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5100 | https://pmc.ncbi.nlm.nih.gov/articles/PMC1152624/ | Anomeric specificity of hexokinase and glucokinase activities in liver and insulin-producing cells - PMC
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Biochem J
. 1985 Sep 1;230(2):345–351. doi: 10.1042/bj2300345
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Anomeric specificity of hexokinase and glucokinase activities in liver and insulin-producing cells.
A Sener
A Sener
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, M H Giroix
M H Giroix
Find articles by M H Giroix
, S P Dufrane
S P Dufrane
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, W J Malaisse
W J Malaisse
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PMCID: PMC1152624 PMID: 3902008
Abstract
Conflicting data have been reported concerning the anomeric specificity of glucokinase. In the present study, liver hexokinase (Km for D-glucose 0.4 mM) displayed a higher affinity for but lower Vmax. with alpha- than with beta-D-glucose. The velocity of the reaction catalysed by liver glucokinase was higher with with beta- than with alpha-D-glucose, whatever the glucose concentration. The apparent Km of glucokinase was somewhat lower, however, with alpha- than with beta-D-glucose. Comparable results were obtained for the high-Km glucokinase-like enzymic activity present in normal pancreatic islets or insulin-producing tumoral cells. These results suggest that the anomeric specificity of glucokinase cannot account for the higher rate of glycolysis found in islets exposed to alpha- as distinct from beta-D-glucose.
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5101 | https://amesweb.info/section/second-moment-of-area-calculator.aspx | SECOND MOMENT OF AREA (AREA MOMENT OF INERTIA) CALCULATOR
Second Moment of Area Calculator for I beam, T section, rectangle, c channel, hollow rectangle, round bar and unequal angle.
Second Moment of Area is defined as the
capacity of a cross-section to resist bending.
| | | |
---
| Sectional properties of I-beam | | |
| Beam Cross Section | I Beam Hollow Rectangle Rectangle C Channel T Section Circular Hollow Section Round Bar Angle | |
| Unit System (Quick selection) | | | | Metric | | Inch | | |
| INPUT PARAMETERS | | |
| Parameter | Value | |
| Flange-flange inner face height [H] | | mm cm m inch ft |
| Width [B] | |
| Flange thickness [h] | |
| Web thickness [b] | |
| | | |
| |
| Metric |
| Inch |
| | | |
---
| RESULTS | | |
| Parameter | Value | |
| Second moment of area [Ixx] mm^4 inch^4 cm^4 ft^4 |
| Second moment of area [Iyy]
Note: Use dot "." as decimal separator.
Second Moment of Area Formula:
| | |
--- |
| I Beam Area Moment of Inertia Formula | |
| Sectional properties of I-beam | |
| Parameter | Equation |
| Area moment of inertia | Ixx = H3b/12 + 2[h3B/12 + hB(H+h)2/4] |
| Area moment of inertia | Iyy = b3H/12 + 2(B3h/12) |
| | |
--- |
| Hollow Rectangle Area Moment of Inertia Formula | |
| Sectional properties of hollow rectangle | |
| Parameter | Equation |
| Area moment of inertia | Ixx = BH3/12 - bh3/12 |
| Area moment of inertia | Iyy = HB3/12 - hb3/12 |
| | |
--- |
| Rectangle Area Moment of Inertia Formula | |
| Sectional properties of hollow rectangle | |
| Parameter | Equation |
| Area moment of inertia | Ixx = BH3/12 |
| Area moment of inertia | Iyy = HB3/12 |
| | |
--- |
| C Channel Area Moment of Inertia Formula | |
| Sectional properties of C-beam | |
| Parameter | Equation |
| Area moment of inertia | Ixx = H3b/12 + 2[h3B/12 + hB(h+H)2/4] |
| Area moment of inertia | Iyy = b3H/12 + bH(xc-b/2)2+ 2B3h/12+ 2Bh(xc - B/2)2 |
| Centroid distance | xc = (2hB2/2 + b2H/2)/A |
| Centroid distance | yc= H/2 + h |
| | |
--- |
| T Section Area Moment of Inertia Formula | |
| Section properties of T-beam | |
| Symbol | Equation |
| Area moment of inertia | Ixx = bH(yc-H/2)2 + bH3/12 + hB(H + h/2 - yc)2 + h3B/12 |
| Area moment of inertia | Iyy = b3H/12 + B3h/12 |
| Centroid distance | xc = B/2 |
| Centroid distance | yc= [(H+h/2)hB+H2b/2]/A |
| | |
--- |
| Hollow Circle Area Moment of Inertia Formula | |
| Section properties of hollow circle (shaft) | |
| Parameter | Equation |
| Area moment of inertia | Ixx = πd4/64 - πd14/64 |
| Area moment of inertia | Iyy = πd4/64 - πd14/64 |
| | |
--- |
| Round Bar Area Moment of Inertia Formula | |
| Section properties of round bar | |
| Parameter | Equation |
| Area moment of inertia | Ixx = πd4/64 |
| Area moment of inertia | Iyy = πd4/64 |
| | |
--- |
| Angle with Unequal Legs Area Moment of Inertia Formula | |
| Sectional properties of angle with unequal legs | |
| Symbol | Equation |
| Area moment of inertia | Ixx = 1/3[bd3 - (b-t) (d-t)3] - A (d - yc)2 |
| Area moment of inertia | Iyy = 1/3[db3 - (d-t) (b-t)3] - A (b - xc)2 |
| Centroid distance | xc=(b2+dt-t2)/(2(b+d-t)) |
| Centroid distance | yc=(d2+bt-t2)/(2(b+d-t)) |
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5102 | https://askfilo.com/user-question-answers-smart-solutions/11-23-61-a-table-has-4-legs-a-chair-has-4-legs-and-a-stool-3335303430323036 | Question asked by Filo student
11:23 61. A table has 4 legs, a chair has 4 legs, and a stool has 3 legs. If a room has 5 tables, 10 chairs, and 7 stools, how many legs are there in total? 87 93 97 102 62. Which word does not belong? Chair Table Bed Dog 63. If a train travels at 80 km/h and a car travels at 100 km/h, how much farther will the car travel in 3 hours? 20 km 40 km 60 km
Views: 5,514 students
Updated on: May 28, 2025
Text SolutionText solutionverified iconVerified
Explanation
Below are the step-by-step solutions for each problem. Each question covers basic reasoning, arithmetic, or logic skills suitable for elementary/primary grade level.
Question 61
Concepts
arithmetic, multiplication, addition
Explanation
To find the total number of legs, multiply the number of each object by its number of legs, then add them all together.
Step-By-Step Solution
Step 1
Calculate the number of legs for tables:
Number of tables = 5
Legs per table = 4
5×4=20
Step 2
Calculate the number of legs for chairs:
Number of chairs = 10
Legs per chair = 4
10×4=40
Step 3
Calculate the number of legs for stools:
Number of stools = 7
Legs per stool = 3
7×3=21
Step 4
Add up all the legs:
20+40+21=81
Step 5
Double check options. Since 81 is not an option, let's re-calculate:
5×4=20
10×4=40
7×3=21
20+40=60
60+21=81
There seems to be a mismatch. Let's check for misreading: all the values are correct, but 81 is not one of the options.
Given the options 87, 93, 97, 102, the closest is 81, but since none are present, but mathematically:
Step 6
Confirm calculations:
5×4=20
10×4=40
7×3=21
20+40+21=81
Final Answer
81 (Closest option is 87, but this seems to be a mistake in the options. The correct computed answer is 81 )
Question 62
Concepts
classification, logic
Explanation
Three of the words are types of furniture; one is an animal. The word that does not belong is the one that is not furniture.
Step-By-Step Solution
Step 1
Chair, Table, and Bed are all furniture.
Step 2
Dog is an animal, not furniture.
Final Answer
Dog
Question 63
Concepts
speed, distance, subtraction
Explanation
Find the distance each vehicle travels in 3 hours, and subtract to find how much farther the car travels.
Step-By-Step Solution
Step 1
Distance travelled by train in 3 hours:
80 km/h×3 h=240 km
Step 2
Distance travelled by car in 3 hours:
100 km/h×3 h=300 km
Step 3
Difference in distance:
300 km−240 km=60 km
Final Answer
60 km
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| | |
--- |
| Question Text | 11:23 61. A table has 4 legs, a chair has 4 legs, and a stool has 3 legs. If a room has 5 tables, 10 chairs, and 7 stools, how many legs are there in total? 87 93 97 102 62. Which word does not belong? Chair Table Bed Dog 63. If a train travels at 80 km/h and a car travels at 100 km/h, how much farther will the car travel in 3 hours? 20 km 40 km 60 km |
| Updated On | May 28, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 11 |
| Answer Type | Text solution:1 |
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5103 | https://homepages.abdn.ac.uk/h.tan/pages/teaching/computation2/demo/Wikipedia,%20the%20free%20encyclopedia.htm | Bisection method - Wikipedia, the free encyclopedia
Bisection method
From Wikipedia, the free encyclopedia
Jump to: navigation, search
A few steps of the bisection method applied over the starting range [a 1;b 1]. The bigger red dot is the root of the function.
In mathematics, the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists.
Suppose we want to solve the equation f(x) = 0. Given two points a and b such that f(a) and f(b) have opposite signs, we know by the intermediate value theorem that f must have at least one root in the interval [a, b] as long as f is continuous on this interval. The bisection method divides the interval in two by computing c = (a+b) / 2. There are now two possibilities: either f(a) and f(c) have opposite signs, or f(c) and f(b) have opposite signs. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs.
The bisection method is less efficient than Newton's method but is not prone to the same odd behavior.
Contents
[hide]
1 Analysis
2 Pseudo-code
3 See also
4 References
5 External links
[edit]Analysis
If f is a continuous function on the interval [a, b] and f(a)f(b) < 0, then the bisection method converges to a root of f. In fact, the absolute error is halved at each step. Thus, the method converges linearly, which is quite slow.We can find the asymptotic error constant is 1/2. This is because .On the other aspect, since the absolute error of bisection convergence is , the rate of convergence is which means the decrease of error is exponential. This means the convergence speed actually is not so bad.On the positive side, the method is guaranteed to converge if f(a) and f(b) have different signs.
The bisection method gives only a range where the root exists, rather than a single estimate for the root's location. Without using any other information, the best estimate for the location of the root is the midpoint of the range. In that case, the absolute error after n steps is at most
If either endpoint of the interval is used, then the maximum absolute error is
the entire length of the interval.
If f has several roots in the interval [a, b], then the bisection method finds the odd-numbered roots with equal, non-zero probability and the even-numbered roots with zero probability. More precisely, suppose that f has 2 k + 1 simple roots x 1<x 2< … <x 2 k+1 in the interval [a, b] (the number of roots is odd because f(a) and f(b) have opposite signs). Assume that the roots are distributed independently and uniformly in this interval. Then, the probability that the bisection method converges to the root x i with i = 1, 2, …, 2 k + 1 is zero if i is even and 1 / (k + 1) if i is odd (Corliss 1977).
[edit]Pseudo-code
Here is a representation of the bisection method in Visual Basic code. The variables left and right correspond to a and b above. The initial left and right must be chosen so that f(left) and f(right) are of opposite sign (they 'bracket' a root). The variable epsilon specifies how precise the result will be.
'Bisection Method
'Start loop
Do While (abs(right - left) > 2epsilon)
'Calculate midpoint of domain
midpoint = (right + left) / 2
'Find f(midpoint)
If ((f(left) f(midpoint)) > 0) Then
'Throw away left half
left = midpoint
Else
'Throw away right half
right = midpoint
End If
Loop
Return (right + left) / 2
[edit]See also
Binary search algorithm
Lehmer-Schur algorithm, generalization of the bisection method in the complex plane
[edit]References
Burden, Richard L. & Faires, J. Douglas (2000), Numerical Analysis (7th ed.), Brooks/Cole, ISBN 978-0-534-38216-2.
Corliss, George (1977), "Which root does the bisection algorithm find?", SIAM Review19(2): 325–327, ISSN1095-7200.
[edit]External links
Bisection Method on Mathcad Application Server.
Bisection Method Notes, PPT, Mathcad, Maple, Matlab, Mathematica from Holistic Numerical Methods Institute
Module for the Bisection Method by John H. Mathews
Java Code for Bisection Method by Behzad Torkian
True example of using bisection method in computer programming free program to isoelectric point calculation
Retrieved from "
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5104 | https://www.youtube.com/watch?v=_Ffcr98c7EE | Multi-variable Optimization & the Second Derivative Test
Dr. Trefor Bazett
518000 subscribers
3623 likes
Description
121842 views
Posted: 24 Nov 2019
Finding Maximums and Minimums of multi-variable functions works pretty similar to single variable functions. First,find candidates for maximums/minimums by finding critical points. Critical Points are where the partial derivatives with respect to x and y are both zero. Then we classify each critical point using the second derivative test. In the multivariable case, there is a new option beyond max/min/neither, there is also the case of the saddle point. The second derivative test involves computing the Hessian, the determinant of a matrix that helps decide whether points are maximums/minimums/saddle or inconclusive. We sketch the geometric intuition behind the Hessian.
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118 comments
Transcript:
Introduction in this video we're gonna study optimization if I have some particular function that depends on an x and y how do I find where this function is a Mac so how do I find where it has a minimum and sometimes how do I find where has this interesting new phenomenon called a saddle in the single variable case there were actually three different major possibilities the first of them was you have a minimum so some point for example a function x squared and it would go up at the value of x equal to 0 you would have a minimum location the derivative was 0 you had a horizontal tangent line but function appear to sort of bought an out and flatten off at that point likewise we had maximums and also a third case where it appears at the derivative was 0 you had a horizontal tangent line at the value of x equal to 0 but in this third case an example of which is someone like Y it would be X cubed but it was neither a maximum or a minimum it went up and I leveled off and then went other day no the first thing to say is all three of these scenarios are peep themselves in the multi variable calculus case you can have a minimum you can have a maximum and that neither situation can also occur you can be going not flattening off and then going up again but there's a new situation that can happen in the multi variable calculus case that couldn't happen before and that's the idea of a saddle when I have a saddle when I look at one of those directions for example I've plugged away if I go along the x axis I get this yellow curve then it appears to be a minimum according to the yellow curve but if you go parallel now to the y axis at the point 0 0 you get this blue curve which is going downwards and then it appears to be a maximum the idealist saddle is that in one direction it appears to be a maximum the other appears to be a minimum when you both those properties you get a saddle and single variable isn't relevant cuz you only ever had one direction that you could go now these two curves the yellow and blue ones be yellow one is what happens when I just plug in the value of y it would is around if I have a multivariable function f of X & Y and I plug in y producer I know these single variable function and you can see its graph likewise for the blue one if I plug in now this time X equal to zero I also get a single variable function f of zero because there are single variable functions I can use all the analysis from first or calculus but where to find that something was a minimum you took the derivative set it equal to 0 and those gave you candidates to be minimums maximums or 9 so in both of these cases I can pick the derivative with respect to X of the yellow curve or the derivative with respect to Y of the blue curve and both of those are derivatives of single variable functions now one of the big features of single variable calculus was just because the derivative was 0 didn't tell you it was a max or a minimum or neither Q had to do more and else beyond that but setting the derivatives equal to 0 gives you candidates to be maximums and candidates three minimums in the same exact story it may be true here is just that when I take the derivatives of these yellow and blue curves the derivatives say of f of X comma 0 that is going to turn into a partial derivative in the multivariable case so this is my statement of the First Derivative Test so-called first derivative test it says if you have a local Max or min at a point then the following must be true it must be the true that both of the partials are 0 in that particular point so this is the exact analog of the previous theorem in single variable calculus saying that if they function head of maximum at one point the derivative had to be 0 at that particular point now let's see an example of this so I need to give the parabola x squared plus y squared so I want to go and figure out well where is the parcel throat is equal to 0 where is my candidates to be a maximum or a minimum well you take the first part so privileged is to exit that serum like you do the next is equal to 0 taking the second partial derivative this is going to do to wife take it respect to Y it said that it would is really good wide loser just because I tell you that the partials are both so doesn't tell you it was a maximum or a minimum or neither we have to investigate a little bit further but in this particular case if I just look at what the function is well the function is always positive or equal to zero of this one spot zero zero so that one spot zero zero discovered has to be a minimum because the thumb just always bigger around it but notice how I'm doing a bit of an ad arguing here I had to go and investigate the certain specifics of the function you changed function and this analysis might be harder or smaller it's easy to find candidates it's easy to find candidates to be maxims or minimums that the partial is equal to zero but how do I really know that they're going to be maximums or minimums in a general case where I can't just do this easy little bit of analysis but for that we have the analog of this second derivative test and Europeans it's a lot too much so we Second Derivative Test have to unpack it basically what it says in the following first of all assume that your functions are nice so what an Imagi first and second partial derivatives all are going to exist in there it would exist in some open disk around your point so you've got some nicely functions then it gives you conditions to tell you whether you've got a max a min a saddle or when the test fails and is inconclusive and you cannot use the test so the test is not completed tells you sometimes you can use this test and sometimes it's inconclusive but a good number of times we can use the test so it's worthwhile knowing okay the conditions here are a little bit weird there's something about the second derivative at X X being negative being positive but then there's also this weird expression in pink the second partial respect to x times the second partial respect to Y minus the square of the mixed partials F XY and whether this is greater than zero or less what's going on so let's try to break it down piece by piece first I'm gonna study that's an example of a minimum here now what I've highlighted in yellow here is the curve when I plug in y equal to zero so I have f of X zero then the second derivative of that curve tells me the concavity of it if the second derivative is positive it tells me that this yellow curve is facing off it's concave if the yellow curve here is concave down the second derivative will be negative indeed we use this analysis of concavity in single variable calculus to tell whether or not our particular point that was a candidate remax or a minimum was a maxima or minima and so indeed it makes a lot of sense that we demand that the second partial is going to be positive at this point down at the bottom is going to be a minimum and likewise if it's going to be a maximum you'd expect the exact opposite that the second partial has to be negative which means it's concave down however studying just the person with the X is not enough you might think okay well why don't you go and also look at the second partial factor Y so if you go the other direction if you fix your x value or Y is changing look at the second derivative that is a concave up concave down so that is actually done a little bit in the middle of the pink expression the pink expression is where FY y first appears but it's a little bit complex so let's switch to the third exam area that could happen to see why we might have this let's look now at the stat well in the case of the saddle what we have is that my our second partial respective X is positive but my second partials but 2y is negative so this means it's sort of concave up in one way concave down the other consistent with my notion of a saddle point then if I go and look at what the product of these two things are the partial of F respective X twice the partial of Y with about two y twice the product of those things is sort of a code for whether the two partial derivatives have the same sign or opposite sign so for instance in the case when they have opposite sign like one is concave up in the one direction and concave down in the other so positive second derivative the one Russian negative second derivative the other then the product FX and at YY will be negative since we're then subtracting something square is always positive in the case of a saddle what you have is that if these mixed partials or opposite directions for sure it's going to be a saddle so we think it's so far we justify why we have the FX X being less than zero and greater than zero with differential between maxims minimums and we've also seen the sort of first part of my pig expression the F xx times FY why why the sign of that might be important for telling us whether or not we have a saddle but what about the other part of this expression consider the following weird example ok so this is a new example it is a weird type of saddle it's actually the function x squared plus y squared minus 3x y know if I go along here and I specify that I am on the x axis I plug in Y for the 0 then what I just get is the curve x squared that's what I've highlighted in yellow and x squared its second partial derivative is just 2 which is positive it's concave up as you go along the x-axis alright but what happens when you go along the y-axis well I lose the exact same thing if you're going along the y axis of this now it's also concave up indeed the function is y squared along the y axis and its second derivative expected Y is 2 so whether you all on the X whether you go on the Y is concave up in both directions which might make you think that it should be a minimum and yet if I choose this somewhat clever third what if I go along the path y equal to X map so I'm here I'm restricting to y equal to X and if I plug it back in then what I get out of this is x squared plus x squared minus 3x squared I get minus x squared minus x squared is concave down so basically when I want to be x axis with concave up going on the y axis is concave up but going along y equal to X it was concave down now watch what happens to the mixed partial the f sub x y well first of all I'm going to say that that's taking the first version but the Y of I pick or soap derivatives for 2x 2x minus 3y I take that parcel and I get the value of minus 3 when you swear that you're gonna get nine which is just much bigger than the two times two you have for FXX times FY y so in this case considering that FX y the so-called mixed partial squared even allowed us to see that this was a saddle even though the analysis along x and y respectively could not see that with a saddle indeed it can be going down in some direction that's not the X and the y axis so the way to think about the mixed partial is that as I change my access I in for example parts of respect to X partial derivative with respect to Y now as I studied that it changes as I move off of the x axis and indeed you might recall that F X Y was equal to my ex so likewise you can study the partial respect to Y and then see how that changes if you move the X which is now moving off of the y axis so the mixed partials are a way of figuring out what happens when we're not just constrained to the axes that's what FXX I have to why why does the mixed partials tell us well how big are these changes if I go off a little on X and then I globe along Y that captures changes for example in y equal to X which is not along either of the axes and so if this you change X and then as you change Y which is picking you off the axes if that change is very large then the FX Y squared is gonna be very large and indeed then you can have a saddle even way along both axes it appears to be a maximum or appear to be a minimum now exactly proving this that the grotesque feel on the intuition that I've given you is beyond the scope of this particular video I can tell you that it comes from the generalization of Taylor's theorem we've seen in single variable calculus to the multivariable calculus case and indeed we're gonna have this expression is related to something called the hestian and that will be our way to formally demonstrate that these conditions are valid ways to decide maximum or minimum or a sample from the Conclusion computational perspective our method is now relatively straightforward you find all the candidates by the first derivative test that is you go along and you figure out where is the partial present at X equal to 0 partial reified to y 0 and when both of those occur then the second derivative test gives us the conditions to classify those when is it the case is gonna be a maximum a minimum a saddle or unfortunately sometimes the test fails and we have to go further we have to do more analysis that we can get from just this test if you have a question about this video leave it down in the comments below real mathematicians here we appreciate algorithms so let's just hope the YouTube l know about by giving this video a like and finally if you want to watch more multivariable calculus videos this video is part of a larger playlist a multivariable calculus so you can check out those videos here and we'll do some more math in the next video |
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5106 | https://www.youtube.com/watch?v=Rw70zkvqEiE | Calculus 3 - Vector Projections & Orthogonal Components
The Organic Chemistry Tutor
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This calculus 3 video tutorial explains how to find the vector projection of u onto v using the dot product and how to find the vector component of u orthogonal to v. W1 is the component of u parallel to v and w2 is the component of u perpendicular to v.
3D Coordinate System:
3D Distance Formula:
Equation of a 3D Sphere:
Calculus 3 - Intro to Vectors:
Calculus 3 - The Dot Product:
Angle Between Two Vectors:
Parallel & Orthogonal Vectors:
Direction Cosines and Vectors:
Calculus 3 - Vector Projections:
Cross Product of 2 Vectors:
Area - Vector Cross Product:
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Vector Equations of Lines:
The Equation of a Plane:
Planar Equation - 3 Points:
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let's say we have two vectors vector u and vector v how can we find two components of vector u one of which is parallel to vector v and the other is perpendicular to vector v so let's draw a picture so let's say this is w1 and this perpendicular to it is w2 how can we find those two components of vector u where w one is parallel to vector v and uh w two a component of vector u is perpendicular or orthogonal to vector v how can we find these two components now these two components they're not necessarily along the x or y axis they're simply parallel and perpendicular to another vector the question is how do we find these two components well let's redraw the picture a different way so once again this is going to be vector u and we're going to break it up into two components let's call this w1 and this one w2 and here we have the angle theta now we're going to say this is vector v it's not as long as w1 but notice that it's parallel to it which means that it's perpendicular to w2 what you need to know is this w1 which i'm going to write in red a component of vector u is the projection of vector u onto vector v because w 1 is the component of vector u that travels along vector v w 2 is the component of vector v that is orthogonal to vector u and the way you calculate it it's vector u minus the component w1 so this here is the component of vector u that is parallel to the other vector vector v and this right here is the component of vector u that is perpendicular or orthogonal to vector v so using these formulas that's how we can find that is these two's formulas that's how we could find a component that is parallel to another vector and another component of vector u that is perpendicular to another vector so now let's work on some practice problems now let's work on some practice problems let's start with part a find w1 the projection of u onto v and let's say that vector u is three comma five and vector v is two comma four so with this information how can we find the projection of u onto v so here's the formula w 1 which is the projection of u onto v it's equal to the dot product of u and v divided by the square of the magnitude of vector v times v i didn't give that formula to you before but now you have it so using this formula how can we find w1 so the dot product of u and v we need to multiply the x components together so that's going to be 3 times 2 and then we need to multiply the y components together so that's five times four now the magnitude of vector v is the square is the sum of the squares of the x and y components all within a square root but since that is squared the square will cancel the square root and then times the vector v which is two comma four now let's get rid of this and let's perform the operation so we have three times two which is six five times four is twenty and we could cancel the square root and the square two squared is four four squared is sixteen and then six plus twenty is twenty six four plus sixteen is twenty and we can reduce that fraction if we divide the numerator and the denominator by 2 so 26 over 20 becomes 13 over 10. now 13 times 2 is 26 4 times 13 is 52 and then we can also reduce those fractions 26 over 10 reduces to 13 over 5 if you divide both numbers by 2 and the same is true for 52 over 10. if you divide the top and bottom by 2 you get 26 over 5. so this vector here is the vector that is parallel to or rather it is the component of vector u that is parallel to vector v so that is the projection of u onto v now the last thing we need to do is find w 2 which is the vector component of u orthogonal or perpendicular to vector v and w 2 is going to be vector u minus w 1 so it's 3 comma 5 minus thirteen over five comma twenty six over five now let's get common denominators so three is the same as fifteen over five and five is the same as twenty five over five so fifteen minus thirteen is two and 25 minus 26 is negative one so this is w2 so this is the component of vector u that is orthogonal or perpendicular to vector v now let's work on another example so let's say that vector u is 6 i minus 3 j plus 9 k and let's say that vector v is 4i minus j plus 8k so let's go ahead and find the two components of vector u w1 and w2 so let's start with w1 so just like before w1 is the projection of vector u onto v and so it's the dot product of u and v divided by the square of the magnitude of v times vector v so let's start with the dot product of u and v so let's multiply the x components together so that's going to be 6 times 4 and then let's multiply the y components together so that's negative three times negative one and then multiply the z components so nine times eight and on the bottom we're going to have the square root of vector v so the magnitude of v is going to be the square root of 4 squared plus negative 1 squared plus 8 squared and we're going to square this times vector v and now let's simplify 6 times 4 is 24 negative 3 times negative 1 is 3 9 times 8 is 72 these two will cancel on the bottom we have 4 squared which is 16 plus 1 squared plus 8 squared which is 64. 24 plus 3 plus 72 that's 99 16 plus 64 is 80 plus 1 that's 81. now 99 over 81. if we divide both numbers by 9 99 divided by 9 is 11 81 divided by 9 is 9. so that gives us this and so w 1 is going to be 11 over 9 times 4 so that's 44 over 9 i and then minus 11 over 9 j plus 88 over 9 k so this is w1 the projection of u onto v now let's move on to part b so let's calculate w 2 which is u minus w 1. so we have 6i minus 3j plus 9k minus w1 so let's begin by getting common denominators just like we did before so 6 times 9 is 54. so let's write 6 as 54 over 9. negative 3 we can write that as negative 27 over 9 and 9 times 9 is 81. so let's write 9 as 81 over 9. and now let's distribute the negative sign so this is going to be negative 44 over 9 and then positive 11 over 9 and then negative 88 over 9. now let's combine like terms so let's combine these two first 54 minus 44. that's 10. so we have 10 over 9i next we have negative 27 plus 11 so that gives us negative 16. and then 81 minus 88 that's negative 7. so this is w2 the component of vector u that is orthogonal to vector v that's it |
5107 | https://www.quora.com/What-is-the-electronic-geometry-of-SO2 | What is the electronic geometry of SO2? - Quora
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What is the electronic geometry of SO2?
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Harini Ram
Studied at The High School Experience (Graduated 2023)
·5y
SO2 has 18 valence electrons so, keeping formal charge in mind, the lewis structure would look like this.
The central atom, S, has three electron domains so the electron geometry must be trigonal planar.
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SO2 has 18 valence electrons so, keeping formal charge in mind, the lewis structure would look like this.
The central atom, S, has three electron domains so the electron geometry must be trigonal planar.
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Michael Mombourquette
Retired Chemistry Prof, Church member, Knight of Columbus, · Author has 6.7K answers and 17.5M answer views
·Updated 1y
The S in SO2 has two double bonds one set of double bonds to each O and it has a lone pair. That makes three domains with electrons. 2 bonding regions and a lone pair. That means trigonal planar electron geometry and a bent molecular shape. The lone pair takes more space than the bonding pairs so the bond angle is slightly less than the perfect trigonal planar angle of 120 degrees.
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The S in SO2 has two double bonds one set of double bonds to each O and it has a lone pair. That makes three domains with electrons. 2 bonding regions and a lone pair. That means trigonal planar electron geometry and a bent molecular shape. The lone pair takes more space than the bonding pairs so the bond angle is slightly less than the perfect trigonal planar angle of 120 degrees.
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The electronic geometry of sulfur dioxide (SO₂) is bent or angular.
In SO₂, the sulfur atom is the central atom bonded to two oxygen atoms. The molecule has a total of 18 valence electrons (6 from sulfur and 6 from each of the two oxygen atoms). The sulfur atom uses one lone pair and forms double bonds with the two oxygen atoms, leading to a bent shape due to the repulsion between the lone pair and the bonding pairs of electrons.
The bond angle in SO₂ is approximately 120 degrees, which is characteristic of a trigonal planar arrangement of electron groups, but the presence of the lone pair caus
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The electronic geometry of sulfur dioxide (SO₂) is bent or angular.
In SO₂, the sulfur atom is the central atom bonded to two oxygen atoms. The molecule has a total of 18 valence electrons (6 from sulfur and 6 from each of the two oxygen atoms). The sulfur atom uses one lone pair and forms double bonds with the two oxygen atoms, leading to a bent shape due to the repulsion between the lone pair and the bonding pairs of electrons.
The bond angle in SO₂ is approximately 120 degrees, which is characteristic of a trigonal planar arrangement of electron groups, but the presence of the lone pair causes the molecular shape to be bent.
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How is the molecular geometry for SO2 determined?
What is VSEPR notation of SO2?
Why is the atom geometry of SO2 bent?
What is the electron geometry for H2S?
What is the electron geometry for CH2O?
Graham Foster
Works at University of Auckland · Author has 365 answers and 431.1K answer views
·7y
Do you mean the shape of a sulfur dioxide molecule?
Please re-write the question. If that is the question it is bent linear
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Do you mean the shape of a sulfur dioxide molecule?
Please re-write the question. If that is the question it is bent linear
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Nirosha Ameer
BSc sp in Chemistry in Chemistry, South Eastern University of Sri Lanka (Graduated 2015)
·5y
The center atom sulfur is bonded by two oxygen atoms with two double bonds and one lone pair electrons too. According to the electron geometry it gets trigonal planer and sp2 hybridization on central atom.
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What are the different errors and mistakes in English?
Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional.
Ambiguous (“Squinting”) modifiers
Incorrect: Listening to loud music slowly gives me a headache.
Correct: When I listen to loud music, I slowly develop a headache.
A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly gett
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Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional.
Ambiguous (“Squinting”) modifiers
Incorrect: Listening to loud music slowly gives me a headache.
Correct: When I listen to loud music, I slowly develop a headache.
A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly getting a headache? To correct a squinting modifier, move its position in the sentence to clarify to the reader which word you intend to modify.
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Incorrect: He was laying on the couch.
Correct: He was lying on the couch.
If you plan to place or put an object somewhere, such as a plate on a table, you should use “lay.” If you intend to stretch out on a bed for a nap, you should use “lie.” The verb “lie” is an intransitive verb, which means it does not need an object. The transitive verb “lay” requires an object.
It may take some getting used to this “lay” or “lie” business; after all, misuse of these verbs is common. But if you remember to lay down your fork before you’re full, then you won’t have to lie down later from overeating.
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Correct: He was very hungry. He ate a whole pizza.
He was very hungry, so he ate a whole pizza.
To splice means to connect or join. When a writer joins two independent sentences with a comma instead of separating them with a period or coordinating conjunction, that’s a comma splice.
The comma has its jobs to do, but connecting two independent sentences isn’t one of them. Besides, the period gets testy when his sister, the comma, steals his thunder. Periods have their jobs, and so do commas, but never the twain shall meet—unless it’s in the form of a semicolon. Semicolons can also take the place of a coordinating conjunction, such as “and,” “but,” or “so,” among others.
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Correct: Lila enjoyed the bouquet of tulips John gave her on prom night; however, she prefers roses.
Run-on sentences, also known as fused sentences, occur when two complete sentences are squashed together without using a coordinating conjunction or proper punctuation, such as a period or a semicolon. Run-on sentences can be short or long. A long sentence isn’t necessarily a run-on sentence.
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Example: She prefers roses.
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Correct: Sam could have received an A on his essay, but he made too many grammatical mistakes.
“Could have” is always correct; “could of” never is. Writers probably make this grammar gaffe because, when we speak, the contraction “could’ve” sounds an awful lot like “could of.”
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Correct: Jack made a water pail for Jill.
Tautologies express the same thing twice with different words. In our example, the word “made” implies that Jack used his own two hands to create the pail. The prepositional phrase “with his own hands” creates a redundancy. Once you know what they are, it’s fun to discover tautologies: dilapidated ruins, close proximity, added bonus, large crowd...The list goes on and on!
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Greg Burns
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What is the molecular structure of SO2?
It is a bent molecule with C 2 v C 2 v symmetry. Due to the two resonance forms, it has a bond order of 1.5.
Image courtesy Talk:Sulfur dioxide - Wikipedia
Image courtesy Sulfur dioxide - Wikipedia
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It is a bent molecule with C 2 v C 2 v symmetry. Due to the two resonance forms, it has a bond order of 1.5.
Image courtesy Talk:Sulfur dioxide - Wikipedia
Image courtesy Sulfur dioxide - Wikipedia
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What is the name of the electronic geometry of H2S?
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What is SO2 molecular geometry?
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What is the basicity of SO2?
Albert Straub
B.S. from University of Colorado Denver (Graduated 1986)
·7y
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What is the molecular structure of SO2?
Something like this.
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Something like this.
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Charles Seymour
, PhD Chemistry & Inorganic Chemistry, The University of Texas at Austin (1986)
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What is the molecular shape of SO2?
Even though the answer to this is easily ‘google-able’ I'll take this opportunity to introduce the concept Of VSEPR (pronounced vess-per) theory. This is a theory that provides shapes for all molecules by counting the number of electron domains. Or areas in which there are bonded electrons (Double/triple bonds count as one. And lone pairs of electrons also count). There is a very simple image one can find using a simple image search, whigh will tell you exactly the name of the shape, and the bond angle.
Therefore, by counting the electron domains of SO2, which is 3, and looking at the the numbe
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Even though the answer to this is easily ‘google-able’ I'll take this opportunity to introduce the concept Of VSEPR (pronounced vess-per) theory. This is a theory that provides shapes for all molecules by counting the number of electron domains. Or areas in which there are bonded electrons (Double/triple bonds count as one. And lone pairs of electrons also count). There is a very simple image one can find using a simple image search, whigh will tell you exactly the name of the shape, and the bond angle.
Therefore, by counting the electron domains of SO2, which is 3, and looking at the the number of lone pairs, I can tell that SO2 is bent. Which also means that the bond angle between the oxygen atoms should be 120 degrees But is actually closer to 117.5 degrees because the lone pair on the sulfur atom causes a stronger repultion, ‘squeezing’ the bond.
So to find the shape of a molecule, look at the number of electron domains, take into consideration whether or not any of those domains are a lone pair, and make the decision of molecular shape based off of that.
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Arkopaul Das
B.Sc. & M.Sc.( Inorganic) from Presidency University, Kolkata (Graduated 2020)
·4y
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What is the molecular geometry of O2F2?
O2F2 has the open book like structure like Hydrogen Peroxide. Here the Oxygen have +1 oxidation state which is very much unusual for it. The angle between the O and F is 109.5°. Two O-F bonds stay in two almost perpendicular planes.
Hope it will help you!
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O2F2 has the open book like structure like Hydrogen Peroxide. Here the Oxygen have +1 oxidation state which is very much unusual for it. The angle between the O and F is 109.5°. Two O-F bonds stay in two almost perpendicular planes.
Hope it will help you!
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Maria Goldade
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Is the CLion IDE good for learning C or C++?
Indeed, IDEs are sometimes more of a hindrance than a help when you're learning a new programming language. However, this isn't true of CLion, which can be very helpful right off the bat. Let's run through some of the advantages of using CLion:
CLion makes it easy to start a new project. The wizard will generate a simple project structure with stub code that you can explore and run right away.
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For many
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Indeed, IDEs are sometimes more of a hindrance than a help when you're learning a new programming language. However, this isn't true of CLion, which can be very helpful right off the bat. Let's run through some of the advantages of using CLion:
CLion makes it easy to start a new project. The wizard will generate a simple project structure with stub code that you can explore and run right away.
Then, as you start writing code, CLion will highlight its structure and suggest improvements. This allows you to learn both the language itself and the best code practices from the very beginning.
For many error cases, CLion suggests quick-fixes, which means you can start memorizing the proper solutions right off the bat. For example, the IDE catches typical errors like dangling pointers – the types of errors that might pop up a lot when you’re a newbie but can also be hard to debug.
The building, running, and debugging processes are completely transparent, as are VCS operations. You can control every step, see the exact commands, parameters, and flags that are being used, and experiment with them. If necessary, you can always perform the same actions in the built-in terminal.
CLion allows you to become familiar with a variety of compilers and build systems. It works with gcc, Clang, MSVC compilers, and project formats, such as CMake, Makefile, Meson, and others.
Understanding the underlying assembly is important when you’re learning a language like C++. CLion allows you to examine the assembly of a file without having to build the entire project. You can change the compiler flags, refresh the assembly view, and see the effect immediately.
And, of course, the IDE debugger is very convenient for learning. The IDE will help you investigate any runtime problem, including those that require memory analysis and disassembly.
To summarize, CLion will not overwhelm you with a complex set-up process and will not hide the essentials in the background. You’ll be able to get up and running with a project in no time, while learning best code practices and efficient debugging right out of the gate.
And if you’re a student, you can get it all for free!
Upvote ·
999 159
9 4
9 3
Ahad B.
Lives in Hafizabad, Punjab, Pakistan (2003–present)
·3y
Related
What is the electron geometry for H2S?
H2S is a non-linear molecule.
The bond angle between the two H-S bonds is about 92%.
Each two 3p-orbitals of sulphur containing one electron can overlap with 1s orbitals of hydrogen atoms.
Thus, the VBT requires the idea of overlap to explain the geometry of the hydrogen
Bonding in H2S showing overlap of orbitals.
Continue Reading
H2S is a non-linear molecule.
The bond angle between the two H-S bonds is about 92%.
Each two 3p-orbitals of sulphur containing one electron can overlap with 1s orbitals of hydrogen atoms.
Thus, the VBT requires the idea of overlap to explain the geometry of the hydrogen
Bonding in H2S showing overlap of orbitals.
Upvote ·
9 2
9 1
Muhammad Attaullah
BS Industrial Chemistry from GC University, Faisalabad. (Graduated 2018) · Author has 875 answers and 1.4M answer views
·4y
Related
What is the electron pair geometry of O3?
Ozone has three electron groups around the central oxygen, so has the trigonal planar electron geometry.
Or
Based on the VSEPR (valance shell electron pair repulsion) theory, electrons will repel the electron cloud of the two oxygen atoms on each end. This will result in the end O groups being pushed down giving the O3 molecule a bent molecular geometry or V shape.
Continue Reading
Ozone has three electron groups around the central oxygen, so has the trigonal planar electron geometry.
Or
Based on the VSEPR (valance shell electron pair repulsion) theory, electrons will repel the electron cloud of the two oxygen atoms on each end. This will result in the end O groups being pushed down giving the O3 molecule a bent molecular geometry or V shape.
Upvote ·
9 3
9 1
Jonathan Boden
8y
Related
What is the electron geometry for H2S?
Ok electron geometry of H2S.
S (sulfur) is the central atom. H (hydrogen) bonds twice with each S atom to form a H2S molecule. S is in group 6 - it has 6 electrons on its outer shell. H has 1. This forms a molecule with 8 electrons on its outer shell. This means there are 2 lone pairs on S (as one electron from S bonds with each H). This forms a V shaped or bent molecule. This has 2 bonding pairs and 2 lone pairs - like a water molecule. In this case, replace O with S.
Thanks for the A2A - means a lot!!
Continue Reading
Ok electron geometry of H2S.
S (sulfur) is the central atom. H (hydrogen) bonds twice with each S atom to form a H2S molecule. S is in group 6 - it has 6 electrons on its outer shell. H has 1. This forms a molecule with 8 electrons on its outer shell. This means there are 2 lone pairs on S (as one electron from S bonds with each H). This forms a V shaped or bent molecule. This has 2 bonding pairs and 2 lone pairs - like a water molecule. In this case, replace O with S.
Thanks for the A2A - means a lot!!
Upvote ·
9 5
Ernest Leung
M.Phil. in Chemistry, The Chinese University of Hong Kong · Author has 11.4K answers and 5.5M answer views
·4y
Related
What is the electron-pair geometry for S in SF4?
