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https://www.mechamath.com/precalculus/equation-of-the-circumference-with-center-outside-the-origin/ | 1,653,090,990,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00588.warc.gz | 1,033,195,307 | 41,881 | Select Page
Circles are formed by a set of points that are located at the same distance from a fixed point. The fixed point is called the center of the circumference and the distance between the points is called the radius. The equation of a circumference with center outside the origin is found using the equation of a circumference that has a center at the origin and applying vertical and horizontal translations.
Here, we will learn to find equations for these types of circumferences. Then, we will look at some practice problems.
##### PRECALCULUS
Relevant for
Finding the equation of the circumference centered outside the origin.
See examples
##### PRECALCULUS
Relevant for
Finding the equation of the circumference centered outside the origin.
See examples
## Determining the equation of a circumference with center outside the origin
To find the equation of a circumference centered outside the origin, we use the equation of a circumference that has a center at the origin, and then we apply vertical and horizontal translations. Recall that the equation of a circumference with the center at the origin is .
This equation was derived using the Pythagorean theorem. If we rewrite this equation using the center, we would have .
Now, let’s consider the following circumference:
We can see that this circle has its center located at the point (h, k). Therefore, if we use the equation of the circumference with this center, we have:
This is the equation of the circle centered outside the origin, where r is the radius, (x, y) is any point that is located on the circle, and (h, k) are the coordinates of the center of the circle.
## Circumference center outside the origin – Examples with answers
The following exercises facilitate the understanding of the application of the equation of the circumference with the center outside the origin. Try to solve the exercises yourself before looking at the answer.
### EXAMPLE 1
Find the radius and center of the circle .
The general equation for the circumference is , where (h, k) is the center and r is the radius. Comparing this equation with the given equation, we have:
Therefore, the radius of the circumference is 3 and the center is (2, 3).
### EXAMPLE 2
What is the radius and center of a circle that has the equation ?
We recall that the general equation of the circumference is . This equation tells us that (h, k) is the center and r is the radius of the circumference. Therefore, we can determine the following:
Therefore, the radius of the circle is 4 and the center is (-4, 5).
### EXAMPLE 3
Find the equation of the circumference that has the center at the point (2, -3) and has a radius of 4.
We plug the values , and into the general equation . Therefore, we have:
### EXAMPLE 4
Find the equation of the circumference that has the center at the point (-1, 2) and in which the point (2, 6) is part of the circumference.
In this case, we know the values . However, we do not know the radius of the circumference. For this we can use the formula for the distance between two points since this distance represents the radius. Therefore, we have:
The radius of the circumference is . We use the general equation of the circumference with these values:
## Circumference with center outside the origin – Practice problems
Solve the following problems using what you have learned about the equation of the circumference with the center outside the origin. You can look at the solved examples above in case you need help. | 741 | 3,519 | {"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} | 4.84375 | 5 | CC-MAIN-2022-21 | longest | en | 0.929574 |
http://www.ehow.com/about_5470825_careers-combine-biology-mathematics.html | 1,484,637,330,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00270-ip-10-171-10-70.ec2.internal.warc.gz | 439,494,973 | 21,564 | # Careers That Combine Biology & Mathematics
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Biology and mathematics seems to be mutually exclusive fields but in reality the two disciplines have crossected throughout and have become intertwined even more with the avdvent of computers in biological research. These are five fields that combine biology and mathematics on a regular basis.
An agricultural sales manager, Galynn Beer, presented some of the ways that the agricultural industry depends on math to meet business goals. Examples where math and biology would intersect are soil analysis in which the percentages of certain elements like nitrogen are measured to determine how well a crop would grow. Also, math is used in analyzing fertilizer blends before being used as well as acre shaping and projections of crops.
Biomechanics is the study of how things move in biological processes, such as how far your leg can bend back and forth. Biomechanics are used by podiatrists and orthopedic surgeons to determine treatment options according to course in biomechanics taught at the University of Ottawa. Physics and biology play a role in biomechanics but quite a bit of math is used. To be successful in biomechanics, a strong knowledge of vectors and angles is needed. In addition to trigonometry, linear algebra and vector calculus are used in the field.
This field uses statistics to measure biological and medical interactions well as studies in research. Biostatistics is prevalent in clinical research to measure if a certain drug or device is successful in treating a randomized patient population. The type of math would fall under statistics, and most biology degrees require at least a course in biostatistics. Advanced degrees in biostatistics are available as well.
As its name suggests, biochemistry studies the interactions of biology and chemistry. However, even the biochemist uses math in routine tasks such as titrations of acid and base illustrated by Curtright et al, logarithms in determining how strong an acid or base is. Also, biochemists use math to interpret data from research results. For a degree in biochemistry, most students would need math up to the level of calculus.
The field of bioinformatics is another field where the merging of biology and math can happen. Bioinformatics is the use of computers to solve biological problems. Specifically, mathematic calculations are used to reach conclusions on various biological mysteries. Nair Achuthsankar, of the Computer Society of India, illustrates that the study of bioinformatics has the most relevance in the discipline of molecular biology, using math to explain different the phenomena of DNA. The type of math used for bioinformatics would fit under the umbrella of applied mathematics.
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Is DIY in your DNA? Become part of our maker community. | 567 | 2,934 | {"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-2017-04 | latest | en | 0.950739 |
https://nuspaceaudio.com/2017/02/15/geometric-audio-4-taxi-cab-distance-bounds-on-integer-lattice-maps/?like_comment=25&_wpnonce=28499d71ed | 1,632,664,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00613.warc.gz | 464,270,261 | 28,531 | # Geometric Audio 4: Taxi-Cab Distance Bounds on Integer Lattice Maps
Recall from the introductory post, the taxi-cab distance metric on the integer lattice map relates the max-order of image-sources to the lattice volume or the total number of image-sources contained within a $k$ radius $L_1$ “ball”. This quantity is useful for determining the total costs of processing individual image-sources in cases where $k$ can be known or estimated beforehand (e.g. $k$ selected according to an estimate of a room’s RT60 using Sabine’s Equation).
In $D$ dimensional space, the $L_1$ norm is given by $||\nu||_1 = \sum_{d=1}^D |\nu_d|$ where the lattice volume is defined by the number of integer lattice coordinates $||\nu||_1 \leq k$. The visual equal-distance curve resembles a diamond in $D=2$ and the growth rate of the lattice volume seems obvious if we introduce a lattice-shell term given by the number of integer coordinates that satisfies $||\nu||_1 = k$ (see the animation below). Such ease is not the case for dimensions $D>2$.
We start with a counting argument that relates the lattice shell and lattice volumes across dimensions via a recurrence relation:
The lattice shell relation $C(k,D)$ follows from observation that all lattice coordinates on the boundary can be formed by augmenting $||\hat{\nu} ||_1 \leq k, \hat{\nu} \in Z^{D-1}$ with $\tilde{\nu}_D = k-||\hat{\nu}||_1$ and $||\tilde{\nu} ||_1 \leq k-1, \tilde{\nu} \in Z^{D-1}$ with $\tilde{\nu}_D = -(k-||\tilde{\nu}||_1)$. The lattice volume relation $S(k,D)$ follows from the definition of a lattice shell which through substitution can be re-written as self-referential difference equation between successive dimensions. The counting solution is practical for small $k$ and $D$ but can be improved by the following ansatz.
Ansatz: Suppose that the lattice volume can be expressed as a polynomial equation given by $S(k, D) = 1 + \sum_{d=1}^D P_{d,D} k^d$. Substitution into the difference equation gives
where $P_{d,D}$ are unknown polynomial coefficients of the powers of radius $k$. Using the binomial theorem, the powers of $k$ rearranged as to form a generalized square matrix-system given by
where upper triangular matrices $B, \bar{B}$ contain the binomial coefficients and $P_D$ is the vector of unknown polynomial coefficients to be solved for in terms of the polynomial coefficients $P_{D-1}$ in the preceding dimension. Carrying the recursion out from the base case $S(k, D=1) = 1+2k$, the higher-dimensional lattice volumes are given by
where by inspection, the highest order coefficients decrease towards $0$ as $D$ increases. This is expected as the majority of the volume moves towards the origin as $D$ grows. For some context, we can compare the lattice volumes bounded under larger p-norms such as $L_2$ or Euclidean (see post on Gauss circle problem, volume appx. hypersphere) and the $L_{\infty}$ norm given by $||\nu||_{\infty} = \max \{ |\nu_1|, \hdots, |\nu_D|\}$ (lattice volume trivially $(2k+1)^D$). By inspection, the lattice volume gap between $L_1$ and $L_2$ grows wider as the leading polynomial terms of $L_1$ decay to $0$ (see plot below).
This concludes our four-part foray into image-source models within high-dimensional integer lattice maps. If any new or interesting results come up or any errors found, I will update accordingly.
Notes: Animations generated in GeoGebra, equations and figures were from my draft paper.
## 3 thoughts on “Geometric Audio 4: Taxi-Cab Distance Bounds on Integer Lattice Maps”
1. […] to process individual image-sources within DSP pipelines (e.g. updating a tapped delay-line). See post. Related: linear algebra and dynamic […]
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2. […] With the distance bound and Gauss circle problem out of the way, we will continue our investigation of lattice volume bounds in terms of the or taxi-cab distance in the next post. […]
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3. […] direct computation in these spaces is expensive but some clever maths [see tutorial (parts 1, 2, 3, 4)] reduced the asymptotic costs to the point of practical use (e.g. a Reverb plugin). Parameterizing […]
Like | 1,098 | 4,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 39, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.786109 |
https://caicedoteaching.wordpress.com/2011/09/19/414514-metric-spaces/ | 1,521,759,835,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00219.warc.gz | 542,860,525 | 23,995 | ## 414/514 – Metric spaces
This is homework 2, due Monday September 26 at the beginning of lecture.
Let $(X,d)$ be a metric space.
• Show that if $d_1:X\times X\to{\mathbb R}$ is defined by $\displaystyle d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$, then $d_1$ is also a metric on $X$.
• Show that if $U$ is open in $(X,d)$ then it is open in $(X,d_1)$, and viceversa.
Recall that $U$ is open iff it is a union of open balls. Use this to explain why it suffices to show that if $U$ is open in $(X,d)$ then for any $x\in U$ there is an $\epsilon>0$ such that
$B_\epsilon^{d_1}(x):=\{y\mid d_1(x,y)<\epsilon\}\subseteq U$,
and similarly, if $V$ is open in $(X,d_1)$ then for any $z\in V$ there is a $\delta>0$ such that
$B_\delta^d(z):=\{w\mid d(z,w)<\delta\}\subseteq V$.
In turn, explain why to show this it suffices to prove that for any $x\in X$ and any $\eta>0$ there is a $\rho>0$ such that
$B^{d_1}_\eta(x)\supseteq B^d_\rho(x)$
and, similarly, for any $\tau>0$ there is a $\mu>0$ such that
$B_\tau^d(x)\supseteq B^{d_1}_\mu(x)$.
Finally, prove this by showing that we can take $\rho=\epsilon$ (no matter what $x$ is) and similarly, find an appropriate $\mu$ that works for $\tau$ (again, independently of $x$).
• Illustrate the above in ${\mathbb R}^2$ as accurately as possible.
• Suppose that a sequence $(x_n)_{n\in{\mathbb N}}$ converges to $x$ in $(X,d)$ and to $x'$ in $(X,d_1)$. Show that $x=x'$.
• Is it true that a sequence $(x_n)_{n\in{\mathbb N}}$ is Cauchy in $(X,d)$ iff it is Cauchy in $(X,d_1)$? (Give a proof or else exhibit a counterexample, with a proof that it is indeed a counterexample.)
• $(*)$ Show that any dense $G_\delta$ subset of ${\mathbb R}$ has the same size as ${\mathbb R}$. | 602 | 1,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 45, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.86084 |
https://nl.pinterest.com/hariveeraghattam/maths-worksheets-for-kids/ | 1,618,988,211,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00366.warc.gz | 491,920,985 | 56,210 | # Maths Worksheets for Kids
## Collection by Hari • Last updated 8 weeks ago
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Fun learning online worksheets for pre-k, online math printable worksheets
These kindergarten worksheets are dynamically created which allow you to select different variables to customize for your needs.
Download and print Turtle Diary's Identify Rectangle Shaped Objects worksheet. Our large collection of math worksheets are a great study tool for all ages.
These kindergarten worksheets are dynamically created which allow you to select different variables to customize for your needs.
The Two-Digit Subtraction with No Regrouping -- 49 Questions (A) Math Worksheet from the Subtraction Worksheets Page at Math-Drills.com.
Challenge your kid to fill in the missing numbers from 1 to 50. This is an essential skill for understanding numbers beyond just counting them from memory.
eine Arbeitsblattsammlung zum Kopfrechnen wird auch zu den Minusaufgaben entstehen (wie immer zur Auswahl gedacht) LG Gille Schrift: Grundschrift Will Software Bild: Joa Pabst Rechenblattgenerator hier eine Ansicht und hier der Link
This worksheet serves up plenty of time practice for kids who need that extra nudge.
Time worksheets for learning to tell time. Produce clock faces for lesson plans or use for extra practice.
The Two-Digit Subtraction with Some Regrouping -- 100 Questions (A) Math Worksheet from the Subtraction Worksheets Page at Math-Drills.com.
hier noch die Arbeitsblätter mit Plus- und Minusaufgaben und ich werde schauen, wie ich sie noch anpassen muss.... Schrift: Grundschrift Will Software Bild: Joa Pabst Rechenblattgenerator LG Gille hier eine Ansicht und hier der Link
The 2-Digit Minus 2-Digit Subtraction (A) Math Worksheet from the Subtraction Worksheets Page at Math-Drills.com.
The Two-Digit Addition -- No Regrouping -- 100 Questions (B) Math Worksheet from the Addition Worksheets Page at Math-Drills.com.
Olá, pessoal! Trouxemos aqui 60 atividades de matemática para crianças do 2º Ano do Ensino Fundamental para baixar e
Here is our range of Subtraction Facts to 20 Worksheets that will help your child learn their subtraction facts.
The 100 Two-Digit Addition and Subtraction Questions with Sums/Minuends to 99 (A) Math Worksheet from the Mixed Operations Worksheets Page at Math-Drills.com. | 526 | 2,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.381625 |
https://www.physicsforums.com/threads/trig-question-eq-in-the-form-y-sin-k-x.349805/ | 1,532,217,254,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00297.warc.gz | 976,480,543 | 12,866 | # Homework Help: Trig question eq in the form y = sin k x
1. Oct 28, 2009
### Matt1234
1. The problem statement, all variables and given/known data
http://img62.imageshack.us/img62/7677/30639229.jpg [Broken]
2. Relevant equations
y = sin k x
3. The attempt at a solution
Looking for help with part 12b in particular so i can try 13c
Im not quite sure how the coefficent of x (K) in the function y=sin kx relate to peroid and frequency. I know that if k is less then 1 the graph is stretched horizontally and if k > 1 its compressed horizontally but i cant come up with the equation.
Last edited by a moderator: May 4, 2017
2. Oct 28, 2009
### rock.freak667
y=Acos(ωt)
OR
y=Asin(ωt)
ω is the angular frequency. (In your question, A=1)
3. Oct 28, 2009
### Matt1234
i got it the formula i could not recall was that period = 2Pi / K (constant in my question) | 267 | 872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | latest | en | 0.891182 |
https://tnomlaccm.com/lakes-creek/interest-rate-parity-theory-example.php | 1,638,776,110,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00058.warc.gz | 639,440,973 | 12,463 | # Lakes Creek Interest Rate Parity Theory Example
## Interest Rate Parity and Purchasing Power Parity SlideShare
### CHAPTER 5 INTERNATIONAL PARITY CONDITIONS INTEREST
What is Interest Rate Parity? Financial Web - finweb.com. Covered and uncovered interest parities should not be confused with each other. For example, suppose you have PPP is purchasing power parity exchange rate,, The Interest Rate Parity When interest rates are highly manipulated by the governments, USD serves as a good example in this matter..
### (3.1) FX Theory Interest Rate Parity snbchf.com
Exchange Rates and Interest Parity SSCC. from Covered Interest Rate Parity . cause their prices to deviate from levels implied by CIP.the For example, theory. 1 For instance, during, Purchasing power parity (PPP) is a theory of exchange rate determination This contrast with the interest rate parity theory which assumes For example, US.
Theory One: Interest Rate Parity (IRP) According to IRP, at equilibrium, the forward rate of a foreign currency will differ (in %) from the current 1/11/2018В В· Perhaps the most famous example of purchasing power parity was given This indexed exchange rate would then What Is Purchasing Power Parity Theory?
Interest rate parity is a financial theory that connects forward exchange rates, spot exchange rates, and nations' individual interest rates. It is the theory with The basic concept of Purchasing Power Parity theory or PPP, For example, if the interest rate in the U.S. is 5 percent and the interest rate in Japan is 8
How to calcuate forward exchange rate or interest rate parity? using price interest rate divided by base interest rate in its examples. Theory; Worldbuilding 2! 1 Introduction Foreign exchange trading gave rise to the theory of interest rate parity, which relates the difference between foreign and domestic interest rates
16/07/2014В В· Hi John, In the interest rate parity formula which is denoted by Spot rate x (1+ih)/(1+if). Could you please explain me how the domestic currency depreciates when the Interest Rate Parity and Purchasing Power Parity 1. Interest Rate Parity & Purchasing power parity Presented by Danish Hasan Ramiz Junaid Zamir
The purchasing-power parity (PPP) theory states that the amount of purchasing between nominal prices and nominal exchange rates so that, for example, 19/04/2016В В· This course is presented in simple language with examples. Interest Rate Parity Theory Case study 1 - Duration: Interest rate parity
The basic concept of Purchasing Power Parity theory or PPP, For example, if the interest rate in the U.S. is 5 percent and the interest rate in Japan is 8 Covered and uncovered interest parities should not be confused with each other. For example, suppose you have PPP is purchasing power parity exchange rate,
Exchange Rate Theories: Purchasing Power Parity 17 Exchange Rate Theories: Purchasing Power Parity Purchasing power parity (PPP) is a theory of exchange rate Understand how inflation and the exchange rate between 2 countries are linked through Purchasing Power Parity (PPP). Example scenarios are provided.
The international Fisher effect is an extension of the Fisher effect hypothesized by American Relation to interest rate parity To check this example, Exchange Rate Theories: Purchasing Power Parity 17 Exchange Rate Theories: Purchasing Power Parity Purchasing power parity (PPP) is a theory of exchange rate
The purchasing-power parity (PPP) theory states that the amount of purchasing between nominal prices and nominal exchange rates so that, for example, Interest rate parity is a no-arbitrage condition representing an equilibrium state under which investors will i \$ is the interest rate in one country (for example
16/07/2014В В· Hi John, In the interest rate parity formula which is denoted by Spot rate x (1+ih)/(1+if). Could you please explain me how the domestic currency depreciates when the Interest Rate Parity theory; A simple example may be a situation, where interest rates in the United Kingdom are, say, 2%, while interest rates in Japan are,
Purchasing power parity (PPP) is a theory of exchange rate determination This contrast with the interest rate parity theory which assumes For example, US Interest Rate Parity theory; A simple example may be a situation, where interest rates in the United Kingdom are, say, 2%, while interest rates in Japan are,
The Interest Rate Parity When interest rates are highly manipulated by the governments, USD serves as a good example in this matter. Interest rate parity connects the interest rates, spot exchange rates and forward exchange rates in a single comparison. The theory is that the differential between
from Covered Interest Rate Parity . cause their prices to deviate from levels implied by CIP.the For example, theory. 1 For instance, during CHAPTER 5 INTERNATIONAL PARITY CONDITIONS: INTEREST RATE It then develops the theory and reviews the empirical evidence of the Interest Rate Parity:
Interest Rate Parity and Purchasing Power Parity 1. Interest Rate Parity & Purchasing power parity Presented by Danish Hasan Ramiz Junaid Zamir The most ideal example of risk-free arbitrage would be buying a security on one exchange and simultaneously selling it on the other for a profit.
... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53 Interest rate parity is a financial theory that connects forward exchange rates, spot exchange rates, and nations' individual interest rates. It is the theory with
Interest Rate Parity and Purchasing Power Parity 1. Interest Rate Parity & Purchasing power parity Presented by Danish Hasan Ramiz Junaid Zamir The purchasing-power parity (PPP) theory states that the amount of purchasing between nominal prices and nominal exchange rates so that, for example,
Purchasing power parity (PPP) is a theory of exchange rate determination This contrast with the interest rate parity theory which assumes For example, US Learn the basics of forward exchange rates and hedging strategies to understand interest rate parity a higher interest rate. In the example theory had held
5.1 Overview of Interest Rate Parity. For example, back in 1997, short-term interest rates were 60 percent per year in Russia and 75 Interest Rate Parity Theory. 2! 1 Introduction Foreign exchange trading gave rise to the theory of interest rate parity, which relates the difference between foreign and domestic interest rates
Credit Migration and Covered Interest Rate Parity Credit Migration and Covered Interest Rate Parity a cornerstone of п¬Ѓnance theory stating that assets with from Covered Interest Rate Parity . cause their prices to deviate from levels implied by CIP.the For example, theory. 1 For instance, during
16/07/2014В В· Hi John, In the interest rate parity formula which is denoted by Spot rate x (1+ih)/(1+if). Could you please explain me how the domestic currency depreciates when the Purchasing power parity (PPP) is a theory of exchange rate determination This contrast with the interest rate parity theory which assumes For example, US
### Interest Rate Parity International Economics
Interest Rate Parity Essay Example for Free. 19/04/2016 · This course is presented in simple language with examples. Interest Rate Parity Theory Case study 1 - Duration: Interest rate parity, The parity conditions Interest parity conditions are no-arbitrage profit (and in this example, This condition is called “covered interest rate parity.
Three Types of Purchasing Power Parity Bizfluent. Interest rate parity is a theory that suggests a strong relationship between interest rates and the movement of currency values. In fact, you can predict what a, Theory: Expected Interest rate parity 1 1 t t t t t no risk –eg, violation of covered interest rate parity of expected interest rate parity • Examples..
### Evidence on Financial Globalization and Crises Interest
Evidence on Financial Globalization and Crises Interest. Learn the basics of forward exchange rates and hedging strategies to understand interest rate parity a higher interest rate. In the example theory had held https://en.m.wikipedia.org/wiki/Purchasing_power_parity Monetary Policy and the Uncovered Interest Rate Parity Puzzle For example, when the Fed sharply lowered rates in 2001 like much of the modern theory.
• MAN 322 PROJECT I
• Interest Rate Parity (IRP) Theory (With Criticisms
• Evidence on Financial Globalization and Crises: Interest Rate Parity by the theory of interest rate parity was formalized by John 1 For example, in An example. Let's say the current exchange rate between the US dollar and theUK pound is \$1.60 / £. Interest Rate Parity Theory (IRPT)
What is Interest Rate Parity? Interest Rate Parity (IRP) is a theory in which the differential between the interest rates of two countries remains equal to the ... Currency of Payment (Managing Transaction Risks) This is an example of a floating Under the Interest Rate Parity theory of exchange rate
16/07/2014В В· Hi John, In the interest rate parity formula which is denoted by Spot rate x (1+ih)/(1+if). Could you please explain me how the domestic currency depreciates when the The Interest Rate Parity (IPR) theory is used to analyze the relationship between at the spot rate and a corresponding forward (future) rate of currencies.
Interest rate parity is a no-arbitrage condition representing an equilibrium state under which investors will i \$ is the interest rate in one country (for example Interest rate parity (IRP) A condition in which the rates of return on comparable assets in two countries are equal. is a theory used to explain the value and
CHAPTER 5 INTERNATIONAL PARITY CONDITIONS: INTEREST RATE It then develops the theory and reviews the empirical evidence of the Interest Rate Parity: Costas Arkolakis teaching fellow: Federico Esposito Economics 407, Nominal and Real Exchange Rate A Theory of Purchasing Power Parity Interest Rate
Estimation of Interest Rate Parity Theory (Recitation: Due: March 10th, 2006) In this project, you will do econometric analysis of (covered) For example, an The interest rate parity theory is a powerful idea with real implications. This theory argues that the difference between the risk free interest rates offered for
What is Interest Rate Parity? Interest Rate Parity (IRP) is a theory in which the differential between the interest rates of two countries remains equal to the Costas Arkolakis teaching fellow: Federico Esposito Economics 407, Nominal and Real Exchange Rate A Theory of Purchasing Power Parity Interest Rate
2! 1 Introduction Foreign exchange trading gave rise to the theory of interest rate parity, which relates the difference between foreign and domestic interest rates Learn the basics of forward exchange rates and hedging strategies to understand interest rate parity a higher interest rate. In the example theory had held
International Finance Theory and Policy Numerical Examples Using the Rate this page provides the interest rate parity condition when interest is Purchasing power parity (PPP) is a neoclassical economic theory that states that the exchange rate between two countries is equal to the ratio of the currencies
19/04/2016В В· This course is presented in simple language with examples. Interest Rate Parity Theory Case study 1 - Duration: Interest rate parity As Palkesh has rightly mentioned that we get to see a net effect of all the theories in reality. However, if I were to specifically talk about the Interest rate
CHAPTER 5 INTERNATIONAL PARITY CONDITIONS: INTEREST RATE It then develops the theory and reviews the empirical evidence of the Interest Rate Parity: Understand how inflation and the exchange rate between 2 countries are linked through Purchasing Power Parity (PPP). Example scenarios are provided.
## Interest Rate Parity (IRP) Theory of Exchange Rate
Which theory will you use to predict future foreign. can be called the deviation from uncovered interest parity,2 the expected excess return, interest rates, real and nominal, and expected future inflation., How to calcuate forward exchange rate or interest rate parity? using price interest rate divided by base interest rate in its examples. Theory; Worldbuilding.
### Overview of Interest Rate Parity GitHub Pages
CHAPTER 5 INTERNATIONAL PARITY CONDITIONS INTEREST. As Palkesh has rightly mentioned that we get to see a net effect of all the theories in reality. However, if I were to specifically talk about the Interest rate, Learn the basics of forward exchange rates and hedging strategies to understand interest rate parity a higher interest rate. In the example theory had held.
... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53 from Covered Interest Rate Parity . cause their prices to deviate from levels implied by CIP.the For example, theory. 1 For instance, during
The parity conditions Interest parity conditions are no-arbitrage profit (and in this example, This condition is called “covered interest rate parity Theory: Expected Interest rate parity 1 1 t t t t t no risk –eg, violation of covered interest rate parity of expected interest rate parity • Examples.
The purchasing-power parity (PPP) theory states that the amount of purchasing between nominal prices and nominal exchange rates so that, for example, 1/11/2018В В· Perhaps the most famous example of purchasing power parity was given This indexed exchange rate would then What Is Purchasing Power Parity Theory?
... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53 Interest rate parity is a financial theory that connects forward exchange rates, spot exchange rates, and nations' individual interest rates. It is the theory with
19/04/2016В В· This course is presented in simple language with examples. Interest Rate Parity Theory Case study 1 - Duration: Interest rate parity Interest Rate Parity and Purchasing Power Parity 1. Interest Rate Parity & Purchasing power parity Presented by Danish Hasan Ramiz Junaid Zamir
11/02/2018В В· Which theory will you use to predict future foreign exchange rates? theory, if, for example, exchange rates is the theory of interest rate parity Learn the basics of forward exchange rates and hedging strategies to understand interest rate parity a higher interest rate. In the example theory had held
The basic concept of Purchasing Power Parity theory or PPP, For example, if the interest rate in the U.S. is 5 percent and the interest rate in Japan is 8 Interest Rate Parity theory; A simple example may be a situation, where interest rates in the United Kingdom are, say, 2%, while interest rates in Japan are,
Credit Migration and Covered Interest Rate Parity Credit Migration and Covered Interest Rate Parity a cornerstone of п¬Ѓnance theory stating that assets with Its equivalent in the financial markets is a theory called the Interest Rate Parity For example, if the interest rate in India is higher than that in the US,
Interest rate parity is a theory that suggests a strong relationship between interest rates and the movement of currency values. In fact, you can predict what a The most ideal example of risk-free arbitrage would be buying a security on one exchange and simultaneously selling it on the other for a profit.
The parity conditions Interest parity conditions are no-arbitrage profit (and in this example, This condition is called “covered interest rate parity As Palkesh has rightly mentioned that we get to see a net effect of all the theories in reality. However, if I were to specifically talk about the Interest rate
How to calcuate forward exchange rate or interest rate parity? using price interest rate divided by base interest rate in its examples. Theory; Worldbuilding This is in a sense an extension of the covered interest rate parity we just when for some reason, the interest rate goes in Computer Science and Game Theory.
The interest rate parity gives a mathematical explanation for the purchasing power parity and real effective interest rates Credit Migration and Covered Interest Rate Parity Credit Migration and Covered Interest Rate Parity a cornerstone of п¬Ѓnance theory stating that assets with
Interest rate parity is a no-arbitrage condition representing an equilibrium state under which investors will i \$ is the interest rate in one country (for example This is in a sense an extension of the covered interest rate parity we just when for some reason, the interest rate goes in Computer Science and Game Theory.
Interest Rate Parity Exchange Rates, Interest Rates, Interest Rate Parity An investor has ВҐ1 invest in Japan have Example of IRP Interest rate parity connects the interest rates, spot exchange rates and forward exchange rates in a single comparison. The theory is that the differential between
The parity conditions Interest parity conditions are no-arbitrage profit (and in this example, This condition is called “covered interest rate parity The basic concept of Purchasing Power Parity theory or PPP, For example, if the interest rate in the U.S. is 5 percent and the interest rate in Japan is 8
Monetary Policy and the Uncovered Interest Rate Parity Puzzle For example, when the Fed sharply lowered rates in 2001 like much of the modern theory International Finance Theory and Policy Numerical Examples Using the Rate this page provides the interest rate parity condition when interest is
PURCHASING POWER PARITY & INTERNATIONAL FISHER EFFECT Did Purchasing Power Parity theory hold in these Interest Rate Local 62.88% 67.08% 54.93% 55.38% ... Currency of Payment (Managing Transaction Risks) This is an example of a floating Under the Interest Rate Parity theory of exchange rate
Credit Migration and Covered Interest Rate Parity Credit Migration and Covered Interest Rate Parity a cornerstone of п¬Ѓnance theory stating that assets with CHAPTER 5 INTERNATIONAL PARITY CONDITIONS: INTEREST RATE It then develops the theory and reviews the empirical evidence of the Interest Rate Parity:
Interest Rate Parity Exchange Rates, Interest Rates, Interest Rate Parity An investor has ¥1 invest in Japan have Example of IRP Theory: Expected Interest rate parity 1 1 t t t t t no risk –eg, violation of covered interest rate parity of expected interest rate parity • Examples.
Theory: Expected Interest rate parity 1 1 t t t t t no risk –eg, violation of covered interest rate parity of expected interest rate parity • Examples. What is Interest Rate Parity? Interest Rate Parity (IRP) is a theory in which the differential between the interest rates of two countries remains equal to the
Interest Rate Parity International Economics. This is in a sense an extension of the covered interest rate parity we just when for some reason, the interest rate goes in Computer Science and Game Theory., Interest rate parity is a financial theory that connects forward exchange rates, spot exchange rates, and nations' individual interest rates. It is the theory with.
### What is Interest Rate Parity? Financial Web - finweb.com
Exchange Rates and Interest Parity SSCC. Understand how inflation and the exchange rate between 2 countries are linked through Purchasing Power Parity (PPP). Example scenarios are provided., Interest Rate Parity theory; A simple example may be a situation, where interest rates in the United Kingdom are, say, 2%, while interest rates in Japan are,.
Evidence on Financial Globalization and Crises Interest. Real Interest Rate Parity: Evidence from Industrialized known as “real interest rate parity,” example, Meese and Rogoff,, Costas Arkolakis teaching fellow: Federico Esposito Economics 407, Nominal and Real Exchange Rate A Theory of Purchasing Power Parity Interest Rate.
### (3.1) FX Theory Interest Rate Parity snbchf.com
Interest Rate Parity and Purchasing Power Parity SlideShare. When Purchasing Power Parity (PPP) Theory applies to product markets, Interest Rate Parity (IRP) condition applies to financial markets. Interest Rate Parity (IRP https://en.m.wikipedia.org/wiki/Fixed_exchange_rate Real interest rate parity hypothesis in post-Soviet countries: Evidence from unit root tests. Real interest parity theory rests on Uncovered Interest Parity.
Interest and Price Parity and Foreign Exchange Market Efficiency: The For example, purchasing power parity rates. In Section IV, interest parity theory is Costas Arkolakis teaching fellow: Federico Esposito Economics 407, Nominal and Real Exchange Rate A Theory of Purchasing Power Parity Interest Rate
... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53 ... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53
from Covered Interest Rate Parity . cause their prices to deviate from levels implied by CIP.the For example, theory. 1 For instance, during In theory were the nominal interest rates the same, Therefore interest rate parity hold since it shows how exchange rate of currencies If, for example,
... including interest rate parity, the international Fisher effect, For example, the interest rate in the UK is different from Definition & Parity Theory 4:53 Purchasing power parity (PPP) is a neoclassical economic theory that states that the exchange rate between two countries is equal to the ratio of the currencies
Costas Arkolakis teaching fellow: Federico Esposito Economics 407, Nominal and Real Exchange Rate A Theory of Purchasing Power Parity Interest Rate Credit Migration and Covered Interest Rate Parity Credit Migration and Covered Interest Rate Parity a cornerstone of п¬Ѓnance theory stating that assets with
When Purchasing Power Parity (PPP) Theory applies to product markets, Interest Rate Parity (IRP) condition applies to financial markets. Interest Rate Parity (IRP How to calcuate forward exchange rate or interest rate parity? using price interest rate divided by base interest rate in its examples. Theory; Worldbuilding
Real interest rate parity hypothesis in post-Soviet countries: Evidence from unit root tests. Real interest parity theory rests on Uncovered Interest Parity International Finance Theory and Policy Numerical Examples Using the Rate this page provides the interest rate parity condition when interest is
16.1 Overview of Interest Rate Parity. For example, back in 1997, short-term interest rates were 60 percent per year in Russia and 75 Interest Rate Parity Theory. Purchasing power parity (PPP) is a neoclassical economic theory that states that the exchange rate between two countries is equal to the ratio of the currencies
An example. Let's say the current exchange rate between the US dollar and theUK pound is \$1.60 / £. Interest Rate Parity Theory (IRPT) Covered Interest Rate Parity (IRP) it is based on the risk-free interest rates for the currencies involved, For example, suppose you are
An example. Let's say the current exchange rate between the US dollar and theUK pound is \$1.60 / £. Interest Rate Parity Theory (IRPT) International Finance Theory and Policy Numerical Examples Using the Rate this page provides the interest rate parity condition when interest is
The international Fisher effect is an extension of the Fisher effect hypothesized by American Relation to interest rate parity To check this example, Covered and uncovered interest parities should not be confused with each other. For example, suppose you have PPP is purchasing power parity exchange rate,
View all posts in Lakes Creek category | 4,943 | 23,909 | {"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-2021-49 | longest | en | 0.909272 |
http://howtoloseweight.com.pk/exactly-much-water-need-drink-per-day-lose-weight/ | 1,660,097,277,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00151.warc.gz | 24,345,447 | 17,730 | # Here Is Exactly How Much Water You Need To Drink Per Day to Lose Weight
Here Is Exactly How Much Water You Need To Drink Per Day
Our body’s texture is consists of almost 71% of water. And the disturbed balance of water in the body can lead to severe and unhealthy consequences such as heart strokes, blood thickness, kidney issues, lungs issues and weight gain. And if you drink enough water, it not only rescues your from different chronic diseases but also freshens your skin, boost up your metabolism, smoothens the digestion, keeps you fresh and energetic for the whole of the day and helps you lose weight. Water is the only thing in this world that is easily available and carries magical qualities for all aspects of healthy life. But the question is should you keep drinking water 24/7? If not, how much water you should consume and on what basis? Here is a holistic way to know how much water you should drink in order to see better and healthy results.
# Go and weigh yourself
Drinking water always feels refreshing but if drunk in the right amount at right time, nothing can compensate it for weight losing tricks. To know what is your exclusive targeted amount of water, you must be aware of your current body weight and if you know your BMI (Body Mass Index) that’s gold. Water makes a lot of your body weight so to know how much intake of water you need on daily basis, knowing your current body weight is important. Because a person with 200 pounds can never be compared to a person with 70 pounds in weight losing journey or drinking water battle.
# Now, do some math
This is not about algebra, honestly. Just divide your body weight by 2 like ½. For instances, if your current body weight is 80 pounds, the half is 40 obviously. Now, here is your result. If you weigh 80 pounds you require 40 ounces of water per day. This is as simple as that. | 403 | 1,869 | {"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-2022-33 | latest | en | 0.948168 |
http://ask.metafilter.com/231710/The-use-of-Linear-Algebra-in-Programming | 1,508,442,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00038.warc.gz | 27,738,255 | 18,357 | # The use of Linear Algebra in Programming.December 23, 2012 10:46 AM Subscribe
I just finished my first course in Linear Algebra. I know a fair amount of Python. Please help me come up with a project that combines the two.
This past Fall semester, I took Linear Algebra I, a 200-level Math class at my school. (The next class, sequentially, is Linear Algebra II -- a 300-level class which is much more proof based.) I don't want to list everything that I learned, but I will say that the course covered a lot of material, and was quite rigorous.
Through the lectures, my professor would frequently reference Linear Algebra's application to computer science/computer graphics (e.g. how each pixel in an image can effectively be manipulated as an entry in a matrix).
Likewise, I want come up with a project that combines my newly acquired knowledge of Linear Algebra, with my ability to program.
A couple of notes with regards to the type of project that I am looking for:
1. I'm not really interested in game design or computer animation (and I know that both are predicated on Linear Algebra).
2. I really like databases. In addition to Python, I know/am learning SQL -- I would love something of the sort that combines matrix manipulation with querying a database (if this is even a thing.)
3. I'm learning Django, so any idea that involves building a (not too complex) web app would be great.
Thanks a lot!
posted by lobbyist to Technology (11 answers total) 13 users marked this as a favorite
Well, I reviewed Khan Academy's Linear Algebra videos. I don't know if they're less rigorous than the course you took (probably), but then I started playing around with this in Python.
I didn't really understand what I was doing but more or less got it to work. Perhaps it's a starting point.
posted by dfriedman at 10:51 AM on December 23, 2012 [1 favorite]
3d graphics is most easily done with 4x1, 4x4 and 1x4 matrices, and implementing a simple ray tracer is a great way to play with algebra and transforms. You can do it with high school algebra, but it's easier with 4x4 matrices. The only problem is that there are already a gazillion implementations.
But if you start to look at matrices as collections of data, there are a lot of analysis techniques in genomics, things like protein expression analysis, that work on matrics with a thousand or ten thousand rows and columns. Finding a data set that'd be useful for computational biology that you can explore Principal Component Analysis or similar on could be a lot of fun!
posted by straw at 10:54 AM on December 23, 2012
Unsupervised machine learning is basically all linear algebra with a coating of stats for reliability statistics, and like all languages python has its share of machine learning packages.
Two things I personally have been poking around with recently are non-negative matrix factorization (NMF) and hidden markov models (HMM). Scipy and numpy get you pretty far. I understand orange is the best-regarded generalized machine learning packages, though I'm still at pretty preliminary stages in my fiddlings. Natural language toolkit NLTK is good if you want to look at corpus/document based things.
Edit to add: The reason I mention corpus/document based things in particular is that depending on scale you might have some reliance on databases/database-like things. Some general problems are topic identification/classification, author identification, and preference classification.
posted by PMdixon at 11:16 AM on December 23, 2012 [1 favorite]
There are a number of network analysis techniques that use linear algebra - Google's PageRank, for instance, is computed as an eigenvector. One of the standard ways of representing a network is the adjacency matrix.
posted by obvious at 11:29 AM on December 23, 2012
How about implementing something like this online matrix calculator?
And, if you do, I'd appreciate if you could let me know.
posted by aroberge at 11:37 AM on December 23, 2012
Face recognition algorithms are all about eigenvectors. OpenCV is a great computer vision library with pretty good Python support. Here are two tutorials on face recognition in OpenCV. It's a lot of fun to feed in your webcam video and detect your own face, but it works on static images too. Building a web app login system that uses face recognition (like face unlock on Android devices) would be a pretty cool project.
posted by ecmendenhall at 11:47 AM on December 23, 2012
1. I'm not really interested in game design or computer animation (and I know that both are predicated on Linear Algebra).
2. I really like databases. In addition to Python, I know/am learning SQL -- I would love something of the sort that combines matrix manipulation with querying a database (if this is even a thing.)
There's an infant startup in Estonia working on putting together a database of math problems for school teachers to use for classwork and homework. I didn't fully understand what they're exactly trying to do but I did understand they are building a database of math problems. Can I put you in touch with them ?
posted by infini at 1:50 PM on December 23, 2012
"tensor algebra" is related to JOIN in SQL as in this patent.
i'm pretty sure the "relationship" between tensor algebra and relational databases is standard stuff but it's not my field and i'm too lazy to dig something up. tensor algebra is a logical continuation of what you learned in linear algebra, where you define "products" on vectors which produce a new construction called a tensor rather than a vector. so vector spaces are subsumed in tensor spaces.
anyway, tensors are pretty standard mathematics which you could learn from a ton of books (you may have had an intro in your linear algebra class) and a logical continuation of linear algebra and i bet in some advanced DB book you can find the relationship between them and relational databases explained...
posted by ennui.bz at 2:56 PM on December 23, 2012
Get up on a kaggle challenge! You could use scikit-learn which makes use of numpy, scipy, and matplotlib, all of which are things that combine Python with Linear Algebra, and much more.
posted by oceanjesse at 2:19 AM on December 24, 2012
I came in to suggest a simple raytracer too.
posted by aeighty at 8:14 AM on December 24, 2012
You may also wish to explore the APL family of languages (APL, J, and K) - they're based on linear algebra. K's killer app, kdb+, combines vector/matrix operations with databases for a high-performance finance database. (K/kdb+ is rather expensive, though there is a demo version of q, the new commercial K, at Kx.)
Two free, open-source options are J (recently open-sourced) and kona, a sort-of clone of K that I've contributed to.
posted by silentbicycle at 9:21 AM on December 24, 2012
« Older Shower runs hot, then cold, then hot. | t-mobile unlocking in the uk for iphone Newer » | 1,547 | 6,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-43 | latest | en | 0.971078 |
https://www.w3resource.com/php-exercises/basic-algorithm/index.php | 1,723,078,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00845.warc.gz | 839,921,365 | 34,717 | # PHP Basic Algorithm: Exercises, Practice, Solution
## PHP Basic Algorithm [136 exercises with solution]
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1. Write a PHP program to compute the sum of the two given integer values. If the two values are the same, then returns triple their sum.
Sample Input
1, 2
3, 2
2, 2
Sample Output:
3
5
12
Click me to see the sample solution
2. Write a PHP program to get the absolute difference between n and 51. If n is greater than 51 return triple the absolute difference.
Sample Input:
53
30
51
Sample Output:
6
21
0
Click me to see the sample solution
3. Write a PHP program to check two given integers, and return true if one of them is 30 or if their sum is 30.
Sample Input:
30, 0
25, 5
20, 30
20, 25
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
4. Write a PHP program to check a given integer and return true if it is within 10 of 100 or 200.
Sample Input:
103
90
89
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
5. Write a PHP program to create a new string where 'if' is added to the front of a given string. If the string already begins with 'if', return the string unchanged.
Sample Input:
"if else"
"else"
"if"
Sample Output:
if else
if else
if
Click me to see the sample solution
6. Write a PHP program to remove the character in a given position of a given string. The given position will be in the range 0..string length -1 inclusive.
Sample Input:
"Python", 1
"Python", o
"Python", 4
Sample Output:
Pthon
ython
Pythn
Click me to see the sample solution
7. Write a PHP program to exchange the first and last characters in a given string and return the new string.
Sample Input:
"abcd"
"a"
"xy"
Sample output:
dbca
a
yx
Click me to see the sample solution
8. Write a PHP program to create a new string which is 4 copies of the 2 front characters of a given string. If the given string length is less than 2 return the original string.
Sample Input:
"C Sharp"
"JS"
"a"
Sample Output:
C C C C
JSJSJSJS
a
Click me to see the sample solution
9. Write a PHP program to create a new string with the last char added at the front and back of a given string of length 1 or more.
Sample Input:
"Red"
"Green"
"1"
Sample Output:
dRedd
nGreenn
111
Click me to see the sample solution
10. Write a PHP program to check if a given positive number is a multiple of 3 or a multiple of 7.
Sample Input
3
14
12
37
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
11. Write a PHP program to create a new string taking the first 3 characters of a given string and return the string with the 3 characters added at both the front and back. If the given string length is less than 3, use whatever characters are there.
Sample Input:
"Python"
"JS"
"Code"
Sample Output:
PytPythonPyt
JSJSJS
CodCodeCod
Click me to see the sample solution
12. Write a PHP program to check if a given string starts with 'C#' or not.
Sample Input:
"PHP"
"C#"
"C++"
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
13. Write a PHP program to check if one given temperatures is less than 0 and the other is greater than 100.
Sample Input:
120, -1
-1, 120
2, 120
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
14. Write a PHP program to check two given integers whether either of them is in the range 100..200 inclusive.
Sample Input:
100, 199
250, 300
105, 190
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
15. Write a PHP program to check whether three given integer values are in the range 20..50 inclusive. Return true if 1 or more of them are in the said range otherwise false.
Sample Input:
11, 20, 12
30, 30, 17
25, 35, 50
15, 12, 8
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
16. Write a PHP program to check whether two given integer values are in the range 20..50 inclusive. Return true if 1 or other is in the said range otherwise false.
Sample Input:
20, 84
14, 50
11, 45
25, 40
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
17. Write a PHP program to check if a string 'yt' appears at index 1 in a given string. If it appears return a string without 'yt' otherwise return the original string.
Sample Input:
"Python"
"ytade"
"jsues"
Sample Output:
Phon
ytade
jsues
Click me to see the sample solution
18. Write a PHP program to check the largest number among three given integers.
Sample Input:
1,2,3
1,3,2
1,1,1
1,2,2
Sample Output:
3
3
1
2
Click me to see the sample solution
19. Write a PHP program to check which number nearest to the value 100 among two given integers. Return 0 if the two numbers are equal.
Sample Input:
78, 95
95, 95
99, 70
Sample Output:
95
0
99
Click me to see the sample solution
20. Write a PHP program to check whether two given integers are in the range 40..50 inclusive, or they are both in the range 50..60 inclusive.
Sample Input:
78, 95
25, 35
40, 50
55, 60
Sample Output:
bool(false)
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
21. Write a PHP program to find the larger value from two positive integer values that is in the range 20..30 inclusive, or return 0 if neither is in that range.
Sample Input:
78, 95
20, 30
21, 25
28, 28
Sample Output:
0
30
25
28
Click me to see the sample solution
22. Write a PHP program to check if a given string contains between 2 and 4 'z' character.
Sample Input:
"frizz"
"zane"
"Zazz"
"false"
Sample Output:
bool(true)
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
23. Write a PHP program to check if two given non-negative integers have the same last digit.
Sample Input:
123, 456
12, 512
7, 87
12, 45
Sample Output:
bool(false)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
24. Write a PHP program to convert the last 3 characters of a given string in upper case. If the length of the string has less than 3 then uppercase all the characters.
Sample Input:
"Python"
"Javascript"
"js"
"PHP"
Sample Output:
PytHON
JavascrIPT
JS
PHP
Click me to see the sample solution
25. Write a PHP program to create a new string which is n (non-negative integer) copies of a given string.
Sample Input:
"JS", 2
"JS", 3
"JS", 1
Sample Output:
JSJS
JSJSJS
JS
Click me to see the sample solution
26. Write a PHP program to create a new string which is n (non-negative integer) copies of the the first 3 characters of a given string. If the length of the given string is less than 3 then return n copies of the string.
Sample Input:
"Python", 2
"Python", 3
"JS", 3
Sample Output:
PytPyt
PytPytPyt
JSJSJS
Click me to see the sample solution
27. Write a PHP program to count the string "aa" in a given string and assume "aaa" contains two "aa".
Sample Input:
"bbaaccaag"
"jjkiaaasew"
"JSaaakoiaa"
Sample Output:
2
2
3
Click me to see the sample solution
28. Write a PHP program to check if the first appearance of "a" in a given string is immediately followed by another "a".
Sample Input:
"caabb"
"babaaba"
"aaaaa"
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
29. Write a PHP program to create a new string made of every other character starting with the first from a given string.
Sample Input:
"Python"
"PHP"
"JS"
Sample Output:
Pto
PP
J
Click me to see the sample solution
30. Write a PHP program to create a string like "aababcabcd" from a given string "abcd".
Sample Input:
"abcd"
"abc"
"a"
Sample Output:
aababcabcd
aababc
a
Click me to see the sample solution
31. Write a PHP program to count a substring of length 2 appears in a given string and also as the last 2 characters of the string. Do not count the end substring.
Sample Input:
"abcdsab"
"abcdabab"
"abcabdabab"
"abcabd"
Sample Output:
1
2
3
0
Click me to see the sample solution
32. Write a PHP program to check a specified number is present in a given array of integers.
Sample Input:
{1,2,9,3}, 3
{1,2,2,3}, 2
{1,2,2,3}, 9
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
33. Write a PHP program to check if one of the first 4 elements in an array of integers is equal to a given element.
Sample Input:
{1,2,9,3}, 3
{1,2,3,4,5,6}, 2
{1,2,2,3}, 9
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
34. Write a PHP program to check whether the sequence of numbers 1, 2, 3 appears in a given array of integers somewhere.
Sample Input:
{1,1,2,3,1}
{1,1,2,4,1}
{1,1,2,1,2,3}
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
35. Write a PHP program to compare two given strings and return the number of the positions where they contain the same length 2 substring.
Sample Input:
"abcdefgh", "abijsklm"
"abcde", "osuefrcd"
"pqrstuvwx", "pqkdiewx"
Sample Output:
1
1
2
Click me to see the sample solution
36. Write a PHP program to create a new string of the characters at indexes 0,1,4,5,8,9 ... from a given string.
Sample Input:
"Python"
"JavaScript"
"HTML"
Sample Output:
Pyon
JaScpt
HT
Click me to see the sample solution
37. Write a PHP program to count the number of two 5's are next to each other in an array of integers. Also count the situation where the second 5 is actually a 6.
Sample Input:
{ 5, 5, 2 }
{ 5, 5, 2, 5, 5 }
{ 5, 6, 2, 9}
Sample Output:
1
2
1
Click me to see the sample solution
38. Write a PHP program to check if a triple is presents in an array of integers or not. If a value appears three times in a row in an array it is called a triple.
Sample Input:
{ 1, 1, 2, 2, 1 }
{ 1, 1, 2, 1, 2, 3 }
{ 1, 1, 1, 2, 2, 2, 1 }
Sample Output:
bool(false)
bool(false)
bool(true)
Click me to see the sample solution
39. Write a PHP program to compute the sum of the two given integers. If the sum is in the range 10..20 inclusive return 30.
Sample Input:
12, 17
2, 17
22, 17
20, 0
Sample Output:
29
30
39
30
Click me to see the sample solution
40. Write a PHP program that accept two integers and return true if either one is 5 or their sum or difference is 5.
Sample Input:
5, 4
4, 3
1, 4
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
41. Write a PHP program to test if a given non-negative number is a multiple of 13 or it is one more than a multiple of 13.
