&
(\$row8['score'] < 40){
}else{
}
}
``````
`````` \$grade_point = 0;
while (\$row4 = mysql_fetch_assoc(\$query4)) {
if (\$row4['score'] >= 70) {
\$score = 5;
}
elseif (\$row4['score'] >= 60) {
\$score = 4;
}elseif (\$row4['score'] >= 50) {
\$score = 3;
}elseif (\$row4['score'] >= 45) {
\$score = 2;
}elseif(\$row4['score'] >= 40) {
\$score = 1;
}elseif(\$row4['score'] >= 0 AND \$row4['score'] < 40) {
\$score = 0;
}else{
\$score = 0;
}
}
``````
`````` if ( \$grade == 'AR' )
{
continue;
}
``````
But the calculations are still the same. It adds the course_unit value of any course having
``````\$grade == 'AR' .
``````
I'll be most delighted with you answers. Thanks very much.
UPDATE
I have being able to solve the grade piont part by adding
`````` elseif(\$row4['score'] >= 0 AND \$row4['score'] < 40) {
\$score = 0;
}else{
\$score = 0;
}
``````
This sets both the occurences of a score between 0 and 39 to zero and also the default score of <0 (i.e AR) to zero. But it still set's the value of the courses having a grade of AR and a score of -1 to the default respective values of the course_unit.
I think this problem is being cause due to the fact that the course_unit are preloaded from the database. Any help?
``````Courses Table Stucture
=================
course_id
course_code
course_title
course_unit
``````
Is it as simple as adding "AND NOT 'AR'" to your SELECT SUM statement?
Or... if your DB values are coming in as AR, why can't you use PHP is_int() in your loop? That would allow you to still assign 0 for F, and just skip over any non integer values being sent from your DB.
## Detecting emails in a text
I'm trying to create a function that translates every occurrence of a plain text email address in a given string into it's htmlized version.
Let's say I have the following code, where `htmlizeEmails` is the function I'm looking for:
``````\$str = "Send me an email to bob@example.com.";
echo htmlizeEmails(\$str); // Echoes "Send me an email to <a href="mailto:bob@example.com">bob@example.com</a>."
``````
If possible, I'd like this function to use the `filter_var` function to check if the email is valid.
Does anyone know how to do this? Thanks!
### Edit:
Thanks for the answers, I used Shocker's regex to match potential email addresses and then, only if the `filter_var` validates it, it gets replaced.
``````function htmlizeEmails(\$text)
preg_match_all('/([a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\.[a-zA-Z]{2,6})/', \$text, \$potentialEmails, PREG_SET_ORDER);
\$potentialEmailsCount = count(\$potentialEmails);
for (\$i = 0; \$i < \$potentialEmailsCount; \$i++) {
if (filter_var(\$potentialEmails[\$i][0], FILTER_VALIDATE_EMAIL)) {
\$text = str_replace(\$potentialEmails[\$i][0], '<a href="mailto:' . \$potentialEmails[\$i][0] .'">' . \$potentialEmails[\$i][0] .'</a>', \$text);
}
}
}
``````
``````\$str = preg_replace('/([a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\.[a-zA-Z]{2,6})/', '<a href="mailto:\$1">\$1</a>', \$str);
``````
where `([a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\.[a-zA-Z]{2,6})` is the regular expression used for detecting an email address (this is a general example, email addresses may be more complicated than this and not all addresses may be covered, but finding the perfect regex for emails is up to you)
## Securely send a Plain Text password?
I'm working on an application for iOS which will have the user fill out their password. The password will then be posted to a PHP page on my site using either POST or GET. (It must be plaintext because it is used in a script.)
Besides HTTPS, is there any way to secure the password? Encrypt it in Obj-C and then decrypt it in PHP?
NOTE: The username is not sent... only the password is posted to the server.
EDIT: To clarify, David Stratton is correct... I'm trying to prevent malicious sniffers in public locations from simply reading clear text passwords as they are posted to the server.
Challenge response outline
Lets assume you have one-way hash function `abc` (in practice use `md5` or `sha1`).
The password you store in your database is `abc(password + salt)` (store the `salt` separately)
The server generates a random challenge `challenge` and sends it to the client (with the `salt`) and calculates the expected response: `abc(challenge + abc(password + salt))`
The client then calculates: `abc(user_password + salt)` and applies the `challenge` to get `abc(challenge + abc(user_password + salt))`, that is sent to the server and the server can easily verify validity.
This is secure because:
• The password is never sent in plaintext, or stored in plaintext
• The hash value that is sent changes every time (mitigates replay attack)
There are some issues:
How do you know what salt to send? Well, I've never really found a solution for this, but using a deterministic algorithm to turn a username into a salt solves this problem. If the algorithm isn't deterministic an attacker could potentially figure out which username exists and which do not. This does require you to have a username though. Alternatively you could just have a static salt, but I don't know enough about cryptography to assess the quality of that implementation.
## How do facebook update content when somebody has posted something
I'm really interested on how facebook loads only content when someone else has posted something. The only thing that I can think of is using something like the one below to constantly update the page without reloading the page.
``````setInterval(ajax_stuff, 1000);
``````
I was watching the console and indeed the request occurs and another new content is added to the page.
I want to be enlightened on how is this done. It would really be awesome if I can use this on a project. I mean doing setInterval every second really consumes much resource. Making a request only when its needed would be the best way to do things. Specifically I want to use it on this project:
https://github.com/anchetaWern/ChatRo
It's basically just a chat box, currently it still uses the setInterval(). I want to update only the content when someone else on the chat session has actually entered something.
I cannot speak directly to how FaceBook does this, but in general, you should be looking at WebSockets.
WebSockets allow the JavaScript on your page to maintain an open connection with a server whereby you can push data out in near-realtime to all of the clients connected to the server.
Take a look at http://pusher.com | 12,189 | 48,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-20 | latest | en | 0.894317 |
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