The central S atom has 6 valence electrons. In SF₄, 4 of them make 4 S-F bonds and the rest 2 make an unshared pair. Hence, the central S atom has 5 electron pairs. According to VSEPR theory, the electron-pair geometry for S atom in SF₄ is trigonal bipyramidal.
Continue Reading
The central S atom has 6 valence electrons. In SF₄, 4 of them make 4 S-F bonds and the rest 2 make an unshared pair. Hence, the central S atom has 5 electron pairs. According to VSEPR theory, the electron-pair geometry for S atom in SF₄ is trigonal bipyramidal.
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5108 | https://www.resuscitationjournal.com/article/S0300-9572(01)00484-1/fulltext | Electrocardiographic changes in occult pneumothorax - Resuscitation
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CorrespondenceVolume 52, Issue 3p306-307 March 2002
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Electrocardiographic changes in occult pneumothorax
Julián Ortega-Carnicer
Julián Ortega-Carnicer
Correspondence
jortegacar@nexo.es
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Intensive Care Unit, Hospital Alarcos, Los Alisos 10, Av. Pı́o XII s/n., 13002 Ciudad Real, Spain
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jortegacar@nexo.es ∙ Francisco Ruı́z-Lorenzo
Francisco Ruı́z-Lorenzo
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∙ Miguel Angel Zarca
Miguel Angel Zarca
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∙ José Villanueva
José Villanueva
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Radiology Service, Hospital Alarcos, Los Alisos 10, Av. Pı́o XII s/n., 13002 Ciudad Real, Spain
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Affiliations & Notes Article Info
1 Intensive Care Unit, Hospital Alarcos, Los Alisos 10, Av. Pı́o XII s/n., 13002 Ciudad Real, Spain
2 Radiology Service, Hospital Alarcos, Los Alisos 10, Av. Pı́o XII s/n., 13002 Ciudad Real, Spain
DOI: 10.1016/S0300-9572(01)00484-1 External LinkAlso available on ScienceDirect External Link
Copyright: © 2002 Elsevier Science Ireland Ltd.
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Pneumothoraces in critically ill patients may be unrecognized on a supine chest radiograph since they could develop anterior to the lung. In such circumstances, a chest CT scan may identify these anterior (‘occult’) pneumothoraces
1.
Tocino, I.M. ∙ Miller, M.H. ∙ Fairfax, W.R.
Distribution of pneumothorax in the supine and semirecumbent critically ill adult
Am. J. Roentgenol. 1985; 144:901-905
Crossref
Scopus (215)
PubMed
Google Scholar
. We report a patient with both apparent and occult left pneumothoraces who had electrocardiography (ECG) changes consistent in generalized low QRS voltage and precordial QS pattern simulating anterior myocardial infarction.
A 70-year-old man was admitted to the hospital because of an acute inferior myocardial infarction complicated by pulseless ventricular tachycardia requiring immediate resuscitation with defibrillation, endotracheal intubation and left subclavian vein cannulation. Following these maneuvres, the patient was awake, responsive, breathing spontaneously, and he was extubated. His vital signs revealed a blood pressure of 135/100 mmHg and a pulse of 100 beats/min. Examination of the chest showed normal breath and heart sounds. The ECG revealed sinus tachycardia at 105 beats/min, a qR pattern with ST-segment elevation in leads III and aVF, and a ST-segment depression in leads I, aVL, and V 5–6. A supine chest radiograph did not show abnormalities in the lungs and in the pleura space. Four hours after admission, the patient referred pleuritic chest pain radiating to the left shoulder. He had diminished breath sounds in left lung. In comparison to the previous tracing, the ECG showed generalized low QRS voltage and a QS pattern without concomitant ST-segment abnormalities in leads V 1–V 4 simulating anterior myocardial infarction (Fig. 1A). A portable chest radiograph revealed a left-sided pneumothorax with more than 40% of collapse, and a lateral chest tube was inserted. Although the following supine chest radiograph showed complete reinflation of the left lung, the ECG did not show significant change. A residual anterior (occult) left pneumothorax was suspected, and a chest CT scan confirmed the ECG diagnosis (Fig. 1B). After anterior chest tube insertion, the ECG showed a marked increase in the QRS amplitude and return of the precordial R-waves (Fig. 1C). A chest CT scan showed total resolution of the anterior left pneumothorax (Fig. 1D). The patient was transferred to the Nephrology Service for hemodialysis therapy.
Figure viewer
Fig. 1 Twelve-lead ECG shows generalized low QRS voltage and QS configuration in leads V 1–V 4 (A), coinciding with an anterolateral left pneumothorax seen on chest CT scan (B). Twelve-lead ECG revealing marked increase in the QRS amplitude and return of precordial R-waves (C), coinciding with a total reinflation of the left lung observed on chest CT scan (D).
The ECG manifestations associated with pneumothorax have been well described. Right tension pneumothorax may rarely cause anterior
2.
Slay, R.D. ∙ Slay, L.E. ∙ Luehrs, J.G.
Transient ST elevation associated with tension pneumothorax
JACEP. 1979; 8:16-18
Abstract
Full Text (PDF)
Scopus (34)
PubMed
Google Scholar
or inferolateral
3.
Janssens, U. ∙ Koch, C. ∙ Graf, J. ...
Severe transmyocardial ischemia in a patient with tension pneumothorax
Crit. Care Med. 2000; 28:1638-1641
Crossref
Scopus (29)
PubMed
Google Scholar
ST-segment elevation resembling an acute myocardial infarction due to severe transmyocardial ischemia. The commonly described ECG abnormalities in left pneumothorax, with or without tension, include a right QRS axis deviation, low QRS voltage, reduced precordial R-wave voltage, and anterior T-wave inversions
4.
Walston, A. ∙ Brewer, D.L. ∙ Ritchens, C.S. ...
The electrocardiographic manifestations of spontaneous left pneumothorax
Ann. Intern. Med. 1974; 80:375-379
Crossref
Scopus (63)
PubMed
Google Scholar
. Other ECG findings in patients with left pneumothorax have been described, including phasic variation in the QRS amplitude
5.
Kuritzky, P. ∙ Goldfarb, A.L.
Unusual electrocardiographic changes in spontaneous pneumothorax
Chest. 1976; 70:535-537
Crossref
Scopus (28)
PubMed
Google Scholar
, precordial QS pattern
6.
Diamond, J.R. ∙ Estes, N.M.
ECG changes associated with iatrogenic left pneumothorax simulating anterior myocardial infarction
Am. Heart J. 1982; 103:303-305
Full Text (PDF)
Scopus (31)
PubMed
Google Scholar
, and inferior PR-segment elevation
7.
Strizik, B. ∙ Forman, R.
New ECG changes associated with a tension pneumothorax
Chest. 1999; 115:1742-1744
Crossref
Scopus (39)
PubMed
Google Scholar
. The mechanisms proposed to explain the ECG abnormalities observed in left pneumothorax include cardiac rotation or displacement, acute right ventricular dilation, and air within the thoracic cavity [4,8]
4.
Walston, A. ∙ Brewer, D.L. ∙ Ritchens, C.S. ...
The electrocardiographic manifestations of spontaneous left pneumothorax
Ann. Intern. Med. 1974; 80:375-379
Crossref
Scopus (63)
PubMed
Google Scholar
8.
Master, A.M.
The electrocardiographic changes of pneumothorax in which the heart has been rotated; similarly of some of these changes to those indicating myocardial involvement
Am. Heart J. 1928; 3:472-483
Abstract
Full Text (PDF)
Scopus (30)
Google Scholar
. Our case convincingly demonstrates that the interposition of air between the electrodes and cardiac muscle may cause low QRS voltage and precordial QS pattern in left pneumothorax setting.
In conclusion, generalized low QRS voltage and precordial QS pattern may be due to an anterior (‘occult’) left-sided pneumothorax.
References
1.
Tocino, I.M. ∙ Miller, M.H. ∙ Fairfax, W.R.
Distribution of pneumothorax in the supine and semirecumbent critically ill adult
Am. J. Roentgenol. 1985; 144:901-905
Crossref
Scopus (215)
PubMed
Google Scholar
2.
Slay, R.D. ∙ Slay, L.E. ∙ Luehrs, J.G.
Transient ST elevation associated with tension pneumothorax
JACEP. 1979; 8:16-18
Abstract
Full Text (PDF)
Scopus (34)
PubMed
Google Scholar
3.
Janssens, U. ∙ Koch, C. ∙ Graf, J. ...
Severe transmyocardial ischemia in a patient with tension pneumothorax
Crit. Care Med. 2000; 28:1638-1641
Crossref
Scopus (29)
PubMed
Google Scholar
4.
Walston, A. ∙ Brewer, D.L. ∙ Ritchens, C.S. ...
The electrocardiographic manifestations of spontaneous left pneumothorax
Ann. Intern. Med. 1974; 80:375-379
Crossref
Scopus (63)
PubMed
Google Scholar
5.
Kuritzky, P. ∙ Goldfarb, A.L.
Unusual electrocardiographic changes in spontaneous pneumothorax
Chest. 1976; 70:535-537
Crossref
Scopus (28)
PubMed
Google Scholar
6.
Diamond, J.R. ∙ Estes, N.M.
ECG changes associated with iatrogenic left pneumothorax simulating anterior myocardial infarction
Am. Heart J. 1982; 103:303-305
Full Text (PDF)
Scopus (31)
PubMed
Google Scholar
7.
Strizik, B. ∙ Forman, R.
New ECG changes associated with a tension pneumothorax
Chest. 1999; 115:1742-1744
Crossref
Scopus (39)
PubMed
Google Scholar
8.
Master, A.M.
The electrocardiographic changes of pneumothorax in which the heart has been rotated; similarly of some of these changes to those indicating myocardial involvement
Am. Heart J. 1928; 3:472-483
Abstract
Full Text (PDF)
Scopus (30)
Google Scholar
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5109 | https://periodictable.com/Isotopes/005.14/index.html | Isotope data for boron-14 in the Periodic Table
HHomeBackground Color:He LiBeBoron Main PageBlackWhiteGrayBCNOFNe NaMgBoron Pictures PageAlSiPSClAr KCaBoron Technical DataScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe CsBaLaCePrNdPmSmEuGdTbDyHoErTmYbLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn FrRaAcThPaUNpPuAmCmBkCfEsFmMdNoLrRfDbSgBhHsMtDsRgCnNhFlMcLvTsOg
Isotopes of Boron (click to see decay chain): 6 B7 B8 B9 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B
14 B Half-life Fermion, 5p 9n 12.5 ms Spin 2 Parity -1 Show Decay Modes Show Ultimate Decay ProductsAtomic Weight 14.025404009 Abundance None Mass Excess 23.663683MeV Binding Energy 6.101645MeV Magnetic Moment N/A Quadrupole Moment N/A Show Element Pictures. Also show decay chains leading to this isotope. Show only decays with probability >0.1%.
Click any isotope in diagram to see its data.
Decay chain image generated by Mathematica's GraphPlot and IsotopeData functions from Wolfram Research, Inc.. |
5110 | https://www.cliffsnotes.com/study-notes/28449223 | Week 3 Company Law (docx) - CliffsNotes
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Company Law Week 3 Chapter 5 - The Company's Relations with Outsiders Organic Theory A company is an artificial legal entity - it cannot act on its own. The law recognises certain persons (directors, officers, agents) as organs of the company. Their actions are treated as the actions of the company itself. Two main aspects: o Directors and officers act as organs, representing the company in dealings. o Actions of these individuals are legally attributed to the company itself. The Common Seal Historically, companies executed documents under a common seal (company stamp) Today, use of a common seal is optional (s127 CA). Corporations Act 2001 (Cth) s127(1): A company may execute a document without using a common seal if the document is signed by: o(a) 2 directors of the company; or o(b) a director and a company secretary of the company; or o(c) for a proprietary company that has a sole director who is also the sole company secretary—that director. Agency A company, being artificial, must act through human agents. Agency law governs the relationship between o Principal (the company), and o Agent (director, officer, employee). Company is bound if the agent acts within their authority Types of Authority 1.Actual Express Authority a.Directly conferred by the company constitution, resolution, or contract of employment.
2.Actual Implied Authority a.Authority inferred from position held. b.E.g., a company secretary has implied authority to enter into routine contracts 3.Apparent (or Ostensible) Authority a.Arises when company, by words or conduct, leads a third party to reasonably believe an agent has authority. b.Even if no actual authority, the company may be bound. c.Case example:Freeman & Lockyer v Buckhurst Park Properties (Mangal) Ltd 2 QB 480 - Company bound by director's actions because his authority was represented to outsiders. Outsiders' Rights of Assumption Corporations Act 2001 (Cth) s128-129: Outsiders may make statutory assumptions when dealing with a company. Protects third parties acting in good faith S128(1): A person is entitled to make the assumptions in s129 in relation to dealings with a company, and a company is not entitled to assert otherwise. S128(4): Assumptions can't be relied on if the person knew/suspected they were incorrect. Key Assumptions under s129: s129(1): Company's constitution and replaceable rules have been complied with. s129(2): Officers are duly appointed and have authority to exercise customary powers. s129(3): Agent/Officer properly performs their duties. s129(4): Company's common seal is valid if it appears genuine. s129(5)-(8): Covers execution of documents under s127, and that officers named in ASIC records are properly appointed. Chapter 6 - Promoters and Pre-Registration Contracts Promoters
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Definition: Persons who undertake to form a company, set it up, and secure necessary capital/resources. Duties of Promoters: Fiduciary in nature, owed to: The potential company, and Potential investors. Main fiduciary duties: No conflict of interest - must act in the company's best interest. No secret profit. Full disclosure of all material facts. Case Law on Promoters' Duties Erlanger v New Sombrero Phosphate Co (1878) 3 App Cas 1218 -Facts: Syndicate bought island, sold to company they promoted an inflated price without full disclosure. -Held: Contract rescinded - disclosure must be full and made to an independent board or shareholders. Gluckstein v Barnes AC 240 -Facts: Promoters failed to disclose $20,000 profit from discount on mortgage. -Held: Secret profit recoverable - disclosure to board comprised of syndicate members insufficient. Must disclose to independent board or ultimate shareholders. Pre-Registration Contracts Promoters often enter contracts before company is formed. At common law: company cannot ratify contracts made before its existence. Statutory modification in Corporations Act 2001 (Cth) s131-s133. S131 Corporations Act: 1.If a person enters into a contract on behalf of or for the benefit of a company before incorporation: a.Company becomes bound if it is later registered and ratifies within agreed or reasonable time. 2.If company not registered or fails to ratify: a.Person who entered contract is liable for damages. 3.Court may order company to pay damages if fair (discretionary)
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4.Even if company ratifies but does not perform, person who signed may still be liable. S132 Corporations Act: 1.Outsider may release promoter from liability 2.Company cannot indemnify promoter Case Example: Bay v Illawarra Stationery Supplies Pty Ltd (1986) 4 ACLC 429 -Facts: Accountant/promoter signed contract for office supplies before registration. -Company failed to ratify. Outsider sued all promoters. -Held: Only the signing promoter liable - personal liability attaches only to signatory under s131 Remedies for Breach of Promoters' Duties -Rescission of contract (if disclosure not made). -Recovery of secret profit (company may claim profits not disclosed). -Damages under s131 if pre-registration contracts not ratified.
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5111 | http://www.coolmath.com/reference/polygons-06-hexagons | Polygons - Hexagons
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Polygons - Hexagons
Properties of hexagons, interior angles of hexagons
Polygons: Properties of Hexagons
Sum of the Interior Angles of a Hexagon:
To find the sum of the interior angles of a hexagon, divide it up into triangles... There are four triangles... Because the sum of the angles of each triangle is 180 degrees... We get So, the sum of the interior angles of a hexagon is 720 degrees.
Regular Hexagons:
The properties of regular hexagons:
All sides are the same length (congruent) and all interior angles are the same size (congruent).
To find the measure of the interior angles, we know that the sum of all the angles is 720 degrees (from above)... And there are six angles...
So, the measure of the interior angle of a regular hexagon is 120 degrees.
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To find the measure of the central angle of a regular hexagon, make a circle in the middle... A circle is 360 degrees around... Divide that by six angles...
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5112 | https://sites.math.rutgers.edu/~zeilberg/calc2/ho72.pdf | Dr. Z’s Math152 Handout #7.2 [Trigonometric Integrals] By Doron Zeilberger Problem Type 7.2a: Integrate an odd power of a sine or cosine.
Z sin2n+1 x dx OR Z cos2n+1 x dx (n = 1, 2, . . .) Example Problem 7.2a: Evaluate the integral Z sin3 x dx Steps Example 1. Rewrite sin2n+1 x dx as (−sin2n x)(−sin x dx), and cos2n+1 x dx as (cos2n x)(cos x dx). Get ready to make the substition u = TheOtherGuy (i.e.
if you have an odd power of sine, then u = cos x and if you have an odd power of cosine, then u = sin x.) 1.
Z sin3 x dx = Z (−sin2 x)(−sin x dx) .
Get ready to make the substitution u = cos x, that implies du = (−sin x)dx since (cos x)′ = −sin x .
2. Use the famous trig-identity sin2 x + cos2 = 1 in one of the equivalent forms sin2 x = 1−cos2 x , cos2 x = 1−sin2 x , to have everything in terms of the other guy, also replacing (−sin x dx) by d(cos x) or (cos x dx) by d(sin x), as the case may be.
Then make the substitution. u = cos x or u = sin x (as the case may be).
Z (−sin2n x)(−sin x dx) = Z (−sin2n x) d(cos x) = Z −(sin2 x)n d(cos x) = Z −(1−cos2 x)n d(cos x) = Z −(1−u2)n du and analogously for the other case.
2. u = cos x, so Z sin3 x dx = Z (−sin2 x)(−sin x dx) = = Z −sin2 x d(cos x) = Z −(1−cos2 x)d(cos x) = Z −(1 −u2) du 1 3. Evaluate the u-integral by first using algebra to simplify the integrand (if nec-essary). Then replace u by cos x or sin x (as the case may be). Add +C at the very end.
3.
= Z −(1−u2) du = −u+u3/3 = −cos x+cos3 x 3 +C Ans: −cos x + cos3 x 3 + C.
Problem Type 7.2b: Integrate a product of sine- and cosine-powers where at least one of the powers is odd Z coswhatever x sin2n+1 x dx OR Z sinwhatever x cos2n+1 x dx (n = 1, 2, . . .) Example Problem 7.2b: Z cos2 x sin5 x dx Steps Example 1. Locate the trig-function that has the odd-power.
Make the substition u = TheOtherGuy. If they both have odd pow-ers, then go by the lower one. If they are identical then it does not matter which one you pick. Otherwise do the same as in 7.2a. Make sure you convert everything into the u-language.
1.
Z cos2 x sin5 x dx = Z −(cos2 x)(sin2 x)2(−sin x dx) .
Get ready to make the substitution u = cos x, that implies du = (−sin x)dx since (cos x)′ = −sin x .
2. Use the famous trig-identity sin2 x + cos2 = 1 in one of the equivalent forms sin2 x = 1−cos2 x , cos2 x = 1−sin2 x , to have everything in terms of the other guy, also replacing (−sin x dx) by d(cos x) or (cos x dx) by d(sin x), as the case may be.
Then make the substitution u = cos x or u = sin x (as the case may be).
2. u = cos x, so = Z −(cos2 x)(1 −cos2 x)2 d(cos x) = Z −u2(1 −u2)2 du 2 3. Evaluate the u-integral by first using algebra to simplify the integrand (if nec-essary). Then replace u by cos x or sin x (as the case may be). Add +C at the very end.
3.
= Z −u2(1−2u2+u4) du = Z (−u2+2u4−u6) du = −u3/3+2u5/5−u7/7 = −cos3 x 3 +2 cos5 x 5 −cos7 x 7 +C Ans: −cos3 x 3 + 2 cos5 x 5 −cos7 x 7 + C.
Problem Type 7.2c: Integrate an even power of sine or cosine Z sin2n x dx OR Z cos2n x dx (n = 1, 2, . . .) Example Problem 7.2c: Evaluate the integral Z cos4 x dx Steps Example 1. Use the trig-identities sin2 θ = 1 −cos 2θ 2 cos2 θ = 1 + cos 2θ 2 possibly more than once, with θ either x, 2x or whatever shows up.
1. Focus on the trig (and algebra)! There is no integration or calculus at this step.
cos4 x = (cos2 x)2 = 1 + cos 2x 2 2 = (1 + cos 2x)2 4 = 1 + 2 cos 2x + cos2 2x 4 = 1 4+2 cos 2x 4 +cos2 2x 4 = 1 4+cos 2x 2 +(1 + cos 4x)/2 4 = 3 8+cos 2x 2 +cos 4x 8 2. Integrate the simplified trig-expression using repeatedly R sin Ax dx = −(cos Ax)/A and/or R cos Ax dx = (sin Ax)/A.
Add +C at the very end.
2.
Z cos4 x dx = 3 8 + cos 2x 2 + cos 4x 8 dx 3x 8 + sin 2x 4 + sin 4x 32 + C Ans.: 1 4( 3x 2 + sin 2x + sin 4x 8 ) + C 3 Powers of tangents: R tan x dx = ln | sec x| + C. To do R tan2 x dx you use the trig-identity tan2 x = sec2 x −1 so R tan2 x dx = R (sec2 x −1) dx = tan x −x + C. To do higher powers of tan x, you separate tan2 x and relpace it by sec2 x −1, and then use, if neccessary the substitution u = tan x. e.g.
Z tan4 x dx = Z (tan2 x)(tan2 x) dx = Z (tan2 x)(sec2 x −1) dx = Z tan2 x sec2 x dx − Z tan2 x dx.
To do R (tan2 x)(sec2 x) dx use u = tan x, giving du = (sec2 x) dx, making it R u2 du = u3/3 = (tan3 x)/3, and R tan2 x dx we did above. So R tan4 x dx = (tan3 x)/3 −tan x + x + C.
Mixtures of powers of tan and sec: If the power of sec is even use u = tan x, if the power of tan is odd, use u = sec x. If the power of sec is odd and the power of tan is even, bad luck! Use trig-identities and/or integration-by-parts.
4 |
5113 | https://math.stackexchange.com/questions/1074100/if-x-and-y-are-not-both-0-then-x2-xy-y2-0 | calculus - If $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$ - Mathematics Stack Exchange
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If x x and y y are not both 0 0 then x 2+x y+y 2>0 x 2+x y+y 2>0 [duplicate]
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How can I prove that x y≤x 2+y 2 x y≤x 2+y 2? [closed] (25 answers)
Closed 6 years ago.
Can't quite finish this proof:
Prove that if x x and y y are not both 0 0 then x 2+x y+y 2>0 x 2+x y+y 2>0
x 2+x y+y 2+x y−x y>0 x 2+x y+y 2+x y−x y>0
(x+y)2−x y>0(x+y)2−x y>0 Without loss of generality define x≥y x≥y
Firstly the case where y<0 y<0 and x>0 x>0 is trivial as both (x+y)2(x+y)2 and −(x y)>0−(x y)>0.
I want to say by transitivity as each are >0 that there sum must be >0>0 can I do that?
The second trivial case is y=0 y=0 as we simply have x 2>0 x 2>0 where x>0 x>0.
Lastly we have define c=x+y c=x+y since both x x and y y must have the same sign we know |c|>x|c|>x since x≥y x≥y we know that x 2≥x y x 2≥x y finally since |c|2=c 2|c|2=c 2∀c∈R∀c∈R
We have c 2>x 2≥x y c 2>x 2≥x y hence c 2−x y>0 c 2−x y>0
I feel like I have all the pieces but it doesn't feel finished how do I fix it?
Edit: I am really glad i asked this question love the variety of different good answers no idea what a discriminant is but im going to go look that up. I wanted to give the accepted answer to Edward as he actually answered my question but Anurag s proof was just so much better then my attempt. ^^
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edited Dec 19, 2014 at 9:29
Chris Culter
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asked Dec 19, 2014 at 2:52
FaustFaust
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2 One way is x 2+x y+y 2=3 4(x+y)2+1 4(x−y)2>0 x 2+x y+y 2=3 4(x+y)2+1 4(x−y)2>0 of not both x,y x,y are 0 0. Another way is looking at the discriminant of the quadratic.r9m –r9m 2014-12-19 02:58:18 +00:00 Commented Dec 19, 2014 at 2:58
EDIT: never mind.MackTuesday –MackTuesday 2014-12-19 08:34:39 +00:00 Commented Dec 19, 2014 at 8:34
Either x 2≥|x||y|x 2≥|x||y| or y 2≥|x||y|y 2≥|x||y|, so x 2+x y+y 2≥x 2−|x||y|+y 2>0 x 2+x y+y 2≥x 2−|x||y|+y 2>0.MackTuesday –MackTuesday 2014-12-19 08:49:20 +00:00 Commented Dec 19, 2014 at 8:49
Almost the same question: math.stackexchange.com/questions/920605/inequality-x2y2xy-ge-0Chris Culter –Chris Culter 2014-12-19 09:31:44 +00:00 Commented Dec 19, 2014 at 9:31
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Consider T=x 2+x y+y 2 T=x 2+x y+y 2. then 2 T=x 2+y 2+(x+y)2≥0 2 T=x 2+y 2+(x+y)2≥0. The only way T=0 T=0 is when x=0,y=0 x=0,y=0, but that has already been ruled out by the conditions given.
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answered Dec 19, 2014 at 2:59
Anurag AAnurag A
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Here's another way:
If x=y x=y, then obvious. Otherwise:
x 2+x y+y 2=x 3−y 3 x−y>0 x 2+x y+y 2=x 3−y 3 x−y>0
as f(x)=x 3 f(x)=x 3 is monotonically increasing.
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answered Dec 19, 2014 at 3:25
SaibalSaibal
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You can simply note that
x 2+x y+y 2=(x+y/2)2+3 y 2/4≥0.x 2+x y+y 2=(x+y/2)2+3 y 2/4≥0.
Equality is iff x=y=0 x=y=0. So if x,y x,y are not both 0 0, then x 2+x y+y 2>0 x 2+x y+y 2>0.
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answered Dec 19, 2014 at 2:59
Kim Jong UnKim Jong Un
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Let y y a fixed non zero real. The discriminant of the quadratic expression x 2+x y+y 2 x 2+x y+y 2 is Δ=y 2−4 y 2=−3 y 2<0 Δ=y 2−4 y 2=−3 y 2<0 hence the expression doesn't change the sign and we have
x 2+x y+y 2>0 x 2+x y+y 2>0
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answered Dec 19, 2014 at 2:58
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Your proof is correct. The first two cases are obvious. The third case follows directly from the AM-GM inequality:
x+y 2≥x y−−√x+y 2≥x y
⇒x+y>x y−−√⇒x+y>x y
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answered Dec 19, 2014 at 2:59
Edward JiangEdward Jiang
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Whats the full name of this inequality so i can look it up?Faust –Faust 2014-12-19 03:12:39 +00:00 Commented Dec 19, 2014 at 3:12
@Faust7: Inequality of arithmetic and geometric meansDennis –Dennis 2014-12-19 03:51:45 +00:00 Commented Dec 19, 2014 at 3:51
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We only need to consider 2 cases:-
(1) Both x and y are negative
Then, x 2+x y+y 2=(x–y)2+3 x y=(+v e)+(+v e)>0 x 2+x y+y 2=(x–y)2+3 x y=(+v e)+(+v e)>0
(2) Only one is negative
Then, x 2+x y+y 2=(x+y)2−x y=(+v e)–(−v e)>0 x 2+x y+y 2=(x+y)2−x y=(+v e)–(−v e)>0
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answered Dec 19, 2014 at 3:09
MickMick
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what's (+ve) stand for/mean?Faust –Faust 2014-12-19 03:10:28 +00:00 Commented Dec 19, 2014 at 3:10
@Faust7 A short form to mean the value of a quantity is positive.Mick –Mick 2014-12-19 03:12:06 +00:00 Commented Dec 19, 2014 at 3:12
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Switching to polar coordinates (x=r cos θ,y=r sin θ)x=r cosθ,y=r sinθ), we need to prove
1+1 2 sin 2 θ>0,1+1 2 sin2 θ>0,
which is obvious.
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answered Dec 19, 2014 at 9:32
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3Does Continuity of Composite Functions hold for both ways?
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5114 | https://www.youtube.com/watch?v=Ou8H_2HpK58 | Linear Programming Problem: Maximizing Profit Plant Example Using the Desmos Graphing Calculator
The Math Dawg
16 subscribers
Description
39 views
Posted: 21 Apr 2022
In tis video I go through a linear programming word problem. I write the constraint inequalities and the profit function. I then enter the constraints into the Desmos Graphing Calculator and find the potential points that will maximize profit. I enter the points into the profit function to find the mix of plants that gives the most profit
Transcript:
hello the math dog here in this video i'm going to look at a linear programming problem using the desmos graphing calculator and we'll be maximizing profit growing and selling plants so desmos is an online calculator that is pretty amazing has great functionality and it allows you to get good visuals of functions and so forth so let's get started so a nursery sells two products rose bushes and maple trees it wants to figure out how many maple trees and rose bushes to grow and sell to maximize its profit on the two items so here is some information about the maple trees and the bushes and what the nursery has available so square footage needed per plant the maple tree needs seven square feet the rose bush four square feet per plant cubic feet of soil needed per plant is 14 cubic feet for the maple tree two for a rose bush two cubic feet from rosebush and the cost per plant is fourteen dollars to grow a maple plant five dollars to grow a rose bush so now the nursery has the following things available these are called constraints because we have limited resources so it needs 168 square feet of planning so nursery has square feet of planting area available has 210 cubic feet of soil available the nursery has a budget of 252 dollars for these two plants the nursery then plans to sell the maple plant trees at 28 each in the rose bushes at eight dollars each so we're going to let we're going to use x and y here as our unknowns so whereas our variables let x equal the number of maple trees grown and sold let y equal number of rose bushes grown and sold so we're gonna have we're gonna write out now our constraint inequalities there's gonna be five of them so the first one is we cannot have a negative number of maple trees and rose bushes and or rose bushes so we can state x is greater than zero and y has to be greater than equal to zero okay then we have our planning area constraint 168 so this is x maple trees are x rose bushes are y so we know that seven x this is going to be 7x plus 4y has to be less than or equal to 168. okay now for the cubic feet of soil we have 210 cubic feet we need 14 here two here so this is to be 14 times the number of maple trees plus two times the number of rays rose bushes so that's going to be 14x plus 2y less than or equal to 210 and finally we have a cost constraint we only have 252 dollars available fourteen dollars per maple tree five dollars per rose bush so we can write 14x plus 5y must be less than or equal to 252. so i'm going to copy this because i'm going to need this in a few minutes so copy these constraints i'm going to send it to the clipboard and we'll get to that in just a second so now i'm going to write a profit function i'm going to call it p for profit and it's it's a function actually of two variables so we could write as p of x y equals so we're selling the maple trees for 28 each and the rose bushes for eight each but you gotta remember profit is is revenue minus cost so it's costing 14 per maple tree so 28 minus 14 the profit is 14 per maple tree and then rose bush is eight dollars you're selling at but it's cost you five so that's going to be eight minus five which is three so the profit function p of x y is going to be 14 x plus three y all right so now we have all the information we need to answer the question we're going to graph it now using desmos so first of all i'm going to paste this information here because i can't remember it so i'm now going to open the calculator okay so i'm going to put in my first two constraints i really don't need to put in these but i'm going to so x greater than or equal to zero y greater than equal to zero and then we've got and i had to write these on a piece of paper because i just can't remember these so i have seven x plus four y less than or equal to 168. seven x plus 4y less than or equal to 168. okay so we're going to get these and then we'll change our grid on our grid so we can see what we need to see then i have 14 x plus 2h 2y i'm sorry less than or equal to 210 and then finally i have 14 x plus 5y is less than or equal to 252 now the beauty of decimals is i can actually enter these equations like these inequalities as is i don't have to solve say for y so that saves us a lot of work so now let's um do is i'm going to first of all zoom out a bit okay and we got we're zoomed out too much let's just manually change our grids so we're going to go try i'm going to go 0 to go 0 to 80 here for x if i can't see what i need i'll change it some go 0 to 80 and then for y i'm going to go 0 to 120 i can get it in there okay let's try that again zero to 120. all right so now this is what i want to see all right so this is what you want to see but i'm going to move this a little bit to help it i think make it a little easier to see what we're doing so the area we're going to look at actually is what i'm outlining here so we're going to go here to here to here now why do i do that i go with the lowest line segment at every as i go left to right so first of all i'm going to click on this point here that's 042 okay then there's a i'm going to go here click on this it's because this is too high now there's nothing below this so i that's why i click on that that's gonna be 828. now i keep going this is the lowest point there's nothing below it here so i go there then here now what i'm going to do is these are the values i'm going to plug into the profit function and the one that gives me the grace value will be our final answer so i'm going to copy this graph now and i'm going to go back to the worksheet and actually i'm going to take it to the next page so i'm going to copy this cut this go to my next page there's my profit function okay so let's uh go up extend the page if we need to so now we're going to plug in these values to this function and evaluate it so p let's use gold that works pretty well so p i don't know gold works so well let's try uh more contrast i'm going to use reds more contrasting more contrast so p of use 0 42 for starters that's to be 14 times 0 plus 3 times 42 well 3 times 42 is 126. so that's 126 dollars there so now i'm going to go so i've done this one now i'm going to do 828 p of 8 28 is 14 times eight plus three times twenty eight okay fourteen times eight is 112 that's 112 plus 28 times three it's 84. if you add those up you get 196 dollars so now we've done that now we're going to do 1314 p of 13 14. is 14 times 13 plus 3 times 14 and that's going to equal 14 times 13 12 carry 1 that's 4 0 four times one is four one times one is one that's one eighty two that's a hundred eighty two plus fourteen times three is forty two so that's going to give us um 200 and well i think it's a 226. i mean 4. that's 224 dollars and then finally we have to check p of 18 0 has to be 14 times 18 plus 7 times zero well fourteen times eighteen that's a hundred fifty-two so our max profit is achieved at 13 14 so let's write a little summary the profit is maximized when 13 maple trees and 14 rose bushes are grown and sold so this concludes the work on this particular problem so if you found this video informative and you liked it please give it a thumbs up and consider subscribing to my youtube channel the math dog you may share the contents video with others thank you and have a good day you |
5115 | https://veterans.gc.ca/en/mental-and-physical-health/mental-health-and-wellness/compensation-illness-or-injury/disability-benefits/entitlement-eligibility-guidelines/az-index/acoustic-neuroma | Vestibular Schwannoma (Acoustic Neuroma) | Veterans Affairs Canada
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Entitlement Eligibility Guidelines
AZ Index
Acoustic Neuroma
Vestibular Schwannoma (Acoustic Neuroma)
Entitlement Eligibility Guideline (EEG)
Date reviewed: 22 January 2025
Date created: February 2005
ICD-11 code: 2A02.3
VAC medical code:38850 Vestibular schwannoma (Acoustic neuroma)
This publication is available upon request in alternate formats.
Full document – PDF Version
On this page
Definition
Diagnostic standard
Anatomy and physiology
Clinical features
Entitlement considerations
Links
References as of 22 January 2025
Definition
Vestibular schwannoma is a benign tumor formed from Schwann cells of the vestibular division (or rarely on the cochlear division), of the vestibulocochlear (Cranial nerve VIII)/acoustic nerve.
For the purpose of this entitlement eligibility guideline (EEG), equivalent diagnoses for vestibular schwannoma include:
acoustic neuroma
acoustic schwannoma
acoustic neurinoma
acoustic neurilemmoma
vestibular neurilemmoma.
For Veterans Affairs Canada (VAC) purposes, vertigo, hearing loss (HL), and/or tinnitus may present as part of the symptom complex of a diagnosed medical condition, or they may present as a primary stand-alone diagnosed medical condition. In those presenting with symptoms of vertigo, HL, and/or tinnitus, but with a known diagnosed cause (e.g. Meniere’s disease), these symptoms are included in entitlement and assessment of the medical condition. Prior to adjudicating the entitlement and assessment of vertigo, HL, and/or tinnitus, or a diagnosed medical condition that may cause these symptoms, a close review of previously entitled medical conditions with potentially overlapping symptoms is required.
Diagnostic standard
A diagnosis from a qualified medical practitioner (ear, nose and throat specialist [ENT]/otolaryngologist, neuro-surgeon, neurologist, family physician) is required.
The diagnosis is based on magnetic resonance imaging (MRI) findings. Additional investigations may include the following:
audiogram
computed tomography (CT) scan.
Anatomy and physiology
Vestibular schwannoma is a benign slow growing tumor arising from the myelin- forming Schwann cells which form a sheath or coating over the vestibulocochlear nerve fibers.
A vestibular schwannoma can grow and expand into the cerebellopontine angle.
Clinical features
Early symptomatology may be subtle and easily escape detection. The growth rate is generally slow. The history often reveals symptoms going back many years. Signs and symptoms develop primarily because of pressure effects that result from increasing tumour size.
Clinical features may include:
Unilateral hearing loss is a primary indicator of vestibular schwannoma. Any pattern of hearing loss may develop, including sudden hearing loss. Routine audiometry may show a hearing loss of which the person is unaware.
Tinnitus in the affected ear is a common complaint and may occur prior to onset of hearing loss.
Vertigo is an uncommon complaint. Minor disturbances of balance are common, with episodes of rotary vertigo being less common. These episodes may last several seconds or minutes to hours, with associated marked visceral autonomic symptoms.
Other than the neurologic deficit associated with the vestibulocochlear nerve (Cranial Nerve VIII), the most common neurologic deficits are changes in sensation within the distribution of either or both of the following:
trigeminal nerve (Cranial Nerve V): facial numbness (paresthesia), pain
facial nerve (Cranial Nerve VII): facial paresis and, less often, taste disturbances, dry eyes.
When comparing males and females, there are no differences in rates of occurrence, presentation or symptoms of vestibular schwannoma.
Entitlement considerations
In this section
Section A: Causes and/or aggravation
Section B: Medical conditions which are to be included in entitlement/assessment
Section C: Common medical conditions which may result, in whole or in part, from vestibular schwannoma and/or its treatment
Section A: Causes and/or aggravation
For VAC entitlement purposes, the following factors are accepted to cause or aggravate the conditions included in the Definition section of this EEG, and may be considered along with the evidence to assist in establishing a relationship to service. The factors have been determined based on a review of up- to-date scientific and medical literature, as well as evidence-based medical best practices. Factors other than those listed may be considered, however consultation with a disability consultant or medical advisor is recommended.
The timelines cited below are for guidance purposes. Each case should be adjudicated on the evidence provided and its own merits.
Factors
Therapeutic radiation to the head at least 10 years before clinical onset or aggravation of vestibular schwannoma. The latency time for the development of vestibular schwannoma following therapeutic radiation to the head has been recognized to be at least 10 years, and up to 30 years.
Neurofibromatosis prior to clinical onset or aggravation of vestibular schwannoma. Neurofibromatosis (Von Recklinghausen’s Disease) is an autosomal dominant genetic disorder with distinctive features. The features may be present at birth, as in café au lait lesions, or may develop decades later. One of these delayed features is vestibular schwannoma. Bilateral vestibular schwannoma is considered characteristic of neurofibromatosis.
Inability to obtain appropriate clinical management of vestibular schwannoma.
Note: At the time of publication of this EEG, the medical literature indicates the following:
Head trauma does not cause or permanently aggravate vestibular schwannoma.
Occupational noise exposure does not cause or permanently aggravate vestibular schwannoma.
Mobile/cellular phone use does not cause or permanently aggravate vestibular schwannoma.
Section B: Medical conditions which are to be included in entitlement/assessment
Section B provides a list of diagnosed medical conditions which are considered for VAC purposes to be included in the entitlement and assessment of vestibular schwannoma.
Hearing loss in affected ear
Tinnitus in affected ear
Vertigo
Section C: Common medical conditions which may result, in whole or in part, from vestibular schwannoma and/or its treatment
Section C is a list of conditions which can be caused or aggravated by vestibular schwannoma and/or its treatment. Conditions listed in Section C are not included in the entitlement and assessment of vestibular schwannoma. A consequential entitlement decision may be considered where the individual merits and the medical evidence of the case support a consequential relationship. Conditions other than those listed in Section C may be considered; consultation with a disability consultant or medical advisor is recommended.
Trigeminal nerve injury
Facial nerve injury
Links
Related VAC guidance and policy:
Hearing Loss – Entitlement Eligibility Guidelines
Tinnitus – Entitlement Eligibility Guidelines
Vertiginous Disorders – Entitlement Eligibility Guidelines
Pain and Suffering Compensation – Policies
Royal Canadian Mounted Police Disability Pension Claims – Policies
Dual Entitlement – Disability Benefits – Policies
Establishing the Existence of a Disability – Policies
Disability Benefits in Respect of Peacetime Military Service – TheCompensation Principle – Policies
Disability Benefits in Respect of Wartime and Special Duty Service – The Insurance Principle – Policies
Disability Resulting from a Non-Service Related Injury or Disease – Polices
Consequential Disability – Policies
Benefit of Doubt – Policies
References as of 22 January 2025
Australian Government, Repatriation Medical Authority. (1996). Statement of principles concerning acoustic neuroma, reasonable hypothesis, No 67 of 1996. SOPs - Repatriation Medical Authority
Australian Government, Repatriation Medical Authority. (1996). Statement of principles concerning acoustic neuroma, balance of probabilities, No 68 of 1996. SOPs - Repatriation Medical Authority
Australian Government, Repatriation Medical Authority. (2019). Statement of principles concerning acoustic neuroma, reasonable hypothesis, No 96 of 2019. SOPs - Repatriation Medical Authority
Australian Government, Repatriation Medical Authority. (2019). Statement of principles concerning acoustic neuroma, balance of probabilities, No 97 of 2019. SOPs - Repatriation Medical Authority
Biswas, R., Genitsaridi, E., Trpchevska, N., Lugo, A., Schlee, W., Cederroth, C. R., Gallus, S., & Hall, D. A. (2022). Low evidence for tinnitus risk factors: A systematic review and Meta-analysis. Journal of the Association for Research in Otolaryngology, 24(1), 81–94.