Sample Input:
13
14
27
41
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
42. Write a PHP program to check if a given non-negative given number is a multiple of 3 or 7, but not both.
Sample Input:
3
7
21
Sample Output:
int(1)
int(1)
int(0)
Click me to see the sample solution
43. Write a PHP program to check if a given number is within 2 of a multiple of 10.
Sample Input:
3
7
8
21
Sample Output:
bool(false)
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
44. Write a PHP program to compute the sum of the two given integers. If one of the given integer value is in the range 10..20 inclusive return 18.
Sample Input:
3, 7
10, 11
10, 20
21, 220
Sample Output:
10
18
18
241
Click me to see the sample solution
45. Write a PHP program to check whether a given string starts with "F" or ends with "B". If the string starts with "F" return "Fizz" and return "Buzz" if it ends with "B" If the string starts with "F" and ends with "B" return "FizzBuzz". In other cases return the original string.
Sample Input:
"FizzBuzz"
"Fizz"
"Buzz"
"Founder"
Sample Output:
Fizz
Fizz
Buzz
Fizz
Click me to see the sample solution
46. Write a PHP program to check if it is possible to add two integers to get the third integer from three given integers.
Sample Input:
1, 2, 3
4, 5, 6
-1, 1, 0
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
47. Write a PHP program to check if y is greater than x, and z is greater than y from three given integers x,y,z.
Sample Input:
1, 2, 3
4, 5, 6
-1, 1, 0
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
48. Write a PHP program to check if three given numbers are in strict increasing order, such as 4 7 15, or 45, 56, 67, but not 4 ,5, 8 or 6, 6, 8.However,if a fourth parameter is true, equality is allowed, such as 6, 6, 8 or 7, 7, 7.
Sample Input:
1, 2, 3, false
1, 2, 3, true
10, 2, 30, false
10, 10, 30, true
Sample Output:
bool(true)
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
49. Write a PHP program to check if two or more non-negative given integers have the same rightmost digit.
Sample Input:
11, 21, 31
11, 22, 31
11, 22, 33
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
50. Write a PHP program to check three given integers and return true if one of them is 20 or more less than one of the others.
Sample Input:
11, 21, 31
11, 22, 31
10, 20, 15
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
51. Write a PHP program to find the larger from two given integers. However if the two integers have the same remainder when divided by 7, then the return the smaller integer. If the two integers are the same, return 0.
Sample Input:
11, 21
11, 20
10, 10
Sample Output:
21
20
0
Click me to see the sample solution
52. Write a PHP program to check two given integers, each in the range 10..99. Return true if a digit appears in both numbers, such as the 3 in 13 and 33.
Sample Input:
11, 21
11, 20
10, 10
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
53. Write a PHP program to compute the sum of two given non-negative integers x and y as long as the sum has the same number of digits as x. If the sum has more digits than x then return x without y.
Sample Input:
4, 5
7, 4
10, 10
Sample Output:
9
7
20
Click me to see the sample solution
54. Write a PHP program to compute the sum of three given integers. If the two values are same return the third value.
Sample Input:
4, 5, 7
7, 4, 12
10, 10, 12
12, 12, 18
Sample Output:
16
23
12
18
Click me to see the sample solution
55. Write a PHP program to compute the sum of the three integers. If one of the values is 13 then do not count it and its right towards the sum.
Sample Input:
4, 5, 7
7, 4, 12
10, 13, 12
13, 12, 18
Sample Output:
16
23
10
0
Click me to see the sample solution
56. Write a PHP program to compute the sum of the three given integers. However, if any of the values is in the range 10..20 inclusive then that value counts as 0, except 13 and 17.
Sample Input:
4, 5, 7
7, 4, 12
10, 13, 12
17, 12, 18
Sample Output:
16
11
13
17
Click me to see the sample solution
57. Write a PHP program to check two given integers and return the value whichever value is nearest to 13 without going over. Return 0 if both numbers go over.
Sample Input:
4, 5
7, 12
10, 13
17, 33
Sample Output:
5
12
13
0
Click me to see the sample solution
58. Write a PHP program to check three given integers (small, medium and large) and return true if the difference between small and medium and the difference between medium and large is same.
Sample Input:
4, 5, 6
7, 12, 13
-1, 0, 1
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
59. Write a PHP program to create a new string using two given strings s1, s2, the format of the new string will be s1s2s2s1.
Sample Input:
"Hi", "Hello"
"whats", "app"
Sample Output:
HiHelloHelloHi
whatsappappwhats
Click me to see the sample solution
60. Write a PHP program to insert a given string into middle of the another given string of length 4.
Sample Input:
"[[]]","Hello"
"(())", "Hi"
Sample Output:
[[Hello]]
((Hi))
Click me to see the sample solution
61. Write a PHP program to create a new string using three copies of the last two character of a given string of length atleast two.
Sample Input:
"Hello"
"Hi"
Sample Output:
lololo
HiHiHi
Click me to see the sample solution
62. Write a PHP program to create a new string using first two characters of a given string. If the string length is less than 2 then return the original string.
Sample Input:
"Hello"
"Hi"
"H"
" "
Sample Output:
He
Hi
H
Click me to see the sample solution
63. Write a PHP program to create a new string of the first half of a given string of even length.
Sample Input:
"Hello"
"Hi"
Sample Output:
He
H
Click me to see the sample solution
64. Write a PHP program to create a new string without the first and last character of a given string of length atleast two.
Sample Input:
"Hello"
"Hi"
"Python"
Sample Output:
ell
ytho
Click me to see the sample solution
65. Write a PHP program to create a new string from two given string one is shorter and another is longer. The format of the new string will be long string + short string + long string.
Sample Input:
"Hello", "Hi"
"JS", "Python"
Sample Output:
HelloHiHello
PythonJSPython
Click me to see the sample solution
66. Write a PHP program to concat two given string of length atleast 1, after removing their first character.
Sample Input:
"Hello", "Hi"
"JS", "Python"
Sample Output:
elloi
Sython
Click me to see the sample solution
67. Write a PHP program to move the first two characters to the end of a given string of length at least two.
Sample Input:
"Hello"
"JS"
Sample Output:
lloHe
JS
Click me to see the sample solution
68. Write a PHP program to move the last two characters to the start of a given string of length at least two.
Sample Input:
"Hello"
"JS"
Sample Output:
loHel
JS
Click me to see the sample solution
69. Write a PHP program to create a new string without the first and last character of a given string of any length.
Sample Input:
"Hello"
"JS"
''
Sample Output:
ell
Click me to see the sample solution
70. Write a PHP program to create a new string using the two middle characters of a given string of even length (at least 2).
Sample Input:
"Hell"
"JS"
Sample Output:
el
JS
Click me to see the sample solution
71. Write a PHP program to check if a given string ends with "on".
Sample Input:
"Hello"
"Python"
"on"
"o"
Sample Output:
string(2) "lo"
string(2) "on"
string(2) "on"
string(1) "o"
Click me to see the sample solution
72. Write a PHP program to create a new string using the first and last n characters from a given string of length at least n.
Sample Input:
"Hello", 1
"Python", 2
"on", 1
"o", 1
Sample Output:
Ho
Pyon
on
oo
Click me to see the sample solution
73. Write a PHP program to create a new string of length 2 starting at the given index of a given string.
Sample Input:
"Hello", 1
"Python", 2
"on", 1
Sample Output:
el
th
on
Click me to see the sample solution
74. Write a PHP program to create a new string taking 3 characters from the middle of a given string at least 3.
Sample Input:
"Hello"
"Python"
"abc"
Sample Output:
ell
yth
abc
Click me to see the sample solution
75. Write a PHP program to create a new string of length 2, using first two characters of a given string. If the given string length is less than 2 use '#' as missing characters.
Sample Input:
"Hello"
"Python"
"a"
""
Sample Output:
He
Py
a#
#
Click me to see the sample solution
76. Write a PHP program to create a new string taking the first character from a given string and the last character from another given string. If the length of any given string is 0, use '#' as its missing character.
Sample Input:
"Hello", "Hi"
"Python", "PHP"
"JS", "JS"
"Csharp", ""
Sample Output:
Hi
PP
JS
C#
Click me to see the sample solution
77. Write a PHP program to concat two given strings (lowercase). If there are any double character in new string then omit one character.
Sample Input:
"abc", "cat"
"python", "php"
"php", "php"
Sample Output:
abcat
PythonPHP
phphp
Click me to see the sample solution
78. Write a PHP program to create a new string from a given string after swapping last two characters.
Sample Input:
"Hello"
"Python"
"PHP"
"JS"
"C"
Sample Output:
Helol
Pythno
PPH
SJ
C
Click me to see the sample solution
79. Write a PHP program to check if a given string begins with 'abc' or 'xyz'. If the string begins with 'abc' or 'xyz' return 'abc' or 'xyz' otherwise return the empty string.
Sample Input:
"abc"
"abcdef"
"C"
"xyz"
"xyzsder"
Sample Output:
abc
abc
xyz
xyz
Click me to see the sample solution
80. Write a PHP program to check whether the first two characters and last two characters of a given string are same.
Sample Input:
"abab"
"abcdef"
"xyzsderxy"
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
81. Write a PHP program to concat two given strings. If the given strings have different length remove the characters from the longer string.
Sample Input:
"abc", "abcd"
"Python", "Python"
"JS", "Python"
Sample Output:
abcbcd
PythonPython
JSon
Click me to see the sample solution
82. Write a PHP program to create a new string using 3 copies of the first 2 characters of a given string. If the length of the given string is less than 2 use the whole string.
Sample Input:
"abc"
"Python"
"J"
Sample Output:
ababab
PyPyPy
JJJ
Click me to see the sample solution
83. Write a PHP program to create a new string from a given string. If the two characters of the given string from its beginning and end are same return the given string without the first two characters otherwise return the original string.
Sample Input:
"abcab"
"Python"
Sample Output:
cab
Python
Click me to see the sample solution
84. Write a PHP program to create a new string from a given string without the first and last character if the first or last characters are 'a' otherwise return the original given string.
Sample Input:
"abcab"
"python"
"abcda"
"jython"
Sample Output:
bcab
Python
bcd
jython
Click me to see the sample solution
85. Write a PHP program to create a new string from a given string. If the first or first two characters is 'a', return the string without those 'a' characters otherwise return the original given string.
Sample Input:
"abcab"
"python"
"aacda"
"jython"
Sample Output:
bcab
Python
cda
jython
Click me to see the sample solution
86. Write a PHP program to check a given array of integers of length 1 or more and return true if 10 appears as either first or last element in the given array.
Sample Input:
{ 10, 20, 40, 50 }
{ 5, 20, 40, 10 }
{ 10, 20, 40, 10 }
{ 12, 24, 35, 55 }
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
87. Write a PHP program to check a given array of integers of length 1 or more and return true if the first element and the last element are equal in the given array.
Sample Input:
{ 10, 20, 40, 50 }
{ 10, 20, 40, 10 }
{ 12, 24, 35, 55 }
Sample Output:
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
88. Write a PHP program to check two given arrays of integers of length 1 or more and return true if they have the same first element or they have the same last element.
Sample Input:
{[10, 20, 40, 50], [10, 20, 40, 50]}
{[10, 20, 40, 10], [10, 20, 40, 5]}
{[12, 24, 35, 55], [1, 20, 40, 5]}
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
89. Write a PHP program to compute the sum of the elements of an given array of integers.
Sample Input:
{ 10, 20, 30, 40, 50 }
{ 10, 20, -30, -40, 50 }
Sample Output:
150
10
Click me to see the sample solution
90. Write a PHP program to rotate the elements of a given array of integers (length 4 ) in left direction and return the new array.
Sample Input:
{ 10, 20, -30, -40 }
Sample Output:
Rotated array: 20,-30,-40,10
Click me to see the sample solution
91. Write a PHP program to reverse a given array of integers and length 5.
Sample Input:
{ 10, 20, -30, -40, 50 }
Sample Output:
Reverse array: 50,-40,-30,20,10
Click me to see the sample solution
92. Write a PHP program to find out the maximum element between the first or last element in a given array of integers ( length 4), replace all elements with maximum element.
Sample Input:
{ 10, 20, -30, -40 }
Sample Output:
New array with maximum values: 20,20,20,20
Click me to see the sample solution
93. Write a PHP program to create a new array containing the middle elements from the two given arrays of integers, each length 5.
Sample Input:
{ 10, 20, -30, -40, 30 }, { 10, 20, 30, 40, 30 }
Sample Output:
New array: -30, 30
Click me to see the sample solution
94. Write a PHP program to create a new array taking the first and last elements of a given array of integers and length 1 or more.
Sample Input:
{ 10, 20, -30, -40, 30 }
Sample Output:
New array: 10,30
Click me to see the sample solution
95. Write a PHP program to check if a given array of integers and length 2, contains 15 or 20.
Sample Input:
{ 12, 20 }
{ 14, 15 }
{ 11, 21 }
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
96. Write a PHP program to check if a given array of integers and length 2, does not contain 15 or 20.
Sample Input:
{ 12, 20 }
{ 14, 15 }
{ 11, 21 }
Sample Output:
bool(false)
bool(false)
bool(true)
Click me to see the sample solution
97. Write a PHP program to check a given array of integers and return true if the array contains 10 or 20 twice. The length of the array will be 0, 1, or 2.
Sample Input:
{ 12, 20 }
{ 20, 20 }
{ 10, 10 }
{ 10 }
Sample Output:
bool(false)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
98. Write a PHP program to check a given array of integers, length 3 and create a new array. If there is a 5 in the given array immediately followed by a 7 then set 7 to 1.
Sample Input:
{ 1, 5, 7 }
Sample Output:
New array with maximum values: 1,5,1
Click me to see the sample solution
99. Write a PHP program to compute the sum of the two given arrays of integers, length 3 and find the array which has the largest sum.
Sample Input:
{ 10, 20, -30 }, { 10, 20, 30 }
Sample Output:
New array with maximum values: 10,20,30
Click me to see the sample solution
100. Write a PHP program to create an array taking two middle elements from a given array of integers of length even.
Sample Input:
{ 1, 5, 7, 9, 11, 13 }
Sample Output:
New array: 7 9
Click me to see the sample solution
101. Write a PHP program to create a new array from two give array of integers, each length 3.
Sample Input:
{ 10, 20, 30 }, { 40, 50, 60 }
Sample Output:
New array: 10,20,30,40,50,60
Click me to see the sample solution
102. Write a PHP program to create a new array swapping the first and last elements of a given array of integers and length will be least 1.
Sample Input:
{ 1, 5, 7, 9, 11, 13 }
Sample Output:
New array, after swapping first and last elements: 13,5,7,9,11,1
Click me to see the sample solution
103. Write a PHP program to create a new array length 3 from a given array (length atleast 3) using the elements from the middle of the array.
Sample Input:
{ 1, 5, 7, 9, 11, 13 }
Sample Output:
New array: 7,9,11
Click me to see the sample solution
104. Write a PHP program to find the largest value from first, last, and middle elements of a given array of integers of odd length (atleast 1).
Sample Input:
{1}
{1,2,9}
{1,2,9,3,3}
{1,2,3,4,5,6,7}
{1,2,2,3,7,8,9,10,6,5,4}
Sample Output:
1
9
9
7
8
Click me to see the sample solution
105. Write a PHP program to create a new array taking the first two elements from a given array. If the length of the given array is less than 2 then return the give array.
Sample Input:
{ 1, 5, 7, 9, 11, 13 }
Sample Output:
New array: 1,5
Click me to see the sample solution
106. Write a PHP program to count even number of elements in a given array of integers.
Sample Input:
{ 1, 5, 7, 9, 10, 12 }
Sample Output:
Number of even elements: 2
Click me to see the sample solution
107. Write a PHP program to compute the difference between the largest and smallest values in a given array of integers and length one or more.
Sample Input:
{ 1, 5, 7, 9, 10, 12 }
Sample Output:
Difference between the largest and smallest values: 11
Click me to see the sample solution
108. Write a PHP program to compute the sum of values in a given array of integers except the number 17. Return 0 if the given array has no integer.
Sample Input:
{ 1, 5, 7, 9, 10, 17 }
Sample Output:
Sum of values in the array of integers except the number 17: 32
Click me to see the sample solution
109. Write a PHP program to compute the sum of the numbers in a given array except those numbers starting with 5 followed by atleast one 6. Return 0 if the given array has no integer.
Sample Input:
{ 1, 5, 7, 9, 10, 17 }
{ 1, 5, 6, 9, 10, 17 }
{ 5, 6, 7, 9, 10, 17, 1 }
{ 11, 9, 10, 17, 5, 6 }
Sample Output:
Sum of the numbers of the said array except those numbers starting with 5 followed by atleast one 6:
49
Sum of the numbers of the said array except those numbers starting with 5 followed by atleast one 6:
37
Sum of the numbers of the said array except those numbers starting with 5 followed by atleast one 6:
44
Sum of the numbers of the said array except those numbers starting with 5 followed by atleast one 6:
47
Click me to see the sample solution
110. Write a PHP program to check if a given array of integers contains 5 next to a 5 somewhere.
Sample Input:
{ 1, 5, 6, 9, 10, 17 }
{ 1, 5, 5, 9, 10, 17 }
{ 1, 5, 5, 9, 10, 17, 5, 5 }
Sample Output:
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
111. Write a PHP program to check whether a given array of integers contains 5's and 7's.
Sample Input:
{ 1, 5, 6, 9, 10, 17 }
{ 1, 4, 7, 9, 10, 17 }
{ 1, 1, 2, 9, 10, 17}
Sample Output:
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
112. Write a PHP program to check if the sum of all 5' in the array exactly 15 in a given array of integers.
Sample Input:
{ 1, 5, 6, 9, 10, 17 }
{ 1, 5, 5, 5, 10, 17 }
{ 1, 1, 5, 5, 5, 5}
Sample Output:
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
113. Write a PHP program to check if the number of 3's is greater than the number of 5's.
Sample Input:
{ 1, 5, 6, 9, 3, 3 }
{ 1, 5, 5, 5, 10, 17 }
{ 1, 3, 3, 5, 5, 5}
Sample Output:
bool(true)
bool(false)
bool(false)
Click me to see the sample solution
114. Write a PHP program to check if a given array of integers contains a 3 or a 5.
Sample Input:
{ 5, 5, 5, 5, 5 }
{ 3, 3, 3, 3 }
{ 3, 3, 3, 5, 5, 5}
{ 1, 6, 8, 10}
Sample Output:
bool(true)
bool(true)
bool(true)
bool(false)
Click me to see the sample solution
115. Write a PHP program to check if a given array of integers contains no 3 or a 5.
Sample Input:
{ 5, 5, 5, 5, 5 }
{ 3, 3, 3, 3 }
{ 3, 3, 3, 5, 5, 5}
{ 1, 6, 8, 10}
Sample Output:
bool(true)
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
116. Write a PHP program to check if an array of integers contains a 3 next to a 3 or a 5 next to a 5 or both.
Sample Input:
{ 5, 5, 5, 5, 5 }
{ 1, 2, 3, 4 }
{ 3, 3, 5, 5, 5, 5}
{ 1, 5, 5, 7, 8, 10}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
117. Write a PHP program to check a given array of integers and return true if the given array contains two 5's next to each other, or two 5 separated by one element.
Sample Input:
{ 5, 5, 1, 5, 5 }
{ 1, 2, 3, 4 }
{ 3, 3, 5, 5, 5, 5}
{ 1, 5, 5, 7, 8, 10}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
118. Write a PHP program to check a given array of integers and return true if there is a 3 with a 5 somewhere later in the given array.
Sample Input:
{ 3, 5, 1, 3, 7 }
{ 1, 2, 3, 4 }
{ 3, 3, 5, 5, 5, 5}
{ 2, 5, 5, 7, 8, 10}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
119. Write a PHP program to check a given array of integers and return true if the given array contains either 2 even or 2 odd values all next to each other.
Sample Input:
{ 3, 5, 1, 3, 7 }
{ 1, 2, 3, 4 }
{ 3, 3, 5, 5, 5, 5}
{ 2, 4, 5, 6}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(true)
Click me to see the sample solution
120. Write a PHP program to check a given array of integers and return true if the value 5 appears 5 times and there are no 5 next to each other.
Sample Input:
{ 3, 5, 1, 5, 3, 5, 7, 5, 1, 5 }
{ 3, 5, 5, 5, 5, 5, 5}
{ 3, 5, 2, 5, 4, 5, 7, 5, 8, 5}
{ 2, 4, 5, 5, 5, 5}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
121. Write a PHP program to check a given array of integers and return true if every 5 that appears in the given array is next to another 5.
Sample Input:
{ 3, 5, 5, 3, 7 }
{ 3, 5, 5, 4, 1, 5, 7}
{ 3, 5, 5, 5, 5, 5}
{ 2, 4, 5, 5, 6, 7, 5}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
122. Write a PHP program to check a given array of integers and return true if the specified number of same elements appears at the start and end of the given array.
Sample Input:
{ 3, 7, 5, 5, 3, 7 }, 2
{ 3, 7, 5, 5, 3, 7 }, 3
{ 3, 7, 5, 5, 3, 7, 5 }, 3
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
123. Write a PHP program to check a given array of integers and return true if the array contains three increasing adjacent numbers.
Sample Input:
{ 1, 2, 3, 5, 3, 7 }
{ 3, 7, 5, 5, 3, 7 }
{ 3, 7, 5, 5, 6, 7, 5 }
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
124. Write a PHP program to shift an element in left direction and return a new array.
Sample Input:
{ 10, 20, -30, -40, 50 }
Sample Output:
New array: 20,-30,-40,50,10
Click me to see the sample solution
125. Write a PHP program to create a new array taking the elements before the element value 5 from a given array of integers.
Sample Input:
{ 1, 2, 3, 5, 7 }
Sample Output:
New array: 1,2,3
Click me to see the sample solution
126. Write a PHP program to create a new array taking the elements after the element value 5 from a given array of integers.
Sample Input:
{ 1, 2, 3, 5, 7, 9, 11 }
Sample Output:
New array: 7,9,11
Click me to see the sample solution
127. Write a PHP program to create a new array from a given array of integers shifting all zeros to left direction.
Sample Input:
{ 1, 2, 0, 3, 5, 7, 0, 9, 11 }
Sample Output:
New array: 0,0,1,3,5,7,2,9,11
Click me to see the sample solution
128. Write a PHP program to create a new array after replacing all the values 5 with 0 shifting all zeros to right direction.
Sample Input:
{ 1, 2, 0, 3, 5, 7, 0, 9, 11 }
Sample Output:
New array: 1,2,3,7,9,11,0,0,0
Click me to see the sample solution
129. Write a PHP program to create new array from a given array of integers shifting all even numbers before all odd numbers.
Sample Input:
{ 1, 2, 5, 3, 5, 4, 6, 9, 11 }
Sample Output:
New array: 2,4,6,3,5,1,5,9,11
Click me to see the sample solution
130. Write a PHP program to check if the value of each element is equal or greater than the value of previous element of a given array of integers.
Sample Input:
{ 5, 5, 1, 5, 5 }
{ 1, 2, 3, 4 }
{ 3, 3, 5, 5, 5, 5}
{ 1, 5, 5, 7, 8, 10}
Sample Output:
bool(false)
bool(true)
bool(true)
bool(true)
Click me to see the sample solution
131. Write a PHP program to check a given array (length will be atleast 2) of integers and return true if there are two values 15, 15 next to each other.
Sample Input:
{ 5, 5, 1, 15, 15 }
{ 15, 2, 3, 4, 15 }
{ 3, 3, 15, 15, 5, 5}
{ 1, 5, 15, 7, 8, 15}
Sample Output:
bool(true)
bool(false)
bool(true)
bool(false)
Click me to see the sample solution
132. Write a PHP program to find the larger average value between the first and the second half of a given array of integers and minimum length is atleast 2. Assume that the second half begins at index (array length)/2.
Sample Input:
{ 1, 2, 3, 4, 6, 8 }
{ 15, 2, 3, 4, 15, 11 }
Sample Output:
6
10
Click me to see the sample solution
133. Write a PHP program to count the number of strings with given length in given array of strings.
Sample Input:
{"a", "b", "bb", "c", "ccc" }, 1
Sample Output:
Number of array:
3
Click me to see the sample solution
134. Write a PHP program to create a new array using the first n strings from a given array of strings. (n>=1 and <=length of the array).
Sample Input:
{"a", "b", "bb", "c", "ccc" }, 3
Sample Output:
a b bb
Click me to see the sample solution
135. Write a PHP program to check a positive integer and return true if it contains a number 2.
Sample Input:
123
13
222
Sample Output:
bool(true)
bool(false)
bool(true)
Click me to see the sample solution
136. Write a PHP program to create a new array of given length using the odd numbers from a given array of positive integers.
Sample Input:
{1,2,3,5,7,9,10},3
Sample Output:
New array: 1,3,5
Click me to see the sample solution
PHP Code Editor:
More to Come !
Do not submit any solution of the above exercises at here, if you want to contribute go to the appropriate exercise page.

Follow us on Facebook and Twitter for latest update. | 11,195 | 36,779 | {"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-2024-33 | latest | en | 0.7705 |
https://timesrussia.com/fixed-spring-hinge-character-configurable-joint-explained-unity-tutorial/ | 1,585,999,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370521876.48/warc/CC-MAIN-20200404103932-20200404133932-00077.warc.gz | 722,915,414 | 19,775 | # Fixed, Spring, Hinge, Character & Configurable Joint explained – Unity Tutorial
In this video we will talk about all unity joins, so we will talk about fixed joint spring joints hinge joints character joints and configurable joints Yes we’re talking about all of them and you can find the time codes in the Descriptions so that you can skip all the parts. You don’t need at the time of the recording. You can’t find any Tutorial on YouTube that covers up all the joints of unity. So this is very exciting Let’s start. So we’re creating a plane position the plane at 0 0 0 and Yes, there we go we create one 3d object cube and We create another video. You object cube So let’s move up this a little bit more up and We create a rigid body This is kinematic so it won’t be affected by anything. So no gravity No other coli or no other objects will ever move this box Except I move it. So and This one is the other widget or e It uses gravity. Everything can be ever this by default and We start with the first joint, which is a fixed joint. So the fixed joint is very simple you can just Connect it to any other rigid body. So I Maybe I call this the connected rigid body So then it’s a little bit more clear Here we have added the joint and this is a connected rigid body there’s a break and a Force a break force and a break talk and you can enable collision what that means is Let’s create another cube and call this tube three cube and This has also a rigid body and has no gravity as you can see nothing happens So but this is a connect wicked rigid body. And as soon as I move the connected rigid body the other rigid body will also be moved by unity because it is connected to the connected rigidbody and Here is my free rigid body. I can’t move it into the other box but what I could do is I can just Set this break and talk And this break force at work force for example to 100? Maybe in edit mode 100 100 and this means if the force of 100 is applied to this box then this Complete joint will break so now it’s double I can move it. But as soon as I hit this very hard like this. Oh Okay, it wasn’t hard enough or the connection is too strong let’s set it down to 10 and try it again There we go, we break it now it’s completely broken The complete connection on the complete joint is gone now and you can do this with every joint so I’ll skip this in the next joint Next up is spring joint You do not have to set a connected which body but it makes things a little bit clearer Let’s start with a rigid body. Let’s remove the ground and I think we can start right there the Initial values I could enough for us and here you can see it’s connected to this one and Here is a button you can see it’s anchored right there. So when I move it As an anchor is right there. I can set the anchor. This is the first one to the middle and As you can see the anchor is in the middle It behaves a little bit different So it rotates a little bit more if I set the anchor back to 0.5 Its Again anchored at the top and you see the bounce pop there is that it’s an anchor So it goes a little bit down every time I change it here because it’s also configure the anchor I can just uncheck the box and then it should be a little bit easier to Change these values during one time. Yes. There we go And I can move this rigid body as well because this is the connected one if I do not have connected which body Then everything looks like this It’s still a joint but it’s not connected to this rigid body anymore Let’s be connected and Let’s talk about the other values if I increase the spring you can see it’s a little bit stronger I can increase damper decreases doing spring again. And you see it’s moving way slow work You see its damping. It’s not that It bounces so often So if I set the damper very high Let the cube fall, oh it’s not falling anymore. It’s it’s So and now You see There’s a high damper so these are the values booing and damper say They are returning and occurring in all the other joints, so keep in mind this is a spring as the force that is applied to the rigid body to reach a Specific position and the damper is a damper so it dams it on the way to the position if you have a spring and the damper there’s often a minute max distance and this indicates where the force is applied and where as a position or the target position of the rigid body is for example now, it’s Directly in this brown spot if I set it to one the target position is in the circle around a Brown dot so it’s either here or up here The force is applied always to the nearest position and the box twice to be around in this shape next up hinge joint as you can see it’s just this axis and the complete widget body is moving allowances access and you can use a connected rigid body to Just move it with a rigid body so, for example, I can connect it to our connected rigid body and now I can move the connected rigid body and The widget body underneath it or connected to it will move sim simultaneously and I can change the axis now. It’s X-axis I can set it to the z-axis and now it’s turning around this z axis I Can change the anchor? Which is maybe interesting, for example, I can set it to zero now the anchor is Directly in the middle of the box And if I rotate or move the box, maybe I have to potate it If I use the other cube to have some rotation There you go. You see it’s rotating around Itself because the anchor is set in the middle. Let’s move the anchor up a bit now It’s hanging Around here. And now we have the same concepts coming over and over again. Here’s the spring again For example, I can enter a spring value of 100 say ok. I will use the spring and avataric position and This tube tries to go to the target position with the force of the spring I can set it to 90 to rotate it around 90 degrees. I can set it to 180 I can we as a forest so That it tries to go to this position, but with a lower force maybe a little bit lower and then You see it tries to get to a certain position What’s really interesting is also if I sets anchor to zero again so that it’s rotating I can use the motor and the motor is just Too Force a rotation so you have sus insured and the motor always and constantly Adds force to it so that it rotates around this hinge axis, and I can use limit for example – 90 degrees and 90 degrees Check use limit and even if I use the motor to Rotate it. It will stop Directly here where the limit is I can say? Okay bounce a little bit at the limit then it bounces as you can see here and That’s everything. I got to say about the hinge joint next up the character joint This is a little bit more complex joint. It’s used for Characters and specific for ragdolls so you can use it on arm or something if you started it just Looks like a normal hinge joint But there’s one difference as soon as I collide with another object It’s in each joint and it can rotate. So this is a really real difference between those two Let’s see what you can do we can set a connected body as always so that it follows the connected body It behaves a little bit different you see it right here We have an access this is the main axis where this inch rotation or Galatian from the hinge joint is applied You can say twist swing and Twist limit and The same with a swing so the swing is everything that the hinge joint is also doing and the twist is just the rotation about this axis and you see The limit that already there I can’t rotate over a certain amount of degrees one last thing that is really helpful is Joined angular limit button here. You can set the limits here. It’s a little bit more visual for example as soon as I Decrease the rotation here It does not rotate anymore that much for example, I try to rotate it. It will always Reach the limited immediately so we can set the lab a little as a limit a little bit higher Or I can just change this limit or this one okay, the last one and the most interesting one is a configurable joint because this joint is very complex and unity Itself says you can do everything with it. You can do with other joints. So it’s a hinge joint It’s a spring joint it two-character joint. It’s a fixed joint. You can do everything you want So and this is a reason why it’s so interesting so you can see Everything you can know already from the character joint. You can have a connected rigidbody. Let’s started right away For example We can lock everything x motion Y motion that motion for example now it can lock the Y motion and We will see that the box is not falling down anymore as soon as we set it free it goes down so we can do it with every motion and also With the rotation we can set limits in every single way So spring limits or linear limits angular spring limits low angular X limits Y limits that limits wise at limits and one of the most interesting things is here the target position for example, the target position is now 0 0 0 that’s where the object is right now and You can use these drive values to add some force to the object so that it always goes back to the target position and the target philosophy velocity and Let’s add a hundred To everything and let’s see what will happen and as you can see as soon as I Drag it out of the position it always twice to get there as soon as possible always a force of 100 This is really interesting. But the rotation is a little bit strange It does not try to get back to the desired rotation and as you can see it Already uses an anchor for the force. I can set the anchor to zero now. It’s rotating Around every single axis we can do the same thing for the rotation. So the target rotation is here zero zero zero and We can Set the angular drive for X Y Z and No x y&z not this one And now the cube tries to go back to this position From everywhere. So I Can’t rotate the cube. I can move it it always twice to go back to this position and This is very interesting. Especially if you have a connected rigidbody It tries to go back to the position underneath the rigidbody and What we can do now is to use a configurable joint on The other widget body as well Connected to our cube And says is not Cannon Matic anymore and we can do as an exact same thing So we set the forces and now this cube tries to stay above this cube and this cubes that tries to stay above this cube, and they all using is spring and Maybe the spring value is a little bit hard Let’s lower the values a little bit maybe a long 10 and this is interesting. So they try to Avoid it each other and they are connected in some ways so that This is like a little jumping character instead of the angular drive you could use Slope drive, and this means this loved wife always tries to rotate around the axis with the least amount of degrees to the target rotation and It’s basically a little bit smoother and a little bit faster to move around this axis We could divide out. It should be the same result in general. Yes, as you can see It’s a little bit more stable even though I rotate the cubes a little bit so here are so many properties and If you have questions to one of these in specific, I can really recommend use a help page You can reach it via the questionnaire on the pork right here You click it and here are all the things in detail in the unity documentation with some graphics this is really one of the Components and said that has a really big documentation in contrast to the others So I hope you liked the video It was a little bit longer, but this is one of the videos I really wanted to make because it covers up all the joints that are and unity
## 33 thoughts on “Fixed, Spring, Hinge, Character & Configurable Joint explained – Unity Tutorial”
1. Nitin Joseph says:
Good Video Dude! Thanks!
2. Paul Robinson says:
Nice tutorial again. I am new to game dev and was wondering if you could do a tutorial based on a mobile multi level game, explaining setting up of where the touch controller screen goes and a bit of explanation about correct use of 'game manager'. Is it possible to have the main character teleport into the level start and teleport out at the end of each level. Any help on these matters would be amazing, thanks, Paul
3. Anthony der says:
Why is every video on spring joint 100% the same? I don't think anyone really understand them…
4. storm says:
Please make another tutorial
5. W N says:
how would I attach my character(which has a fixed joint) to a object like a cube through script? for instance I want to have a interaction script on the cube I want to pick up and have a interaction radius of about 0.4 so that when the character enters the interaction zone the fixed joint automatically enables between the two objects and he can move around with it
6. Esben Rasmussen says:
Nice tutorial, thanks mate
7. Arnaud B says:
Nice tutorial thanks! what kind of joint can be rigged with more than 1 other objets?
8. Roy Meredith says:
How would connect a spring or hinged joint using c# script, ?
9. Gendalf Gray says:
Hellow, nice tutorial!
But can you explain how ancor axis in configurable joint works? I suppose it needs for turning limit area.
I'm making body with 2 limbs attached to it. And those limbs must have assimetrical angle limits on 2 direction.
1) how rotate areas with this axis? Only x angle axis have min max setting.
2) also areas on both limbs looking in one direction, but i need them to be mirrored. How to mirror them?
I tried to change axis values and areas moved in weird way.
May be i should use several hinge joints, but i think it will be harder to script.
If so, i still want to know how axis work.
10. Tr Gamer says:
Hello, why my hinge joint is looks like 2d ? i need to make like you but it looks circle, i need spherical hinge joint ? its only effected horizontal
11. singleton pattern says:
Thank you mery much ,good tutorial .
12. 26PM says:
Thank you! You really helped me understand configurable joints.
13. Mikkyle Matthews says:
You think you could help me out: https://answers.unity.com/questions/1579139/hinge-joints-interfering-with-each-other.html
14. Olivia Cobb says:
just what I needed, thank you
15. Andrea Maiellaro says:
So, for having a Punching Bag effect should I use Spring Joint??I think It makes sense….what do you think about ?
16. Redman One says:
Дякую(Ukraine)
17. xcogames says:
GREAT UPLOAD!!
18. Ossobuko says:
doesn't explain all the variables.
19. Lewis Scrivens says:
These are so much better than the physic constraints in ue4 they're so limited
20. Ryutaku Zaki says:
Tactic like. If the video is bad I’ll be back to dislike
21. Ryutaku Zaki says:
Naaah, I hate improvised tutorials. You aren’t explaining
22. Necronomicron says:
16:30 Shouldn't you have changed Rotation Drive Mode to Slerp as well? You basically just disabled springs here.
23. Ziggy's Theorem says:
I can't afford health care
24. Dan Opris says:
slider joint ?
25. AZURE BATMAN says:
Can you guide me where should a beginner like me start from i don't know about any of these like why to use this ?
26. Ben Zen says:
great video! I caught a really good second lesson and that is your rapid prototyping technique on a test project. This will save me a ton of time!
27. Dance Hall says:
Newbie here. Would a spring joint approach make sense for trying to create a pole vaulting mechanic? Or what would be a good approach for creating pole vaulting?
28. Drew Mileham says:
One of the best Unity tuts ive seen. Thanks
29. sword111000 says:
Hi you seem to be the guys who may be able to help regarding, physics and configurable joints !! Im trying to build a real time bouncy blade for my fencing game www.ufencevr.com where the blade bends using physics both by gravity at its end, and bending uniformly against a collider , like the real thing, I have faked this using animations and physics combined, but its not the ideal solution. Is it possible to set up this type of action using configrerabe joints ? Ive tried every combination but I just don’t know enough to get to the bottom of it any ideas https://youtu.be/saWmHK4XETI ?? Best Hugo
HOME | ufencevr
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30. juhohyv says:
thank you and have a nice day 🙂
31. PepsiFruit says:
"Heres the time codes except you can press them"
Setup 0:26
Fixed joint 1:15
Spring joint 3:33
Hinge joint 6:54
Character joint 9:52
Configurable joint 11:55
32. Thayakorn Rakwetpakorn says:
2:11 notice that he did that in play mode
33. two cee says:
All this talk about joints is making me wanna roll one | 3,742 | 16,527 | {"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-2020-16 | latest | en | 0.934382 |
https://www.aqua-calc.com/calculate/food-calories/substance/weis-blank-quality-coma-and-blank-small-blank-tender-blank-sweet-blank-peas-coma-and-blank-upc-column--blank-041497181049 | 1,576,122,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540536855.78/warc/CC-MAIN-20191212023648-20191212051648-00356.warc.gz | 620,023,357 | 7,379 | # Calories of WEIS QUALITY, SMALL TENDER SWEET PEAS, UPC: 041497181049
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#### See how many calories in0.1 kg (0.22 lbs) of WEISQUALITY, SMALL TENDERSWEET PEAS, UPC:041497181049
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 56 234.3 Total 56 234.3
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 56 Protein g 3.2 Total lipid (fat) g 0.4 Carbohydrate,bydifference g 9.6 Fiber,totaldietary g 2.4 Sugars, total g 4.8 Minerals Calcium, Ca mg 16 Iron, Fe mg 1.15 Potassium, K mg 84 Sodium, Na mg 240 Vitamins Vitamin C,totalascorbic acid mg 9.6 Vitamin A, IU IU 240 Lipids Fatty acids,totalsaturated g 0 Cholesterol mg 0
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
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• WEIS QUALITY, SMALL TENDER SWEET PEAS, UPC: 041497181049 weigh(s) 264.17 gram per (metric cup) or 8.82 ounce per (US cup), and contain(s) 56 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
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• Reference (ID: 10801)
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#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Bahamas Oolite weighs 1 537.8 kg/m³ (96.00172 lb/ft³) with specific gravity of 1.5378 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Coal naphtha [C6H6] weighs 878.65 kg/m³ (54.85233 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 30 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
watt hour (Wh) is a derived unit of energy equal to 3600 joules, and expressed as power [watt (W)] multiplied by time [hour (h)].
The online number conversions include conversions between binary, octal, decimal and hexadecimal numbers.
troy/US qt to dwt/in³ conversion table, troy/US qt to dwt/in³ unit converter or convert between all units of density measurement.
#### Calculators
Compute cost of oils, fuels, refrigerants at different temperatures | 977 | 3,498 | {"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-2019-51 | latest | en | 0.650502 |
http://slides.com/michaelfreeman/simulation | 1,638,537,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00390.warc.gz | 68,715,940 | 14,668 | # How?
## Predicting health burden...
### How much sickness is experienced by a pop. of 1000 people?
Number Flu = Pop. * Flu Probability = 100
$Number Flu = Pop. * Flu Probability = 100$
Flu Burden = Number Flu * Flu Disability = 5
$Flu Burden = Number Flu * Flu Disability = 5$
## Predicting health burden...
### Flu is .05 disabling, pneumonia is .1 disabling
Disability = 1000 * (Flu Prob * Flu Disability + Pneu. Prob. * Pneu. Disability)
$Disability = 1000 * (Flu Prob * Flu Disability + Pneu. Prob. * Pneu. Disability)$
### How much sickness is experienced by a pop. of 1000 people?
Disability = 25
$Disability = 25$
### Pneu.
DW(combined) = 1 - (1 - flu) * (1 - pneu)
$DW(combined) = 1 - (1 - flu) * (1 - pneu)$
DW(combined) = .145
$DW(combined) = .145$
# Simulation!
## Random number generators
### Randomly sample out of a vector
# Given a 10% probability, simulate 1000 people getting the flu
people <- rbinom(n = 1000, size = 1, prob = .2)
# Generate a normal distribution of grades for 20 students
grades <- rnorm(20, mean = 80, sd = 10)
# Get the eye color for 100 people: blue, green, or hazel
colors <- c('blue', 'green', 'hazel')
eye_colors <- sample(colors, 100, replace = TRUE)
There is a very long, straight highway with some number of cars (N) placed somewhere along it, randomly. The highway is only one lane, so the cars can’t pass each other. Each car is going in the same direction, and each driver has a distinct positive speed at which she prefers to travel. Each preferred speed is chosen at random. Each driver travels at her preferred speed unless she gets stuck behind a slower car, in which case she remains stuck behind the slower car. On average, how many groups of cars will eventually form? (A group is one or more cars traveling at the same speed.)
For example, if the car in the very front happens to be slowest, there will be exactly one group — everybody will eventually pile up behind the slowpoke. If the cars happen to end up in order, fastest to slowest, there will be groups — no car ever gets stuck behind a slower car.
## Initial Steps
### Write a function to test if a car creates a new group
# Simulate 100 cars w/mean speed 50
cars <- rnorm(n = 100, mean = 50, sd = 5)
# A function to determine if a car is slower than all of the cars
# in front of it (which createa a new group of cars **behind** it)
slower_than <- function(index) {
return(cars[index] < min(cars[1:index - 1]))
}
### Apply the function to the list of indicies
# Apply the slower_than function to all of the cars
new_group <- lapply(2:length(cars), slower_than)
## Initial Steps
# Simulate 100 cars w/mean speed 50
cars <- rnorm(n = 1000, mean = 50, sd = 5)
# A function to determine if a car is slower than all of the cars
# in front of it (which createa a new group of cars **behind** it)
slower_than <- function(index) {
return(cars[index] < min(cars[1:index - 1]))
}
# Apply the slower_than function to all of the cars
new_group <- lapply(2:length(cars), slower_than)
# Determine number of groups created
groups <- length(new_group[new_group == TRUE]) + 1
## Better yet....
simulate_groups <- function() {
# Simulate 100 cars w/mean speed 50
cars <- rnorm(n = 1000, mean = 50, sd = 5)
# A function to determine if a car is slower than all of the cars
# in front of it (which createa a new group of cars **behind** it)
slower_than <- function(index) {
return(cars[index] < min(cars[1:index - 1]))
}
# Apply the slower_than function to all of the cars
new_group <- lapply(2:length(cars), slower_than)
# Determine number of groups created
groups <- length(new_group[new_group == TRUE]) + 1
return(groups)
}
## Loops
### Pass a different element into the block of code in each iteration
items <- 1:10
# Loop through each element in your vector items
for(i in items) {
print(2*i) # i takes on the identity of each element in the vector items
}
## Repeating your simulation
repeat_simulation <- function(num_sims) {
# Create an empty vector to store your results
results <- vector()
# Run your simulation 100 times, and track your results
for(i in 1:num_sims) {
results <- c(results, simulate_groups())
}
# Work with your results
hist(results)
return(mean(results))
}
## Parameterizing your simulation
simulate_groups <- function(mean, sd, num_cars) {
# Simulate 100 cars w/mean speed 50
cars <- rnorm(n = num_cars, mean = mean, sd = sd)
# A function to determine if a car is slower than all of the cars
# in front of it (which createa a new group of cars **behind** it)
slower_than <- function(index) {
return(cars[index] < min(cars[1:index - 1]))
}
# Apply the slower_than function to all of the cars
new_group <- lapply(2:length(cars), slower_than)
# Determine number of groups created
groups <- length(new_group[new_group == TRUE]) + 1
return(groups)
}
## Create an interface: ui.R
# Create UI
shinyUI(fluidPage(
# UI for the traffic simulation
titlePanel('Traffic Simulation'),
# Controls
sidebarLayout(
sidebarPanel(
sliderInput("num_cars", "Number of Cars:",
min = 10, max = 1000, value = 100, step = 10),
sliderInput("num_sims", "Iterations of Simulation",
min = 10, max = 1000, value = 100, step= 10),
sliderInput("speed", "Average Speed",
min = 10, max = 150, value = 40, step= 5),
sliderInput("deviation", "Speed Deviation",
min = 1, max = 20, value = 5, step= 1)
),
# Render plot
mainPanel(
plotOutput("histogram")
)
)
))
## Run your simulation: server.R
# Run the traffic simulation
source('traffic_sim.R')
shinyServer(function(input, output) {
output$histogram <- renderPlot({ repeat_simulation( num_sims = input$num_sims,
mean = input$speed, sd = input$deviation,
num_cars = input\$num_cars
)
})
})
## Assignments
### Keep working on your final projects!
#### simulation
By Michael Freeman
• 1,007 | 1,598 | 5,782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.877046 |
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Anonymous
Not applicable
## Tough One: Margin Mix Variance and solving wrong totals
First, this is the second post. My first one disappeared. I recommend keeping a copy of your posts in a text file as this has happened before.
I want to automate the impact of product mix change on Margin in BI. I get the wrong totals for the critical calculations and it is the total that matters.
Below is a snip from Excel showing the formulas and with A,B,C column type references too.
I get correct answers for all but E,F
Calculations
A From Data Table of detailed sales
B From Data Table
C CY Mix = _Sales[Revenue] / CALCULATE( _Sales[Revenue] , ALLSELECTED() )
D=AxC Sales PY x CY MIX = _Sales[Revenue PY] * CY Mix
E=BxD Margin% PY x (Sales PY x CY MIX) = _Sales[Margin% PY] * Sales PY x CY MIX
F=E/D but made with virtual table that incorporates the above measures
GM Rate =
SUMX(
SUMMARIZE( VALUES( 'Product'[ProductCategory1ID] ) ,
" Sales PY x CY MIX " , [Sales PY x CY MIX] ,
" Margin% PY x (Sales PY x CY MIX)" , [Margin% PY x (Sales PY x CY MIX)] ) ,
( [Margin% PY x (Sales PY x CY MIX)] / [Sales PY x CY MIX] ) )
It could be the ALLSELECTED() in Colum C calculation or/and column F virtual table calculation
Thyank you brainiacs!
Excel Visual of the data:
DATA DATA DATA DATA DATA Correct rows, wrong total Correct rows, wrong total A B C = A Row /A Total D = A x C E = B x D F = E / D Product Sales Margin% Sales PY Margin% PY CY Mix Sales PY x CY MIX Margin% PY x (Sales PY x CY MIX) GM Rate One 22,135,134 26.4% 20,481,961 28.6% 19.0% 20,093,108 5,739,249 28.6% Two 17,637,550 32.9% 20,627,738 33.2% 15.2% 16,010,439 5,309,906 33.2% Three 18,845,322 28.2% 20,558,646 29.0% 16.2% 17,106,790 4,955,420 29.0% Four 22,765,600 20.2% 10,038,249 21.5% 19.6% 20,665,412 4,451,544 21.5% Five 11,881,163 23.6% 13,680,710 23.7% 10.2% 10,785,094 2,560,293 23.7% Six 6,366,515 32.1% 5,473,459 33.0% 5.5% 5,779,187 1,908,000 33.0% Seven 3,448,453 32.3% 3,712,855 31.6% 3.0% 3,130,324 989,455 31.6% Eight 3,894,096 21.3% 4,918,662 22.1% 3.4% 3,534,855 781,763 22.1% nine 6,350,536 8.7% 3,136,442 10.7% 5.5% 5,764,682 614,478 10.7% Ten 2,079,378 29.9% 2,101,299 32.2% 1.8% 1,887,549 607,044 32.2% Eleven 836,071 51.8% 787,161 33.6% 0.7% 758,941 255,368 33.6% Twelve (567) 160.1% 0.0% Thirteen 256 0.0% 0.0% 233 Wrong Total Wrong Total 116,240,074 25.8% 105,516,615 27.9% 100% 105,516,614 29,399,456 27.9% Total Should Be Total Should Be 28,172,520 26.7%
2 ACCEPTED SOLUTIONS
Microsoft
Hi @Anonymous
You may refer to below measure: If it is not your case, please share your data sample file which could reproduce your scenario and your desired output, you can upload it to OneDrive or Dropbox and post the link here. Show a simplified sample as below:
```C =
CALCULATE ( SUM ( Table[Sales PY] ) )
/ CALCULATE ( SUM ( Table[Sales PY] ), ALL ( Table ) )```
```E =
SUMX (
SUMMARIZE ( Table, Table[Product] ),
CALCULATE ( SUM ( Table[Margin%PY] ) ) * [D]
)```
```F =
SUMX ( SUMMARIZE ( Table, Table[Product] ), [E] / [D] )```
Regards,
Cherie
Community Support Team _ Cherie Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
Hi Cherie,
We solved this together!
I need ALLSELECTED in [ C ] “Product Revenue Percent of Total” as that will calculate correctly with filters applied.
This will deliver correct rows and totals with filters applied. Your help is new measure E-TOTAL, thank you.
This stream of measures can be consolidated and I most likely will do that.
Cheers!
A = Prior Year Revenue
B = Prior Year Margin %
C = [Current Year Revenue] / CALCULATE( [Current Year Revenue] ,ALLSELECTED())
D = CALCULATE( [ A ], ALLSELECTED() ) * [ C ]
E = [ D ] * [ B ] (This is the Line Item Values, the column actually comes from the virtual table below)
New Measure: This is your formula for the Total of the Margin Mix Column and comes out of the virtual table below:
E-TOTAL = SUMX(
SUMMARIZE( Table3, [Product] ) ,
CALCULATE( [B] ) * [D] )
Virtual Table:
MIX Margin PY at CY Mix =
VAR Margin_PY_at_CY = [ E ]
VAR Margin_PY_TOTAL = CALCULATE( [ E-Total ] , FILTER( Table3 , [ Product ] <> BLANK() ) )
RETURN
IF( HASONEVALUE( Table3[Product] ) ,
Margin_PY_at_CY ,
Margin_PY_TOTAL )
F = New Margin % = MIX Margin PY at CY Mix / [ D ]
4 REPLIES 4
Microsoft
Hi @Anonymous
You may refer to below measure: If it is not your case, please share your data sample file which could reproduce your scenario and your desired output, you can upload it to OneDrive or Dropbox and post the link here. Show a simplified sample as below:
```C =
CALCULATE ( SUM ( Table[Sales PY] ) )
/ CALCULATE ( SUM ( Table[Sales PY] ), ALL ( Table ) )```
```E =
SUMX (
SUMMARIZE ( Table, Table[Product] ),
CALCULATE ( SUM ( Table[Margin%PY] ) ) * [D]
)```
```F =
SUMX ( SUMMARIZE ( Table, Table[Product] ), [E] / [D] )```
Regards,
Cherie
Community Support Team _ Cherie Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
Oh so close.