Blettner, M., Schlehofer, B., Samkange-Zeeb, F., Berg, G., Schlaefer, K., & Schüz, J. (2007). Medical exposure to ionising radiation and the risk of brain tumours: Interphone study group, Germany. European Journal of Cancer, 43(13), 1990– 1998.
Braganza, M. Z., Kitahara, C. M., Berrington de Gonzalez, A., Inskip, P. D., Johnson, K. J., & Rajaraman, P. (2012). Ionizing radiation and the risk of brain and central nervous system tumors: A systematic review. Neuro-Oncology, 14(11), 1316–1324.
Chen, M., Fan, Z., Zheng, X., Cao, F., & Wang, L. (2016). Risk Factors of Acoustic Neuroma: Systematic Review and Meta-Analysis. Yonsei Medical Journal, 57(3), 776.
Clinical Key. (2022). Vestibular Schwannoma Clinical Overview. Elsevier.
Durham, A. R., Tooker, E. L., Patel, N. S., & Gurgel, R. K. (2023). Epidemiology and Risk Factors for Development of Sporadic Vestibular Schwannoma.Otolaryngologic Clinics of North America, 56(3), 413–420.
Fauci, A. S., Braunwald, E., Isselbacher, K.J., Wilson, J.D., Martin, J.B., Kasper, D. Hauser, S.L., & Longo, D.L. (1998). Har rison’s principles of internal medicine. (14 th ed.). McGraw-Hill.
Illis, L. S. (1998). Harrison’s Principles of Internal Medicine 14th Edition: Edited by: A Fauci, E Braunwald, K Isselbacher, J Wilson, J Martin, D Kasper, S Hauser, D Longo. Spinal Cord, 36(9), 665–665.
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Tag Archives: 3-regular graph
Lecture 10 – the configuration model
Posted on January 15, 2019 by dominicyeo
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I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.
As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. Although we will not get far into the details during this course, the overall goal is to develop models which are close to Erdos-Renyi in terms of ease of analysis, while also allowing more of the features characteristic of networks observed in the real world.
One of the more obvious deficiencies of the sparse regime of Erdos-Renyi random graphs for modelling ‘real-world phenomena’ concerns the degree sequence. Indeed, the empirical degree distribution of G(n,c/n) converges to Poisson(c). By contrast, in real-world networks, a much wider range of degrees is typically observed, and in many cases it is felt that these should follow a power law, with a small number of a very highly connected agents.
One way around this problem to construct random graphs where we insist that the graph has a given sequence of degrees. The configuration model, which is the subject of this lecture and this post (and about which I’ve written before), offers one way to achieve this.
Definition and notes
Let and let be a sequence of non-negative integers such that is even. Then the configuration model with degree sequence d is a random multigraph with vertex set [n], constructed as follows:
To each vertex , assign half-edges;
Then, take a uniform matching of these half-edges;
Finally, for each pair of half-edges in the matching, replace the two half-edges with a genuine edge, to obtain the multigraph , in which, by construction, vertex i has degree .
One should note immediately that although the matching is uniform, the multigraph is not uniform amongst multigraphs with that degree sequence. Note also that the condition on the sums of the degrees is necessary for any graph, and in this context means that the number of half-edges is even, without which it would not be possible to construct a matching.
This effect is manifest in the simplest possible example, when n=2 and d=(3,3). There are two possible graphs, up to isomorphism, which are shown below:
For obvious reasons, we might refer to these as the handcuffs and the theta , respectively. It’s helpful if we, temporarily, assume the half-edges are distinguishable at the moment we join them up in the configuration model construction. Because then there are 3×3=9 ways to join them up to form the handcuffs (think of which half-edge ends up forming the edge between the two vertices) while there are 3!=6 ways to pair up the half-edges in the theta.
In general, for multigraphs H with the correct degree sequence, we have
where is the multiplicity with which a given edge e appears in H.
Note: it might seem counterintuitive that this procedure is biased against multiple edges and self-loops, but it is really just saying that there are more ways to form two distinct edges than to form two equal edges (ie a multiedge pair) when we view the half-edges as distinguishable. (See this post for further discussion of this aspect in the 3-regular setting.)
However, a consequence of this result is that if we condition on the event that is simple, then the resulting random graph is uniform on the set of simple graphs satisfying the degree property. Note that the same example as above shows that there’s no guarantee that there exists a simple graph whose degrees are some given sequence.
d-regular configuration model
In general, from a modelling point of view, we are particularly interested in simple, connected graphs, and so it is valuable to study whether the large examples of the configuration model are likely to have these properties. In this lecture, I will mainly focus on the case where the multigraphs are d-regular, meaning that all the vertices have degree equal to d. For the purposes of this lecture, we denote by , the d-regular configuration model .
d=1: to satisfy the parity condition on the sums of degrees, we must have n even. But then will consist of n/2 disjoint edges.
d=2: will consist of some number of disjoint cycles, and it is a straightforward calculation to check that when n is large, with high probability the graph will be disconnected.
In particular, I will focus on the case when d=3, which is the first interesting case. Most of the results we prove here can be generalised (under various conditions) to more general examples of the configuration model. The main goal of the lecture is revision of some techniques of the course, plus one new one, in a fresh setting, and the strongest possible versions of many of these results can be found amongst the references listed at the end.
Connectedness
In the lecture, we showed that is connected with high probability. This is, in fact, a very weak result, since in fact is d-connected with high probability for [Bol81, Wor81]. Here,d-connected means that one must remove at least d vertices in order to disconnect the graph, or, equivalently, that there are d disjoint paths between any pair of vertices. Furthermore, Bollobas shows that for , is a (random) expander family [Bol88].
Anyway, for the purposes of this course, the main tool is direct enumeration. The matching number satisfies
and so Stirling’s approximation gives the asymptotics
although it will be useful to use the true bounds
instead in some places. Anyway, in , there are 6n half-edges in total, and so the probability that the graph may be split into two parts consisting of vertices, with , and with no edges between the classes is Continue reading →
Posted inCombinatorics, Probability Theory, Random Graphs, Uncategorized | Tagged3-regular graph, configuration model, d-regular graph, degree distribution, Erdos Renyi, k-connected, matching, multigraph, Poisson approximation lemma, power law, random graph, random regular graph, regular graph, simple graph, uniform matching | Leave a reply
Kernels of critical graph components
Posted on November 9, 2017 by dominicyeo
1
This post is motivated by G(N,p), the classical Erdos-Renyirandom graph, specifically its critical window, when .
We start with the following observation, which makes no restriction on p. Suppose a component of G(N,p) is a tree. Then, the graph geometry of this component is that of a uniform random tree on the appropriate number of vertices. This is deliberately informal. To be formal, we’d have to say “condition on a particular subset of vertices forming a tree-component” and so on. But the formality is broadly irrelevant, because at the level of metric scaling limits, if we want to describe the structure of a tree component, it doesn’t matter whether it has or vertices, because in both cases the tree structure is uniform. The only thing that changes is the scaling factor.
In general, when V vertices form a connected component of a graph with E edges, we define the excess to be E-V+1. So the excess is non-negative, and is zero precisely when the component is a tree. I’m reluctant to say that the excess counts the number of cycles in the component, but certainly it quantifies the amount of cyclic structure present. We will sometimes, in a mild abuse of notation, talk about excess edges. But note that for a connected component with positive excess, there is a priori no way to select which edges would be the excess edges. In a graph process, or when there is some underlying exploration of the component, there sometimes might be a canonical way to classify the excess edges, though it’s worth remarking that the risk of size-biasing errors is always extremely high in this sort of situation.
Returning to the random graph process, as so often there are big changes around criticality. In the subcritical regime, the components are small, and most of them, even the largest with high probability, are trees. In the supercritical regime, the giant component has excess , which is qualitatively very different.
It feels like every talk I’ve ever given has begun with an exposition of Aldous’s seminal paper [Al97] giving a distributional scaling limit of the sizes of critical components in the critical window, and a relation between the process on this time-scale and the multiplicative coalescent. And it remains relevant here, because the breadth-first exploration process can also be used to track the number of excess edges.
In a breadth-first exploration, we have a stack of vertices we are waiting to explore. We pick one and look its neighbours restricted to the rest of the graph, that is without the vertices we have already fully explored, and also without the other vertices in the stack. That’s the easiest way to handle the total component size. But we can simultaneously track how many times we would have joined to a neighbour within the stack, which leads to an excess edge, and Aldous derives a joint distributional scaling limit for the sizes of the critical components and their excesses. (Note that in this case, there is a canonical notion of excess edge, but it depends not just on the graph structure, but also on the extra randomness of the ordering within the breadth-first search.)
Roughly speaking, we consider the reflected exploration process, and its scaling limit, which is a reflected parabolically-drifting Brownian motion (though the details of this are not important at this level of exposition, except that it’s a well-behaved non-negative process that hits zero often). The component sizes are given by the widths of the excursions above zero, scaled up in a factor . Then conditional on the shape of the excursion, the excess is Poisson with parameter the area under the excursion,with no rescaling. That is, a critical component has excess.
So, with Aldous’s result in the background, when we ask about the metric structure of these critical components, we are really asking: “what does a uniformly-chosen connected component with fixed excess look like when the number of vertices grows?”
I’ll try to keep notation light, but let’s say T(n,k) is a uniform choice from connected graphs on n vertices with excess k.
[Note, the separation of N and n is deliberate, because in the critical window, the connected components have size , so I want to distinguish the two problems.]
In this post, we will mainly address the question: “what does the cycle structure of T(n,k) look like for large n?” When k=0, we have a uniform tree, and the convergence of this to the Brownian CRT is now well-known [CRT2, LeGall]. We hope for results with a similar flavour for positive excess k.
2-cores and kernels
First, we have to give a precise statement of what it means to study just the cycle structure of a connected component. From now on I will assume we are always working with a connected graph.
There are several equivalent definitions of the 2-core C(G)of a graph G:
When the excess is positive, there are some cycles. The 2-core is the union of all edges which form part of some cycle, and any edges which lie on a path between two edges which both form part of some cycle.
C(G) is the maximal induced subgraph where all degrees are at least two.
If you remove all the leaves from the graph, then all the leaves from the remaining graph, and continue, the 2-core is the state you arrive at where there are no leaves.
It’s very helpful to think of the overall structure of the graph as consisting of the 2-core, with pendant trees‘hanging off’ the 2-core. That is, we can view every vertex of the 2-core as the root of a (possibly size 1) tree. This is particular clear if we remove all the edges of the 2-core from the graph. What remains is a forest, with one tree for each vertex of the 2-core.
In general, the k-core is the maximal induced subgraph where all degrees are at least k. Thecoreis generally taken to be something rather different. For this post (and any immediate sequels) I will never refer to the k-core for k>2, and certainly not to the traditional core. So I write ‘core’ for ‘2-core’.
As you can see in the diagram, the core consists of lots of paths, and topologically, the lengths of these paths are redundant. So we will often consider instead the kernel, K(G), which is constructed by taking the core and contracting all the paths between vertices of degree greater than 2. The resulting graph has minimal degree at least three. So far we’ve made no comment about the simplicity of the original graphs, but certainly the kernel need not be simple. It will regularly have loops and multiple edges. The kernel of the graph and core in the previous diagram is therefore this:
Kernels of critical components
To recap, we can deconstruct a connected graph as follows. It has a kernel, and each edge of the kernel is a path length of some length in the core. The rest of the graph consists of trees hanging off from the core vertices.
For now, we ask about the distribution of the kernel of a T(n,K). You might notice that the case k=1 is slightly awkward, as when the core consists of a single cycle, it’s somewhat ambiguous how to define the kernel. Everything we do is easily fixable for k=1, but rather than carry separate cases, we handle the case .
We first observe that fixing k doesn’t confirm the number of vertices or edges in the kernel. For example, both of the following pictures could correspond to k=3:
However, with high probability the kernel is 3-regular, which suddenly makes the previous postrelevant. As I said earlier, it can introduce size-biasing errors to add the excess edges one-at-a-time, but these should be constant factor errors, not scaling errors. So imagine the core of a large graph with excess k=2. For the sake of argument, assume the kernel has the dumbbell / handcuffs shape. Now add an extra edge somewhere. It’s asymptotically very unlikely that this is incident to one of the two vertices with degree three in the core. Note it would need to be incident to both to generate the right-hand picture above. Instead, the core will gain two new vertices of degree three.
Roughly equivalently, once the size of the core is fixed (and large) we have to make a uniform choice from connected graphs of this size where almost every vertex has degree 2, and of the rest have degree 3 or higher. But the sum of the degrees is fixed, because the excess is fixed. If there are n vertices in the core, then there are more graphs where all the vertices have degree 2 or 3, than graphs where a vertex has degree at least 4. Let’s state this formally.
Proposition:The kernel of a uniform graph with n vertices and excess is, with high probability as , 3-regular.
This proved rather more formally as part of Theorem 7 of [JKLP], essentially as a corollary after some very comprehensive generating function setup; and in [LPW] with a more direct computation.
In the previous post, we introduced the configuration model as a method for constructing regular graphs (or any graphs with fixed degree sequence). We observe that, conditional on the event that the resulting graph is simple, it is in fact uniformly-distributed among simple graphs. When the graph is allowed to be a multigraph, this is no longer true. However, in many circumstances, as remarked in (1.1) of [JKLP], for most applications the configuration model measure on multigraphs is the most natural.
Given a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges, and K a uniform choice from the configuration model with these parameters, we have
where t(H) is the number of loops in H, and mult(e) the multiplicity of an edge e. This might seem initially counter-intuitive, because it looks we are biasing against graphs with multiple edges, when perhaps our intuition is that because there are more ways to form a set of multiple edges we should bias in favour of it.
I think it’s most helpful to look at a diagram of a multigraph as shown, and ask how to assign stubs to edges. At a vertex with degree three, all stub assignments are different, that is 3!=6 possibilities. At the multiple edge, however, we care which stubs match with which stubs, but we don’t care about the order within the multi-edge. Alternatively, there are three choices of how to divide each vertex’s stubs into (2 for the multi-edge, 1 for the rest), and then two choices for how to match up the multi-edge stubs, ie 18 in total = 36/2, and a discount factor of 2.
We mention this because in fact K(T(n,k)) converges in distribution to this uniform configuration model. Once you know that K(T(n,k)) is with high probability 3-regular, then again it’s probably easiest to think about the core, indeed you might as well condition on its total size and number of degree 3 vertices. It’s then not hard to convince yourself that a uniform choice induces a uniform choice of kernel. Again, let’s state that as a proposition.
Proposition:For any H a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges as before,
As we said before, the kernel describes the topology of the core. To reconstruct the graph, we need to know the lengths in the core, and then how to glue pendant trees onto the core. But this final stage depends on k only through the total length of paths in the core. Given that information, it’s a combinatorial problem, and while I’m not claiming it’s easy, it’s essentially the same as for the case with k=1, and is worth treating separately.
It is worth clarifying a couple of things first though. Even the outline of methods above relies on the fact that the size of the core diverges as n grows. Again, the heuristic is that up to size-biasing errors, T(n,k) looks like a uniform tree with some uniformly-chosen extra edges. But distances in T(n,k) scale like (and thus in critical components of G(N,p) scale like ). And the core will be roughly the set of edges on paths between the uniformly-chosen pairs of vertices, and so will also have length .
Once you have conditioned on the kernel structure, and the (large) number of internal vertices on paths in the core (ie the length of the core), it is natural that the assignment of the degree-2 vertices to core paths / kernel edges is uniform. A consequence of this is that if you record the lengths of paths in the core, where m=3(k-1), then
This is stated formally as Corollary 7 b) of [ABG09]. It’s worth noting that this confirms that the lengths of core paths are bounded in probability away from zero after the appropriate rescaling. In seeking a metric scaling limit, this is convenient as it means there’s so danger that two of the degree-3 vertices end up in ‘the same place’ in the scaling limit object.
To recap, the only missing ingredients now to give a complete limiting metric description of T(n,k) are 1) a distributional limit of the total core length; 2) some appropriate description of set of pendant trees conditional on the size of the pendant forest. [ABG09] show the first of these. As remarked before, all the content of the second of these is encoded in the unicyclic k=1 case, which I have written about before, albeit slightly sketchily, here. (Note that in that post we get around size-biasing by counting a slightly different object, namely unicyclic graphs with an identified cyclic edge.)
However, [ABG09] also propose an alternative construction, which you can think of as glueing CRTs directly onto the stubs of the kernel (with the same distribution as before). The proof that this construction works isn’t as painful as one might fear, and allows a lot of the other metric distributional results to be read off as corollaries.
References
[ABG09] – Addario-Berry, Broutin, Goldschmidt –Critical random graphs: limiting constructions and distributional properties
[CRT2] – Aldous – The continuum random tree: II
[Al97] – Aldous – Brownian excursions, critical random graphs and the multiplicative coalescent
[JKLP] – Janson, Knuth, Luczak, Pittel – The birth of the giant component
[LeGall] – Le Gall – Random trees and applications
[LPW] – Luczak, Pittel, Wierman – The structure of a random graph at the point of the phase transition
Posted inCombinatorics, Probability Theory, Random Graphs, Random Trees | Tagged2-core, 3-regular graph, Brownian continuum random tree, configuration model, continuum random tree, critical window, Dirichlet distribution, Erdos Renyi, excess edges, exploration process, generating function, graph core, graph kernel, k-core, kernel, multigraph, multiplicative coalescence, pendant trees, random graph, random graph process, random regular graph, regular graph, scaling limit, size-biased, unicyclic graph, uniform tree | 1 Reply
Random 3-regular graphs
Posted on November 1, 2017 by dominicyeo
1
A graph is d-regular if every vertex has degree d. Probably the easiest examples of d-regular graphs are the complete graph on (d+1) vertices, and the infinite d-ary tree. A less trivial example is the Petersen graph, which is 3-regular. 3-regular graphs will be the main focus for some of this post, but initially we lose nothing by considering general d.
Throughout, a necessary condition for the existence of a d-regular graph with N vertices is that at least one of d and N is even, as the sum of the degrees of a graph must be even. We will always assume that this holds, so that when d=3, we are always taking N to be even.
A natural pair of questions for a probabilist is ‘can we sample a d-regular graph with N vertices uniformly at random?’ and ‘what does a typical large d-regular graph look like?’
In a rather old post, I addressed some aspects of the first question, but revisit it briefly here. A good idea, due to Bollobas [B80] is to assign to all the vertices d stubs (or half-edges), and choose a matching of the Nd stubs uniformly at random. This works as a method to generate a random graph with any fixed degree sequence.
If you want your graphs to be simple, this can go wrong, because there’s a chance you get loops (that is, an edge from a vertex v to itself) and multiple edges between the same pair of vertices. It would be nice the graph formed in this fashion was simple with high probability when . Unfortunately that’s not the case, however the probability that the graph is simple remains asymptotically bounded away from 0 and 1. Indeed, because the presence of a loop / multiple edge is asymptotically independent of the presence of a loop / multiple edge elsewhere, it’s unsurprising we have a Poisson limit for the number of such occurences. So from a sampling point of view, it’s reasonable to sample a graph in this way until you find a simple one. This takes O(1) steps, and it’s O(N) steps to check whether a given multigraph is simple.
It’s clear that conditional on the graph generated in this fashion being simple, its distribution is uniform on the set of simple graphs with the correct degree distribution. If you are happy for your graphs to have loops, then it’s a little bit more complicated, because if an edge has multiplicity k, these can appear in k! ways in the configuration construction.
Other asymptotic properties
Loops and multiple edges can be thought of as cycles of length 1 and 2 respectively if you want. We might ask about other small cycles. A calculation in expectation is relatively straightforward. Given three vertices, the probability they form a triangle (in at least one way) is , and there are ways to choose three vertices. Thus the expected number of triangles is . Finally, the edge structure induced on disjoint triples is asymptotically independent, and hence a Poisson limit. (See [J06] for details, including more detail on the general configuration construction.) The same result holds for the same reasons for cycles of any fixed finite length.
We might also ask about connectivity. At a heuristic level, there are two ways for the graph to be disconnected: it could have some small components; or it could have two components of size . The smallest possible component is , and an argument like for the cycles above shows that the number of copies of vanishes in expectation. Now, consider having two components of size roughly N/2. There are ways to make this choice. However, given such a choice, we can handle the probability that all the stubs from one class match within that class by going through the class one stub at a time:
We approximate this as
and this dominates the number of choices powerfully enough that we might believe it remains valid for a broader range of class sizes. In fact we have a much stronger statement, namely that G(N,3) is 3-connected with high probability. This means that the graph cannot be disconnected by removing two vertices, or equivalently that there are three vertex-disjoint paths between any pair of vertices in the graph, essentially one emerging from each stub. See this note by David Ellis for a quick proof. We might return to this later.
You might ask about planarity. It’s clear from degree consideration that there are no induced copies of in any random 3-regular graph, and since contains a cycle of length 4, and with high probability G(N,3) doesn’t, that takes care of that possibility too. However, there might be minors of this form. This seemed a good example of the Kuratowski criterion not actually being that useful, since I certainly don’t find the minors of the 3-regular graph an obvious structure to handle.
However, we can use Euler’s formula V – E + F = 2 for planar graphs. Here V = N, E = 3N/2. Faces are described by (a subset of the) cycles, and we there are asymptotically O(1) small cycles, so most faces include a large number of edges. But each edge corresponds to at most two faces. So we have , and so with high probability Euler’s formula can’t hold in G(N,3) for large N.
We can also ask about the local limit of G(N,3). Since the vertices are exchangeable, we don’t need to worry about whether we choose the root uniformly at random (often referred to as the Benjamini-Schramm sense) or by some other method.
The root has up to three neighbours, and with high probability it has exactly three neighbours. These neighbours have at most two other neighbours themselves. However, we’ve already seen that there are asymptotically O(1) cycles, and so with high probability there are no small cycles near a fixed root vertex. So the six neighbours-of-neighbours are with high probability different to the root and the root’s neighbours and to each other. We can make this argument at arbitrary finite radius from the root, to conclude that the local limit of G(N,3) is the infinite 3-ary tree.
Spectral expansion
[Caveat – this is something I read about and wanted to mention, but I really don’t know much at all about any of this theory, and it’s definitely not certain that what follows wouldn’t be better replaced by a set of links.]
This straightforward local limit offers good heuristics on some of the more global properties. Almost by definition, the d-ary tree expands as rapidly as is possible away from the root among infinite d-regular graphs. There are a number of ways to measure the expansion of a graph, and some methods transfer better to the infinite setting than others. The adjacency matrix of an infinite graph can be defined similarly to that of a finite graph, and it remains possible to talk about eigenfunctions and spectrum. As for the finite setting, d is an eigenvalue because the tree is d-regular, and -d is an eigenvalue because it is also bipartite.
The next largest eigenvalue governs the spectral gap which is a measure of the expansion of a graph. A graph is a good (spectral) expander if all the non-trivial eigenvalues are close to zero. A priori, all we know is that . For the infinite d-ary tree, we have . This blog postby Luca Trevisan gives a very readable proof.
A key result is that finite graphs can have , but not asymptotically. That is, taking N to be the number of vertices:
This is the content of the Alon-Boppana theorem [Al86]. In fact the error can be quantified as – the diamater of the graph is relevant here. A finite d-regular graph for which is called a Ramanujan graph. The existence of Ramanujan graphs has been much studied, and various constructions often rely on number theoretic properties of N, and lie at the interface of disparate branches of mathematics where my understanding is zero rather than epsilon.
Now return to our view of the d-ary tree as the local limit of a d-regular graph on N vertices for large N. We might expect from everything above that the uniform d-regular graph is a good expander. Bollobas shows that in the sense of edge-expansion, asymptotically almost all d-regular graphs have edge-expansion bounded away from zero. (See Section 2 of [Ell], including history of the d=3 case.) Friedman [Fri08] proves the conjecture of Alon that for every , a.a.s. for G(N,d) is at most . In this sense, G(N,d) is asymptotically ‘almost Ramanujan’. (See also [Bor17] for another proof and an introduction including history, context and references.)
Some other links: The Wikipedia page on expanders, which includes a discussion of the different descriptions of expansion, and the Cheeger inequalities and other relations between them; slides for a talk by Spielman on spectra and Ramanujan graphs; a survey by Murty on Ramanujan graphs;.
What next?
This post took a slightly different direction from what I had intended, and rather than make a halting U-turn back to my planned finale, I’ll postpone this. However, a short overture is that I’m interested in the structure of critical components of random graphs during the critical window. This is the window during which the largest components first have cycles with probability . Indeed, the critical components have size and surplus edges. Conditional on their size, and number of surplus edges, the choice of the graph structure on the component is uniform among such (connected) graphs.
Addario-Berry, Broutin and Goldschmidt [ABG09] study scaling limits of such components. Central to this analysis is the 2-core of such components, which can be described in terms of 3-regular (multi)graphs. Various processes we are now interested in running on the critical components of critical RGs can then be studied in terms of related processes on random 3-regular graphs.
References
[ABG09] – Addario-Berry, Broutin, Goldschmidt – Critical random graphs: limiting constructions and distributional properties
[Al86] – Alon – Eigenvalues and expanders
[B80] – Bollobas – A probabilistic proof of an asymptotic formula for the number of labelled regular graphs
[B88] – Bollobas – The isoperimetric number of random regular graphs
[Bor17] – Bordenave – A new proof of Friedman’s second eigenvalue theorem and its extension to random lifts. Arxiv.
[Ell] – Ellis – The expansion of random regular graphs
[Fri08] – Friedman – A proof of Alon’s second eigenvalue conjecture and related problems
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5117 | https://motto.tc/siteler/www.tiaftturkey2018.org/topic/Christianity.html | Christianity | Description, History, Doctrines, & Traditions | Britannica
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Contents
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Introduction
The church and its history
The essence and identity of Christianity
Historical views of the essence
Early views
Medieval and Reformation views
Modern views
The question of Christian identity
The history of Christianity
The primitive church
The relation of the early church to late Judaism
The relation of the early church to the career and intentions of Jesus
The Gentile mission and St. Paul
The contemporary social, religious, and intellectual world
The internal development of the early Christian church
The problem of jurisdictional authority
The problem of scriptural authority
The problem of theological authority
Early heretical movements
Relations between Christianity and the Roman government and the Hellenistic culture
Church-state relations
Christianity and Classical culture
The Apologists
The early liturgy, the calendar, and the arts
The alliance between church and empire
Theological controversies of the 4th and 5th centuries
Western controversies
Eastern controversies
Liturgy and the arts after Constantine
New forms of worship
Historical and polemical writing
Political relations between East and West
Literature and art of the “Dark Ages”
Missions and monasticism
The Photian schism and the great East-West schism
The Photian schism
The great East-West schism
From the schism to the Reformation
Papacy and empire
Medieval thought
Reformation
Christianity from the 16th to the 21st century
Contemporary Christianity
Roman Catholicism
The Eastern churches
Eastern Orthodoxy
Oriental Orthodoxy
Protestantism
Lutheranism
Anglicanism
Presbyterian and Reformed churches
Other Protestant churches
Christian doctrine
The nature and functions of doctrine
Scripture and tradition: the apostolic witness
Evangelism: the first teaching about the God of Jesus Christ
Catechesis: instructing candidates for baptism
Liturgy: the school and feast of faith
Ethics: obeying the truth
Aversion of heresy: the establishment of orthodoxy
Apologetics: defending the faith
Restatement: respecting language and knowledge
Inculturation: respecting places and peoples
Dogma: the most authoritative teaching
Consensus: patterns of agreement
Theology: loving God with the mind
Symbolics: creeds and confessions
Development: the maturation of understanding
Schism: division over substantial matters
Controversy: fighting over the faith
Ecumenism: speaking the truth in love
God the Father
Characteristic features of the Christian concept of God
The specific concept of God as Father
The belief in the oneness of the Father and the Son
The revelatory character of God
God as Creator, Sustainer, and Judge
The view that God is not solitary
Modern views of God
Satan and the origin of evil
God the Son
Different interpretations of the person of Jesus
The Christological controversies
Messianic views
The doctrine of the Virgin Mary and holy Wisdom
God the Holy Spirit
Contradictory aspects of the Holy Spirit
Conflict between order and charismatic freedom
The operations of the Holy Spirit
The Holy Trinity
The basis for the doctrine of the Trinity
Introduction of Neoplatonic themes
Attempts to define the Trinity
Anthropology
What it is to be human
The human as a creature
The human as the image of God
Human redemption
The problem of suffering
The resurrection of the body
Progressive human perfection
The “new man”: The human being in the light of Christ
The “reborn human”
Human liberation
Joy in human existence
The charismatic believer
Christian perfection
Fellow humans as the present Christ
The church
Normative defenses in the early church
Evolution of the episcopal office
Authority and dissent
Organization
Church polity
Liturgy
New liturgical forms and antiliturgical attitudes
Church tradition
The sacraments
Scriptural traditions
Veneration of places, objects, and people
Monasticism
The saintly life
Art and iconography
Theology of icons
Eschatology
Expectations of the Kingdom of God in early Christianity
Expectations of the Kingdom of God in the medieval and Reformation periods
Expectations of the Kingdom of God in the post-Reformation period
The role of imminent expectation in missions and emigrations
Eschatological expectations and secularization
Concepts of life after death
Aspects of the Christian religion
Christian philosophy
History of the interactions of philosophy and theology
Influence of Greek philosophy
Emergence of official doctrine
Aristotle and Aquinas
Other influences
Faith and reason
Christian philosophy as natural theology
Arguments for the existence of God
The design (or teleological) argument
The cosmological argument
The ontological argument
Moral arguments
Arguments from religious experience and miracles
The immortality of the soul
20th-century discussions
Influence of logical positivism
Evidentialist approach
Christian mysticism
History of Christian mysticism
Early church
Eastern Christianity
Western Catholic Christianity
Protestant Christianity
Stages of Christian mysticism
The dying to self
The union with God
The readjustment
Forms of Christian mysticism
Christ-mysticism
Trinitarian mysticism
Negative mysticism: God and the Godhead
Significance of Christian mysticism
Christian myth and legend
Characteristics of Christian myth and legend
History of Christian myth and legend
The early church
The ages of the world
Messianic secrets and the mysteries of salvation
The Magi and the Child of Wondrous Light
Relics and saints
The Middle Ages
Renaissance magic and science
Christian practice in the modern world
The Christian community and the world
The relationships of Christianity
Historical views
Church, sect, and mystical movement
Church and state
The history of church and state
The church and the Roman Empire
The church and the Byzantine, or Eastern, Empire
The church and Western states
Separation of church and state
Church and state in Eastern and Western theology
The views of Eusebius of Caesarea
The views of Augustine
Later developments
Church and society
The problem of slavery and persecution
Theological and humanitarian motivations
Church and education
Intellectualism versus anti-intellectualism
Forms of Christian education
Church and social welfare
Curing and caring for the sick
Healing the sick
Care for the sick
Care for widows and orphans
Property, poverty, and the poor
Pastoral care
Church and minorities
Church and family
The tendency to spiritualize and individualize marriage
The tendency toward asceticism
Church and the individual
Love as the basis for Christian ethics
Freedom and responsibility
Christian missions
Biblical foundations
The history of Christian missions
First transition, to ad 500
Second transition, to ad 1500
Western mission
Papal mission
Eastern and Nestorian missions
The rise of Islam
Third transition, to ad 1950
Roman Catholic mission, 1500–1950
Protestant missions, 1500–1950
Early Protestant missions
Missions to Asia
Missions to South East Asia and the Pacific
Missions to Africa and South America
Missionary associations
Orthodox and nondenominational missions
Fourth transition, from 1950
Scripture translations
Ecumenism
The biblical perspective
The history of ecumenism
Early controversies
The Schism of 1054
The Reformation
Ecumenism in the 17th and 18th centuries
19th-century efforts
Ecumenism since the start of the 20th century
Christianity and world religions
Conflicting Christian attitudes
Contemporary views
HomePhilosophy & ReligionScriptures
Christianity
WRITTEN BY
Matt StefonSee All Contributors
Matt Stefon was a religion editor at Encyclopaedia Britannica. He earned B.A. degrees in English and American studies from the Pennsylvania State University and an M.A. in religion and literature and an...
Last Updated: Aug 26, 2020See Article History
Christianity, major religion stemming from the life, teachings, and death of Jesus of Nazareth (the Christ, or the Anointed One of God) in the 1st century ce. It has become the largest of the world’s religions and, geographically, the most widely diffused of all faiths. It has a constituency of more than two billion believers. Its largest groups are the Roman Catholic Church, the Eastern Orthodox churches, and the Protestant churches. The Oriental Orthodox churches constitute one of the oldest branches of the tradition but had been out of contact with Western Christianity and Eastern Orthodoxy from the middle of the 5th century until the late 20th century because of a dispute over Christology (the doctrine of Jesus Christ’s nature and significance). Significant movements within the broader Christian world and sometimes transcending denominational boundaries are Pentecostalism, Charismatic Christianity, Evangelicalism, and fundamentalism. In addition, there are numerous independent churches throughout the world. See alsoAnglicanism; Baptist; Calvinism; Congregationalism; Evangelical church; Lutheranism; Oriental Orthodoxy; presbyterian; Reformed and Presbyterian churches.
mosaic; ChristianityChrist as Ruler, with the Apostles and Evangelists (represented by the beasts). The female figures are believed to be either Santa Pudenziana and Santa Práxedes or symbols of the Jewish and Gentile churches. Mosaic in the apse of Santa Pudenziana basilica, Rome, 401–417 ce.De Antonis
Britannica Quiz Christianity Quiz Which of the following philosophical concepts suggesting both “Word” and “cosmic order” was associated with Jesus Christ?
This article first considers the nature and development of the Christian religion, its ideas, and its institutions. This is followed by an examination of several intellectualmanifestations of Christianity. Finally, the position of Christianity in the world, the relations among its divisions and denominations, its missionary outreach to other peoples, and its relations with other world religions are discussed. For supporting material on various topics, seeangel and demon; Bible; biblical literature; canon law; creed; Christology; doctrine and dogma; ecumenism; eschatology; exegesis; faith; grace; heaven; hell; heresy; Jesus Christ; liturgical movement; millennialism; miracle; monasticism; monotheism; New Testament; Old Testament; original sin; papacy; prayer; priesthood; purgatory; sacrament; salvation; schism; scripture; theism; theology; and worship.
The church and its history
The essence and identity of Christianity
At its most basic, Christianity is the faith tradition that focuses on the figure of Jesus Christ. In this context, faith refers both to the believers’ act of trust and to the content of their faith. As a tradition, Christianity is more than a system of religious belief. It also has generated a culture, a set of ideas and ways of life, practices, and artifacts that have been handed down from generation to generation since Jesus first became the object of faith. Christianity is thus both a living tradition of faith and the culture that the faith leaves behind. The agent of Christianity is the church, the community of people who make up the body of believers.
To say that Christianity “focuses” on Jesus Christ is to say that somehow it brings together its beliefs and practices and other traditions in reference to a historical figure. Few Christians, however, would be content to keep this reference merely historical. Although their faith tradition is historical—i.e., they believe that transactions with the divine do not occur in the realm of timeless ideas but among ordinary humans through the ages—the vast majority of Christians focus their faith in Jesus Christ as someone who is also a present reality. They may include many other references in their tradition and thus may speak of “God” and “human nature” or of the “church” and the “world,” but they would not be called Christian if they did not bring their attentions first and last to Jesus Christ.
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While there is something simple about this focus on Jesus as the central figure, there is also something very complicated. That complexity is revealed by the thousands of separate churches, sects, and denominations that make up the modern Christian tradition. To project these separate bodies against the background of their development in the nations of the world is to suggest the bewildering variety. To picture people expressing their adherence to that tradition in their prayer life and church-building, in their quiet worship or their strenuous efforts to change the world, is to suggest even more of the variety.
Given such complexity, it is natural that throughout Christian history both those in the tradition and those surrounding it have made attempts at simplification. Two ways to do this have been to concentrate on the “essence” of the faith, and thus on the ideas that are integral to it, or to be concerned with the “identity” of the tradition, and thus on the boundaries of its historical experience.
Modern scholars have located the focus of this faith tradition in the context of monotheistic religions. Christianity addresses the historical figure of Jesus Christ against the background of, and while seeking to remain faithful to, the experience of one God. It has consistently rejected polytheism and atheism.
A second element of the faith tradition of Christianity, with rare exceptions, is a plan of salvation or redemption. That is to say, the believers in the church picture themselves as in a plight from which they need rescue. For whatever reason, they have been distanced from God and need to be saved. Christianity is based on a particular experience or scheme directed to the act of saving—that is, of bringing or “buying back,” which is part of what redemption means, these creatures of God to their source in God. The agent of that redemption is Jesus Christ.
It is possible that through the centuries the vast majority of believers have not used the term essence to describe the central focus of their faith. The term is itself of Greek origin and thus represents only one part of the tradition, one element in the terms that have gone into making up Christianity. Essence refers to those qualities that give something its identity and are at the centre of what makes that thing different from everything else. To Greek philosophers it meant something intrinsic to and inherent in a thing or category of things, which gave it its character and thus separated it from everything of different character. Thus, Jesus Christ belongs to the essential character of Christianity and gives it a unique identity.
If most people are not concerned with defining the essence of Christianity, in practice they must come to terms with what the word essence implies. Whether they are engaged in being saved or redeemed on the one hand, or thinking and speaking about that redemption, its agent, and its meaning on the other, they are concentrating on the essence of their experience. Those who have concentrated from within the faith tradition have also helped to give it its identity. It is not possible to speak of the essence of a historical tradition without referring to how its ideal qualities have been discussed through the ages. Yet one can take up the separate subjects of essence and identity in sequence, being always aware of how they interrelate.
Christianity
key people
St. Paul the Apostle
Martin Buber
Jan Hus
St. Albertus Magnus
Nikolaus Ludwig, count von Zinzendorf
Pseudo-Dionysius the Areopagite
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Theōdūrus Abū Qurrah
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5118 | https://repair.wiki/w/Inductors_-_Repair_Basics | Inductors - Repair Basics - Repair Wiki
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| Inductors - Repair Basics |
| Type | Component |
| Device(s) | Inductors |
| Difficulty | ◉◌◌◌ Easy |
This article aims to provide a comprehensive overview of inductors from a repair perspective, detailing their function, types, common issues, testing methods, and replacement considerations.
Inductor schematic symbol and common packages (Figure 1)
What is an Inductor?
An inductor is an essential passive electronic component that stores energy in a magnetic field when an electric current flows through it. Inductors resist changes in current flow and are widely used in various electronic circuits.
Inductors are measured in "Henry" but a Henry is a large unit so you'll typically find them in milli-Henry "mH", nano-Henry "nH", or micro-Henry "uH"
In essence, an inductor consists of a wire coiled around a core, with the choice of core material and the number of turns influencing its inductance and magnetic field strength.
Inductors are typically identified by the symbol shown in Figure 1 and are usually denoted with the letter L followed by an identifier number. (E.g., L3)
Function
Inductors serve various functions in electronic circuits, including:
Voltage Smoothing
Inductors in the output stage of voltage converter circuits serve to filter and stabilize the output voltage. By smoothing the pulsating DC output, inductors reduce voltage ripples, contributing to a stable and continuous power supply. In almost every device that has DC-DC voltage converters you will most likely find an inductor at the output.
Energy Storage
Inductors store energy in their magnetic fields, and this property is crucial in applications like boost and buck voltage converters, which are essential in power supplies and voltage regulators.
Electro-magnet
When an electric current flows through the coil, it generates a magnetic field, turning the inductor into a temporary magnet. This principle finds practical application in relays, where an inductor serves as the electromagnet to control the switch mechanism.
Filtering
Inductors are used in conjunction with capacitors to create low-pass and high-pass filters (LC Filters), allowing certain frequencies to pass while attenuating others. By themselves, Inductors block AC, and let DC pass through.
Resonance Circuits
Inductors are crucial in resonance circuits, where they interact with capacitors to establish resonance at a specific frequency. In series resonance, inductors limit current increase, while in parallel resonance, they contribute to maintaining a high impedance at the resonant frequency. The strategic use of inductors in resonance circuits is fundamental for designing efficient filters, antennas, and oscillators in electronic systems.
Inductive Sensing
Inductors are used in sensors for inductive sensing applications, such as proximity sensors and metal detectors, where changes in inductance are used to detect the presence or absence of certain materials.
Types
Air Core Inductors
Air core inductors have a coil wound around a non-magnetic form, such as plastic or ceramic. They are used in radio-frequency applications and where magnetic interference must be minimized.
Iron Core Inductors
Iron core inductors have a coil wound around a magnetic core, typically made of iron or other magnetic materials. They are used in power applications and transformers.
Ferrite Core Inductors
Ferrite core inductors use a core made of ferrite, a ceramic material with high magnetic permeability. They are common in high-frequency applications and EMI suppression.
Toroidal Inductors
Toroidal inductors have a coil wound around a ring-shaped core. They offer high inductance and are often used where compact size and low magnetic interference are critical.