When I use these I get incorrect rows and correct total. That's the opposite oif my formuls which is correct at rows and incorrect total.
I'm trying to incorporate an IF( HASONEVALUE() to combine them but am failing.
For C = My formula incorporates ALLSELECTED(). I could not get ALL() to work.
I'll get back to you my friend with what I ended up with or I will create a data set if i fail.
Cheers!
Microsoft
Hi @Anonymous
I've tried to use ALLSELECTED() for C with my test data. Show a sample as below for you to check if it could help you.
`D =IF ( HASONEVALUE ( Table3[Product] ), CALCULATE ( SUM ( Table3[Sales PY] ) ) * [C], SUMX ( SUMMARIZE ( Table3, Table3[Product] ), CALCULATE ( SUM ( Table3[Sales PY] ) ) * [C] ))`
```E =
IF (
HASONEVALUE ( Table3[Product] ),
CALCULATE ( SUM ( Table3[Margin%PY] ) ) * [D],
SUMX (
SUMMARIZE ( Table3, Table3[Product] ),
CALCULATE ( SUM ( Table3[Margin%PY] ) ) * [D]
)
)```
`F = SUMX(SUMMARIZE(Table3,Table3[Product]),[E]/[D])`
Regards,
Cherie
Community Support Team _ Cherie Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
Hi Cherie,
We solved this together!
I need ALLSELECTED in [ C ] “Product Revenue Percent of Total” as that will calculate correctly with filters applied.
This will deliver correct rows and totals with filters applied. Your help is new measure E-TOTAL, thank you.
This stream of measures can be consolidated and I most likely will do that.
Cheers!
A = Prior Year Revenue
B = Prior Year Margin %
C = [Current Year Revenue] / CALCULATE( [Current Year Revenue] ,ALLSELECTED())
D = CALCULATE( [ A ], ALLSELECTED() ) * [ C ]
E = [ D ] * [ B ] (This is the Line Item Values, the column actually comes from the virtual table below)
New Measure: This is your formula for the Total of the Margin Mix Column and comes out of the virtual table below:
E-TOTAL = SUMX(
SUMMARIZE( Table3, [Product] ) ,
CALCULATE( [B] ) * [D] )
Virtual Table:
MIX Margin PY at CY Mix =
VAR Margin_PY_at_CY = [ E ]
VAR Margin_PY_TOTAL = CALCULATE( [ E-Total ] , FILTER( Table3 , [ Product ] <> BLANK() ) )
RETURN
IF( HASONEVALUE( Table3[Product] ) ,
Margin_PY_at_CY ,
Margin_PY_TOTAL )
F = New Margin % = MIX Margin PY at CY Mix / [ D ]
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# Is xy=21 linear or nonlinear
0
1. 6 September, 00:16
0
I believe xy＝21 is nonlinear.
An example of a linear equation would be, y＝5x+4 or 7x＝8+2y.
2. 6 September, 01:11
0
Xy=21 is nonlinear.
I think that because the definitions for nonlinear are:
not denoting, involving, or arranged in a straight line.
and
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Advanced Analysis 1: Metric Spaces and Applications (MATH3116)
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Analysis 1 is a foundational course in Mathematics, leading on to other areas of analysis, such as topology and measure theory, complex analysis, functional analysis, and harmonic analysis. It also provides important tools for application areas such as theoretical computer science, physics and engineering.
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## Indicative Assessment
1. Assignments (20) [LO 1,2,3,4]
2. Midsemester exam (30-40%) (30) [LO 1,2,3,4]
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## Inherent Requirements
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To enrol in this course you must have successfully completed MATH2305 and MATH2306 with a mark of 80 and above or MATH 2405 with a mark of 60 and above. You are not able to enrol in this course if you have previously completed MATH2320.
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Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### First Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
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http://saslist.com/blog/2020/02/05/finding-possible-data-errors-using-the-auto_outliers-macro/ | 1,627,770,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00452.warc.gz | 33,035,939 | 11,390 | One of the first and most important steps in analyzing data, whether for descriptive or inferential statistical tasks, is to check for possible errors in your data. In my book, Cody's Data Cleaning Techniques Using SAS, Third Edition, I describe a macro called %Auto_Outliers. This macro allows you to search for possible data errors in one or more variables with a simple macro call.
### Example Statistics
To demonstrate how useful and necessary it is to check your data before starting your analysis, take a look at the statistics on heart rate from a data set called Patients (in the Clean library) that contains an ID variable (Patno) and another variable representing heart rate (HR). This is one of the data sets I used in my book to demonstrate data cleaning techniques. Here is output from PROC MEANS:
The mean of 79 seems a bit high for normal adults, but the standard deviation is clearly too large. As you will see later in the example, there was one person with a heart rate of 90.0 but the value was entered as 900 by mistake (shown as the maximum value in the output). A severe outlier can have a strong effect on the mean but an even stronger effect on the standard deviation. If you recall, one step in computing a standard deviation is to subtract each value from the mean and square that difference. This causes an outlier to have a huge effect on the standard deviation.
### Macro
Let's run the %Auto_Outliers macro on this data set to check for possible outliers (that may or may not be errors).
Here is the call:
```%Auto_Outliers(Dsn=Clean.Patients,
Id=Patno,
Var_List=HR SBP DBP,
Trim=.1,
N_Sd=2.5)```
This macro call is looking for possible errors in three variables (HR, SBP, and DBP); however, we will only look at HR for this example. Setting the value of Trim equal to .1 specifies that you want to remove the top and bottom 10% of the data values before computing the mean and standard deviation. The value of N_Sd (number of standard deviations) specifies that you want to list any heart rate beyond 2.5 trimmed standard deviations from the mean.
### Result
Here is the result:
After checking every value, it turned out that every value except the one for patient 003 (HR = 56) was a data error. Let's see the mean and standard deviation after these data points are removed.
Notice the Mean is now 71.3 and the standard deviation is 11.5. You can see why it so important to check your data before performing any analysis.
You can download this macro and all the other macros in my data cleaning book by going to support.sas.com/cody. Scroll down to Cody's Data Cleaning Techniques Using SAS, and click on the link named "Example Code and Data." This will download a file containing all the programs, macros, and data files from the book. By the way, you can do this with any of my books published by SAS Press, and it is FREE!
Let me know if you have questions in the comments section, and may your data always be clean! To learn more about SAS Press, check out up-and-coming titles, and to receive exclusive discounts make sure to subscribe to the newsletter.
Finding Possible Data Errors Using the %Auto_Outliers Macro was published on SAS Users. | 706 | 3,195 | {"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} | 3.265625 | 3 | CC-MAIN-2021-31 | longest | en | 0.890704 |
https://drtecho.com/operations-with-integers-worksheet/ | 1,624,628,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00146.warc.gz | 204,558,800 | 8,849 | # Operations with integers worksheet Latest
Written by Ines Apr 28, 2021 ยท 10 min read
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Operations With Integers Worksheet. Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Once you find your worksheet click on pop-out icon or print icon to worksheet. Some of the worksheets for this concept are Integer operations review name Order of operations with integers work Order of operations work order of operations with All operations with integers a Ooperations with integersperations with integers Integer order of operations Integer. Math integers add subtract multiply divide negative positive operations.
Https Www Dadsworksheets Com Negative Numbers Addition Worksheet And Subtraction Worksheet 3 Negative Numbers Worksheet Negative Numbers Integers Worksheet From pinterest.com
Math integers add subtract multiply divide negative positive operations. Some of the worksheets for this concept are Word problems with integers Word problem practice workbook Math review packet Addingsubtracting integers date period Integer addition subtraction multiplication and division Operations with integers Basic integral representations and absolute value state Unit 4 integers. Integers Worksheet – All Operations with Integers Range -50 to 50 with All Integers in Parentheses Author. Chapter 3 Lesson 15 Transcript. Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Order of operations with integers and exponents worksheet.
### Some of the worksheets for this concept are Grade 8 integers Basic integral representations and absolute value state Integer operations review name Word problems with integers Adding and subtracting integers Unit 1 grade 8 integers and algebraic expressions 8th to 9th grade summer math.
6 5 7. Displaying top 8 worksheets found for - Integer For Grade 8. 413 6 5. Once you find your worksheet click on pop-out icon or print icon to worksheet. Integer For Grade 8. 8 42 52.
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Integers Worksheet – All Operations with Integers Range -50 to 50 with All Integers in Parentheses Author. 6 5 7. Operations with Integers Worksheets for 6th Grade and 7th Grade. Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Operations with integers interactive worksheet Integers online activity for Grade 7.
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Exclusive pages to compare and order integers and representing integers on a number line are given here with a variety of activities and exercises. Operations with integers interactive worksheet Integers online activity for Grade 7. Some of the worksheets for this concept are Integer operations review name Order of operations with integers work Order of operations work order of operations with All operations with integers a Ooperations with integersperations with integers Integer order of operations Integer. Some of the worksheets for this concept are Word problems with integers Word problem practice workbook Math review packet Addingsubtracting integers date period Integer addition subtraction multiplication and division Operations with integers Basic integral representations and absolute value state Unit 4 integers. Mixed Operations with Integers.
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Some of the worksheets for this concept are Order of operations work order of operations with Order of operations work order of operations with Working with integers adding rules Name operations with integers Operations with integers Adding positive and negative numbers date period Addsubtracting fractions and mixed numbers Exponents and order of operations. Once you find your worksheet click on pop-out icon or print icon to worksheet. 8 42 52. 6 2 15. Parenthesis brackets exponents division multiplication addition subtraction.
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Some of the worksheets displayed are Name operations with integers All operations with integers a Integer operations review name Basic integral representations and absolute value state Addingsubtracting integers date period Integers Integer addition subtraction multiplication and division Order of operations with integers work. Operations with Integers Worksheets for 6th Grade and 7th Grade. Some of the worksheets for this concept are Grade 8 integers Basic integral representations and absolute value state Integer operations review name Word problems with integers Adding and subtracting integers Unit 1 grade 8 integers and algebraic expressions 8th to 9th grade summer math. Offering three levels of difficulty the printable order of operations worksheets provide practice in using DMAS on expressions with 4 integers and 3 operators. You miss the.
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6 5 7. 413 6 5. Order of operations worksheet order of operations with whole numbers and no exponents three steps author. You miss the. Order of operations with integers 2-Step Order of Operations with Integers 3-Step Order of Operations with Integers 4-Step Order of Operations with Integers 5-Step Order of Operations with Integers 6-Step Order of Operations with Integers.
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Working with integers integer worksheets. Math integers add subtract multiply divide negative positive operations. You can do the exercises online or download the worksheet as pdf. 6 2 15. Mixed Operations with Integers.
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Chapter 3 Lesson 15 Transcript. Integer For Grade 8. 6 5 7. Integer worksheets contain a huge collection of practice pages based on the concepts of addition subtraction multiplication and division. Try it risk-free for 30 days.
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Parenthesis brackets exponents division multiplication addition subtraction. These grade 6 worksheets cover addition subtraction multiplication and division of integers integers are whole numbers no fractional or decimal part and can be negative or positive. Worksheet - Integer Operations. Some of the worksheets for this concept are Integer operations review name Order of operations with integers work Order of operations work order of operations with All operations with integers a Ooperations with integersperations with integers Integer order of operations Integer. Operations with Integers Worksheets for 6th Grade and 7th Grade.
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Order of operations with integers 2-Step Order of Operations with Integers 3-Step Order of Operations with Integers 4-Step Order of Operations with Integers 5-Step Order of Operations with Integers 6-Step Order of Operations with Integers. Some of the worksheets displayed are Performing combined operations on integers Order of operations pemdas practice work All matrix operations Operations with decimals review work Addition and subtraction when adding Function operations date period Exercise work Order of operations basic. Chapter 3 Lesson 15 Transcript. 8 42 52. You can do the exercises online or download the worksheet as pdf.
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Some of the worksheets displayed are Name operations with integers All operations with integers a Integer operations review name Basic integral representations and absolute value state Addingsubtracting integers date period Integers Integer addition subtraction multiplication and division Order of operations with integers work. These grade 6 worksheets cover addition subtraction multiplication and division of integers integers are whole numbers no fractional or decimal part and can be negative or positive. Some of the worksheets for this concept are Word problems with integers Word problem practice workbook Math review packet Addingsubtracting integers date period Integer addition subtraction multiplication and division Operations with integers Basic integral representations and absolute value state Unit 4 integers. Integer For Grade 8. You can do the exercises online or download the worksheet as pdf.
Source: pinterest.com
Integers Worksheet – All Operations with Integers Range -50 to 50 with All Integers in Parentheses Author. Mixed Operations with Integers. Some of the worksheets for this concept are Integer operations review name Order of operations with integers work Order of operations work order of operations with All operations with integers a Ooperations with integersperations with integers Integer order of operations Integer. Offering three levels of difficulty the printable order of operations worksheets provide practice in using DMAS on expressions with 4 integers and 3 operators. 32 6 2.
Source: pinterest.com
Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Offering three levels of difficulty the printable order of operations worksheets provide practice in using DMAS on expressions with 4 integers and 3 operators. Order of operations with integers and exponents worksheet. Displaying top 8 worksheets found for - Integer For Grade 8. 413 6 5.
Source: pinterest.com
6 2 15. Some of the worksheets for this concept are Order of operations work order of operations with Order of operations work order of operations with Working with integers adding rules Name operations with integers Operations with integers Adding positive and negative numbers date period Addsubtracting fractions and mixed numbers Exponents and order of operations. Worksheet - Integer Operations. Add Subtract Multiply. 8 42 52.
Source: pinterest.com
Displaying top 8 worksheets found for - Integer For Grade 8. Add Subtract Multiply. Order of operations with integers 2-Step Order of Operations with Integers 3-Step Order of Operations with Integers 4-Step Order of Operations with Integers 5-Step Order of Operations with Integers 6-Step Order of Operations with Integers. Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Integer worksheets contain a huge collection of practice pages based on the concepts of addition subtraction multiplication and division.
Source: pinterest.com
Parenthesis brackets exponents division multiplication addition subtraction. Try it risk-free for 30 days. Operations with integers interactive worksheet Integers online activity for Grade 7. Exclusive pages to compare and order integers and representing integers on a number line are given here with a variety of activities and exercises. Some of the worksheets displayed are Name operations with integers All operations with integers a Integer operations review name Basic integral representations and absolute value state Addingsubtracting integers date period Integers Integer addition subtraction multiplication and division Order of operations with integers work.
Source: pinterest.com
Mixed Operations with Integers. Some of the worksheets for this concept are Grade 8 integers Basic integral representations and absolute value state Integer operations review name Word problems with integers Adding and subtracting integers Unit 1 grade 8 integers and algebraic expressions 8th to 9th grade summer math. Order of operations with integers 2-Step Order of Operations with Integers 3-Step Order of Operations with Integers 4-Step Order of Operations with Integers 5-Step Order of Operations with Integers 6-Step Order of Operations with Integers. Working with integers integer worksheets. 6 2 15.
Source: pinterest.com
Mixed Operations with Integers. This page includes Integers worksheets for comparing and ordering integers adding subtracting multiplying and dividing integers and order of operations with integers. 8 42 52. 16112020 Order Of Operations Worksheets 8th Grade Math Worksheets Pemdas Worksheets Order Of Operations. Offering three levels of difficulty the printable order of operations worksheets provide practice in using DMAS on expressions with 4 integers and 3 operators.
Source: pinterest.com
Working with integers integer worksheets. Order of Operations with Integers Worksheet Circle the part of the expression that you would complete first. Integer For Grade 8. Subtracting Integers 1 1 3 _____ 2 2 -5 _____ 3 6 -9 _____ 4 -7 -1 _____ 5 -7 4 _____ 6 3 -2 _____ 7 -1 9 _____ 8 2 9 _____ 9 -8 -1 _____ Multiplying Integers. Operations with integers interactive worksheet Integers online activity for Grade 7.
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Feb 03 . 7 min read | 2,991 | 14,517 | {"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} | 3.359375 | 3 | CC-MAIN-2021-25 | latest | en | 0.819928 |
https://socratic.org/questions/how-do-you-solve-0-5-1-8-3-3 | 1,576,451,969,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00037.warc.gz | 538,763,636 | 5,877 | # How do you solve -0.5-1.8=-3.3?
By declaring that the given equation is not valid; since $\left(- 0.5 - 1.8 = - 2.3\right)$
$\textcolor{w h i t e}{\text{XXX}} - 0.5 x - 1.8 = - 3.3$
$\textcolor{w h i t e}{\text{XXX}} - 0.5 x = - 1.5$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = 3$ | 135 | 290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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} | 3.984375 | 4 | CC-MAIN-2019-51 | longest | en | 0.472029 |
https://www.genbays.com/cpl17-array-operation-person-above-average-height/ | 1,674,902,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00572.warc.gz | 771,111,318 | 6,335 | # Array Operation: Person above average height
#### Program 7: person above average height
##### Intro Text
This exercise can be considered as the real life example. Where you get number of peoples height and calculate the average height of person. With a simple comparison between peoples average height and height of individuals, we can find the number of peoples above average height. That’s all.
##### Programming Procedure
###### Get Peoples height
we need to declare an array of size 100. Here 100 is considered as the maximum. First we need to get the number of persons using a simple scanf statement. Then using a or loop we will get the height of each persons. here is he code snippet for that.
```printf("\nEnter number of persons...");
scanf("%d",&n);
for(i=0;i<n;i++){
person[i]=0.0f;
}
printf("\nEnter the height for each person");
for(i=0;i<n;i++){
printf("\nEnter the height for person%d...",i+1);
scanf("%f",&person[i]);
}```
###### Finding the average
We will find the average height of person is by summing up the height of individual person then divided by the total number of person.
`avg height = (sum of individual weight / no. of person)`
###### People above average height
we can find the people above average height by comparing the average height with height of individuals. If the height of the person is above average height the we will add 1 to count.
##### Program Code
The final program for finding people above average height is….
``` /*
Experiment:
1.Get number of persons
2.Get height for each person
3.take avg height
4.display person whose height is more than avg
*/
#include<stdio.h>
#include<conio.h>
#define MAX 100
void main(){
int i,n=0,noOfPerson=0;
float person[MAX],avg=0.0f,sum=0.0f;
clrscr();
printf("\nPEOPLE WHOES HEIGHT ABOVE AVG");
printf("\n*****************************");
printf("\nEnter number of persons...");
scanf("%d",&n);
for(i=0;i<n;i++){
person[i]=0.0f;
}
printf("\nEnter the height for each person");
for(i=0;i<n;i++){
printf("\nEnter the height for person%d...",i+1);
scanf("%f",&person[i]);
}
for(i=0;i<n;i++){
sum +=person[i];
}
avg = sum/n;
printf("Average Height is...%f",avg);
for(i=0;i<n;i++){
if(person[i]>avg){
noOfPerson+=1;
}
}
printf("\nNo. of Person above average...%d",noOfPerson);
getch();
}``` | 561 | 2,285 | {"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-2023-06 | latest | en | 0.718699 |
http://de.metamath.org/mpeuni/xnegneg.html | 1,721,499,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00442.warc.gz | 7,196,128 | 5,812 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > xnegneg Structured version Visualization version GIF version
Theorem xnegneg 11919
Description: Extended real version of negneg 10210. (Contributed by Mario Carneiro, 20-Aug-2015.)
Assertion
Ref Expression
xnegneg (𝐴 ∈ ℝ* → -𝑒-𝑒𝐴 = 𝐴)
Proof of Theorem xnegneg
StepHypRef Expression
1 elxr 11826 . 2 (𝐴 ∈ ℝ* ↔ (𝐴 ∈ ℝ ∨ 𝐴 = +∞ ∨ 𝐴 = -∞))
2 rexneg 11916 . . . . 5 (𝐴 ∈ ℝ → -𝑒𝐴 = -𝐴)
3 xnegeq 11912 . . . . 5 (-𝑒𝐴 = -𝐴 → -𝑒-𝑒𝐴 = -𝑒-𝐴)
42, 3syl 17 . . . 4 (𝐴 ∈ ℝ → -𝑒-𝑒𝐴 = -𝑒-𝐴)
5 renegcl 10223 . . . . 5 (𝐴 ∈ ℝ → -𝐴 ∈ ℝ)
6 rexneg 11916 . . . . 5 (-𝐴 ∈ ℝ → -𝑒-𝐴 = --𝐴)
75, 6syl 17 . . . 4 (𝐴 ∈ ℝ → -𝑒-𝐴 = --𝐴)
8 recn 9905 . . . . 5 (𝐴 ∈ ℝ → 𝐴 ∈ ℂ)
98negnegd 10262 . . . 4 (𝐴 ∈ ℝ → --𝐴 = 𝐴)
104, 7, 93eqtrd 2648 . . 3 (𝐴 ∈ ℝ → -𝑒-𝑒𝐴 = 𝐴)
11 xnegmnf 11915 . . . 4 -𝑒-∞ = +∞
12 xnegeq 11912 . . . . . 6 (𝐴 = +∞ → -𝑒𝐴 = -𝑒+∞)
13 xnegpnf 11914 . . . . . 6 -𝑒+∞ = -∞
1412, 13syl6eq 2660 . . . . 5 (𝐴 = +∞ → -𝑒𝐴 = -∞)
15 xnegeq 11912 . . . . 5 (-𝑒𝐴 = -∞ → -𝑒-𝑒𝐴 = -𝑒-∞)
1614, 15syl 17 . . . 4 (𝐴 = +∞ → -𝑒-𝑒𝐴 = -𝑒-∞)
17 id 22 . . . 4 (𝐴 = +∞ → 𝐴 = +∞)
1811, 16, 173eqtr4a 2670 . . 3 (𝐴 = +∞ → -𝑒-𝑒𝐴 = 𝐴)
19 xnegeq 11912 . . . . . 6 (𝐴 = -∞ → -𝑒𝐴 = -𝑒-∞)
2019, 11syl6eq 2660 . . . . 5 (𝐴 = -∞ → -𝑒𝐴 = +∞)
21 xnegeq 11912 . . . . 5 (-𝑒𝐴 = +∞ → -𝑒-𝑒𝐴 = -𝑒+∞)
2220, 21syl 17 . . . 4 (𝐴 = -∞ → -𝑒-𝑒𝐴 = -𝑒+∞)
23 id 22 . . . 4 (𝐴 = -∞ → 𝐴 = -∞)
2413, 22, 233eqtr4a 2670 . . 3 (𝐴 = -∞ → -𝑒-𝑒𝐴 = 𝐴)
2510, 18, 243jaoi 1383 . 2 ((𝐴 ∈ ℝ ∨ 𝐴 = +∞ ∨ 𝐴 = -∞) → -𝑒-𝑒𝐴 = 𝐴)
261, 25sylbi 206 1 (𝐴 ∈ ℝ* → -𝑒-𝑒𝐴 = 𝐴)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∨ w3o 1030 = wceq 1475 ∈ wcel 1977 ℝcr 9814 +∞cpnf 9950 -∞cmnf 9951 ℝ*cxr 9952 -cneg 10146 -𝑒cxne 11819 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-op 4132 df-uni 4373 df-br 4584 df-opab 4644 df-mpt 4645 df-id 4953 df-po 4959 df-so 4960 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-sub 10147 df-neg 10148 df-xneg 11822 This theorem is referenced by: xneg11 11920 xltneg 11922 xnegdi 11950 xnpcan 11954 xposdif 11964 xrsxmet 22420 xrhmeo 22553 xaddeq0 28907 xrge0npcan 29025 carsgclctunlem2 29708
Copyright terms: Public domain W3C validator | 2,253 | 3,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.133526 |
http://spmath7310.blogspot.com/2010/10/part-1-two-term-ratio-compares-2.html | 1,532,088,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591596.64/warc/CC-MAIN-20180720115631-20180720135631-00385.warc.gz | 342,853,395 | 20,818 | ## Friday, October 29, 2010
### Sesame Street final post
Part 1
Two Term Ratio: compares 2 quantities measured in the same units.
ex.
or 1:2 or 1/2 or 1 to 2
Three Term Ratio: compares three quantities measured in the same units.
ex.
or 1:2:3 or 1 to 2 to 3
Part to Part Ratio: compares different parts of a group to each other.
ex. 1:1 or 1 to 1
Part to Whole Ratio: compares one part of a group to the whole group.
ex.
1/2 or 50% or 0.5 - A part to whole ratio can be written as a fraction, a decimal, and a percent.
ates: compares two quantities measured in different units.
ex.
\$2.00 per 10g or \$2.00/10g- rates can be used when buying food.
Unit Rate- a rate in which the second term is one.
ex.
20km an hour or 20km/hr
Unit Price- a unit rate used when shopping.
ex.
per 100g or per 100ml
Proportion- a relationship that says that two ratios or two rates are equal.
ex. | 268 | 887 | {"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-2018-30 | latest | en | 0.905345 |
https://www.athletisme-privas.fr/29881_steam_sparger_calculations.html | 1,656,509,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00638.warc.gz | 707,624,717 | 8,945 | ### Perforated Pipe Distributor Sizing Calculations ...
May 16, 2013· Perforated Pipe Distributors have been discussed many times on "Cheresources". How to design a perforated pipe distributor also known as a sparger has been a frequent question on the forum. Some general replies without providing in-depth methodology of sizing a …
### Online calculation of properties of water and steam
Wikipedia -> steam Advanced calculations and graphical presentation, even in russion language, by Valery Ochkov Steamcalculation: if you found an error, please mail to: [email protected] No garanty for correctness. Calculation based on formulas of IAPWS-IF97 by Dr. Bernhard Spang.
### Process Heating by Direct Steam Injection
Oct 01, 1998· Figure 1. The oldest direct steam-injection method of heating, the sparger incorporates tubes at the bottom of the vessel that inject steam into the fluid to be heated. Two heat processing methods use steam to heat liquids, solutions and slurries: direct, using direct steam-injection (DSI) heaters, and indirect, employing heat exchangers.
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Steam injection: Elimination of steam hammers and for direct heating through sparging steam. When ordering the porous sparger for your application, you should select the sparger keeping in mind the type of process application you intend to use the sparger on.
### Spargers - starchjectcooker.com
Companies around the globe rely on ProSonix steam injection heating to support their water heating, starch jet cooker, slurry heating, & sludge heating needs. From inline fluid heating to tank steam sparging, we have the experience and we don't just supply a steam injector valve or steam jet heater, we deliver an integrated process heating ...
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Sparger ‐ The simplest system is the steam sparger, which "bubbles" steam into a tank of cold water (see Figure 1). The sparger does not have the advantages of other direct injection systems. Sparging systems can be simple and relatively inexpensive to install, but the
### Steam sparger design - Chemical process engineering - Eng-Tips
Feb 21, 2016· The north sparger is 3" 125# steam. I need to get 12500 # per hour through these spargers. I designed two 3" spargers with spray tips on them. The "old" north sparger had 6 3/4" holes cut in with no nozzles. The south sparger had roughly 30 - 40 holes cut in it with no pattern. From what I saw the 4" sparger was plugging in the bottom half ...
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Units for gas calculations ... Hot air or steam injection – enhances volatility Horizontal wells – can be more efficient than ... Groundwater Capture Zone Aerated Zone Treatment System 75 0 50 100 150 50 25 0. Design considerations for air sparging Pilot test is critical – needed to determine R I, which is the key design parameter ...
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2 David R. Shonnard Michigan Technological University 3 Sterilization Agents 1. Thermal-preferred for economical large-scale sterilizations of liquids and equipment. 2. …
### Steam Flow - Orifices
Steam flow through orifices - for steam pressures ranging 2 - 300 psi. Steam flow rates can be measured by using orifice plates. The diagram below indicates steam flow rate (lbs/hr) through orifices ranging diameter 1/32" to 1/2" and 2 - 300 psig steam line pressure. Downstream pressure …
### Steam Heating Process - Load Calculating
In general steam heating is used to. change a product or fluid temperature; maintain a product or fluid temperature; A benefit with steam is the large amount of heat energy that can be transferred. The energy released when steam condenses to water is in the range 2000 - 2250 kJ/kg (depending on the pressure) - compared to water with 80 - 120 kJ/kg (with temperature difference 20 - 30 o C).
### STEAM SPARGER BY SIGMAC PROCESS - BlueSinga
Steam is presented into the liquid flow through the designed nozzle and the inlet to the diffuser section of the heater. Advantage of using a steam sparger which gives less hammer shock. Well designed steam sparger can be also used to heat liquid in the tanks by providing a control loop with steam …
### Calculator: Steam Flow Rate through an Orifice | TLV - A ...
Online calculator to quickly determine Steam Flow Rate through an Orifice. Includes 53 different calculations. Equations displayed for easy reference.
### Sparger Design Guide - Mott Corporation
sparger hardware is 316 stainless steel. Other materials are available on special order, including 304L SS, 347 SS, 430 SS, Inconel® 600, Monel® 400, Nickel 200; Hastelloy® C276, C22 and X; and Alloy 20. Sparging Media Grades For most gas sparging applications, Mott Media Grade 2* is recommended. For steam sparging, Media Grade 10 is ...
### Sparger: design - Chemical plant design & operations - Eng ...
Dec 24, 2008· The steam INSIDE the sparger will likely be collapsing very rapidly as well. Calculate your heat transfer and exit quality before you really commit to having such a high velocity. 80 m/sec is a lot higher than you want to have for liquid flow in a stainless line. Published guidelines I have seen are closer to 15 m/sec max.
### Heating Liquids by Steam Sparging - Accendo Reliability
Steam sparging can generate a great amount of vibration. Especially if there is a big pressure difference between the steam pressure and the process pressure. When pressure drops steam expands. In order for steam to flow through the pipe to the outlet at the same rate it is turning from high pressure to low pressure steam it speeds up.
### Engineering Calculator | TLV - A Steam Specialist Company ...
Engineering Calculator. For Steam, Water, Air and Gas Systems. Includes 50+ different calculations. Equations displayed for easy reference.
### Steam Control and Condensate Drainage for Heat Exchangers
smallest size steam regulator. The steam regulator should be sized based on the maximum lb./hr. of steam required by the heat exchanger. To prop-erly size the regulator, the available inlet steam pressure and the heat exchanger design operating pressure must be known. The steam regulator should not be over-sized. Oversizing the regulator may cause
### Eductor Operating Principles - BEX
A steam sparger is any device used to inject steam into a tank liquid for the purpose of heating the tank liquid. Eductors make very effective steam spargers. When steam is used to heat a tank liquid, there is the risk of damaging the tank due to cavitation of the steam bubbles.
### Steam Calculator · SteamDB
Steam Calculator Calculate the value of your Steam account Get disappointed in your life. Any SteamID format is accepted, enter anything you want and we will convert it on the fly for you. ...
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4 Determine the surface area of the sparger 5 Determine appropriate sparger configuration to accommodate the required area Design Example for Static Sparger Application Process: 500 gallons of H2O in a non-agitated tank, 48 inch diameter and 5 foot head at the sparger Heat from 60-100°F in 10 minut 2 psi across element. Weld Joint calculators
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### How to Calculate Steam Velocities - Heating Help
May 09, 2018· How to calculate steam velocity. Lbs/Hr x Cubic Volume of Steam divided by 25 x Internal area of pipe = Steam Velocity. Steam Velocity = Lbs Hour x Cubic Volume of Steam/25 x Internal Area of Pipe. Lbs. Steam/Hr = Btuh/960 Lbs. Steam/Hr = Boiler HP x 34.5 Another formula to calculate steam velocity is:
### Modeling Choked Flow Through an Orifice - AFT
The following notes apply to each modeling/calculation method: Applied Flow Technology 2955 Professional Place, Colorado Springs, CO 80904 USA (719) 686-1000 / …
### What Is Steam Sparging - Komax
Oct 15, 2018· A steam sparger is a type of direct contact heating device. Steam sparging is considered a traditional heating method, where the input temperature is much higher than the process temperature. Although steam injection is considered more efficient, steam spargers are also used in several industries. Some of the most important benefits of the ...
### Steam sparger design - Chemical process engineering - Eng-Tips
Feb 20, 2016· The north sparger is 3" 125# steam. I need to get 12500 # per hour through these spargers. I designed two 3" spargers with spray tips on them. The "old" north sparger had 6 3/4" holes cut in with no nozzles. The south sparger had roughly 30 - 40 holes cut in it with no pattern. From what I saw the 4" sparger was plugging in the bottom half ... | 2,186 | 9,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.883214 |
http://uk.ask.com/question/which-is-greater-63-cm-or-6-m | 1,386,674,522,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164017049/warc/CC-MAIN-20131204133337-00038-ip-10-33-133-15.ec2.internal.warc.gz | 193,077,914 | 14,860 | # What is greater 63cm or 6m?
The latter.
Q&A Related to "What is greater 63cm or 6m?"
Actually 6m is greater than 63cm :P http://wiki.answers.com/Q/Which_is_greater_63cm_or...
6 meter is greater than 63 cm .The reason is 6meter =600 cm.So,it is well known that 600>63. http://wiki.answers.com/Q/What_is_greater_63_cm_or...
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There are 100 centimeters in a meter, when referring to a basic unit of length. However, there are meters of various types, such as a flow meter, (measurement ...
One meter is equal to 100 centimeters. To convert centimeters to meters, all you need to do is divide the number of centimeters by 100. Therefore, 6000 centimeters ... | 231 | 870 | {"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-2013-48 | longest | en | 0.914448 |
https://jeopardylabs.com/print/multiplication-review-group-1 | 1,611,804,435,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00687.warc.gz | 393,329,278 | 3,368 | Multiplication Properties
Mixed Facts A
Mixed Facts B
Mixed Facts C
Mixed Facts D
100
4x0=0 is an example of what property?
Zero Property of Multiplication
100
2x2
4
100
10x3
30
100
4x10
40
100
7x2
14
200
2x1=2 is an example of what property
Identity Property of Multiplication
200
3x2
6
200
8x2
16
200
6x2
12
200
5x3
15
300
2+2+2=6 is what as a multiplication sentence?
2x3 OR 3x2
300
5x10
50
300
2x9
18
300
5x7
35
300
4x6
24
400
2x3=3x2 is an example of what property
Commutative Property of Multiplication
400
4x3
12
400
4x5
20
400
8x4
32
400
7x3
21
500
(2x3)x4=(4x3)x2 is an example of what property
Associative Property of Multiplication
500
5x5
25
500
10x10
100
500
5x9
45
500
12x3
36
Click to zoom | 301 | 779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-04 | latest | en | 0.613247 |
https://ask.sagemath.org/question/33055/find-all-n-torsion-of-an-elliptic-curve/?sort=oldest | 1,726,494,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00844.warc.gz | 97,474,140 | 13,924 | # Find all n-torsion of an elliptic curve
I'm sorry if this is a really easy question. I found out how to compute the n-torsion of an elliptic curve $E$ over a given field. But what if I would like to do is find a field $K$ such that all n-torsion points of $E$ are defined over $K$.
In effect compute the n-torsion part of the Tate module. Is this possible with Sagemath?
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this would get you close:
E.torsion_polynomial(n).splitting_field()
You're just some quadratic extensions away from the desired field (you just need to make sure that roots of the division polynomial are indeed x-coordinates of points).
Be careful, though: generally such explicit representations of splitting fields are entirely unworkable because they tend to have very high degrees.
more
The exact answer to your question is to use the mehod division_field, for example
sage: E = EllipticCurve([0,-1,1,-10,-20])
sage: E.division_field(5,'a')
Number Field in a with defining polynomial x^4 - x^3 + x^2 - x + 1
so that
sage: len(E.change_ring(K).torsion_points())
25
but as Nils said the degree is in general very large -- I chose an example where the degree of the 5-division field is only 4, which is as small as possible over QQ.
more | 330 | 1,295 | {"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} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.927754 |
https://www.learnfk.com/python-machine-learning/machine-learning-with-python-confusion-matrix.html | 1,659,996,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00594.warc.gz | 778,195,163 | 8,871 | # Confusion Matrix函数
• True Positives (TP) − It is the case when both actual class & predicted class of data point is 1.
• True Negatives (TN) − It is the case when both actual class & predicted class of data point is 0.
• False Positives (FP) − It is the case when actual class of data point is 0 & predicted class of data point is 1.
• False Negatives (FN) − It is the case when actual class of data point is 1 & predicted class of data point is 0.
from sklearn.metrics import confusion_matrix
[[ 73 7]
[ 4 144]]
## 准确性
$$准确性 =\frac{TP+TN}{TP+FP+FN+TN}$$
Hence, 准确性=217/228=0.951754385965 which is same as we have calculated after creating our binary classifier.
## 精确
$$精确 =\frac{TP}{TP+FP}$$
Hence, 精确=73/80=0.915
## 召回或敏感性
$$Recall =\frac {TP} {TP + FN}$$
Hence, 精确=73/77=0.94805
## 特异性
$$特异性 =\frac{TN}{TN+FP}$$
Hence, 精确=144/151=0.95364
## 相关推荐
10x程序员工作法 -〔郑晔 - 〕
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Python机器学习 - 10-逻辑回归原理 | 349 | 923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.52705 |
https://stats.stackexchange.com/questions/348807/find-probability-of-rejecting-a-true-null-hypothesis | 1,718,925,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00180.warc.gz | 474,179,437 | 41,018 | # Find probability of rejecting a true null hypothesis
A binomial experiment tests $H_0$ : p = 0.5 against p $\neq$ 0.5 using significance level 0.05. Only 5 observations are available. Show that the probability to reject a true null hypothesis is 1/16.
The steps for the solution:
$H_0$ : p = 0.5
$H_a$: p $\neq$ 0.5
$N$ = 5
$\alpha$ = 0.05
Pr(Type I error) = Pr(reject $H_0$ when $H_0$ is true)
The formula implemented from Hypothesis testing, page 12
$\sum_{x=1}^5 b(x, n = 5, p = 0.5) = \sum_{x=1}^5 \binom{5}{n} 0.5^x0.95^{5-x}$
but it doesn't get the answer 1 / 16.
Any suggestion what's wrong ?
• Is this an exact binomial test or are you using the normal approximation to the sample proportion? Commented May 29, 2018 at 15:32
– Bor
Commented Jun 1, 2018 at 8:17
For calculating the probability of a Type I Error, we start with: $$$$\label{eql} \begin{split} \text{Pr}(\text{Type I Error}) & = \text{Pr}(\text{reject }H_0 | H_0 \text{ is true}) \\ & = \text{Pr}(\text{reject }H_0 | p=.5, n=5) \end{split}$$$$
The probability mass function $\text{Pr}(X=x)=\binom{5}{x}.5^x .5^{5-x}$ (note that your pmf incorrectly uses $1-p=.95$) for a binomial random variable $X$ given our $H_0$ ($p=.5,n=5$) is: $$\begin{split} \text{Pr}(X=0) = \frac{1}{32} = .03125 \\ \text{Pr}(X=1) = \frac{5}{32} = .15625 \\ \text{Pr}(X=2) = \frac{5}{16} = .31250 \\ \text{Pr}(X=3) = \frac{5}{16} = .31250 \\ \text{Pr}(X=4) = \frac{5}{32} = .15625 \\ \text{Pr}(X=5) = \frac{1}{32} = .03125 \end{split}$$
Noting above that only $\text{Pr}(X=0)$ and $\text{Pr}(X=5)$ are below our $\alpha=.05$ threshold, and therefore that $H_0$ may only be rejected if a sample results in $X=0$ or $X=5$, we can move forward as follows:
$$$$\label{eql1} \begin{split} \text{Pr}(\text{Type I Error}) & = \text{Pr}(\text{reject }H_0 | p=.5,n=5) \\ & = \text{Pr}(X=0| p=.5,n=5) + \text{Pr}(X=5| p=.5,n=5) \\ & =2\cdot.03125=.0625=\frac{1}{16} \end{split}$$$$
• But this doesn't meet the goal of having the false alarm probability being at most $0.05$; the false alarm probability is $0.0625 > 0.05$. Commented May 29, 2018 at 20:45
• is missing the 0 in .5 some kind of convention here ?
– Bor
Commented May 30, 2018 at 7:48
• also why are 6 observations made(0-5) ?
– Bor
Commented May 30, 2018 at 11:59
• 1) Not really - some people do it, some don't. It's a matter of preference, and I find it to be more easily readable. 2) A total of 0, 1, 2, 3, 4, or 5 successes can possibly be observed, because there are 5 observations in the experiment. An example would be being able to flip anywhere from 0 to 5 heads when flipping n=5 times. Commented May 30, 2018 at 12:09
• In other words, there are 6 possible outcomes of the experiment when you make 5 observations. Commented May 30, 2018 at 12:18
I have no idea what the significance level is doing in this hypothesis test or what its meaning might be, but stripped of all the malarkey and statistical verbiage, on 5 tosses of a fair coin, the probability of tossing 5 heads or 5 tails is $\displaystyle \frac{1}{32}+ \frac{1}{32}= \frac{1}{16}$. Any reasonable test of the hypothesis $H_0$ that $p=0.5$ versus the hypothesis $H_1$ that $p\neq 0.5$ must reject this event (all heads or all tails) giving a false alarm probability of $\frac{1}{16}=0.0625$ on the reasonable assumption that the only observation that causes rejection of $H_0$ is the all heads or all tails event. If you insist that the false alarm probability should not exceed 0.05, then you either need to use an asymmetric test such as rejecting $H_0$ when you observe 5 heads but not rejecting $H_0$ when you observe 5 tails, or, more simply, never reject $H_0$ thereby achieving a false alarm probability of $0$ which, when I last checked, is smaller than the maximum allowable false alarm probability of $0.05$.
• No more clarifications are given on the task, just to prove it's 1/16
– Bor
Commented May 30, 2018 at 7:15 | 1,354 | 3,922 | {"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": 2, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.71098 |
https://speakerdeck.com/hayaosuzuki/effective-python-in-python-3-dot-6 | 1,726,172,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00458.warc.gz | 500,537,305 | 15,385 | HayaoSuzuki
January 31, 2017
3.8k
# Effective Python in Python 3.6
Lightning talk at Python 3.6 Release Party
January 31, 2017
## Transcript
1. ### Effective Python in Python 3.6 Hayao Suzuki Python 3.6 Release
Party at Yahoo Japan January 31, 2017
2. ### Who Are You? About Me Name Hayao Suzuki (@CardinalXaro) Blog
http://xaro.hatenablog.jp/ Major Mathematics (Combinatorics, Number Theory) Work Python Programmer (Not Mathematics!) Reviewed Books Effective Python (O’Reilly Japan) Algorithms in A Nutshell 2nd ed. (O’Reilly Japan) And more...? 2 / 9
3. ### Today’s Theme Metaprogramming without Metaclasses Metaclass can create extremely bizarre
behaviors that are unapproachable to newcomers. (Effective Python, P.87) 3 / 9
4. ### Metaprogramming without Metaclasses Items about Metaclass in Effective Python Item
33 Validate Subclasses with Metaclasses Item 34 Register Class Existence with Metaclasses Item 35 Annotate Class Attributes with Metaclasses Using Python 3.6, we can write metaprogramming without using metaclass. 4 / 9
5. ### Item 33 With Metaclasses class ValidatePolygon(type): def __new__(meta, name, bases,
class_dict): if bases != (object,): if class_dict["sides"] < 3: raise ValueError("polygons need 3+ sides.") return type.__new__(meta, name, bases, class_dict) class Polygon(object, metaclass=ValidatePolygon): sides = None @classmethod def interior_angles(cls): return (cls.sides - 2) * 180 class Triangle(Polygon): sides = 3 5 / 9
6. ### Item 33 Without Metaclasses class Polygon(object): def __init_subclass__(cls, sides, **kwargs):
cls.sides = sides if cls.sides < 3: raise ValueError("polygons need 3+ sides.") @classmethod def interior_angles(cls): return (cls.sides - 2) * 180 class Triangle(Polygon, sides=3): pass 6 / 9
7. ### Item 35 with Metaclasses class Meta(type): def __new__(meta, name, bases,
class_dict): for key, value in class_dict.items(): if isinstance(value, Field): value.name = key value.internal_name = '_' + key return type.__new__(meta, name, bases, class_dict) class Field(object): def __init__(self): self.name = None self.internal_name = None def __get__(self, instance, instance_type): if instance is None: return self return getattr(instance, self.internal_name, '') def __set__(self, instance, value): setattr(instance, self.internal_name, value) class DatabaseRow(object, metaclass=Meta): pass class Customer(DatabaseRow): first_name = Field() 7 / 9
8. ### Item 35 Without Metaclasses class Field(object): def __init__(self): self.name =
None self.internal_name = None def __get__(self, instance, instance_type): if instance is None: return self return getattr(instance, self.internal_name, '') def __set__(self, instance, value): setattr(instance, self.internal_name, value) def __set_name__(self, owner, name): self.name = name self.internal_name = '_' + name class Customer(object): first_name = Field() 8 / 9
9. ### Summary Summary Using Python 3.6, we can write metaprogramming without
using metaclass. Item 33 to Item 35 in Effective Python no longer have to use metaclass. Python 3.6 is good I think. 9 / 9 | 847 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.527827 |
https://us.metamath.org/nfegif/elrabsf.html | 1,716,205,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00758.warc.gz | 550,683,022 | 4,257 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > elrabsf Unicode version
Theorem elrabsf 3084
Description: Membership in a restricted class abstraction, expressed with explicit class substitution. (The variation elrabf 2993 has implicit substitution). The hypothesis specifies that must not be a free variable in . (Contributed by NM, 30-Sep-2003.) (Proof shortened by Mario Carneiro, 13-Oct-2016.)
Hypothesis
Ref Expression
elrabsf.1
Assertion
Ref Expression
elrabsf
Proof of Theorem elrabsf
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 dfsbcq 3048 . 2
2 elrabsf.1 . . 3
3 nfcv 2489 . . 3
4 nfv 1619 . . 3
5 nfsbc1v 3065 . . 3
6 sbceq1a 3056 . . 3
72, 3, 4, 5, 6cbvrab 2857 . 2
81, 7elrab2 2996 1
Colors of variables: wff setvar class Syntax hints: wb 176 wa 358 wcel 1710 wnfc 2476 crab 2618 wsbc 3046 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-rab 2623 df-v 2861 df-sbc 3047 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 576 | 1,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.31629 |
https://hextobinary.com/unit/angularacc/from/degps2/to/arcminpd2 | 1,722,685,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00433.warc.gz | 239,078,478 | 16,891 | # Degree/Square Second to Arcmin/Square Day Converter
Angular Acceleration
Degree/Square Second
Arcmin/Square Day
1 Degree/Square Second = 447897600000 Arcmin/Square Day
## How many Arcmin/Square Day are in a Degree/Square Second?
The answer is one Degree/Square Second is equal to 447897600000 Arcmin/Square Day and that means we can also write it as 1 Degree/Square Second = 447897600000 Arcmin/Square Day. Feel free to use our online unit conversion calculator to convert the unit from Degree/Square Second to Arcmin/Square Day. Just simply enter value 1 in Degree/Square Second and see the result in Arcmin/Square Day.
## How to Convert Degree/Square Second to Arcmin/Square Day (deg/s2 to arcmin/day2)
By using our Degree/Square Second to Arcmin/Square Day conversion tool, you know that one Degree/Square Second is equivalent to 447897600000 Arcmin/Square Day. Hence, to convert Degree/Square Second to Arcmin/Square Day, we just need to multiply the number by 447897600000. We are going to use very simple Degree/Square Second to Arcmin/Square Day conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Degree/Square Second} = 1 \times 447897600000 = \text{447897600000 Arcmin/Square Day}$$
## What is Degree/Square Second Unit of Measure?
Degree per square second is a unit of measurement for angular acceleration. By definition, if an object accelerates at one degree per square second, its angular velocity is increasing by one degree per second every second.
## What is the symbol of Degree/Square Second?
The symbol of Degree/Square Second is deg/s2. This means you can also write one Degree/Square Second as 1 deg/s2.
## What is Arcmin/Square Day Unit of Measure?
Arcmin per square day is a unit of measurement for angular acceleration. By definition, if an object accelerates at one arcmin per square day, its angular velocity is increasing by one arcmin per day every day.
## What is the symbol of Arcmin/Square Day?
The symbol of Arcmin/Square Day is arcmin/day2. This means you can also write one Arcmin/Square Day as 1 arcmin/day2.
## Degree/Square Second to Arcmin/Square Day Conversion Table
Degree/Square Second [deg/s2]Arcmin/Square Day [arcmin/day2]
1447897600000
2895795200000
31343692800000
41791590400000
52239488000000
62687385600000
73135283200000
83583180800000
94031078400000
104478976000000
10044789760000000
1000447897600000000
## Degree/Square Second to Other Units Conversion Table
Degree/Square Second [deg/s2]Output
1 degree/square second in degree/square millisecond is equal to0.000001
1 degree/square second in degree/square microsecond is equal to1e-12
1 degree/square second in degree/square nanosecond is equal to1e-18
1 degree/square second in degree/square minute is equal to3600
1 degree/square second in degree/square hour is equal to12960000
1 degree/square second in degree/square day is equal to7464960000
1 degree/square second in degree/square week is equal to365783040000
1 degree/square second in degree/square month is equal to6915848040000
1 degree/square second in degree/square year is equal to995882117760000
1 degree/square second in radian/square second is equal to0.017453292519943
1 degree/square second in radian/square millisecond is equal to1.7453292519943e-8
1 degree/square second in radian/square microsecond is equal to1.7453292519943e-14
1 degree/square second in radian/square nanosecond is equal to1.7453292519943e-20
1 degree/square second in radian/square minute is equal to62.83
1 degree/square second in radian/square hour is equal to226194.67
1 degree/square second in radian/square day is equal to130288130.53
1 degree/square second in radian/square week is equal to6384118395.95
1 degree/square second in radian/square month is equal to120704318865.6
1 degree/square second in radian/square year is equal to17381421916646
1 degree/square second in gradian/square second is equal to1.11
1 degree/square second in gradian/square millisecond is equal to0.0000011111111111111
1 degree/square second in gradian/square microsecond is equal to1.1111111111111e-12
1 degree/square second in gradian/square nanosecond is equal to1.1111111111111e-18
1 degree/square second in gradian/square minute is equal to4000
1 degree/square second in gradian/square hour is equal to14400000
1 degree/square second in gradian/square day is equal to8294400000
1 degree/square second in gradian/square week is equal to406425600000
1 degree/square second in gradian/square month is equal to7684275600000
1 degree/square second in gradian/square year is equal to1106535686400000
1 degree/square second in arcmin/square second is equal to60
1 degree/square second in arcmin/square millisecond is equal to0.00006
1 degree/square second in arcmin/square microsecond is equal to6e-11
1 degree/square second in arcmin/square nanosecond is equal to6e-17
1 degree/square second in arcmin/square minute is equal to216000
1 degree/square second in arcmin/square hour is equal to777600000
1 degree/square second in arcmin/square day is equal to447897600000
1 degree/square second in arcmin/square week is equal to21946982400000
1 degree/square second in arcmin/square month is equal to414950882400000
1 degree/square second in arcmin/square year is equal to59752927065600000
1 degree/square second in arcsec/square second is equal to3600
1 degree/square second in arcsec/square millisecond is equal to0.0036
1 degree/square second in arcsec/square microsecond is equal to3.6e-9
1 degree/square second in arcsec/square nanosecond is equal to3.6e-15
1 degree/square second in arcsec/square minute is equal to12960000
1 degree/square second in arcsec/square hour is equal to46656000000
1 degree/square second in arcsec/square day is equal to26873856000000
1 degree/square second in arcsec/square week is equal to1316818944000000
1 degree/square second in arcsec/square month is equal to24897052944000000
1 degree/square second in arcsec/square year is equal to3585175623936000000
1 degree/square second in sign/square second is equal to0.033333333333333
1 degree/square second in sign/square millisecond is equal to3.3333333333333e-8
1 degree/square second in sign/square microsecond is equal to3.3333333333333e-14
1 degree/square second in sign/square nanosecond is equal to3.3333333333333e-20
1 degree/square second in sign/square minute is equal to120
1 degree/square second in sign/square hour is equal to432000
1 degree/square second in sign/square day is equal to248832000
1 degree/square second in sign/square week is equal to12192768000
1 degree/square second in sign/square month is equal to230528268000
1 degree/square second in sign/square year is equal to33196070592000
1 degree/square second in turn/square second is equal to0.0027777777777778
1 degree/square second in turn/square millisecond is equal to2.7777777777778e-9
1 degree/square second in turn/square microsecond is equal to2.7777777777778e-15
1 degree/square second in turn/square nanosecond is equal to2.7777777777778e-21
1 degree/square second in turn/square minute is equal to10
1 degree/square second in turn/square hour is equal to36000
1 degree/square second in turn/square day is equal to20736000
1 degree/square second in turn/square week is equal to1016064000
1 degree/square second in turn/square month is equal to19210689000
1 degree/square second in turn/square year is equal to2766339216000
1 degree/square second in circle/square second is equal to0.0027777777777778
1 degree/square second in circle/square millisecond is equal to2.7777777777778e-9
1 degree/square second in circle/square microsecond is equal to2.7777777777778e-15
1 degree/square second in circle/square nanosecond is equal to2.7777777777778e-21
1 degree/square second in circle/square minute is equal to10
1 degree/square second in circle/square hour is equal to36000
1 degree/square second in circle/square day is equal to20736000
1 degree/square second in circle/square week is equal to1016064000
1 degree/square second in circle/square month is equal to19210689000
1 degree/square second in circle/square year is equal to2766339216000
1 degree/square second in mil/square second is equal to17.78
1 degree/square second in mil/square millisecond is equal to0.000017777777777778
1 degree/square second in mil/square microsecond is equal to1.7777777777778e-11
1 degree/square second in mil/square nanosecond is equal to1.7777777777778e-17
1 degree/square second in mil/square minute is equal to64000
1 degree/square second in mil/square hour is equal to230400000
1 degree/square second in mil/square day is equal to132710400000
1 degree/square second in mil/square week is equal to6502809600000
1 degree/square second in mil/square month is equal to122948409600000
1 degree/square second in mil/square year is equal to17704570982400000
1 degree/square second in revolution/square second is equal to0.0027777777777778
1 degree/square second in revolution/square millisecond is equal to2.7777777777778e-9
1 degree/square second in revolution/square microsecond is equal to2.7777777777778e-15
1 degree/square second in revolution/square nanosecond is equal to2.7777777777778e-21
1 degree/square second in revolution/square minute is equal to10
1 degree/square second in revolution/square hour is equal to36000
1 degree/square second in revolution/square day is equal to20736000
1 degree/square second in revolution/square week is equal to1016064000
1 degree/square second in revolution/square month is equal to19210689000
1 degree/square second in revolution/square year is equal to2766339216000
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 2,677 | 9,765 | {"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} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.904467 |
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Contributor III
## Calculate sales percentage based on a group of selected products: denominator issue
Hi Everyone
I'd need some help with the following expression. I need to select some products and then calculate the percentage of sales based on the current selection --> sales of selected products / Total sales of selected product lines. Nothing difficult. But every product belongs to a different product line.
product 1 --> product line 1
product 2 --> product line 2
product 3 --> product line 3
So, if I select only product 1 and product 2, I'd need in the denominator only the total of product line 1 and product line 2.
Formula: sum({\$<year={2019}>} [quantity product]) / sum({<year={2019}, Product=>} total [quantity product]) doesn't work because it returns to me a denominator with ALL the values including the product line 3.
Is there anyone that can help me?
Labels (3)
• ### Set Analysis
1 Solution
Accepted Solutions
MVP
May be this
``````Sum({\$<year = {2019}>} [quantity product])
/
Sum({\$<year = {2019}, Product, ProductLine = p(ProductLine)>} [quantity product])``````
Or this if you are doing this in a chart where ProductLine is a dimension
``````Sum({\$<year = {2019}>} [quantity product])
/
Sum({\$<year = {2019}, Product, ProductLine = p(ProductLine)>} TOTAL <ProductLine> [quantity product])``````
2 Replies
MVP
May be this
``````Sum({\$<year = {2019}>} [quantity product])
/
Sum({\$<year = {2019}, Product, ProductLine = p(ProductLine)>} [quantity product])``````
Or this if you are doing this in a chart where ProductLine is a dimension
``````Sum({\$<year = {2019}>} [quantity product])
/
Sum({\$<year = {2019}, Product, ProductLine = p(ProductLine)>} TOTAL <ProductLine> [quantity product])``````
Contributor III
Author
it worked perfectly.
Thanks a lot!
Tags
Community Browser | 550 | 2,087 | {"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-2022-49 | latest | en | 0.82042 |
http://masteringelectronicsdesign.com/page/5/ | 1,519,140,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812978.31/warc/CC-MAIN-20180220145713-20180220165713-00292.warc.gz | 204,795,337 | 16,527 | ## Differential Amplifier Output Common-Mode Voltage Calculator
A differential amplifier frequent use is the amplification of the voltage difference between its inputs, while rejecting the common-mode level. However, the output common-mode level cannot be zero. The operational amplifier technological limitations, as well as the outside resistor tolerances let the common-mode voltage to make it to the amplifier output as an output error. As a consequence, the amplifier output voltage is the input signal difference times gain, plus the output common-mode voltage.
Read moreDifferential Amplifier Output Common-Mode Voltage Calculator
## How to Design a Circuit from its Transfer Function Graph
Sometimes all we know about a circuit is its transfer function graph. The transfer function might look like the one in Figure 1. How can we design a circuit so that its input-output behavior will match the graph?
Figure 1
The design starts with the mathematical form of the transfer function. This is a linear function, with the general form of a first order polynomial function.
Read moreHow to Design a Circuit from its Transfer Function Graph
## Differential Amplifier Calculator
### Unipolar to Bipolar Converter Example
If you need to design a differential amplifier, here is a handy calculator. All you need to define are the input range, the output range and a choice of voltage reference.
The differential amplifier was explained in different articles on this website. Solving the Differential Amplifier – Part 1, Part 2 and Part 3 shows a numerical example and how to design such an amplifier. Also, the common mode voltage level and the common mode output error were explained in the series of articles The Differential Amplifier Common-Mode Error – Part 1 and Part 2.
Enter the input range, Vin1 to Vin2, the output range, Vout1 to Vout2 and a reference voltage Vref. You need to choose two resistors, R2 and R3. The calculator will compute R1 and R4.
## Design a Bipolar to Unipolar Converter with a 3-input Summing Amplifier
Since the publication of Design a Bipolar to Unipolar Converter to Drive an ADC, several readers contacted me with requests to help in solving their particular converter. The common problem they had was the fact that the components’ calculation resulted in a negative value for at least one resistor.
To provide a solution, first we need to understand the root cause of the problem. Let’s take one of the circuits I received and analyze it.
The reader wrote that he would like to drive an ADC with the input range of 0 to 2.5V from a signal with the range of –5V to +5V, connected at V1 (see Figure 1).
Read moreDesign a Bipolar to Unipolar Converter with a 3-input Summing Amplifier
## Summing Amplifier Calculator
### Bipolar to Unipolar Converter Example
The calculator solves the summing amplifier resistors based on the input and output voltage range requirements. It is a great tool to design a bipolar to unipolar converter, as an example and other circuits.
Enter the input range, Vin1 to Vin2, the output range, Vout1 to Vout2 and a reference voltage Vref which helps in adjusting the common-mode level of the amplifier. Since the 2-input summing amplifier has 4 resistors, you need to choose two resistors, R1 and R3, and calculate R2 and R4. For more details about this calculator read How to Design a Summing Amplifier Calculator.
## How to Design a Summing Amplifier Calculator
Several articles in this website describe the Summing Amplifier. In one of these articles, Solving the Summing Amplifier, I showed a numeric method to design a non-inverting summing amplifier based on its input and output voltage range requirements.
This article shows how to design a summing amplifier calculator and the mathematical relations it uses. You can find the calculator here:
JavaScript Summing Amplifier Calculator
Type the input voltage range, output range, a reference voltage and a choice of two resistors. The calculator gives you the answer for the remaining two resistors. The default values are for a bipolar to unipolar converter, which is explained in Design a Bipolar to Unipolar Converter to Drive an ADC.
What are the underlying equations?
Read moreHow to Design a Summing Amplifier Calculator
## The Virtual Ground
In my articles I talked about the op amp virtual ground and sometimes I wrote a brief explanation of this concept. In this article I will show you why an op amp input can be considered at a zero potential, without being galvanically connected to ground. Let’s take a simple circuit, the inverting amplifier.
Figure 1
In MasteringElectronicsDesign.com : How to Derive the Inverting Amplifier Transfer Function I showed the proof of its formula by using the virtual ground. The inverting input is at a zero potential, therefore virtual ground, which is a direct consequence of the feedback provided by R2 and the op amp high gain. Let’s see why. | 1,052 | 4,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-09 | latest | en | 0.853196 |
https://www.neetprep.com/questions?courseId=8&testId=403-JIPMER-MBBS-&questionId=94195-charged-particle-accelerated-potential--V-its-velocity--ms-value-em-particle------------ | 1,721,135,830,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00324.warc.gz | 783,410,536 | 61,251 | # A charged particle is accelerated by a potential of 200 V. If its velocity is 8.4 × 108 m/s, then the value of e/m for that particle is: 1. 17.6 × 1016 2. 14.5 × 1012 3. 1.76 × 1015 4. 1.45 × 1015
Subtopic: Electric Field |
61%
Hints
When an open organ is dipped in water up to half of its height, then its frequency will become:
1. half
2. double
3. remain same
4. four times
Subtopic: Standing Waves |
58%
Hints
A sound source producing waves of frequency 300 Hz and wavelength 1 m observer is stationary, while the source is going away with the velocity 30 m/s, then apparent frequency heard by the observer is:
1. 270 Hz
2. 273 Hz
3. 383 Hz
4. 300 Hz
Subtopic: Doppler's Effect (OLD NCERT) |
67%
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A particle moves towards east for 2 s with velocity 15 m/s and move towards north for 8 s with velocity 5 m/s. Then, the average velocity of the particle is:
1. 2 m/s
2. 5 m/s
3. 7 m/s
4. 10 m/s
Subtopic: Average Speed & Average Velocity |
60%
Hints
Relation between a wavelength of photon and electron of same energy is:
1. $$\lambda_{p h}>\lambda_e$$
2. $$\lambda_{p h}<\lambda_e$$
3. $$\lambda_{p h}=\lambda_e$$
4. $$\frac{\lambda_e}{\lambda_{p h}}=\text { constant }$$
Subtopic: De-broglie Wavelength |
52%
Hints
Match the following.
A. Angular momentum 1 [M-1L2T-2] B. Torque 2 [M1T-2] C. Gravitational constant 3 [M1L2T-2] D. Tension 4 [M1L2T-1]
1. C-2, D-1
2. A-4, B-3
3. A-3, C-2
4. B-2, A-1
Subtopic: Dimensions |
72%
Hints
If we increase the kinetic energy of a body by 300%, then percent increase in its momentum is:
1. 50%
2. 300%
3. 100%
4. 150%
Subtopic: Concept of Work |
56%
Hints
Change in acceleration due to gravity is same at height h above the Earth surface and depth $$x ,$$ then relation between $$x$$ and $$h$$ is: ($$h$$ and $$x<<R_e$$)
1. $$x=h$$
2. $$x=2h$$
3. $$x=\frac h2$$
4. $$x=h^2$$
Subtopic: Acceleration due to Gravity |
77%
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A mass of $$1~\mathrm{kg}$$ is suspended from a spring of force constant $$400$$ N, executing SHM total energy of the body is $$2$$ J, then the maximum acceleration of the spring will be:
1. $$4 \mathrm{~m} / \mathrm{s}^2$$
2. $$40 \mathrm{~m} / \mathrm{s}^2$$
3. $$200 \mathrm{~m} / \mathrm{s}^2$$
4. $$400 \mathrm{~m} / \mathrm{s}^2$$
Subtopic: Energy of SHM |
64%
Two capacitors of capacities $$C_1$$ and $$C_2$$ are charged upto the potential $$V_1$$ and $$V_2,$$ then the condition for not flowing the charge between them when connected the capacitors, in parallel is:
1. $$C_1=C_2$$
2. $$C_1V_1=C_2V_2$$
3. $$V_1V_2$$
4. $$\frac{C_1}{V_1}=\frac{C_2}{V_2}$$ | 941 | 2,579 | {"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} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.759657 |
http://www.edhelper.com/Logarithms56.htm | 1,432,938,709,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930443.64/warc/CC-MAIN-20150521113210-00196-ip-10-180-206-219.ec2.internal.warc.gz | 440,714,584 | 2,165 | edHelper subscribers - Create a new printable
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Exponent and Logarithm Applications
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1 A translucent plastic paper reduces the intensity of light that passes through it by eight percent. Amy wants to combine a number of these papers to only allow at the most forty percent of light to pass through. How many translucent plastic papers should Amy use?
2 Jill's investment in a bank account will be worth \$3,746.42 in six months. The current value of Jill's account is \$3,600. If the bank pays a fixed rate that is compounded monthly, how much will the account be worth in two years?
3 Brad spent \$1,200 on a credit card this month. The credit card charges 24.77% interest compounded continuously. If Brad does not have to pay any monthly minimums and makes no payments, how long will it take until Brad owes the credit card company \$2,400?
4 The half-life of a radioactive substance is one hundred fifty-three days. How many days will it take for seventy percent of the substance to decay?
Sample
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edHelper subscribers - Create a new printable | 257 | 1,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2015-22 | longest | en | 0.912267 |
https://math.stackexchange.com/questions/3207154/calculate-double-integral-for-a-segment-of-a-sphere | 1,568,649,990,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572879.28/warc/CC-MAIN-20190916155946-20190916181946-00315.warc.gz | 573,077,320 | 29,550 | # Calculate double integral for a segment of a sphere
I have one problem and the exact answer for it. But did not figure out the solution. $$A_{P-A} = \large R^2\int_{\theta=\frac{\pi}{2}-\beta}^{\alpha_{eq}}\int_{\varphi=\arcsin\left(\frac{1}{\tan\beta\tan\theta}\right)}^{\pi-\arcsin\left(\frac{1}{\tan\beta\tan\theta}\right)}\sin\theta\ d\theta\ d\varphi$$
I have tried to do that with mathematica but it do not provide the same solution rather it provides very big non realistic solution. I can solve up to the following but do not know how to proceed later.
Can anyone help me to continue? | 181 | 597 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-39 | latest | en | 0.834439 |
https://www.classace.io/learn/math/4thgrade/divide-four-digit-numbers-with-remainder | 1,680,036,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00389.warc.gz | 790,697,259 | 36,570 | Divide Four-Digit Numbers With Remainder
## How to Divide Four-Digit Numbers With a Remainder
Division is splitting a number into equal groups.
What if we can't split a number exactly?
That's when we get a remainder.
A remainder is a number that's left over from division.
### Dividing Four-Digit Numbers
In an earlier lesson, you learned how to divide 4-digit numbers without remainders.
Now, let's learn how to divide 4-digit numbers with remainders. It's mostly the same.
Let's learn with an example:
9317 ÷ 5 = ?
The first step is to write this in long division form.
We write the dividend, or the big number we want to split, inside the division bar.
We write the divisor on the outside.
Tip: the answer to our division equation is called the quotient.
We're all set! 🙂
👉 Start dividing from the first digit of the dividend.
What's the first digit in the dividend?
Yes! It's 9.
Let's divide 9 by 5.
To get the answer, we think about how many 5's can fit in 9. 🤔
You're right! Only 1.
We write that on top of 9.
Now let's multiply 1 × 5.
What product do we get?
Good! It's 5.
Let's write that below 9. Then we subtract.
Now, we bring down 3, the second digit in the dividend.
We have 43.
How many 5's are in 43?
Correct! There are 8.
Let's write that above 3.
Now we multiply 8 by 5.
We get 40. We write that below 43.
Then we subtract.
Now, we bring down the third digit in the dividend.
Now we divide 31 by 5.
How many 5's can fit in 31?
That's right! 6.
We write 6 on top of 1.
Now let's multiply 6 with our divisor.
What do we get when we multiply 6 x 5?
You got it! 30
We write 30 below 31. Then we subtract.
What's the next step? 🤔
Correct! We bring down 7, the fourth digit in the dividend.
Now we have 17.
How many 5's can fit in 17? 🤔
You're right! 3
We write 3 on top of 7.
Now we multiply 3 by 5, our divisor.
3 x 5 is equal to 15.
We write 15 below 17.
Then we subtract.
The difference is 2.
Are there any digits left to bring down?
There are none left! 🤷♂️
That means 2 is our remainder.
Let's complete our division by writing 2 as our remainder.
So this is our final answer:
9317 ÷ 5 = 1863 R2
Great work! 🌟
Did you get all those steps?
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Duplicate | 711 | 2,530 | {"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} | 4.90625 | 5 | CC-MAIN-2023-14 | longest | en | 0.903623 |
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.14400.1/holomorphic/4.4.14400.1-20.1-k | 1,607,137,807,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141746033.87/warc/CC-MAIN-20201205013617-20201205043617-00179.warc.gz | 122,762,659 | 5,662 | # Properties
Label 4.4.14400.1-20.1-k Base field $$\Q(\sqrt{5}, \sqrt{6})$$ Weight $[2, 2, 2, 2]$ Level norm $20$ Level $[20, 10, \frac{2}{19}w^{3} - \frac{3}{19}w^{2} + \frac{1}{19}w - 1]$ Dimension $4$ CM no Base change no
# Related objects
• L-function not available
## Base field $$\Q(\sqrt{5}, \sqrt{6})$$
Generator $$w$$, with minimal polynomial $$x^{4} - 2x^{3} - 13x^{2} + 14x + 19$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2, 2]$ Level: $[20, 10, \frac{2}{19}w^{3} - \frac{3}{19}w^{2} + \frac{1}{19}w - 1]$ Dimension: $4$ CM: no Base change: no Newspace dimension: $20$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{4} - 50x^{2} + 56x + 184$$
Norm Prime Eigenvalue
4 $[4, 2, -\frac{2}{19}w^{3} + \frac{3}{19}w^{2} + \frac{37}{19}w - 3]$ $\phantom{-}1$
5 $[5, 5, \frac{3}{19}w^{3} + \frac{5}{19}w^{2} - \frac{27}{19}w - 1]$ $\phantom{-}1$
5 $[5, 5, -\frac{3}{19}w^{3} + \frac{14}{19}w^{2} + \frac{8}{19}w - 2]$ $-\frac{1}{172}e^{3} - \frac{7}{86}e^{2} + \frac{13}{86}e + \frac{77}{43}$
9 $[9, 3, -\frac{2}{19}w^{3} + \frac{3}{19}w^{2} + \frac{37}{19}w - 4]$ $\phantom{-}\frac{1}{172}e^{3} + \frac{7}{86}e^{2} - \frac{13}{86}e - \frac{120}{43}$
19 $[19, 19, -w]$ $\phantom{-}e$
19 $[19, 19, \frac{4}{19}w^{3} - \frac{6}{19}w^{2} - \frac{55}{19}w + 1]$ $-\frac{1}{172}e^{3} - \frac{7}{86}e^{2} - \frac{73}{86}e + \frac{77}{43}$
19 $[19, 19, \frac{4}{19}w^{3} - \frac{6}{19}w^{2} - \frac{55}{19}w + 2]$ $\phantom{-}\frac{2}{43}e^{3} + \frac{13}{86}e^{2} - \frac{52}{43}e - \frac{57}{43}$
19 $[19, 19, w - 1]$ $-\frac{3}{86}e^{3} + \frac{1}{86}e^{2} + \frac{39}{43}e - \frac{54}{43}$
29 $[29, 29, -\frac{5}{19}w^{3} - \frac{2}{19}w^{2} + \frac{64}{19}w + 2]$ $-\frac{7}{172}e^{3} - \frac{3}{43}e^{2} + \frac{91}{86}e + \frac{195}{43}$
29 $[29, 29, \frac{10}{19}w^{3} - \frac{34}{19}w^{2} - \frac{71}{19}w + 9]$ $\phantom{-}\frac{7}{172}e^{3} + \frac{3}{43}e^{2} - \frac{91}{86}e + \frac{192}{43}$
29 $[29, 29, \frac{4}{19}w^{3} - \frac{6}{19}w^{2} - \frac{55}{19}w + 4]$ $-\frac{1}{43}e^{3} + \frac{15}{86}e^{2} + \frac{69}{43}e - \frac{122}{43}$
29 $[29, 29, -w + 3]$ $\phantom{-}\frac{9}{172}e^{3} + \frac{10}{43}e^{2} - \frac{203}{86}e - \frac{48}{43}$
49 $[49, 7, \frac{5}{19}w^{3} - \frac{17}{19}w^{2} - \frac{64}{19}w + 11]$ $-\frac{21}{172}e^{3} - \frac{9}{43}e^{2} + \frac{359}{86}e + \frac{198}{43}$
49 $[49, 7, \frac{6}{19}w^{3} - \frac{9}{19}w^{2} - \frac{92}{19}w - 1]$ $\phantom{-}\frac{5}{43}e^{3} + \frac{11}{86}e^{2} - \frac{173}{43}e + \frac{266}{43}$
71 $[71, 71, \frac{3}{19}w^{3} + \frac{5}{19}w^{2} - \frac{46}{19}w - 1]$ $-\frac{13}{172}e^{3} + \frac{19}{43}e^{2} + \frac{341}{86}e - \frac{461}{43}$
71 $[71, 71, \frac{6}{19}w^{3} - \frac{9}{19}w^{2} - \frac{73}{19}w]$ $\phantom{-}\frac{27}{172}e^{3} + \frac{30}{43}e^{2} - \frac{523}{86}e - \frac{316}{43}$
71 $[71, 71, \frac{6}{19}w^{3} - \frac{9}{19}w^{2} - \frac{73}{19}w + 4]$ $\phantom{-}\frac{11}{172}e^{3} - \frac{9}{86}e^{2} - \frac{229}{86}e + \frac{486}{43}$
71 $[71, 71, -\frac{3}{19}w^{3} + \frac{14}{19}w^{2} + \frac{27}{19}w - 3]$ $-\frac{4}{43}e^{3} - \frac{13}{43}e^{2} + \frac{147}{43}e + \frac{415}{43}$
101 $[101, 101, \frac{7}{19}w^{3} - \frac{1}{19}w^{2} - \frac{63}{19}w - 3]$ $\phantom{-}\frac{2}{43}e^{3} + \frac{28}{43}e^{2} - \frac{95}{43}e - \frac{616}{43}$
101 $[101, 101, -\frac{9}{19}w^{3} + \frac{23}{19}w^{2} + \frac{81}{19}w - 5]$ $\phantom{-}\frac{7}{43}e^{3} - \frac{19}{86}e^{2} - \frac{354}{43}e + \frac{338}{43}$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$4$ $[4, 2, -\frac{2}{19}w^{3} + \frac{3}{19}w^{2} + \frac{37}{19}w - 3]$ $-1$
$5$ $[5, 5, \frac{3}{19}w^{3} + \frac{5}{19}w^{2} - \frac{27}{19}w - 1]$ $-1$ | 2,000 | 3,798 | {"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} | 3.046875 | 3 | CC-MAIN-2020-50 | latest | en | 0.296572 |
https://www.kaysonseducation.co.in/questions/p-nbsp-p-p_1479 | 1,708,802,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00451.warc.gz | 892,499,764 | 12,022 | Equation of a line passing through the intersection of the lines x + 2y – 10 = 0 and 2x + y + 5 = 0 is : Kaysons Education
# Equation Of A Line Passing Through The Intersection Of The Lines X + 2y – 10 = 0 And 2x + y + 5 = 0 Is
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## Question
### Solution
Correct option is
#### SIMILAR QUESTIONS
Q1
The straight line is x + y = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 from a triangle which is
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If the line x + 2ay + a = 0, x + 3by + b = 0 and x + 4cy + c = 0 are concurrent, then abc are in
Q3
If the line 2 (sin a + sin bx – 2 sin (a – by = 3 and 2 (cos a + cos bx + 2 cos (a – by = 5 are perpendicular, then sin 2a+ sin2b is equal to
Q4
If p1p2 denote the lengths of the perpendiculars from the origin on the lines x sec α + y cosec α = 2a and
x cos α + y sin α = a cos 2α respectively, then is equal to
Q5
The locus of the point of intersection of the lines x sin θ + (1 – cos θ) y = a sin θ and x sin θ – (1 + cos θ) y + a sin θ = 0 is
Q6
The straight lines 4x – 3y – 5 = 0, x – 2y – 10 = 0, 7x + 4y – 40 = 0 and x + 3y + 10 = 0 form the sides of a
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If two vertices of a triangle are (5, –1) and (–2, 3), and the orthocenter lies at the origin, the coordinate of the third vertex are
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The lengths of the perpendicular from the points (m2, 2m), (mmm +m) and (m2, 2m) to the line x + y + 1 = 0 form
Q9
The sine of the angle between the pair of lines represented by the equation x2 – 7xy + 12y2 = 0 is
Q10
The square of the differences of the slopes of the lines represented by the equation x2(sec2θ – sin2θ) – (2xy tan θ + y2 sin2θ = 0) is | 801 | 2,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.562174 |
https://forwardonclimate.org/popular-articles/how-do-you-evaluate-a-camera-calibration/ | 1,669,913,300,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00170.warc.gz | 309,195,651 | 11,974 | # How do you evaluate a camera calibration?
## How do you evaluate a camera calibration?
To evaluate calibration accuracy, you look at the distances between the detected points and the reprojected points. This is what a calibration algorithm typically tries to minimize. Depending on the calibration software you use, you may also be able to look at the uncertainty of the estimated camera parameters.
What is distortion calibration?
The goal of the distortion calibration is to find the transformation that maps the actual camera image plane onto an image following the perspective camera model. So far, there are many lens distortion models. Brown proposed a radial and decentering distortion model which are widely used in this field.
### Why Do cameras distort?
Barrel distortion often occurs when using wide-angle lenses. This is because the field of view of wide-angle lenses is wider than the image sensor on a digital camera and therefore the image looks like it has been squeezed and constricted to fit in the edges of the frame.
What is the camera calibration problem?
The result of camera calibration is an explicit transformation that maps a 3D world point M=(X,Y,Z,1)T into a 2D pixel m=(u,v,1)T.
## What is the need for camera calibration?
Camera calibration is the process of finding the true parameters of the camera that took your photographs. Some of these parameters are focal length, format size, principal point, and lens distortion. If you need high accuracy though, completing a camera calibration is very important.
What is distortion coefficients?
Tangential distortion occurs when the lens and the image plane are not parallel. The tangential distortion coefficients model this type of distortion. Normalized image coordinates are calculated from pixel coordinates by translating to the optical center and dividing by the focal length in pixels.
### What is camera calibration used for?
Geometric camera calibration, also referred to as camera resectioning, estimates the parameters of a lens and image sensor of an image or video camera. You can use these parameters to correct for lens distortion, measure the size of an object in world units, or determine the location of the camera in the scene.
What does lens 18 55mm mean?
A reading of 18-55mm is a focal length range. It means that you can change your focal length. The widest angle is 18mm, and you can zoom in to 55mm.
## What are the types of distortion?
Distortion occurs in six main forms:
• Longitudinal shrinkage.
• Transverse shrinkage.
• Angular distortion.
• Bowing and dishing.
• Buckling.
• Twisting.
What is camera calibration explain it with all its methods?
Camera calibration is the process of estimating intrinsic and/or extrinsic parameters. Intrinsic parameters deal with the camera’s internal characteristics, such as, its focal length, skew, distortion, and image center. Extrinsic parameters describe its position and orientation in the world.
### What is a distortion model?
Distortion models describe the mathematical deviation of a camera from the pinhole model.
What is distortion correction in camera?
When distortion control is applied to an image either in-camera or through software it corrects barrel or pin cushion distortion that occurs in images due to the characteristics of specific lenses. With Manual Distortion Control in the Retouch Menu you can adjust the image with a fine-tuning slider. | 678 | 3,437 | {"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} | 3.390625 | 3 | CC-MAIN-2022-49 | latest | en | 0.91655 |
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growth graph equity line
gmfx
May 25 2011 at 22:58
posts 8
I don't understand how the equity line is supposed to be calculated in the growth graph.
Can anyone help?
Thanks
speki
May 28 2011 at 19:30
posts 735
Suppose you have 1K balance and +100 profit on open trades (that's 10% of the balance).
Equity would show 110%.
Savvy?
Surround yourself with people whose eyes light up when they see you and who have no agenda for your reform.
gmfx
May 29 2011 at 12:57
posts 8
The concept of equity is very clear.
But that's not my problem.
My problem is with the growth chart.
It shows the line of growth of the account (in red) and the line of growth of the equity held for that account (in yellow).
The problem is that the equity growth is definitely very different from what this graph seems to indicate.
For instance, my account shows a growth of about 110% (which makes perfect sense, since the account started at \$50k and has now reached about \$105k so +\$55k), however it also shows a growth of equity of about 60% when my equity is next to \$60k. Now this doesn't make any sense: in this case the equity growth is no more than 20%. So where does that value of 60% come from? One might think, ok that's not the growth in equity, it's only how much equity you're holding against your balance. But that's also not possible because it would have to start at 100%, instead it starts at 0% just like the balance growth and it grows along with it although following the ups and down of the equity (but how exactly?).
In conclusion, the growth chart only gives an indication about the balance growth, while the equity growth line simply doesn't make any sense (to me now).
I'd be grateful if anyone form myfxbook could explain this.
Many thanks
Ethan (Staff)
May 29 2011 at 14:37
posts 1400
The growth's equity line is consistent with the balance's equity line in percent. For your last point, equity is at 56.29%, so if referring to the current growth, it is the 106.59%*56.29%=~60% which is what is seen in the chart.
Let me know if it's still unclear.
gmfx
May 30 2011 at 12:03
posts 8
Hi Ethan,
Thanks a lot for answering this.
However, I have to say this doesn't make a lot of sense to me: if this was right the equity line should start at 100%, right?
Why does it start at 0%, then?
I would expect any growth chart to start at 0%.
But what's the purpose if it plots a figure of ~60% when the actual increase is only ~20%?
That's a very large difference and the value that would be most informative about the system is the ~20% growth in equity, rather than the fact that equity is at ~60% of account balance in that moment (while it's still unclear why it starts at 0% instead of 100%)
Even worse: when the equity growth went in the negative (meaning my system had open trades losing more than it had earnt) the equity growth indicates a positive value of about 10% while in truth I had an equity of ~\$35k, thus making a loss of ~30% on \$50k.
I understand the growth line must be progressive and independent of the amount actually in the account, but it should also be indicative of the actual increase or decrease of the amount it tracks, be it the account balance or the equity held for that account.
I'm not expecting the MyFXbook team to change the way charts are displayed, but I must say I'd appreciate much more a growth chart that plots the actual growth of the equity as well as that of the balance of a given account.
I think this would give better indication to all traders.
Please correct me if I'm wrong.
Thank you :)
Ethan (Staff)
May 30 2011 at 12:10
posts 1400
That's simply a different visual plot of the same data 😄
The equity line is plotted as a percentage of the growth at each data point which would explain why it starts at 0% - the growth starts at 0% and 100% (equity) of 0% is still 0%.
In any case, I will forward your request to the development team for review.
Thanks.
gmfx
May 30 2011 at 14:41
posts 8
Ok, thanks for making this clear, Ethan, but still how meaningful is it?
It leads us to compare an increment (that of the account balance) not with the increment in equity for that account, but with the portion of equity held for that account scaled down by the increment of the account!! (perverse!)
What information do we gather from such comparison?
It's not just useless, it's misleading ... think about it ... (ultimately it's dimensionally wrong: we must compare increments with increments)
If you want to check the growth of a system as an indication of its performance you want to look at the increment (positive or negative) it has produced on the account balance AND on the equity.
If I have a 20% increment in equity, I don't want to read a chart that shows my equity at 60% of the increment on balance!
If my trading system produces a 30% loss in equity, I don't want to see a chart that plots a positive figure of 10% for my equity 'growth'!
Do you see my point?
Thanks for your response and for submitting this to the development team
Ethan (Staff)
May 30 2011 at 15:02
posts 1400
gmfx posted:
Ok, thanks for making this clear, Ethan, but still how meaningful is it?
It leads us to compare an increment (that of the account balance) not with the increment in equity for that account, but with the portion of equity held for that account scaled down by the increment of the account!! (perverse!)
What information do we gather from such comparison?
It's not just useless, it's misleading ... think about it ... (ultimately it's dimensionally wrong: we must compare increments with increments)
If you want to check the growth of a system as an indication of its performance you want to look at the increment (positive or negative) it has produced on the account balance AND on the equity.
If I have a 20% increment in equity, I don't want to read a chart that shows my equity at 60% of the increment on balance!
If my trading system produces a 30% loss in equity, I don't want to see a chart that plots a positive figure of 10% for my equity 'growth'!
Do you see my point?
Thanks for your response and for submitting this to the development team
I see your point, but the chart is a cumulative growth/balance plot.
The equity plot is relative to the growth and cumulative, so it's useful to see over time how it diverges/converges from the growth/balance.
I'm not sure why you think it's useless, as what you're asking is the same plot, only in absolute terms, not relative 😄.
In any case, as mentioned before I've already forwarded your suggestion to the development team. Your feedback is much appreciated!
speki
May 30 2011 at 16:17
posts 735
Man, for a year and a half, I thought this was useful and informative. Especially the convergences / divergences, equity crossing / touching the main growth curve. Now I have to learn it's perverse and useless. I need my head checked ASAP.
gmfx, could we keep shtum about this? It's gonna be fun to watch the other few thousand members fooling themselves with those useless equity curves 😁 😁 😁
Surround yourself with people whose eyes light up when they see you and who have no agenda for your reform.
gmfx
May 30 2011 at 16:22
posts 8
It remains that the current growth chart instead of comparing balance growth with equity growth, compares balance growth with the total value of equity proportion of balance scaled down by the balance growth ... meaningful?
I'm really not sure of the use of this, of one thing I'm sure: it's misleading to plot these two lines together in what's called a growth chart.
From a growth chart I expect to see how a trading system has performed on a given account in terms of (relative) balance growth and (relative) equity growth, if either goes in the negative it must show so.
With a bit of (tedious) math, just to make thing crystal clear:
b = balance [\$] = b(t)
e = equity [\$] = e(t)
t = time [days]
d = increment (delta): dt = t2-t1, db = b(t2)-b(t1), de = e(t2)-e(t1)
daily balance growth: b' = db/dt
daily equity growth: e' = de/dt
relative balance growth: b'/b
relative equity growth: e'/e
equity proportion of balance: e/b
Why not plot b'/b with e'/e instead of b'/b with e/b*b'/b?
Note that b is very different from b'/b and e is very different from e'/e.
Besides, in the chart of b with e, deposits and withdrawals are also taken into account, while they wouldn't show in the chart of b'/b with e'/e.
Thanks again and look forward to the response from the development team
gmfx
May 30 2011 at 16:32
posts 8
speki posted:
Man, for a year and a half, I thought this was useful and informative. Especially the convergences / divergences, equity crossing / touching the main growth curve. Now I have to learn it's perverse and useless. I need my head checked ASAP.
gmfx, could we keep shtum about this? It's gonna be fun to watch the other few thousand members fooling themselves with those useless equity curves 😁 😁 😁
Speki,
I'm simply pointing out how in the growth chart my equity appears to perform way better than it actually does if the equity line is accepted as equity growth, as the title of the chart suggests.
I have a system giving an equity increasing by 20% and a chart plotting it at 60%.
Worse, when equity goes into a loss of 30% this chart still shows it in the positive by 10%.
I do find this misleading, if you don't good on you, but please respect my voicing my doubts.
speki
May 30 2011 at 17:46
posts 735
Ah-ha. I think we're looking at things from a slightly different angle. But it's diversity that makes the world beautiful, so no problem with that. How about the new curve(s) beside the existing ones? Not instead of the old, well-known ones.
I think I understand the difference between your suggestion and the current e-b composite graph, just I don't get where would this extra information help us with trading? I'm just asking, not saying it wouldn't. I can't withdraw my equity, I can't pay the bills with it, it's purely virtual, that's why (I tend to think that) equity itself, without the balance / growth component, is hard to interpret. In other words, the goal is to have the balance increasing with equity as close to 100% as possible.
Then again, I might be mistaken!
Surround yourself with people whose eyes light up when they see you and who have no agenda for your reform.
gmfx
May 30 2011 at 20:41
posts 8
Hey speki,
I'm only a newbie trader trying to understand the use of these tools and I give my feedback hoping that can help improve things.
I think myfxbook offers a fantastic platform and we can only be grateful they made it freely available.
Surely they make it for money and I hope they can keep making more by offering even better tools.
So, obviously I understand thousands have been and are using the charts as they are and making the most of them.
In my small experience I found that the growth chart in particular offers some reasons to doubt the way it's designed.
Lately, I've found myself looking increasingly to the balance chart because I didn't know what to think about what was actually being plot as equity in the growth chart. Finally, I decided to ask the community, and the answer has been revealing.
From Ethan explanation, the growth chart plots equity in such a distorted way, that I find it really difficult to make much sense out of it.
Basically, equity is plot as a percentage of the account balance scaled by its growth (!?!).
So if you have zero growth you have equity at zero (!), if you have 100% growth you see your equity at the rate of its portion of your account balance, if you have 10% growth you have equity at one tenth of that value and so on.
Meaningful? what indication does this give us? how do we interpret this in terms of efficiency of a trading system?
Besides, the chart title being growth, leads one to think that the equity line should also represent some sort of growth of the equity. Absolutely not so! if you do, you may think of a trading system as way more reliable than it actually is.
As I said previously, I'm not expecting these charts to be changed only because I find it awkward to interpret them.
But from what Ethan said, it sounds as if this chart is indeed plotting two things that I wouldn't dream to put one next to the other in my wildest dreams ... (ok, let's not dig here ...)
I think if the chart will eventually get redesigned (which I hope, but I doubt), the current design won't be missed much at all.
In fact, because of how it's built, the equity line is often so close to the balance growth line, it's not really showing any interesting information. The crossings and convergences you can see them magnified and in more detail (because not dampened by the balance growth factor) in the balance chart. What you'd compare in the new growth chart would be the daily growth in balance and in equity, so much more informative and indeed quite different from the balance vs equity amounts chart.
If you want to track a history of the equity as a portion of an account balance, that can have it's own dedicated chart (starting at 100%), just like the profit does (starting at \$0).
About your point regarding equity being 'purely virtual', I have to disagree (sorry).
My understanding (please do correct me if I'm wrong) is that we can withdraw funds from an account up to a certain portion of the equity held (provided this is not greater than the account balance).
So I regard equity as one of the most important values to monitor in a trading system, because I think it's there that accounts get blown down. No matter how much the system has been able to accumulate, if it gets too exposed with open trades, it'll be forced to a halt.
And that's what's worrying me in the system I'm testing.
And that's why I want to monitor more closely the performance of this system in terms of equity and why ultimately the equity line in the growth chart doesn't give me any valuable information.
Of course, I will be more than happy to be shown ways to make use of such a strange indicator as the one plotted by the equity line in the current growth chart.
speki
May 30 2011 at 21:49
posts 735
No worries, gmfx. I think it's a good thing that you share your observations and that you suggest improvements. 😄
How do we interpret the current equity curve? Simple. If it's close to the growth curve, it's good. If it's going up and down like a pro's underwear, then the system is risky. Like this one:
https://www.myfxbook.com/members/bpmapan/bp-trend3/93756
I wouldn't say it's bad, but it's definitely risky. He's a lucky guy!
Yes, I agree, equity is a very important thing. You must keep your exposure in mind because if you are overleveraged, the usual wild swings could finish you in no time. But you can see your equity % on the current chart (or better, updated on a tick by tick basis, in your trading program). Equity here gets updated only when you have some activity on your account (trades opened, closed). But, the unrealised profit or loss, is virtual, until you close the trades.
This is what I don't fully understand:
<i>What you'd compare in the new growth chart would be the daily growth in balance and in equity, so much more informative and indeed quite different from the balance vs equity amounts chart.</i>
Suppose you have a trading logic or system or any set of rules. You wait for a signal, open the trade then the market either goes your way and you make a profit or it goes against you and you take a loss (well this was a bit of oversimplification since you can finish at breakeven too). Anyway, equity returns to 100% once the trade is done. So basically there's no 'equity growth', because you want to bank your profits. Again, this was a bit of simplification, if you're a position trader trading off the weeklies, you can monitor your daily equity progress, but again, you can see the same information from the current curve.
Sorry if I fail to understand your logic, I know I'm not the quickest of cats and you definitely have a unique approach. I will give it some time to think it through again.
Surround yourself with people whose eyes light up when they see you and who have no agenda for your reform.
speki
May 31 2011 at 06:18
posts 735
gmfx, I gave some thought to this matter but alas I still don't see any advantages of a separate equity curve. This does not mean that you're wrong or your idea is wrong, I might be missing something - so if you can give some examples how do you think this extra information would be useful, then thank you very much.
laters!
Surround yourself with people whose eyes light up when they see you and who have no agenda for your reform.
gmfx
Jun 01 2011 at 00:17
posts 8
Hey speki,
A day away brought some advice.
Indeed, I must say I have neglected some important details when I wrote down the math.
And recognising that means I now understand why the MyFxBook team has designed the Growth chart this way.
On the other hand, I must say I still regard it as potentially misleading.
So what have I realised that wasn't right in the math?
Well let's have a little repeat for sake of clarity:
gmfx posted:
b = balance [\$] = b(t)
e = equity [\$] = e(t)
t = time [days]
d = increment (delta): dt = t2-t1, db = b(t2)-b(t1), de = e(t2)-e(t1)
daily balance growth: b' = db/dt
daily equity growth: e' = de/dt
relative balance growth: b'/b
relative equity growth: e'/e
equity proportion of balance: e/b
Now, what the Growth chart plots, is not b'/b but the cumulative amount of b'/b since t0 (start) up to a given day t: sum(b'/b).
So this was my mistake.
cumulative balance growth to date: sum(b'/b) = sum(b'(t0)/b(to), ... , b'(t)/b(t))
Now it's easy to see why they decided to plot this cumulative growth in the account balance towards the portion of equity of the balance and even why they decided to scale this by that amount.
In fact, while the account balance can reach (hopefully) many times above 100%, the portion of equity held by the balance will supposedly hardly ever go too far higher than that. Hence, why they must have decided to scale the equity portion of balance by the balance cumulative growth. This dampens the equity line when balance cumulative growth is below 100% (<1) and amplifies it when it's higher (>1).
But I now see why (and that's thanks to you, speki and Ethan too).
If there were no funds added or withdrawn in the course of trading (like in demo accounts), the Growth chart would be nothing else but a distorted view of the Balance chart. Qualitatively they're the same, but the Balance chart is quantitatively more accurate.
When the trader adds and withdraws funds from the account, however, the Balance chart becomes confused and it's difficult to gather a clear picture of the performace of the trading system.
The Growth chart (however 'perverse' and potentially misleading it may be) screens away all these ups and downs and renders a picture qualitatively similar to what the Balance chart would be without the funds added and withdrawn.
Yes, ok we knew that before, but the point is that I now have figured out what's being plot and why.
And this solves my initial confusion and doubt about potentially misleading interpretation of the chart.
Which is a good result.
I still would find useful a chart plotting the equity portion of balance undeformed (e/b tout court, starting at 100%), and perhaps compare that with the cumulative equity portion of balance, sum(e'/e), as that would give us an understanding of how the system has performed in terms of equity: we would see when the equity has reached a certain portion (maybe 100%, maybe 10%, maybe 110%) of the account balance and we would be able to compare that with how the system has managed the open trades and the gain/loss in equity (proper actual equity growth).
This chart would be independent of funds added and withdrawn too and would also help avoiding being misled by interpreting the 'equity line' in the Growth chart as equity growth.
In my case, I would see the equity portion line hit +60% while the equity growth hits +20% or more extremely, when the latter hits -35%, the former is hitting +10%.
Yes, I believe that might help.
But I suppose, there would also be a myriad of other charts that surely many traders would find useful.
So I guess, I'll have to just be content of my gained understanding of what the current charts display.
Once again, thank you speki and Ethan.
I hope other readers found this topic of some interest too and who knows maybe someone in the development team also thinks this is worth considering.
engineofdeath
Apr 13 2018 at 07:12
posts 1
Hello!
Can you please tell me, is Growth chart working fine for me?
https://www.myfxbook.com/members/engineofdeath/mt5-17016249/2479290
I lost my first deposit (105 usd), and then I made another 100 usd deposit.
Now I have 290 usd in the account, but 'Equity Growth' and 'Growth' shows -98%.
Thank You!
Cloudballs
Apr 13 2018 at 12:24
posts 12
I don't think it takes into account deposits
Veuillez vous connecter pour commenter . | 5,008 | 21,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.968867 |
http://www.mathworks.com/matlabcentral/answers/54705-nonlinear-system-has-anyone-seen-this-equation-form?requestedDomain=www.mathworks.com&nocookie=true | 1,481,003,530,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541883.3/warc/CC-MAIN-20161202170901-00124-ip-10-31-129-80.ec2.internal.warc.gz | 539,999,419 | 13,520 | ## Nonlinear system (has anyone seen this equation form?)
### Jeff (view profile)
on 25 Nov 2012
Has anyone seen this form of a nonlinear equation with respect to X, but linear with respect to Y & Z? I provided a contour plot within the region for all 3 variables between -2 & 2. The plot is actually Z*conjugate(Z) so that the magnitude is above ZERO. If I am correct I may have seen this before in orbital mechanics, but my undergrad years are a bit behind me. I generated this eq. during a parametric analysis of a 2nd order oscillator. I would like to come up to speed about the physical phenomena this equation might describe.
I would provide a PDF of my results, but I do not see that I can upload a file. So below is my MATLAB code. If you can provide info how to attach or upload a file please respond.
```alpha=-2:0.004:2;
rho=10^-6;
for k=1:length(alpha)-1
rho(k+1)=rho(k)*10^(12/(length(alpha)-1));
end
ones(1:k+1)=1;
X=fliplr(-1*ones'*rho);
X=[X ones'*rho];
Y=alpha'*ones;
Y=[Y Y];
z=((1-Y).*((1-Y).^2+1).*((1+Y).^2-1).^2.*X-(1+Y).*((1+Y).^2+1).*((1-Y).^2-1).^2)./(4*(1-Y).*(1+Y).*((1-Y).*(1+(1+Y).^2)-(1+Y).*((1-Y).^2+1).*X));
Z=z.*conj(z);
mesh(X,Y,Z);
[m n]=size(Z);
axis([X(1,1) X(m,n) Y(1,1) Y(m,n) 0 2 0 2])
```
Walter Roberson
on 25 Nov 2012
bym
### bym (view profile)
on 25 Nov 2012
```ones()
```
is built in function which you are overwriting in your code, not a good idea
## Products
No products are associated with this question.
#### Join the 15-year community celebration.
Play games and win prizes!
MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi | 506 | 1,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-50 | latest | en | 0.817215 |
http://mathhelpforum.com/algebra/9720-general-exponents-question.html | 1,524,786,755,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00304.warc.gz | 216,458,272 | 9,492 | 1. ## General Exponents Question
The question is:
2^x = 512
I know how to solve this problem through just multiplying by two. I found that it was 9, but how would one solve this equation without going the long way? Such as:
5^x = 625
Thanks
2. [QUOTE=Xavier20;34022]
5^x = 625
Thanks
Take the logarithm of both sides,
$\displaystyle \log 5^x = \log 625$
$\displaystyle x\log 5=\log 625$
$\displaystyle x=\frac{\log 625}{\log 5}$
Use calculator to compute the answer.
3. Thanks a bunch. | 152 | 493 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-17 | latest | en | 0.886436 |
https://www.indiabix.com/civil-engineering/theory-of-structures/103005 | 1,708,752,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00709.warc.gz | 830,449,073 | 8,338 | # Civil Engineering - Theory of Structures
Exercise : Theory of Structures - Section 4
21.
For a strongest rectangular beam cut from a circular log, the ratio of the width and depth, is
0.303
0.404
0.505
0.606
0.707
Explanation:
No answer description is available. Let's discuss.
22.
The moment of inertia of a circular section about any diameter D, is
Explanation:
No answer description is available. Let's discuss.
23.
The ratio of maximum and average shear stresses on a rectangular section, is
1
1.25
1.5
2.0
2.5
Explanation:
No answer description is available. Let's discuss.
24.
The locus of the moment of inertia about inclined axes to the principal axis, is
straight line
parabola
circle
ellipse. | 180 | 708 | {"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-2024-10 | latest | en | 0.777801 |
https://socratic.org/questions/how-do-you-solve-x-2-y-8-and-x-2y-6 | 1,580,115,109,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00418.warc.gz | 668,551,290 | 6,298 | # How do you solve x^2+y=8 and x-2y=-6 ?
Jan 28, 2017
$\left(- \frac{5}{2} , \frac{7}{2}\right) , \left(2 , 4\right)$
#### Explanation:
${x}^{2} + y = 8$
$x - 2 y = - 6$
In the first equation solve for y in terms of x:
${x}^{2} + y = 8$ => $y = 8 - {x}^{2}$
Substitute for $y$ in the second equation:
$x - 2 \left(8 - {x}^{2}\right) = - 6$ => simplify:
$x - 16 + 2 {x}^{2} = - 6$ => add 6 to both sides and rewrite as:
$2 {x}^{2} + x - 10 = 0$
Solve by quadratic formula or by factoring,
$\left(2 x + 5\right) \left(x - 2\right) = 0$
$x = - \frac{5}{2} \mathmr{and} x = 2$
Solve for y:
$y = 8 - {\left(- \frac{5}{2}\right)}^{2} \mathmr{and} y = 8 - {2}^{2}$
$y = 8 - \frac{25}{4} \mathmr{and} y = 8 - 4$
$y = \frac{7}{4} \mathmr{and} y = 4$
Hence the solutions are:
$\left(- \frac{5}{2} , \frac{7}{2}\right) , \left(2 , 4\right)$ | 389 | 833 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-05 | latest | en | 0.592391 |
https://electricalandelectronicsengineering.com/current-divider-rule/ | 1,726,243,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00474.warc.gz | 209,065,261 | 13,563 | # Current Divider Rule: All Important Things you need to Know about CDR
In series circuits electric current always remains the same. However in parallel circuits, the current is not the same, instead, it is different based on the resistance of components. The Current Divider rule is used to determine how current splits at the node and how does it divide among parallel components. In this post, you’ll learn all about current divider.
## The formula
In figure above, two resistors Rx and R1 are connected in parallel.
The equation Ix is the CDR formula.
Mathematically: Current divider rule formula Ix = (Rt/Rx) * It
where Rt is the parallel equivalent resistance of Rx and R1.