Measuring an Inductor (Figure 2)
Testing an Inductor
Make sure the inductor does not have any physical damage first! That's the quickest way of identifying a faulty inductor.
To measure an inductor, use a digital multimeter and check for continuity. Remember, an inductor is just a wire!
Set your multimeter to resistance mode.
Connect the multimeter's leads across the inductor
The multimeter should display around 1 Ohm since the inductor is just a wire.
If you measure high resistance or OL that means the inductor is blown open.
The inductor is the most reliable electrical component. If you find a damaged inductor, that means it is a symptom and not the cause and you most likely have a short circuit somewhere.How to find short circuits?
If an inductor fails, it most likely fails in a "blown open" way. Caused by excessive flow of current, leading to overheating and subsequent failure.
You can also use an LCR meter to measure the inductance the same way but it is usually not necessary.
Replacement Considerations
When replacing an inductor, consider the following factors:
Inductance Value: Choose a replacement inductor with the same or very close inductance value to the faulty inductor.
Current Rating: Ensure the replacement inductor can handle the same or higher current than the original, especially in power applications.
Type of Core: Match the type of core (air core, iron core, ferrite core, etc.) to the original inductor, as it can affect the inductor's performance.
Physical Size: Choose a replacement with a similar or compatible physical size.
Replacing an inductor with the wrong specifications can lead to it breaking again.
Retrieved from "
Last modified
17 July 2024
Contents
Back to top
Contents
1 What is an Inductor?
1.1 Function
1.1.1 Voltage Smoothing
1.1.2 Energy Storage
1.1.3 Electro-magnet
1.1.4 Filtering
1.1.5 Resonance Circuits
1.1.6 Inductive Sensing
1.2 Types
1.2.1 Air Core Inductors
1.2.2 Iron Core Inductors
1.2.3 Ferrite Core Inductors
1.2.4 Toroidal Inductors
2 Testing an Inductor
3 Replacement Considerations
Categories:
Explanatory guide
Explanatory guides for Inductors
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5119 | http://faculty.ung.edu/mgoodroe/Math_1001/PP_pdf_Files/3.2_Truth_Tables(02-10-2019).pdf | Slide 1 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Logic 3.2 Slide 2 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables 3.2 • State the truth tables for the five fundamental connectives.
• Compute truth tables for compound statements.
• Determine when statements are logically equivalent.
• State and apply DeMorgan’s laws.
Slide 3 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables We want to know if a pair of similar statements such as and mean the same thing.
We use truth tables to determine when compound statements are true and when they are false, and whether a pair of statements have the same meaning.
( ) p q ∧ p q ∧ Slide 4 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables - Negation If p is true, then ~p is false.
If p is false, then ~p is true.
Slide 5 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Finding the Truth Value of Negations Determine whether each statement is true or false and then state its negation. Notice how the negation of each statement has the opposite truth value of the original statement.
a) Maya Angelou is a well-known American poet.
b) The Bill of Rights begins the Constitution of the United States.
Slide 6 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Finding the Truth Value of Negations (cont) a) Maya Angelou is a well-known American poet.
This is true. Therefore, its negation, “Maya Angelou is not a well-known American poet.,” is false.
Slide 7 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Finding the Truth Value of Negations (cont) b) The Bill of Rights begins the Constitution of the United States.
This is false. So, “The Bill of Rights does not begin the Constitution of the United States.” is true.
Slide 8 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables - Conjunction A conjunction is true only when both of its components are true.
Slide 9 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables - Disjunction A disjunction is false only when both p and q are false.
Slide 10 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Analyzing the Logic of an Advertisement Consider the following help-wanted ad: “Management trainee wanted. Applicant must have four-year degree in accounting or three years of experience working in a financial institution.” Which applicants should be considered for the position if we interpret or according to the table?
Slide 11 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Analyzing the Logic of an Advertisement (cont) a) Aya has a four-year degree in accounting and has worked two years for a loan company.
b) Heidi has studied accounting in college (but did not graduate) and has worked for five years selling electronics.
c) Monte earned a four-year degree in accounting and has five years of experience working for a credit card company.
Slide 12 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Analyzing the Logic of an Advertisement (cont) Solution a) Aya’s background corresponds to line 2 of the table, so she should be considered for the job.
b) Heidi’s credentials are reflected in line 4 of the table, so she is not qualified.
c) Monte’s background corresponds to line 1 of the table, so he should be considered.
Slide 13 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Truth Tables Because of their similar structures, there are parallels that occur in logic and set theory. Often if a result is true for set theory, then a similar result also holds for logic (and vice versa).
Slide 14 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Compound Statements We use truth tables to find the logical values of complex statements. Slide 15 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Finding the Truth Table for a Compound Statement Compute a truth table for a statement of the form Solution So we can solve this problem easily by breaking it into three simpler problems: Step 1 Calculate Step 2 Calculate Step 3 Use the results from steps 1 and 2 to calculate ( ) ( ).
p q p q ∧ ∨ ∧ ( ).
∧ p q ( ).
∧ p q ( ) ( ).
∧ ∨ ∧ p q p q Slide 16 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Finding the Truth Table for a Compound Statement (cont) We will next perform each of these steps in the following truth table: If p is false and q is true, line 3 tells us that is true.
( ) ( ) p q p q ∧ ∨ ∧ Slide 17 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Compound Statements Definition: If the final column of a truth table contains all T’s, then the statement is always true. Such a statement is called a tautology. Slide 18 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Compound Statements How many lines should your truth table have?
If the statement has k variables, then its truth table will have 2k lines.
Slide 19 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Logically Equivalent Statements Two statements are logically equivalent if they have the same variables and, when their truth tables are computed, the final columns in the tables are identical.
Slide 20 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Determining When Statements Mean the Same Thing Assume that you have an entertainment book containing discount coupons for movies, restaurants, and other leisure activities. You are considering eating at either the Pasta Bar or the Deli. Do the following two statements say the same thing?
a) Both the Pasta Bar and the Deli accepts coupons is false.
b) The Pasta Bar does not accept coupons or the Deli does not accept coupons.
Slide 21 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Determining When Statements Mean the Same Thing (cont) Solution If we let p represent “The Pasta Bar accepts coupons” and let d represent “The Deli accepts coupons,” then we can write a) and b) symbolically as To decide whether these statements say the same thing, all we have to do is make two truth tables.
( ) ( ) ( ) ) ) ∧ ∨ a p d b p d Slide 22 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Determining When Statements Mean the Same Thing (cont) Because the final columns (shaded) in the two truth tables are identical, the two statements are logically equivalent and therefore express exactly the same information.
Slide 23 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G DeMorgan’s Laws for Logic If p and q are statements, then a) b) ( ) is logically equivalent to ( ) ( ).
∧ ∨ p q p q ( ) is logically equivalent to ( ) ( ).
∨ ∧ p q p q Slide 24 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Applying Logic to Legal Documents Use DeMorgan’s laws to rewrite the following statement, which is based on instructions for filing Form 1040-A with the U.S. Internal Revenue Service.
You received interest from a seller-financed mortgage and the buyer used the property as a personal residence is false.
Slide 25 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: Applying Logic to Legal Documents (cont) Solution It is easy to rewrite this statement if first we represent it in symbolic form. Let r represent “You received interest from a seller-financed mortgage” and let b represent “The buyer used the property as a personal residence.” This statement has the form By DeMorgan’s laws this statement is equivalent to We can now rewrite this in English as “You did not receive interest from a seller-financed mortgage or the buyer did not use the property as a personal residence.” ( ).
∧ r b ( ) ( ).
∨ r b Slide 26 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: An Alternative Method to Construct Truth Tables a) We will construct a truth table for which you saw in a previous example.
Recall that we constructed a table by performing the following steps.
Step 1 Calculate Step 2 Calculate Step 3 Use the results from steps 1 and 2 to calculate ( ) ( ), ∧ ∨ ∧ p q p q ( ).
∧ p q ( ).
∧ p q ( ) ( ).
∧ ∨ ∧ p q p q Slide 27 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: An Alternative Method to Construct Truth Tables (cont) Slide 28 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: An Alternative Method to Construct Truth Tables (cont) b) We will construct a truth table for by performing the calculations as follows: Step 1: Calculate Step 2: Calculate Step 3: Calculate ( ) ( ) ∨ ∧ p q r ( ).
∨ p q .
r ( ) ( ).
∨ ∧ p q r Slide 29 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Example: An Alternative Method to Construct Truth Tables (cont) Slide 30 Copyright © 2018, 2014, 2010 Pearson Education Inc. A L W A Y S L E A R N I N G Assignment Do 3.2.38 and turn in next class. Show how you came to your conclusion by building a Truth Table. |
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Symmetric sum
The symmetric sum of a function of variables is defined to be , where ranges over all permutations of .
More generally, a symmetric sum of variables is a sum that is unchanged by any permutation of its variables.
Any symmetric sum can be written as a polynomial of elementary symmetric sums.
A symmetric function of variables is a function that is unchanged by any permutation of its variables. The symmetric sum of a symmetric function therefore satisfies
Given variables and a symmetric function with , the notation is sometimes used to denote the sum of over all subsets of size in .
See also
Cyclic sum
Muirhead's Inequality
PaperMath’s sum
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5121 | http://www.sciencemadness.org/smwiki/index.php/Bromoform | Bromoform - Sciencemadness Wiki
Bromoform
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Bromoform | Names |
| IUPAC name Tribromomethane |
| Other names Methane tribromide Methenyl tribromide Methyl tribromide |
| Properties |
| Chemical formula | CHBr 3 |
| Molar mass | 252.731 g/mol |
| Appearance | Clear colorless liquid |
| Odor | Sweet, typical halocarbon |
| Density | 2.8912 g/cm 3 (20 °C) |
| Melting point | 8.3°C (46.9°F; 281.4 K) |
| Boiling point | 149.5°C (301.1°F; 422.6 K) |
| Solubility in water | 0.1 g/100 ml (20 °C) |
| Solubility | Miscible with acetone, benzene, chloroform, DCM, diethyl ether, ethanol, hexane, methanol, oils, petroleum ether, pyridine, toluene, xylene |
| Vapor pressure | 5.025 mmHg at 20 °C |
| Acidity (p K a) | 13.7 |
| Thermochemistry |
| Std enthalpy of formation (Δ f H~~o~~298) | 6.1–12.7 kJ/mol |
| Hazards |
| Safety data sheet | Sigma-Aldrich |
| Lethal dose or concentration (LD, LC): |
| LD 50 (Median dose) | 933 mg/kg (rat, oral) |
| LC 50 (Median concentration) | 1,151 ppm (mammal) |
| Related compounds |
| Related compounds | Chloroform Iodoform |
| Except where otherwise noted, data are given for materials in their standard state (at 25°C [77°F], 100 kPa). |
| Infobox references |
| | |
Bromoform is a brominated organic solvent, colorless liquid at room temperature, with a high refractive index, very high density, and sweet odor is similar to that of chloroform. It is one of the four haloforms, the others being fluoroform, chloroform and iodoform. Bromoform has the chemical formula CHBr 3.
Contents
1 Properties
1.1 Chemical
1.2 Physical
2 Availability
3 Preparation
4 Projects
5 Handling
5.1 Safety
5.2 Storage
5.3 Disposal
6 References
6.1 Relevant Sciencemadness threads
Properties
Chemical
Partial reduction of bromoform yields dibromomethane.
Physical
Bromoform is a dense colorless liquid at standard conditions, with a high refractive index, with a sweet odor typical of halocarbons and immiscible with water and polar solvents, but miscible with most non-polar solvents.
Availability
Bromoform is sold by chemical suppliers.
Small amounts of bromoform are produced by seaweed.
Preparation
Bromoform can be prepared by the haloform reaction using a methyl ketone like acetone and sodium hypobromite.
(CH 3)2 C=O + 3 NaOBr → CHBr 3 + 2 NaOH + CH 3 COONa
Sodium hypobromite can be prepared by adding bromine to sodium hydroxide.
Other routes involve the electrolysis of potassium bromide in ethanol, or by treating chloroform with aluminium bromide.
Projects
Extraction solvent
Separation of minerals by density
Compound collecting
Handling
Safety
Bromoform is harmful and exposure may cause irritation to the respiratory tract, the eyes and the skin, and may cause effects on the central nervous system and liver, resulting in impaired functions. It can be absorbed into the body by inhalation and through the skin.
The International Agency for Research on Cancer (IARC) concluded that bromoform is not classifiable as to human carcinogenicity, however the EPA classifies bromoform as a probable human carcinogen.
Storage
In closed amber glass bottles with a suitable cap, in a dark place away from heat and light. Ethanol is sometimes used as stabilizer.
Disposal
Bromoform should be first diluted, before destroyed with a solution of sodium hydroxide.
References
Relevant Sciencemadness threads
Bromoform
"Large" Scale Bromoform Synthesis
Ultrahigh-purity CHBr3 preparation (interesting use, 1atm synthetic ultrananocrystalline hexagonal diamond
Tribromomethane (bromoform) purity
Bromine from bromoform
Retrieved from "
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Tools & Reference>Gastroenterology
Ascites Treatment & Management
Updated: May 30, 2023
Author: Rahil Shah, MD; Chief Editor: Praveen K Roy, MD, MSc more...;)
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Sections
Ascites
Sections Ascites
Overview
Background
Pathophysiology
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Epidemiology
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Treatment
Medical Care
Sodium restriction (20-30 mEq/d) and diuretic therapy constitute the standard medical management for ascites and are effective in approximately 95% of patients. Note the following:
Water restriction is used only if persistent hyponatremia is present (see Diet, below).
Patients with ascites and hyponatremia may be treated with aquaretics—vasopressin V2-receptor antagonists that promote excretion of electrolyte-free water. Tolvaptan has been approved by the FDA for treating hypervolemic hyponatremia in ascites. Conivaptan has been approved by the FDA for treating hypervolemic hyponatremia in syndrome of inappropriate antidiuretic hormone; however, studies specific to efficacy in cirrhosis are lacking.
In a multicenter study that assessed the safety and efficacy of an automated pump system for the treatment of refractory ascites in 40 patients at 9 centers, Bellot et al reported the automated pump was an efficacious tool to remove ascites from the peritoneal cavity to the bladder. During the 6-month follow-up period, 90% of the ascites was removed with the pump system; there was also a significant reduction in the monthly median number of large volume paracentesis as well as a reduction in the number of cirrhosis-related adverse events.
Therapeutic paracentesis may be performed in patients who require rapid symptomatic relief for refractory or tense ascites. When small volumes of ascitic fluid are removed, saline alone is an effective plasma expander. The removal of 5 L of fluid or more is considered large-volume paracentesis. Total paracentesis, that is, removal of all ascites (even >20 L), can usually be performed safely.
Supplementing 5 g of albumin per each liter over 5 L of ascitic fluid removed decreases complications of paracentesis, such as electrolyte imbalances and increases in serum creatinine levels secondary to large shifts of intravascular volume. Note: The AASLD indicates that postparacentesis albumin infusion may not be necessary for a single paracentesis of less than 4 to 5 L ((class I, level C recommendation); however, for large-volume paracenteses, an albumin infusion of 6-8 g per liter of fluid removed appears to improve survival and is recommended (class IIa, level C recommendation).
To avoid exposing patients to blood products, the use of terlipressin (eg, 1 mg every 4 hours for 48 hours) rather than albumin has been proposed for prevention of circulatory dysfunction after large-volume paracentesis. Initial studies suggest that terlipressin is as effective as albumin for this purpose. [20, 21]
Repeated therapeutic paracentesis can be used to treat refractory ascites (class I, level C recommendation). For palliative care in patients with advanced cancer, an alternative to serial paracenteses is placement of an indwelling peritoneal catheter; ascitic fluid can then be removed by continuous drainage or intermittent drainage with a proprietary system utilizing vacuum bottles, which can be performed in the patient's home. Preservation of good nutrition status is important.
The transjugular intrahepatic portosystemic shunt (TIPS) is an interventional radiologic technique that reduces portal pressure and may be the most effective treatment for patients with diuretic-resistant ascites. In the procedure, which is performed with the patient under conscious sedation or general anesthesia, an interventional radiologist places a stent percutaneously from the right jugular vein into the hepatic vein, thereby creating a connection between the portal and systemic circulations. TIPS is gradually becoming the standard of care in patients with diuretic-refractory ascites.
In a systematic review and meta-analysis of 10 trials comprising 462 patients with cirrhotic ascites, Guo et al reported that midodrine, a vasopressor, used as a novel therapy for the ascites caused by cirrhosis did not improve survival but potentially improved response rates and reduced plasma renin activity. However, when midodrine was used as an alternative to albumin in large-volume paracentesis, the mortality was higher for those receiving midodrine than for those receiving albumin; midodrine and albumin had a similar association with the development of paracentesis-induced circulatory dysfunction.
Conservative management appears to be the treatment of choice for patients with chylous ascites.
Patients can actually be maintained free of ascites if sodium intake is limited to 10 mmol/d. However, this is not practical outside a metabolic ward.
Twenty-four–hour urinary sodium measurements are useful in patients with ascites related to portal hypertension in order to assess the degree of sodium avidity, monitor the response to diuretics, and assess compliance with diet.
For grade 3 or 4 ascites, therapeutic paracentesis may be necessary intermittently.
Consultation with a gastrointestinal specialist and/or hepatologist should be considered for all patients with ascites, particularly if the ascites is refractory to medical treatment.
Next: Surgical Care
Surgical Care
The peritoneovenous shunt is an alternative for patients with medically intractable ascites. This is a megalymphatic shunt that returns the ascitic fluid to the central venous system. Beneficial effects of these shunts include increased cardiac output, renal blood flow, glomerular filtration rate, urinary volume, and sodium excretion and decreased plasma renin activity and plasma aldosterone concentration. Although it has largely been supplanted by transjugular intrahepatic portosystemic shunt (TIPS), peritoneovenous shunting has been shown to improve short-term survival (compared with paracentesis) in cancer patients with refractory malignant ascites. The AASLD suggests considering peritoneovenous shunting for patients with refractory ascites who are not candidates for paracentesis, transplant, or TIPS (class I, level A recommendation).
Ascites. Transjugular intrahepatic portosystemic shunt (TIPS).
View Media Gallery)
The AASLD recommends that patients with cirrhosis and ascites be considered for liver transplantation (class I, level B recommendation).
Previous
Next: Surgical Care
Diet
Sodium restriction of 500 mg/d (22 mmol/d) is feasible in a hospital setting; however, it is unrealistic in most outpatient settings. A more appropriate sodium restriction is 2000 mg/d (88 mmol). Indiscriminate fluid restriction is inappropriate. Fluids need not be restricted unless the serum sodium level drops below 120 mmol/L.
Previous
Next: Surgical Care
Long-Term Monitoring
The best method of assessing the effectiveness of diuretic therapy is by monitoring body weight and urinary sodium levels.
In general, the goal of diuretic treatment of ascites should be to achieve a weight loss of 300-500 g/d in patients without edema and 800-1000 g/d in patients with edema.
Once ascites has disappeared, diuretic treatment should be adjusted to the minimum effective dose to avoid recurrences.
Previous
Medication
References
Chiejina M, Kudaravalli P, Samant H. Ascites. StatPearls [Internet]. 2023 Jan. [QxMD MEDLINE Link].[Full Text].
Pericleous M, Sarnowski A, Moore A, Fijten R, Zaman M. The clinical management of abdominal ascites, spontaneous bacterial peritonitis and hepatorenal syndrome: a review of current guidelines and recommendations. Eur J Gastroenterol Hepatol. 2016 Mar. 28(3):e10-8. [QxMD MEDLINE Link].
Weniger M, D'Haese JG, Angele MK, Kleespies A, Werner J, Hartwig W. Treatment options for chylous ascites after major abdominal surgery: a systematic review. Am J Surg. 2016 Jan. 211(1):206-13. [QxMD MEDLINE Link].
CDC/National Center for Health Statistics. Chronic liver disease and cirrhosis. Centers for Disease Control and Prevention. January 17, 2023. Available at Accessed: May 30, 2023.
Raza S, Rajak S, Upadhyay A, Tewari A, Sinha RA. Current treatment paradigms and emerging therapies for NAFLD/NASH. Front Biosci (Landmark Ed). 2021 Jan 1. 26(2):206-37. [QxMD MEDLINE Link].[Full Text].
Asrani SK, Devarbhavi H, Eaton J, Kamath PS. Burden of liver diseases in the world. J Hepatol. 2019 Jan. 70(1):151-71. [QxMD MEDLINE Link].
Fede G, D'Amico G, Arvaniti V, et al. Renal failure and cirrhosis: a systematic review of mortality and prognosis. J Hepatol. 2012 Apr. 56(4):810-8. [QxMD MEDLINE Link].
Wallerstedt S, Olsson R, Simren M, et al. Abdominal tenderness in ascites patients indicates spontaneous bacterial peritonitis. Eur J Intern Med. 2007 Jan. 18(1):44-7. [QxMD MEDLINE Link].
Tapper EB, Bonder A, Cardenas A. Preventing and treating acute kidney injury among hospitalized patients with cirrhosis and ascites: a narrative review. Am J Med. 2016 May. 129(5):461-7. [QxMD MEDLINE Link].
Han CM, Lee CL, Huang KG, et al. Diagnostic laparoscopy in ascites of unknown origin: Chang Gung Memorial Hospital 20-year experience. Chang Gung Med J. 2008 Jul-Aug. 31(4):378-83. [QxMD MEDLINE Link].[Full Text].
Wong CL, Holroyd-Leduc J, Thorpe KE, Straus SE. Does this patient have bacterial peritonitis or portal hypertension? How do I perform a paracentesis and analyze the results?. JAMA. 2008 Mar 12. 299(10):1166-78. [QxMD MEDLINE Link].
[Guideline] Biggins SW, Angeli P, Garcia-Tsao G, et al. Diagnosis, evaluation, and management of ascites, spontaneous bacterial peritonitis and hepatorenal syndrome: 2021 practice guidance by the American Association for the Study of Liver Diseases. Hepatology. 2021 Aug. 74(2):1014-48. [QxMD MEDLINE Link].[Full Text].
[Guideline] Aithal GP, Palaniyappan N, China L, et al. Guidelines on the management of ascites in cirrhosis. Gut. 2021 Jan. 70(1):9-29. [QxMD MEDLINE Link].[Full Text].
Gallo A, Dedionigi C, Civitelli C, Panzeri A, Corradi C, Squizzato A. Optimal management of cirrhotic ascites: a review for internal medicine physicians. J Transl Int Med. 2020 Dec. 8(4):220-36. [QxMD MEDLINE Link].[Full Text].
Gines P, Cardenas A. The management of ascites and hyponatremia in cirrhosis. Semin Liver Dis. 2008 Feb. 28(1):43-58. [QxMD MEDLINE Link].
Facciorusso A, Amoruso A, Neve V, Antonino M, Prete VD, Barone M. Role of vaptans in the management of hydroelectrolytic imbalance in liver cirrhosis. World J Hepatol. 2014 Nov 27. 6(11):793-9. [QxMD MEDLINE Link].[Full Text].
Bellot P, Welker MW, Soriano G, et al. Automated low flow pump system for the treatment of refractory ascites: a multi-center safety and efficacy study. J Hepatol. 2013 May. 58(5):922-7. [QxMD MEDLINE Link].
Sola-Vera J, Minana J, Ricart E, et al. Randomized trial comparing albumin and saline in the prevention of paracentesis-induced circulatory dysfunction in cirrhotic patients with ascites. Hepatology. 2003 May. 37(5):1147-53. [QxMD MEDLINE Link].[Full Text].
[Guideline] Runyon BA, American Association for the Study of Liver Diseases. Management of adult patients with ascites due to cirrhosis: update 2012. National Guideline Clearinghouse. Available at Accessed: Nov 6, 2014.
Lata J, Marecek Z, Fejfar T, et al. The efficacy of terlipressin in comparison with albumin in the prevention of circulatory changes after the paracentesis of tense ascites--a randomized multicentric study. Hepatogastroenterology. 2007 Oct-Nov. 54(79):1930-3. [QxMD MEDLINE Link].
Singh V, Kumar R, Nain CK, Singh B, Sharma AK. Terlipressin versus albumin in paracentesis-induced circulatory dysfunction in cirrhosis: a randomized study. J Gastroenterol Hepatol. 2006 Jan. 21(1 pt 2):303-7. [QxMD MEDLINE Link].
Mercadante S, Intravaia G, Ferrera P, Villari P, David F. Peritoneal catheter for continuous drainage of ascites in advanced cancer patients. Support Care Cancer. 2008 Aug. 16(8):975-8. [QxMD MEDLINE Link].
Courtney A, Nemcek AA Jr, Rosenberg S, et al. Prospective evaluation of the PleurX catheter when used to treat recurrent ascites associated with malignancy. J Vasc Interv Radiol. 2008 Dec. 19(12):1723-31. [QxMD MEDLINE Link].
Sorrentino P, Castaldo G, Tarantino L, et al. Preservation of nutritional-status in patients with refractory ascites due to hepatic cirrhosis who are undergoing repeated paracentesis. J Gastroenterol Hepatol. 2012 Apr. 27(4):813-22. [QxMD MEDLINE Link].
Guo TT, Yang Y, Song Y, Ren Y, Liu ZX, Cheng G. Effects of midodrine in patients with ascites due to cirrhosis: Systematic review and meta-analysis. J Dig Dis. 2016 Jan. 17(1):11-9. [QxMD MEDLINE Link].
Seike M, Maetani I, Sakai Y. Treatment of malignant ascites in patients with advanced cancer: peritoneovenous shunt versus paracentesis. J Gastroenterol Hepatol. 2007 Dec. 22(12):2161-6. [QxMD MEDLINE Link].
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Liu YB, Chen MK. Epidemiology of liver cirrhosis and associated complications: current knowledge and future directions. World J Gastroenterol. 2022 Nov 7. 28(41):5910-30. [QxMD MEDLINE Link].[Full Text].
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Ascites. This computed tomography scan demonstrates free intraperitoneal fluid due to urinary ascites.
Ascites. Transjugular intrahepatic portosystemic shunt (TIPS).
Ascites. Massive ascites.
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Rahil Shah, MD Consulting Staff, Lebanon Endoscopy Center
Rahil Shah, MD is a member of the following medical societies: American College of Gastroenterology, American Society for Gastrointestinal Endoscopy
Disclosure: Received consulting fee from Takeda for speaking and teaching.
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Janice M Fields, MD, FACG, FACP Assistant Professor of Internal Medicine, Oakland University William Beaumont School of Medicine; Consulting Staff, Department of Internal Medicine, Section of Gastroenterology, Providence Hospital, St John Macomb-Oakland Hosptial
Janice M Fields, MD, FACG, FACP is a member of the following medical societies: American College of Gastroenterology, American College of Physicians-American Society of Internal Medicine, American Gastroenterological Association, American Medical Association, American Society for Gastrointestinal Endoscopy, National Medical Association
Disclosure: Nothing to disclose.
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Disclosure: Received salary from Medscape for employment.
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For `0ltxlt(pi)/(2), (1+4cosecx)(1+8secx...
For 0<x<π 2,(1+4 cos e c x)(1+8 sec x), is
A
≥81
B
81
C
83
D
≥83
Video Solution
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Text Solution
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The correct Answer is:B
To solve the inequality (1+4 csc x)(1+8 sec x) for 0<x<π 2, we will follow these steps:
Step 1: Rewrite the expressions
We start by rewriting the cosecant and secant functions in terms of sine and cosine:
csc x=1 sin x and sec x=1 cos x
Thus, we can rewrite the expression as:
(1+4 csc x)(1+8 sec x)=(1+4 sin x)(1+8 cos x)
Step 2: Simplify the expression
Now, we can expand the expression:
=1+4 sin x+8 cos x+32 sin x cos x
Step 3: Combine terms
We can combine the terms into a single fraction:
=sin x cos x+4 cos x+8 sin x+32 sin x cos x
Step 4: Use the identity for sine and cosine
Recall the identity 2 sin x cos x=sin 2 x:
=2 sin x cos x+8 sin x+4 cos x+32 sin x cos x
Step 5: Find the maximum of the denominator
To minimize the entire expression, we need to maximize the denominator sin x cos x. The maximum value occurs at x=π 4:
sin x cos x=1 2 when x=π 4
Step 6: Substitute x=π 4
Now, substituting x=π 4:
sin π 4=1√2 and cos π 4=1√2
Thus,
=2⋅1 2+8⋅1√2+4⋅1√2+32 1 2
=1+8√2+4√2+32 1 2
=2(1+12√2+32)
=2(33+12√2)
Step 7: Calculate the value
Now we can calculate the approximate value:
=66+24√2
Using √2≈1.414:
=66+24×1.414≈66+33.936≈99.936
Conclusion
Thus, the value of (1+4 csc x)(1+8 sec x) is approximately 99.936, which is greater than 81.
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5125 | https://www.thesaurus.com/browse/progeny | 192 Synonyms & Antonyms for PROGENY | Thesaurus.com
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noun (1)
as in offspring
Example SentencesRelated WordsQuiz
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View definitions for progeny
progeny
noun as in offspring
Synonyms Antonyms
Strong matches
breed
children
descendants
family
get
issue
kids
lineage
posterity
race
scions
seed
stock
young
Weak matches
begats
progeniture
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The series wisely taps some familiar faces from Bravo shows of yore, tossing “Housewives” progenies Ariana Biermann, Riley Burruss, Gia Giudice and Brooks Marks into the batter before adding a few special ingredients.
FromSalon
In any other sport, the progeny of a former star player ascending to the highest level would be especially noteworthy.
FromLos Angeles Times
Rich men can afford to support lots of progeny they never have to see.
FromSalon
Conscious robots, he believes, "are our progeny. Down the road, machines like these will be entities that will be on Earth and maybe on other planets when we are no longer around".
FromBBC
But there’s an attitude, a worldview and a fundamental set of principles that guide the tech industry and its progeny, like a secular catechism.
FromLos Angeles Times
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Related Words
Words related to progeny are not direct synonyms, but are associated with the word progeny. Browse related words to learn more about word associations.
breed
noun as in kind, class
brand
character
extraction
family
feather
genus
ilk
likes
line
lineage
lot
nature
number
pedigree
progeny
race
sort
species
stamp
stock
strain
stripe
type
variety
brood
noun as in cluster of young
begats
breed
chicks
clutch
descendants
family
flock
hatch
infants
issue
litter
offspring
posterity
progeniture
progeny
scions
seed
young
descendant
noun as in person in line of ancestry
brood
child
children
chip off old block
get
heir
issue
kin
offshoot
offspring
posterity
product
progeniture
progeny
scion
seed
spin-off
families
noun as in kin, offspring; classification
ancestors
ancestry
birth
blood
brood
children
clan
class
descendants
descent
dynasty
extraction
folk
forebears
genealogy
generations
genre
group
heirs and assigns
house
household
in-laws
inheritance
issue
kind
kindred
kith and kin
line
lineage
menage
ménage
network
parentage
pedigree
people
progenitors
progeny
race
relations
relationship
relatives
siblings
strain
subdivision
system
tribe
family
noun as in kin, offspring; classification
ancestors
ancestry
birth
blood
brood
children
clan
class
descendants
descent
dynasty
extraction
folk
forebears
genealogy
generations
genre
group
heirs and assigns
house
household
in-laws
inheritance
issue
kind
kindred
kith and kin
line
lineage
menage
ménage
network
parentage
pedigree
people
progenitors
progeny
race
relations
relationship
relatives
siblings
strain
subdivision
system
tribe
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Q: Instead of the term COTTON CANDY, British English speakers are more likely to use the word …
candycotton.
candyfloss.
candyfluff.
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5126 | https://center.ef.com.cn/blog/article/speaking-conversation/ | 英语口语情景对话 | EF成人英语
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5127 | https://www.quora.com/What-are-some-examples-of-a-real-life-situation-illustrating-each-of-the-following-inductive-reasoning-deductive-Reasoning-analogy-and-intuition | Something went wrong. Wait a moment and try again.
Cognitive Processes
Logic (philosophy)
Real Life Examples of Mea...
Analogy Method
Inductive Inference
Real Life Case Studies
5
What are some examples of a real-life situation illustrating each of the following inductive reasoning, deductive Reasoning, analogy, and intuition?
Dennis Rohatyn
Author has 3.3K answers and 608.2K answer views
·
4y
Glad to oblige.
Deductive: All men are mortal. Donald Trump is a man. Therefore, despite
Regeneron, and a team of doctors at Walter Reed, Donald Trump is mortal.
Inductive: Donald Trump cheated on his taxes. Most people who cheat on
their taxes are caught and punished within a few years. Therefore, there is
a good chance that Donald Trump will be caught and punished for failing to
pay his taxes to the federal government, of which he is no longer a member.
Intuitive: Something inside tells me that Donald Trump ain’t gonna make it.
Got it? Now collect your fee from Quora, and don’t ask questions to whic
Glad to oblige.
Deductive: All men are mortal. Donald Trump is a man. Therefore, despite
Regeneron, and a team of doctors at Walter Reed, Donald Trump is mortal.
Inductive: Donald Trump cheated on his taxes. Most people who cheat on
their taxes are caught and punished within a few years. Therefore, there is
a good chance that Donald Trump will be caught and punished for failing to
pay his taxes to the federal government, of which he is no longer a member.
Intuitive: Something inside tells me that Donald Trump ain’t gonna make it.
Got it? Now collect your fee from Quora, and don’t ask questions to which
anyone can find the answers in a logic textbook, or by learning to read, and
write, and think. [Disclaimer: The opinions expressed here are not those of
Donald Trump, Elise Stefanik or even (so far as I know) Quora—thank God].
Dan Schmidt
Full-Triangle Builder
·
Author has 124 answers and 1.2M answer views
·
12y
Related
What are the best real examples of deductive, inductive or abductive reasoning? I mean non-mathematical deduction, with a logical chain of reasoning, leading to a compelling conclusion that convincingly demonstrated that the reasoning was correct.
John Snow's work to trace the source of a cholera outbreak in 1854 in Soho, England seems to be a good example of this. Here's a description from the BBC web site (
At the time, it was assumed that cholera was airborne. However, Snow did not accept this 'miasma' (bad air) theory, arguing that in fact entered the body through the mouth. He published his ideas in an essay 'On the Mode of Communication of Cholera' in 1849. A few years later, Snow was able to prove his theory in dramatic circumstances. In August 1854, a cholera outbreak
John Snow's work to trace the source of a cholera outbreak in 1854 in Soho, England seems to be a good example of this. Here's a description from the BBC web site (
At the time, it was assumed that cholera was airborne. However, Snow did not accept this 'miasma' (bad air) theory, arguing that in fact entered the body through the mouth. He published his ideas in an essay 'On the Mode of Communication of Cholera' in 1849. A few years later, Snow was able to prove his theory in dramatic circumstances. In August 1854, a cholera outbreak occurred in Soho. After careful investigation, including plotting cases of cholera on a map of the area, Snow was able to identify a water pump in Broad (now Broadwick) Street as the source of the disease. He had the handle of the pump removed, and cases of cholera immediately began to diminish. However, Snow's 'germ' theory of disease was not widely accepted until the 1860s.
According to the Wikipedia page dedicated to this topic ( "This discovery came to influence public health and the construction of improved sanitation facilities beginning in the 19th century."
I learned of this example in Edward Tufte's book The Visual Display of Quantitative Information. Tufte uses Snow's work to demonstrate the power of information visualization in problem solving. Here's the map Snow used to connect instances of the disease to the location of a water pump.
Assistant
Bot
·
1y
Here are real-life situations illustrating each type of reasoning:
Inductive Reasoning
Example: A teacher notices that every time she gives a quiz on Fridays, her students perform better than on quizzes given on Mondays. She concludes that quizzes on Fridays lead to better performance.
Explanation: This reasoning is based on specific observations (students performing better on Fridays) and leads to a general conclusion (Fridays are better for quiz performance).
Deductive Reasoning
Example:
- Premise 1: All humans are mortal.
- Premise 2: Socrates is a human.
- Conclusion: Therefore, Socrate
Here are real-life situations illustrating each type of reasoning:
Inductive Reasoning
Example: A teacher notices that every time she gives a quiz on Fridays, her students perform better than on quizzes given on Mondays. She concludes that quizzes on Fridays lead to better performance.
Explanation: This reasoning is based on specific observations (students performing better on Fridays) and leads to a general conclusion (Fridays are better for quiz performance).
Deductive Reasoning
Example:
- Premise 1: All humans are mortal.
- Premise 2: Socrates is a human.
- Conclusion: Therefore, Socrates is mortal.
Explanation: This reasoning starts with a general statement and applies it to a specific case to arrive at a conclusion.
Analogy
Example: A doctor compares the human brain to a computer. She explains that just as a computer processes information and can crash if overloaded, the brain processes thoughts and can become overwhelmed if too much information is presented at once.
Explanation: This reasoning uses a comparison between two different things (brain and computer) to draw a conclusion about their similarities in function.
Intuition
Example: A seasoned firefighter enters a burning building and immediately feels that the structure is unstable. Based on his years of experience, he trusts his gut feeling and decides to evacuate his team.
Explanation: This situation illustrates intuition, where the firefighter relies on his instinctive understanding and experience rather than a formal analysis of the situation.
These examples show how different forms of reasoning can be applied in real-life scenarios to draw conclusions and make decisions.
Related questions
What are some deductive reasoning examples found in everyday life?
What are some examples of inductive reasoning in everyday life?
What are the best real examples of deductive, inductive or abductive reasoning? I mean non-mathematical deduction, with a logical chain of reasoning, leading to a compelling conclusion that convincingly demonstrated that the reasoning was correct.
What is a real life situation where we use inductive reasoning?
What are the best examples of deductive reasoning?
Daniel Chen
Works at Uber (company)
·
Author has 131 answers and 642.1K answer views
·
13y
Related
What is inductive and deductive reasoning?
Deductive reasoning refers to the act of reaching a conclusion by showing that such a conclusion must follow from a set of premises. In contrast, inductive reasoning refers to the act of reaching a conclusion by abstracting or generalizing a premise.
One of the most famous examples of deductive reasoning is the argument that Socrates is a mortal. It goes something like this:
All men are mortal.
Socrates is a man.
Thus, Socrates is mortal.
If you accept that premise 1 and premise 2 hold, then you must also accept that premise 3 holds.
There are many forms of inductive reasoning (I recommend
Deductive reasoning refers to the act of reaching a conclusion by showing that such a conclusion must follow from a set of premises. In contrast, inductive reasoning refers to the act of reaching a conclusion by abstracting or generalizing a premise.
One of the most famous examples of deductive reasoning is the argument that Socrates is a mortal. It goes something like this:
All men are mortal.
Socrates is a man.
Thus, Socrates is mortal.
If you accept that premise 1 and premise 2 hold, then you must also accept that premise 3 holds.
There are many forms of inductive reasoning (I recommend to learn about them), but a simple example of inductive reasoning is as follows: My cat is meaner than my dog, so all cats are meaner than all dogs.
Aishwa Patnaik
Environment & Economics Enthusiast
·
Updated 1y
Related
What is inductive and deductive reasoning?
INDUCTIVE REASONING
Consider a basket full of mangoes. We have to check here whether the mangoes are raw or ripe. We can find out this by observing the mangoes individually.
First, we pick up a mango from the basket and observe it. Suppose we find the mango to be raw. Then we pick up another mango from the basket and find out that’s raw as well. Based on this, many of us would conclude that all the mangoes in the basket are raw. So, here we examined a couple of mangoes in the basket and accordingly, arrived at a general conclusion. We generalized the idea by saying that all the mangoes in the ba
INDUCTIVE REASONING
Consider a basket full of mangoes. We have to check here whether the mangoes are raw or ripe. We can find out this by observing the mangoes individually.
First, we pick up a mango from the basket and observe it. Suppose we find the mango to be raw. Then we pick up another mango from the basket and find out that’s raw as well. Based on this, many of us would conclude that all the mangoes in the basket are raw. So, here we examined a couple of mangoes in the basket and accordingly, arrived at a general conclusion. We generalized the idea by saying that all the mangoes in the basket are raw. So by observing specific outcomes of the experiments, we concluded the observation in a generalized form. This approach of reasoning from specific to general is called as Inductive Reasoning.
Inductive Reasoning is logically true but may or may not be realistically true.
SPECIFIC ARGUMENT to GENERAL CONCLUSION
For example-
Statement 1- Mango is a fruit ( specific statement)
Statement 2- The box is full of fruits (specific statement)
Conclusion- The box is full of mangoes (this is a generalized conclusion)
Here, statements 1 and 2 are true but the conclusion drawn is wrong although it is logically true. It can be false if the basket contains any other fruit other than mangoes. The conclusion is logically true but not definitely true( realistically may or may not be true).
DEDUCTIVE REASONING
GENERAL ARGUMENT to SPECIFIC CONCLUSION
Logically True
Realistically True
Unlike Inductive reasoning, Deductive reasoning is always true.
For example-
Statement 1- All mangoes are fruits ( general statement)
Statement 2-All fruits have seeds( general statement)
Conclusion- Mangoes have seeds (specific conclusion)
Hence, the conclusion is both Logically true and Realistically true.
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Grammarly
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Aug 18
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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Samuel Muldoon
Bs in Mathematics, University of Colorado Denver (Graduated 2017)
·
Author has 1K answers and 1.3M answer views
·
5y
Related
What are some deductive reasoning examples found in everyday life?
Occasionally, I engage in what I call “imposition minimization.”
I try to impose other people as little as possible. Equivalently, I sometimes try to negatively impact other people as much as possible, or reduce their “utlity” as little as possible. The word “utility” comes from the field of economics; it is a numerical quantification of happiness. For example, 50 units of utility is 50 units of happiness.