## An Example
Consider a circuit having 10 Amps current source and two resistors R1 and R2. Using CDR formula we can find the current flowing through both resistors:
## Some Quick Points to remember
• Two equal resistors in parallel will have equal current flowing across them
• The higher the amount of resistance is, the lower will be the current (This is in accordance with Ohm’s law)
• A very small resistance in parallel with a large resistance will have almost all the current dissipated across it
• For any number of equal resistances, the overall current will be equally dissipated among the individual resistors
Categories EEE | 284 | 1,325 | {"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} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.927533 |
http://www.reddit.com/r/nba/comments/15o95q/the_worst_shot_in_basketball_and_some_ideas_for/c7oof32?context=3 | 1,429,618,646,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246641393.36/warc/CC-MAIN-20150417045721-00311-ip-10-235-10-82.ec2.internal.warc.gz | 767,022,806 | 14,261 | you are viewing a single comment's thread.
[–]Spurs[S] 1 point2 points (3 children)
sorry, this has been archived and can no longer be voted on
Well, I can tell you the overall average for those players :D
Transition: 38.87%
Spot up: 39.98%
But the real question is what is the conversion rate for transition layups that you are forgoing...
[–][WAS] John Wall 0 points1 point (2 children)
sorry, this has been archived and can no longer be voted on
and transition free throws
[–]Spurs[S] 2 points3 points (1 child)
sorry, this has been archived and can no longer be voted on
Exactly.
40% 3P works out to about 1.2 ppp
The Thunder are the no.1 Transition team at 1.26 ppp, but 3s are part of that but not all that much (< 20% of fga).
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Making a marginal argument, where you compare two sets of specifics shots, assumes that a shot is taken INSTEAD OF another. Can you show that link here? Is a T3 taken instead of a layup, or instead of starting the half court offence?
We should start with the simplest answer (the T3 is better than standard offence) until we can show that the T3 is taken instead of some other shot. | 332 | 1,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-18 | latest | en | 0.957058 |
https://ramblingsofaneuroticwriter.com/?post=99 | 1,627,058,725,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00035.warc.gz | 487,980,356 | 5,714 | # Can a wave propagate in 1D
## Mechanical waves¶
In the following, waves are considered that have a spatially sinusoidal propagation pattern. The wave starts at the origin of the coordinates with the deflection , the result is a wave propagation as shown in the following figure.
At a distance of an integral multiple of the wavelength The wave pattern is repeated in each case. From a spatial point of view, the wave has a period of length ; at the same time every sine function has a period of . The wave can thus be characterized by the following formula:
Here referred to the amplitude of the wave. Is an integral multiple of , the argument of the sine function becomes an integral multiple of . The wave begins at with the value , the above equation is sufficient to describe the wave, otherwise an initial phase angle must be added to the argument of the sine function to be added.
With the exception of standing waves, wave patterns do not stay in place, but move on over time. For example, if the wave moves in a positive direction -Direction, so the wave pattern moves in time about the length further.
For "shifting" the wave around applies:
This relationship is useful for determining the state of deflection of a sinusoidal wave at any given location and to be determined at any time: At the time got the shaft in place namely exactly the same deflection as it was at the time at the point would have. The following applies:
This has been simplified for the difference between the time and the starting point written. The equation can be further transformed by looking for the wave relationship uses:
In the second calculation step, the factor was multiplied into the inner bracket. If you write in this form for the frequency , the spatial and temporal period of the wave becomes clear:
The wave starts over again and again when a multiple of the wavelength is (spatial period), or if a multiple of the period of oscillation is (time period).
For practical calculations it is even more “handy”, including the factor to multiply in the argument of the sine function into the brackets. One obtains:
In this representation, the term just the angular frequency the wave; this indicates the speed at which the shaft oscillates in the pointer representation. The term is called accordingly as so-called "circular wave number" . This results in the following "simple" form of the equation for the state of deflection of a shaft:
The circular wavenumber indicates how many waves in a certain unit of length (for example or ) fit in. So the shorter the wavelength of a wave, the bigger it is -Value. For microwaves, for example on the order of about ever , with light waves in the order of magnitude of over ever .
Remarks: | 558 | 2,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-31 | latest | en | 0.922936 |
benchr267.github.io | 1,576,280,678,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00283.warc.gz | 267,646,974 | 4,107 | # Parsel
Parsel is a parser combinator library that makes it easy to write parsers. Parser combinators lets you create simple parses that can be combined together to very complex ones. Take for example a parser that parses a digit from a given String: (You can use the pre-defined lexical parser for digit: `L.digit`)
``````let digit = Parser<String, Int> { input in
guard let first = input.first, let number = Int(String(first)) else {
return .fail(/* some pre-defined error */)
}
return .success(result: number, rest: String(input.dropFirst()))
}
``````
We can now simply extend this to create a parser that parses an addition of two digits from a string:
``````let addition = (digit ~ L.plus ~ digit).map { a, _, b in a + b } // `L.plus` is a predefined parser that parses the `+` sign
try! result.unwrap() // Int: 6
``````
# Why should I use this?
Parsing is a very common task, it does not always mean to parse source code or JSON strings. Parsing means to transform an unstructured input to a structured output. In case of source code this means to parse a raw string to an AST (abstract syntax tree), in case of an addition it means to parse the result of adding two numbers out of a string. Parsing can always fail, if the input does not match the needed grammer. If the input string in the above example would have been `1+`, it would have been failed because the second number is missing. The advantage of parser combinators is that you start with a very basic parser. In the above example `digit` parses only one digit. But it is not hard, to add a parser that parses more than one digit. A number is a repetition of mulitple digits. For repetition, we can use `rep`, which tries to apply the parser until it fails and collects the result as an array. Parsing an integer addition is as easy as
``````func intFromDigits(_ digits: [Int]) -> Int {
return digits.reduce(0) { res, e in
return res * 10 + e
}
}
let number = digit.rep.map(intFromDigits)
let addition = number ~ L.plus ~ number ^^ { a, _, b in // ^^ is convenience for map
return a + b
}
try! result.unwrap() // Int: 579
``````
(There is also a pre-defined lexical parser for numbers `L.number` that is able to parse a number in different formats [binary, octal, hexadecimal, decimal])
Since parser combinators are very high level, they abstract the whole process of parsing a lot. That means it is easier to use but it also means that the performance is not as good as an optimized, handwritten parser.
# Installation
Parsel is currently available via Swift Package Manager and Cocoapods. Support for Carthage will come later.
## Swift Package Manager
To use Parsel in your project, just add it as a dependency in your `Package.swift` file and run `swift package update`.
``````import PackageDescription
let package = Package(
name: "MyAwesomeApp",
dependencies: [
.package(url: "https://github.com/BenchR267/Parsel", from: "2.2.0")
]
)
``````
## Cocoapods
To use Parsel with Cocoapods, just add this entry to your Podfile and run `pod install`:
``````target 'MyAwesomeApp' do
use_frameworks!
pod 'Parsel'
end
``````
# Requirements
• Swift 4.0
• Parsel is written in Swift 4.0 development snapshots and will be available when Swift 4.0 is released
# Example
Calculator is a small example I wrote with Parsel.
# Documentation
Check out the documentation on the Github page. | 827 | 3,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-51 | latest | en | 0.80277 |
https://web2.0rechner.de/mitglieder/reddragon1/ | 1,716,857,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00830.warc.gz | 519,667,216 | 4,907 | # RedDragon1
Benutzername RedDragon1 Punkte 36 Membership Stats Fragen 11 Antworten 1
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### Triangle DEF is inscribed in circle G. In circle G, angle EDF=29 and arc DF has a measure of 132. If the circumference of circle G is 24
RedDragon1 vor 6 Stunden
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### In triangle ABC, cos A=sqrt(7/10) and cos B=sqrt(3/10) Find cos C
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### The line x=k bisects the area of triangle ABC. Find the value of k. A on the coodinate grid is (3,2), B is (-5,0), and C is (1,0)
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### The point (8,2) is on an asyptote of the hyperbola x^2/a^2-y^2/b^2=1. What is |a/b|?
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### Four points form the vertices of a square. Find the area of the square if the points are (2,b), (0,a), (5,c), and (7,d).
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### A translation maps A ti A' and B to B'. We know AA'=6, AB=5, and AB'=5 find BA'
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RedDragon1 13.04.2024 | 392 | 977 | {"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} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.496829 |
https://forum.image.sc/t/circularity-roundness-din-iso-1101/5597 | 1,601,365,911,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401632671.79/warc/CC-MAIN-20200929060555-20200929090555-00208.warc.gz | 347,319,572 | 8,810 | # Circularity/ roundness (DIN ISO 1101)
Dear all,
I’m trying to find the minimum and maximum distance of center of ROI to permeter of ROI, see picture attached.
I’d really appreciate if you could help me
Katharina
1 Like
Good day Kath,
please tell us if the centerpoint coordinates are known?
Best
Herbie
Hi Herbie,
We get the centerpoint of the roi with the roi-manager.
The other two circles should have the same centerpoint coordinates.
Geetings
Katharina
Katharina,
here is a macro that does what you want for selections (ROIs) that are not rectangular:
/////////////////////// macro start ///////////////////////
requires(“1.51o”);
if (Roi.getType == “rectangle”) { exit(“Rectangular ROIs are not allowed!”); }
Roi.getCoordinates( roiX, roiY );
//Array.show( “Coordinates”, roiX, roiY );
List.setMeasurements;
cX = List.getValue(“X”);
cY = List.getValue(“Y”);
print( "Center X = " + cX + "; Center Y = " + cY + “;” );
w = getWidth();
h = getHeight();
dSqmin = w * w + h * h;
dSqmax = 0;
for ( i=0; i<roiX.length; i++ ) {
dSq = (roiX[i]-cX) * (roiX[i]-cX) + (roiY[i]-cY) * (roiY[i]-cY);
if ( dSq > dSqmax ) { dSqmax = dSq; }
if ( dSq < dSqmin ) { dSqmin = dSq; }
}
print( "Max Distance = " + sqrt(dSqmax) + "; Min Distance = " + sqrt(dSqmin) + “;” );
exit();
/////////////////////// macro end ///////////////////////
The code uses the Centroid as center.
Paste the code in a new macro window:
Plugins > New > Macro
You run the macro by Cmd-R (Mac) or Ctrl-R (PC).
HTH
Herbie
Edit: Updated script to include calibrated calculations.
Hello Katharina, and welcome to the forum. I see that @anon96376101 has already provided a macro solution to your question. I was playing around with this before I saw his post, just for practicing, so I will post my solution as well.
Edit: I added a lot of comments to the code, in case you are not familiar with programming (or just python).
It is very simple, but you can build on it if you are using this as part of a greater analysis protocol. It currently assumes that the ROI you want to analyse is the only (or first) ROI in the Roimanager.
To run it (in Fiji) copy-paste the code below to:
File >> New >> Script.
In the script window, select the language at the top (Python).
Click Run.
``````from ij import IJ
from ij.gui import Roi
from ij.plugin.frame import RoiManager
from java.awt import Polygon
import math
from ij import ImagePlus
from ij.measure import Calibration
def Roi_Dist():
""" Distance from center calculator. """
# Gets active image and it's calibration.
imp = IJ.getImage()
calib = imp.getCalibration()
# Get RoiManager.
rm = RoiManager.getInstance()
# Get the desired region, I only have the one in rm so index = 0.
region = rm.getRoi(0)
# Find the center [x,y] however you prefer, this gets the contourcentroid.
center_pixels = region.getContourCentroid()
center_calib = [calib.getX(center_pixels[0]), calib.getY(center_pixels[1])]
# Gets the pixel coords of all the points in the ROI polygon.
poly = region.getPolygon()
poly_xy = zip(poly.xpoints, poly.ypoints)
# Calibrated coordinates.
calib_x = [calib.getX(poly.xpoints[i]) for i in range (len(poly.xpoints))]
calib_y = [calib.getY(poly.ypoints[i]) for i in range (len(poly.ypoints))]
poly_xy_calib = zip(calib_x, calib_y)
# Calculates distance from the center to each point in the polygon.
# sqrt( (x2 - x1)^2 + (y2 - y1)^2 )
distances = [ math.sqrt((poly_xy[i][0] - center_pixels[0])**2
+ (poly_xy[i][1] - center_pixels[1])**2) for i in range(len(poly_xy)) ]
distances_calib = [ math.sqrt((poly_xy_calib[i][0] - center_calib[0])**2
+ (poly_xy_calib[i][1] - center_calib[1])**2) for i in range(len(poly_xy)) ]
# To round the distances to 2 decimals, include the following line.
#distances = [ round(i, 2) for i in distances ]
# Prints distances.
print "List of distances from center: ", distances
print "List of calibrated distances from center: ", distances_calib
# Prints (and IJ.logs) longest and shortest distances.
print "Longest distance: ", max(distances), "Shortest distance: ", min(distances)
IJ.log( "Longest distance: " + str(max(distances) ))
IJ.log( "Shortest distance: " + str(min(distances) ))
print "Longest calibrated distance: ", max(distances_calib)
print "Shortest calibrated distance: ", min(distances_calib)
IJ.log( "Longest calibrated distance: " + str(max(distances_calib) ))
IJ.log( "Shortest calibrated distance: " + str(min(distances_calib) ))
Roi_Dist()
``````
Hi you two,
thanks a lot for your help!
It seems, that both solutions work. My only problem now is, that I set a scale to change pixel for mm before, but I think the macros don’t consider that. The values appear too high.
Katharina
HTH Katharina.
No, none of these approaches consider the image calibration/scale. I updated my post to include this, try it out. Note that it targets the active image, so the image you selected before you hit run. There are better ways to do this, so if you need to improve it just ask!
Best,
Sverre
Hi Sverre,
I tried it with some examples and I just don’t get it. Perhaps some of the things that I do before I use your code are wrong.
1. I take the big white round circles major distance to calibrate the picture (set scale --> 40 mm) (there e.g. results 28,975 pixel/mm)
2. Calculate the area (with roi manager) of the white circle with the black hole, calculate the area of the black hole to get the share of the black whole
3. Calculate the centerpoints of white circle and black hole (with roi manager) to get displacement of the black whole in the white circle
4. get the roundess of the black whole (what we try to solve since yesterday ;-))
To show you two examples of the roundness-calculation:
The values that I get don’t seem right to me.
Hello Kath, let me try to clarify this.
This is what the script does, and to me it looks very accurate. Why do you feel these values are off?
I ran your bottom image with a scale of 40x40 mm and the script told me the longest distance from the center is ~15mm. Sounds correct in the current scale. A manual test with the line tool (see below) gives me 14.96 mm.
So your scale is defined as: the longest distance in the white circle = 40 mm? This sounds off. If the circle is not rigid, how can you know the real value of the longest distance?
Can you specify what the big white round circles are? Do you mean the big black circle? Or the white area between the black hole and the big black circle?
Aha, so what you actually want is the distance from the centerpoint of the black hole to the perimeter of the white circle?
This script does not measure roundness! It simply measure the distance of the center of the ROI to the perimeter. It is in no way a roundness-calculation. If you want roundness all you need is the feret diameter and the area… There is also a built-in roundness measurement in Fiji.
Analyze >> Set measurements >> Shape descriptors >>>> select ROI and click measure.
Good day Katharina,
here is a version of my macro that respects the scale setting of the image:
/////////////////////// macro start ///////////////////////
requires(“1.51o”);
if (Roi.getType == “rectangle”) { exit(“Rectangular ROIs are not allowed!”); }
Roi.getCoordinates( roiX, roiY );
toScaled( roiX, roiY );
List.setMeasurements;
cX = List.getValue(“X”);
cY = List.getValue(“Y”);
print( "Center X = " + cX + "; Center Y = " + cY + “;” );
w = getWidth();
h = getHeight();
dSqmin = w * w + h * h;
dSqmax = 0;
for ( i=0; i<roiX.length; i++ ) {
dSq = (roiX[i]-cX) * (roiX[i]-cX) + (roiY[i]-cY) * (roiY[i]-cY);
if ( dSq > dSqmax ) { dSqmax = dSq; }
if ( dSq < dSqmin ) { dSqmin = dSq; }
}
print( "Max Distance = " + sqrt(dSqmax) + "; Min Distance = " + sqrt(dSqmin) + “;” );
exit();
/////////////////////// macro end ///////////////////////
Regarding some other issues I think they may be related to making reasonable selections for your analyses but I may be wrong.
Regards
Herbie | 2,142 | 7,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-40 | latest | en | 0.80883 |
https://www.scia.net/en/scia-engineer/fact-sheets/concrete-design/concrete-2d-all-one-design-uls-sls | 1,726,118,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00599.warc.gz | 935,395,249 | 118,333 | # Concrete 2D All-in-one design (ULS+ SLS)
• Software
• SCIA Engineer
• Category Concrete Design
Highlights
• Design of 2D concrete elements for ULS
• Design of 2D concrete elements for ULS + SLS (cracks)
• Design of 2D concrete elements for ULS + SLS (reinforcement stress limitation)
• All in one design (ULS+SLS (cracks + stress limitation)).
• Calculation of Delta reinforcement due to SLS requirement
• Detailed output with transparent results
• Support the standard EN 1992-1-1:2004 and the national annexes.
In version of SCIA Engineer 19.1, the design of 2D members has been improved a lot. Thanks to our overall design check, the user with this version can calculate the required reinforcement for the ultimate limit state as well as for the serviceability limit state in one click.
## Design of 2D concrete elements for ULS+SLS based on EN 1992-1-1:2004
The design of concrete slabs and walls for ULS has been introduced since version 16 of SCIA Engineer.
The calculations for the ULS design has been improved with time. In the beginning, the calculation was only for the required shear and longitudinal reinforcement including the requirement from the code regarding the detailing of the reinforcement.
After that, SCIA has introduced a new methodology to simplify the results of the designed reinforcement by using a smart template to assume a basic amount of reinforcement As basic and let the software calculate the additional amount of reinforcement where it is needed As additional.
## The procedure of the design:
### Step #1: Calculation of As,req for ULS = As,ult.
The first step in the procedure of the design is to calculate the required reinforcement for the ultimate limit state. The new procedure is quite similar to the procedure in the previous version of SCIA Engineer. In this step, SCIA Engineer will calculate in fact two values:
As,ult = which the required amount of reinforcement to resist the applied forces.
As,req = which the required amount of reinforcement including the detailing provision form EN
### Step #2: Calculation of the required amount of reinforcement for ULS+SLS.
In this step and after the calculation of As, ult., the user has two possibilities to include the effect of SLS design on the calculation of the long. reinforcement.
The user can combine the calculation of ULS+SLS (based on cracks)
The user can combine the calculation of ULS+SLS (based on stress limitation)
The user can combine the calculation of ULS + SLS( based on cracks and stress limitation).
### Step # 3: Calculation of Delta As based on cracks
In this step and if the user activates the calculation of Cracks, SCIA Engineer will calculate the required reinforcement for the ultimate limit state and use this reinforcement for the calculation of cracks size wk based on 7.3.4 of the EN 1992-1-1:2004
To calculate the final required reinforcement, SCIA ENGINEER is doing the following:
• Calculation of principal forces MEd, Ch & MEd,QP
In this step, the user has to select a class of combination which should include at least one ULS and one SLS combination. It will make no difference which SLS combination is there in the background. Thanks to our smart combinator which will generate all the necessary SLS combination in the background for the calculation of the cracks and stresses.
• Re-Calculation of area of reinforcement in the direction of the principal forces.
In order to calculate the appearance of the crack, we need the amount of reinforcement calculated in the direction of the principal forces.
• Calculation of cracks.
Based on chapter 7.3.4 from EN 1992-1-1:2004, SCIA Engineer will calculate the size of the cracks.
As shown in the images below
In this case, SCIA Engineer is checking if the cracks are within the limits or not.
IF yes, then As,ult is good enough to fulfil the reinforcement for ULS + SLS (Cracks).
IF not, then SCIA Engineer will start the iteration processing to increase the As,ult by an extra amount of reinforcement to make the size of the crack within the allowable limits.
After this calculation, the user will be able to see the table below:
In this table, the user can be sure that the calculation of cracks did not require any additional longitudinal reinforcement because the value of Delta As,serv. is equal to zero.
So, the user can be sure also from that by reviewing the detailed output where the value of wk should be less than the limit.
• Calculation of stresses.
If the user activate the stress limitation design, then SCIA Engineer will calculate the amount of reinforcement for the ULS and use this reinforcement to calculate the actual stresses in the reinforcement and finally, compare that with the allowable limit which is located inside the concrete setting menu.
The user has three possibilities to define the limit of the stresses:
Auto= Based on definition in the national annexes 7.2(5)
Yield Strength= the limit is determined based on fyk (Characteristic yield strength of reinforcement)
User input= thee limit must be decided by the user. | 1,102 | 5,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.907562 |
https://softmath.com/algebra-software-4/worksheet-for-linear-equasions.html | 1,718,667,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00334.warc.gz | 474,821,802 | 9,489 | worksheet for linear equasions
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Author Message
dorh/Raem
Registered: 19.03.2007
From:
Posted: Monday 15th of Apr 07:41 Hi! We just started tackling a new lesson in math regarding worksheet for linear equasions and I did well for most assignments we had but the latest one my professor gave really hard so I'd love if somebody will assist me to understand it! It’s a problem solving assignment my math professor gave out this day and it’s due next week and I tried answering it but still can’t get it right. I can’t finish it easily unlike the other assignments. I had an easy time solving my past assignments but this particular assignment with specific topic of adding matrices gives me a hard time just figuring out how to begin. I’m desperately in need of help. I’ll really appreciate if somebody help me in discussing the steps and how to solve it in a organized and clear way.
Vofj Timidrov
Registered: 06.07.2001
From: Bulgaria
Posted: Monday 15th of Apr 16:55 I have been in your situation some time agowhen I was learning worksheet for linear equasions. What part of radical equations and equivalent fractions poses more problems ? Because I am sure that what you really need is a good program to help you understand the basic concepts and ways of solving the exercises. Did you ever use a program like that? I have tried many but I have to say that Algebrator is the best and the easiest to use. It's not like those other programs because it teaches you how to solve, it doesn't just give you the solutions.
Gog
Registered: 07.11.2001
From: Austin, TX
Posted: Tuesday 16th of Apr 20:26 I remember I faced similar difficulties with adding exponents, quadratic formula and graphing parabolas. This Algebrator is rightly a great piece of algebra software program. This would simply give step by step solution to any algebra problem that I copied from homework copy on clicking on Solve. I have been able to use the program through several Pre Algebra, Basic Math and Pre Algebra. I seriously recommend the program.
3Di
Registered: 04.04.2005
From: 45°26' N, 09°10' E
Posted: Thursday 18th of Apr 15:07 A great piece of math software is Algebrator. Even I faced similar problems while solving rational inequalities, converting fractions and simplifying fractions. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several math classes - Remedial Algebra, Intermediate algebra and Intermediate algebra. I highly recommend the program.
ov9db
Registered: 22.04.2003
From: Swindon, UK
Posted: Friday 19th of Apr 07:33 I hope this software has an easy interface . Can I have a look at it? | 674 | 2,956 | {"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-2024-26 | latest | en | 0.945778 |
https://therightmentor.com/tag/interact/page/2/ | 1,686,384,247,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00697.warc.gz | 648,246,561 | 40,163 | # Tag: Interact
• ## Quick Study Template
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error: Content is protected !! | 852 | 4,246 | {"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-2023-23 | longest | en | 0.63534 |
https://publisher.uthm.edu.my/periodicals/index.php/rpmme/article/view/3765 | 1,725,732,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00280.warc.gz | 456,221,606 | 5,988 | # Empirical Mathematical Model of Multiple Interaction Cracks
## Authors
• Muhammad Najmi Che Ariffin Universiti Tun Hussien Onn Malaysia
• Muhammad Khairudin Awang
## Keywords:
stress intensity factor, empirical mathematical model, multiple cracks, independent variable, dependent variable
## Abstract
As Stress Intensity Factors (SIF) may be used to measure the fracture force of the faulty structure quantitatively, the engineering application of fracture mechanics relies largely on knowledge of stress intensity factors. Many researcher methods have so far been proposed to measure stress intensity factors. Over the past few years, because of their reducing structural integrity, the issue of extended operations of aging aircraft structures is becoming pressing. Therefore, this study aims to investigate the SIF for multiple cracks growth by using software such Microsoft excel. It detailed about the existing study of multiple cracks and how to formulate the empirical mathematical model to predict the SIF. The findings of relationship between independent variable and dependent variable of multiple cracks with different load types. The parameters of the model are a/D and a/b which are crack depth ratio and cracks aspect ratio respectively. By using different cracks shapes of 0.1,0.2,0.3 and 0.4. For a/b, the range are from 0.2 and 1.2 with the increment of 0.2. The variable of c/l is 0.005, 0.01, 0.02,0.04, 0.08,0.16, 0.32. In conclusion, the equation has been made for SIF. The objective was achieved with seven equations are shown in this thesis from different C/L, a/D, a/b values
17-01-2022
Articles
## How to Cite
Che Ariffin, M. N., & Awang, M. K. (2022). Empirical Mathematical Model of Multiple Interaction Cracks. Research Progress in Mechanical and Manufacturing Engineering, 2(2), 311-317. https://publisher.uthm.edu.my/periodicals/index.php/rpmme/article/view/3765 | 436 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.880563 |
http://pillars.che.pitt.edu/student/slide.cgi?course_id=10&slide_id=19.0 | 1,506,209,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689806.55/warc/CC-MAIN-20170923231842-20170924011842-00618.warc.gz | 265,596,726 | 2,530 | # ITP: Forces Acting on Fluids
Before we identify forces acting on fluids it is important to recall that we are interested (ultimately) in performing a momentum balance. Recall that momentum is mass X velocity; therefore if we are performing a differential balance on momentum, we will look at rates at which momentum flows in or out of our system/control volume (i.e., a momentum per unit time). Clearly momentum per unit time is equivalent to mass X acceleration, and therefore any differential momentum balance will necessarily balances forces on the fluid!
### Stresses at a Point
Stresses in fluids can be of two types: normal and shear.
##### DEFINITION:
A normal stress is force per unit area acting normal (perpendicular) to a fluid area element $\Delta$A, and is usually denoted as $\sigma_{ij}$.
$\displaystyle{\sigma_{ij} = \frac{dF_n}{dA} \equiv \lim_{\Delta A\to \delta A} \frac{\Delta F_n}{\Delta A}}$
##### DEFINITION:
A shear stress is force per unit area acting tangential (parallel) to a fluid element $\Delta$A, and is usually denoted as $\tau_{ij}$.
$\displaystyle{\tau_{ij} = \frac{dF_s}{dA} \equiv \lim_{\Delta A\to \delta A} \frac{\Delta F_s}{\Delta A}}$
### Fluid Pressure at a Point
The fluid pressure is one of the components of the normal stress. In fact, if the fluid is at rest, the pressure is identically the magnitude of the compressive normal stress.
It is important to note that the pressure always acts compressively on a fluid element so that while the force magnitude due to the pressure is given as:
$dF_P = PdA$
the direction of the force (i.e., the force vector) will always be in the direction opposite of the fluid element's normal vector ( vector which points "outward" from a surface):
$d\vec{F_P} = - P\vec{n}dA$
##### NOTE:
Even in moving fluids the primary normal stress will be the pressure. Other normal stresses will be ignored except in Non-Newtonian fluids (we will define this term soon).
### Forces Acting on Fluid in Control Volume
In and open system (i.e., one with flow into and out of the system), the forces acting on the fluid comprise three types:
• Pressure force at inlets and outlets (a surface force): $\displaystyle{\vec{F_P} = - \int_{inlet}^{outlet} P\vec{n}dA}$
• Pressure and shear force exerted by wall surfaces on fluid in CV (surface forces): $\displaystyle{\vec{R}\equiv - \int_{walls}P\vec{n}dA + \int_{walls}\vec{\tau}dA}$
• Gravitational force exerted on all fluid within CV (a body or volumetric force): $\displaystyle{\vec{W} \equiv \int_{CV}\rho\vec{g}dV = \rho V\vec{g}}$
##### OUTCOME:
Name and explain the origin of forces acting on fluids in a control volume. | 680 | 2,664 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-39 | longest | en | 0.857203 |
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Math Corrections I understand where they get this answer but they did the division wrong, for example 21/7. You take 7 and the long way would to multiply it by 1 more till you get 21, 7x1=7,7x2=14, and 7x3=21. To make it even longer you can add. 7+7=14, and 7+7=14+7=21.
Article posted October 3, 2011 at 09:49 PM • comment • Reads 1685 • see all articles | 207 | 716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2013-48 | latest | en | 0.923042 |
https://music.stackexchange.com/questions/80917/must-a-tritone-substitution-use-a-dominant-functioning-seventh-chord/80923 | 1,576,322,917,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540586560.45/warc/CC-MAIN-20191214094407-20191214122407-00105.warc.gz | 468,466,830 | 35,974 | # Must a tritone substitution use a dominant functioning seventh chord?
The tritone substitution has been discussed in several questions here, perhaps most notably What is tritone substitution? and Why is bII 7(b5) considered a dominant? (I believe my question is different, but I'd understand and respect the urge to mark this as a duplicate.)
My understanding has always been that "tritone substitution" means that we use a dominant seventh chord that also has the same tritone found in the regular V7 chord. In C, `G–B–D–F` is our normal V7 chord. But we can instead play `D♭–F–A♭–C♭`, because the `B–F` tritone of the G7 becomes the `F–C♭` tritone of the D♭7.
But the answers in the linked questions all emphasize that the chords have a root a tritone away from each other, and that these "usually" involve dominant seventh chords; it's only mentioned later (although not always) that they also happen to share the same tritone.
I ask because—and here's why I think my question is not a duplicate—if "tritone substitution" means the first definition above, then it must be used with dominant seventh functioning chords. But if the definition is the second definition, then we can use any chord qualities. Needless to say, a proper understanding of this definition could prevent some major confusion regarding (and incorrect usage of) tritone substitutions, and I don't believe the current questions clarify this distinction.
Another way of asking this question would be: does the term "tritone substitution" refer to the interval shared between two chords, or the root distance between them?
• A related question that I think falls into the same umbrella: can tritone substitution be applied to diminished and half-diminished 7th chords? If so, must they have dominant functions, or can tritone substitution also be applied to common-tone diminished 7th chords? – Dekkadeci Mar 4 at 16:40
• @Dekkadeci, I wondered the same with `B F` found in `F` half diminished 7 `F Ab Cb (B) Eb`. But that would clearly not be the common jazz tritone sub. This could be an interesting new question. – Michael Curtis Mar 4 at 17:27
• Interesting, that, writing the chromatic scale out round a circle, the tritone is as far away from its 'oppo' as possible. Whereas either side of it, are the two most important notes in relation to the original I - IV and V. Also interesting is that the tritone on I is a semitone one way from its IV tritone, and a semitone the other way from its V tritone. – Tim Mar 4 at 18:13
• A rare example of a tritone sub on a chord with no tritone in it: Once in a concert I was playing a song with the final cadence Gm7 - Bb/C - F(6) and did a tritone sub on Bb/C (aka C9sus or C11), which has a dominant function but no 3rd. The resulting substitute is a GbM7#5(#11) (a nice chord if you play the root in a low register). – Mirlan Mar 5 at 4:26
• @jdjazz Yes, I should have clarified "dominant functioning chords." I thus include altered and extended dominants like V♭13, yes. – Richard Mar 6 at 20:42
My understanding has been that 'tritone' in tritone substitution refers to both preserving the shared tritone (`B` and `F` in this example, the chord 3rds and 7ths) and the fact that the roots will be a tritone apart.
So, when we put those two points together the sub will be a minimum of...
`Db F ? Cb (enharm. B)`
If we use either `Ab` or `Abb` for the chord fifth we get either `Db7` or `Db7b5`.
I suppose we could use `A` natural for a `Db7#5`.
So they aren't all dominant seventh chords, but include altered versions of a dominant seventh.
I've never tried the `#5` but I have tried the `b5`, but I imagine it would work just fine. The `A` natural could be held for resolving to a `C6`.
Anyway, my understanding is a two part meaning: common tritone & roots a tritone apart.
It was not part of the original question, but it may be worth adding that the tritone substitution does not only apply to the `V` chord in `ii V I` it could be used for any progression by descending fifth to a major or minor based chord. In practical application jazz likes to 'back cycle' meaning a chain of dominant sevenths leading to some (probably diatonic) chord. Like this: `VI7 II7 V7 I6` as `A7 D7 G7 C6`. Tritone subs could happen in various points of the chain `Eb7 D7 Db7 C6` or `A7 Ab7 G7 C6` etc.
• This seems like the right answer to me, since the OP is asking about the use of the term, as opposed to the use of the chord. – Ben Crowell Mar 4 at 20:08
The textbook answer actually can differ from how this is approached in practice. I'll give the textbook answer first. A tritone substitution is defined to have these characteristics:
1. you're substituting in one dominant 7th chord for an existing dominant 7th chord
2. the original chord and the substituted chord have roots that are a tritone away
3. the 3rd and 7th of the original chord and the 3rd and 7th of the substituted chord are enharmonically equivalent (e.g., B and F form the 3rd and 7th of both G7 and D♭7).
Most textbook definitions I've seen include stipulation #1 alongside either #2 or #3. However, I have actually encountered a definition that only cited stipulation #3! (It essentially listed #1 and #2 as corollaries!) Stipulation #3 is perhaps the most important factor of them all, because it's the reason why a tritone substitution works so well and sounds so good. This emphasis on the two chords sharing the same 3rd & 7th is well-justified because the 3rd and 7th notes define the chord quality:
• M3 + M7 → major 7th chord
• m3 + m7 → minor 7th chord
• M3 + m7 → dominant 7th chord
This certainly isn't a comprehensive list, but it covers the vast majority of chord qualities that one encounters in jazz. (And you can think about other chords like half-diminished or altered dominant as fitting into one of these three chord qualities.)
So for this question, G7, G7♯5, G7♯9, G7♭9, G7♭13, etc. can all be treated the same, because (a) they all contain the same 3rd and 7th, (b) their roots are all a tritone away from D♭7, and (c) they all share the same 3rd and 7th as D♭7.
What about other chord qualities besides dominant 7th--like minor 7th or major 7th? The textbook definition says that these don't qualify as tritone substitutions. For example, you can't substitute Dmin7 for A♭min7, since their 3rd's and 7th's aren't enharmonically the same. (Dmin7 has F and C as its 3rd and 7th, while A♭min7 has C♭ and G♭ as its 3rd and 7th. Those are not enharmonically equivalent, and thus this isn't a textbook tritone substitution.)
But in practice, this requirement isn't always obeyed, and it's not terribly uncommon to hear a tritone substitution that only follows stipulation #2 from above--that the roots are a tritone away. Here's the general evolution that occurred:
``````| Dmin7 | Dmin7 | G7 | G7 | Cmaj | Cmaj |
| Dmin7 | Dmin7 | Db7 | Db7 | CMaj | CMaj |
| Dmin7 | Dmin7 | Abmin7 | Db7 | CMaj | CMaj |
| Dmin7 | Dmin7 | Abmin7 | Abmin7 | CMaj | CMaj |
``````
The final version essentially performs the tritone substitution only using the ii chords, and entirely bypasses the V7 chords. This sort of thing can allude to a V7sus sound, or it can serve as something unique in its own right. I believe the final version is more frequently found in fusion jazz (a la Chick Corea). Perhaps because of this evolution, the emphasis has shifted from stipulation #3 to stipulation #2 when defining the tritone substitution.
• As a side note, just stipulating that a tritone is a dominant 7th chord and shares a tritone interval isn't sufficient because G7b9 and Bb7b9 share a tritone interval (they share two, in fact--Ab & D, and B & F). We need to specify that the 3rd's and 7th's are the same notes. – jdjazz Mar 5 at 11:26
No, it doesn't have to be a dominant 7th. But it does have to have a dominant function.
A tritone substitution substitutes the tritone found in one chord for the same tritone in another chord. C7 contains the tritone E-Bb; F#7 contains the enharmonic tritone A#-E. When you make a tritone substitution, the roots are also a tritone apart (F# is a tritone from C), but you couldn't use an F#m7 instead of F#7, because you will have changed the chord's function in the progression.
But retaining the function (and the internal tritone) does not require the substitution also be a dominant 7th chord. It could be Gb7+, or F#7b9, or any other altered or extended chord with a dominant function and a tritone between its third and seventh.
• Why change the key from the question in your answer examples? Richard gave `C`, but your answer is for `F`. It just makes things a little harder to follow IMO. – Michael Curtis Mar 4 at 17:21
• But retaining the function (and the internal tritone) does not require the substitution also be a dominant 7th chord. It could be Gb7+, or F#7b9, or any other altered or extended chord with a dominant function and a tritone between its third and seventh. To me, a 7+ or 7b9 is a dominant 7th chord. Additional colors, tensions, or alterations doesn't change the fact that it's a dominant 7th. What makes it a dominant 7th is that it has the 3 and 7 in it. I guess this is just terminology, but the OP's question is a question about terminology. – Ben Crowell Mar 4 at 20:10
• @BenCrowell - looks like that could become an answer! But with the internal tritone, won't the substitution be a dominant something anyway? What happens on the 5th or 9th won't affect that tritone, which is needed for tts anyway. – Tim Mar 5 at 8:12
• @BenCrowell - What makes it a dominant chord is that it has a 3 and b7, but what makes it a seventh chord is not having a 9, 11, or 13. That's why I distinguished function from "seventh" – Tom Serb Mar 5 at 12:35
• @TomSerb: What makes it a dominant chord is that it has a 3 and b7, but what makes it a seventh chord is not having a 9, 11, or 13. You don't need a 7 to make it a dominant chord. A triad can be a dominant, e.g., a G triad in the key of C. – Ben Crowell Mar 5 at 22:11
Tritone substitution is one chord swapped for another, whose root is a tritone away. As in C7 and F♯7.
In C7 there is a tritone between E and B♭, (M3 and m7), which get reversed in the tts of F♯7,, where it's A♯ and E, (M3 and m7).
So, if I understand correctly, the chords themselves are a tritone apart, and both contain a tritone which changes round (M3>m7 and m7>M3), making a full tts.
I understand the development of the tritonus substition as a derivation of the following chord construction: (my explanation is in major-C)
• V7b5 (G,B,Db,F) and its 2nd inversion (Db,F,G,B) - the so called augmented 34 chord (relating to the 4th Db-G)
• the augmented 34 Db,F,G,B is enharmonic identically with Db7b5, the tritonus substitution of G7b5, (namely Db,F,Abb,Cb)
• the 2 chords G7b5 and its tritonus substitution Db7b5 are distanced by definition a tritonus and are built by definition of the same tones.
• Nod in jazz also the chord of the normal dominant G7b9 (G,B,D,F,Ab) or natural Dom.7b9 (natural because of the natural fifth in G7) has been substituted by Db7b9 (Db,F,Ab,Cb,Ebb) - the Dom.7b9 of the tritonus. (Cb,Eb => B,D)
• The substitution of natural dominants (V7) allows a chromatic downwards progression of secondary dominants (V7) in steps of a minor 2nd instead of chains of secondary 5ths.
• This will lead to all kind of variations of steps by a minor 2nd or a 5th. But the tetrades are basically all (V7) or extended chords of (V7) or V7b5 and their function is one of a secondary dominante as far they follow the rule above (chromatic down-step or circle of 5th.
Now I think that Richard proposes that we can substitute any chord by (a same chord?) in distance of a tritonus.
Why not? This would be nothing else but the logical development of the process described above!
If we imagine a progression of a normal chain of (ii7-V7) progression following the circle of 5ths it would be obvious that also the IIm7 chords can be replaced by their tritonus substitution. All jazz chaps can tell whether this is usual. I would be surprised if it’s not. | 3,269 | 12,094 | {"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-2019-51 | latest | en | 0.945488 |
http://www.mathworks.com/help/symbolic/mupad_ref/fresnels.html?requestedDomain=www.mathworks.com&nocookie=true | 1,508,772,723,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00140.warc.gz | 488,152,003 | 14,028 | # Documentation
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# `fresnelS`
The Fresnel sine integral function
MuPAD® notebooks are not recommended. Use MATLAB® live scripts instead.
MATLAB live scripts support most MuPAD functionality, though there are some differences. For more information, see Convert MuPAD Notebooks to MATLAB Live Scripts.
## Syntax
```fresnelS(`z`)
```
## Description
`fresnelS(z)` = $\underset{0}{\overset{z}{\int }}\mathrm{sin}\left(\frac{\pi {t}^{2}}{2}\right)dt$.
The function S = `fresnelS` is analytic throughout the complex plane. It satisfies ```fresnelS(-z) = -fresnelS(z)```, `fresnelS(conjugate(z)) = conjugate(fresnelS(z))`, ```fresnelS(I*z) =-I*fresnelS(z)``` for all complex values of z.
`fresnelS(z)` returns special values for ```z = 0```, `z = ±∞`, and ```z = ±i∞```. Symbolic function calls are returned for all other symbolic values of `z`. In a MuPAD® notebook `fresnelC(z)` appears in a typeset notation as $S\left(z\right)$.
For floating-point arguments, `fresnelS` returns floating-point values.
`simplify` and `Simplify`, `fresnelS` uses the reflection rule `fresnelS(-z) = -fresnelS(z)` to create a “normal form” of symbolic function calls. See Example 3.
## Environment Interactions
When called with floating-point arguments, these functions are sensitive to the environment variable `DIGITS` which determines the numerical working precision.
## Examples
### Example 1
Call the Fresnel sine integral function with various arguments:
```fresnelS(0), fresnelS(1), fresnelS(PI + I), fresnelS(z), fresnelS(infinity)```
For floating-point arguments, `fresnelS` returns floating-point values:
```fresnelS(1.0), fresnelS(float(PI)), fresnelS(-3.45 + 0.75*I)```
### Example 2
`diff`, `float`, `limit`, `series`, and other functions handle expressions involving the Fresnel sine integral function:
`diff(fresnelS(x), x)`
`float(fresnelS(-100))`
`limit(fresnelS(x), x = -infinity)`
`series(fresnelS(x), x = infinity, 4)`
### Example 3
`simplify` use the reflection rule `fresnelS(-z) = -fresnelS(z)` to create a “normal form” of symbolic function calls:
`simplify(3*fresnelS(z) + 2*fresnelS(-z))`
## Parameters
`z`
## Return Values
Arithmetical expression.
`z`
Watch now | 706 | 2,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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-2017-43 | latest | en | 0.592591 |
https://math.stackexchange.com/questions/3464261/finding-the-nearest-power-of-2-with-a-formula | 1,716,235,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00311.warc.gz | 335,688,784 | 36,141 | # Finding the nearest power of 2 with a formula
Is there any simple formula that can be used to find the nearest power of 2 of a number?
So, let's say I got the number $$15$$, it has to give $$16$$, If I got the number $$55$$, it gives me $$64$$, etc.
• Nearest in what sense? is $12$ "nearer" to $8$ or is it "nearer" to $16$? What about $11.5$? Whatever your final interpretation, play with $\log_2$ and floor/ceiling functions. Dec 5, 2019 at 14:56
• indeed, if it is the same distance to two power of 2's, it can use any of the two. I was wondering if their is a closed form formula, without using floor/ceiling functions... Dec 5, 2019 at 14:57
• Given the very nature of what you are describing as being a step function of sorts... a floor or ceiling is almost guaranteed in some form. Dec 5, 2019 at 14:58
• If you convert you number into binary, you can easily do this Dec 5, 2019 at 14:59
• @Steven31415 Not true. That would make $2^{n+1/2}$ the dividing line between $2^n$ and $2^{n+1}$, but it should be $3 \cdot 2^{n-1}$. Dec 5, 2019 at 15:09
$$a=\log_2 x- \lfloor \log_2 x \rfloor$$
$$b=\lceil \log_2 x \rceil -\log_2 x$$
If $$min\{a,b\}=a$$ then the nearest power of $$2$$ is $$2^{\lfloor \log_2 x \rfloor}$$,
Else it is $$2^{\lceil \log_2 x \rceil }$$ | 434 | 1,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | latest | en | 0.905065 |
https://byjus.com/question-answer/60-g-of-ice-at-0-0-is-mixed-with-60-g-of-steam-at/ | 1,643,023,341,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00352.warc.gz | 206,928,186 | 23,459 | Question
# $$60 g$$ of ice at $$0^0$$ is mixed with $$60 g$$ of steam at $$100^0$$. At thermal equilibrium, the mixture contains:(Latent heats of steam and ice are $$540 cal { g }^{ -1 }$$ and $$80 cal { g }^{ -1 }$$ respectively, specific heat of water $$= 1 cal { g }^{ -1 } { }^{ -1 }$$)
A
80g of water and 40g of steam at 1000
B
120g of water at 900
C
120g of water at 1000
D
40g of steam and 80g of water at 00
Solution
## The correct option is A $$80 g$$ of water and $$40 g$$ of steam at $$100 ^0$$Heat energy required to melt $$60 g$$ of ice $$= 60\times 80=4800 Cal$$.Heat energy required to raise the temprature of water at $$0$$ degree centigrade to $$100$$ degree centigrade $$= 100\times 1\times 60=6000 Cal$$Total energy required $$= 4800+6000=10800 Cal$$This energy is given by the steam by getting converted into water.Amount of steam getting converted into water is $$\dfrac{10800}{540}=20g$$Thus, finally total $$80 gm$$ of water will be there and $$40 gm$$ of steam. Physics
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1. ## Matrix Browser
Hello all, I need a help regarding Matrixbrowser. I am not able to get the area of the surfaces in Matrixbrowser. I see that area can be calculated only based on elements. Any inputs or alternative options available on this. Please help me. Thanks and regards, Punith
2. ## Synergy between Compose and Matrix Browser
I'd like to share the nice and slick synergy Compose has with Matrix Browser, a feature in HyperMesh to handle model data in table format. This is the typical use case when the model data will lead the process and Compose will play as the Math/Automation engine. Let’s assume that we have the model ExampleRib.fem (attached) and that we would like to get the central coordinates X, Y and Z of a few elements: In this example, I’d like to calculate the central distance of these elements to the point (0,0,0), which would be simply the square root of x²+y²+z². For that, I created a custom Compose function distZero (also attached): Which will receive as inputs from HyperMesh the coordinates X, Y and Z that I queried with Matrix Browser. In order to expose the Compose function to other HyperWorks applications, we need to register the function. This is done clicking on the right mouse button on Register Function in Compose: After this step, we may now see that the function has been registered under the same mouse click shown above under Show Registered Functions button. Once it is registered, we may now use it inside HyperMesh + Matrix Browser: The function will be available without the need for closing HyperMesh and reopening. Then we give the columns that will serve as arguments: And it's done! The custom Compose function calculated everything and the results are shown in a new column: Regards, Roberta ExampleRib.fem distZero.oml
3. ## 要素の持つ板厚情報をCSVで出力
各要素のもつ板厚情報をCSVで出力したいです。 Midmesh Thicknessesで関連付けした要素の板厚情報を出力しようと 試みていますがうまくいきません・・・ なにか良い方法があればご教授のほどよろしくお願いいたします。
×
×
• Create New... | 814 | 2,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-16 | longest | en | 0.518507 |
https://mechanicalland.com/how-to-create-points-in-siemens-nx/ | 1,696,021,855,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00860.warc.gz | 419,484,826 | 19,019 | # Creating Points in Siemens NX(Illustrated Expression)
In some cases, there could be a requirement about specifying a point in a CAD project to use later for some purpose in Siemens NX®. In this article, we will show you creating points ins Siemens NX by using the Point command. So, you could create your points efficiently if you follow the steps that we introduced to you. Siemens NX® provides lots of tools to create the desired geometry features in general, also you will understand that this fact is true for this topic also.
Our website includes lots of useful information about various CAD programs, so visit the main page to see them…
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## How a Point is Created in Siemens NX®?
Firstly, to create a point in Siemens NX®, you need to click on the Point command as shown by the blue arrow.
### Point Creating Methods in Siemens NX
Secondly, in the Point command interface, there are lots of point creating methods in Siemens NX as above.
• Inferred Point: You could create your point by selecting a point in the drawing scheme in Siemens NX®.
• Cursor Location: You could click on the drawing scheme to obtain points in Siemens NX®.
• Existing Point: You could create points on existing points in Siemens NX®.
• End Point: Simply selecting of en points of sketches in Siemens NX®.
• Control Point: You could place your point to a control point of a Spline in Siemens NX®.
• Intersection Point: You can select an intersection point of sketches to place your point in Siemens NX®.
• Arc/Ellipse/Sphere Center: You can place your point in the center of an arc, an ellipse is a sphere in Siemens NX®.
• Angle on Arc/Ellipse: You could place your point on arc or ellipse by entering an angle value to place your point o arc or ellipse at an angle in Siemens NX®.
• Quadrant Points: You can place your point in quadrants of an arc.
• Point On Curve: Edge: You could place your point on an edge or face by clicking on Siemens NX®.
• Point On Face: You could place your point on a selected face or plane.
• Between Two Points: You place your point by selecting two points to place your point between them by entering a value.
• Spline Pole: You could place your points on the selected Spline pole in Siemens NX®.
• Spline Defining Point: You can place your point on selected Spline defining points.
### Point Creating in Siemens NX
For example, we selected an edge of the geometry to place our point as shown by the blue arrow. Also, you can enter your X, Y, and Z coordinates according to the selected reference to place your point in the blue box in Siemens NX®. You could finish your creating points Siemens NX® by clicking on the OK or Apply buttons as shown by the green box.
Finally, our point is created as shown by the green arrow above.
There are lots of other commands in Siemens NX® like creating points in Siemens NX. Check the related posts about Siemens NX® to learn about other useful referencing commands.
## Conclusion
So, creating points in Siemens NX is very simple like this.
Do not forget to leave your comments and questions below about the Point command in Siemens NX®!
Your precious feedbacks are very important to us.
NOTE: All the screenshots and images are used for educational and informative purposes. Images used courtesy of Siemens NX®.
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http://mathhelpforum.com/calculus/121127-derivative.html | 1,481,433,350,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00123-ip-10-31-129-80.ec2.internal.warc.gz | 174,857,136 | 10,463 | 1. ## Derivative
Use the radius and tangent theorem to find the derivatives of:
$f(x)=7-\sqrt{2x-x^2}$
2. Originally Posted by deltaxray
Use the radius and tangent theorem to find the derivatives of:
$f(x)=7-\sqrt{2x-x^2}$
$\frac{d}{dx} f(g(x))=g'(x)f'(g(x))$
3. I believe that the "radius and tangent theorem" is the theorem from geometry that a line tangent to a circle is perpendicular to the radius of the circle to the point of tangency. Since the function $f(x)=7-\sqrt{2x-x^2}$ is the lower half of the circle $(x-1)^2+(y-7)^2=1$, the circle of radius 1 and center (1,7), a point (x,f(x)) is the endpoint of a radius whose other endpoint is the center (1,7), and thus with slope $\frac{7-f(x)}{1-x}$. Therefore the line tangent to our circle at that point is perpendicular to that radius, and so has opposite reciprocal slope; the tangent line at (x,f(x)) has slope
$-\frac{1-x}{7-f(x)}=\frac{x-1}{7-f(x)}$. But the slope of the tangent line is f'(x), so one just plugs $f(x)=7-\sqrt{2x-x^2}$ into this $f'(x)=\frac{x-1}{7-f(x)}$ to get one's answer. | 331 | 1,062 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.862759 |
https://socratic.org/questions/5875d2b2b72cff3e2ba877d7 | 1,623,706,093,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00100.warc.gz | 476,065,127 | 5,584 | # Question #877d7
Jan 11, 2017
$12 < 2 \sqrt{40} < 13$
#### Explanation:
Since
$2 \sqrt{40} = \sqrt{{2}^{2} \cdot 40} = \sqrt{160}$
and
$\sqrt{144} < \sqrt{160} < \sqrt{169}$
then
$12 < \sqrt{160} < 13$ | 97 | 211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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} | 3.390625 | 3 | CC-MAIN-2021-25 | latest | en | 0.364726 |
https://docs.analytica.com/index.php?title=User-Defined_Functions&oldid=56382 | 1,686,313,650,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00417.warc.gz | 238,011,844 | 9,779 | # User-Defined Functions
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
You can create your own functions to perform calculations you use frequently. A function has one or more parameters; its definition is an expression that uses these parameters. You can specify that the function check the type or dimensions of its parameters, and control their evaluation by using various parameter qualifiers.
A Library is a a collection of functions grouped in a library file to extend Analytica’s built-in functions for use for particular types of applications. Analytica is distributed with an initial set of libraries, accessed from the File menu with the Add Library... option. If you add a Library to a model, it will appear with its functions in the Definition menu, and these functions will appear almost the same as the built-in functions.
You may want to look at these libraries to see if they provide functions useful for your applications. You may also look at library functions as a starting point or inspiration for writing your own functions.
Analytica experts may create their own function libraries for particular domains. Other Analytica users can benefit from these libraries.
One of the nice things about defining functions in Analytica is that the function definition uses the same expression syntax used to define variables, so that defining a new function is just like defining a variable, along with an additional parameters attribute.
## Example function
The following function, Capm(), computes the expected return for a stock under the capital asset pricing model.
Parameters: (Rf, Rm, Beta: Numeric)
Definition: Rf + Beta * (Rm - Rf)
Sample usage: You use the Capm() function in a definition in the same way you would use Analytica’s built-in functions. For example, if the risk free rate is 5%, the expected market return is 8%, and StockBeta is defined as the beta value for a given stock, we can find the expected return according to the capital asset pricing model as:
Stock_return: Capm(5%, 8%, StockBeta)
This definition functions equally well when StockBeta is an array of beta values. In this case, the result will be an array of expected returns.
## Creating a function
To define a function:
1. Make sure the Edit tool () is selected and you can see the node palette.
2. Drag the Function node icon from the node palette into the Diagram area.
3. Title the node, and double-click on it to open its Object window.
4. Fill in the new function's attributes.
## Attributes of a function
Like other objects, a Function is defined by a set of attributes. Many of these attributes are the same as the attributes of Variables, including identifier, title, units, description, and definition, inputs, and outputs. It possesses one unique Attribute, Parameters, which specifies the parameters available to the function.
Identifier
If you are creating a library of functions, make a descriptive identifier. This identifier appears in the function list for the library under the Definition menu, and is used to call the function. Analytica makes all characters except the first one lower case.
Title
The title appears in the Function node. It is usually good to have the title match the identifier, although for clarity you can also include a brief parameter description as part of the title. For example, we might title our Capm function as Capm(Rf, Rm, Beta).
Units
If desired, use the units field to document the units of the function’s result. The units are not used in any calculation.
Parameters
The parameters to be passed to the function must be enclosed in parentheses, separated by commas. For example: (x, y, z)
The parameters may have type qualifiers (see the next section).
If you are creating a library of functions, use descriptive abbreviations for the parameters and give them a logical sequence. The parameters will appear in the Object Finder dialog box and they will be pasted when the function is pasted from its library in the Definition menu.
Note: Earlier releases of Analytica (pre-4.0) limited the number of total parameters and local variables to 32. No such limit exists any more -- you can have as many parameters as you desire.
Description
The Description should first document what the function returns, and explain each of its parameters. If the Definition is not immediately obvious, the second part of the Description should explain how it works. The Description text for a function in a Library will also appears in a scrolling box in the bottom half of the Object Finder dialog.
Definition
The definition of a function is an expression or compound list of expressions. It should usually use all of its parameters. When you select the definition field of a function in Edit mode, it shows the Inputs pull-down menu that lists the parameters as well as any other Variables or functions that have been specified as inputs to the function. You can specify the inputs to a function in the same way as for a Variable—by drawing arrows from each input node into the function node.
Recursive
If your function needs to call itself, or needs to call another function that will ultimately call your function again, then it is a recursive function. By default, Analytica will issue an error message indicating that you have a dependency cycle if you attempt a recursive definition without setting this attribute to 1 (true). Since recursive functions are fairly rare, this helps to catch mistakes.
If you really intend a recursive function, then you should set this parameter to true. You can do this from the attribute pane (below the diagram), or you can configure Analytica to show the recursive attribute in the object window. To keep things simple, the attribute is hidden by default. To configure it to show, select Attributes... from the Object menu. Select Function in the pulldown, and set a checkmark next to Recursive in the attribute list. Then to indicate that your function should allow recursion, set the attribute to 1.
## Parameter qualifiers
Parameter qualifiers are keywords you may use in parameters to specify for each parameter how, or whether, it should be evaluated when the function is used (called), and whether it should have a particular type of value, such as number, text value, or other. Other qualifiers specify whether a parameter should be an array, and if so, which indexes it expects.You can also specify whether a parameter is optional. By using qualifiers properly, you can help make functions easier to use, more flexible, and more reliable.
For example, consider this Parameters Attribute:
(A: Number Array[I, J]; I, J: Index; C; D: Optional Atom Text)
It defines five parameters, A, I, J, C, and D.A should be an array of numbers, indexed by parameters I and J. I and J, being separated by commas "," rather than semicolons ";" are subject to the same qualifier, IndexType. C has no qualifiers, and so can be of any type, or dimensions. The semicolon ";" between C and D means that the qualifiers following D do not apply to C. D has three qualifiers, specifying that it is Optional, Atomic, and a text value.
The page Function Parameter Qualifiers defines all parameter qualifiers in detail.
Function Parameter Qualifiers provide very rich control over Evaluation Modes and Array Abstraction, which can contribute a tremendous amount of programming power alone.
## Libraries
To do:
• Library icons, node shapes
• Creating a library | 1,512 | 7,468 | {"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} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.872392 |
https://www.mrexcel.com/board/threads/recalculate-expiry-date-based-on-two-factors.751457/ | 1,695,808,004,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00264.warc.gz | 992,501,956 | 19,624 | # Recalculate expiry date based on two factors
#### michelleha
##### New Member
Using excel 2013 (2010 at work) I have the following...
D F O P Q S T 2 Date Received Date Due Delay required New Due Date Notify Manager Date Closed Days Overdue 3 11/12/2013 =IF(ISBLANK(D3),"",D3+14) 25/12/2013 N/A N/A N/A 26/12/2013 =DATEDIF(F3,S3,"D") 1 4 12/12/2013 =IF(ISBLANK(D4),"",D4+14) 26/12/2013 Yes =IF(ISBLANK(O4),"",F4+7) 02/01/2014 =IF(F4=TODAY(),HYPERLINK("mailto:?subject="&SetUp!\$B\$25&"&body=" &SUBSTITUTE(SUBSTITUTE(SetUp!\$E\$26,"\$",J6),"@",B6), "Notify Manager"),"") Notify Manager ?? 5 12/12/2013 =IF(ISBLANK(D5),"",D5+14) 26/12/2013 Yes =IF(ISBLANK(O5),"",F5+7) 02/01/2014 =IF(F5=TODAY(),HYPERLINK("mailto:?subject="&SetUp!\$B\$25&"&body=" &SUBSTITUTE(SUBSTITUTE(SetUp!\$E\$26,"\$",J6),"@",B6), "Notify Manager"),"") Notify Manager 05/01/2014 ??
<tbody>
</tbody>
Explanation:
Row 4 - Date Received is entered in D, this calculate the required due date in F. If there is going to be a delay in response then "yes" is selected in 0, which calculates the New Due Date in P.
Here is were I get stuck...
1. Q "Notify manager" works when Today=F but how do I get it to take into consideration if a new due date is calculated in P?
2. When the "Notify Manager" is active Admin personal go in and send the email off... but how do I take into consideration weekends.. I want a => function I think...? Thoughts?
3. T "Days Overdue" considers F & S but how do I get it to take into consideration if a new due date is calculated in P?
Thanks heaps!!
### Excel Facts
Back into an answer in Excel
Use Data, What-If Analysis, Goal Seek to find the correct input cell value to reach a desired result
First observation: It's good practice not to have hard-coded numbers like 14 and 7 scattered throughout your spreadsheet. What happens if in future the extension period is changed from 7 days to 10 days, say? If you try a global replace of 10 for 7, you'll make all sorts of unintended changes to your spreadsheet.
Instead, it's better to set your parameters once, e.g. set \$A\$1=14 and \$A\$2 = 7, and refer to these cells in all your formulae. You can also give these cells meaningful range names e.g. ProcessingTime, and ExtensionTime.
I suggest then that you only need one column for Due Date. Formula in F3, for example, can be
=IF(ISBLANK(D3),"",D3 + ProcessingTime + If(O3="YES",ExtensionTime,0))
Finally, yes, if you test F4<=TODAY() it will identify all orders due today or on previous days (so if today is the first day back after a weekend or public holiday, it will include orders due on those dates and now overdue). But presumably you'll want to turn this off if the DateClosed is non-blank, i.e. you don't want "Notify Manager" for all historical orders, filled and unfilled.
Very good point - I am just about to show this spreadsheet to my "wizz" brother-in-law to pick up any flaws of that nature... So taking your advice I have added parameters in my "setup" tab and using the suggested formula - we have lift off!
I have added a conditional format so when "Notify Manager" action has been completed the next cell =yes it will strike through "Notify Manager"..
Thanks so much for your help!
That all sounds good! Thanks for the feedback.
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Go back | 1,218 | 4,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.85018 |
https://convertoctopus.com/34-2-feet-per-second-to-miles-per-hour | 1,624,107,233,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00132.warc.gz | 185,726,637 | 8,111 | ## Conversion formula
The conversion factor from feet per second to miles per hour is 0.68181818181818, which means that 1 foot per second is equal to 0.68181818181818 miles per hour:
1 ft/s = 0.68181818181818 mph
To convert 34.2 feet per second into miles per hour we have to multiply 34.2 by the conversion factor in order to get the velocity amount from feet per second to miles per hour. We can also form a simple proportion to calculate the result:
1 ft/s → 0.68181818181818 mph
34.2 ft/s → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 34.2 ft/s × 0.68181818181818 mph
V(mph) = 23.318181818182 mph
The final result is:
34.2 ft/s → 23.318181818182 mph
We conclude that 34.2 feet per second is equivalent to 23.318181818182 miles per hour:
34.2 feet per second = 23.318181818182 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.042884990253411 × 34.2 feet per second.
Another way is saying that 34.2 feet per second is equal to 1 ÷ 0.042884990253411 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-four point two feet per second is approximately twenty-three point three one eight miles per hour:
34.2 ft/s ≅ 23.318 mph
An alternative is also that one mile per hour is approximately zero point zero four three times thirty-four point two feet per second.
## Conversion table
### feet per second to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from feet per second to miles per hour
feet per second (ft/s) miles per hour (mph)
35.2 feet per second 24 miles per hour
36.2 feet per second 24.682 miles per hour
37.2 feet per second 25.364 miles per hour
38.2 feet per second 26.045 miles per hour
39.2 feet per second 26.727 miles per hour
40.2 feet per second 27.409 miles per hour
41.2 feet per second 28.091 miles per hour
42.2 feet per second 28.773 miles per hour
43.2 feet per second 29.455 miles per hour
44.2 feet per second 30.136 miles per hour | 600 | 2,191 | {"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-2021-25 | latest | en | 0.780683 |
https://ask.learncbse.in/t/suppose-you-start-an-antique-car-by-exerting-a-force-of-300-n/57399 | 1,708,674,739,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00124.warc.gz | 107,698,692 | 3,464 | # Suppose you start an antique car by exerting a force of 300 N
Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What is the angular momentum given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time? | 82 | 321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.925091 |
https://republicofsouthossetia.org/question/what-is-the-slope-of-the-line-represented-by-the-equation-y-1-2-1-4-1-2-1-4-1-4-1-2-16359209-59/ | 1,632,514,921,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00527.warc.gz | 523,103,339 | 13,181 | ## What is the slope of the line represented by the equation y=-1/2x+1/4? -1/2 -1/4 1/4 1/2
Question
What is the slope of the line represented by the equation y=-1/2x+1/4?
-1/2
-1/4
1/4
1/2
in progress 0
2 weeks 2021-09-10T20:20:59+00:00 1 Answer 0 | 112 | 251 | {"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-2021-39 | latest | en | 0.904432 |
https://id.scribd.com/document/367169458/Week-9-Ultimate-Shear-Capacity | 1,569,316,849,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00273.warc.gz | 510,691,810 | 72,178 | Anda di halaman 1dari 10
# Topic Outcome:
## At the end of this topic, students should be able to
determine the ultimate shear capacity of the
prestressed concrete beam
Shear in PSC is considered at ULS.
The behaviour of PSC beam at failure in
shear is distinctly different from their
behaviour in flexure.
The beams fail abruptly without
sufficient warning, and the diagonal
cracks that develop are considerably
wider than the flexural cracks.
Shear forces result in shear stress.
Such a stress can result in principal
tensile stresses at the critical section
which can exceed the tensile strength
of the concrete.
When the tensile strength of the
concrete is exceeded, cracks will form.
The response of a PSC member in
resisting shear is similar to RC
the compression due to prestressing
force need to take into account.
Based on Variable Strut Inclination
Method of shear design of RC design,
EC2 recommends few slight
modifications to equation to compute
shear capacity (VRd,c) for PSC design.
In calculating the design ultimate
shear force, VEd, it is permissible to
take into account the vertical
component of force in any inclined
tendons which will tend to act in a
direction that resists shear.
Prestressing force should be
multiplied by the partial factor of
safety, p= 0.9
In calculating the ultimate shear
capacity:
Sections do not require shear
reinforcement
Sections that require shear reinforcement
Shear strength without shear
reinforcement for uncracked regions in
bending (special case)
Cracking in prestressed concrete beams at ultimate
load depends on the local magnitudes of moment
and shear.
In regions where the moment is large and the
shear is small, vertical flexural cracks appear
after the normal tensile stress in the extreme
concrete fibres exceeds the tensile strength of
concrete. (These are the cracks shown as type
A in figure).
Where both the moment and shear force are
relatively large, flexural cracks which are
vertical at the extreme fibres become inclined
as they extend deeper into the beam owing to
the presence of shear stresses in the beam
web. These inclined cracks, which are often
quite flat in a prestressed beam, are called
flexure-shear cracks and are designated crack
(illustrated as type B in figure).
If adequate shear reinforcement is not
provided, a flexure-shear crack may lead to a
so-called shear-compression failure, in which
the area of concrete in compression above
the advancing inclined crack is so reduced as
to be no longer adequate to carry the
compression force resulting from flexure.
A second type of inclined crack sometimes
occurs in the web of a prestressed beam in
the regions where moment is small and shear
is large, such as the cracks designated type C
adjacent to the discontinuous support and
near the point of contraflexure in the Figure.
In such locations, high principal tensile stress
may cause inclined cracking in the mid-depth
region of the beam before flexural cracking
occurs in the extreme fibres. These cracks are
known as web-shear cracks and occur most
often in beams with relatively thin webs. | 721 | 3,119 | {"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} | 2.921875 | 3 | CC-MAIN-2019-39 | latest | en | 0.913948 |
https://oeis.org/A158449 | 1,653,575,946,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00573.warc.gz | 505,153,618 | 4,357 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A158449 The number of sigma-admissible subsets of {1,2,...,n} as defined by Marzuola-Miller. 1
1, 0, 1, 0, 2, 0, 3, 1, 7, 3, 17, 7, 43, 24, 118, 74, 330, 206, 888, 612, 2571, 1810, 7274, 5552, 21099, 16334, 61252, 49025, 179239, 146048, 523455, 440980 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS a(n), or Asigma(n), equals the number of sigma-admissible subsets of {1,2,...,n}. Alternate description: (1) Asigma(k) is the same as the number of additive 2-bases for k which are not additive 2-bases for k+1. (2) Asigma(n) is the number of vertices at height n in the rooted tree in figure 5 of [Marzuola-Miller] which spawn only one vertex at height n+1. [Jeremy L. Marzuola (marzuola(AT)math.uni-bonn.de), Aug 08 2009] The number of symmetric numerical sets S with atom monoid A(S) equal to {0,n+1,2n+2,2n+3,2n+4,2n+5,...} LINKS S. R. Finch, Monoids of natural numbers S. R. Finch, Monoids of natural numbers, March 17, 2009. [Cached copy, with permission of the author] J. Marzuola and A. Miller, Counting Numerical Sets with No Small Atoms, arXiv:0805.3493 [math.CO], 2008. J. Marzuola and A. Miller, Counting numerical sets with no small atoms, J. Combin. Theory A 117 (6) (2010) 650-667. FORMULA Recursively related to A164047 by the formula Asigma(2k+1)' = 2Asigma(2k)'-Asigma(k) EXAMPLE a(1)=a(3)=1 since {0,2,4,5,6,7,...} and {0,1,4,5,8,9,10,11,...} are the only sets satisfying the required conditions. CROSSREFS Cf. A066062, A164047. Sequence in context: A350962 A241644 A241640 * A106533 A192421 A035223 Adjacent sequences: A158446 A158447 A158448 * A158450 A158451 A158452 KEYWORD nonn,more AUTHOR Steven Finch, Mar 19 2009 EXTENSIONS Definition rephrased by Jeremy L. Marzuola (marzuola(AT)math.uni-bonn.de), Aug 08 2009 Edited by R. J. Mathar, Aug 31 2009 STATUS approved
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Last modified May 26 10:17 EDT 2022. Contains 354086 sequences. (Running on oeis4.) | 788 | 2,300 | {"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-2022-21 | latest | en | 0.741964 |
http://www.slidesearchengine.com/slide/representative-sets | 1,508,332,517,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00806.warc.gz | 554,449,695 | 11,638 | Representative Sets
50 %
50 %
Education
Published on March 7, 2014
Author: ASPAK2014
Source: slideshare.net
Parameterized Algorithms using Matroids Lecture II: Representative Sets Saket Saurabh The Institute of Mathematical Sciences, India ASPAK, IMSc, March 3–8, 2014
Problems we would be interested in... Vertex Cover Input: A graph G = (V, E) and a positive integer k. Parameter: k Question: Does there exist a subset V 1 Ď V of size at most k such that for every edge (u, v) P E either u P V 1 or v P V 1 ? Hamiltonian Path Input: A graph G = (V, E) Question: Does there exist a path P in G that spans all the vertices? Path Input: A graph G = (V, E) and a positive integer k. Parameter: k Question: Does there exist a path P in G of length at least k?
Representative Sets Why, What and How.
Representative Sets Why, What and How.
Ham-Path Dynamic Programming for Hamiltonian Path
Ham-Path . 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n
Ham-Path . 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ n − 1 n
Ham-Path . vj 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ n − 1 n
Ham-Path . vj 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ n − 1 n
Ham-Path . vj 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ n − 1 n
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn V[Paths of length i ending at vj ] n
Ham-Path . Example: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn V[Paths of length i ending at vj ] n
Ham-Path . Example: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn V[Paths of length i ending at vj ] n
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn SETS, NOT SEQUENCES. V[Paths of length i ending at vj ] n
Ham-Path . Example: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn SETS, NOT SEQUENCES. V[Paths of length i ending at vj ] n
Ham-Path . Example: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn SETS, NOT SEQUENCES. V[Paths of length i ending at vj ] n
Ham-Path . Example: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 v1 .. . vj .. . vn SETS, NOT SEQUENCES. V[Paths of length i ending at vj ] n
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . vj .. . vn SETS, NOT SEQUENCES. V[Paths of length i ending at vj ] Two paths that use the same set of vertices but visit them in different orders are equivalent.
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . vj V[Paths of length i ending at vj ] .. . = V[Paths of length (i − 1) ending at u, avoiding vj .] vn
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . vj V[Paths of length i ending at vj ] .. . = V[Paths of length (i − 1) ending at u, avoiding vj .] vn u P N(vj )
Ham-Path . Valid: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . vj V[Paths of length i ending at vj ] .. . = V[Paths of length (i − 1) ending at u, avoiding vj .] vn u P N(vj )
Ham-Path . Invalid: vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . vj V[Paths of length i ending at vj ] .. . = V[Paths of length (i − 1) ending at u, avoiding vj .] vn u P N(vj )
Ham-Path . vj 1 2 3 ¨¨¨ i ¨¨¨ n − 1 n v1 .. . Potentially storing (n) i sets. vj V[Paths of length i ending at vj ] .. . = V[Paths of length (i − 1) ending at u, avoiding vj .] vn u P N(vj )
K-Path Let us now turn to k-Path. To find paths of length at least k, we may simply use the DP table for Hamiltonian Path restricted to the first k columns.
K-Path . 1 2 3 ¨¨¨ i ¨¨¨ k−1 v1 .. . vj .. . vn Worst case running time: O⋆ ((n)) k k
K-Path . 1 2 3 ¨¨¨ i ¨¨¨ k−1 v1 .. . vj .. . vn Worst case running time: O⋆ (nk ) k
Do we really need to store all these sets?
Do we really need to store all these sets? In the ith column, we are storing paths of length i.
Do we really need to store all these sets? In the ith column, we are storing paths of length i. Let P be a path of length k.
Do we really need to store all these sets? In the ith column, we are storing paths of length i. Let P be a path of length k. There may be several paths of length i that “latch on” to the last (k − i) vertices of P.
Do we really need to store all these sets? In the ith column, we are storing paths of length i. Let P be a path of length k. There may be several paths of length i that “latch on” to the last (k − i) vertices of P. We need to store just one of them.
Example. .
Example. Suppose we have a path P on seven edges. .
Example. Suppose we have a path P on seven edges. Consider it broken up into the first four and the last three edges. .
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. A Fixed Future (vi+1 − ¨ ¨ ¨ − vk ).
The Possibilities for Partial Solutions Compatible with vi+1 − ¨ ¨ ¨ − vk . . A Fixed Future (vi+1 − ¨ ¨ ¨ − vk ).
Let’s try a different example. .
The Possibilities for Partial Solutions Compatible with vi+1 − ¨ ¨ ¨ − vk . . A Fixed Future (vi+1 − ¨ ¨ ¨ − vk ).
Here’s one more example: .
The Possibilities for Partial Solutions Compatible with vi+1 − ¨ ¨ ¨ − vk . . A Fixed Future (vi+1 − ¨ ¨ ¨ − vk ).
For any possible ending of length (k − i), we want to be sure that we store at least one among the possibly many “prefixes”.
For any possible ending of length (k − i), we want to be sure that we store at least one among the possibly many “prefixes”. This could also be ( ) n k−i .
For any possible ending of length (k − i), we want to be sure that we store at least one among the possibly many “prefixes”. This could also be ( ) n k−i . The hope for “saving” comes from the fact that a single path of length i is potentially capable of being a prefix to several distinct endings.
For example...
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Representative Sets Why, What and How.
Partial solutions: paths of length j ending at vi .
Partial solutions: paths of length j ending at vi . A “small” representative family.
If: Partial solutions: paths of length j ending at vi . A “small” representative family.
If: vi Partial solutions: paths of length j ending at vi . A “small” representative family.
j vertices If: vi (k − j) vertices Partial solutions: paths of length j ending at vi . A “small” representative family.
j vertices If: vi (k − j) vertices Partial solutions: paths of length j ending at vi . A “small” representative family.
j vertices If: vi (k − j) vertices Partial solutions: paths of length j ending at vi . A “small” representative family. Then:
j vertices If: vi (k − j) vertices Partial solutions: paths of length j ending at vi . A “small” representative family. Then:
j vertices If: vi (k − j) vertices Partial solutions: paths of length j ending at vi . A “small” representative family. We would like to store at least one path of length j that serves the same purpose. Then:
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that:
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. The “second half” of a solution — can be any subset.
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. The “second half” of a solution — can be any subset.
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. This is a valid patch into X.
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. This is a guaranteed replacement for S.
Given: A (BIG) family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. The “second half” of a solution — can be any subset.
Given: A ď (n) p family F of p-sized subsets of [n]. S1 , S2 , . . . , St p Want: A (small) subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. The “second half” of a solution — can be any subset.
Given: A ď (n) p family F of p-sized subsets of [n]. S1 , S2 , . . . , St (k) p Known: D p subfamily F of F such that: For any X Ď [n] of size (k − p), if there is a set S in F such that X X S = H, p p p then there is a set S in F such that X X S = H. Bolobás, 1965.
Given: A a matroid (M, I), and a family of p-sized subsets from I: S1 , S2 , . . . , St
Given: A a matroid (M, I), and a family of p-sized subsets from I: S1 , S2 , . . . , St p Want: A subfamily F of F such that:
Given: A a matroid (M, I), and a family of p-sized subsets from I: S1 , S2 , . . . , St p Want: A subfamily F of F such that: For any X Ď [n] of size at most q, if there is a set S in F such that X X S = H and X Y S P I, p p p p then there is a set S in F such that X X S = H and X Y S P I. Lovász, 1977
Given: A a matroid (M, I), and a family of p-sized subsets from I: S1 , S2 , . . . , St p There is a subfamily F of F of size at most (p+q) p such that: For any X Ď [n] of size at most q, if there is a set S in F such that X X S = H and X Y S P I, p p p p then there is a set S in F such that X X S = H and X Y S P I. Lovász, 1977
Given: A a matroid (M, I), and a family of p-sized subsets from I: S1 , S2 , . . . , St p There is an efficiently computable subfamily F of F of size at most (p+q) p such that: For any X Ď [n] of size at most q, if there is a set S in F such that X X S = H and X Y S P I, p p p p then there is a set S in F such that X X S = H and X Y S P I. Márx (2009) and Fomin, Lokshtanov, Saurabh (2013)
Summary. We have at hand a p-uniform collection of independent sets, F and a number q. Let X be any set of size at most q. For any set S P F, if: a X is disjoint from S, and b X and S together form an independent set, p p then a q-representative family F contains a set S that is: a disjoint from X, and b forms an independent set together with X. Such a subfamily is called a q-representative family for the given family.
Representative Sets Back to Why.
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 [RECALL] Worst case running time: O⋆ ((n)) k k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 [RECALL] (n) (( )) Worst case running time: O⋆ n k k k
. 1 v1 .. . 2 3 ¨¨¨ i ¨¨¨ k−1 [RECALL] (n) (( )) Worst case running time: O⋆ n k k vj .. . vn Representative Set Computation k
. 1 v1 .. . 2 3 ¨¨¨ i ¨¨¨ k−1 [RECALL] (n) (( )) Worst case running time: O⋆ n k k vj .. . vn Representative Set Computation (k ) i k
. 1 v1 .. . 2 3 ¨¨¨ i ¨¨¨ k−1 Not so fast! (n) (( )) Worst case running time: O⋆ n k k vj .. . vn Representative Set Computation (k ) i k
. 1 v1 .. . 2 3 ¨¨¨ i ¨¨¨ k−1 Not so fast! (n) (( )) Worst case running time: O⋆ n k is too big! k vj .. . vn Representative Set Computation (k ) i k
We are going to compute representative families at every intermediate stage of the computation.
We are going to compute representative families at every intermediate stage of the computation. For instance, in the ith column, we are storing i-uniform families. Before moving on to column (i + 1), we compute (k − i)-representative families. This keeps the sizes small as we go along.
. 1 v1 .. . vn 3 ¨¨¨ i ¨¨¨ k−1 RECALL .. . vj 2 Worst case running time: O⋆ Blah blah. ((n)) k n Representative Set Computation (k) i k
. 1 v1 .. . vn 3 ¨¨¨ i ¨¨¨ k−1 RECALL .. . vj 2 (k ) n 1 Worst case running time: O⋆ Blah blah. ((n)) k Representative Set Computation (k) i k
. 1 2 v1 .. . vn ¨¨¨ i ¨¨¨ k−1 RECALL .. . vj 3 (k) (k) n 1 1 n Worst case running time: O⋆ Blah blah. ((n)) k Representative Set Computation (k) i k
. 1 2 v1 .. . vn ¨¨¨ i ¨¨¨ k−1 RECALL .. . vj 3 (k) ((k) ) k n 1 12 n Worst case running time: O⋆ Blah blah. ((n)) k Representative Set Computation (k) i k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 RECALL Worst case running time: O⋆ Blah blah. (k) ((k) (k) ) k n 1 12 n 2 n ((n)) k Representative Set Computation (k) i k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 RECALL Worst case running time: O⋆ Blah blah. (k) ((k) ((k) ) ) k k n 1 12 n 23 n ((n)) k Representative Set Computation (k) i k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 RECALL Worst case running time: O⋆ Blah blah. ( k ) (k) ((k) ((k) ) ) k k n ¨ ¨ ¨ i−1 n 1 12 n 23 n ((n)) k Representative Set Computation (k) i k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 RECALL Worst case running time: O⋆ Blah blah. ( (k) ) (k) ((k) ((k) ) ) k k k n ¨ ¨ ¨ i−1 n 1 12 n 23 n i ((n)) k Representative Set Computation (k) i k
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 k RECALL (( )) Worst case running time: O⋆ n Blah blah. k ( (k) ) (k) ((k) ((k) ) ) k k k n ¨ ¨ ¨ i−1 n ¨¨¨ 1 12 n 23 n i Representative Set Computation (k) i 2k n
. 1 v1 .. . vj .. . vn 2 3 ¨¨¨ i ¨¨¨ k−1 k RECALL (( )) Worst case running time: O⋆ n Blah blah. k ( (k) ) (k) ((k) ((k) ) ) k k k n ¨ ¨ ¨ i−1 n ¨¨¨ 1 12 n 23 n i Representative Set Computation (k) i kn 22k
Let Pj be the set of all paths of length i ending at vj . i It can be shown that the families thus computed at the ith column, jth row are j indeed (k − i)-representative families for Pi . The correctness is implicit in the notion of a representative family.
Representative Sets A Different Why.
Vertex Cover Can you delete k vertices to kill all edges? .
Vertex Cover Can you delete k vertices to kill all edges? .
Let (G = (V, E), k) be an instance of Vertex Cover. Note that E can be thought of as a 2-uniform family over the ground set V .
Let (G = (V, E), k) be an instance of Vertex Cover. Note that E can be thought of as a 2-uniform family over the ground set V . Goal: Kernelization. In this context, we are asking if there is a small subset X of the edges such that G[X] is a YES-instance ↔ G is a YES-instance.
Note: If G is a YES-instance, then G[X] is a YES-instance for any subset X Ď E.
Note: If G is a YES-instance, then G[X] is a YES-instance for any subset X Ď E. We get one direction for free!
Note: If G is a YES-instance, then G[X] is a YES-instance for any subset X Ď E. We get one direction for free! It is the NO-instances that we have to worry about preserving.
Note: If G is a YES-instance, then G[X] is a YES-instance for any subset X Ď E. We get one direction for free! It is the NO-instances that we have to worry about preserving. What is a NO-instance?
. If G is a NO-instance: For any subset S of size at most k, there is an edge that is disjoint from S.
. If G is a NO-instance: For any subset S of size at most k, there is an edge that is disjoint from S. Ring a bell?
Recall. We have at hand a p-uniform collection of independent sets, F and a number q. Let X be any set of size at most q. For any set S P F, if: a X is disjoint from S, and b X and S together form an independent set, p then a q-representative family contains a set S that is: a disjoint from X, and b forms an independent set together with X. Such a subfamily is called a q-representative family for the given family.
Claim: A k-representative family for E is in fact an O(k2 ) kernel for vertex cover.
E(G) = {e1 ,.e2 , . . . , em } Is there a Vertex Cover of size at most k?
E(G) = {e1 ,.e2 , . . . , em } k-Representative Family Is there a Vertex Cover of size at most k?
E(G) = {e1 ,.e2 , . . . , em } k-Representative Family {f1 , f2 , . . . , fr } Is there a Vertex Cover of size at most k?
E(G) = {e1 ,.e2 , . . . , em } k-Representative Family {f1 , f2 , . . . , fr } O(k2 ) Is there a Vertex Cover of size at most k?
E(G) = {e1 ,.e2 , . . . , em } ” {f1 , f2 , . . . , fr } O(k2 ) Is there a Vertex Cover of size at most k?
Let us show that if G[X] is a YES-instance, then so is G.
Let us show that if G[X] is a YES-instance, then so is G. This time, by contradiction.
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. Try the solution for G[X] on G.
. Suppose there is an uncovered edge.
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k:
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H,
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, e then there is a set p in X such that p X S = H. e Note that the green edges denote G[X].
. Since X is a k-representative family, for ANY S Ď V , where |S| ď k: if there is a set e in E such that e X S = H, then there is a set p in X such that p X S = H. e e Note that the green edges denote G[X]. Contradiction!
A k-representative family for E(G) is in fact an O(k2 ) instance kernel for Vertex Cover!
Representative Sets Why, What and How.
Notation Det(M) : M Let M be a m ˆ n matrix, and let I Ď [m], J Ď [n]. M[I, J] : M restricted to rows indexed by I and columns indexed by J M[⋆, J] : M restricted to all rows and columns indexed by J M[I, ⋆] : M restricted to rows indexed by I and all columns
Standard Laplace Expansion
a11 . a21 . a31 . a41 . a51 . a61 . . . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns.
a11 . a21 . a31 . a41 . a51 . a61 . . . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns.
a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns.
a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 .
a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 .
a11 . a21 . a31 . a41 . a51 . a61 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 .
a11 . a21 . a31 . a41 . a51 . a61 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 .
a11 . a21 . a31 . a41 . a51 . a61 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . . a13 . a23 . a33 . a43 . a53 . a63 . a11 . a21 . a31 . a41 . a51 . a61 . . a12 . a22 . a32 . a42 . a52 . a62 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a row and expand along the columns.
Generalized Laplace Expansion
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+3) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+4) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+2+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+3+4) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+3+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+3+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+4+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+4+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(1+5+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+3+4) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+3+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+3+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+4+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+4+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(2+5+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(3+4+5) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(3+4+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(3+5+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . Fix a set of columns, J Ď [6]. Iterate over all I Ď [6] such that |I| = |J|. a11 . a21 . a31 . a41 . a51 . a61 . a12 . a22 . a32 . a42 . a52 . a62 . a13 . a23 . a33 . a43 . a53 . a63 . a14 . a24 . a34 . a44 . a54 . a64 . a15 . a25 . a35 . a45 . a55 . a65 . a16 . a26 . a36 . a46 . a56 . a66 . (−1)(1+3+6)+(4+5+6) Det(A[I, J]). Det(A[I, J]). . . Det(A) = ÿ IĎ[n],|I|=|J| Det(A[I, J]) ¨ Det(A[I, J]) ¨ (−1) ř ř I+ J
Recall: A Linear (or Representable) Matroid
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E ...are linearly independent.
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . a41 . a51 . a61 . a71 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . a47 . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns corresponding to S P I AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns corresponding to S P I AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns that are linearly independent... AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns that are linearly independent... AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...correspond to sets in I. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 .
M = (E, I), where E = {e1 , . . . , en } and I Ď 2E Columns indexed by elements of E AM = a11 . a21 . a31 . x41 ae1 . a51 . a61 . a71 . a12 . a22 . a32 . x42 ae2 . a52 . a62 . a72 . a13 . a23 . a33 . a¨. ¨ ¨ 43 a53 . a63 . a73 . a14 . a24 . a34 . xei a44 . a54 . a64 . a74 . a15 . a25 . a35 . a¨. ¨ ¨ 45 a55 . a65 . a75 . a16 . a26 . a36 . xan−1 . e46 a56 . a66 . a76 . ...are linearly independent. a17 . a27 . a37 . x 47 aen . a57 . a67 . a77 . rk(M)
Given: A collection of p-sized independent sets1 : S = {S1 , . . . , St }. 1 The rank of the underlying matroid is (p + q).
Given: A collection of p-sized independent sets1 : S = {S1 , . . . , St }. Want: A q-representative subfamily p of size ď S 1 The rank of the underlying matroid is (p + q). (p+q) p .
. ZPS
. ZPS YĎE
. ZPS PI YĎE
. ZPS PI YĎE p S ZPp
. ZPS PI YĎE PI p S ZPp
. ZPS Given PI YĎE PI p S ZPp
. ZPS Given PI YĎE PI p S ZPp
. | Z| = p Given PI YĎE PI p S ZPp
Given . | Z| = p PI |Y | = q PI p S ZPp
Given . | Z| = p PI |Y | = q PI p | Z| = p
AM = a11 . a21 . a31 . a41 . a51 . a61 . a71 . a81 . a91 . a101 . a111 . a121 . a131 . a141 . a151 . a161 . a171 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a82 . a92 . a102 . a112 . a122 . a132 . a142 . a152 . a162 . a172 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a83 . a93 . a103 . a113 . a123 . a133 . a143 . a153 . a163 . a173 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a84 . a94 . a104 . a114 . a124 . a134 . a144 . a154 . a164 . a174 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a85 . a95 . a105 . a115 . a125 . a135 . a145 . a155 . a165 . a175 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . a86 . a96 . a106 . a116 . a126 . a136 . a146 . a156 . a166 . a176 . a17 . a27 . a37 . a47 . a57 . a67 . a77 . a87 . a97 . a107 . a117 . a127 . a137 . a147 . a157 . a167 . a177 . a18 . a28 . a38 . a48 . a58 . a68 . a78 . a88 . a98 . a108 . a118 . a128 . a138 . a148 . a158 . a168 . a178 . a19 . a29 . a39 . a49 . a59 . a69 . a79 . a89 . a99 . a109 . a119 . a129 . a139 . a149 . a159 . a169 . a179 . a110 . a210 . a310 . a410 . a510 . a610 . a710 . a810 . .a910 . a1010 . a1110 . a1210 . a1310 . a1410 . a1510 . a1610 . a1710 . . a111 . a211 . a311 . a411 . a511 . a611 . a711 . a811 . a911 . a1011 . a1111 . a1211 . a1311 . a1411 . a1511 . a1611 . a1711 . a112 . a212 . a312 . a412 . a512 . a612 . a712 . a812 . a912 . a1012 . a1112 . a1212 . a1312 . a1412 . a1512 . a1612 . a1712 . a113 . a213 . a313 . a413 . a513 . a613 . a713 . a813 . a913 . a1013 . a1113 . a1213 . a1313 . a1413 . a1513 . a1613 . a1713 . a114 . a214 . a314 . a414 . a514 . a614 . a714 . a814 . a914 . a1014 . a1114 . a1214 . a1314 . a1414 . a1514 . a1614 . a1714 . a115 . a215 . a315 . a415 . a515 . a615 . a715 . a815 . a915 . a1015 . a1115 . a1215 . a1315 . a1415 . a1515 . a1615 . a1715 . a116 . a216 . a316 . a416 . a516 . a616 . a716 . a816 . a916 . a1016 . a1116 . a1216 . a1316 . a1416 . a1516 . a1616 . a1716 . a117 . a217 . a317 . a417 . a517 . a617 . a717 . a817 . a917 . a1017 . a1117 . a1217 . a1317 . a1417 . a1517 . a1617 . a1717 . (p + q)
p Columns corresponding to Z AM = a11 . a21 . a31 . a41 . a51 . a61 . a71 . a81 . a91 . a101 . a111 . a121 . a131 . a141 . a151 . a161 . a171 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a82 . a92 . a102 . a112 . a122 . a132 . a142 . a152 . a162 . a172 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a83 . a93 . a103 . a113 . a123 . a133 . a143 . a153 . a163 . a173 . a14 . a24 . a34 . a44 . a54 . a64 . a74 . a84 . a94 . a104 . a114 . a124 . a134 . a144 . a154 . a164 . a174 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a85 . a95 . a105 . a115 . a125 . a135 . a145 . a155 . a165 . a175 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . a86 . a96 . a106 . a116 . a126 . a136 . a146 . a156 . a166 . a176 . a17 . a27 . a37 . a47 . a57 . a67 . a77 . a87 . a97 . a107 . a117 . a127 . a137 . a147 . a157 . a167 . a177 . a18 . a28 . a38 . a48 . a58 . a68 . a78 . a88 . a98 . a108 . a118 . a128 . a138 . a148 . a158 . a168 . a178 . a19 . a29 . a39 . a49 . a59 . a69 . a79 . a89 . a99 . a109 . a119 . a129 . a139 . a149 . a159 . a169 . a179 . a110 . a210 . a310 . a410 . a510 . a610 . a710 . a810 . .a910 . a1010 . a1110 . a1210 . a1310 . a1410 . a1510 . a1610 . a1710 . . a111 . a211 . a311 . a411 . a511 . a611 . a711 . a811 . a911 . a1011 . a1111 . a1211 . a1311 . a1411 . a1511 . a1611 . a1711 . a112 . a212 . a312 . a412 . a512 . a612 . a712 . a812 . a912 . a1012 . a1112 . a1212 . a1312 . a1412 . a1512 . a1612 . a1712 . a113 . a213 . a313 . a413 . a513 . a613 . a713 . a813 . a913 . a1013 . a1113 . a1213 . a1313 . a1413 . a1513 . a1613 . a1713 . a114 . a214 . a314 . a414 . a514 . a614 . a714 . a814 . a914 . a1014 . a1114 . a1214 . a1314 . a1414 . a1514 . a1614 . a1714 . a115 . a215 . a315 . a415 . a515 . a615 . a715 . a815 . a915 . a1015 . a1115 . a1215 . a1315 . a1415 . a1515 . a1615 . a1715 . a116 . a216 . a316 . a416 . a516 . a616 . a716 . a816 . a916 . a1016 . a1116 . a1216 . a1316 . a1416 . a1516 . a1616 . a1716 . a117 . a217 . a317 . a417 . a517 . a617 . a717 . a817 . a917 . a1017 . a1117 . a1217 . a1317 . a1417 . a1517 . a1617 . a1717 . (p + q)
p AM = a11 . a21 . a31 . a41 . a51 . a61 . a71 . a81 . a91 . a101 . a111 . a121 . a131 . a141 . a151 . a161 . a171 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a82 . a92 . a102 . a112 . a122 . a132 . a142 . a152 . a162 . a172 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a83 . a93 . a103 . a113 . a123 . a133 . a143 . a153 . a163 . a173 . q Columns corresponding to Z Columns corresponding to Y a14 . a24 . a34 . a44 . a54 . a64 . a74 . a84 . a94 . a104 . a114 . a124 . a134 . a144 . a154 . a164 . a174 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a85 . a95 . a105 . a115 . a125 . a135 . a145 . a155 . a165 . a175 . a16 . a26 . a36 . a46 . a56 . a66 . a76 . a86 . a96 . a106 . a116 . a126 . a136 . a146 . a156 . a166 . a176 . a17 . a27 . a37 . a47 . a57 . a67 . a77 . a87 . a97 . a107 . a117 . a127 . a137 . a147 . a157 . a167 . a177 . a18 . a28 . a38 . a48 . a58 . a68 . a78 . a88 . a98 . a108 . a118 . a128 . a138 . a148 . a158 . a168 . a178 . a19 . a29 . a39 . a49 . a59 . a69 . a79 . a89 . a99 . a109 . a119 . a129 . a139 . a149 . a159 . a169 . a179 . a110 . a210 . a310 . a410 . a510 . a610 . a710 . a810 . .a910 . a1010 . a1110 . a1210 . a1310 . a1410 . a1510 . a1610 . a1710 . . a111 . a211 . a311 . a411 . a511 . a611 . a711 . a811 . a911 . a1011 . a1111 . a1211 . a1311 . a1411 . a1511 . a1611 . a1711 . a112 . a212 . a312 . a412 . a512 . a612 . a712 . a812 . a912 . a1012 . a1112 . a1212 . a1312 . a1412 . a1512 . a1612 . a1712 . a113 . a213 . a313 . a413 . a513 . a613 . a713 . a813 . a913 . a1013 . a1113 . a1213 . a1313 . a1413 . a1513 . a1613 . a1713 . a114 . a214 . a314 . a414 . a514 . a614 . a714 . a814 . a914 . a1014 . a1114 . a1214 . a1314 . a1414 . a1514 . a1614 . a1714 . a115 . a215 . a315 . a415 . a515 . a615 . a715 . a815 . a915 . a1015 . a1115 . a1215 . a1315 . a1415 . a1515 . a1615 . a1715 . a116 . a216 . a316 . a416 . a516 . a616 . a716 . a816 . a916 . a1016 . a1116 . a1216 . a1316 . a1416 . a1516 . a1616 . a1716 . a117 . a217 . a317 . a417 . a517 . a617 . a717 . a817 . a917 . a1017 . a1117 . a1217 . a1317 . a1417 . a1517 . a1617 . a1717 . (p + q)
p AM = a11 . a21 . a31 . a41 . a51 . a61 . a71 . a81 . a91 . a101 . a111 . a121 . a131 . a141 . a151 . a161 . a171 . a12 . a22 . a32 . a42 . a52 . a62 . a72 . a82 . a92 . a102 . a112 . a122 . a132 . a142 . a152 . a162 . a172 . a13 . a23 . a33 . a43 . a53 . a63 . a73 . a83 . a93 . a103 . a113 . a123 . a133 . a143 . a153 . a163 . a173 . q Columns corresponding to Z Columns corresponding to Y a14 . a24 . a34 . a44 . a54 . a64 . a74 . a84 . a94 . a104 . a114 . a124 . a134 . a144 . a154 . a164 . a174 . a15 . a25 . a35 . a45 . a55 . a65 . a75 . a85 . a95 . a105 . a115 . a125 . a135 . a145 . a155 . a165 . a175 . a16 . a2
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# There are 32 students in Jamie’s eighth-grade class. Each st
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There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 08 Dec 2017, 20:41
Expert's post
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Question Stats:
75% (01:32) correct 25% (02:12) wrong based on 4 sessions
There are 32 students in Jamie’s eighth-grade class. Each student took a 50-point test; the class average (arithmetic mean) was 82% correct. The teacher has assigned one 4-point, extra-credit question. How many students will need to answer the extra-credit question correctly in order to bring the class average to 86% correct?
A. 15
B. 16
C. 17
D. All of the students
E. It will not be possible for the class to reach an average of 86% correct.
Drill 3
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Re: There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 08 Dec 2017, 22:26
We know that 82 % of 50 is 41. Multiplied by the number of students 1312.
The new desired average would be 86% of 50 which is 43. Multiplied by the number of students 1376.
Difference between the two: 1376 - 1312 = 64/4 = 16
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Re: There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 29 Dec 2017, 14:14
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Expert's post
Explanation
If the class average is 82% on a 50-point test, the average score was 41 points out of 50.
Use the Average Pie to find the sum of the class’s scores: (41)(32) = 1,312. To reach a class average of 86%, each student will need to average 43 points out of 50 points. Use the Average Pie to find the desired sum of the class’s scores: (43)(32) = 1,376. The difference is 1,376 − 1,312 = 64, so the class needs to make up 64 points;$$\frac{64}{4} =16$$, so 16 students need to answer the extra credit question correctly. The answer is B.
Alternatively, notice that the class’s average needs to increase by 4%, or 2 points on average for a 50-question test. But the extra credit is
worth 4 points, so to average half of a 4-point increase, only half the students (16) need to get the extra credit correct.
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Re: There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 28 May 2018, 18:30
Expert's post
sandy wrote:
There are 32 students in Jamie’s eighth-grade class. Each student took a 50-point test; the class average (arithmetic mean) was 82% correct. The teacher has assigned one 4-point, extra-credit question. How many students will need to answer the extra-credit question correctly in order to bring the class average to 86% correct?
A. 15
B. 16
C. 17
D. All of the students
E. It will not be possible for the class to reach an average of 86% correct.
We can create the equation:
(0.82(50)(32) + 4x)/32 = 0.86(50)
0.82(50)(32)/32 + 4x/32 = 0.86(50)
41 + x/8 = 43
x/8 = 2
x = 16
Alternate Solution:
Each test point is worth 2%. Thus, the extra credit question is worth 4 points, or 8%. If all 32 students answered it correctly, then the class average would increase by 8%. Since we want to increase the class average by only 4%, then only half of the students (i.e., 16 students) need to answer it correctly.
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Re: There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 11 Jan 2019, 10:09
sandy wrote:
Explanation
If the class average is 82% on a 50-point test, the average score was 41 points out of 50.
Use the Average Pie to find the sum of the class’s scores: (41)(32) = 1,312. To reach a class average of 86%, each student will need to average 43 points out of 50 points. Use the Average Pie to find the desired sum of the class’s scores: (43)(32) = 1,376. The difference is 1,376 − 1,312 = 64, so the class needs to make up 64 points;$$\frac{64}{4} =16$$, so 16 students need to answer the extra credit question correctly. The answer is B.
Alternatively, notice that the class’s average needs to increase by 4%, or 2 points on average for a 50-question test. But the extra credit is
worth 4 points, so to average half of a 4-point increase, only half the students (16) need to get the extra credit correct.
I was confused because i thought some will get 1 out of 4, some 2 out of 4, some 3 and some 4 out of 4.
Re: There are 32 students in Jamie’s eighth-grade class. Each st [#permalink] 11 Jan 2019, 10:09
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https://openpress.usask.ca/physics155/chapter/12-1-ac-sources/ | 1,716,543,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00260.warc.gz | 384,860,628 | 25,775 | # 12.1 AC Sources
#### LEARNING OBJECTIVES
By the end of the section, you will be able to:
• Explain the differences between direct current (dc) and alternating current (ac)
• Define characteristic features of alternating current and voltage, such as the amplitude or peak and the frequency
Most examples dealt with so far in this book, particularly those using batteries, have constant-voltage sources. Thus, once the current is established, it is constant. Direct current (dc) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit.
Most well-known applications, however, use a time-varying voltage source. Alternating current (ac) is the flow of electric charge that periodically reverses direction. An ac is produced by an alternating emf, which is generated in a power plant, as described in Induced Electric Fields. If the ac source varies periodically, particularly sinusoidally, the circuit is known as an ac circuit. Examples include the commercial and residential power that serves so many of our needs.
The ac voltages and frequencies commonly used in businesses and homes vary around the world. In a typical house, the potential difference between the two sides of an electrical outlet alternates sinusoidally with a frequency of or and an amplitude of or depending on whether you live in North America or Europe, respectively. Most people know the potential difference for electrical outlets is or in North America or Europe, but as explained later in the chapter, these voltages are not the peak values given here but rather are related to the common voltages we see in our electrical outlets. shows graphs of voltage and current versus time for typical dc and ac power in North America.
(Figure 12.1.1)
Suppose we hook up a resistor to an ac voltage source and determine how the voltage and current vary in time across the resistor. shows a schematic of a simple circuit with an ac voltage source. The voltage fluctuates sinusoidally with time at a fixed frequency, as shown, on either the battery terminals or the resistor. Therefore, the ac voltage, or the “voltage at a plug,” can be given by
(12.1.1)
where is the voltage at time is the peak voltage, and is the angular frequency in radians per second. For a typical house in North America, and whereas in Europe, and
For this simple resistance circuit, so the ac current, meaning the current that fluctuates sinusoidally with time at a fixed frequency, is
(12.1.2)
where is the current at time and is the peak current and is equal to For this example, the voltage and current are said to be in phase, meaning that their sinusoidal functional forms have peaks, troughs, and nodes in the same place. They oscillate in sync with each other, as shown in(b). In these equations, and throughout this chapter, we use lowercase letters (such as ) to indicate instantaneous values and capital letters (such as ) to indicate maximum, or peak, values.
(Figure 12.1.2)
Current in the resistor alternates back and forth just like the driving voltage, since If the resistor is a fluorescent light bulb, for example, it brightens and dims times per second as the current repeatedly goes through zero. A flicker is too rapid for your eyes to detect, but if you wave your hand back and forth between your face and a fluorescent light, you will see the stroboscopic effect of ac. | 752 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.919051 |
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Evaluating Composite Functions
```
Date: 1/31/96 at 10:13:35
From: Anonymous
Subject: Calculus
What is the domain of f(x) = 3? Also, if f(x) = 3, and g(x) = x^2 + 1
divided by the square root of (x^2 -1), how do you evaluate the
composite function of f composed with g (f circle g)?