Sometime I enjoy impacting other people very little. It makes me feel good about myself.
An example of “imposition minimization” occurred recently in my life. My father often goes to the gym. M
Occasionally, I engage in what I call “imposition minimization.”
I try to impose other people as little as possible. Equivalently, I sometimes try to negatively impact other people as much as possible, or reduce their “utlity” as little as possible. The word “utility” comes from the field of economics; it is a numerical quantification of happiness. For example, 50 units of utility is 50 units of happiness.
Sometime I enjoy impacting other people very little. It makes me feel good about myself.
An example of “imposition minimization” occurred recently in my life. My father often goes to the gym. My father invited me to go to the gym with him. I said yes.
I wanted to leave the gym whenever my father wanted to leave. That puts the smallest imposition on him: we leave when he wants to.
The first thing I asked myself was:
In order to minimize imposing on Dad,
(1) should Dad come get me when he is done working out,
or
(2) should I go get him when I am done working out?
So, I explored the implications.
If I go get dad when I am done working out, then it is possible that…
my father will be ready to leave before I am done
my father will not yet be ready to leave yet when I am done.
If I go get Dad when I am done, then it’s possible I would “impose” on my father. For example, if my father wants to leave, but I do not go find him until 45 minutes later, then he will have been waiting for me for 45 minutes. He does not want to wait. That is an imposition.
So, I decided Dad should find me when he was finished.
Secondly, I don’t want for him to have to search all over hell’s half acre to find me. That would be an imposition. So, I chose a spot I would work out. I made sure to choose a highly visible, easy to see location.
I told my father where I would be. I did not gesture vaguely, there was a big sign on the wall. I told my father “I’ll be in that room with the sign on the wall.” I told him that when he was ready to go home, he should come get me.
Even if I wanted to use different machines, etc… I would not do so. I would not move from the spot I told him about. I would stay working out in the same spot the whole time, so he didn’t have to search for me. That way, when he wanted to leave, he could find me quickly, and we leave as close to when he wanted to leave as possible.
Example #2.
My father has a vegetable garden. When the first snow of winter comes, the tomato plants often still have many green unripe fruit. Sometimes, we have picked the green tomatoes, pureed them in a blender, and frozen them. You can make green tomato and beef stew, and things like that.
We have also shredded and frozen zucchini too. Frozen zucchini mush is fine for making zucchini bread or pumpkin pie.
For many years, my father could not tell the difference between frozen zucchini and frozen green tomato.
So I explained:
The inside of a zucchini: BRIGHT GREEN
The inside of a green tomatoe: also BRIGHT GREEN
However, the outside colors are different.
green tomatoes outside: BRIGHT GREEN
zucchini outside: DARK GREEN.
Thus, there exist flecks of dark green color inside of the bags of frozen zucchini.
The bags of green tomato are pure light green.
There do not exist dark green flecks inside of the bags of green tomato.
Related questions
In your daily life, give 1 example of each situation that can be compared to inductive and deductive reasoning?
What are some true examples of inductive reasoning?
What is an example of inductive reasoning that you have used in your life? How did it help you?
What is an example of a real life situation illustrating each of the following analogy reasoning fundamentals?
Does deductive reasoning give an example of a real life situation?
Dan Stillit
Former Former Managing Director, Research at UBS
·
Author has 90 answers and 328.3K answer views
·
12y
Related
What is inductive and deductive reasoning?
It's worth distinguishing inductive from deductive reasoning. Sherlock Holmes' reasoning was inductive, notwithstanding his famous comment "Simple deduction Watson."
Valid deductive arguments are ones that cannot be false if the premises from which they follow are true. That is, the conclusion of the argument is true and is entailed by the premises. It should be viewed as a mechanical relationship between premises and conclusion. If it turns out that the premises are actually true, then the argument is called 'sound'.
Deduction is found in two strands of logical reasoning: propositional logic a
It's worth distinguishing inductive from deductive reasoning. Sherlock Holmes' reasoning was inductive, notwithstanding his famous comment "Simple deduction Watson."
Valid deductive arguments are ones that cannot be false if the premises from which they follow are true. That is, the conclusion of the argument is true and is entailed by the premises. It should be viewed as a mechanical relationship between premises and conclusion. If it turns out that the premises are actually true, then the argument is called 'sound'.
Deduction is found in two strands of logical reasoning: propositional logic and categorical logic. Propositional logic looks at reasoning using simple or complex propositions or statement. For example, "If p, then q" is a premise and "p" is a second premise, then the two premises together if true imply "q". With categorical logic, the propositions or statements relate to sets or groups or categories of people or objects or subjects. An example might be:
All diplomats are charming All D is C
Gordon is a diplomat g is L
Thus, Gordon is charming g is C
In inductive reasoning (in contrast to deductive reasoning) we are concerned with the probability of the conclusion given the premises. All inductive arguments are vulnerable to contrary evidence, which can weaken them. This 'vulnerability' is referred to as “defeasibility” – in contrast to indefeasibility of deductive logic. Thus, as Ian Hacking has put it ('Probability and Inductive Logic') "Inductive logic is about risky arguments." Not coincidently, induction lies close to the field of probability, statistics and decision theory.
Inductive logic employs various tools, many of which use the premises-plus-conclusion structure: Inference to the best explanation, statistical generalisation, and statistical application, argument by analogy and causal reasoning (much rooted in the work of John Stuart Mill). The first of these, inference to the best explanation was referred to as "abduction" by the 19th century philosopher Charles Seymour Peirce.
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Mark Bradley
Economist
·
Updated Aug 14
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Greg Burns
Systems Engineer - Safety at Various Companies (2008–present)
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Author has 4.1K answers and 9.7M answer views
·
8y
Related
Is inductive or deductive reasoning more scientific?
Deductive reasoning is more logically certain, however the vast majority of science fits within what would be called inductive reasoning.
The reason science is mostly inductive is because science is based on empiricism. Empiricism states that things are only considered ‘real’ or ‘true’ if they are measurable. Science, with its main tool being experiment, is reliant on phenomena being measurable in order to make its findings.
Deductive reasoning enables us to establish things with certainty, since it starts with axioms (that are taken to be self evident) and forms deductive conclusions that do no
Deductive reasoning is more logically certain, however the vast majority of science fits within what would be called inductive reasoning.
The reason science is mostly inductive is because science is based on empiricism. Empiricism states that things are only considered ‘real’ or ‘true’ if they are measurable. Science, with its main tool being experiment, is reliant on phenomena being measurable in order to make its findings.
Deductive reasoning enables us to establish things with certainty, since it starts with axioms (that are taken to be self evident) and forms deductive conclusions that do not require measurement.
The best inductive reasoning can do is establish correlation and determine the statistical probability something is true. We can take those findings and perform deductive logic to find other outcomes, but we need to assume the findings are ‘true’ based on inductive reasoning.
Inductive reasoning isn’t possible to prove deductively without circular logic. It might be possible to compare some inductive evidence with a deductive proof, but that will only demonstrate a single case, not all possible cases, which is where science sits.
The story of Russell’s turkey gives a rationale as to why we should hold a little distrust of logic which is purely based on induction.
A turkey, in an american nurture, decide to shape its vision of the world scientifically well founded (a wissenschaftliche Weltauffassung, according to the Logical Positivism by the Wiener Kreis).
The turkey found that, on his first morning at the turkey farm, he was fed at 9 a.m. Being a good inductivist turkey he did not jump to conclusions. He waited until he collected a large number of observations that he was fed at 9 a.m. and made these observations under a wide range of circumstances, on Wednesdays, on Thursdays, on cold days, on warm days. Each day he added another observation statement to his list. Finally he was satisfied that he had collected a number of observation statements to inductively infer that “I am always fed at 9 a.m.”.
However on the morning of Christmas eve he was not fed but instead had his throat cut.
You haven’t mentioned the rare cousin to deduction and induction, which is logical abduction.
This is logic which goes against the flow and generates the principles that can help simplify and explain the details provided by inductive examination of evidence. If I also go strictly tongue in cheek, then you could consider when this is done by a foreigner that you will have concrete evidence of alien abduction.
Footnotes
Bertrand Russell’s Inductivist Turkey
Abductive reasoning - Wikipedia
Kevin Wayne Pledger
Math and Computer Science Teacher (2016–present)
·
7y
Related
What are the best real examples of deductive, inductive or abductive reasoning? I mean non-mathematical deduction, with a logical chain of reasoning, leading to a compelling conclusion that convincingly demonstrated that the reasoning was correct.
Deductive reasoning is when you apply one or more properties or rules in order to reach a conclusion. An example: “I had two cookies on the counter a minute ago. Now, there is only one cookie on the counter. My son is the only other person in the house. I did not take the missing cookie. Therefore, my son must have taken it.” That’s obviously not the best example in the world, but it’s the easiest to see. The best real examples would probably be the detective work used in solving cold cases. The case against the Golden State Killer is one of the more prominent, recent examples.
Abductive reason
Deductive reasoning is when you apply one or more properties or rules in order to reach a conclusion. An example: “I had two cookies on the counter a minute ago. Now, there is only one cookie on the counter. My son is the only other person in the house. I did not take the missing cookie. Therefore, my son must have taken it.” That’s obviously not the best example in the world, but it’s the easiest to see. The best real examples would probably be the detective work used in solving cold cases. The case against the Golden State Killer is one of the more prominent, recent examples.
Abductive reasoning comes in where there is more than one possible explanation for the conclusion, but we apply Occam’s Razor to find the most likely explanation. (The one with the fewest assumptions.) I try not to get political, but here’s a humorous example of abductive reasoning (source):
Finally, inductive reasoning is a weaker form of reasoning in which you assume that one particular piece of evidence supports a rule or property that can be applied to other pieces of evidence. Examples of this could happen when you look at a situation and attempt to infer a rule from it that you can apply to another, similar situation. One I can think of off the top of my head (as I look out my window): “We have watering restrictions in my area. My lawn is yellowing due to the watering restrictions in place and the summer heat. My neighbor’s lawn is lush and green. He must be violating the watering restrictions.”
Keep in mind that the results of inductive reasoning are considered weak because the conclusion is based on an assumption that may or may not hold for the second situation. [Note: This is different from (though similar to) inductive reasoning in mathematics, where you show that the property is true for and then show that if it is true for , then it must also be true for . The question asked about real-life examples.]
A good real-life example is the search for life, past or present, on Mars. There is evidence that water flowed on the Martian surface sometime in the past. Water is essential to life on Earth. Fixed nitrogen and carbon monoxide have been found on the planet as well as an as-yet unexplained source of methane. These are essential ingredients and waste products of microbes on Earth. Therefore, Mars may well have life on it currently or have had it sometime in the past.
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Which is usually more effective: inductive or deductive reasoning?
I think of it like pie.
Inductive is like having a slice of pie and drawing conclusions about what the whole of the pie may be.
Deductive reasoning is like having the whole pie (the surface of it at least) and trying to determine the ingredients or character of a single piece of it.
I would never say that one is more effective than the other because it would be like asking, “which is more effective when driving cars, taking right turns or taking left turns?” Depending on available data and circumstance, you often need one or both, to make useful conclusions.
Charles Gray
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Mikkal VanPelt
, PhD Musicology & Philosophy, Florida State University (1983) · Author has 1.9K answers and 1.9M answer views
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What is a good example of inductive reasoning?
Here is an example from chapter VI of Bertrand Russell’s The Problems of Philosophy:
And this kind of association [induction] is not confined to men; in animals also it is very strong. A horse which has been often driven along a certain road resists the attempt to drive him in a different direction. Domestic animals expect food when they see the person who feeds them. We know that all these rather
Here is an example from chapter VI of Bertrand Russell’s The Problems of Philosophy:
And this kind of association [induction] is not confined to men; in animals also it is very strong. A horse which has been often driven along a certain road resists the attempt to drive him in a different direction. Domestic animals expect food when they see the person who feeds them. We know that all these rather crude expectations of uniformity are liable to be misleading. The man who has fed the chicken every day throughout its life at last wrings its ...
Nagarajan Ramachandran
Non-dualist at Mata Amritanandamayi Math (1996–present)
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What is intuition with different real examples?
The question is:
What is intuition with different real examples?
Intuition is a thought that emerges instanteneously whenever it is required, and it cannot but emerge from our own knowledge base and convictions.
Therefore intuition is subjective and fallible.
A toddler sees a helium balloon for the first time. You can see the wonder and disbelief in his face. His little intuition tells him, “Objects usually fall to the ground and never float!!”.
For millions of years people have seen objects fall to the ground. However it was only one man who had the intuition in the 1700s that there must n
The question is:
What is intuition with different real examples?
Intuition is a thought that emerges instanteneously whenever it is required, and it cannot but emerge from our own knowledge base and convictions.
Therefore intuition is subjective and fallible.
Example:
A toddler sees a helium balloon for the first time. You can see the wonder and disbelief in his face. His little intuition tells him, “Objects usually fall to the ground and never float!!”.
For millions of years people have seen objects fall to the ground. However it was only one man who had the intuition in the 1700s that there must necessarily be an invisible cause that is responsible for the object to be attracted towards the center of our planet Earth.
If you look up the dictionary for the definition of any word or concept, you will invaraibly see the definition mentioning several other concepts to define this one!
Example: Fact.
a thing that is known or proved to be true.
"he ignores some historical and economic facts"
If you wish to understand the definition of fact, then you better already know the definitions of “proof”, “true” and “knowing”!
It is your intuition that makes sense of any word or concept.
K.F. Wally
Investor, small business operator, stock and futures trader
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What are some examples of deductive logic?
All men are mortal. Socrates is a man. Therefore, Socrates is a mortal.
If it is raining, then it is cloudy. It is not cloudy. Therefore, not raining.
If it is true that all US presidents are intelligent or well informed, and if it is true that Trump is US president, but it is also true that Trump is not intelligent, then it is true that Trump is well informed.
However, if it is true that Trump is neither intelligent nor well informed, then it is false that all US presidents are intelligent or well informed.
If it is true that an intelligent and informed person will not repeatedly commit crass blu
All men are mortal. Socrates is a man. Therefore, Socrates is a mortal.
If it is raining, then it is cloudy. It is not cloudy. Therefore, not raining.
If it is true that all US presidents are intelligent or well informed, and if it is true that Trump is US president, but it is also true that Trump is not intelligent, then it is true that Trump is well informed.
However, if it is true that Trump is neither intelligent nor well informed, then it is false that all US presidents are intelligent or well informed.
If it is true that an intelligent and informed person will not repeatedly commit crass blunders, and Trump repeatedly commits crass blunders, then either Trump is not intelligent or not informed or both.
Related questions
What are some deductive reasoning examples found in everyday life?
What are some examples of inductive reasoning in everyday life?
What are the best real examples of deductive, inductive or abductive reasoning? I mean non-mathematical deduction, with a logical chain of reasoning, leading to a compelling conclusion that convincingly demonstrated that the reasoning was correct.
What is a real life situation where we use inductive reasoning?
What are the best examples of deductive reasoning?
In your daily life, give 1 example of each situation that can be compared to inductive and deductive reasoning?
What are some true examples of inductive reasoning?
What is an example of inductive reasoning that you have used in your life? How did it help you?
What is an example of a real life situation illustrating each of the following analogy reasoning fundamentals?
Does deductive reasoning give an example of a real life situation?
Which is usually more effective: inductive or deductive reasoning?
What are some examples for each type of reasoning, intuition, inductive, and inductive?
What's the difference between "inductive", "deductive" and "abductive" reasoning?
What are real life examples of sensing vs intuition?
What is deductive reasoning and examples?
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5128 | https://www.ck12.org/flexi/chemistry/molar-mass/what-is-the-molar-mass-of-o2/ | Flexi answers - What is the molar mass of O2? | CK-12 Foundation
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What is the molar mass of O2?
Flexi Says:
To find the molar mass of O 2, you need to know the atomic mass of oxygen. The atomic mass of a single oxygen atom is approximately 16.00 atomic mass units (amu).
Since O 2 is a molecule made up of two oxygen atoms, you multiply the atomic mass of oxygen by 2:
16.00,amu×2=32.00,amu
So, the molar mass of O 2 is 32.00 grams per mole.
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5129 | https://pubmed.ncbi.nlm.nih.gov/11766876/ | The structure and function of nuclear lamins: implications for disease - PubMed
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. 2001 Nov;58(12-13):1748-57.
doi: 10.1007/PL00000814.
The structure and function of nuclear lamins: implications for disease
R D Moir1,T P Spann
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Affiliation
1 Department of Cell and Molecular Biology, Northwestern University Medical School, Chicago, Illinois 60611, USA. r-moir@northwestern.edu
PMID: 11766876
PMCID: PMC11337319
DOI: 10.1007/PL00000814
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The structure and function of nuclear lamins: implications for disease
R D Moir et al. Cell Mol Life Sci.2001 Nov.
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. 2001 Nov;58(12-13):1748-57.
doi: 10.1007/PL00000814.
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R D Moir1,T P Spann
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1 Department of Cell and Molecular Biology, Northwestern University Medical School, Chicago, Illinois 60611, USA. r-moir@northwestern.edu
PMID: 11766876
PMCID: PMC11337319
DOI: 10.1007/PL00000814
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The nuclear lamins polymerize to form the nuclear lamina, a fibrous structure found on the inner face of the nuclear membrane. The lamins also appear to form structures within the nucleoplasm. These various lamin structures help to establish and maintain the shape and strength of the interphase nucleus, but recent work also suggests that the lamins have a role in nuclear processes such as DNA replication. Furthermore, mutations in the human lamin A/C gene have recently been linked to several diseases, including Emery-Dreifuss muscular dystrophy. This review discusses the nature of these mutations and the possible effects of lamin mutations on nuclear function.
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The nuclear envelope: emerging roles in development and disease.Wolfner MF, Wilson KL.Wolfner MF, et al.Cell Mol Life Sci. 2001 Nov;58(12-13):1737-40. doi: 10.1007/PL00000812.Cell Mol Life Sci. 2001.PMID: 11766874 Free PMC article.
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The A-type lamins: nuclear structural proteins as a focus for muscular dystrophy and cardiovascular diseases.Mounkes LC, Burke B, Stewart CL.Mounkes LC, et al.Trends Cardiovasc Med. 2001 Oct;11(7):280-5. doi: 10.1016/s1050-1738(01)00126-8.Trends Cardiovasc Med. 2001.PMID: 11709282 Review.
Novel and recurrent mutations in lamin A/C in patients with Emery-Dreifuss muscular dystrophy.Brown CA, Lanning RW, McKinney KQ, Salvino AR, Cherniske E, Crowe CA, Darras BT, Gominak S, Greenberg CR, Grosmann C, Heydemann P, Mendell JR, Pober BR, Sasaki T, Shapiro F, Simpson DA, Suchowersky O, Spence JE.Brown CA, et al.Am J Med Genet. 2001 Sep 1;102(4):359-67. doi: 10.1002/ajmg.1463.Am J Med Genet. 2001.PMID: 11503164
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5131 | https://gogeometry.com/school-college/6/p1529-triangle-interior-cevian-30-100-degree-congruence-find-missing-angle.htm | Geometry Problem 1529: Unlock the Mystery of Triangles: Solving for the Missing Angle with 100-50-30 Degree Angles and Cevian Lengths - A High School Challenge. Thematic Poem. Difficulty Level: High School.
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Geometry Problem 1529: Unlock the Mystery of Triangles: Solving for the Missing Angle with 100-50-30 Degree Angles and Cevian Lengths - A High School Challenge
In a triangle ABC, angle A measures 30 degrees, angle C measures 100 degrees. BD is an interior cevian such that BD is equal to AC. Calculate the measure of angle BDC.
View or post a solution
Discover Even More: See Also...
Problem 1530
Problem 1528
Problem 1527
Problem 1526
Problem 1525
Problem 1524
Thematic Poem:
Unraveling the Secrets of a Mysterious Triangle
Amidst the angles sharp and bold,
A triangle with secrets untold,
Degrees of 100, 50, and 30,
A riddle waiting to ensue.
But as we search, we may discover,
A shape with sides so equal in measure,
An equilateral, hidden within,
A hint to the mystery therein.
And as we ponder, with our minds so bright,
An auxiliary line may come in sight,
A tool to unlock the triangle's hold,
And reveal its secrets, pure as gold.
And lo and behold, what do we see?
An isosceles triangle, waiting to be set free,
With sides and angles that match so fine,
A missing angle, no longer left behind.
So let us solve this puzzle, with grace and wit,
And uncover the mysteries, bit by bit,
With equilateral, auxiliary, and isosceles,
Our journey ends in triumph, oh so seamless.
If you're interested in finding more poems with a focus on geometry, you may enjoy this collection: More geometry thematic poems.
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5132 | https://www.aatbio.com/resources/faq-frequently-asked-questions/what-are-the-differences-between-ribosomes-and-endoplasmic-reticulum | What are the differences between ribosomes and endoplasmic reticulum? | AAT Bioquest
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What are the differences between ribosomes and endoplasmic reticulum?
Posted January 10, 2023
Chemical ReagentsCellular Structures and OrganellesEndoplasmic Reticulum (ER)OrganellesPhysiological ProbesProtein
Answer
Endoplasmic reticulum and ribosomes are two types of organelles that work together to synthesize proteins and transport them to their various destinations. There are several differences between these two organelles.
Endoplasmic reticulum
Endoplasmic reticulum is a membrane-bound organelle. It is made up of tubes and sacs, called cisternae, enclosed in a complex network of membranes inside a cell. The endoplasmic reticulum extends from the exterior surface of the nuclear membrane into the body of the cell. It is continuous with the nuclear membrane and plays a key role in protein synthesis, lipid metabolism, and calcium storage.
There are two types of endoplasmic reticulum – smooth and rough. The rough endoplasmic reticulum (rough ER) acts as a host for ribosomes that continuously attach to and detach from its outer surface. The main function of the rough ER is to facilitate the synthesis and storage of proteins. Many of the proteins synthesized by the attached ribosomes pass into the rough endoplasmic reticulum, from where they are transported to other parts of the cell for immediate use or to be stored or transported out of the cell to other parts of the organism.
Ribosomes
Ribosomes are small, round, membrane-less organelles. They are composed of ribosomal RNA and proteins and make up of a large and a small subunit. Ribosomes may be found floating freely in the cell cytosol or attached to the outer surface of the rough endoplasmic reticulum through tiny, complex docking stations called translocons.
Ribosomes are the main site of protein synthesis. They synthesize proteins through a process called translation, which occurs in three distinct phases. The newly synthesized protein is transported through the translocon into the endoplasmic reticulum. From there it may be transported to other parts of the cell to be used immediately or it may be transported out the cell to other parts of the organism. Proteins that are not needed for immediate use are stored in the rough ER.
Additional resources
Heterogeneous translational landscape of the endoplasmic reticulum revealed by ribosome proximity labeling and transcriptome analysisCell Navigator® Live Cell Endoplasmic Reticulum (ER) Staining Kit Blue FluorescenceReadiUse™ TCA Deproteinization Sample Preparation Kit
Related questions
What are the similarities between free and attached ribosomes?What types of proteins are produced by attached ribosomes?What are the differences between free and attached ribosomes?What is the significance of free ribosomes?What converts mRNA into a protein?
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5133 | https://www.youtube.com/watch?v=YYbr2j2wQWM | Linear Maps | Linear Algebra
Kyle Broder
15000 subscribers
261 likes
Description
20713 views
Posted: 10 Feb 2020
If you're interested in personal help, I've posted my tutoring services on Fiverr:
I have not had the opportunity to teach mathematics as much lately, given the amount of focus I have given to my research. I enjoy the process of teaching and interacting with students and so have decided to do some online tutoring.
In this lecture, we discuss the definition of a linear map, relating it to the definition of a vector space. We also give an elementary example, illustrating how to verify that a map is linear or not linear.
Video Notes:
👍 If you like the video, hit the like button and subscribe to the channel if you want to show your support/want to learn some mathematics.
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These videos are separate from my research and teaching roles at the Australian National University, University of Sydney, and Beijing University.
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Transcript:
okay so the purpose of this lecture is to discuss the definition of a linear map between vector spaces and give an example computation so let's begin by recalling the definition so a map we'll call it T between two vector spaces V and W so these vector spaces and for the moment we'll just assume that they're real vector spaces is let's keep the color consistent is linear if the following two conditions are satisfied if T of U plus V is T of U plus T of V and secondly that T of so this is for for all u V in V and T of lambda V is lambda T of V for all real numbers lambda and for all vectors V ok so let's give some intuition for this definition here and to do this we we recall the definition of a vector space so remember that a vector space has two conditions the first condition is that it's closed under vector addition so if I take two vectors in the vector space and I add them together I get another vector now whenever we define a linear define a map between two sets with some structure we need to make sure that this actually makes sense so we know that we can add in a vector space and the first condition pretty much says only uses that property so here for example we have an addition vectors but this is an addition of vectors in V U and V are elements of the vector space V so I can add them so what this is saying is that when I add the vectors and then apply the map that's the same as applying the map to each vector and then adding the result but notice that this addition occurs in W ok so the next condition of a vector space was that if I took a vector V in V and a scalar in this case a a real scalar since we're assuming the vector spaces are real then lambda times V is also in V so I can scale vectors and if we look at condition two we're only using the scaling property so here we're scaling a vector in V and here we're scaling a vector in W the other thing to point out here is that this is precisely the definition you want if you want the range of this map to be a vector subspace of the codomain which in this case is w so recall that a the subspace the definition of a vector subspace is simply a subset of W which is a vector space in and of itself okay so let's suppose you get an exam question or an assignment problem that asks you to verify that something is a linear map between vector spaces so let's consider a very simple example we'll consider other examples in later tutorials but let's just consider the map that takes a real number to a real number and is defined by the formula MX plus B ok so let's have a look at what we can do here so we need to check two conditions we need to check that T of U plus V is T of U plus T of V so let's look at what T of U plus V is well that's in this case F of U plus V which is M of U plus V plus B and we can expand that out to get M of u mu plus MV plus B now we need to compare this with T of U plus T of V so T of U plus T of V that's just F of U plus F of V now F of U is just mu plus B and F of V is M V plus B now if we add these together what do we get we get mu plus M V plus there's a B term here and a B term here so I get to be now if the map is linear meaning it satisfies this T of U plus V it was the same as T of U plus T or V and these should be the same but notice they're not the same this implies that B equals to B and in fact they are equal if B equals zero so for condition one to be true we need B equals zero okay so now let's work with the map f of X is M X and we need to check that T of lambda V is lambda T of V okay so T of lambda V is f of lambda V and that's simply M lambda V and if we look at lambda T of V then we see that this is just lambda F of V which is lambda M V and these are equal so if you want the map f of X defined by M of X plus B is linear to be linear then that's true if and only if B equals zero as a side note if you have this for be nonzero this type of map is is called affine okay so to briefly sum up the definition of a linear map means the following two conditions that if you add two vectors and then apply the map that's the same as applying the map to each vector and then adding and similarly if you scale the vectors and then apply the map that's the same as applying the map and then scaling this is the precise definition you want if you want the range of T to be a vector subspace of W and this mimics the definition or uses the two conditions of a vector space and this is how you verify that a map is indeed linear or check if it is nonlinear we'll cover more examples in later tutorials |
5134 | https://dermnetnz.org/topics/granuloma-inguinale | Are you a healthcare professional
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Granuloma inguinale
Granuloma inguinale — extra information
Synonyms:
Calymmatobacterium granulomatis infection, Donovanosis, Klebsiella granulomatis infection
Categories:
Infections
ICD-10:
A58
ICD-11:
1A91
SNOMED CT:
28867007, 417443008
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Infections
Granuloma inguinale
Author: Vanessa Ngan, Staff Writer, 2003. Updated by Dr Natalie Renaud, Registrar, Sexual Health Services, Hamilton, New Zealand. April 2018.
Introduction Prevalence Demographics Signs and symptoms Complications Diagnosis Treatment Prevention
What is granuloma inguinale?
Granuloma inguinale is a sexually transmitted infection (STI) caused by the bacteria previously classified as Calymmatobacterium granulomatosis and now reclassified as Klebsiella granulomatis.
It is also called donovanosis after ‘Donovan bodies’, which are cellular components that are seen when scrapings from the genital skin lesions are viewed under a microscope.
How common is granuloma inguinale?
Granuloma inguinale is most commonly found in tropical or subtropical areas such as Papua New Guinea, Southern India, Southern Africa, the Caribbean and South America. In Australia, it is largely confined to Aboriginal and Torres Strait Islander population in the remote northern region. A successful elimination program in Australia targeting this population was started in the mid-1990s. In countries where it is less frequently seen, cases are most probably related to foreign travel. Granuloma inguinale is extremely rare in New Zealand.
Who is at risk of granuloma inguinale?
Sexually active people are at risk of getting granuloma inguinale. The peak incidence is between 20 and 40 years of age. Vertical transmission during childbirth can also occur from contact with the bacteria in the genital tract. A small minority of people (mainly children) may be infected from direct skin-to-skin contact. Klebsiella granulomatis is of low pathogenicity and it may require repeated exposure to develop skin lesions.
What are the signs and symptoms of granuloma inguinale?
The time from exposure to the bacteria to the development of signs and symptoms of granuloma inguinale has not been well defined. The reported median is 50 days, but it can vary from a few days to up to a year.
The initial lesion is usually a firm papule or subcutaneous nodule at the site of contact, that later ulcerates. The areas most commonly affected are the penis (corona, glans and prepuce), the vulva (labia minora and fourchette) and anal area (which is more common in men).
Ulcerovegetative type
The ulcerovegetative type of granuloma inguinale is the most common clinical presentation.
Single or multiple painless firm papules and nodules (granulomas) arise at the site of contact.
They then erode into soft, painless ulcers that bleed easily.
In skin folds, they have a linear ‘knife cut’ appearance.
The ulcer base consists of beefy red granulation tissue and the margins are sharp with rolled or heaped-up borders.
Without treatment, these ulcers continue to enlarge.
Nodular type
The nodular type of granuloma inguinale is less common than the ulcerovegetative type and consists of a mixture of papules and nodules that are soft and red, also with some granulation tissue.
Hypertrophic (verrucous) type
The hypertrophic type of granuloma inguinale consists of large vegetating masses, which has been described as a ‘walnut' appearance.
Necrotic type
The necrotic type of granuloma inguinale consists of a deep foul-smelling ulcer.
Cicatricial (sclerotic) type
Cicatricial granuloma inguinale consists of extensive plaques of scar tissue.
What are the complications of granuloma inguinale?
The infection can spread to the inguinal region where it can cause swelling and ulceration. This is called pseudobubo. Complications of chronic granuloma inguinale include bleeding, secondary infection, swelling from lymphoedema, and significant scarring causing destruction of the genital architecture.
Granuloma inguinale
Granuloma inguinale
Credit. Images supplied by Dr Shahbaz A. Janjua
How is granuloma inguinale diagnosed?
Due to the variation in presentation and significant overlap with other genital ulcers, granuloma inguinale can be difficult to diagnose clinically.
In the early stages, it may be difficult to differentiate it from chancroid. In the later stages, it may look like lymphogranuloma venereum. It may also be confused with other conditions, such as penile or vulvar cancer, genital herpes or syphilis.
Laboratory diagnosis is made by direct microscopy of Giemsa-stained crushed tissue smear looking for characteristics Donovan bodies or by histological examination of a tissue biopsy. Culture is very difficult. Nucleic acid amplification tests are not routinely available and serology testing is not useful.
What is the treatment for granuloma inguinale?
Granuloma inguinale is treated with antibiotics. In settings where this infection is very rare, it is recommended that management involve a sexual health or infectious diseases specialist.
The suggested preferred antibiotic is 1 g of weekly azithromycin for 3 or more weeks. Other treatment options include trimethoprim and sulfamethoxazole, doxycycline, erythromycin, with parenteral gentamicin as a treatment adjunct. Treatment should be continued for a minimum of 3 weeks or until the lesions heal completely; provided there is an adequate clinical response.
Sexual activity should not take place until all lesions are completely healed. A follow-up examination is necessary to ensure a complete recovery. Sexual contacts should be notified and undergo clinical examination for possible lesions. Antibiotic prophylaxis should be considered for a baby born to an infected mother with genital lesions.
A full sexual health screen should be done, in particular tests for genital herpes, syphilis and human immunodeficiency virus (HIV). Lesions that bleed easily increase the risk of HIV transmission and multiple infections can be present.
Can granuloma inguinale be prevented?
Like other sexually transmitted infections, the risk of acquiring this infection is reduced by safe-sex practices including limiting the number of sexual partners, avoiding sexual contact with individuals from a high-risk population, and using condoms.
If you think you are infected, stop all sexual contact and see your usual doctor or a specialist clinician at a sexual health clinic immediately.
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References
Rishi, A. Granuloma inguinale. PathologyOutlines.com website. Accessed March 27th, 2018.
Susan Tuddenham, Khalil G Ghanem. Approach to the patient with genital ulcers. UpToDate. Topic last updated Feb 2018.
UK National guideline for the management of Donovanosis 2018. Bashh guidelines. Nigel 0’Farrel et al.
Bowden FJ. Donovanosis in Australia: going, going… Sexually Transmitted Infections 2005; 81:365-366
Genital Dermatology. Libby Edwards, Peter J. Lynch. Second edition. 2011.
NZSHS STI management guidelines for use in primary care 2017. Genital ulcer disease guideline.
Sexually Transmitted Diseases. King K. Holmes. 4th edition. 2008.
On DermNet
Bacterial skin infections
Genital skin problems
Other websites
STD Transmitted Infections — CDC
Granuloma Inguinale (Donovanosis) — Medscape Reference
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Dermatology Made Easy - second edition
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5135 | https://www.khanacademy.org/math/cc-third-grade-math/equivalent-fractions-and-comparing-fractions/imp-comparing-fractions/v/comparing-fractions-with-greater-than-and-less-than-symbols | Comparing fractions with > and < symbols (video) | Khan Academy
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3rd grade math
Course: 3rd grade math>Unit 6
Lesson 1: Comparing fractions
Comparing fractions with > and < symbols
Comparing fractions visually
Compare fractions with fraction models
Compare fractions on the number line
Comparing fractions with the same denominator
Compare fractions with the same denominator
Comparing unit fractions
Comparing fractions with the same numerator
Compare fractions with the same numerator
Compare fractions with the same numerator or denominator
Math>
3rd grade math>
Equivalent fractions and comparing fractions>
Comparing fractions
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When comparing fractions, remember that the numerator is the top number and the denominator is the bottom number. With the same denominator, the larger numerator means a larger fraction. With the same numerator, the smaller denominator means a larger fraction. Two fractions can be equivalent even with different numerators and denominators.Created by Sal Khan.
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Lilly Brown 7 years ago Posted 7 years ago. Direct link to Lilly Brown's post “How can 3/4 be larger tha...” more How can 3/4 be larger than 3/7? 3/4 has a smaller denominator. Can someone help me?
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ABHISAB 7 years ago Posted 7 years ago. Direct link to ABHISAB's post “If you have a bigger deno...” more If you have a bigger denominator it is smaller because then you have more parts, and that makes it smaller. fractions can also be turned into decimals, so 3/7= 3 divided by seven, which is about 0.4285. but 3 divided by 4 is 0.75.
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Joanna Ni 9 years ago Posted 9 years ago. Direct link to Joanna Ni's post “I understand what <,>, & ...” more I understand what <,>, & = equals, but what does the < and > mean with a line under it?
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Elysia Du 9 years ago Posted 9 years ago. Direct link to Elysia Du's post “Do you mean ≤ and ≥? ...” more Do you mean ≤ and ≥?
≥ is equal to or greater than. For example, you could use 2 ≥ 1 or 4 ≥ 4.
≤ is equal to or less than. For example, you could use 5 ≤ 8 or 8 ≤ 8.
Hope this helped! <3
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Stephen White 5 years ago Posted 5 years ago. Direct link to Stephen White's post “Fractions and division ha...” more Fractions and division have some things in common. For instance, if you want to convert a fraction into a decimal, you would divide the two numbers.
Example: 1/4 (spoken one quarter) To convert 1/4 into a decimal, you would take 1 divided by 4 and you would get 0.25
If you look at that as money, it would be $0.25 and that is also what we call "one quarter." Looking at it another way, one quarter is 1/4 of a dollar.
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When you write a fraction, there are actually words for the top number and the bottom number. And the words are a lot more fancy than the word "top number" and "bottom number." What mathematicians typically use is the word "numerator" for the top number and "denominator" for the bottom number. And what I want to do now that we know that the top number is the numerator of the fraction and the bottom number is the denominator, I want to compare pairs of fractions that have either the same denominator or the same numerator. So let's look at this first pair. I want to compare 4/7 to 3/7. And I have two wholes right over here. They're the same hole, and I've divided them into sevenths. I've divided them into 7 equal chunks. And I want to see what's larger, 4/7 or 3/7. So what I can do is, I can fill in 4/7. Let me select 4 out of the 7. So, that's 1, 2, 3, 4. And the fact that trying to even get to 4/7 I had to have 3/7 first gives you good clue that 4/7 is probably larger, or it is larger. But now let's color in 3/7, just so we can compare. So 1, 2, 3/7. And so it's pretty clear that on the left-hand side, we are shading in more of the whole than on the right-hand side. So, 4/7 represents a larger fraction, more of the whole than 3/7. And the way that we can state that comparison mathematically is with the greater than symbol. We can write 4/7 is greater than 3/7. Now, the greater than and less than symbols can sometimes be confusing. This is greater than. This is less than. And the way that I remember it is that the greater than symbol, either symbol, the small pointy side is always on the side of the smaller number, and the big open side is always on the side of the larger number. So here, big open side is opening towards the 4/7, small pointy side opening to the 3/7. 4/7 is greater than 3/7. Now, what about 3/7 and 3/4? So, here I have different denominators, but I have the same numerator. And so I encourage you to pause this video and draw maybe little boxes like this and try to judge for yourself which of these is a larger fraction, represents a larger number. Well, let's color them in. So, let's think about 3/7 first. And we actually already drew it here, but I can do it really fast right over here. So that is 3/7. I've colored in 3 of the 7 equal groups. And what would 3/4 be? Well, that's 1/3, 2/4, and 3/4. And so it's pretty clear that 3/4 represents a larger fraction of the whole, that 3/4 is larger, or that 3/7 is smaller. So we could write that 3/7 is less than 3/4. So, notice, same exact numerator. When I divided it-- because this fraction symbol could also be viewed as division. When I have it as more equal groups, so it's a fraction of more equal groups, so 3 out of 7 versus 3 out of 4, this is a smaller number, which which makes sense. Now, let's compare these two. We have the same denominator, different numerators-- 3/4 versus 2/4. Well, 3/4 we've already looked at. We can just shade in 3 of these. So 3 of these fourths. So that's 3/4 right over there. And then 2/4, well, we're going to only have 2 of the fourths, 1, 2. So 2/4 is clearly the smaller number. 3/4 is the greater number. So, once again, we could write 3/4 is greater than 2/4. And then finally, I encourage you to pause the video. Try to come up with whether 2/4 or 3/6 is a larger number. Well, let's color it in again. We've already seen 2/4. We just have to color in 2 of our fourths. So let's just color in 2 of our fourths right over there. And then 3/6, we've split our whole into 6 equal sections-- 1, 2, 3, 4, 5, 6. We need to color in 3 of them. And as you see, we're coloring in the exact same amount of the whole. These two fractions are equivalent. These are equivalent fractions. 2/4 is equal to 3/6. And as you see here, they're both filling in half of the whole. If we were to just draw the whole and split it only-- let me do this in a different color. If we have our whole, and we were to split it only into two sections, we are shading in exactly 1 out of the 2 sections. So you could say that 2/4 is equal to 3/6, and they both equal 1/2. So 1/2 is equal to 2/4 is equal to 3/6.
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5136 | https://www.geeksforgeeks.org/dsa/number-of-leaf-nodes-in-a-perfect-n-ary-tree-of-height-k/ | Number of leaf nodes in a perfect N-ary tree of height K
Find the number of leaf nodes in a perfect N-ary tree of height K.
Note: As the answer can be very large, return the answer modulo 109+7.
Examples:
Input: N = 2, K = 2
Output: 4
Explanation: A perfect Binary tree of height 2 has 4 leaf nodes.
Input: N = 2, K = 1
Output: 2
Explanation: A perfect Binary tree of height 1 has 2 leaf nodes.
Approach: This problem can be solved based on the observation that the number of leaf nodes in a perfect N-ary tree with height K will be equal to NK. Use Modular exponentiation to calculate the power modulo 109+7.
Follow the steps mentioned below to implement the above idea.