The constant really threw me off in these problems (am I making a simple
thing hard?).
```
```
Date: 6/18/96 at 0:48:20
From: Doctor Paul
Subject: Re: Calculus
The domain of a function is the set of points on the x-axis that can be
plugged into the equation with a finite result returned. In other
words, if a particular x leads to an undefined point, then that point is
not in the domain of f(x).
For example:
x
if f(x) = -----
(x-3)
Then the domain would be the set of all x's (from -infinity to positive
infinity) with the exception of x=3 (b/c x=3 leads to dividing by zero).
Since your function, f(x) = 3 doesn't have any x's in it (and is hence
just a horizontal line) the domain is the set of all x's (from -infinity
to infinity)
The composite function, f(g(x)) [pronounced "f of g of x"], is very
simple in this case. Let me explain the concept first.
If f(x) = x^3+1 and g(x) = x^2-4
Then f(g(x)) is calculated by plugging g(x) directly into every occurring
x in f(x). In this case, f(g(x)) = (x^2-4)^3 +1.
In your case, since the function f(x) is just a constant, there is
nowhere to plug g(x) into f(x). Thus, f(g(x)) is just 3. In the case
where f(x) is a constant, f(g(x)) is always a constant regardless of how
complicated g(x) is.
-Doctor Paul, The Math Forum
```
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## Introduction and Background
Recently I decided to implement image based lighting in BGFX, since I had never done so before and it’s a great way to produce realistic renders with assets authored for PBR. As I started reading I realized that approaches had gotten a lot more involved than they had been even 10 years ago, and that it might be useful to others to have a reference for implementation from start to finish. I’ve documented the steps involved in implementation, starting with some background, then moving onto Karis’s 2014 paper explaining Unreal’s IBL implementation, and then finally new approaches from Fdez-Agüera and others to address the error present in existing models.
Let’s start by defining the equation which defines the outgoing radiance from a point $\mathbf{p}$ in the viewing direction $\mathbf{v}$:
$L_o(\mathbf{v}) = \int_{\Omega} f_{\text{brdf}}(\mathbf{v}, \mathbf{l}) L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l}$
Where $\mathbf{v}$ is our viewing angle, $f_{\text{brdf}}(\mathbf{v}, \mathbf{l})$ is our bidirectional reflectance distribution function (BRDF), $L_i(\mathbf{l})$ is the radiance incident on point $\mathbf{p}$ from direction $-\mathbf{l}$, and $\mathbf{n}$ is the normal vector at point $\mathbf{p}$. The hemisphere $\Omega$ is aligned with $\mathbf{n}$, and $\mathbf{h}$ is the is the halfway vector, halfway between $\mathbf{l}$ and $\mathbf{v}$.
I’ve omitted $\mathbf{n}$ and $\mathbf{h}$ as function arguments, but in reality they are also required to evaluate the BRDF. Here’s a diagram of our different vectors:
For our BRDF, we will take the popular approach of using a Lambertian diffuse lobe and Cook-Torrance microfacet model for our specular lobe:
\begin{aligned} f_{\text{brdf}}(\mathbf{v}, \mathbf{l}) &= f_{\text{specular}}(\mathbf{v}, \mathbf{l}) + f_{\text{diffuse}}, \\\\ f_{\text{specular}}(\mathbf{v}, \mathbf{l}) &= \frac{ D(\mathbf{h}) F(\mathbf{v}\cdot\mathbf{h}) G(\mathbf{v},\mathbf{l},\mathbf{h}) }{ 4 \langle\mathbf{n}\cdot\mathbf{l}\rangle \langle\mathbf{n}\cdot\mathbf{v}\rangle }, \\ f_{\text{diffuse}} &= \frac{c_{\text{diff}}}{\pi}, \end{aligned}
Where $D$ is our normal distribution function (NDF), which tells us what proportion of our perfectly reflecting microfacets will have normals aligned with $\mathbf{h}$, therefore reflecting the light from $-\mathbf{l}$ in the direction $\mathbf{v}$. $F$ is the Fresnel term which defines what proportion of light is reflected as opposed to refracted into the surface. $G$ is the “self shadowing term”, which defines what proportion of our microfacets will be occluded by the surrounding surface along the direction $\mathbf{l}$.
For more detail on this model, there are many resources but Naty Hoffman’s talk from this 2013 Siggraph workshop is very helpful in explaining this BRDF if you’ve never seen it before. I’ll also go ahead and define our functions explicitly, to dispell any ambiguity as to which variation we’re using here.
One extremely important detail is that the $f_\text{specular}$ lobe only accounts for a single scattering of energy. In reality, rougher surfaces will produce multiple scattering events. This is especially important for metals, where reflection dominates. We will see the consequences of only including a single scattering event in our BRDF model later.
For our NDF $D$, we’re using the GGX lobe:
$D(\mathbf{h}) = \frac{ \alpha^2 }{ \pi ((\mathbf{n}\cdot\mathbf{h})^2(\alpha^2 - 1) + 1)^2 }$
where $\alpha = \text{roughness}^2$. Here, $\text{roughness}$, AKA perceptually linear roughness, will be an input into our shading model, stored as a channel in our textures.
Meanwhile, for our geometric shadowing/attenuation term $G$ we’re using the height correlated Smith function for GGX:
\begin{aligned} V(\mathbf{v}, \mathbf{l}) &= \frac{ G(\mathbf{v},\mathbf{l},\mathbf{h}) }{ 4 \langle\mathbf{n}\cdot\mathbf{l}\rangle \langle\mathbf{n}\cdot\mathbf{v}\rangle }, \\ V(\mathbf{v}, \mathbf{l}) &= \frac{0.5}{ \langle \mathbf{n}\cdot\mathbf{l} \rangle \sqrt{(\mathbf{n}\cdot\mathbf{v})^2 (1 - \alpha^2) + \alpha^2} + \langle \mathbf{n}\cdot\mathbf{v} \rangle \sqrt{(\mathbf{n}\cdot\mathbf{l})^2 (1 - \alpha^2) + \alpha^2} } \end{aligned}
Note that we’ve expressed this in a way that incorporates the denominator of the BRDF into our $V$ term. For more detail on this term, I’ve found this section of the Filament documentation to be helpful.
Finally, our Fresnel term will use the Schlick approximation:
\begin{aligned} F(\mathbf{v}, \mathbf{l}) &= F_0 + (1 - F_0)(1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5 \\ &= F_0 (1 - (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5) + (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5 \end{aligned}
Where $F_0$ is a material property representing the specular reflectance at incident angles ($\mathbf{l} = \mathbf{n}$). The second form will come in handy later. For dielectrics, this is taken to be a uniform 4% as per the GLTF spec, while for metals we can use the albedo channel in the texture:
const float DIELECTRIC_SPECULAR = 0.04;
F_0 = lerp(vec3(DIELECTRIC_SPECULAR), albedo.rgb, metalness);
For the constant diffuse term $f_{\text{diffuse}}$, we’ll be using the GLTF spec for determining $c_{\text{diff}}$, which allows for a single albedo texture used by both metals and dielectrics:
diffuseColor = albedo.rgb * (1 - DIELECTRIC_SPECULAR) * (1.0 - metalness)
## Image Based Lighting Challenges
Okay, so that’s our BRDF and the reflectance equation we need to evaluate for every pixel covering our mesh. Let’s turn our attention now to the hemisphere $\Omega$ we need to integrate across. In order to do so, we need to be able to evaluate $L_i(\mathbf{l})$ for any given $\mathbf{l}$. In real time applications, we have two approaches, often used together:
1. Describe $L_i(\mathbf{l})$ using a set of $N$ analytical lights. Then, we only need to evaluate the integral $N$ times, for each corresponding light direction $\mathbf{l}_i$.
2. Describe $L_i(\mathbf{l})$ using an environment map, often using a cube map.
We’ll be focused on item 2 in this post. While an environment map can provide much more realistic lighting, the naive evaluation of our hemispherical integral is not going to cut it for interactive computer graphics. To gain a quick understanding of the challenges involved, consider the following:
• With a typical microfacet model we have both specular and diffuse lobes.
• The Lambertian lobe is dependent on light incoming from every part of our hemisphere. We must sample as much of the hemisphere as possible.
• Sampling the hemisphere naively, we’ll need potentially thousands of texture reads.
• For the specular lobe, our roughness will determine what parts of the hemisphere are most important. Other parts will not be as important and will not help us converge on a solution.
• Additionally, each sample will require evaluation of the BRDF, which for the specular component, is not “cheap”.
Performing all of this work with every mesh isn’t feasible for anything but the simplest of scenes. Additionally, sometimes we’ll be redoing work — for instance, calculating the diffuse irradiance for each direction really only needs to happen once for each environment map!
Instead, we’re going to try to pre-compute as many parts of the integral as possible. In order to do so, we’ll need to identify what we can precalculate and have to make some approximations.
## Lambertian Diffuse Component
First, since it’s easier, let’s look at the diffuse part of our reflectance integral:
\begin{aligned} L_{\text{diffuse}}(\mathbf{v}) &= \int_{\Omega} \frac{c_{\text{diff}}}{\pi} L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \\&= c_{\text{diff}} \int_{\Omega} \frac{1}{\pi} L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \end{aligned}
Since our integral is not dependent on any material properties or view direction, and instead dependent only on $\mathbf{n}$, we can solve this integral for every direction $\mathbf{n}$ and store the results in a cube map, allowing us to later look up the irradiance for any normal direction. Note that since we’re not using an NDF, we can get away with simple uniform sampling of our hemisphere.
In BGFX, I implemented this using compute shaders, using thread groups with a third dimension corresponding to the faces of the cube. I’m not certain this is actually a good idea as it may result in poor coherency, but I didn’t have time to benchmark against seperate dispatches for each face. This is left as an exercise for the reader.
#define TWO_PI 6.2831853071795864769252867665590
#define HALF_PI 1.5707963267948966192313216916398
SAMPLERCUBE(s_source, 0);
IMAGE2D_ARRAY_WR(s_target, rgba16f, 1);
void main()
{
const float imgSize = 64.0;
ivec3 globalId = gl_GlobalInvocationID.xyz;
vec3 N = normalize(toWorldCoords(globalId, imgSize));
vec3 up = abs(N.z) < 0.999 ? vec3(0.0, 0.0, 1.0) : vec3(1.0, 0.0, 0.0);
const vec3 right = normalize(cross(up, N));
up = cross(N, right);
vec3 color = vec3_splat(0.0);
uint sampleCount = 0u;
float deltaPhi = TWO_PI / 360.0;
float deltaTheta = HALF_PI / 90.0;
for (float phi = 0.0; phi < TWO_PI; phi += deltaPhi) {
for (float theta = 0.0; theta < HALF_PI; theta += deltaTheta) {
// Spherical to World Space in two steps...
vec3 tempVec = cos(phi) * right + sin(phi) * up;
vec3 sampleVector = cos(theta) * N + sin(theta) * tempVec;
color += textureCubeLod(s_source, sampleVector, 0).rgb * cos(theta) * sin(theta);
sampleCount++;
}
}
imageStore(s_target, globalId, vec4(PI * color / float(sampleCount), 1.0));
}
Note that you may not need a such a small step size. I was struggling with ringing artifacts with certain environment maps, so I opted to use a smaller delta which helped. Since the operation is done outside the render loop and only once per environment map, I didn’t want to spend too much time tweaking performance.
## Importance Sampling
In order to integrate the specular part of our radiance equation, we’ll need to use importance sampling, using the following equation:
$\int_{\Omega} f_{\text{brdf}}(\mathbf{v}, \mathbf{l}) L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \approx \frac{1}{N} \sum_{k=1}^{N} \frac{f_{\text{brdf}}(\mathbf{v}, \mathbf{l}_k) L_i(\mathbf{l}_k) \langle\mathbf{n} \cdot \mathbf{l}_k\rangle}{\text{pdf}(\mathbf{v}, \mathbf{l}_k)}$
Where $\text{pdf}(\mathbf{v}, \mathbf{l}_k)$ is the probability distribution function we use for sampling our hemisphere. Intuitively, the idea behind having the PDF in the denominator is that if a direction is more likely to be sampled than others, it should contribute less to the sum than a direction that is very unlikely to be sampled.
We could naively sample our hemisphere randomly, but at low roughness our specular component will converge to the actual solution very slowly. The same is true if we attempt to sample our hemisphere uniformly. To reason about this, consider a roughness close to zero, which corresponds to a perfect reflector. In this case, the only light direction that matters is the direction of perfect reflection $\mathbf{l}_{\text{reflection}}$ from $\mathbf{v}$ based on $\mathbf{n}$. Every other direction will contribute nothing to our sum. And if we do manage to sample $\mathbf{l}_{\text{reflection}}$, we still won’t converge due to the $\frac{1}{N}$ term. At hight roughness, this is less of a factor as the distribution of our microfacet normals will increase, and we’ll have to consider most parts of our hemisphere.
Instead, we’ll choose a PDF that resembles our BRDF in order to decrease the variance. We can use our Normal Distribution Function $D$ as our PDF, the reasoning being that the NDF determines which directions contribute the most to the light leaving our larger surface patch. More correctly, we’ll actually be sampling $D(\mathbf{h})\langle \mathbf{n} \cdot \mathbf{h} \rangle$, since this is how the normal distribution is actually defined:
$\int_\Omega D(\mathbf{h}) \langle \mathbf{n} \cdot \mathbf{h} \rangle d\mathbf{l}s = 1$
If we want our PDF to be used for $\mathbf{l}$ instead of the half vector $\mathbf{h}$ then we’ll need to include the Jacobian of the transformation from half vector to $\mathbf{l}$, $J(\mathbf{h})$:
$J(\mathbf{h}) = \frac{1}{4\langle \mathbf{v} \cdot \mathbf{h} \rangle}$ $\text{pdf}(\mathbf{v}, \mathbf{l}_k) = \frac{D(\mathbf{h}) \langle \mathbf{n} \cdot \mathbf{h} \rangle}{4\langle \mathbf{v} \cdot \mathbf{h} \rangle}$
You can also read section 5.3 of Walter et al. 2007, as well as Legarde et al’s course notes for more detail.
It’s also worth noting that we won’t be sampling $\mathbf{l}$ directly and instead we’ll sample our microfacet normal $\mathbf{h}$ and use it to find the relevant $\mathbf{l}$. To do so, we’ll need to map two uniformly random variables in the interval $[0, 1)$, let’s call them $\xi_1, \xi_2$, to $\phi$ and $\theta$ in spherical coordinates. Then, we can turn our $\phi, \theta$ into a cartesian direction $\mathbf{h}_i$ in world space. We can then use this to find $\mathbf{l}$ to sample our environment map to evaluate $L_i$.
The mapping from $\xi_1, \xi_2$ to $\phi$ and $\theta$ won’t be derived here, but Tobias Franke and A Graphics Guy have good blog posts that break this down step by step. Ultimately the mapping is as follows:
\begin{aligned} \phi &= 2 \pi \xi_1, \\ \theta &= \arccos \sqrt{\frac{1 - \xi_2}{\xi_2(\alpha^2 - 1) + 1}} \end{aligned}
Note that this mapping assumes we’re restricting ourselves to an isotropic version of the GGX, which is why the $\phi$ term is just randomly sampled. You may also notice the $\alpha$ term in our equation for $\theta$, but it may not be obvious how this dependency causes the equation to behave. I created a Desmos demo that you can play around with to get a sense of how this maps our $\xi_2$ to $\theta$. Observe that at low roughness, most of the $[0, 1)$ range will map to a small $\theta$, while for greater roughness we approach a standard cosine distribution, and we’re much more likely to get directions “further” from $\mathbf{n}$.
With this mapping we now have a method of importance sampling our hemisphere that will drastically improve convergence. Here’s some GLSL code that shows how we could sample a quasi-random direction $\mathbf{h}$:
// Taken from https://github.com/SaschaWillems/Vulkan-glTF-PBR/blob/master/data/shaders/genbrdflut.frag
// Based on http://holger.dammertz.org/stuff/notes_HammersleyOnHemisphere.html
vec2 hammersley(uint i, uint N)
{
uint bits = (i << 16u) | (i >> 16u);
bits = ((bits & 0x55555555u) << 1u) | ((bits & 0xAAAAAAAAu) >> 1u);
bits = ((bits & 0x33333333u) << 2u) | ((bits & 0xCCCCCCCCu) >> 2u);
bits = ((bits & 0x0F0F0F0Fu) << 4u) | ((bits & 0xF0F0F0F0u) >> 4u);
bits = ((bits & 0x00FF00FFu) << 8u) | ((bits & 0xFF00FF00u) >> 8u);
float rdi = float(bits) * 2.3283064365386963e-10;
return vec2(float(i) /float(N), rdi);
}
// Based on Karis 2014
vec3 importanceSampleGGX(vec2 Xi, float roughness, vec3 N)
{
float a = roughness * roughness;
// Sample in spherical coordinates
float phi = 2.0 * PI * Xi.x;
float cosTheta = sqrt((1.0 - Xi.y) / (1.0 + (a * a - 1.0) * Xi.y));
float sinTheta = sqrt(1.0 - cosTheta * cosTheta);
// Construct tangent space vector
vec3 H;
H.x = sinTheta * cos(phi);
H.y = sinTheta * sin(phi);
H.z = cosTheta;
// Tangent to world space
vec3 upVector = abs(N.z) < 0.999 ? vec3(0.0, 0.0, 1.0) : vec3(1.0, 0.0, 0.0);
vec3 tangentX = normalize(cross(upVector, N));
vec3 tangentY = cross(N, tangentX);
return tangentX * H.x + tangentY * H.y + N * H.z;
}
One thing in the above code I did not touch on at all is the Hammersley sequence, which is the method by which we create our random numbers $\xi_1, \xi_2$. It’s a low-discrepancy sequence that I won’t do any justice describing, so here’s a wikipedia link.
We can further improve the rate of convergence by filtering our samples as well. The idea is described in detail in Křivánek and Colbert’s GPU Gems article which I will briefly summarize. When using our importance sampling routine, we evaluate the PDF value for the sample and if it is “small” then this corresponds to a direction we sample infrequently. To improve convergence, we can consider this equivalent to sampling a larger solid angle in our hemisphere. The inverse is also true: samples with higher PDF values will correspond to smaller solid angles. But how do we sample a larger or smaller solid angle in our environment without having to integrate? The solution is to approximate this by sampling the mip chain of our environment map! It’s a very clever trick, and drastically improves convergence. You’ll see it demonstrated in some shader code later on.
Now that we can importance sample our hemisphere, we can turn our attention back to the radiance equation and see which parts we can calculate and store to cut down on the amount of work we need to do for each frame.
## Split Sum Approximation
In Karis’s presentation of the Unreal Engine PBR pipeline, he explains that we can actually approximate the specular part of our radiance sum by splitting it in two:
$\frac{1}{N} \sum_{k=1}^{N} \frac{ f_s(\mathbf{v}, \mathbf{l}_k) L_i(\mathbf{l}_k) \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \text{pdf}(\mathbf{v}, \mathbf{l}_k) } \approx \left( \frac{1}{N} \sum_{k=1}^{N} L_i(\mathbf{l}_k) \right) \left( \frac{1}{N} \sum_{k=1}^{N} \frac{f_s(\mathbf{v}, \mathbf{l}_k) \langle\mathbf{n} \cdot \mathbf{l}_k\rangle}{\text{pdf}(\mathbf{v}, \mathbf{l}_k)} \right)$
As Karis notes, this approximation is exact for a constant $L_i$ term, and reasonable for “common environments”. This approximation enables us to store the two different sums separately. Let’s examine each sum:
### Pre-filtered Environment Map
The first sum is what’s commonly referred to as the “pre-filtered environment map” or just the “radiance map” in other papers. When we say “pre-filtering”, what we’re actually doing is evaluating and storing a value such that when we go to render our mesh, we can simply sample this stored value to obtain $L_i$. The idea is that we can skip having to integrate across our hemisphere in order to figure out the radiance reflected in $\mathbf{v}$, and instead just look it up in a prefiltered cube map.
Interestingly, Karis and others convolve this sum with the GGX by still using the GGX based importance sampling. This improves convergence, which is important if you’re planning on having your environment map change over time and need to re-evaluate it outside the context of an offline pipeline. So really, we’ll be performing the following sum instead of the one above:
$\frac{4}{\sum_{k=1}^{N} \langle \mathbf{n} \cdot \mathbf{l}_k \rangle } \left( \sum_{k=1}^{N} \langle \mathbf{n} \cdot \mathbf{l}_k \rangle \frac{ L_i(\mathbf{l}_k) \langle \mathbf{v} \cdot \mathbf{h} \rangle } { D(\mathbf{h}) \langle \mathbf{n} \cdot \mathbf{h} \rangle } \right)$
Karis doesn’t provide any mathematic justification for the additional summation in the denominator, or why we should evaluate $L_i$ by importance sampling GGX, but as noted by Sebastian Legarde (pg. 64) these empirical terms seem to provde the best correction for our split sum approximation for a constant $L_i$.
However, one big problem with this is that the GGX NDF is dependent on $\mathbf{v}$ — which creates “stretchy” reflections. Karis approximates the NDF as isotropic where $\mathbf{n} = \mathbf{v}$. This is a much larger source of error than the split sum, and is demonstrated by the figure below, which was taken from the previously mentioned Moving Frostbite to PBR course notes:
Notice how we lose the stretch reflections visible in the ground truth render on the right once we assume $\mathbf{n} = \mathbf{v}$. While the error is large, it’s the price we must pay to be able to perform this sum outside of our render loop when $\mathbf{v}$ is known.
When it comes to storing the pre-filtered environment map, it’s important to note once again that we’re creating a different map for different roughness levels of our choosing. Since we’ll lose high frequency information as roughness increases, we can actually use mip map layers to store the filtered map, with higher roughnesses corresponding to higher mip levels.
Once again we can compute shaders, with a different pass for each roughness/mip level. Here’s the loop used in the application code to issue each dispatch call:
// width is the width/length of a single face of our cube map
float maxMipLevel = bx::log2(float(width));
for (float mipLevel = 0; mipLevel <= maxMipLevel; ++mipLevel)
{
uint16_t mipWidth = width >> uint16_t(mipLevel);
float roughness = mipLevel / maxMipLevel;
float params[] = { roughness, mipLevel, float(width), 0.0f };
bgfx::setUniform(u_params, params);
// Bind the source using a Sampler, so we don't have to perform the cube mapping ourselves
bgfx::setTexture(0, u_sourceCubeMap, sourceCubeMap);
// Set the image using a specific mipLevel
bgfx::setImage(1, filteredCubeMap, uint8_t(mipLevel), bgfx::Access::Write, bgfx::TextureFormat::RGBA16F);
// Dispatch enough groups to cover the entire _mipped_ face
}
Then our compute shader looks like the following:
#define THREADS 8
#define NUM_SAMPLES 64u
// u_params.x == roughness
// u_params.y == mipLevel
// u_params.z == imageSize
uniform vec4 u_params;
SAMPLERCUBE(s_source, 0);
IMAGE2D_ARRAY_WR(s_target, rgba16f, 1);
// From Karis, 2014
vec3 prefilterEnvMap(float roughness, vec3 R, float imgSize)
{
// Isotropic approximation: we lose stretchy reflections :(
vec3 N = R;
vec3 V = R;
vec3 prefilteredColor = vec3_splat(0.0);
float totalWeight = 0.0;
for (uint i = 0u; i < NUM_SAMPLES; i++) {
vec2 Xi = hammersley(i, NUM_SAMPLES);
vec3 H = importanceSampleGGX(Xi, roughness, N);
float VoH = dot(V, H);
float NoH = VoH; // Since N = V in our approximation
// Use microfacet normal H to find L
vec3 L = 2.0 * VoH * H - V;
float NoL = saturate(dot(N, L));
// Clamp 0 <= NoH <= 1
NoH = saturate(NoH);
if (NoL > 0.0) {
// Based off https://developer.nvidia.com/gpugems/GPUGems3/gpugems3_ch20.html
// Typically you'd have the following:
// float pdf = D_GGX(NoH, roughness) * NoH / (4.0 * VoH);
// but since V = N => VoH == NoH
float pdf = D_GGX(NoH, roughness) / 4.0 + 0.001;
// Solid angle of current sample -- bigger for less likely samples
float omegaS = 1.0 / (float(NUM_SAMPLES) * pdf);
// Solid angle of texel
float omegaP = 4.0 * PI / (6.0 * imgSize * imgSize);
// Mip level is determined by the ratio of our sample's solid angle to a texel's solid angle
float mipLevel = max(0.5 * log2(omegaS / omegaP), 0.0);
prefilteredColor += textureCubeLod(s_source, L, mipLevel).rgb * NoL;
totalWeight += NoL;
}
}
return prefilteredColor / totalWeight;
}
// Notice the 6 in the Z component of our NUM_THREADS call
// This allows us to index the faces using gl_GlobalInvocationID.z
void main()
{
float mipLevel = u_params.y;
float imgSize = u_params.z;
float mipImageSize = imgSize / pow(2.0, mipLevel);
ivec3 globalId = ivec3(gl_GlobalInvocationID.xyz);
if (globalId.x >= mipImageSize || globalId.y >= mipImageSize)
{
return;
}
vec3 R = normalize(toWorldCoords(globalId, mipImageSize));
// Don't need to integrate for roughness == 0, since it's a perfect reflector
if (u_params.x == 0.0) {
vec4 color = textureCubeLod(s_source, R, 0);
imageStore(s_target, globalId, color);
return;
}
vec3 color = prefilterEnvMap(u_params.x, R, imgSize);
// We access our target cubemap as a 2D texture array, where z is the face index
imageStore(s_target, globalId, vec4(color, 1.0));
}
I also wrote a small ShaderToy example that lets you visualize this for different levels of roughness. The mouse allows you to change the roughness amount. Let’s take a look at the resulting output for the Pisa environment map:
Note that this is in linear space and will appear darker than it would otherwise when rendered properly with the appropriate gamma transformation applied. The size of the mips has been scaled to match the original image size. Also note that the banding is due to GIF encoding, and you shouldn’t see anything like that in your output.
### Environment BRDF
Okay, so that’s one sum taken care of, let’s look at the second sum. Once again, our goal here is to try and pre-calculate as much of this sum as possible and store it, allowing us to resolve the radiance equation without having to perform hundreds of texture lookups.
$\frac{1}{N} \sum_{k=1}^{N} \frac{ f_s(\mathbf{v}, \mathbf{l}_k) \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \text{pdf}(\mathbf{v}, \mathbf{l}_k) }$
Let’s take a look at it with our specular BRDF and the $\text{pdf}$ subbed in:
$\frac{1}{N} \sum_{k=1}^{N} D(\mathbf{h}) F(\mathbf{v}\cdot\mathbf{h}) V(\mathbf{v},\mathbf{l},\mathbf{h}) \frac{ 4 \langle \mathbf{v} \cdot \mathbf{h} \rangle }{ D(\mathbf{h}) \langle \mathbf{n} \cdot \mathbf{h} \rangle } \langle\mathbf{n} \cdot \mathbf{l}_k\rangle = \frac{4}{N} \sum_{k=1}^{N} F(\mathbf{v}\cdot\mathbf{h}) \frac{ V(\mathbf{v},\mathbf{l},\mathbf{h}) \langle \mathbf{v} \cdot \mathbf{h} \rangle \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \langle \mathbf{n} \cdot \mathbf{h} \rangle }$
At this point, we’ve managed to factor out the NDF but our sum is still dependent on both $\mathbf{v}$, $\text{roughness}$ (through the $G$ term) and $F_0$ (through the Fresnel term). The $F0$ term is particularly inconvenient, because each material will have a potentially different $F0$ term but we don’t want to have to store a different LUT for each material. However, if we substitute our Schlick’s Fresnel equation and rearrange our terms, we get:
\begin{aligned} = &\frac{4}{N} \sum_{k=1}^{N} \frac{ V(\mathbf{v},\mathbf{l},\mathbf{h}) \langle \mathbf{v} \cdot \mathbf{h} \rangle \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \langle \mathbf{n} \cdot \mathbf{h} \rangle } \Big(F_0(1 - (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5) + (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5\Big) \\= &F_0 \left(\frac{4}{N} \sum_{k=1}^{N} \frac{ V(\mathbf{v},\mathbf{l},\mathbf{h}) \langle \mathbf{v} \cdot \mathbf{h} \rangle \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \langle \mathbf{n} \cdot \mathbf{h} \rangle } (1 - (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5) \right) \\ & + \left( \frac{4}{N} \sum_{k=1}^{N} \frac{ V(\mathbf{v},\mathbf{l},\mathbf{h}) \langle \mathbf{v} \cdot \mathbf{h} \rangle \langle\mathbf{n} \cdot \mathbf{l}_k\rangle }{ \langle \mathbf{n} \cdot \mathbf{h} \rangle } (1 - \langle \mathbf{v} \cdot \mathbf{h} \rangle)^5 \right) \\ = &F_0 f_a + f_b \end{aligned}
Where $f_a, f_b$ are the scale and the bias terms applied to $F_0$. You may ask yourself what we have accomplished by doing all this, but notice that the only terms outside the sum $f_a, f_b$ are dependent on are $\text{roughness}$ (through $V$) and $\mathbf{n} \cdot \mathbf{v}$. For $\mathbf{n}$ we’ll set it to some constant (like the positive z-direction). We can then create a two dimensional LUT where the x-axis is $\langle \mathbf{n} \cdot \mathbf{v} \rangle$ and the y-axis is $\text{roughness}$. The red channel of this LUT will be $f_a$ while the green channel will be $f_b$.
I used a compute shader that runs once and the resulting LUT is stored in a RG16F texture. Here’s the compute shader code to perform the calculation:
#include "bgfx_compute.sh"
#include "pbr_helpers.sh"
#define GROUP_SIZE 256
IMAGE2D_WR(s_target, rg16f, 0);
// From the filament docs. Geometric Shadowing function
float V_SmithGGXCorrelated(float NoV, float NoL, float roughness) {
float a2 = pow(roughness, 4.0);
float GGXV = NoL * sqrt(NoV * NoV * (1.0 - a2) + a2);
float GGXL = NoV * sqrt(NoL * NoL * (1.0 - a2) + a2);
return 0.5 / (GGXV + GGXL);
}
// Karis 2014
vec2 integrateBRDF(float roughness, float NoV)
{
vec3 V;
V.x = sqrt(1.0 - NoV * NoV); // sin
V.y = 0.0;
V.z = NoV; // cos
// N points straight upwards for this integration
const vec3 N = vec3(0.0, 0.0, 1.0);
float A = 0.0;
float B = 0.0;
const uint numSamples = 1024;
for (uint i = 0u; i < numSamples; i++) {
vec2 Xi = hammersley(i, numSamples);
// Sample microfacet direction
vec3 H = importanceSampleGGX(Xi, roughness, N);
// Get the light direction
vec3 L = 2.0 * dot(V, H) * H - V;
float NoL = saturate(dot(N, L));
float NoH = saturate(dot(N, H));
float VoH = saturate(dot(V, H));
if (NoL > 0.0) {
// Terms besides V are from the GGX PDF we're dividing by
float V_pdf = V_SmithGGXCorrelated(NoV, NoL, roughness) * VoH * NoL / NoH;
float Fc = pow(1.0 - VoH, 5.0);
A += (1.0 - Fc) * V_pdf;
B += Fc * V_pdf;
}
}
return 4.0 * vec2(A, B) / float(numSamples);
}
void main()
{
// Normalized pixel coordinates (from 0 to 1)
vec2 uv = vec2(gl_GlobalInvocationID.xy + 1) / vec2(imageSize(s_target).xy);
float mu = uv.x;
float a = uv.y;
// Output to screen
vec2 res = integrateBRDF(a, mu);
// Scale and Bias for F0 (as per Karis 2014)
imageStore(s_target, ivec2(gl_GlobalInvocationID.xy), vec4(res, 0.0, 0.0));
}
This shader is a bit simpler than the last, since we don’t have to write to a cube map this time. I’ve written a ShaderToy for this example as well, which shows you what the output looks like. Note that these outputs will be in RGBA8, and aren’t suitable for direct use (i.e. you can’t use the screen cap as an LUT). The two values vary smoothly allowing us to get away with a fairly small LUT. In my BGFX code I store it in 128x128 texture.
## Single Scattering Results
Now that we have a prefiltered environment map and a LUT for most parts of our BRDF, and our diffuse irradiance map, all we have to do is put it together.
$\int_{\Omega} f_{\text{brdf}}(\mathbf{v}, \mathbf{l}) L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \approx \left( F_0 f_a + f_b \right) \text{radiance} + \frac{c_{\text{diff}}}{\pi} \text{irradiance}$
Where $f_a, f_b$ are the values stored in our lookup table, $\text{radiance}$ is the prefiltered environment map, $\text{irradiance}$ is the irradiance map, $F_0$ and $c_{\text{diff}}$ are material properties for specular color and diffuse color respectively. Here’s the fragment shader code as well:
\$input v_position, v_normal, v_tangent, v_bitangent, v_texcoord
#include "../common/common.sh"
#include "pbr_helpers.sh"
#define DIELECTRIC_SPECULAR 0.04
// Scene
uniform vec4 u_envParams;
#define numEnvLevels u_envParams.x
uniform vec4 u_cameraPos;
// Material
SAMPLER2D(s_diffuseMap, 0);
SAMPLER2D(s_normalMap, 1);
SAMPLER2D(s_metallicRoughnessMap, 2);
// IBL Stuff
SAMPLER2D(s_brdfLUT, 3);
SAMPLERCUBE(s_prefilteredEnv, 4);
void main()
{
mat3 tbn = mat3FromCols(
normalize(v_tangent),
normalize(v_bitangent),
normalize(v_normal)
);
vec3 normal = texture2D(s_normalMap, v_texcoord).xyz * 2.0 - 1.0;
normal = normalize(mul(tbn, normal));
vec3 viewDir = normalize(u_cameraPos.xyz - v_position);
vec3 lightDir = reflect(-viewDir, normal);
vec3 H = normalize(lightDir + viewDir);
float VoH = clampDot(viewDir, H);
float NoV = clamp(dot(normal, viewDir), 1e-5, 1.0);
vec4 baseColor = texture2D(s_diffuseMap, v_texcoord);
vec3 OccRoughMetal = texture2D(s_metallicRoughnessMap, v_texcoord).xyz;
float roughness = max(OccRoughMetal.y, MIN_ROUGHNESS);
float metalness = OccRoughMetal.z;
// From GLTF spec
vec3 diffuseColor = baseColor.rgb * (1.0 - DIELECTRIC_SPECULAR) * (1.0 - metalness);
vec3 F0 = mix(vec3_splat(DIELECTRIC_SPECULAR), baseColor.xyz, metalness);
vec2 f_ab = texture2D(s_brdfLUT, vec2(NoV, roughness)).xy;
float lodLevel = roughness * numEnvLevels;
vec3 radiance = textureCubeLod(s_prefilteredEnv, lightDir, lodLevel).xyz;
vec3 FssEss = F0 * f_ab.x + f_ab.y;
}
Let’s take a look at the results! Here’s a render of spheres of increasing roughness (from left to right) that uses the Pisa courtyard environment map:
We can see how as the roughness increases, we’re using our blurrier prefiltered environment maps. However, you may also be able to notice that as roughness increases, the metal spheres become darker. One way we can demonstrate this fully is to render the spheres in a totally uniform environment, where $L_i(\mathbf{l}) = C$ for all directions $\mathbf{l}$. This is also commonly referred to as a “furnace test”:
Here the darkening is much more pronounced. About 40% of the energy is being lost, but since our spheres have a “white” albedo, this is violating conservation of energy. Recall that our model is simply a first order approximation which includes only the single scattering events. In reality, light will scatter several times across the surface, as this diagram from Heitz et al, 2015 illustrates:
As the material becomes rougher, the multiscattering events account for a larger proportion of the reflected energy, and our approximation breaks down. So we need some way of recovering this lost energy if we want to have more plausible results.
## Accounting for Multiple-Scattering
Luckily for us, there’s been a significant amount of work done on improving this approximation by accounting for the additional scattering events. Heitz et al. presented their work on modelling the problem as a free-path problem, validating their results by simulating each successive scattering event across triangular meshes using ray tracing. However, their work is too slow for production path tracing, nevermind real time interactive graphics.
Following this, Kulla and Conty presented their own alternative to recovering the energy based on the furnace test, but it requires additional 2D and 3D LUTs and was written in the context of path tracing. Additionally, Emmanuel Turquin built upon the Heitz paper to model the additional scattering events by noting that Heitz’s paper showed the additional scattering events to have lobes resembling smaller versions of the first event’s lobe.
Unfortunately none of these solutions were designed as real time techniques, but just this year Fdez-Agüera published a paper outlining how we could extend the methods outlined by Karis without having to introduce any additional LUTs or parameters. The paper provides a full mathematical derivation, but I’ll try to provide the highlights.
### Metals
First, let’s consider a perfect reflector with no diffuse lobe, where $F = 1$. In this case, the total amount of energy reflected regardless of number of bounces must always be equal to the incident energy (due to conservation of energy):
$1 = E_{ss} + E_{ms} \Rightarrow E_{ms} = 1 - E_{ss}$
Where $E_{ss}$ is our directional albedo, but with $F = 1$:
$E_{ss} = \int_{\Omega} \frac{ D(\mathbf{h}) G(\mathbf{v},\mathbf{l},\mathbf{h}) }{ 4 \langle\mathbf{n}\cdot\mathbf{l}\rangle \langle\mathbf{n}\cdot\mathbf{v}\rangle } \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l}$
To define $E_{ms}$, we can express it in the same fashion, but with an unknown BRDF:
$E_{ms} = \int_{\Omega} f_{ms} \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} = 1 - E_{ss}$
So we could effectively add this additional lobe to our reflectance equation:
\begin{aligned} L_o(\mathbf{v}) &= \int_{\Omega} f_{\text{ss}} + f_{\text{ms}}) L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \\ &= \int_{\Omega} f_{\text{ss}} L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} + \int_{\Omega} f_{\text{ms}} L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \end{aligned}
We’ve already discussed the integral on the left, so let’s focus instead on the second integral. We’ll once again use the split sum introduced by Karis, but here Fdez-Agüera makes the assumption that we can consider the secondary scattering events to be mostly diffuse, and therefore use irradiance as a solution to the $L_i$ integral:
\begin{aligned} \int_{\Omega} f_{\text{ms}} L_i(\mathbf{l}) \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} &= \int_{\Omega} f_{\text{ms}} \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \int_{\Omega} \frac{L_i(\mathbf{l})}{\pi} \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \\ &= (1 - E_{ss}) \int_{\Omega} \frac{L_i(\mathbf{l})}{\pi} \langle\mathbf{n} \cdot \mathbf{l}\rangle d\mathbf{l} \end{aligned}
Fdez-Agüera further notes that this approximation fails for narrow lights, such as analytical point or directional lights. Interestingly, Stephen McAuley made the clever observation that this multiscattering term will only dominate at higher roughnesses, where our radiance map is going to be highly blurred and fairly diffuse. His comparison shows little difference, so if you weren’t previously using irradiance you could potentially skip it here as well. However we’ve already created our irradiance map, so we will be using it again here.
If we use the approximation from earlier for the single scattering event, and bring in this approximation for the multiscattering event, we end up with the following:
$L_o(\mathbf{v}) = \left( f_a + f_b \right) \text{radiance} + (1 - E_{ss}) \text{irradiance}$
Where $f_a, f_b$ are the scale and bias terms from Karis that we’ve stored in our LUT.
Recall, however, that so far we’ve constrained ourselves to a perfectly reflecting metal. In order to extend it to generic metals, we need to re-introduce $F$. Like others, Fdez-Agüera splits $F$ into two terms, $F_{ss}$ and $F_{ms}$ such that:
$E = F_{ss} E_{ss} + F_{ms} E_{ms}$
However, unlike previously, we cannot simply set $E=1$. Instead, Fdez-Agüera models $F_{ms}$ as a geometric series such that:
$E = F_{ss} E_{ss} + \sum_{k=1}^{\inf} F_{ss} E_{ss} (1 - E_{\text{avg}})^k F_{\text{avg}}^k$
Where $F_{\text{avg}}$ and $E_{\text{avg}}$ are defined as:
$E_{\text{avg}} = E_{ss} \\ F_{\text{avg}} = F_0 + \frac{1}{21} (1 - F_0),$
You’ll have to check the papers for the details on this I’m afraid, as I don’t want to entirely reproduce the paper in this blog post. The conclusion is that we can therefore write our equation for $L_o$ as the following:
$L_o(\mathbf{v}) = \left( F_0 f_a + f_b \right) \text{radiance} + \frac{(F_0 f_a + f_b)F_{\text{avg}}}{1 - F_{\text{avg}}(1 - E_{ss})} (1 - E_{ss}) \text{irradiance}$
Let’s take a second and look at some renders of metals with this new formulation. For the furnace test, we should not be able to see the spheres at all. Below is a comparison between the single scattering BRDF and multiple scaterring BRDF inside the furnace:
Sure enough, using our multiple scattering BRDF we can’t see the spheres at all! The approximation is perfect for constant environments like the furnace, so let’s take a look at a less optimal case:
The improvement is considerable, as roughness does not darken the spheres.
One important thing to mention is that for colored metals, you’ll see an increase in saturation at the higher roughnesses not unlike what Heitz, Kulla and Conty and Hill see. Whether this is desired behaviour is up to you. It appears Turqin’s method does not produce these shifts in saturation, so that is worth investigation if increased saturation is to be avoided.
### Dielectrics
So far, we haven’t talked about how our diffuse component fits into this at all. Let’s first examine how dielectrics behave in the furnace test for single scattering by revisting our render from earlier:
At lower roughnesses the Fresnel effect is too strong, and there is still some darkening at higher roughnesses.
$E = 1 = F_{ss} E_{ss} + F_{ms} E_{ms} + E_{\text{diffuse}}$
This approximation ignores some of the physical reality of diffuse transmission and re-emission, but Fdez-Agüera makes a good argument for why we can keep ignoring them:
It doesn’t explicitly model the effects of light scattering back and forth between the diffuse and specular interfaces, but since specular reflectance is usually quite low and unsaturated for dielectrics, radiated energy will eventually escape the surface unaltered, so the approximation holds. (p. 52)
To extend to non-white dielectrics, we can simply multiply it by the diffuse albedo, $c_{\text{diff}}$.
\begin{aligned} F_{ss}E_{ss} &= \left( F_0 f_a + f_b \right) \times \text{radiance}, \\ F_{ms}E_{ms} &= \frac{E_{ss} F_{\text{avg}}}{1 - F_{\text{avg}}(1 - E_{ss})} (1 - E_{ss}) \times \text{irradiance}, \\ K_d &= c_{\text{diff}} (1 - F_{ss} E_{ss} + F_{ms} E_{ms}) \times \text{irradiance}, \\ L_o &= F_{ss}E_{ss} + F_{ms}E_{ms} + K_d \end{aligned}
Let’s look at our final results using the same comparison as before, but with dielectrics this time:
Interestingly, I actually get a darker result with multiple scattering than with single scattering. I’m not totally convinced that this isn’t some subtle bug in my implementation of the dielectric part of the BRDF. However, the significant excess energy at lower roughnesses that we observed with the single scattering BRDF is not present with our new BRDF. Here’s a final render using the Pisa environment, as before, but this time with our new BRDF:
#### Roughness Dependent Fresnel
In the Fdez-Agüera paper’s sample GLSL code, the author includes a modification to the Fresnel term:
vec3 Fr = max(vec3_splat(1.0 - roughness), F0) - F0;
vec3 k_S = F0 + Fr * pow(1.0 - NoV, 5.0);
// Typically, simply:
// vec3 FssEss = F_0 * f_ab.x + f_ab.y;
vec3 FssEss = k_S * f_ab.x + f_ab.y;
Unfortunately, I haven’t been able to track down the origin of this modification, and it’s not expanded upon in the paper. It doesn’t make much difference when rendering our spheres, especially in the furnace test, but when rendering more complex objects the difference is noticable. To reason about what this term does exactly, I’ve made a little desmos graph demo. You can adjust the $F_0$ term to see how the modification deviates from the Schlick approximation for different levels of roughness. Interestingly, this makes the ramp start earlier for lower roughness levels. I can see this change being plausible: for rougher surfaces, the proprtion of microfacets contributing to the angle dependent Fresnel will decrease. At least that’s the best explanation I could come up with. If you know the reasoning or source for this modification, please let me know!
Here’s a series of renders for you to compare, using the Flight Helmet model. The roughness dependent Fresnel displays the increase in reflections for smoother surfaces like the leather cap and wooden base.
You can also see the brightening of the metal goggles and base’s metal plaque, but otherwise this model is mostly composed of dielectric materials and so the difference is not as large.
Here’s the multiscattering code, which can be appended to the shader code we used for single scattering above. We are still only using the BRDF LUT, prefitlered environment map, and irradiance map as uniforms; there are no additional uniforms introduced by multiple scattering.
// Same code as from earlier... no extra uniforms
void main()
{
// Same code from earlier ...
// Roughness dependent fresnel, from Fdez-Aguera
vec3 Fr = max(vec3_splat(1.0 - roughness), F0) - F0;
k_S = F0 + Fr * pow(1.0 - NoV, 5.0);
vec3 FssEss = k_S * f_ab.x + f_ab.y;
// Multiple scattering, from Fdez-Aguera
float Ems = (1.0 - (f_ab.x + f_ab.y));
vec3 F_avg = F0 + (1.0 - F0) / 21.0;
vec3 FmsEms = Ems * FssEss * F_avg / (1.0 - F_avg * Ems);
vec3 k_D = diffuseColor * (1.0 - FssEss - FmsEms);
vec3 color = FssEss * radiance + (FmsEms + k_D) * irradiance;
gl_FragColor = vec4(color, baseColor.w);
}
Notice how little code has been added for calculating the new terms! And yet the results for metals are very noticable.
## Future Work
• The multiscattering BRDF I’ve implemented here is only presented for image based lighting. McAuley goes into some detail on how it could be extended for area lights using Heitz’s linearly transformed cosine approach.
• My model doesn’t fully support the GLTF material spec. I still need to add better support for Occlusion, Emissive, scaling factors and non-texture uniforms (baseColor for instance can be encoded as a single vec4). I’m trying to accomplish this piecemeal while implementing other algorithms.
• Since writing this post, I’ve added support for all the material properties that the GLTF spec provides in the “roughness metallic” workflow. There is no skinning or animation support yet though.
• For our prefiltered environment map, at lower roughness values we are likely to encounter substantial aliasing in the reflections as we can no longer use the cube map’s mip map chain as we typically would. I’ve seen some implementations like Babylon.js actually just use the higher roughness (e.g. blurrier) parts of the map anyway, to reduce aliasing.
• If you have any feedback, please let me know either using the comments below, my email or my twitter account
## Source Code
While I’ve included most of the shader code used, the entire example is available here as 04-pbl-ibl. Note however that it uses non-standard GLTF files, that have had their web-compatible textures swapped out with corresponding DDS files that contain mip maps, generated using the texturec tool provided as part of BGFX. BGFX does not provide an easy way to invoke the equivalent of glGenerateMipmap, so I’ve created a python script to pre-process the GLTF files. It can be found in the scripts folder of the repo. | 12,863 | 45,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 159, "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} | 3.171875 | 3 | CC-MAIN-2021-25 | longest | en | 0.821794 |
https://math.answers.com/natural-sciences/How_many_square_feet_is_12_feet_times_17_feet | 1,679,774,139,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00711.warc.gz | 424,273,143 | 49,662 | 0
# How many square feet is 12 feet times 17 feet?
Wiki User
2013-07-08 20:15:00
The square feet of a room that is 12 feet times 17 feet is 204. Square feet is calculated by multiplying the length of the room by the width of the room.
Wiki User
2013-07-08 20:15:00
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132 Reviews | 106 | 328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.900502 |
https://www.markedbyteachers.com/international-baccalaureate/maths/the-segments-of-a-polygon.html | 1,716,035,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00326.warc.gz | 784,235,806 | 12,388 | # The segments of a polygon
INTERNATIONAL BACCALAUREATE
II. GIMNAZIJA MARIBOR
Portfolio mathematics HL
Assignment 1
The segments of a polygon
Author: Luka Dremelj
Candidate number:
Subject: Mathematics HL
Teacher: Barbara Pećanac
Date written: 25/5/2009
Introduction
First Mathematics HL Portfolio is about investigating the segments of a polygon, using graphical methods and analytical proofs. The task is to find conjecture between ratio of the sides and the ratio of the areas, first of triangles developing it to general conjecture of polygons.
1. In an equilateral triangle ABC, a line segment is drawn from each vertex to a point on the opposite side so that the segment divides the side in the ratio 1:2, crating another equilateral triangle DEF.
(a) What is the ratio of the areas of the two equilateral triangles? To answer this question,
(i) create the above diagram with your geometry package.
(ii) measure the lengths of the sides of the two equilateral triangles
== = 3 units
= = 1.13 units
(iii) find the areas of the two equilateral triangles and the ratio between them.
Formula for area of triangle:
(u=unit)
I got the same results for the areas, as before calculated with program. (see the picture above)
(b) Repeat the procedure above for at least two other side ratios, 1:n.
I decided to draw triangle with ratio: 1:3
== = 3 units
= = 1.67 units
Third ratio: 1:4
== = 3 units
= = 1.96 units
(c) For:
n=2 we get ratio of areas 7
n=3 we get ratio of areas 3.2
n=4 we get ratio of areas 2.3
To see connection between this results the best way is to draw a graph on which n will be represented on ... | 425 | 1,669 | {"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-2024-22 | latest | en | 0.885446 |
http://mingxinglai.com/cn/2012/09/generate-permutation-use-stl/ | 1,657,122,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00263.warc.gz | 34,890,893 | 4,704 | # STL：从N个元素中取出M个的所有组合
``````#include <iostream>
#include <algorithm>
#include <vector>
#include <iterator>
using namespace std;
const size_t M = 3;
void premutate(std::vector<int> &v, int data[], int num = 0)
{
if (num == M)
{//output
copy(data, data + num, ostream_iterator<int>(cout, " "));
cout << endl;
return;
}
else
{
for (vector<int>::iterator it = v.begin(); it != v.end(); it++)
{
//我们注意到,对有序的数排列的时候,排列后面的数都比前面的大
//我们利用这个特性,快速的判断一个数是否已经在排列当中了
if ( num != 0 && data[num - 1] >= *it )
{
continue;
}
else
{
data[num] = *it;
premutate(v, data, num + 1);
}
}
}
}
int main(int argc, char* argv[])
{
int elements[] = {1,2,3,4,5};
const size_t N = sizeof(elements) / sizeof(elements[0]);
std::vector<int> selectors(elements, elements + N);
int data[N] = {0};
sort(selectors.begin(), selectors.end());
premutate(selectors, data);
return 0;
}
``````
``````#include <assert.h>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main()
{
int values[] = { 1, 2, 3, 4, 5, 6, 7 };
int elements[] = { 1, 1, 1, 0, 0, 0, 0 };
const size_t N = sizeof(elements)/sizeof(elements[0]);
assert(N == sizeof(values)/sizeof(values[0]));
std::vector<int> selectors(elements, elements + N);
int count = 0;
do
{
std::cout << ++count << ": ";
for (size_t i = 0; i < selectors.size(); ++i)
{
if (selectors[i])
{
std::cout << values[i] << ", ";
}
}
std::cout << std::endl;
} while (prev_permutation(selectors.begin(), selectors.end()));
}
``````
/
Published under (CC) BY-NC-SA in categories 算法 tagged with STL algorithm 排列 | 525 | 1,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.23619 |
http://www.jiskha.com/display.cgi?id=1290482932 | 1,495,833,523,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608684.93/warc/CC-MAIN-20170526203511-20170526223511-00154.warc.gz | 685,467,382 | 4,082 | # chemistry
posted by on .