Below is the implementation of the above approach:
````
// C++ code to implement the approach
include
using namespace std;
const int mod = 1e9 + 7;
// Find the number of leaf nodes in a
// perfect k-ary tree of height m
long long karyTree(long long N, int K)
{
// Initialize variable
long long res = 1;
// Run until height is positive
while (K > 0) {
if (K & 1)
res = (res N) % mod;
N = (N N) % mod;
K >>= 1;
}
// Return answer
return res;
}
// Driver code
int main()
{
int N = 2, K = 2;
// Function call
cout << karyTree(N, K);
return 0;
}
// Java code to implement the approach
import java.io.;
class GFG {
static final int mod = 1000000007;
public static void main(String[] args)
{
int N = 2, K = 2;
// Function call
System.out.println(karyTree(N, K));
}
public static long karyTree(long N, int K)
{
// Initialize variable
long res = 1;
// Run until height is positive
while (K > 0) {
if (K % 2 == 1)
res = (res N) % mod;
N = (N N) % mod;
K >>= 1;
}
// Return answer
return res;
}
}
// This code is contributed by ishankhandelwals.
python code to implement the approach
mod = 1e9 + 7
Find the number of leaf nodes in a
perfect k-ary tree of height m
def karyTree(N, K):
# Initialize variable
res = 1
# Run until height is positive
while (K > 0):
if (K & 1):
res = (res N) % mod
N = (N N) % mod
K >>= 1
# Return answer
return res
Driver code
N,K = 2,2
Function call
print(karyTree(N, K))
This code is contributed by ishankhandelwals
// C# code to implement the approach
using System;
class GFG {
static int mod = 1000000007;
public static void Main()
{
int N = 2, K = 2;
// Function call
Console.Write(karyTree(N, K));
}
static long karyTree(long N, int K)
{
// Initialize variable
long res = 1;
// Run until height is positive
while (K > 0) {
if (K % 2 == 1)
res = (res N) % mod;
N = (N N) % mod;
K >>= 1;
}
// Return answer
return res;
}
}
// This code is contributed by Samim Hossain Mondal.
// JavaScript code to implement the approach
const mod = 1e9 + 7;
// Find the number of leaf nodes in a
// perfect k-ary tree of height m
function karyTree(N, K)
{
// Initialize variable
let res = 1;
// Run until height is positive
while (K > 0) {
if (K & 1)
res = (res N) % mod;
N = (N N) % mod;
K >>= 1;
}
// Return answer
return res;
}
// Driver code
let N = 2, K = 2;
// Function call
document.write(karyTree(N, K));
// This code is contributed by ishankhandelwals
````
Time Complexity: O(logK)
Auxiliary Space: O(1)
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5137 | https://artofproblemsolving.com/wiki/index.php/Permutation?srsltid=AfmBOorQj3MVVOAXmCun9nLu53xTaR9sO3KHKXMoqcElBtta09TfOrYT | Art of Problem Solving
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Permutation
A permutation of a set of objects is any rearrangement (linear ordering) of the objects. There are (the factorial of ) permutations of a set with distinct objects.
One can also consider permutations of infinite sets. In this case, a permutation of a set is simply a bijection between and itself.
Contents
[hide]
1 Video
2 Notations
3 The Symmetric Group
4 Picking Ordered Subsets of a Set
5 Related Videos
6 See also
Video
This video goes over what Permutations & Combinations are, their various types, and how to calculate each type! It serves as a great introductory video to combinations, permutations, and counting problems in general! Permutations & Combinations Video
This video is a great introduction to permutations, combinations, and constructive counting:
Notations
A given permutation of a finite set can be denoted in a variety of ways. The most straightforward representation is simply to write down what the permutation looks like. For example, the permutations of the set are and . We often drop the brackets and commas, so the permutation would just be represented by .
Another common notation is cycle notation.
The Symmetric Group
The set of all permutations of an -element set is denoted . In fact, forms a group, known as the Symmetric group, under the operation of permutation composition.
A permutation in which no object remains in the same place it started is called a derangement.
Picking Ordered Subsets of a Set
An important question is how many ways to pick an -element subset of a set with elements, where order matters. To find how many ways we can do this, note that for the first of the elements, we have different objects we can choose from. For the second element, there are objects we can choose, for the third, and so on. In general, the number of ways to permute objects from a set of is given by .
Related Videos
Permutation Videos
(Counting Permutations)
(Permutations and Factorials)
See also
Combinations
Combinatorics
Derangement
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5138 | https://stats.stackexchange.com/questions/27682/what-is-the-reason-why-we-use-natural-logarithm-ln-rather-than-log-to-base-10 | Skip to main content
What is the reason why we use natural logarithm (ln) rather than log to base 10 in specifying function in econometrics?
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What is the reason why we use natural logarithm (ln) rather than log to base 10 in specifying functions in econometrics?
econometrics
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asked Mar 27, 2012 at 10:11
rithoritho
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Check this for details youtube.com/watch?v=IXhucU6214M&feature=youtu.be this will tell why natural logs are calculated with reasons and references of the renowned authors
– Amit Kumar
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Oct 16, 2017 at 22:39
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In the context of linear regression in the social sciences, Gelman and Hill write:
We prefer natural logs (that is, logarithms base e) because, as
described above, coefficients on the natural-log scale are directly
interpretable as approximate proportional differences: with a
coefficient of 0.06, a difference of 1 in x corresponds to an
approximate 6% difference in y, and so forth.
Andrew Gelman and Jennifer Hill (2007). Data Analysis using Regression and Multilevel/Hierarchical Models. Cambridge University Press: Cambridge; New York, pp. 60-61.
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answered Aug 6, 2012 at 5:20
fmarkfmark
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+1: For concrete reason to prefer the natural logarithm.
– Neil G
Commented
Aug 6, 2012 at 6:02
3
More generally, the exponential function is the only continuous function that is equal to its derivative.
– user603
Commented
May 11, 2016 at 14:03
1
would this not apply if we apply log10 to the dependent and the independent variable(s)?
– cs0815
Commented
Jul 23, 2017 at 18:02
3
@cs0815 if you apply the Taylor expansion around the point b
f(x)=∑n=0∞f(n)(b)n!(x−b)n
to the exponential function f(x)=ax, with f(n)(x)=ln(a)nax then you get for the first two terms:
f(b+x)=f(b)+ln(a)f(b)x+O(x2)
and the ln(a) term becomes 1 for a=e such that you can use f(b+x)∼f(b)(1+x), which is however only true for small x. Also you can simply try it out exp(1.06)/exp(1) = 1.0618 and 10^1.06/10^1=1.1418154
– Sextus Empiricus
Commented
Oct 17, 2017 at 6:59
Expanding on @SextusEmpiricus. "small" x are definitely within [-0.5, 0.5] which yields 0.64 (an error of 0.14). See a plot of the relationship between x and the error here: stats.stackexchange.com/a/244237/17459.
– Jonas Lindeløv
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Feb 8, 2022 at 9:36
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There is no very strong reason for preferring natural logarithms. Suppose we are estimating the model:
```
ln Y = a + b ln X
```
The relation between natural (ln) and base 10 (log) logarithms is ln X = 2.303 log X (source). Hence the model is equivalent to:
```
2.303 log Y = a + 2.303b log X
```
or, putting a / 2.303 = a:
```
log Y = a + b log X
```
Either form of the model could be estimated, with equivalent results.
A slight advantage of natural logarithms is that their first differential is simpler: d(ln X)/dX = 1/X, while d(log X)/dX = 1 / ((ln 10)X) (source).
For a source in an econometrics textbook saying that either form of logarithms could be used, see Gujarati, Essentials of Econometrics 3rd edition 2006 p 288.
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answered Mar 27, 2012 at 14:29
Adam BaileyAdam Bailey
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The natural log is also useful in a semi-log time series regression since the estimated coefficients can be interpreted as continuously compounded growth rates.
– Jason B
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Mar 27, 2012 at 18:20
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I think that the natural logarithm is used because the exponential is often used when doing interest/growth calculation.
If you are in continuous time and that you are compounding interests, you will end up having a future value of a certain sum equal to F(t)=N.ert (where r is the interest rate and N the nominal amount of the sum).
Since you end up with exponential in the calculus, the best way to get rid of it is by using the natural logarithm and if you do the inverse operation, the natural log will give you the time needed to reach a certain growth.
Also, the good thing about logarithms (be it natural or not) is the fact that you can turn multiplications into additions.
As for mathematical explanations of why we end up using an exponential when compounding interest, you can find it here:
Basically, you need to take the limit to have an infinite number of interest rate payment, which ends up being the definition of exponential
Even thought, continuous time is not widely used in real life (you pay your mortgages with monthly payments, not every seconds..), that kind of calculation is often used by quantitative analysts.
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answered Mar 29, 2012 at 7:09
MesopMesop
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I probably would have given an answer like this. The point that it doesn't matter in modeling is a good one too. We could just as easily use base 2. The difference is only a constant factor
– Michael R. Chernick
Commented
May 3, 2012 at 18:44
Surely, you could write it Nrt (more intuitively), and so are you arguing that the base doesn't matter as Adam Bailey does?
– Neil G
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Aug 6, 2012 at 6:16
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An additional reason why economists like to use regressions with logarithmic functional forms is an economic one: Coefficients can be understood as elasticities of a Cobb-Douglas function. This function is probably the most common one used among economists to analyze issues regarding microeconomic behaviour (consumers´preferences, technology, production functions) and macroeconomic issues (economic growth). The elasticity term is used to describe the degree of response of a change of a variable with respect to another.
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answered Feb 25, 2013 at 21:35
user21259user21259
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The only reason is that the Taylor expansion, gives an intuitive interpretation of the result.
Let's look at a typical variable used in econometrics a lot, the log difference of GDP:
ΔlnYt=lnYt−lnYt−1=lnYtYt−1=ln(1+ΔYtYt−1)
, where ΔYtYt−1 is GDP growth rate now.
Let's apply Taylor expansion of the log:
ΔlnYt≈ΔYtYt−1−12(ΔYtYt−1)2+…
Since, GDP growth rate is usually small, e.g. for USA around 2% lately, we can drop all the higher order terms then we get:
ΔlnYt≈ΔYtYt−1
So, if you're using the log differences of GDP in the right hand side of the equation, e.g. as an explanatory variable in the regression you may have the following:
⋯=⋯+β×ΔlnYt
which can be interpreted as "β times percentage change in GDP."
Economists like the variables that can be interpreted easily. If you plugged the different log base then the interpretability is weaker. For example, see what happens to the log base 10:
⋯=⋯+β×Δlog10Yt≈⋯+β×1ln(10)ΔYtYt−1
This still works, but now you need to divide β by some unintuitive number to get the "percentage change" effect interpretation.
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answered Oct 17, 2017 at 0:07
AksakalAksakal
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Is this unique to economics? The standard normal distribution features an e−12x2 in it, and the normal distribution is only one of the large family of exponential distributions that cover a huge swath of statistics. (See GLM's.) It seems like the natural log would be useful in these cases.
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answered May 3, 2012 at 20:37
WayneWayne
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Yes, but the variance term of the normal distribution is equivalent to a change of base of the exponent. Even the way you've drawn it, the base could be (e√)−x2
– Neil G
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Aug 6, 2012 at 5:59
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There is a good reason to use the log transformation of the variable if you think that the inverse function of logarithm is the exponential function which is a continuous version of conpounding. The economic variable which is growing around 10% at a time can be transformed to the variable with its mean around 10 (plus a constant). You cannot do that with the transformation of logarithm of different base.
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answered Aug 6, 2012 at 1:55
IkuyasuIkuyasu
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Not only in econometrics, using base e is more "natural" in almost every domain, including computer science, where dominated by 0,1 (where log2 may be natural).
I would like to use some experiments to show the base e is very natural.
Consider following three functions f1(x)=2x, f2(x)=10x, f3(x)=ex, which one seems more natural? Many people may say first two seems better, because 2 and 10 are small integers, and e is an irrational number.
However, consider following experiment, we want to investigate the relationship between the function's derivative at x0 and the function value at x0.
We pick two random points, say 1.23 and 2.34 and we will use f1 as example.
Here are some facts:
f′1(1.23)=1.625894, f′1(2.34)=3.509423
f1(1.23)=2.34567, f1(2.34)=5.063026
We see there are some patterns: if we calculate f′1(x0)/f1(x0), it is a constant: 0.6931472=log(2).
For another function f2, this constant is 2.302585=log(10). (I will attach the code for some fun experiments)
So, the natural question to ask is when we can get a simplified results, that do not need to scale with this constant (or this constant is equal to 1.0)?
The answer is when the base is e. Where f′3(x)=f3(x). Now, we may think base e is more natural?
some code for fun experiments
```
suppose we have some exponential functions
because 2 and 10 are small integers f1, f2 seems pretty natural
f1 <- function(x) 2^x
f2 <- function(x) 10^x
simple code to calculate numeric gradient
simpleNumDiff <- function(f, x0){
(f(x0+1e-8)-f(x0))/1e-8
}
we try to investigate derv value and function value pattern
note, they are not equal but have very strong pattern
a1 = simpleNumDiff(f1,1.23)
b1 = f1(1.23)
a2 = simpleNumDiff(f1,2.34)
b2 = f1(2.34)
c(a1/b1,a2/b2) # log(2)
```
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edited May 6, 2020 at 20:14
answered May 6, 2020 at 20:02
Haitao DuHaitao Du
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very nice explain of e in a specific area (computer science)! I mention that in ODE, the equation y'=a y is called one of natural growth rate.
– Black Mild
Commented
Jan 30, 2021 at 0:35
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5139 | https://www.onlinemathlearning.com/multiplicative-comparison-4oa1.html | Multiplicative Comparisons (solutions, examples, videos, worksheets, games, activities)
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Multiplicative Comparison (Grade 4)
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More Lessons for Grade 4
Common Core for Grade 4
Common Core Worksheets for Grade 4
Common Core Worksheets for all Grades
These lessons, with videos, examples, and solutions, help Grade 4 students learn to interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
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Common Core: 4.OA.1
Suggested Learning Targets
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I can explain how a multiple equation can be interpreted as a comparison such as Jason is 4 times as old as Ben.
I can write an equation for a situation involving multiplicative comparison.
The following diagrams give examples of Multiplicative Comparison: Phrases used, Model Diagram, Multiplication Equation for a Word Problem. Scroll down the page for more examples and solutions.
Multiplicative comparison involves comparing two quantities by determining how many times larger or smaller one quantity is than the other. It is often expressed using multiplication or division. For example, if one quantity is 3 times larger than another, it means the first quantity is the product of the second quantity and 3.
Printable Worksheets involving Multiplicative Comparison
Printable math worksheets
Measurement Word Problems
Multiplicative Comparison using writing
How to use models, illustrations and writing to solve multiplicative comparison questions?
Things to remember:
Multiplicative comparison means you are comparing two things together that need to be multiplied.
Multiplicative comparison questions are usually written in word problems that have this format:
Statement, Statement, Question.
3. We use the two statements to determine the number sentence or equation.
4. An equation is just like a number sentence but it includes letters.
Examples:
Darlene picked 7 apples. Juan picked 4 times as many as Darlene. How many apples did Juan pick?
Joelle read 9 books in the fall. George read 7 times as many books. How many books did George read?
Sarah is 12 years old. George is g years old. Sarah is 3 times as old as George. How old is George?
Show Video Lesson
Writing Equations with Multiplicative Comparison Problems
Things to remember:
Multiplicative comparison means you are comparing two things together that need to be multiplied.
An equation is just like a number sentence but it includes letters.
The letter is called a variable.
Examples:
Suzie is 4 feet tall. There is a tree in her yard that is 9 times taller than she is. How tall is the tree in Suzie’s yard?
The tree in Suzie’s yard is 36 feet tall. The tree is 9 times as tall as Suzie. How tall is Suzie?
Suzie is 4 feet tall. The tree in her yard is 36 feet tall. How much taller is the tree than Suzie?
Show Video Lesson
Multiplicative Comparison
Multiplicative comparison problems compare a characteristic of two objects using multiplication. Sometimes we know the value of the object that is bigger/older/faster, sometime we know the value of the object that is smaller/younger/slower and sometimes we know how many times bigger/older/faster one object is over the other?
Use a variable for the value you don’t know and fill in the rest of the equation.
Examples:
Write an equation and solve for the variable.
Mrs. K is five times as old as her son. If she is 35, how old is her son?
Mrs. L has two dogs. Scruffy is 12 inches high at the shoulder and Tuffy is 24 inches high at the shoulders. How much bigger than Scruffy is Tuffy?
Show Video Lesson
Key Ideas:
“Times as many/much”: Multiplicative comparisons often use phrases like “times as many,” “times as much,” or “times the size.” These phrases indicate that one quantity is a multiple of the other.
Multiplication and Division: Solving multiplicative comparison problems involves using multiplication to find a larger quantity or division to find a smaller quantity.
Focus on Scaling: Instead of finding the difference between two quantities (like in additive comparison), multiplicative comparison focuses on how one quantity scales relative to the other.
The following activities are from Howard County Public School System
Best online courses
Activity 1:
Write a multiplication equation to match each comparison statement.
comparison statementmultiplication equation
21 days is 3 times longer than 7 days.
8 pounds is 4 times as heavy as 2 pounds.
72 inches is 12 times the length of 6 inches.
30 fish is 5 times as many as 6 fish.
Write a comparison statement to match the multiplication equation.
comparison statementmultiplication equation
36 = 9 x 4
Activity 2:
Hannah was doing a report on animals’ sleep habits. She made the charts below to show the number of hours certain animals usually sleep each day.
animalbat mouse guinea pig possum gray seal
hours of sleep20 hours 12 hours 9 hours 18 hours 6 hours
animaltiger horse cheetah cow goat
hours of sleep16 hours 3 hours 12 hours 4 hours l5 hours
Fill in the blanks to make the statements true.
A possum sleeps ______ times as many hours a day as a guinea pig.
A bat sleeps ______ times as many hours per day as a cow.
Write a multiplication equation to show the
relationship between the length of time a gray
seal sleeps and the length of time a possum sleeps. _ x _ = __
When Hannah was reading about donkeys, she said, “I can’t believe that goats sleep 5 times as many hours per day as donkeys.” Find the number of hours per day a donkey sleeps. Show your thinking below using words, numbers, and/or pictures.
A donkey sleeps _ hours per day.
Activity 3:
Last weekend, Cassidy, Jefferson, and Braden played three basketball games against their cousins, Sammy, Kara, and Mitchell. The chart below shows how many baskets each were able to make during their three games.
Best video game consoles
PlayerNumber of Baskets
Cassidy 24
Jefferson 18
Braden 8
Sammy 6
Kara 36
Mitchell 3
Fill in each blank with a player’s name or a number to make each comparison statement true. Below each comparison statement, write a multiplication equation to show that the statement is true.
statement: ____ made three times as many baskets as Sammy.
multiplication equation: _____
statement: Cassidy made ______ times as many baskets as Mitchell.
multiplication equation: _____
statement: Jefferson made _ times as many baskets as _______.
multiplication equation: _____
statement: Sammy made double the number of baskets _______ made.
multiplication equation: _____
Try out our new and fun Fraction Concoction Game.
Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level.
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5140 | https://www.quora.com/What-are-the-differences-of-primary-and-secondary-tissue-in-plants | What are the differences of primary and secondary tissue in plants? - Quora
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What are the differences of primary and secondary tissue in plants?
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In plants, primary and secondary tissues refer to different types of growth and development, specifically in how they contribute to the structure and function of the plant.
Primary Tissue
Definition: Primary tissues are the initial tissues formed during the primary growth of a plant. This growth occurs at the tips of roots and stems, allowing for an increase in length.
Origin: They arise from the apical meristems (the growing tips of roots and shoots) and are involved in the formation of new organs.
Types:
Parenchyma: Involved in storage, photosynthesis, and tissue repair.
Collenchyma: Provides fle
Continue Reading
In plants, primary and secondary tissues refer to different types of growth and development, specifically in how they contribute to the structure and function of the plant.
Primary Tissue
Definition: Primary tissues are the initial tissues formed during the primary growth of a plant. This growth occurs at the tips of roots and stems, allowing for an increase in length.
Origin: They arise from the apical meristems (the growing tips of roots and shoots) and are involved in the formation of new organs.
Types:
Parenchyma: Involved in storage, photosynthesis, and tissue repair.
Collenchyma: Provides flexible support, especially in young stems and petioles.
Sclerenchyma: Provides rigid support and is composed of thick-walled cells (e.g., fibers and sclereids).
Xylem: Conducts water and minerals from roots to leaves; consists of tracheids and vessel elements.
Phloem: Transports sugars and nutrients produced by photosynthesis; includes sieve elements and companion cells.
Secondary Tissue
Definition: Secondary tissues are formed during secondary growth, which increases the thickness or girth of the plant. This growth occurs in woody plants and some herbaceous plants.
Origin: They arise from the lateral meristems, namely the vascular cambium and cork cambium.
Types:
Secondary Xylem: Often referred to as wood, it is produced by the vascular cambium and contributes to the plant's structural support and water transport.
Secondary Phloem: Forms part of the bark and is involved in the transport of nutrients; it is produced alongside secondary xylem.
Cork (Phellogen): Produced by the cork cambium, it provides protection and reduces water loss.
Key Differences
Growth Type: Primary tissues are associated with lengthening growth, while secondary tissues are associated with thickening growth.
Meristematic Source: Primary tissues originate from apical meristems, whereas secondary tissues originate from lateral meristems.
Function: Primary tissues are essential for initial plant structure and function, while secondary tissues provide support, protection, and storage as the plant matures.
In summary, primary tissues are crucial for the initial growth and development of the plant, while secondary tissues facilitate further growth in thickness and enhance structural integrity.
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What is the difference between primary and secondary tissues in the root system?
What are primary and secondary growth in plants?
What is the difference between primary and secondary growth in woody plants?
What are the different types of primary tissues and secondary tissues of plants?
What is the difference of primary and secondary growth in plants?
Michael Clayton
Former Science Educator (Retired) at University of Wisconsin - Madison (1986–2018) · Author has 2.5K answers and 1.6M answer views
·Updated 4y
In plants, primary tissues are derived from growth at apical meristems.
Primary growth produces dermal, ground and vascular tissues and results in the elongation of stems and roots, and the production of new leaves.
Secondary tissues are derived from the growth of lateral meristems called cambia. These are sheets of dividing cells. The vascular cambium produces vascular tissue. Another, cambium, the cork cambium, produces dermal tissue. In each case, this growth contributes to the girth of stems and roots.
View of secondary xylem (bottom) and phloem (top) with a layer of vascular cambium between.
Continue Reading
In plants, primary tissues are derived from growth at apical meristems.
Primary growth produces dermal, ground and vascular tissues and results in the elongation of stems and roots, and the production of new leaves.
Secondary tissues are derived from the growth of lateral meristems called cambia. These are sheets of dividing cells. The vascular cambium produces vascular tissue. Another, cambium, the cork cambium, produces dermal tissue. In each case, this growth contributes to the girth of stems and roots.
View of secondary xylem (bottom) and phloem (top) with a layer of vascular cambium between.
Photomicrographs by Michael Clayton
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Gita Ramsay
M.S. in Biology&Ethnobotany, Florida International University (Graduated 2006)
·6y
Plant primary tissue includes: epidermal tissue (like skin, the outer covering of a plant), vascular tissue (xylem and phloem, which conduct water and sap respectively), and ground tissue (all other tissue that makes up a plant). All plants have primary tissue (although bryophytes lack vascular tissue).
Secondary tissue results from lateral (sideways) growth, and some herbaceous plants have hardly any secondary growth. Dicots (woody plants) have secondary growth which makes them increase in girth. Such growth results from a thickening of secondary xylem (which is one of the components of wood),
Continue Reading
Plant primary tissue includes: epidermal tissue (like skin, the outer covering of a plant), vascular tissue (xylem and phloem, which conduct water and sap respectively), and ground tissue (all other tissue that makes up a plant). All plants have primary tissue (although bryophytes lack vascular tissue).
Secondary tissue results from lateral (sideways) growth, and some herbaceous plants have hardly any secondary growth. Dicots (woody plants) have secondary growth which makes them increase in girth. Such growth results from a thickening of secondary xylem (which is one of the components of wood), and from an outgrowth of the cork cambium which produces, in part, what we call bark.
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Ritwik Sunny
Former Customer Support Executive at Ashok Leyland · Author has 39.8K answers and 9.7M answer views
·3y
Originally Answered: What are the different types of primary tissues and secondary tissues of plants? ·
They differentiate into three main tissue types: dermal, vascular, and ground tissue. Each plant organ (roots, stems, leaves) contains all three tissue types Dermal tissue covers and protects the plant, and controls gas exchange and water absorption (in roots)
Continue Reading
They differentiate into three main tissue types: dermal, vascular, and ground tissue. Each plant organ (roots, stems, leaves) contains all three tissue types Dermal tissue covers and protects the plant, and controls gas exchange and water absorption (in roots)
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5141 | https://mathigon.org/course/circles/spheres-cones-cylinders | Share
Select one of the keywords on the left…
Circles and PiSpheres, Cones and Cylinders
Reading time: ~50 min
In the previous sections, we studied the properties of circles on a flat surface. But our world is actually three-dimensional, so lets have a look at some 3D solids that are based on circles:
A cylinder
A cylinder is a three-dimensional solid consisting of two congruent, parallel, circular sides (the bases), joined by a curved surface. You could also think of a cylinder as a “circular prism”.
consists of two congruent, parallel circles joined by a curved surface.
A cone
A cone is a three-dimensional solid that has a circular base joined to a single point (called the vertex) by a curved side. You could also think of a cone as a “circular pyramid”.
A right cone is a cone with its vertex directly above the center of its base.
has a circular base that is joined to a single point (called the vertex).
Every point on the surface of a sphere
A sphere is a three-dimensional solid consisting of all points that have the same distance from a given center. This distance is called the radius of the sphere.
has the same distance from its center.
Notice how the definition of a sphere is almost the same as the definition of a ??? – except in three dimensions!
Cylinders
Here you can see the cylindrical Gasometer in Oberhausen, Germany. It used to store natural gas which was used as fuel in nearby factories and power plants. The Gasometer is 120m tall, and its base and ceiling are two large circles with radius 35m. There are two important questions that engineers might want to answer:
How much natural gas can be stored? This is the ??? of the cylinder.
How much steel is needed to build the Gasometer? This is (approximately) the ??? of the cylinder.
Let’s try to find formulas for both these results!
Gasometer Oberhausen
Volume of a Cylinder
The top and bottom of a cylinder are two congruent circles, called bases. The height h of a cylinder is the perpendicular distance between these bases, and the radius r of a cylinder is simply the radius of the circular bases.
We can approximate a cylinder using a
5
-sided prism
A prism is a three-dimensional solid that has two congruent faces that are polygons (called the bases), whose corresponding vertices are joined by parallel segments. The remaining faces of a prism are all rectangles or parallelograms.
. As the number of sides increases, the prism starts to look more and more like a cylinder:
Even though a cylinder is technically not a prism, they share many properties. In both cases, we can find the volume by multiplying the area of their base with their height. This means that a cylinder with radius r and height h has volume
V=
+
−
×
÷
π
Remember that radius and height must use the same units. For example, if r and h are both in cm, then the volume will be in ???.
In the examples above, the two bases of the cylinder were always directly above each other: this is called a right cylinder. If the bases are not directly above each other, we have an oblique cylinder. The bases are still parallel, but the sides seem to “lean over” at an angle that is not 90°.
The Leaning Tower of Pisa in Italy is not quite an oblique cylinder.
The volume of an oblique cylinder turns out to be exactly the same as that of a right cylinder with the same radius and height. This is due to Cavalieri’s Principle
Cavalieri’s Principle states that if two solids have the same height and the same cross-sectional area at every level, then they both have the same volume.
We can use this fact to derive that the volume of prisms and cylinders is the area of their cross-section multiplied by their height.
, named after the Italian mathematician Bonaventura Cavalieri
Bonaventura Cavalieri (1598 – 1647) was an Italian mathematician and monk. He developed a precursor to infinitesimal calculus, and is remembered for Cavalieri’s principle to find the volume of solids in geometry.
Cavalieri also worked in optics and mechanics, introduced logarithms to Italy, and exchanged many letters with Galileo Galilei.
Timeline
: if two solids have the same cross-sectional area at every height, then they will have the same volume.
Imagine slicing a cylinder into lots of thin disks. We can then slide these disks horizontal to get an oblique cylinder. The volume of the individual discs does not change as you make it oblique, therefore the total volume also remains constant:
Surface Area of a Cylinder
To find the surface area of a cylinder, we have to “unroll” it into its flat net
The net of a polyhedron is what you get when you “unfold” its polygonal faces onto a flat surface.
. You can try this yourself, for example by peeling off the label on a can of food.
There are two ???, one at the top and one at the bottom of the cylinder. The curved side is actually a large ???.
The two circles each have area
+
×
π
.
The height of the rectangle is
and the width of the rectangle is the same as the ??? of the circles:
+
×
π
.
This means that the total surface area of a cylinder with radius r and height h is given by
A=
+
−
×
÷
π
.
Cylinders can be found everywhere in our world – from soda cans to toilet paper or water pipes. Can you think of any other examples?
The Gasometer above had a radius of 35m and a height of 120m. We can now calculate that its volume is approximately m3 and its surface area is approximately m2.
Cones
A cone
A cone is a three-dimensional solid that has a circular base joined to a single point (called the vertex) by a curved side. You could also think of a cone as a “circular pyramid”.
A right cone is a cone with its vertex directly above the center of its base.
is a three-dimensional solid that has a circular base. Its side “tapers upwards” as shown in the diagram, and ends in a single point called the vertex.
The radius of the cone is the radius of the circular base, and the height of the cone is the perpendicular distance from the base to the vertex.
Just like other shapes we met before, cones are everywhere around us: ice cream cones, traffic cones, certain roofs, and even christmas trees. What else can you think of?
Volume of a Cone
We previously found the volume of a cylinder by approximating it using a prism. Similarly, we can find the volume of a cone by approximating it using a pyramid
A pyramid is a polyhedron that has a polygon as base, and triangular faces around the outside, that taper to a vertex.
In a right regular pyramid, the base is a regular polygon and the vertex is directly above the center of the base.
.
Here you can see a
5
-sided pyramid. As the number of sides increases, the pyramid starts to look more and more like a cone. In fact, we could think of a cone as a pyramid with infinitely many sides!
This also means that we can also use the equation for the volume: V=13base×height. The base of a cone is a circle, so the volume of a cone with radius r and height h is
V=
+
−
×
÷
π
Notice the similarity with the equation for the volume of a cylinder. Imagine drawing a cylinder around the cone, with the same base and height – this is called the circumscribed cylinder. Now, the cone will take up exactly ??? of the volume of the cylinder:
Note: You might think that infinitely many tiny sides as an approximation is a bit “imprecise”. Mathematicians spent a long time trying to find a more straightforward way to calculate the volume of a cone. In 1900, the great mathematician David Hilbert
David Hilbert (1862 – 1943) was one of the most influential mathematicians of the 20th century. He worked on almost every area of mathematics, and was particularly interested in building a formal, logical foundation for maths.
Hilbert worked in Göttingen (Germany), where he tutored numerous students who later became famous mathematicians. During the International Congress of Mathematicians in 1900, he presented a list of 23 unsolved problems. These set the course for future research – and four of them are still unsolved today!
Timeline
even named it as one of the 23 most important unsolved problems in mathematics! Today we know that it is actually impossible.
Just like a cylinder, a cone doesn’t have to be “straight”. If the vertex is directly over the center of the base, we have a right cone. Otherwise, we call it an oblique cone.
Once again, we can use Cavalieri’s principle to show that all oblique cones have the same volume, as long as they have the same base and height.
Surface Area of a Cone
Finding the surface area of a cone is a bit more tricky. Like before, we can unravel a cone into its net. Move the slider to see what happens: in this case, we get one circle and one ???.
Now we just have to add up the area of both these components. The base is a circle with radius r, so its area is
ABase=
+
×
π
.
The radius of the sector is the same as the distance from the rim of a cone to its vertex. This is called the slant height s of the cone, and not the same as the normal height h. We can find the slant height using Pythagoras
Pythagoras’ theorem states that in every right-angled triangle, a2+b2=c2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.
Learn more…
:
| | | |
---
| s2 | = | + × π |
| s | = | + × |
The arc length of the sector is the same as the ??? of the base: 2πr. Now we can find the area of the sector using the formula
A sector of a circle is a part of its interior, bounded by two radii and an arc. Its area is proportional to the internal angle, as well as the length of the arc. This means that
ASectorACircle=ArcCircumference=θ2π
Learn more…
we derived in a previous section:
| | | |
---
| ASector | = | ACircle×arccircumference |
| | = | + − × ÷ π |
Finally, we just have to add up the area of the base and the area of the sector, to get the total surface are of the cone:
A=
+
−
×
÷
π
Spheres
A sphere
A sphere is a three-dimensional solid consisting of all points that have the same distance from a given center. This distance is called the radius of the sphere.
is a three-dimensional solid consisting of all points that have the same distance from a given center C. This distance is called the radius r of the sphere.
You can think of a sphere as a “three-dimensional circle
A circle is the set of all points in two dimensions, at a fixed distance (the radius) from a given point (the center).
Learn more…
”. Just like a circle, a sphere also has a diameter d, which is ??? the length of the radius, as well as chords and secants.
In a previous section, you learned how the Greek mathematician Eratosthenes
Eratosthenes of Cyrene (c. 276 – 195 BCE) was a Greek mathematician, geographer, astronomer, historian, and poet. He spent much of his life in Egypt, as head of the library of Alexandria. Among many other achievements, Eratosthenes calculated the circumference of the Earth, measured the tilt of the Earth’s axis of rotation, estimated the distance to the sun, and created some of the first maps of the world. He also invented the “Sieve of Eratosthenes”, an efficient way to calculate prime numbers.
Timeline
calculated the radius of Earth using the shadow of a pole – it was 6,371 km. Now, let’s try to find the Earth’s total volume and surface area.
Volume of a Sphere
To find the volume of a sphere, we once again have to use Cavalieri’s Principle. Let’s start with a hemisphere – a sphere cut in half along the equator. We also need a cylinder with the same radius and height as the hemisphere, but with an inverted cone “cut out” in the middle.
As you move the slider below, you can see the cross-section of both these shapes at a specific height above the base:
Let us try to find the cross-sectional area of both these solids, at a distance height h above the base.
The cross-section of the hemisphere is always a ???.
The radius x of the cross-section is part of a right-angled triangle, so we can use Pythagoras
Pythagoras’ theorem states that in every right-angled triangle, a2+b2=c2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.
Learn more…
:
r2=h2+x2.
Now, the area of the cross section is
| | | |
---
| A | = | + − × ÷ π |
The cross-section of the cut-out cylinder is always a ???.
The radius of the hole is h. We can find the area of the ring by subtracting the area of the hole from the area of the larger circle:
| | | |
---
| A | = | πr2−πh2 |
| | = | πr2−h2 |
It looks like both solids have the same cross-sectional area at every level. By Cavalieri’s Principle, both solids must also have the same ???! We can find the volume of the hemisphere by subtracting the volume of the cylinder
The volume of a cylinder is given by the equation
V=πr2h,
where r is the radius of the circular base, and h is the height of the cylinder (the perpendicular distance between the two bases).
and the volume of the cone
The volume of a cone is given by the equation
V=13πr2h,
where r is the radius of the circular base, and h is the height of the cone (the perpendicular distance from the base to the vertex).
:
| | | |
---
| VHemisphere | = | VCylinder−VCone |
| | = | + − × ÷ π |
A sphere consists of hemispheres, which means that its volume must be
V=43πr3.
The Earth is (approximately) a sphere with a radius of 6,371 km. Therefore its volume is
| | | |
---
| V | = | + − × ÷ π |
| | = | 1 km3 |
The average density of the Earth is 5510kg/m3. This means that its total mass is
Mass=Volume×Density≈6×1024kg
That’s a 6 followed by 24 zeros!
If you compare the equations for the volume of a cylinder, cone and sphere, you might notice one of the most satisfying relationships in geometry. Imagine we have a cylinder with the same height as the diameter of its base. We can now fit both a cone and a sphere perfectly in its inside:
+
This cone has radius r and height 2r. Its volume is
×
π
This sphere has radius r. Its volume is
×
π
This cylinder has radius r and height 2r. Its volume is
×
π
Notice how, if we ??? the volume of the cone and the sphere, we get exactly the volume of the cylinder!
Surface Area of a Sphere
Finding a formula for the surface area of a sphere is very difficult. One reason is that we can’t open and “flatten” the surface of a sphere, like we did for cones and cylinders before.
This is a particular issue when trying to create maps. Earth has a curved, three-dimensional surface, but every printed map has to be flat and two-dimensional. This means that Geographers have to cheat: by stretching or squishing certain areas.
Here you can see few different types of maps, called projections. Try moving the red square, and watch what this area actually looks like on a globe:
To find the surface area of a sphere, we can once again approximate it using a different shape – for example a polyhedron with lots of faces. As the number of faces increases, the polyhedron starts to look more and more like a sphere.
COMING SOON: Sphere Surface Area Proof |
5142 | https://www.teacherspayteachers.com/Product/FREE-Boyles-Law-Worksheet-with-Answer-Key-6505782 | FREE Boyle's Law Worksheet with Answer Key by Keystone Science | TPT
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Boyle's Law investigates the relationship between pressure and volume. This PDF worksheet gives students practice completing word problems in chemistry using these variables. The problems include real-world scenarios for students to relate Boyle's law to their every day life.
Detailed teacher ANSWER KEY IS INCLUDED! All work is shown as well as how to set up each problem!
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5143 | https://www.calculatorsoup.com/calculators/financial/present-value-calculator.php | skip to calculator
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Present Value Calculator
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Calculator Use
Find the present value of a future sum of money. The present value calculator answers the question, "What do I need to invest today to have a specific sum of money at a future date?"
You can think of present value as the amount you need to save now to have a certain amount of money in the future. The present value formula applies a discount to your future value amount, deducting interest earned to find the present value in today's money.
Present Value Formula and Calculator
The present value formula is PV=FV/(1+i)n, where you divide the future value
FV by a factor of 1 + i for each period between present and future dates.
Input these numbers in the present value calculator for the PV calculation:
The future value sum FV
Number of time periods (years) t, which is n in the formula
Interest rate R, which is i in the formula
Compounding frequency
Cash flow payments going out
Annuity growth rate
The present value of an amount of money is worth more in the future when it is invested and earns interest.
The present value is the amount you would need to invest now, at a known interest and compounding rate, so that you have a specific amount of money at a specific point in the future.
You can enter 0 for any variable you'd like to exclude when using this calculator. Our other
present value calculators offer more specialized present value calculations.
What's in the Present Value Calculation
The present value calculator uses the following to find the present value
PV of a future sum plus interest, minus cash flow payments:
Future Value FV
: Future value of a sum of money
Number of time periods t
: • Time periods is typically a number of years
• Be sure all your inputs use the same time period unit (years, months, etc.)
• Enter p or perpetuity for a perpetual annuity
Interest Rate R
: The nominal interest rate or stated rate, as a percentage
Compounding m
: • The number of times compounding occurs per period
• Enter 1 for annual compounding which is once per year
• Enter 4 for quarterly compounding
• Enter 12 for monthly compounding
• Enter 365 for daily compounding
• Enter c or continuous for continuous compounding
Cash flow annuity payments going out PMT
: The payment amount each period
Growth rate G
: The growth rate of annuity payments per period entered as a percentage
Number of payments q per period
: • Payment frequency
• Enter 1 for annual payments which is once per year
• Enter 4 for quarterly payments
• Enter 12 for monthly payments
• Enter 365 for daily payments
When do annuity payments occur T
: • Select end which is an ordinary annuity for payments at the end of the period
• Select beginning for payments at the beginning of the period
Present Value PV
: The result of the PV calculation is the present value of any future value sum plus future cash flows or annuity payments
The sections below show how to derive present value formulas. For a list of the formulas presented here see our
Present Value Formulas page.
Present Value Formula Derivation
The future value (FV) of a present value (PV) sum that accumulates interest at rate i over a single period of time is the present value plus the interest earned on that sum. The mathematical equation is
or
For each period into the future the accumulated value increases by an additional factor (1 + i). Therefore, the future value accumulated over, say 3 periods, is given by
or generally
Equation 1a
Likewise we can solve for PV to get
Equation 1b
The equations we have are (1a) the
future value of a present sum and (1b) the
present value of a future sum at a periodic interest rate i where n is the number of periods in the future. Commonly this equation is applied with periods as years but it is less restrictive to think in the broader terms of periods. Dropping the subscripts from (1b) we have:
Present Value of a Future Sum
Equation 1
Present Value of an Annuity Formula Derivation
An annuity is a sum of money paid periodically, (at regular intervals). Let's assume we have a series of equal present values that we will call payments (PMT) for n periods at a constant interest rate i. We can calculate FV of the series of payments 1 through n using formula (1) to add up the individual future values.
Equation 2a
multiply both sides of this equation by (1 + i) to get
Equation 2b
subtracting the equation for PV (2a) from the equation for
PV(1 + i) (2b) most terms cancel and we are left with
pulling out like terms on both sides
cancelling 1's on the left
and finally, after dividing through by i, the present value of an ordinary annuity, payments made at the end of each period, is
Equation 2c
For an annuity due, payments made at the beginning of each period instead of the end, therefore payments are now 1 period closer to the
PV. We need to discount each future value payment in the formula by 1 period. This could be written on (1b) as
but factoring out the (1 + i)
So, multiplying each payment in equation (2a), or the right side of equation (2c), by the factor (1 + i) will give us the equation of
PV for an annuity due. This can be written more generally as
Present Value of an Annuity
Equation 2
where T represents the type. (similar to Excel formulas) If payments are at the end of the period it is an ordinary annuity and we set T = 0. If payments are at the beginning of the period it is an annuity due an we set T = 1.
Present Value of an Ordinary Annuity
if T = 0, payments are at the end of each period and we have the formula for present value of an
ordinary annuity
Equation 2.1
Present Value of an Annuity Due
if T = 1, payments are at the beginning of each period and we have the formula for present value of an annuity due
Equation 2.2
Present Value Growing Annuity Formula Derivation
In a growing annuity, each payment, after the first, is increased by a factor g such that payment 2 is
PMT(1 + g), payment 3 is
PMT(1 + g)(1 + g), payment 4 is
PMT(1 + g)(1 + g)(1 + g), etc. Modifying equation (2a) to include growth we get
Equation 3a
Multiply PV by (1+i)/(1+g) to get
Equation 3b
subtracting equation (3a) from (3b) most terms cancel and we are left with
with some algebraic manipulation, multiplying both sides by (1 + g) we have
pulling like terms out on both sides
cancelling the 1's on the left then dividing through by (i-g) we finally get
Present Value of a Growing Annuity (g ≠ i)
Similar to equation (2), to account for whether we have a growing annuity due or growing ordinary annuity we multiply by the factor (1 + iT)
Equation 3
Present Value of a Growing Annuity (g = i)
If g = i you'll notice that (1 + g) terms cancel in equation (3a) and we get
since we now have n instances of
PMT/(1+i) we can reduce the equation. Also accounting for an annuity due or ordinary annuity, multiply by (1 + iT), and we get
Equation 4
Present Value of a Perpetuity (t → ∞ and n = mt → ∞)
For a perpetuity, perpetual annuity, time and the number of periods goes to infinity therefore n goes to infinity. As n increases the 1/(1 + i)n term in formula (2) goes to 0 leaving
Equation 5
Present Value of a Growing Perpetuity (g < i) (t → ∞ and n = mt → ∞)
Likewise for a growing perpetuity, where we must have g<i, since (1 + i)n grows faster than (1 + g)n, that term goes to 0 in formula (3) and it reduces to
Equation 6
Present Value of a Growing Perpetuity (g = i) (t → ∞ and n = mt → ∞)
Since n also goes to infinity (n → ∞) as t goes to infinity (t → ∞), we see that
Present Value with Growing Annuity (g = i) also goes to infinity
Equation 7
Present Value Formula for Combined Future Value Sum and Cash Flow (Annuity):
We can combine equations (1) and (2) to have a present value equation that includes both a future value lump sum and an annuity. This equation is comparable to the underlying time value of money equations in Excel.
Present Value
Equation 8
As in formula (2.1) if T = 0, payments at the end of each period, we have the formula for
present value with an ordinary annuity
Equation 8.1
As in formula (2.2) if T = 1, payments at the beginning of each period, we have the formula for
present value with an annuity due
Equation 8.2
Present Value when i = 0
In the case where i = 0 and we look back at equations (1) and (2a) to see that the combined present value formula can reduce to
Present Value with Growing Annuity (g ≠ i)
rewritten from formula (3)
Equation 9
Present Value with Growing Annuity (g = i)
rewritten from formula (4)
Equation 10
Note on Compounding m, Time t, and Rate r
Formula (8) can be expanded to account for compounding (m).
Equation 11
where n = mt and . t is the number of periods, m is the compounding intervals per period and r is rate per period t. (this is easily understood when applied with t in years, r the nominal rate per year and m the compounding intervals per year) When written in terms of i and n, i is the rate per compounding interval and n is the total compounding intervals although this can still be stated as "i is the rate per period and n is the number of periods" where period = compounding interval. "Period" can be a broad term.
Related to the calculator inputs, r = R/100 and g = G/100. If compounding (m) and payment frequencies (q) do not coincide in these calculations, r is converted to an
equivalent rate to coincide with payments then n and i are recalculated in terms of payment frequency, q. The first part of the equation is the
present value of the future sum and the second part is the
present value of an annuity.
Present Value of a Perpetuity (t → ∞ and n = mt → ∞)
For a perpetuity, perpetual annuity, the number of periods t goes to infinity therefore n goes to infinity. The
FV term in equation (11) goes to 0 and the 1/(1 + i)n in the second term also goes to 0 leaving just formula (5)
Equation 5
Present Value of a Growing Perpetuity (g < i) (t → ∞ and n = mt → ∞)
Likewise for a growing perpetuity, where we must have g<i, since (1 + i)n grows faster than (1 + g)n, this term in formula (9) reduces to formula (6)
Equation 6
Present Value of a Growing Perpetuity (g = i) (t → ∞ and n = mt → ∞)
Since n also goes to infinity (n → ∞) as t goes to infinity (t → ∞), we see that
Present Value with Growing Annuity (g = i) (10) goes to infinity and we are back at equation (7)
Equation 7
Continuous Compounding (m → ∞)
We look back to formula (11) for present value where m is the compounding per period t, t is the number of periods and r is the compounded rate with and n = mt.
Equation 11
The
effective rate is ieff = ( 1 + ( r / m ) )m - 1 for a rate r compounded m times per period. It can be proven mathematically that as m → ∞, ieff (the effective rate of r with continuous compounding) reaches the upper limit equal to er - 1.
Present Value with Continuous Compounding (m → ∞)
Removing the m and changing r to the effective rate of r, er - 1, in formula (11), formulas (8) & (11) for Present Value become
cancelling out 1's where possible we get the final formula for present value with continuous compounding
Equation 12
for an ordinary annuity
Equation 12.1
for an annuity due
Equation 12.2
Present Value of a Growing Annuity (g ≠ i) and Continuous Compounding (m → ∞)
We can modify equation (3a) for continuous compounding, replacing i's with er - 1 and we get:
which reduces to
Equation 13a
Multiplying (13a) by er/(1+g)
Equation 13b
subtracting (13a) from (13b) most terms cancel out leaving
multiplying through by (1+g)
solving this equation for
PV and adding on the term to account for whether we have a growing annuity due or growing ordinary annuity we multiply by the factor (1 + (er-1)T)
Equation 13
Present Value of a Growing Annuity (g = i) and Continuous Compounding (m → ∞)
Starting with equation (4) replacing i's with er - 1 and simplifying we get:
Equation 14
Present Value of a Perpetuity (t → ∞) and Continuous Compounding (m → ∞)
As t → ∞, ert → ∞ and formula (12) becomes
Equation 15
Present Value of a Growing Perpetuity (g < i) (t → ∞) and Continuous Compounding (m → ∞)
As t → ∞, n → ∞ and enr in formula (13) grows fastest causing this term to go to 0 and we are left with:
Equation 16
Present Value of a Growing Perpetuity (g = i) (t → ∞) and Continuous Compounding (m → ∞)
From our equation for
Present Value of a Growing Perpetuity (g = i) (7) replacing i with er-1 we end up with the following formula but since n → ∞ for a perpetuity this will also always go to infinity.
Equation 17
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5144 | https://www.themathdoctors.org/three-times-larger-idiom-or-error/ | Skip to content
Three Times Larger: Idiom or Error?
November 14, 2018 January 15, 2024 / AQOTW, Statistics / Ambiguity, Words / By Dave Peterson
Having just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does “three times larger” mean? How about “300% more”? We’ll focus on one discussion that involved several of us, and referred back to other answers we’ve given.
Percent increase vs. factor
The question, from 2006, started with the idea of a percent increase:
Percent Increase and "Increase by a Factor of ..." A math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase. But can't I also say there was an increase by a factor of 2? Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent? This is what is confusing me! I'd never been confused about saying "increased by a factor of" and "increased by percent of" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know! Thank you for any help.
Unfortunately, Joseph didn’t directly quote from the page he had in mind, and we have never said exactly what he said we did; so we couldn’t be sure which page it was. Everything he said, however, was correct.
Doctor Rick was the first to answer:
Hi, Joseph. Yes, this can be very confusing, because some statements about increases are ambiguous. When we say "increased by a factor of 2," the word "factor" makes it clear that we mean "multiplied by 2." When we say "increased by 10%," there is only one reasonable interpretation: the amount of the increase is 10% of the original amount. If we meant multiplication by 10%, that would be a decrease -- not an increase! Even when we say "increased by 100%," there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase. When we want to speak of an increase that is greater than the original amount, then ambiguity can arise. In that situation, I much prefer "increase by a factor of 3" or "by a factor of 2.5," etc. I don't know what page you saw -- but have you seen this one? Percent Greater Than vs. Increased See also the page linked there, about the even more confusing phrase "___ times more than" and the like. I am on the side of avoiding the confusing phrases, as a basic principle of communication. If you saw another page and you are still confused by it, please tell me the URL of that page so I can review it with you.
Ambiguity in percent increase
Before we get back to this conversation, we should take a look at the page he referred to, which is a good starting point:
Percent Greater Than vs. Increased What is the difference between the following statements: My profits are 200% bigger than they were last year. and My profits from last year have increased 200%. This is one of the questions we have to answer in my Middle school methods course and I have looked everywhere for the answer. I hope you can help.
I answered this one:
As far as I can see, they mean the same thing; in fact, both are similarly ambiguous. Taken literally, "200% bigger" (or, more formally, larger or greater) and "increased 200%" (or, more completely, increased _by_ 200%) both mean that the increase from one year to the next is 200% of the first year's value, so that the second year's profit is 3 times the first. But both statements are more likely to have been made with the intention of saying that this year's profit is twice last years. English is not very clear in cases like this.
So there is a literal meaning (which mathematicians tend to see as best), and an idiomatic meaning (which ordinary people are more likely to have intended to say). I referred to a page we’ll be looking at below, and then quoted a favorite book of mine that gives a lexicographer’s perspective:
Since writing that, I found a good reference on "two times greater," although it doesn't mention your "200% greater." It is in Merriam Webster's _Dictionary of English Usage_, which under "times" writes The argument in this case is that _times more_ (or _times larger_, _times stronger_, _times brighter_, etc.) is ambiguous, so that "He has five times more money than you" can be misunderstood as meaning "He has six times as much money as you." It is, in fact, possible to misunderstand _times more_ in this way, but it takes a good deal of effort. If you have $100, five times that is $500, which means that "five times more than $100" can mean (the commentators claim) "$500 more than $100," which equals "$600," which equals "six times as much as $100." The commentators regard this as a serious ambiguity, and they advise you to avoid it by always saying "times as much" instead of "times more." Here again, it seems that they are paying homage to mathematics at the expense of language. The fact is that "five times more" and "five times as much" are idiomatic phrases which have - and are understood to have - exactly the same meaning. The "ambiguity" of _times more_ is imaginary: in the world of actual speech and writing, the meaning of _times more_ is clear and unequivocal. It is an idiom that has existed in our language for more than four centuries, and there is no real reason to avoid its use. I think the same applies to "X percent bigger" and "increased [by] X%." There is just enough ambiguity in a technical context that I would want to ask what was intended before assuming anything, but there is no reason to say that they definitely mean different things, or mean something different than "X percent of" or "increased to X percent." I myself would avoid saying these things, just because there are enough people who have heard that they are ambiguous, and would therefore take them the wrong way (whichever that is!).
As a result of my side interest in linguistics, I recognize that human speech is not as logical as we might wish; what a word means is a matter of actual usage in a culture, rather than pure logic. So rather than state that either understanding of “times bigger” is “correct”, I just recognize that people take it in two ways, so you have to ask, or use contextual cues, in order to decide on what is meant.
Confusion about “three times larger”
Back to the original discussion: Joseph responded with specific references, the first of which was that link of mine that Doctor Rick said to “see also”:
I'm sorry, I should have specified the site. In fact, there were two -- and I still don't see the difference between them. Here is the first example, from Larger Than and As Large As 1) "Three times as large as N" means "3 N." 2) "Three times larger than N" means "4 N" -- but only if you stop to think about it, as many people do not. Here, I don't understand how something can be 3 times larger and be 4 times N. That sounds really weird to me. If you asked "What is something that is three times as large as N?" then I would say 3N ... but apparently I'd be wrong! I just don't see where my thinking is wrong.
His thinking isn’t wrong on this point: 3N is three times as large as N! He seems just to be letting what he’s read sow doubt about everything.
Here is the second example, from Percentage of Increase You can choose two ways to express your answer now. One is to say: there will be a 550% increase by the year 2000. Or you can say: in the year 2000 the (new value) -- you didn't say what the numbers represented, so I'm a little confused right here -- will be about five and a half times greater than what it was in 1995. Many people don't quite grasp those phrases, especially the latter one. Instead you might wish to say it this way: in 2000 the (new value) will be 6 and a half times what it was in 1995. The difference in the wording is subtle, of course, but important. The number 6 1/2 comes from 325,000 --------- = 6.5 or 6 1/2 50,000 which is NOT a percent increase situation. In this problem, I don't understand the difference between the way the doctor explains the two different ways you can talk about the increase, and the implications of each. The doctor says that 6.5 times is not a percent increase; but can you still say it's 650 percent OF the original? I'm sorry -- this is all very confusing at this point!
So his specific question is this:
Why wouldn’t “three times as large” and “three times larger” mean the same thing?
How can “5 1/2 times greater” and “6 1/2 times what it was” mean the same thing?
The two pages quoted are by me (1999) and Doctor Terrel (1997). The first is particularly worth reading in its entirety, as there is a lot more there.
The case for a literal interpretation
Doctor Greenie responded, arguing against laxness on the matter, and making the case for the literalistic interpretation:
I'm going to jump in here, because this is one of my pet peeves. Mathematics is commonly called the exact science. Mathematics must be exact; if it is not, it all falls apart. We can't use ambiguous language in mathematics. I agree that the use of the phrase "x times larger than" is best avoided. However, as a mathematician who believes in using unambiguous language, I cannot accept the proposition that we should be able to interpret "5 times larger than 10" as either 50 or 60. It HAS TO BE ONE OR THE OTHER. And grammatically, "5 times larger than" means the "new" number is 5 times larger than the "old" number; this in turn means the difference between the new and old numbers is 5 times the old number, making the new number 6 times the old number. So the number which is 5 times larger than 10 is 10 + 5(10) = 10 + 50 = 60 (The phrase "... larger than ..." implies comparison by subtraction; the phrase "... as large as ..." implies comparison by division. Sixty is 6 times as large as 10, because 60/10 = 6. But 60 is 5 times larger than 10, because [60 - 10]/10 = 50/10 = 5.) Yes, we hear it all the time in everyday life. Sometimes, we even hear it in the supposedly rigorous world of science -- "an earthquake of magnitude 5 is 10 times greater than one of magnitude 4," and such. But the common idiom of using "10 times greater than" -- when the actual meaning is "10 times as great as" -- has no place in mathematics.
He concluded with an accidental overstatement of what the “other side” says:
I disagree with many of the concessions that other math doctors here have made in interpreting the phrase "x times larger than" as being the same as "x times as large as." On one of the pages I saw, a fellow doctor said that "50% larger than" and "50% as large" mean the same thing. But if my weekly salary last year was $1000 and it is 50% LARGER this year, then it is now $1000 + 50%($1000) = $1000 + $500 = $1500 While if it was $1000 last year and it is 50% AS LARGE this year, then it is now 50%($1000) = $500 If something is 50% larger, then it is larger; if it is 50% as large, then it is smaller. They can't be the same; that is nonsense.
In fact, as he admitted in a subsequent private discussion, he had misremembered what others had said; none of us have claimed that “50% larger than” and “50% as large” mean the same thing. What we say is that, when the percentage or multiplier is greater than 100%, we recognize ambiguity in the likely intent. I think we agree that the phrase should not be used in a mathematical context, and that we both grudgingly interpret it as intended elsewhere.
What the literal interpretation means
Then it was my turn to respond, as the author of the first page Joseph had asked about, wanting to make sure he understood both why people take it literally as they do, and how we should think about it.
First, on “Larger Than and As Large As”, I said this:
Joseph, your thinking is RIGHT: if M is three times AS LARGE AS N, then M = 3N. That's what statement (1) above says. But if we break statement (2) apart carefully (some would say TOO carefully :-)), then it means something different from what people usually mean by it. If I said "M is 50 larger than N," I would mean that if you ADD 50 to N, you get M: M = N + 50. And if I said, "M is 50% larger than N," I would mean that if you add 50% OF N to N, you get M; that is, I mean that M is 50% of N added to N: M = N + (0.50)N. Now, though I'm not entirely sure I agree with this, technically minded people often apply the same thinking to (2), for the sake of consistency. The "larger than" means we add something to N. And what do we add? Three times N. So by this thinking, M = N + 3N = 4N. So "three times LARGER THAN N" means the same as "four times AS LARGE AS N."
I then referred to the usage book quoted above, adding:
English usage experts think it is nonsense. My feeling is that this thinking puts a little too much weight on consistency, and is just too weird for the general public to follow. English is not known for consistency! So we need to recognize that in everyday usage, (1) and (2) really mean the same thing. When we accept that, though, we set ourselves up for the opposite confusion: Cases like "50% larger" and "3 times larger" no longer follow the same pattern, and our language becomes inconsistent, which really bothers mathematicians! As Dr. Rick pointed out, this means that there are gray areas where it's hard to be sure what someone means, so it may be best just to avoid using these phrases in mathematical contexts.
Finally, I commented on Joseph’s other quote, from Doctor Terrel:
Joseph, here the doctor was saying that a 550% INCREASE means adding 550% of the original value to the original value, which means 650% OF the original value. In the other terminology, "5 1/2 times greater" (there again, "greater" is taken to refer to the increase) is the same as "6 1/2 times as much." When he says that the 6 1/2 is not a percent increase, he doesn't mean that it hasn't increased, just that he is talking about multiplying by 650% rather than adding 650%. When you think in terms of increase (adding), it is a 550% increase. Now, the "percent increase" case is pretty standard, because it IS technical terminology (though ordinary people reading it can get confused, so it's still risky). The "times greater" case is more disputable, since that sounds less technical. Most people don't demand absolute consistency from language; they are happy to understand "times greater" idiomatically. I hope that clears up some of the confusion. It isn't all cleared up yet at our end. You will definitely get different opinions as to what it all REALLY means!
Increase by a factor
Joseph replied, returning to his initial question:
That really cleared things up for me and I appreciate your time in driving home the differences! The last question I would like to ask is, how do you deal with factors? If someone says something has changed by a factor of ... or is less/greater than by a factor of ..., do we use the same rules that you've discussed above? Or when using the word "factor," are things a bit different?
I answered:
As Dr. Rick pointed out in the first response, "factor" is used to make it clear that multiplication, rather than addition, is the cause of an increase. Just as "increased by a factor of 2" means "twice as large," so does "greater by a factor of 2." And "decreased by a factor of 2" and "less by a factor of 2" both mean "half as much" (divided by 2). I can't think of a context in which that would not be true -- but English is flexible enough that I probably shouldn't guarantee anything!
Isn’t English fun?
Conclusion
In closing, here is a more recent question (2015) where I summarized this complicated issue:
As Big As vs. Bigger Than I'm having difficulty convincing my 5th graders that "as big as" and "bigger than" do not mean the same thing. For example, when asked, "How many times larger is 10,000 than than 100?" they answer "100." Their tests and homework are full of this misunderstanding! How would you suggest telling them they are wrong?
I referred to most of the pages we’ve seen above, and added:
To be honest, my feeling (basically unchanged since the first of those) is that although your understanding is common among thoughtful people, it is a case of excessive consistency. Mathematical people want a certain word structure to always have the same meaning, so we relate "x times bigger" to "x percent bigger," and that to "x bigger," and want to take all in an incremental (additive) sense. But taken on its own, it is perfectly logical to interpret "x times bigger" as "bigger, as a result of multiplication by x." And human language is not completely consistent; we have idioms all over the place that we interpret with no trouble. Having said that, I think that math books should avoid that form, because there is just enough truth to your thinking, and enough extra expectation of careful use of words in a math book, that it can be confusing. What I would do is to make a brief mention of the fact that many people take it as you do, but then point out that your book is using the phrase in the way it is usually intended in the real world. If I were writing the textbook, I would reverse this: almost always use "x times as big," but mention somewhere the fact that many people use "x times bigger" to mean the same thing, and briefly discuss the controversy before moving on to other things.
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5145 | https://math.stackexchange.com/questions/1404609/solving-an-equation-for-a-give-parameter | fundamental solution - Solving an equation for a give parameter. - Mathematics Stack Exchange
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Solving an equation for a give parameter.
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While solving a question , I came across the following situation , we're required to find a value for λ λ , from this equation :
λ=(1.345)(1−e−λ)λ=(1.345)(1−e−λ),
I tried taking logs, but couldn't get through,
The solution on the other hand , starts off with a trial and error method. ( I have no idea how to apply trial and error method here..)
Can anyone provide me a hint ?
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asked Aug 21, 2015 at 5:03
User9523User9523
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1 I suggest you look at Newton's method, setting f(λ)=1.345(1−e−λ)−λ f(λ)=1.345(1−e−λ)−λen.wikipedia.org/wiki/Newton's_method 2015-08-21 05:11:06 +00:00 Commented Aug 21, 2015 at 5:11
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If we consider the equation
x=a(1−e−x)x=a(1−e−x)
the analytical solution is given by
x=a+W(−a e−a)x=a+W(−a e−a)
where W(z)W(z) is Lambert function which, in the real domain, is defined if z≥−1 e z≥−1 e. This function is defined by z=W(z)e W(z)z=W(z)e W(z) (Euler and Lambert worked together).
You will probably learn quite soon about this marvelous function and I am sure that you will enjoy it.
Applied to the case a=1.345 a=1.345, this gives x≈0.625262 x≈0.625262.
If you cannot use this function, the only solution would be a numerical maethod such as Newton which, starting at a "reasonable" guess x 0 x 0 will update it according to
x n+1=x n−f(x n)f′(x n)x n+1=x n−f(x n)f′(x n)
Writing
f(x)=x−a(1−e−x)f(x)=x−a(1−e−x)
f′(x)=1−a e−x f′(x)=1−a e−x
which leads to the iterative scheme
x n+1=a(e x n−x n−1)e x n−a x n+1=a(e x n−x n−1)e x n−a
For sure, there is a trivial solution x=0 x=0; the derivative cancels at x=log(a)x=log(a) which corresponds to a minimum of the function (by the second derivative test). So, if it exists, the non-trivial solution is somewhere >log(a)>log(a).
For the specific value of a=1.345 a=1.345, let us start at x 0=1 x 0=1; Newton method generates the following iterates
x 1=0.7034892906 x 1=0.7034892906
x 2=0.6315135757 x 2=0.6315135757
x 3=0.6253110986 x 3=0.6253110986
x 4=0.6252619129 x 4=0.6252619129
x 5=0.6252619098 x 5=0.6252619098
which is the solution for ten significant figures.
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answered Aug 21, 2015 at 6:01
Claude LeiboviciClaude Leibovici
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Hint: if the 1.345 1.345 mattered it would be a miracle for trial-and-error to work. But it does work...
The other real solution is 1.345+W(−1.345 exp(−1.345))1.345+W(−1.345 exp(−1.345)), where W W is the Lambert W function.
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answered Aug 21, 2015 at 5:27
Robert IsraelRobert Israel
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While I can't provide a general solution, the value of λ λ is approximately 0.6252619098 0.6252619098, which should be sufficient for your question only. Certainly, as @robert-israel suggested, look into the Lambert-W function, which should provide the means to calculate the solution more generally.
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answered Aug 21, 2015 at 5:39
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This is a related question.Dodo –Dodo 2015-08-21 05:42:03 +00:00 Commented Aug 21, 2015 at 5:42
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5146 | https://www.youtube.com/watch?v=NbFW9Nyu0aw | Learn How to Find the nth Term of an Arithmetic Sequence Example with 2, 6, 10, 14, ...
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Learn How to Find the nth Term of an Arithmetic Sequence Example with 2, 6, 10, 14, ...
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Transcript:
hi everyone in this problem we're given an arithmetic sequence and we have to find what's called the nth term or the general term so the formula for the nth term of an arithmetic sequence is the following a sub n equals and it's the first term so a sub 1 plus and then parentheses n minus 1 times d where d here is called the common difference so d is the number that we keep adding over and over to get from one term to the next you'll notice here that if you add four you get to six and if you add four to six you get ten add four you get fourteen add 4 you get 18 etc so d which is called again the common difference is equal to 4. a sub 1 is going to be our first term in this case our first term is 2 so a sub 1 is equal to 2. so now all we have to do is plug in the numbers so a sub n is equal to a sub 1 which is 2 plus parentheses n minus 1 and then d is 4. so we have a sub n this is equal to two and we can distribute this four see four times n is going to give us four n and four times negative one will give us negative four so we have a sub n equals let's see 2 minus 4. it's going to give us a minus 2 and then we still have the 4n so plus 4 times n and that would be the final answer |
5147 | https://stackoverflow.com/questions/24866970/multiplying-real-part-of-a-complex-number-while-imaginary-part-stay-tact-matlab | Stack Overflow
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multiplying real part of a complex number while imaginary part stay tact (Matlab)
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I am trying to multiply real part of a complex number to a scalar while the imaginary part stay the same.
For example x= a + bi after multiplying to c result us y= ca + bi.
Any suggestions on procedure?
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asked Jul 21, 2014 at 14:00
SinaSina
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y=real(x)c+imag(x)?
Daniel
– Daniel
2014-07-21 14:02:08 +00:00
Commented Jul 21, 2014 at 14:02
Nope. didn't worked. h = 1 + 2i >> y=real(h)2+imag(h) y = 4
Sina
– Sina
2014-07-21 14:04:54 +00:00
Commented Jul 21, 2014 at 14:04
As in my answer, don't forget to multiply imag(x) with 'i' !
EJG89
– EJG89
2014-07-21 14:07:18 +00:00
Commented Jul 21, 2014 at 14:07
Are you sure? I tried but it is not giving me the correct form I wrote in question.
Sina
– Sina
2014-07-21 14:10:56 +00:00
Commented Jul 21, 2014 at 14:10
Yes I am sure: h = 1 + 2 i >> y = real(h) 2 + imag(h) i (<- do not forget the i here) results in y = 2 + 2 i
EJG89
– EJG89
2014-07-21 14:15:48 +00:00
Commented Jul 21, 2014 at 14:15
| Show 1 more comment
1 Answer 1
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3
For this you can use the real and imag functions in MatLab. So the multiplication becomes:
y = real(x) c + imag(x)i
real(see documentation) retrieves a from x and imag(see documentation) retrieves b from x. Don't forget to multiply with i again ;)
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5148 | https://www.wyzant.com/resources/answers/915200/combinatorics-password-problem | WYZANT TUTORING
Hannah S.
Combinatorics password problem
a 7-character password can have upper case letters, lower case letters, and numerical digits. How many passwords are possible if it must start with a number and end with a lowercase letter (repetitions allowed)?
1 Expert Answer
Daniel B. answered • 12/08/22
A retired computer professional to teach math, physics
There are 26 upper case letters, 26 lower case letters, and 10 numbers,
for a total of 62 possible characters.
The first position can be chosen in 10 ways.
The last position can be chosen in 26 ways.
Each of the remaining 5 positions can be chosen in 62 ways.
This gives a total of 10×625×26 possible passwords.
The constraint on the first position guarantees a digit,
the constraint on the last position guarantees a lower case letter,
but the above count includes passwords without any upper case letter.
Therefore from the above count we need to subtract those passwords can contain
only lower case letters and digits.
By the same reasoning as above there are 10×365×26 such combinations.
Therefore the number of allowed passwords is
10×625×26 - 10×365×26 = 222,473,330,560
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5149 | https://www.youtube.com/watch?v=VxM58kv8hs0 | How do we simplify powers of i, the imaginary number? Using arithmetic mod 4!
polymathematic
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Posted: 11 May 2022
How are we supposed to approach problems with absolutely gargantuan powers of i? What does it even mean to raise i, the imaginary unit, to a power at all? The geometry of what's happening in the complex plane when we multiply complex numbers helps us better understand why powers of i recur in a cycle of 4. This means taking the remainder of the power when dividing by 4 can help us simplify imaginary numbers.
imaginarynumbers #complexnumbers #complexplane
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have you ever seen a problem like this where we have some large number i to the 79th i to the hundred and first power and we're supposed to find the value of that expression and teacher says something about yeah it's related to dividing by four or mod four and you're like what is this person talking about if we think about complex numbers in terms of the coordinate plane our real axis moving left to right and our imaginary axis moving up and down we can get a better sense of what's going on with problems like i to the 79th power basically what we're seeing in the complex plane is that we have a four-step cycle of possible powers of i we start at one and we say okay if we multiply by our first power of i that's just going to rotate the coordinate plane over to i itself and so of course i to the first power is going to be equal to i and then when we perform another multiplication by i it's going to rotate the coordinate plane 90 degrees once more and that's what we mean when we say that i squared is equal to negative 1. but if we multiply by another i that gives us the third power of i which we can now see through another rotation takes us over to negative i so that's what we mean when we say that i to the third is equal to negative i finally as we multiply one more time we get that i to the fourth is going to be equal to one and the important thing to recognize here is we're back where we started so every other subsequent power of i is just going to follow this same pattern i to the fifth is going to be i in the same way that i to the first was equal to i i to the sixth is going to be negative one in the same way that i squared was equal to negative one this is why your teacher talks to you about powers of i coming in some sort of cycle of force or thinking of them in terms of mod 4 of the power itself when i want to compute something like i to the 79th i only need to figure out where it fits in this cycle i can tell that the cycle begins at one every single time when my exponent actually is a multiple of four and then every number greater than a multiple of four by one is going to be equal to i every number with an exponent greater by two than a multiple of four is going to be negative one every number with an exponent three greater than some multiple of four is going to be equal to negative i and again that takes me back to every multiple of four power which is going to be equal to one for i to the 79th specifically all i need to figure out is that 79 when you divide it by 4 leaves a remainder of 3. and so this is the same thing as i to the third is the same thing as negative i i to the 101st if i divide 101 by four leaves a remainder of one this must be the same thing as i to the first that is it's the same thing as i itself because divisibility by four actually only depends on the final two digits of the number i don't even need to pay attention to the rest of the number if i have i to some really large number ending in 22 that's the same thing as i to the 22nd which has a remainder of 2 when we divide by 4 so that's the same thing as i squared equals negative one but it all goes back to this cycle which results from our geometric understanding of what precisely complex numbers are |
5150 | https://www.ime.usp.br/~peresi/shortcourse2.pdf | REPRESENTATIONS OF THE SYMMETRIC GROUP AND POLYNOMIAL IDENTITIES LUIZ A. PERESI These are the lecture notes from a short course given by the author during the CIMPA Research School on Associative and Nonassociative Algebras and Dialge-bras: Theory and Algorithms - In Honour of Jean-Louis Loday (1946-2012), held at CIMAT, Guanajuato, Mexico, February 17 to March 2, 2013.
Abstract. Let Sn denote the symmetric group on n symbols. When F has characteristic zero or greater than n, the group algebra FSn is a direct sum of p(n) matrix algebras over F, where p(n) is the number of partitions of n.
We present an efficient method due to J. M. Clifton (1981) that calculates the matrix associated to each element of Sn, for each partition of n. In 1950, A. I. Malcev and W. Specht independently used the representation theory of the symmetric group to classify polynomial identities of algebraic structures. Starting in the 1970’s the method was further developed by A. Regev (see the survey paper Regev ). The method was implemented as a computer algebra system by I. R. Hentzel in 1977. We present this computational method and give applications to research on identities of Cayley-Dickson algebras, nuclear elements of the free alternative algebra and special identities of Bol algebras. For some applications of the computational method to other algebraic structures see the papers - by M. R. Bremner and the author.
1. Representations of the symmetric group Let n be a positive integer, Sn the symmetric group on {1, . . . , n} and FSn the group algebra over the field F. We remind that if G is a group then the group algebra FG is the vector space over F with basis G, and the multiplication in FG is given by extending the multiplication in G. The representation theory of FSn was determined in 1900 by Alfred Young (1873-1940). It has since been simplified and exposited by many authors. See for example Boerner and James and Kerber . We give an exposition of this theory based on Clifton . In his review (MR0624907 (82j:20024)) of Clifton’s paper, James wrote the following: “Most methods for working out the matrices for the natural representation are messy, but this paper gives an approach which is simple both to prove and to apply.” 1.1. Frames and tableaus. A partition of n is an ordered sequence of positive integers λ = λ1 . . . λk satisfying λ1 ≥λ2 ≥· · · ≥λk and λ1+· · ·+λk = n. The frame corresponding to λ consists of n boxes in k left-justified rows of lengths λ1, . . . , λk.
For example, if n = 3 then there are 3 partitions 3, 21, 111 with corresponding frames 1 2 LUIZ A. PERESI A tableau corresponding to λ consists of a bijective assignment of the numbers 1, . . . , n to the n boxes. A standard tableau is one in which the assigned numbers increase in each row from left to right and in each column from top to bottom. For example, if n = 3 then the standard tableaus are: 1 2 3 1 2 3 1 3 2 1 2 3 The number of standard tableaus corresponding to partition λ = λ1 . . . λk of n is given by n!
Y i<j (li −lj) l1! l2! . . . lk!
where li = λi + k −i (i = 1, . . . , k).
If T is a tableau we denote by T (i, j) the number in row i and column j. The lexicographical order on tableaus is defined by T < T ′ if and only if T (i, j) < T ′(i, j) where i is the least row index for which T (i, j) ̸= T ′(i, j).
For finite groups we have: Theorem 1.1. (H. Maschke 1899) For any finite group G and any field F, of characteristic 0 or characteristic p not dividing |G|, the group algebra FG is semisimple.
It follows that FG is isomorphic to the direct sum of simple ideals, and each simple ideal is isomorphic to the endomorphism algebra of a vector space over a division ring over F. For G = Sn, these division rings are all equal to F and we have: Corollary 1.2. The Wedderburn decomposition of FSn is given by an iso-morphism φ: F Sn → M λ Mdλ(F), where the summation is over all partitions λ of n, and dλ is the corresponding number of standard tableaus.
A permutation π of {1, 2, . . . , n} is denoted by the sequence π(1)π(2) . . . π(n).
For n = 3 we have the isomorphism φ: FS3 →F ⊕M2(F) ⊕F.
given by φ(123) = 1, 1 0 0 1 , 1 , φ(132) = 1, 0 1 1 0 , −1 , φ(213) = 1, 1 −1 0 −1 , −1 , φ(231) = 1, −1 1 −1 0 , 1 , φ(312) = 1, 0 −1 1 −1 , 1 , φ(321) = 1, −1 0 −1 1 , −1 .
REPRESENTATIONS AND IDENTITIES 3 1.2. Computing the irreducible representations. Let λ be a partition of n.
Let T1, . . . , Tdλ be the standard tableaus for λ in lexicographical order. For π ∈Sn we denote by Rλ π the dλ ×dλ matrix representing π, that is, the projection of π onto the summand Mdλ(F) in the Wedderburn decomposition given by Corollary 1.2.The map Rλ : Sn →Mdλ(F) defined by Rλ(π) = Rλ π is the irreducible representation of Sn corresponding to the partition λ. Let ι denote the identity permutation in Sn.
The matrix Rλ(π) is given by Rλ π = (Aλ ι )−1Aλ π, where Aλ ι and Aλ π are the Clifton matrices corresponding to ι and π.
The Clifton matrix Aλ π is the dλ × dλ matrix defined by the following algorithm for computing each entry (Aλ π)ij: • Apply π to the standard tableau Tj obtaining the (possibly non-standard) tableau πTj.
• If there exist two numbers that appear together both in a column of Ti and in a row of πTj, then (Aλ π)ij = 0.
• Otherwise, there exists a vertical permutation κ ∈Sn which – leaves the columns of Ti fixed as sets, – and takes the numbers of Ti into the rows they occupy in πTj.
In this case, (Aλ π)ij = ϵ(κ), the sign of κ.
For n ≤4 and for all partitions λ, Aλ ι is the identity matrix. For n ≥5 and for some partitions λ, the matrix Aλ ι is not the identity matrix; however this matrix is always invertible.
Example 1.3. Let n = 3 and λ = 21. We have that dλ = 2. For π = 213 we calculate Aπ.
(i, j) Ti π(Tj) (1, 1) T1 = 1 2 3 πT1 = 2 1 3 κ = ι ϵ(κ) = 1 (1, 2) T1 = 1 2 3 πT2 = 2 3 1 κ = 321 ϵ(κ) = −1 (2, 1) T2 = 1 3 2 πT1 = 2 1 3 κ does not exist (2, 2) T2 = 1 3 2 πT2 = 2 3 1 κ = 213 ϵ(κ) = −1 This gives the Clifton matrix Aλ π = 1 −1 0 −1 .
Example 1.4. Let n = 5 and λ = 32. In this case dλ = 5 and the standard tableaus are 1 2 3 4 5 1 2 4 3 5 1 2 5 3 4 1 3 4 2 5 1 3 5 2 4 4 LUIZ A. PERESI To calculate the (1, 5) entry of the Clifton matrix Aλ ι , we consider T1 and ιT5 = T5.
In this case the transposition κ = 15342 interchanging 2 and 5 is the required vertical permutation, so (Aλ ι )15 = −1. In fact, we have Aλ ι = 1 0 0 0 −1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 , (Aλ ι )−1 = 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 .
This is the simplest case in which Aλ ι is not the identity matrix.
To illustrate the difference that can exist between the Clifton matrix Aλ π and the representation matrix Rλ π = (Aλ ι )−1Aλ π, consider the 5-cycle π = 23451; in this case we calculate Rλ π (Aλ ι )−1 Aλ π −1 −1 1 1 0 −1 0 0 0 1 0 −1 0 0 0 −1 0 0 1 0 0 −1 0 1 0 = 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 −1 0 1 0 0 −1 0 0 0 1 0 −1 0 0 0 −1 0 0 1 0 0 −1 0 1 0 The Wedderburn decomposition φ: F Sn →L λ Mdλ(F) has an inverse φ−1 : M λ Mdλ(F) →F Sn given by φ−1(Eλ ij) = U λ ij, where λ is a partition of n and Eλ ij is the dλ × dλ matrix unit. We describe how the elements U λ ij ∈FSn can be calculate.
Let T λ i be the i-th standard tableau in lexicographical order. We denote by Rλ i the subgroup of Sn which permutes the elements of T λ i within each row but leaves the rows fixed as sets. We denote by Cλ i the subgroup of Sn which permutes the elements of T λ i within each column but leaves the columns fixed as sets. We define the elements Dλ ii by Dλ ii = dλ n!
X σ ∈Rλ i X τ ∈Cλ i ϵ(τ) σ τ.
If T λ j is the j-th standard tableau we denote by sλ ij ∈Sn the permutation such that sλ ij T λ i = T λ j . We define the elements Dλ ij by Dλ ij = Dλ ii (sλ ij)−1.
Let (Aλ ι )−1 = (aλ ij). We finally define the elements U λ ij by U λ ij = dλ X k=1 aλ jk Dλ ik.
Theorem 1.5. The elements U λ ij satisfy the orthogonal matrix unit relations. That is, for any two partitions λ and µ of n, we have U λ ij U µ kℓ= δλµ δjk U λ iℓ.
REPRESENTATIONS AND IDENTITIES 5 Furthermore, the isomorphism φ−1 : M λ Mdλ(F) →F Sn is defined on basis elements by φ−1(Eλ ij) = U λ ij.
Example 1.6. For n = 3, φ−1 : F ⊕M2(F) ⊕F →FS3 is given by φ−1 E(3) 11 = 1 6(123 + 132 + 213 + 231 + 312 + 321), φ−1 E(21) 11 = 1 3(123 + 213 −312 −321), φ−1 E(21) 12 = 1 3(132 + 231 −312 −321), φ−1 E(21) 21 = 1 3(132 −213 −231 + 312), φ−1 E(21) 22 = 1 3(123 −213 −231 + 321), φ−1 E(111) 11 = 1 6(123 −132 −213 + 231 + 312 −321).
2. Polynomial identities and representation theory The application of the representation theory of the symmetric group to polyno-mial identities was initiated independently by Malcev and Specht in 1950.
Starting in the 1970’s the method was further developed by Regev (see ). The implementation of this theory in computer algebra was initiated by Hentzel [25, 26] in 1977.
Let X = {x1, x2, . . . } be a set of variables. We construct the set M[X] of (non-commutative and nonassociative) monomials inductively as follows: X ⊂M[X]; if xi, xj ∈X then xixj ∈M[X]; if u, v ∈M[X]−X then xi(u), (u)xi, (u)(v) ∈M[X].
Let F[X] = n r X i=1 αi ui | r ∈N, αi ∈F, ui ∈M[X] o be the vector space over F spanned by M[X]. The elements of F[X] are called nonassociative polynomials in the variables xi. We define in F[X] a multiplication by the following rules: xi.xj = xixj, xi.u = xi(u), u.xi = (u)xi, u.v = (u)(v), ( r X i=1 αi ui) . ( s X j=1 βj vj) = r,s X i,j=1 αiβj ui.vj, where xi, xj ∈X, u, ui, vj ∈M[X] −X.
We obtain then an algebra called free nonassociative algebra generated by X.
We denote this algebra by F[X] or F[x1, x2, . . . ].
6 LUIZ A. PERESI 2.1. Polynomial identities of an algebra. Let A be a nonassociative alge-bra over F.
A polynomial f = f(x1, x2, . . . , xn) ∈F[X] is an identity of A if f(a1, a2, . . . , an) = 0 for all a1, a2, . . . , an ∈A. When f is an identity of A we say also that A satisfies f = 0. Any identity of degree ≤n over a field F of characteris-tic 0 or p > n is equivalent to a finite set of multilinear identities. If f = Pr i=1 αi ui is a multilinear polynomial of degree n then each term αi ui has a coefficient αi ∈F together with a monomial ui which is a permutation of x1 · · · xn with an association type.
An association type is a placement of parentheses.
For n = 3 we have two association types, (xx)x and x(xx). For n = 4 we have five association types: T1 T2 T3 T4 T5 ((xx)x)x (x(xx))x (xx)(xx) x((xx)x) x(x(xx)) Here x is just a place holder. For the general case of a nonassociative algebra, the number of distinct association types in degree n equals the Catalan number, c(n) = 1 n 2n−2 n−1 .