A mixture of 2.0 mol H2 and 1.0 mol of O2 is placed in a sealed evacuated container made of a perfect insulating material at 28C. The mixture is ignited with a spark and it reacts to form liquid water. What is the final temperature of the water
• chemistry - ,
2H2 + O2 ==> 2H2O
2 moles H2 and 1 mol O2 will form 2 moles H2O or 2*18.015 grams. I looked up delta Ho in my text and found that it releases 187.8 kJ/mol which will be 375,600 J/ 2 moles or 36.03 grams H2O.
375,600 = mass water x specific heat water x (Tfinal-Tinitial).
Solve for Tf. Check my thinking. When you finish (be prepared for a shock), check your work and see if
mass water x specific heat water (Tf-28) = the heat released.
I presume that we neglect that this temperature should vaporize the water. | 233 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-22 | latest | en | 0.89276 |
http://jollymaths.com/blog/category/recreational-math/ | 1,675,090,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00207.warc.gz | 26,631,749 | 17,152 | # How to make the Number puzzle
Some of my friends asked about the procedure for doing the above puuzzle.
Just look into the pattern closely for two or three times.
All yellow numbers are written on clockwise and the blue squares are filled with anti clockwise.
Now check the total of two blue squares and in between yellow square.
Adjust some the numbers in the begining to come at the total correctly.
ENHANCING THE TOTAL TO THE REQUIRED NUMBER:
Now let the required total be 2025.
Deduct 369 from the 2025, you will get 1656
Since the total 369 is the addition of 3 squares, divide 1656 by 3, you will get 552.
Now add 552 in all the squares. the total of two blue squares and in between yellow square. you will get 2025
The format 1 to 210 will give a total of 369 and you have added 3 x 552, you will get 369 + 1656 = 2025.
# Circular Number Puzzle from 1 to 1000 A Unique way– Part Three
In continuation of the number puzzles, a new puzzle was done by me on 8-2-2022.
All numbers from 1 to 1000 are utilised in this puzzle and any three continuous squares ( two brown squares and one Blue square in between will give the same total of 1752 at any given point.
I thank the Almighty for giving me the idea, conducted me, guided me to complete the NEW BIGGEST CIRCULAR NUMBER PUZZLE.
# Number puzzle — A unique way (Part Two) Honoring The great Indian Mathematician Srinivasa Ramanujan’s 135th Birth Anniversary in a Unique Jollymaths.com way
The entire world is celebrating the 135th Birth year of the famous Indian Mathematician Srinivasa Ramanaujan. On this occassion we proudly submit a new idea/ puzzle that gives a total of 135
In the above puzzle, count any three successive (two blue squares and one brown square in between) throughout the round, you will get 135
Concept, Design and execution by T.R.Jothilingam, Retired Railway Station Superintendent, Madurai, South India.
Receipient of Ramanujan Award in India in 2016.
Done Five Maths World Records in Sept 2017
Got First place in a worldwide puzzle contest in 2014
Got “Top 100 Maths Genius Award in December 2017
# Number Puzzle – A new way
Recently I had an opportunity to see a Heptagonal Puzzle done by the Famous Mathematician Henry Dudeney (1847 – 1930).
It is a puzzle with seven sides and each meeting point and the middle of the line are placed with some numbers. The numbers are so arranged that any two numbers on two continuous points and one middle number betwwen them will always give a total of 26.
As a puzzle lover it attracted me to expand this puzzle in a bigger way.
In my first attempt after so many trail and errors I made my first puzzle with numbers 1 to 22 with total of 40
Then I expanded it from 1 to 46 with a total of 82
Now I made a Biggest one successfully with all numbers from 1 to 210 with a total of 369, and it is in a zig zag form with the same concept.
Hope you can enjoy and get some idea and enjoy Maths in a easy way.
Best of luck, kindly go through my website www.jollymaths.com
# Finding the DAY for any given DATE
Finding the DAY for any given DATE
1) Let the given date is 14-7-2021.
2) The formula is [ K + { (M-2) x 2.6 – 0.2 } + C/4 + D/4 +D-2C] divided by 7
3) Where K is the date M is the month C is the century D is the last two digit of the year
4) [ 14 + { (7-2) x 2.6 – 0.2} + 20/4 + 21/4 + 21 – 40 }] divided by 7
if M – 2 is zero or -1 ( Feb or Jan) , less one from the year ( D ) and add 12 months to month M.
Drop all the residuals/fractions in all cases and take the whole number only
5) [ 14 + 12.8 + 5 + 5 + 21 – 40] divided by 7
6) {14 + 12 + 5 + 5 + 21 – 40) divided by 7
7) 17 / 7 = 2 and 3/7
If the reminder comes ZERO, it is SUNDAY. 1/7 means MONDAY, 2/7 means Tuesday, 3/7 means WEDNESDAY, 4/7 means THURSDAY, 5/7 means FRIDAY AND 6/7 means SATURDAY.
Hence 14-7-2021 is Wednesday .
9) Try for other dates also.
9442810486
www.jollymaths.com
# CHESS PUZZLES
There are so many other mind boggling games in the Chess Board, other than playing chess.
Some of the puzzles are given below
In a 8 x 8 Chess Board, you can put a maximum of 8 Queens ( each one is to be treated with the power of QUEEN).
One of the position is given below. There are 8 Queens available and check and satisfy by yourself that they are not cutting each other .
Believe. There are 92 different methods available. Try by yourself to find out more methods. Enjoy the new experience
64 knight(Horse) movement in a chess board. It is one of the hardest puzzles that can be solved by the Human Brain.
It is very tough to do, but it is possible.
Draw a 8 x 8 squares in a paper with Ballpoint pen and start writing 1,2,3 etc with a pencil. Start number one anywhere and continue the next number in a knight move method, till you are reaching 64.
See the magical thing in the picture given below. In the outer ring all the diagonally opposite numbers are having a difference of 6. Also in the second inner ring also the diagonally opposite numbers are giving a difference of 6 excepting 10 and 64
Some of my students ( Eighth Standard – 13 years) in Madurai have done more than 10 different methods. You can find infinite ways of doing this. Best wishes.
FIRST ROW PIECES PROBLEM
There are eight pieces available in the first row of the Chess Board.
They are One King, One Queen, two Bishop (camel) and two Knight (Horse) and two Rook (Elephant).
With these Eight pieces, when placed in some proper positions, they will guard all the sixty four squares with their own power. Please check and satisfy by yourself.
Once again, there are so many different methods available. Try to do it more and more methods
64 King move in a chess Board and 8 x 8 Magic square
In the following chess board, one King starts from number one and make 64 moves and completes his journey without any break.
Finally, if you add the numbers it will give a total of 260 in all vertical, Horizontal and both diagonals. Yes. It is a 8 x 8 Magic square also.
Try for other methods also.
# Puzzling number Puzzles
There are many different kinds of puzzles for entertainment as well as to improve our knowledge as well.
In the above You have to write all numbers from 1 to 9 in the same arrangement and you can put any mathematical symbol as you know.
The result of the final format must be 100.
Remember. There are 320 different methods available.
Some of the solutions are given here for easy understanding.
Hope you will be finding this puzzle is very challenging and enjoyable.
# 9×54 Ramanujan sudoku
This is a variation of the popular puzzle Sudoku.
The word “RAMANUJAN” occupied 89 cells and numbers 1 to 9 are used 9 times each and the balance filled with 1 to 8. 8 along with the properties of NORMAL SUDOKU.
1) The entire word “RAMANUJAN” is first written using 119 squares (or cells).
2) Numbers 1 to 9 are filled inside each of these letters.
3) In total, the numbers 1 to 9 are written 13 times and the remaining cells are filled with 1 and 2.
4) We then take each Sudoku of 9 x 9 individually and fill them. The total number of squares that are covered for the word/letter are filled 1 to 9, the adjacent Sudoku (in the right) will commence with the numbers next to the last with 1 to 9 till it fills in the alphabet in that 9 x 9 Sudoku, and the next word/letter will start with the filled number.
For example, the first sudoku contains 20 squares, that are filled with 1 to 9 two times and the balance with 1 & 2. Hence the next Sudoku will commence with 3 to 9, 1 to 9, and so on.
# 9×27 Ramanujan sudoku
This is a variation of the popular puzzle Sudoku.
The word “RAMANUJAN” occupied 89 cells and numbers 1 to 9 are used 9 times each and the balance filled with 1 to 8. 8 along with the properties of NORMAL SUDOKU.
# Srinivasa Ramanujan – 100-by-100 Biography Magic Square
## Ramanujan and Magic Squares
Srinivasa Ramanujan had a special affinity toward numbers. His taxi-cab number (1729) incident is popular. A Mathematician without parallel, he made extraordinary contributions to mathematical analysis, number theory, infinite series, and continued fractions. His works have been collected and analyzed throughout the world
Incidentally, in the opening page of the first Ramanujan’s notebook, there begins by working out a 3 x 3 Magic Square!
Having worked on a variety of special Magic squares ourselves, we could not think of a greater tribute to Srinivasa Ramanujan than this!
### Summary
This is one of the biggest number puzzles we have done so far! This Biography Magic Square summarizes the important events that happened in the life of Sri Srinivasa Ramanujan.
The important dates in the life of Srinivasa Ramanujan were compiled from various sources. These dates were taken two digits at a time, representing either the date of the month or the month or the first/second half of the four-digit year. As an example, Ramanujan’s date-of-birth 22-12-1887, is taken as four separate entries as 22 12 18 and 87. In short, Ramanujan’s entire life history is reproduced here, from his birth to till date in Ramanujan-style.
### Construction
Important dates from Ramanujan’s life were collected and these were then arranged horizontally in a row, from left to right. This row would form the top row of this biography magic square. The rest of the magic square is constructed after assembling this row.
This magic square has the properties of a conventional magic square, namely the sum of the entries along each row/column/diagonal sum up to the same magic-sum 2183.
– starting from left to right, or, from top to bottom, we have embedded magic squares of orders 4 x 4 , 8 x 8, 12 x 12, 16 x 16, 20 x 20, and then in increased orders of 25 x25, 30 x 30, 36 x 36, 42 x 42, 49 x 49, 56 x 56, 64 x 64, 72 x 72, 81 x 81, 90 x 90, and finally 100 x 100. This is thus a cascade of magic-squares-inside-a-magic-sqaure!
Thus the total 100 x 100 Ramanujan Biography Magic square will contain the following 184 smaller magic squares of sizes as listed below:
Size of Magic Square Number of such Magic Squares Total Entries
4 x 4 Magic squares 25 25 ( 4 x 4 ) = 400 squares
5 x 5 Magic squares 20 20 ( 5 x 5 ) = 500 squares
6 x 6 Magic squares 24 24 ( 6 x 6 ) = 864 squares
7 x 7 Magic squares 28 28 ( 7 x 7 ) = 1372 squares
8 x 8 Magic squares 32 32 ( 8 x 8 ) = 2048 squares
9 x 9 Magic squares 36 36 ( 9 x 9 ) = 2916 squares
10 x 10 Magic squares 19 19 (10 x 10 ) = 1900 squares
Total 184 (Different sized squares) 10,000 Squares
### Sidenote
We have constructed a smaller 16 x 16 version of this Biography Magic Square with fewer details, which you can find here.
This was earlier published in an article “A Unique Novel Homage to the Great Indian Mathematician” in the March 2013 (Volume 23, Pg 146-147) Mathematics Newsletter published by the Ramanujan Mathematics Society. (download free). | 2,985 | 11,248 | {"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-2023-06 | longest | en | 0.897548 |
https://de.mathworks.com/matlabcentral/profile/authors/5030971?page=2 | 1,606,439,021,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189030.27/warc/CC-MAIN-20201126230216-20201127020216-00539.warc.gz | 258,387,028 | 3,851 | Solved
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mehr als 6 Jahre ago | 1,416 | 4,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.838331 |
https://ai.stackexchange.com/questions/9751/what-is-the-concept-of-channels-in-cnns/9775 | 1,623,997,138,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635724.52/warc/CC-MAIN-20210618043356-20210618073356-00385.warc.gz | 99,538,411 | 40,873 | # What is the concept of channels in CNNs?
I am trying to understand what channels mean in convolutional neural networks. When working with grayscale and colored images, I understand that the number of channels is set to 1 and 3 (in the first conv layer), respectively, where 3 corresponds to red, green, and blue.
Say you have a colored image that is 200x200 pixels. The standard is such that the input matrix is a 200x200 matrix with 3 channels. The first convolutional layer would have a filter that is size $$N \times M \times 3$$, where $$N,M<200$$ (I think they're usually set to 3 or 5).
Would it be possible to structure the input data differently, such that the number of channels now becomes the width or height of the image? i.e., the number of channels would be 200, the input matrix would then be 200x3 or 3x200. What would be the advantage/disadvantage of this formulation versus the standard (# of channels = 3)? Obviously, this would limit your filter's spatial size, but dramatically increase it in the depth direction.
I am really posing this question because I don't quite understand the concept of channels in CNNs.
Say you have a colored image that is 200x200 pixels. The standard is such that the input matrix is a 200x200 matrix with 3 channels. The first convolutional layer would have a filter that is size $$N×M×3$$, where $$N,M<200$$ (I think they're usually set to 3 or 5).
Would it be possible to structure the input data differently, such that the number of channels now becomes the width or height of the image? i.e., the number of channels would be 200, the input matrix would then be 200x3 or 3x200. What would be the advantage/disadvantage of this formulation versus the standard (# of channels = 3)? Obviously, this would limit your filter's spatial size, but dramatically increase it in the depth direction.
The different dimensions (width, height, and number of channels) do have different meanings, the intuition behind them is different, and this is important. You could rearrange your data, but if you then plug it into an implementation of CNNs that expects data in the original format, it would likely perform poorly.
The important observation to make is that the intuition behind CNNs is that they encode the "prior" assumption or knowledge, the "heuristic", the "rule of thumb", of location invariance. The intuition is that, when looking at images, we often want our Neural Network to be able to consistently (in the same way) detect features (maybe low-level features such as edges, corners, or maybe high-level features such as complete faces) regardless of where they are. It should not matter whether a face is located in the top-left corner or the bottom-right corner of an image, detecting that it is there should still be performed in the same way (i.e. likely requires exactly the same combination of learned weights in our network). That is what we mean with location invariance.
That intution of location invariance is implemented by using "filters" or "feature detectors" that we "slide" along the entire image. These are the things you mentioned having dimensionality $$N \times M \times 3$$. The intuition of location invariance is implemented by taking the exact same filter, and re-applying it in different locations of the image.
If you change the order in which you present your data, you will break this property of location invariance. Instead, you will replace it with a rather strange property of... for example, "width-colour" invariance. You might get a filter that can detect the same type of feature regardless its $$x$$-coordinate in an image, and regarldess of the colour in which it was drawn, but the $$y$$-coordinate will suddenly become relevant; your filter may be able to detect edges of any colour in the bottom of an image, but fail to recognize the same edges in the top-side of an image. This is not an intuition that I would expect to work successfully in most image recognition tasks.
Note that there may also be advantages in terms of computation time in having the data ordered in a certain way, depending on what calculations you're going to perform using that data afterwards (typically lots of matrix multiplications). It is best to have the data stored in RAM in such a way that the inner-most loops of algorithms using the data (matrix multiplication) access the data sequentially, in the same order that it is stored in. This is the most efficient way in which to access data from RAM, and will result in the fastest computations. You can generally safely expect that implementations in large frameworks like Tensorflow and PyTorch will already require you to supply data in whatever format is the most efficient by default.
• Thanks for the detailed answer. To clarify, when I said to "rearrange the data," I mean that the CNN is ALSO trained under the new format. I am not suggesting that we use a CNN trained on the previous format and use it to make a predictions on an image in the new format. Apologies for the miscommunication. As a side note, this question was inspired by my question in datascience.stackexchange.com/questions/43291/… where I am trying to use CNNs on non-image data. – anonuser01 Dec 30 '18 at 19:07
• @Iamanon Yes that's what I assumed in my answer as well, that you'd do both training and evaluation with the "rearranged" data. I still expect that would generally perform less well. The dimensions that your filters are "sliding" along should by the dimensions where you want the "invariance". That should be in the spatial dimensions (width and height), but NOT in other dimensions (e.g. channels for different colours). – Dennis Soemers Dec 30 '18 at 19:46
• @Iamanon So, intuitively you could say that dimensions for which you expect the invariance property to be beneficial are not "channels", and all other dimensions are "channels". For example, TensorFlow has conv2d and conv3d for inputs with invariance in 2D or 3D spaces (and any other input dimensions are "channels"). It also has convolution (tensorflow.org/api_docs/python/tf/nn/convolution) with an arbitrary number input_spatial_shape of "spatial" inputs, and an arbitrary number in_channels of remaining input dimensions. – Dennis Soemers Dec 30 '18 at 19:55
• @Iamanon For your physics application... it's difficult for me to judge where you'd want the property of invariance. I could easily see it being a useful property in all of the dimensions. Spatial and temporal invariance both can make sense. You have to ask yourself: whatever kind of high-level features I expect my NN to "learn inside hidden layers", should the manner in which it detects them be invariant to time and/or space? If yes, filters should slide along those dimensions. If no, treat them as "channels". – Dennis Soemers Dec 30 '18 at 20:01
• Thanks again. The idea that the channels are NOT the dimension(s) in which you expect the invariance property makes sense! What happens if there is more than one dimension that fits under this idea of channels? It seems these deep learning libraries only allow for a single input as the "channels." With one dimension, say, of size 5, you would simply set channels = 5. What if you had some data where you had 2 dimensions of size 5 and 10, respectively, in which you are NOT expecting the invariance property, how would you define the number of channels? – anonuser01 Dec 30 '18 at 20:37
If you have a gray scale image, that means you are getting data from one sensor. If you have an RGB image, that means you are getting data from three sensors. If you have a CMYK image, that means you are getting data from four sensors.
So, channels can be considered as same information seen from different perspective. (here color)
If you see how the kernel (for example 5*5*3) moves, it moves only in XY direction and not in the channel direction. So, you are trying to learn features in XY direction from all the channels together.
But, if you exchange the dimensions like you mentioned your XY dimensions become 200*3 or 3*200 and your channels become 200. In this case you are moving kernel not in the actual XY spatial space of image. So, it doesn't make any sense according to me. You are contradicting the basic concept of CNN by doing so.
The concept of CNN itself is that you want to learn features from the spatial domain of the image which is XY dimension. So, you cannot change dimensions like you mentioned. | 1,894 | 8,430 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.958161 |
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# What does it mean for two lines to be perpendicular? How are the slopes of perpendicular lines related?
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What does it mean for two lines to be perpendicular? How are the slopes of perpendicular lines related?
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### Harper
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The lines intersect at a right angle and have opposite reciprocal slopes.
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The slope of perpendicular lines are the negative reciprocals of each other, and a pair of these lines intersects at 90 degrees.
Option A is the correct answer.
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Help me a lot (92) | 165 | 727 | {"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-2022-49 | latest | en | 0.888992 |
https://www.teachoo.com/3253/692/Ex-4.5--16---Verify-A3---6A2---9A--4I-=-O-and-hence-find-A-1/category/Ex-4.5/ | 1,527,053,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00157.warc.gz | 842,737,379 | 13,672 | 1. Chapter 4 Class 12 Determinants
2. Serial order wise
Transcript
Ex4.5, 16 If A = 2−11−12−11−12 verify that A3 − 6A2 + 9A − 4I = O and hence find A−1 Calculating A2 A2 = A.A = 2−11−12−11−12 2−11−12−11−12 = 2 2+ −1 −1+1(1)2 −1+ −1 2+1(−1)2 1+ −1 −1+1(2)−1 2+2 −1+(−1)(1)−1 −1+2 2+(−1)(−1)−1 1+2 −1+(−1)(2)1 2+ −1 −1+2(1)1 −1+ −1 2+2(−1)1 1+ −1 −1+2(2) = 4+1+1−2−2−12+1+2−2−2−11+4+1−1−2−22+1+2−1−2−21+1+4 = 6−55−56−55−56 Hence A2 = 6−55−56−55−56 Calculating A3 A3 = A2 . A = 6−55−56−55−56 2−11−12−11−12 = 6 2+ −5 −1+5(1)6 −1+ −5 2+5(−1)6 1+ −5 −1+5(2)−5 2+6 −1+(−5)(1)−5 −1+6 2+(−5)(−1)−5 1+6 −1+(−5)(2)5 2+ −5 −1+6(1)5 −1+ −5 2+6(−1)5 1+ −5 −1+6(2) = 12+5+5−6−10−56+5+10−10−6−55+12+5−5−6−1010+5+6−5−10−65+5+12= 22−2121−2122−2121−2122 Hence, A3 = 22−2121−2122−2121−2122 Now, putting values in A3 – 6A2 +9A – 4I = 22−2121−2122−2221−2122 – 6 6−55−56−55−56 + 9 2−11−12−11−12 – 4 100010001 = 22−2121−2122−2221−2122 – 6(6)6(−5)6(5)6(−5)6(6)6(−5)6(5)6(−5)6(6) + 9(2)9(−1)9(1)9(−1)9(2)9(−1)9(1)9(−1)9(2) – 4(1)0004(1)0004(1) = 22−2121−2122−2221−2122 – 36−3030−3036−3030−3030 + 18−99−918−99−918 – 400040004 = 22−36+18+4−21− −30+ −9+021−30+9+0−21− −30+ −96022−36+18+4−22− −30+ −9+021−30+9+0−21− −30+ −9+022−36+18+4 = 36−36−21+30−930−30−21+30−936−36−21+30−930−30−21+30−936−36 = 000000000 = O Hence proved Now calculating A-1 using A3 – 6A2 + 9A – 4I = 0 Post multiplying by A-1 both side (A3 – 6A2 + 9A – 4I ) A-1 = 0.A-1 A3 . A-1 – 6A2 . A + 9AA-1 – 4IA-1 = 0 A2 . AA-1 – 6A. A-1 A + 9AA-1 – 4IA-1 = 0 A2 I – 6AI + 9I – 4IA-1 = 0 A2 – 6A + 9I – 4A-1 = 0 4A-1 = A2 – 6A + 9I A-1 = 14 (A2 – 6A + 9I) Putting value A-1 = 14 6−55−56−55−56 − 6 2−11−12−11−12 + 9 100010001 = 14 6−55−56−55−56 − (6)26(−1)6(1)(6)(−1)6(2)6(−1)6(1)6(−1)6(2)+ 9(1)0009(1)0009(1) = 14 6−55−56−55−56 − 12−66−612−66−612 + 900090009 = 14 6−12+9−5+6+05−6+0−5+6+06−12+9−5+6+05−6+0−5+6+06−12+9 = 14 31−1131−113 Hence, A – 1 = 𝟏𝟒 𝟑𝟏−𝟏𝟏𝟑𝟏−𝟏𝟏𝟑 | 2,858 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-22 | longest | en | 0.362734 |
https://math.stackexchange.com/questions/2149129/lattice-divisors-of-150-ordered-by-divisibility-draw-hasse-diagram-get-comple | 1,623,976,709,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00329.warc.gz | 349,044,238 | 37,597 | # Lattice divisors of 150, ordered by divisibility. Draw Hasse diagram, get complements, check if distributive, check if Boolean.
I have the following exercise:
Having the lattice $$D_{150}$$ of the divisors of 150, ordered by divisibility
1) Draw Hasse Diagram;
2) Find all complements;
3) Check if it's a distributive lattice;
4) Check if it's a Boolean lattice.
My development:
The divisors are $$D_{150} = \left \{1,2,3,5,6,10,15,25,30,50,75,150 \right \}$$
point 1)
I have tried to draw the Hasse diagram in this way:
I have also realized that I can draw two cube structures like the following, and I don't know if this fact can help to resolve the exercise:
point 2)
assuming that I have done right the diagram, I have found some complements, but, not every elements have a complement:
$$\begin{array}{c|c} \, & 1 & 2 & 3 & 5 & 6 & 10 & 15 & 25 & 30 & 50 & 75 & 150 \\ \hline 1 & - & - & - & - & - & - & - & - & - & - & - & * \\ \hline 2 & - & - & - & - & - & - & - & - & - & - & * & - \\ \hline 3 & - & - & - & - & - & - & - & - & - & * & - & - \\ \hline 5 & - & - & - & - & - & - & - & - & - & - & - & - \\ \hline 6 & - & - & - & - & - & - & - & * & - & - & - & - \\ \hline 10 & - & - & - & - & - & - & - & - & - & - & - & - \\ \hline 15 & - & - & - & - & - & - & - & - & - & - & - & - \\ \hline 25 & - & - & - & - & * & - & - & - & - & - & - & - \\ \hline 30 & - & - & - & - & - & - & - & - & - & - & - & - \\ \hline 50 & - & - & * & - & - & - & - & - & - & - & - & - \\ \hline 75 & - & * & - & - & - & - & - & - & - & - & - & - \\ \hline 150 & * & - & - & - & - & - & - & - & - & - & - & - \\ \end{array}$$
point 3)
I think it is distributive because I can't find any sublattice like the following nondistributive twos:
point 4)
based on what I have found, since not every element has a complement, then it isn't a Boolean lattice. Or does exists any isomorphism?
please, can you tell what do you think? Many thanks!
Moreover, that they are Boolean iff the number is square-free. In this case, $25|150$, and $25=5^2$, so $150$ is not square-free and $D_{150}$ is not Boolean.
Thus, if $D_{150}$ is distributive and not Boolean, it is not complemented. | 779 | 2,173 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-25 | latest | en | 0.753962 |
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4-79.
By mistake, Jim graphed $y=x^3−4x$ instead of $y=x^3−4x+6$. What should he do to his graph to get the correct one?
Notice the difference in the equations. What effect will that have on the graph? | 86 | 260 | {"found_math": true, "script_math_tex": 2, "script_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-2024-22 | latest | en | 0.909991 |
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posted by .
A 0.077-kg arrow is fired horizontally. The bowstring exerts an average force of 50 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?
• physics -
The work done by the bow string, F * X = 50 x 0.75 Joules,
equals the kinetic energy (MV^2/2) of the arrow as it leaves the bow.
Solve for V
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A 77 g arrow is fired from a bow whose string exerts an average force of 95 N on the arrow over a distance of 81 cm. What is the speed of the arrow as it leaves the bow?
7. ### Physics
A 0.069-kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.95 m. With what speed does the arrow leave the bow?
8. ### Physics
A 0.069-kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.95 m. With what speed does the arrow leave the bow?
9. ### Physics
A 75 g arrow is fired horizontally. the bow string exerts an average force of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow string?
10. ### Physics
A 75 g arrow is fired horizontally. The bow string exerts an average of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow string?
More Similar Questions | 616 | 2,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-34 | latest | en | 0.886529 |
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## Discussion of Problem 1876. Centipede's Morning
Understanding the solution
Posted by MARAZ MIA 22 Feb 2020 03:06
After so many calculation and math I have solved the problem.....
Here we can have two worst cases...
Case 1:
having all the right shoes first.so here needed time is 2*b and we have now all the left shoes remaining...so total time is 2*b+40...
Case 2:
we may have 39 right shoes so time needed here is (39*2=78)...then we have only one right foot left but we may encounter all the left shoes and here needed time is 40+2*(a-40)....> 40 for the first 40 shoes and 2*(a-40) is for the remaining shoes as they needed to be thrown away...then we have the only one right foot left and it need 1 second...
so total time = 78+40+2*(a-40)+1 = 119+2*a-80 = 2*a-39
ans=max(Case 1,Case 2)
Edited by author 22.02.2020 03:07 | 309 | 1,068 | {"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} | 3.765625 | 4 | CC-MAIN-2020-16 | latest | en | 0.880598 |
https://www.cpalms.org/Public/PreviewStandard/Preview/5451 | 1,695,716,770,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00202.warc.gz | 793,504,116 | 20,234 | # MAFS.6.EE.2.6Archived Standard
Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
General Information
Subject Area: Mathematics
Domain-Subdomain: Expressions & Equations
Cluster: Level 3: Strategic Thinking & Complex Reasoning
Cluster: Reason about and solve one-variable equations and inequalities. (Major Cluster) -
Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters.
Date of Last Rating: 02/14
Status: State Board Approved - Archived
Assessed: Yes
Test Item Specifications
• Assessment Limits :
Numbers in items should not require students to perform operations with negative rational numbers or result in answers with negative rational numbers. Expressions must contain at least one variable.
• Calculator :
No
• Context :
Allowable
Sample Test Items (1)
## Related Courses
This benchmark is part of these courses.
1205010: M/J Grade 6 Mathematics (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1205020: M/J Accelerated Mathematics Grade 6 (Specifically in versions: 2014 - 2015, 2015 - 2020, 2020 - 2022, 2022 and beyond (current))
1204000: M/J Foundational Skills in Mathematics 6-8 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7812015: Access M/J Grade 6 Mathematics (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2022, 2022 and beyond (current))
7912110: Fundamental Explorations in Mathematics 1 (Specifically in versions: 2013 - 2015, 2015 - 2017 (course terminated))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Formative Assessments
Writing Real-World Expressions:
Students are asked to use variables to write expressions that represent quantities described in context.
Type: Formative Assessment
Gavin’s Pocket:
Students are asked to interpret the significance of a variable and its possible values when given a variable expression in a real-world context.
Type: Formative Assessment
Inventing X:
Students are asked to write and explain a real-world situation to accompany an algebraic expression.
Type: Formative Assessment
## Lesson Plans
Gather Data For Distribution by Programming an App:
This lesson allow students to gather, calculate, and plot data using both computer code and mathematical equations. In this lesson students will create a pedometer app to demonstrate the understanding of algorithms, components (such as buttons, textboxes, sensors, etc.), and If/Then statements. This lesson uses algebraic equations and random data to access the needed components to store data in a spreadsheet.
Type: Lesson Plan
Data Sets Represented in Computers:
This lesson shows how data can be represented by computers, in relation to everyday activities we may not be aware that we use computer. It gives an overview of graphing data by creating a histogram based on population data. Using the data collected, students will get a chance to hand write code to show what structure is needed for computers to collect, analyze and distribute such data. This lesson is lesson 1 of the Data Set and Deviation Statistics Unit and bridges statistical concepts of data collection, graphing and analysis with programming a computer using coding language while reinforcing foundational algebraic skills.
Type: Lesson Plan
Different Bodies of Water:
This is the introductory lesson in a 4-lesson unit of study about different bodies of water, their characteristics, and how to translate natural language into computer language.
Type: Lesson Plan
Waste Not, Want Not:
Students will explore the excretory system through a variety of activities that include an inquiry lab, a reading excerpt, an engineering design challenge, and creating an infomercial. Students will build a simulated kidney and explore what factors improve the filtering ability. Students will analyze the data and formulate findings.
Type: Lesson Plan
Students will use a balance scale graphic organizer to solve for the unknown (variable) in addition and subtraction equations with one variable.
Type: Lesson Plan
Going The Distance:
This lesson provides a hands-on activity where students can apply solving one-step multiplication and division equations to a real-world problem. The lesson focuses on the relationship between distance, rate, and time. The students will also represent data on graphs and draw conclusions and make interpretations based on the graphs.
Type: Lesson Plan
How Much was Lunch?:
This lesson explores using substitution to solve one-step equations. Each real world problems involves discussion and students are expected to support their solutions.
Type: Lesson Plan
Bake Sale:
This lesson challenges student to develop and solve equations for mathematical and real world situations. Students will be encouraged to use the work backwards strategy using inverse operations to find a solution.
Type: Lesson Plan
Decoding Word Phrases-Translating verbal phrases to variable expressions:
This lesson is designed to help students decode word phrases and then translate them from word form into numerical form. It provides a resource, in the form of a foldable, that can be kept all year and used anytime the students needs to decode word phrases.
Type: Lesson Plan
Analyzing Polyhedra:
Students will construct several simple polyhedra, then count the number of faces, edges, and vertices. These data should suggest Euler's formula.
Type: Lesson Plan
Gummy vs. Gum (Number Pattern):
In this lesson from the Beacon Learning Center, students use gummy bears and sticks of gum to discover a number pattern and write an representation that describes it.
This lesson can be adapted for several grade levels and instead of writing an equation, students can write a "rule" or expression based on the pattern or relationship between number of pieces of gum and gummy bears. Consider scaffolding this activity to meet your grade-level needs.
Type: Lesson Plan
## Original Student Tutorials
Algebraic Expressions Part 2: Multiplication and Division:
Help Oscar translate written real-world descriptions of multiplication and division into algebraic expressions in this interactive tutorial.
This is part 2 of 3. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Algebraic Expressions Part 1: Addition and Subtraction:
Follow Oscar as he writes algebraic expressions of addition and subtraction about his new puppy Scooter in this interactive tutorial.
Type: Original Student Tutorial
Balancing the Machine:
Use models to solve balance problems on a space station in this interactive, math and science tutorial.
Type: Original Student Tutorial
## Perspectives Video: Professional/Enthusiast
Using Algebra to Program Robots and Microcontrollers:
There are 10 ways to use algebra to program a binary-counting circuit: fun and more fun.
Type: Perspectives Video: Professional/Enthusiast
Smiles:
In this online problem-solving challenge, students apply algebraic reasoning to determine the "costs" of individual types of faces from sums of frowns, smiles, and neutral faces. This page provides three pictorial problems involving solving systems of equations along with tips for thinking through the problem, the solution, and other similar problems.
Triangular Tables:
Students are asked to use a diagram or table to write an algebraic expression and use the expression to solve problems.
Firefighter Allocation:
In this task students are asked to write an equation to solve a real-world problem.
## Teaching Idea
True, False, and Open Sentences:
"Students first explore arithmetic sentences to decide whether they are true or false. The lesson then introduces students to sentences that are neither true nor false but are algebraic equations, also called open sentences, such as x + 3 = 7 or 2 x = 12." from Math Solutions.
Type: Teaching Idea
## Tutorials
How to Write Basic Expressions with Variables:
Learn how to write basic algebraic expressions.
Type: Tutorial
How to Write Expressions with Variables:
Learn how to write simple algebraic expressions.
Type: Tutorial
How to Write Basic Algebraic Expressions from Word Problems:
Learn how to write basic expressions with variables to portray situations described in word problems.
Type: Tutorial
What is a Variable?:
The focus here is understanding that a variable is just a symbol that can represent different values in an expression.
Type: Tutorial
## MFAS Formative Assessments
Gavin’s Pocket:
Students are asked to interpret the significance of a variable and its possible values when given a variable expression in a real-world context.
Inventing X:
Students are asked to write and explain a real-world situation to accompany an algebraic expression.
Writing Real-World Expressions:
Students are asked to use variables to write expressions that represent quantities described in context.
## Original Student Tutorials Science - Grades K-8
Balancing the Machine:
Use models to solve balance problems on a space station in this interactive, math and science tutorial.
## Original Student Tutorials Mathematics - Grades 6-8
Algebraic Expressions Part 1: Addition and Subtraction:
Follow Oscar as he writes algebraic expressions of addition and subtraction about his new puppy Scooter in this interactive tutorial.
Algebraic Expressions Part 2: Multiplication and Division:
Help Oscar translate written real-world descriptions of multiplication and division into algebraic expressions in this interactive tutorial.
This is part 2 of 3. Click below to open the other tutorials in this series.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Original Student Tutorials
Algebraic Expressions Part 2: Multiplication and Division:
Help Oscar translate written real-world descriptions of multiplication and division into algebraic expressions in this interactive tutorial.
This is part 2 of 3. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Algebraic Expressions Part 1: Addition and Subtraction:
Follow Oscar as he writes algebraic expressions of addition and subtraction about his new puppy Scooter in this interactive tutorial.
Type: Original Student Tutorial
Balancing the Machine:
Use models to solve balance problems on a space station in this interactive, math and science tutorial.
Type: Original Student Tutorial
Smiles:
In this online problem-solving challenge, students apply algebraic reasoning to determine the "costs" of individual types of faces from sums of frowns, smiles, and neutral faces. This page provides three pictorial problems involving solving systems of equations along with tips for thinking through the problem, the solution, and other similar problems.
Triangular Tables:
Students are asked to use a diagram or table to write an algebraic expression and use the expression to solve problems.
Firefighter Allocation:
In this task students are asked to write an equation to solve a real-world problem.
## Tutorials
How to Write Basic Expressions with Variables:
Learn how to write basic algebraic expressions.
Type: Tutorial
How to Write Expressions with Variables:
Learn how to write simple algebraic expressions.
Type: Tutorial
How to Write Basic Algebraic Expressions from Word Problems:
Learn how to write basic expressions with variables to portray situations described in word problems.
Type: Tutorial
What is a Variable?:
The focus here is understanding that a variable is just a symbol that can represent different values in an expression.
Type: Tutorial
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark.
Smiles:
In this online problem-solving challenge, students apply algebraic reasoning to determine the "costs" of individual types of faces from sums of frowns, smiles, and neutral faces. This page provides three pictorial problems involving solving systems of equations along with tips for thinking through the problem, the solution, and other similar problems. | 2,527 | 12,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-40 | latest | en | 0.821115 |
http://thelogiccafe.net/logic/2_2SLintro7.htm | 1,558,419,896,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256281.35/warc/CC-MAIN-20190521062300-20190521084300-00086.warc.gz | 207,466,711 | 4,315 | T2.2: 7 of 7
Good, let's review the connectives before moving on to adding names. You can also find this table in the Chapter 2 Reference (so you need not print it out now).
Connective Name Resulting Sentence Type Component Names Typical English Versions English Statement Symbolization in SL & Ampersand Conjunction Conjuncts "and", "both ... and ... " Agnes and Bob will attend law school. A&B > Horseshoe Conditional Antecedent, Consequent "if ... then ... " If Agnes attends, then Bob will. A>B ~ Tilde Negation Negate "it's not the case that", "not" Agnes will not attend law school. ~A v Wedge Disjunction Disjuncts "or", "either... or... " Either Agnes or Bob will attend law school. AvB = Triple Bar Biconditional Bicomponets1 "if and only if", "just in case" Agnes will attend law school just in case Bob will. A=B
Look carefully at this table to make sure you've caught the basics...then on to names and predicates as we try to disect the atomic sentences a bit .
Names and Predicates
We'll continue to think about connectives and compound sentences. But we can also symbolize the names that occur in sentences.
So far in this tutorial, we have symbolized simple sentences (like "Agnes will go to law school") with a single symbol.
A
But at other times we'll want to break an English sentence up to analyze it.
Start...
We'll use lower case letters as proper names of our symbolic language to stand for objects. So, we might take 'a' as a name of Agnes. We'll typically set out our definition of a proper name, an interpretation, this way:
a: Agnes
In such a case, we will say that 'a' refers to Agnes.
And we'll use upper case letters as predicates or to stand for kinds, properties, or just incomplete thoughts (corresponding to incomplete sentences). So, we could use 'L' to stand for "will go to law school". Because predicates are something like incomplete sentences, we'll write their interpretation with a blank:
L_: ___ will go to law school.
Then,
Agnes will go to law school
can be symbolized simply as
La
Yes, it's a little backward from the English. But we'll get used to it! You may want to read this as "Go to law school will Alice". But I suspect it's best to just read 'La' and say to yourself "Alice will go to law school."
In addition to representing predicates, we will somtimes still to use upper case letters to stand for simple English sentences when there are no names or predicates to do the job. You'll be able to tell from context.
Next...
'La' counts as a new sort of atomic sentence. Let's see what we can build with it.
1. Agnes will go to law school and Bob will go to law school.
Before we had names in our symbolic language, we would symbolize 1 simply as 'A&B'. If our interpretation so defines 'A' and 'B', then this is fine.
But when we have 'a' as a name for Agnes and adding 'b' as a name for Bob, we can symbolize 1 as
La&Lb
And
2. If Agnes will go to law school, then she will be miserable for the first year.
can be symbolized as:
La>Ma
(Hint: if this still looks confusing, enter a question mark in a field, i.e., in one of the boxes, then hit the TAB key to get the answer. For this little quiz, you'll be entering things like "La>Mb". AND you can learn alot by making mistakes and trying to correct them. If at first everything seems a little confusing, do the problems, get them wrong, redo and see if the material doesn't get clearer.)
a: Agnes, b: Bob
L_: ___ will go to law school.
M_: ___ will be miserable.
Don't forget that the greater than sign is for the horseshoe AND the '=' sign for the triple-bar.
1. Both Agnes and Bob will be miserable.
2. Bob will be miserable if and only if he goes to law school.
3. Either Bob or Agnes will go to law school.
4. Bob will go to law school only if Agnes will. (hint)
5. Bob will not be miserable.
6. Bob will not be a law student but he will be miserable.(hint)
7. Agnes will not be miserable and she will not be a law student.
= Symbolic (for Topic 3...enter in T3 Cafe Check) | 1,011 | 4,020 | {"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-2019-22 | latest | en | 0.920613 |
http://www.chegg.com/homework-help/questions-and-answers/yes-r-c-response-square-wave-input-t-100t-100-time-constant-states-set-dc-supply-0v-square-q2774020 | 1,472,505,795,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982965886.67/warc/CC-MAIN-20160823200925-00071-ip-10-153-172-175.ec2.internal.warc.gz | 372,674,740 | 13,675 | Yes, this is a R-C response to a square wave input for T = 100t, or 100 * time constant. It states set the dc supply to 0V and the square wave generator to 1000 Hz, and connect scope to point a. Adjust the amplitude to 10V (p-p). Draw the waveform for one period of the applied square wave. Determine the time constant, determine T and T/2 of the applied square wave, and calculate the value of Vc(t) at t=T/2 using the following equation: Vc(t) = Vp(1 - e^-t/RC). Thanks so much for any help. | 138 | 493 | {"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-2016-36 | latest | en | 0.840659 |
https://www.jiskha.com/display.cgi?id=1264442525 | 1,516,215,227,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886952.14/warc/CC-MAIN-20180117173312-20180117193312-00124.warc.gz | 945,782,870 | 3,746 | # math
posted by .
How many different ways can u make change for 75 cents using quarters, nickels and dimes?
• math -
I will give you some examples
7 dimes, one Nickel
6 dimes, three nickels
You take it from here.
• math -
Change OF 75 cents or change from a dollar (25 cents)?
In either case, I suggest you make a list instead of trying to derive a formula, or plugging into one we derive for you.
When you start making the list, perhaps a formula to use will become obvious.
## Similar Questions
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2. ### math
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3. ### math
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4. ### math
how many ways can you make 75 cents with quarters, dimes, and nickels
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7. ### Math
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More Similar Questions | 522 | 2,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-05 | latest | en | 0.943629 |
https://ebs.sakarya.edu.tr/Ders/Detay/557885 | 1,642,746,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00422.warc.gz | 264,585,775 | 21,022 | Ders Adı Kodu Yarıyıl T+U Saat Kredi AKTS
Engıne Thermodynamıcs OTO 535 0 3 + 0 3 6
Ön Koşul Dersleri Önerilen Seçmeli Dersler Dersin Dili Türkçe Dersin Seviyesi YUKSEK_LISANS Dersin Türü Zorunlu Dersin Koordinatörü Dr.Öğr.Üyesi GÖKHAN ERGEN Dersi Verenler Dersin Yardımcıları Dersin Kategorisi Diğer Dersin Amacı Perform thermodynamic analysis of internal combustion engines; integration to thermodynamic concepts and engines Dersin İçeriği 1. Introduction2. Basic thermodynamic concepts and definitions3. Laws of thermodynamic4. Internal combustion engines and thermodynamic evaluation5. Cycle analysis in internal combustion engines6. Fuels and combustion analysis in internal combustion engines
# Ders Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri
1 Knows the basic thermodynamic concepts Lecture, Question-Answer, Discussion, Testing, Homework,
2 Know and analyze the laws of thermodynamics Lecture, Question-Answer, Discussion, Drilland Practice, Problem Solving, Testing, Homework,
3 Calculates changing the basic state Lecture, Question-Answer, Discussion, Drilland Practice, Problem Solving, Testing, Homework,
4 Analize and calculate the thermodynamic cycle Lecture, Question-Answer, Discussion, Problem Solving, Testing, Homework,
5 Knows and practices the fuels used in internal combustion engines Lecture, Question-Answer, Discussion, Testing, Homework,
6 Measure the performance characteristics of internal combustion engines and reviews Lecture, Question-Answer, Discussion, Drilland Practice, Problem Solving, Testing, Homework,
7 knows the exhaust emissions from internal combustion engines and reviews Lecture, Question-Answer, Discussion, Testing, Homework,
Hafta Ders Konuları Ön Hazırlık
1 Introduction
2 Basic Concepts of Thermodynamics
3 Heat, Work, and the First Law of Thermodynamics
4 Second and Third Laws of Thermodynamics
5 Laws and Applications of Thermodynamics
6 Basic State Replacement Procedures and Theoretical Cycles
7 Theoretical Cycle Accounts and Applications
8 Real Engine Cycles
9 Real Cycle Accounts and Applications
10 Midterm Exam
11 Fuels and Combustion
12 Performance Characteristics of Internal Combustion Engines
13 Emission Formation in Internal Combustion Engines
14 Overall Rating
Kaynaklar
Ders Notu
Ders Kaynakları
Sıra Program Çıktıları Katkı Düzeyi
1 2 3 4 5
1 Ability to reach the knowledge expansion and depth by doing scientific research in the field of engineering, knowledge evaluation, interpretation and application skills X
2 To be able to develop strategies, policies and implementation plans in the field of automotive engineering and evaluate the results obtained within the framework of quality processes
3 Ability to observe social, scientific and ethical values at the stages of collecting, interpreting and announcing the data and at all professional activities X
4 Ability to manipulate and apply knowledge through scientific methods using existing data available, and the ability to integrate knowledge from different disciplines X
5 Ability to design engineering problems, to develop methods to solve and to apply innovative methods of solutions X
6 Ability to develop new and original ideas and methods; the ability to develop innovative solutions in system, component or process design X
7 Awareness of the new and developing practices of the profession; the ability to examine and learn when necessary X
8 Understanding the social and environmental dimensions of engineering practices and the ability to adapt to the social environment
9 Comprehensive information on modern techniques and methods applied in engineering and their boundaries X
10 The ability to transcribe the processes and outcomes of their work in a systematic and explicit way, either in writing or verbally, in the national and international contexts
Değerlendirme Sistemi
Yarıyıl Çalışmaları Katkı Oranı
1. Ara Sınav 60
1. Kısa Sınav 10
1. Ödev 20
2. Kısa Sınav 10
Toplam 100
1. Yıl İçinin Başarıya 50
1. Final 50
Toplam 100
AKTS - İş Yükü Etkinlik Sayı Süre (Saat) Toplam İş Yükü (Saat)
Course Duration (Including the exam week: 16x Total course hours) 16 3 48
Hours for off-the-classroom study (Pre-study, practice) 16 4 64
Mid-terms 1 5 5
Quiz 2 10 20
Assignment 1 15 15
Final examination 1 2 2
Toplam İş Yükü 154
Toplam İş Yükü / 25 (Saat) 6,16
Dersin AKTS Kredisi 6 | 1,020 | 4,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | latest | en | 0.681803 |
https://us.metamath.org/mpeuni/sneqrg.html | 1,702,136,147,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00292.warc.gz | 637,930,949 | 3,721 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sneqrg Structured version Visualization version GIF version
Theorem sneqrg 4683
Description: Closed form of sneqr 4684. (Contributed by Scott Fenton, 1-Apr-2011.) (Proof shortened by JJ, 23-Jul-2021.)
Assertion
Ref Expression
sneqrg (𝐴𝑉 → ({𝐴} = {𝐵} → 𝐴 = 𝐵))
Proof of Theorem sneqrg
StepHypRef Expression
1 snidg 4510 . . 3 (𝐴𝑉𝐴 ∈ {𝐴})
2 eleq2 2873 . . 3 ({𝐴} = {𝐵} → (𝐴 ∈ {𝐴} ↔ 𝐴 ∈ {𝐵}))
31, 2syl5ibcom 246 . 2 (𝐴𝑉 → ({𝐴} = {𝐵} → 𝐴 ∈ {𝐵}))
4 elsng 4492 . 2 (𝐴𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
53, 4sylibd 240 1 (𝐴𝑉 → ({𝐴} = {𝐵} → 𝐴 = 𝐵))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1525 ∈ wcel 2083 {csn 4478 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1781 ax-4 1795 ax-5 1892 ax-6 1951 ax-7 1996 ax-8 2085 ax-9 2093 ax-10 2114 ax-11 2128 ax-12 2143 ax-ext 2771 This theorem depends on definitions: df-bi 208 df-an 397 df-or 843 df-tru 1528 df-ex 1766 df-nf 1770 df-sb 2045 df-clab 2778 df-cleq 2790 df-clel 2865 df-nfc 2937 df-sn 4479 This theorem is referenced by: sneqr 4684 sneqbg 4687 preimane 30101 altopth1 33037 altopth2 33038
Copyright terms: Public domain W3C validator | 663 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-50 | longest | en | 0.21208 |
http://forum.allaboutcircuits.com/threads/tank-circuit-equation.13222/ | 1,485,279,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00440-ip-10-171-10-70.ec2.internal.warc.gz | 114,652,497 | 14,559 | # Tank Circuit Equation
Discussion in 'Math' started by The Lightning Stalker, Jul 24, 2008.
1. ### The Lightning Stalker Thread Starter New Member
Jul 24, 2008
2
0
Hello. I'm trying to isolate L to find out the needed inductance based on resonance frequency and capacitance. I can't remember how to get rid of the square root when it's on the bottom of a fraction.
Robert
2. ### Wendy Moderator
Mar 24, 2008
20,772
2,540
You square both sides.
3. ### The Lightning Stalker Thread Starter New Member
Jul 24, 2008
2
0
Okay, but what parts do I square, I'm not sure. Also I have no idea how to get rid of the 1 on top of the fraction. It's been a long time since I had to do an equation like this.
Thank you,
Robert
4. ### Mark44 Well-Known Member
Nov 26, 2007
626
1
I'm assuming you are trying to solve an equation. As Bill said, you can square both sides of the equation, which will get rid of the radical in the denominator. We can't give you any more advice, since we don't know what the equation you're working with is. If you post it, we can give you some more help.
5. ### PinkDalek Member
Dec 18, 2005
10
0
Hi Robert, I've hand drawn the formulas for you, I believe this is probably what you want, but please take care when you use them as it's very easy to make a mistake, I would suggest you do a practice calculation using values for L & C and find the frequency and then apply the frequency value to the other formulas with either known value of L or C and when you get the values correct you will know you've used the correct method of working out.
All the best,
Lorraine
• ###### resonance.jpg
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Last edited: Jul 25, 2008
6. ### PinkDalek Member
Dec 18, 2005
10
0
With regard to my previous post it has occured to me that non members or guests cannot view attached images, so here's a simple text version of the formulas for our guests.
F=1/(2*PI*sqrt(L*C))
L=1/(4*PI*PI*F*F*C)
C=1/(4*PI*PI*F*F*L)
Where F=frequency in Hertz, L=inductance in Henries, C=capacitance in Farads and PI=$\pi$ which is 3.1415927 correct to 7 decimal places.
I hope that helps.
best wishes,
Lorraine
Last edited: Jul 27, 2008 | 614 | 2,164 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/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-2017-04 | latest | en | 0.933029 |
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