The numbers c(n) grow very rapidly; here is a short table: n 1 2 3 4 5 6 7 8 9 10 11 12 c(n) 1 1 2 5 14 42 132 429 1430 4862 16796 58786 We denote the association types of degree n by T1, . . . , Tc(n).
If f = f(x1, . . . , xn) is a multilinear polynomial then f = f1 + · · · + fc(n).
Within each component fk, all the terms have association type k and differ only in the permutation of the variables. So each component fk can be regarded as an element of the group algebra FSn: fk = X π∈Sn αkπ (x1x2 . . . xn)π ∈FSn.
The permutation π applies to the position, not the subscript. Thus (x1x2 . . . xn)π = xπ−1(1)xπ−1(2) . . . xπ−1(n). For example, (x1x2x3)(231) = x3x1x2. We can therefore regard f as an element of the direct sum of c(n) copies of FSn: f = f1 + · · · + fc(n) ∈FSn ⊕· · · ⊕FSn.
Let φλ : F Sn →Mdλ(F) be the λ-component of the Wedderburn decomposition φ: F Sn →L λ Mdλ(F). Applying φλ to f we obtain the representation matrix of f in partition λ given by Table 1.
T1 T2 · · · Tc(n)−1 Tc(n) h φλ(f1) | φλ(f2) | · · · | φλ(fc(n)−1) | φλ(fc(n)) i Table 1. Representation matrix of f in partition λ REPRESENTATIONS AND IDENTITIES 7 The basic problem concerning identities of an algebra is the following: Problem 2.1. Let f 1, . . . , f k, f be a set of multilinear identities of degre n. Is f a consequence of f 1, . . . , f k?
For i = 1, . . . , k we consider f i = f i 1 + f i 2 + · · · + f i c(n) ∈FSn ⊕· · · ⊕FSn and also f = f1 + f2 + · · · + fc(n) ∈FSn ⊕· · · ⊕FSn.
We then apply φλ to f 1, . . . , f k, f to obtain the representation matrix Mλ = φλ(f 1 1 ) φλ(f 1 2 ) · · · φλ(f 1 c(n)−1) φλ(f 1 c(n)) φλ(f 2 1 ) φλ(f 2 2 ) · · · φλ(f 2 c(n)−1) φλ(f 2 c(n)) .
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φλ(f k 1 ) φλ(f k 2 ) · · · φλ(f k c(n)−1) φλ(f k c(n)) φλ(f1) φλ(f2) · · · φλ(fc(n)−1) φλ(fc(n)) Let Nλ be the matrix consisting of the first k rows of blocks of Mλ. We denote by RCF(Mλ) the row canonical form of Mλ and by RCF(Nλ) the row canonical form of Nλ. Let rank(Mλ) be the number of nonzero rows in RCF(Mλ) and rank(Nλ) the number of nonzero rows in RCF(Nλ).
Proposition 2.2. (Solution of Problem 2.1) Let f 1, . . . , f k, f be a set of mul-tilinear identities of degre n. Then f a consequence of f 1, . . . , f k if and only if rank(Mλ) = rank(Nλ) for all partition λ of n.
Rational arithmetic and modular arithmetic. We prefer to use rational arith-metic (F = Q). However, even for a sparse matrix with integer entries, very large rational numerators and denominators can appear during the computation of the row canonical form. Therefore, when it is necessary, we use modular arithmetic (F = Zp where p is a prime and p > n). The ranks are the same over Q and over Zp (see Bremner and Peresi , Lemma 8).
2.2. An example in alternative algebras. In any algebra the associator (a, b, c) and the commutator [a, b] are defined by (a, b, c) = (ab)c −a(bc) and [a, b] = ab − ba. An algebra A is called an alternative algebra if it satisfies the left and right alternative laws (a, a, c) = 0 and (a, b, b) = 0. The most familiar example of an alternative algebra is the octonions which appear in the chain Reals ⊂Complexes ⊂Quaternions ⊂Octonions.
The Cayley-Dickson algebras are generalizations of the octonions (see §2.4). The classification of simple alternative algebras is given by Theorem 2.3. (Kleinfeld 1953) A simple alternative algebra (which is not associative) is a Cayley-Dickson algebra over its center.
A well-known identity satisfied by alternative algebras is given by Proposition 2.4. Let A be any alternative algebra. Then A satisfies the identity f(a, b, c, d) = (ab, c, d) + (a, b, [c, d]) −a(b, c, d) −(a, c, d)b = 0.
8 LUIZ A. PERESI To prove that f(a, b, c, d) = 0 is an identity in any alternative algebra we need to verify that it is a consequence of the alternative laws. We do this using Proposition 2.2.The linerized form of (a, a, b) = 0 is (a, b, c) + (b, a, c) = 0. The linerized form of (a, b, b) = 0 is (a, b, c) + (a, c, b) = 0. From this we obtain the following lifting identities to degree 4: (a, b, c)d + (b, a, c)d = 0, (a, b, c)d + (a, c, b)d = 0, a(b, c, d) + a(c, b, d) = 0, a(b, c, d) + a(b, d, c) = 0, (ab, c, d) + (c, ab, d) = 0, (ab, c, d) + (ab, d, c) = 0, (a, bc, d) + (a, d, bc) = 0, (a, b, cd) + (b, a, cd) = 0.
For each partition λ we calculate the matrix Mλ representing the lifting identities and f(a, b, c, d), and the matrix Nλ representing only the lifting identities. The ranks of the row canonical forms of these matrices are given in Table 2. The ranks of Nλ are given in the column labeled Lift and those of Mλ are given in the column labeled Lift and f.
The ranks are the same in both columns for all partitions.
Therefore, f(a, b, c, d) is a consequence of the lifting identities to degree 4 of the linearized forms of the alternative laws.
Partition λ dλ Lift Lift and f 4 1 4 4 31 3 12 12 22 2 8 8 211 3 10 10 1111 1 2 2 Table 2. Ranks of liftings identities and f(a, b, c, d) 2.3. Computing polynomial identities. If A is an algebra over F we denote by T(A) the set of polynomial identities of A. The set T(A) is an ideal of identities (or T-ideal).
That is, T(A) is an ideal of F[X] and ψ(T(A)) ⊂T(A) for any homomorphism ψ : F[X] →F[X].
Problem 2.5. (Specht 1950) Given a class of algebras, is it true that any algebra A in this class has a finite basis of identities (i.e., the ideal of identities T(A) is generated by a finite number of identities)?
Specht posed this problem for associative algebras over a field of characteristic zero. The complete solution was given by Kemer .
Theorem 2.6. (Kemmer 1987) Any associative algebra over a field of charac-teristic zero has a finite basis of identities.
Similar results were obtained by Vais and Zelmanov in 1989 for finitely generated Jordan algebras, and by Iltyakov [36, 37] for finetely generated alternative algebras (1991) and Lie algebras (1992).
In this subsection we consider the problem of finding polynomial identities of a fixed degree.
REPRESENTATIONS AND IDENTITIES 9 Problem 2.7. Let A be an algebra over a field F of finite dimension d.
Let {v1, . . . , vd} be a basis of A and multiplication table given by vivj = d X k=1 ck ijvk (ck ij ∈F).
Determine all the multilinear polynomial identities of degree n satisfied by A. What is the mininal degree of a polynomial identity?
For the matrix algebra Mn(F) the polynomial identities of minimal degree is given by Theorem 2.8. (Amitsur and Levitzki 1950) The minimal degree of a poly-nomial identity of Mn(F) is 2n. Furthermore, any multilinear polynomial identity of degree 2n is a multiple of the standard polynomial s2n(x1, . . . , x2n) = X σ∈S2n ϵ(σ) xσ(1) . . . xσ(2n).
In 1973 Leron proved that if char(F) = 0 and n > 2 then all polynomial identities of degree 2n+1 of Mn(F) are consequences of s2n(x1, . . . , x2n). In partic-ular, the identities of degree 7 of M3(F) are consequences of s6. When char(F) = 0, Drensky and Kasparian found in 1983 all identities of degree 8 of M3(F) and showed that they are consequences of the standard identity s6.
The computational methods used to study polynomial identities of matrices are surveyed by Benanti et. al. . We remark also that there are in the literature many results about the ideal of identities T(M2(F)).
In 1997 Bondari introduced an algorithm for computing an independent gen-erating set for the multilinear identifies and central identities of Mn(F) over a field F of characteristic zero or large enough prime. Constructed all the multilinear iden-tifies and all the multilinear central identities of degree ≤8 for M3(F). Checked the existing results in the literature and found a new central identity in degree 8 for M3(F).
The algorithm of Bondari can be used to find identities of any finite dimensional algebra, giving a solution for Problem 2.7.
Algorithm 2.9. (Bondari 1997) Let f(x1, . . . , xn) be a multilinear polynomial of degree n and write f = f1 + · · · + fc(n) ∈FSn ⊕· · · ⊕FSn.
For k = 1, . . . , c(n), assume that the dλ × dλ matrix φλ(fk) is given by φλ(fk) = ak 11Eλ 11 + · · · + ak 1dλEλ 1dλ (ak 1j ∈F). In other words, assume that φλ(fk) is a matrix such that the only nonzero row is the first row ak 11ak 12 . . . ak 1dλ. (Any nonzero row can be moved to the first row.) In this case we say that f is an irreducible polynomial for partition λ.
We consider φ−1(Eλ 1j) = U λ 1j ∈FSn as a multilinear associative polynomial of degree n. For k = 1, . . . , c(n) we denote by [U λ 1j]k (x1, . . . , xn) the multilinear nonassociative polynomial obtained by applying association type Tk to every term of U λ 1j. Therefore f(x1, . . . , xn) = c(n) X k=1 dλ X j=1 ak 1j [U λ 1j]k (x1, . . . , xn).
10 LUIZ A. PERESI Assume that f is an identity of A. Then f(a1, . . . , an) = 0 for any choice of a1, . . . , an ∈A. Given n arbitrary elements a1, . . . , an ∈A, we can evaluate [U λ 1j]k (a1, . . . , an) = α1 jkv1 + · · · + αd jkvd := (α1 jk, . . . , αd jk) ∈Fd.
Therefore c(n) X k=1 dλ X j=1 ak 1j (α1 jk, . . . , αd jk) = 0.
This gives the following linear system with d equations and c(n)dλ unknowns ak 1j: c(n) X k=1 dλ X j=1 αi jk ak 1j = 0 (i = 1, . . . , d).
Solving this linear system we find the coeficients ak 1j of the irreducible polynomial identity f(x1, . . . , xn) = 0 of A for partition λ.
2.4. Polynomial identities of Cayley-Dickson algebras. A Cayley-Dickson algebra is an eight-dimensional alternative algebra C = C(α, β, γ) over a field F with three non-zero parameters α, β, γ in F. Over the reals C(−1, −1, −1) is the algebra of octonions. The algebras C can be constructed by the Cayley-Dickson process (see Zhevlakov et. al. , §2.2). Over a field F of characteristic ̸= 2 it is possible to choose a basis 1, e1, . . . , e7 with the multiplication table given below.
C 1 e1 e2 e3 e4 e5 e6 e7 1 1 e1 e2 e3 e4 e5 e6 e7 e1 e1 α e3 αe2 e5 αe4 −e7 −αe6 e2 e2 −e3 β −βe1 e6 e7 βe4 βe5 e3 e3 −αe2 βe1 −αβ e7 αe6 −βe5 −αβe4 e4 e4 −e5 −e6 −e7 γ −γe1 −γe2 −γe3 e5 e5 −αe4 −e7 −αe6 γe1 −αγ γe3 αγe2 e6 e6 e7 −βe4 βe5 γe2 −γe3 −βγ −βγe1 e7 e7 αe6 −βe5 αβe4 γe3 −αγe2 βγe1 αβγ Problem 2.10. Find a basis for the ideal of identities T(C).
Isaev found in 1984 a finite basis of T(C) when F is a finite field. When char(F) = 0, Iltyakov proved in 1985 that T(C) is finitely generated without giving a set of generators. For fields of characteristic ̸= 2, 3, 5, Racine found in 1988 the identities of degree ≤5 of C.
Let A be an algebra over F with identity element 1. Then A is a quadratic algebra if each element x ∈A satisfies x2 −t(x) x + n(x) 1 = 0, where the trace t : A →F is a linear map and the norm n : A →F is a quadratic form. The trace and the norm are uniquely defined.
The Cayley-Dickson algebra C is a quadratic algebra. Let x = a 1+P7 i=1 ai ei ∈ C and x = a 1−P7 i=1 ai ei. The trace and norm of x is given by t(x) = x+x = 2a and n(x) = x x = a2 −αa2 1 −βa2 2 + αβa2 3 −γa2 4 + αγa2 5 + βγa2 6 −αβγa2 7.
REPRESENTATIONS AND IDENTITIES 11 The Jordan product x ◦y is defined by x ◦y = xy + yx. Let Vx and V given by Vx(y) = x ◦y and V = VaVbVc + VcVaVb + VbVcVa −VbVaVc −VaVcVb −VcVbVa.
Proposition 2.11. (Racine 1985) Any quadratic algebra satisfies the identity V (d2) −V (d) ◦d = 0.
Corollary 2.12. The Cayley-Dickson algebra C satisfies the identity V (d2) −V (d) ◦d = 0.
The linearized form of x2−t(x) x+n(x) 1 = 0 is x◦y−t(x) y−t(y) x+q(x, y) 1 = 0, where q(x, y) = n(x + y) −n(x) −n(y). For every a, b, c, d ∈C, if x = [a, b] and y = [c, d] then t(x) = 0, t(y) = 0. Thus [a, b] ◦[c, d] = −q([a, b], [c, d]) 1 ∈F. It follows that for every e ∈C we have [[a, b] ◦[c, d], e] = 0. Therefore C satisfies the identity [[a, b] ◦[c, d], e] = 0.
Using the Algorithm 2.9 we can obtain the identities of degree ≤7 satisfied by C. For each degree we list only the new ones, that is, those that are not implied by identities of lower degree.
Theorem 2.13. (Hentzel and Peresi 1997) The identities of degree ≤6 of the Cayley-Dickson algebras C are given by (char(F) = 0 or > degree of the identity): degree 1 and 2: None.
degree 3: (a, a, b) = 0 and (a, b, b) = 0 (alternative laws).
degree 4: No new ones (just the lifted alternative identities).
degree 5: V (d2) −V (d) ◦d = 0 and [[a, b] ◦[c, d], e] = 0.
degree 6: [ ALTSUM{a,b,c,d,e} { 24 a(b(c(de)))+8 a((b, c, d)e)−11 (a, b, (c, d, e)) } , f ] = 0, where ALTSUM{a,b,c,d,e} denotes the alternating sum over all the permutations of the variables {a, b, c, d, e}.
We remark that although all the calculations done by Racine in are correct, to obtain all degree 5 identities of C, we need the identity [[a, b] ◦[c, d], e] = 0. The weaker form [[a, b] ◦[a, b], e] = 0 used by Racine is not enough.
Theorem 2.14. (Bremner and Peresi 2012) The identities of degree 7 of the Cayley-Dickson algebra C are consequences of identities of degree ≤6.
In 2002, Bremner and Hentzel studied identities for alternative algebras which are restricted in the sense that the terms in the identities must be built out of associators. These authors discovered two new identities in degree 7 satisfied by the associator in every alternative algebra and five new identities in degree 7 satisfied by the associator in the algebra of octonions.
For fields of characteristic zero, Shestakov and Zhukavets found in 2009 a basis consisting of 3 identities (one of degree 5 and two of degree 6) for the skew-symmetric identities of C. One of this identity of degree 6 is two times the one given in Theorem 2.13. In 2011, Shestakov found a basis of identities for a Cayley-Dickson algebra modulo an associator ideal of a free alternative algebra. More precisely, denote by Tas(C) a homomorphic image of the ideal of identities T(C) of the split Cayley-Dickson algebra C, in a free associative algebra. Over a field F of characteristic not 2, 3, or 5, Shestakov found 12 LUIZ A. PERESI a basis for the ideal Tas(C). Using the relations that appear in the Cayley-Dickson process, Henry found in 2012 a basis for the Z2 2 and Z3 2-graded identities for C.
The Z3 2-grading requires characteristic ̸= 2.
3. Nuclear elements in the free alternative algebra The nucleus of an algebra A is the set N(A) = {p ∈A | (p, x, y) = (x, p, y) = (x, y, p) = 0, ∀x, y ∈A}.
The center of A is the set C(A) = {p ∈N(A) | [p, x] = 0, ∀x ∈A}.
Let F[X] be the free nonassociative algebra over the field F in generators X = {x1, x2, . . . , xn}. Let Alt[X] denote the ideal of F[X] generated by the elements (f1, f1, f2), (f2, f1, f1) (f1, f2 ∈F[X]).
We denote by Altn[X] the subspace of multilinear elements of degree n in Alt[X]. The free alternative algebra in generators X is the quotient algebra ALT[X] = F[X]/Alt[X].
In 1953 Kleinfeld showed that for any x and y in an alternative algebra, the element [x, y]4 is in the nucleus and this element is nonzero in the free alternative algebra on two or more generators. This result was used by Kleinfeld to prove Theorem 2.3. Subsequently, other authors have found elements of larger and smaller degree in the nucleus, as well as elements in the center.
In the Dniester Notebook published in 1993, Shestakov posed the folowing prob-lem (see , Problem 2.121): Problem 3.1. Describe the center and the associative center (i. e., the nucleus in our terminology) of a free alternative algebra as completely characteristic subalge-bras. Are they finitely generated?
A subalgebra S of an algebra A is completely characteristic (or T-subalgebra) if ψ(S) ⊂S for all homomorphisms ψ : A →A.
When the free alternative algebra has more then five free generators, Filippov found in 1999 an element of degree 7 in the center. And Filippov conjectured that the minimal degree of nonzero elements in the center of the free alternative algebra is 7. In 2003 (see ) we presented a new central element of degree 7. In 2006 (see ) we proved that Fillippov conjecture is true and found all the central elements of degree 7 in the free alternative algebra over the field Z103. An infinite set {an | n = 4k (k > 1) or n = 4k +1 (k > 0)} of central elements was constructed in 2006 by Shestakov and Zhukavets .
In this section we consider in more details the problem of finding nuclear elements of degree n in the free alternative algebra. We give the calculations for degree 5 and 6. Let p be an element of the free nonassociative algebra F[X] over the field F in generators X = {x1, x2, . . . , xn}. We say that p is an element of the nucleus of the free alternative algebra in generators X if in the free alternative algebra on generators X ∪{xn+1, xn+2} one has that (p, xn+1, xn+2) = 0.
3.1. Nuclear elements of degree n. A multilinear element p ∈F[X] is in the nucleus if (p, xn+1, xn+2) is in Altn+2 [X ∪{xn+1, xn+2}] that is the subspace of multilinear elements of degree n + 2 in Alt[X ∪{xn+1, xn+2}]. This subspace is generated by the liftings identities to degree n + 2 of the alternative laws.
REPRESENTATIONS AND IDENTITIES 13 We consider only permutations of the variables x1, x2, . . . , xn. We do this by using c(n) association types of degree n, T1, . . . , Tc(n), and k(n) = c(n + 2) n+2 2 association types of degree n + 2, T ′ 1, . . . , T ′ k(n). In each type T ′ j we consider only the permutation of x1, . . . , xn and the positions of xn+1 and xn+2. Also we assume that xn+1 and xn+2 are skew-symmetric since (p, xn+1, xn+2) = −(p, xn+2, xn+1).
Let wk be the monomial x1x2 . . . xn associated as in type Tk. We expand the associator (wk, xn+1, xn+2) as −(wkxn+1)xn+2 + wk(xn+1xn+2) + (wk, xn+1, xn+2) = 0 and call −(wkxn+1)xn+2 + wk(xn+1xn+2) the expansion of the associator (wk, xn+1, xn+2). This expansion of associator can be represented as in Table 3, where ιn is the identity permutation in Sn.
T ′ ik T ′ jk Tk −(wk xn+1) xn+2 wk (xn+1 xn+2) (wk, xn+1, xn+2) −ιn ιn ιn Table 3. Expansion of associator Let λ be a partition of n. Using φλ : F Sn →Mdλ(F) we obtain the representation matrix in Table 4, where φλ(ιn) = Idλ is the dλ × dλ identity matrix.
T ′ ik T ′ jk Tk h φλ(−ιn) | φλ(ιn) | φλ(ιn) i Table 4. Representation matrix of expansion of associator We apply φλ to the lifted alternative identities of degree n + 2 obtaining the rep-resentation matrix of these identities. We append to the bottom of this matrix the representation matrix given in Table 4 for all the associators.
We obtain the representation matrix Mλ given in Table 5.
We calculate the row canoni-cal form RCF(Mλ) of Mλ.
The nonzero rows of RCF(Mλ) that involve only the types (wk, xn+1, xn+2) (k = 1, . . . , c(n)) represent the identities of the form (p, xn+1, xn+2) = 0 with p ∈F[X]. Therefore p is an element of degree n in the the nucleus of the free alternative algebra ALT[X]. Some of these nuclear elements p are not new in the sense that they are consequence of the lifted alternative identities of degree n and the lifted nuclear elements of degree n of possible nuclear elements of degree < n. We calculate the matrix Nλ representing these lifted alternative identities and possible lifted nuclear elements. The nonzero rows that appear in the right-hand portion of RCF(Mλ) but do not appear in RCF(Nλ) represent the new nuclear elements of degree n.
14 LUIZ A. PERESI T ′ 1 T ′ 2 · · · T ′ k(n) T1 T2 · · · Tc(n) representation matrix of lifted alternative identities zero matrix of degree n + 2 representation matrix of expantion of associators Idλ 0 · · · 0 0 Idλ · · · 0 .
.
.
.
.
.
...
.
.
.
0 0 · · · Idλ Table 5. Representation matrix for nuclear elements of degree n 3.2. Nuclear elements of degree 5. Let A be any alternative algebra. Then by Proposition 2.4 A satisfies the identity (ab, c, d) + (a, b, [c, d]) = a(b, c, d) + (a, c, d)b.
Therefore, if (x, c, d) = 0 for all x ∈A, then [c, d] is in the nucleus of A.
Since (x, [a, b][a, b], a) = 0 is an identity in A (, §13.4, Lemma 15) it follows that (1) [ [a, b][a, b], a ] is in the nucleus of A. Filippov proved that (x, , a) = 0 is an identity of A. Therefore (2) , a] is in the nucleus of A. These nuclear elements have degree 5. In the next theorem we obtain that 5 is the minimal degree and obtain all the nuclear elements of degree 5 in the free alternative algebra.
Theorem 3.2. (Hentzel and Peresi 2006) In the free alternative algebra over Z103 on generators {a, b, c, d, e} we have: (i) There are no nonzero nuclear elements of degree < 5.
(ii) All the nuclear elements of degree 5 are consequences of the lifted alternative identities of degree 5 and ([a, b][a, c])a −(a[a, b])[a, c].
(3) As described in §3.1 we compare the right-hand portion of RCF(Mλ) with the RCF(Nλ). For degree < 5 they are the same. The ranks for degree 5 are given in Table 6. The nozero rows containing the new nuclear elements of degree 5 are given in Table 7. Modulo the alternative laws the nuclear element in partition 32 is (1) and the one in partition 311 is (2). Further work shows that these two nuclear elements are equivalent (modulo the alternative laws) to the nuclear element (3).
REPRESENTATIONS AND IDENTITIES 15 Partition λ dλ Alternative laws Nuclear elements 5 1 13 13 41 4 52 52 32 5 65 66 • 311 6 75 76 • 221 5 63 63 2111 4 46 46 11111 1 10 10 Table 6. Nuclear elements of degree 5: rank of matrices Type 14 Partition x(x(x(xx))) 32 1 0 1 1 1 311 1 0 −1 −3 −1 2 Table 7. New nuclear elements of degree 5 3.3. Nuclear elements of degree 6. The degree 6 nuclear elements implied by nuclear element (3) are the following lifted nuclear elements of degree 6: ([a, bd][a, c])a −(a[a, bd])[a, c], (4) ([a, b][a, cd])a −(a[a, b])[a, cd], (5) ([de, b][a, c])a + ([a, b][de, c])a + ([a, b][a, c])(de) −((de)[a, b])[a, c] −(a[de, b])[a, c] −(a[a, b])[de, c].
(6) For each partition λ, we compare the right-hand portion of RCF(Mλ) with the RCF(Nλ) (see §3.1). The matrix Nλ represents the lifted alternative identities of degree 6 and the lifted nuclear elements (4), (5) and (6). We also compare Nλ with the matrix representing just the lifted alternative identities of degree 6. The ranks are given in Table 8. The new nuclear elementes of degree 6 are the ones that are not consequences of the alternative laws, (4), (5) and (6). The new nuclear elements occur in partitions 33, 222, 2211 and 21111. The new nuclear element in partition 33 is [[a, b][a, b], b]a (7) and was obtained by a different method not the representation of Sn (see , §2 and 4.3). For the other partitions we give the nonzero rows and the corresponding new nuclear element (modulo the alternative laws, (4), (5) and (6)).
16 LUIZ A. PERESI Partition dλ Alternative Alternative laws Nuclear elements λ laws and nuclear elements (4)-(6) 6 1 41 41 41 51 5 205 205 205 42 9 369 372 372 411 10 406 409 409 33 5 205 206 207 • 321 16 652 658 658 3111 10 400 403 403 222 5 202 202 204 • 2211 9 360 362 363 • 21111 5 194 195 196 • 111111 1 36 36 36 Table 8. Nuclear elements of degree 6: rank of matrices Partition 222. Nuclear elemenets (8) and (9).
x(x(x(x(xx)))) 1 1 1 0 -1 2 −3 [a, c] ◦ −3 [b, c] ◦ (xx)(x((xx)x)) x(x((xx)(xx))) x(x(x((xx)x))) x(x(x(x(xx)))) 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 1 0 1 0 -1 -2 ⟨a, a, b⟩⟨c, c, b⟩+ ⟨a, a, c⟩⟨b, b, c⟩ − ⟨a, b, c⟩⟨a, b, c⟩+ ⟨b, b, a⟩⟨c, c, a⟩ + ⟨b, b, c⟩⟨a, a, c⟩−⟨b, c, a⟩⟨b, c, a⟩ (9) − ⟨c, a, b⟩⟨c, a, b⟩+ ⟨c, c, a⟩⟨b, b, a⟩ + ⟨c, c, b⟩⟨a, a, b⟩−72 (a, b, c)(a, b, c), where ⟨x, y, z⟩is the Jordan associator ⟨x, y, z⟩= (x ◦y) ◦z −x ◦(y ◦z).
Partition 2211. Nuclear elemenet (10).
x(x(x(x(xx)))) 0 2 3 1 -1 1 -5 1 4 [V (d2) −V (d) ◦d, a] (10) REPRESENTATIONS AND IDENTITIES 17 Partition 21111. Nuclear elemenet (11).
(xx)(x(x(xx))) x((xx)(x(xx))) x(x((xx)(xx))) x(x(x((xx)x))) x(x(x(x(xx)))) 0 0 0 0 1 0 0 0 0 -1 1 0 0 1 0 0 -2 0 0 -1 -1 1 1 0 0 ALTSUM{b,c,d,e} n 3 [[[a, b], a], c] [d, e] + [[[a, b], c], a] [d, e] −2 [[[a, b], c], d] [a, e] + 2 [[[b, c], a], a] [d, e] (11) −4 [[[b, c], a], d] [a, e] o where ALTSUM{b,c,d,e} denotes the alternating sum over all the permutations of the variables {b, c, d, e}.
Therefore we have: Theorem 3.3. (Hentzel and Peresi 2008) In the free alternative algebra over Z103 on generators {a, b, c, d, e, f} all the nuclear elements of degree 6 are con-sequences of the lifted alternative identities of degree 6, the lifted nuclear elements (4), (5),(6) and the nuclear elements (7) - (11).
4. Special identities for Bol algebras The Bol loops were introduced by Bol in 1937. In the 1980’s Sabinin and Mikheev constructed the theory of smooth Bol loops (see Sabinin ). In par-ticular, they introduced the notion of Bol algebra in 1982, and proved that any finite-dimensional Bol algebra over R can be realized as the tangent algebra of a local analytic Bol loop .
4.1. Bol algebras. A (left) Bol algebra is a vector space over F equipped with a binary operation [a, b] and a ternary operation {a, b, c} satisfying these identities: [a, b] + [b, a] = 0, (12) {a, b, c} + {b, a, c} = 0, (13) {a, b, c} + {b, c, a} + {c, a, b} = 0, (14) [{a, b, c}, d] −[{a, b, d}, c] + {c, d, [a, b]} (15) −{a, b, [c, d]} + = 0, {a, b, {c, d, e}} −{{a, b, c}, d, e} (16) −{c, {a, b, d}, e} −{c, d, {a, b, e}} = 0.
In 1994 Filippov gave the classification of homogeneous Bol algebras, i.e, the Bol algebras satisfying {a, b, c} = α[[a, b], c] + β[[b, c], a] + γ[[c, a], b] (α, β, γ ∈F).
Kuzmin and Za˘ ıdi investigated the solvability and semisimplicity of Bol algebras in 1993. An envelope for Bol algebras was constructed by P´ erez-Izquierdo in 2005, and it was proved that any Bol algebra is located inside the generalized left alternative nucleus of the envelope.
From the classification of all two-dimensional right Bol algebras over R given in we obtain two examples of (left) Bol algebras.
18 LUIZ A. PERESI Example 4.1. The Bol algebra B1 has basis {e1, e2}, [a, b] = 0 and the nonzero ternary products are given by {e1, e2, e1} = e2, {e1, e2, e2} = −e1, {e2, e1, e1} = −e2, {e2, e1, e2} = e1.
Example 4.2. The Bol algebra B2 has basis {e1, e2} and the nonzero products are given by [e1, e2] = e1, [e2, e1] = −e1, {e1, e2, e1} = −e1, {e1, e2, e2} = e2, {e2, e1, e1} = e1, {e2, e1, e2} = −e2.
Starting with a left altenartive algebra (i.e., an algebra satisfying (x, x, y) = 0) or a right altenartive algebra (i.e., an algebra satisfying (x, y, y) = 0) A, we construct a Bol algebra Ab.
The Bol algebra Ab. Let A be any algebra with multiplication ab. We denote by Ab the algebra having the same underlying vector space as A, the binary operation [a, b] := ab −ba and the ternary operation {a, b, c} := ⟨b, c, a⟩. Then Ab satisfies (12), (13) and (14). If A is a left or right alternative algebra then Ab satisfies also (15) and (16), and therefore Ab is a Bol algebra.
As an example we prove that if A is a right alternative algebra then Ab satisfies (16). The algebra A# = F.1 ⊕A is also a right alternative as proved by Albert (, Lemma 3). Therefore, we may assume that A has an identity element 1. Let Rb : A →A be the map given by Rb(a) = ab. We denote by [Ra, Rb] the operator RaRb −RbRa. We remind that [a, b] := ab −ba and {a, b, c} := ⟨b, c, a⟩. For any a, b, c ∈A we can prove that , Re −[Rc, Rd], [[Ra, Rb], Re] = 0.
The free Bol algebra Bol[X]. Let X = {x1, x2, x3, . . . } be a set of variables.
We construct the set of binary-ternary monomials BTM[X] inductively as follows: X ⊂BTM[X] and we assume that BTM[X] is closed under [a, b] and {a, b, c}. Let BT[X] = n Pn i=1 αi ui | n ∈N, αi ∈F, ui ∈BTM[X] o be the vector space over F spanned by BTM[X]. The elements of BT[X] are called binary-ternary polynomials in the variables xi. We define in BT[X] a multiplication by the following rules: if f = Pl i=1 αi ui, g = Pm j=1 βj vj and h = Pn k=1 γk wk are polynomials in BT[X] then [f, g] = Pl,m i,j=1 αiβj [ui, vj], {f, g, h} = Pl,m,n i,j,k=1 αiβjγk {ui, vj, wk}.
The algebra BT[X] is called the free binary-ternary algebra. The usual notion of ideal of identities (T-ideal) of F[X] can be extended to BT[X]. The free Bol algebra Bol[X] is BT[X] modulo the T-ideal of BT[X] generated by the (binary-ternary polynomial) identities (12) - (16).
The Bol types of degree n are ways to construct a product of degree n in BT[X].
Identities (12) and (13) reduce the number of these types. The Bol types of degree REPRESENTATIONS AND IDENTITIES 19 ≤3 are x, [x, x], , [{x, x, x}, x], [[[x, x], x], x].
Again x is just a place holder. The number of Bol types we use is denoted by b(n), and for n ≤8 this number is given by n 1 2 3 4 5 6 7 8 b(n) 1 1 2 5 13 38 113 354 We denote the Bol types of degree n by B1, . . . , Bb(n).
Let B be a Bol algebra. An element f = f(x1, . . . , xn) ∈BT[X] is an identity of B if f(a1, . . . , an) = 0 for all a1, . . . , an ∈B. Any multilinear identity f of degree n can be written as a linear combination of multilinear monomials, where each monomial has one of the Bol types of degree n. Collecting the monomials which have the same Bol type, f = f1 + · · · + fb(n), where fk is a linear combination of monomials having Bol type k. We can therefore (as in §2.1) identify each fk with an element of the group algebra FSn of the symmetric group Sn. Therefore, we can regard f as an element of the direct sum of b(n) copies of FSn. Applying φλ : F Sn →Mdλ(F) to f we obtain the representation matrix of f in partition λ: B1 B2 · · · Bb(n)−1 Bb(n) h φλ(f1) | φλ(f2) | · · · | φλ(fb(n)−1) | φλ(fb(n)) i 4.2. Special identities. An algebra J is called a Jordan algebra if it satisfies xy = yx and (x2, y, x) = 0. For any algebra A with multiplication ab, we denote by A+ the algebra having the same underlying vector space as A and the Jordan product a◦b = ab+ba as the multiplication. If A is associative then A+ is a Jordan algebra. A Jordan algebra J is special if there is an associative algebra A such that J is isomorphic to a subalgebra of A+. Otherwise, J is called exceptional. The Albert algebra, consisting of the matrices M ∈M3(C(α, β, γ)) such that M t = M, has dimension 27 and is an exemple of a exceptional Jordan algebra. In the 1960’s Glennie [21, 22, 23] discovered a special identity for Jordan algebras, i.e., an identity satisfied by special Jordan algebras that is not satisfied by all Jordan algebras (see also Hentzel ).
Vel´ asquez and Felipe introduced in 2008 a new class of algebras of Jordan type, the quasi-Jordan algebras, defined by two identities. Two years later, Bremner proved that it is more natural to consider the subclass consisting of the algebras that satisfy a third identity. We say that an algebra is a quasi-Jordan algebra if it satisfies the identities x(yz) = x(zy), (yx)x2 = (yx2)x, (y, x2, z) = 2(y, x, z)x.
If (D, +, ⊣, ⊢) is an associative dialgebra and we define the quasi-Jordan product a • b = a ⊣b + b ⊢a we obatin that D+ = (D, +, •) is a quasi-Jordan algebra. A quasi-Jordan algebra is special if it is isomorphic to a subalgebra of D+ for some associative dialgebra D. Bremner and Peresi found in 2011 special identities for quasi-Jordan algebras, i.e., identities satisfied by all special quasi-Jordan algebras but not satisfied by all quasi-Jordan algebras. One of these special identities for 20 LUIZ A. PERESI quasi-Jordan algebras is a noncommutative preimage of the Glennie identity. In both cases (Jordan and quasi-Jordan) the minimal degree of a special identity is 8.
We consider the analogous problem of finding special identities for Bol algebras.
We denote by LALT[X] (RALT[X]) the free left (right) alternative algebra. Since a homogeneous element in BT[X] is an identity of LALT[X]b if and only if it is an identity of RALT[X]b, we consider only right alternative algebras. A special identity is an identity satisfied by Ab for every right alternative algebra A, but not satisfied by the free Bol algebra.
Example 4.3. In the free binary-ternary algebra BT[a, b, c] let d = {a, b, a} + [[b, a], a].
Then H(a, b, c) := {{{{c, d, d}, d, d}, d, d}, d, d} is a special identity. The operations in RALT[a, b, c]b are [x, y] := xy −yx and {x, y, z} := ⟨y, z, x⟩. We have that d = {a, b, a} + [[b, a], a] = −2(a, a, b). In every right alternative algebra the Mikheev’s identity (x, x, y)4 = 0 is valid (see Zhevlakov , §16.1, Theorem 2). Therefore d4 = 0. We obtain H(a, b, c) = {{{{c, d, d}, d, d}, d, d}, d, d} = R{{{{c,d,d},d,d},d,d},d,d}(1) = [[[[[[[Rc, Rd], Rd], Rd], Rd], Rd], Rd], Rd], Rd = 0, since in each term of the expansion of the commutators of operators appears Rd4 = 0. Therefore H(a, b, c) = 0 is an identity of RALT[a, b, c]b. Now, we prove that H(a, b, c) is not zero in the free Bol algebra Bol[a, b, c]. It is enough to give an example of a Bol algebra where H(a, b, c) does not hold. We consider the algebra B1 (see Example 4.1). For a = e2, b = e1 and c = e2, we have d = {e2, e1, e2} + [[e1, e2], e2] = e1 and {c, d, d} = {e2, e1, e1} = −e2. Therefore H(e2, e1, e2) = e2 ̸= 0.
The special identity H(a, b, c, d) has degree 25 which is far from the minimal degree. In §4.3 we obtain that the minimal degree for special identities is 8.
Remark 4.4. Not every Bol algebra is isomorphic to a subalgebra of Ab for some right alternative algebra A. For instance, consider the Bol algebra B1 (see Example 4.1). Assume that B1 can be imbedded into Ab for some right alternative algebra A. Since Ab satisfies the identity H(a, b, c) = 0 then H(a, b, c) evaluates to zero in B1, a contradiction since H(e2, e1, e2) = e2 ̸= 0.
4.3. Degree n identities of RALT [X]b. Let RAltn[X] be the subspace of mul-tilinear elements of degree n in RALT[X] and {I1, . . . , Ig(n)} a generating set of RAltn[X]. We use a generating set with g(n) elements where g(n) (for n ≤8) is given by n 3 4 5 6 7 8 g(n) 1 4 16 61 234 895 Each Ii can be written as Ii = c(n) X j=1 f i j where all monomials in f i j have type Tj.
REPRESENTATIONS AND IDENTITIES 21 T1 T2 · · · Tc(n) B1 B2 · · · Bb(n)−1 Bb(n) φλ(f 1 1 ) φλ(f 1 2 ) · · · φλ(f 1 c(n)) O O · · · O O φλ(f 2 1 ) φλ(f 2 2 ) · · · φλ(f 2 c(n)) O O · · · O O .
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φλ(f g(n)−1 1 ) φλ(f g(n)−1 2 ) · · · φλ(f g(n)−1 c(n) ) O O · · · O O φλ(f g(n) 1 ) φλ(f g(n) 2 ) · · · φλ(f g(n) c(n) ) O O · · · O O φλ(g1 1) φλ(g1 2) · · · φλ(g1 c(n)) −Idλ O · · · O O φλ(g2 1) φλ(g2 2) · · · φλ(g2 c(n)) O −Idλ · · · O O .
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φλ(gb(n)−1 1 ) φλ(gb(n)−1 2 ) · · · φλ(gb(n)−1 c(n) ) O O · · · −Idλ O φλ(gb(n) 1 ) φλ(gb(n) 2 ) · · · φλ(gb(n) c(n)) O O · · · O −Idλ Table 9. Representation matrix Mλ: Bol types of degree n Let bi = bi(x1, . . . , xn) be the monomial x1 . . . xn written using Bol type Bi.
Using [a, b] := ab −ba and {a, b, c} := ⟨b, c, a⟩we obtain c(n) X j=1 gi j −bi = 0 where all monomials in gi j have type Tj.
For each partition λ, we apply φλ : FSn →Mdλ(F) to the expansion of bi to obtain the (direct sum of matrices) c(n) X j=1 φλ(gi j) −Idλ = 0, where Idλ is the dλ × dλ identity matrix. Also applying φλ to each Ii we obtain the (direct sum of matrices) c(n) X j=1 φλ(f i j).
Putting all these dλ × dλ blocks together we obtain the representation matrix Mλ in Table 9. We compute the row canonical form RCF(Mλ) of Mλ. The identities of degree n for RALT[X]b are represented in the lower right block of RCF(Mλ), under Bol types B1, . . . , Bb(n).
22 LUIZ A. PERESI B1 B2 · · · Bb(n) φλ(h1 1) φλ(h1 2) · · · φλ(h1 b(n)) φλ(h2 1) φλ(h2 2) · · · φλ(h2 b(n)) .
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φλ(hk(n)−1 1 ) φλ(hk(n)−1 2 ) · · · φλ(hk(n)−1 b(n) ) φλ(hk(n) 1 ) φλ(hk(n) 2 ) · · · φλ(f k(n) b(n) ) Table 10. Representation matrix Kλ: known identities of degree n Some of these identities of degree n satisfied by RALT[X]b are known, i.e., they are consequences of identities (12) - (16). Therefore we need to find the nonzero rows in RCF(Mλ) that come from the known identities. To do this we lift to degree n all the known identities of degree < n. This gives a generating set {K1, . . . , Kk(n)} of the subspace Boln[X] consisting of multilinear elements of degree n in Bol[X].
We use a generating set with k(n) elements where k(n) (for 4 ≤n ≤8) is given by n 4 5 6 7 8 k(n) 9 39 148 516 1885 Each Ki can be written as Ki = b(n) X j=1 hi j where all monomials in hi j have Bol type Bj. Applying φλ to each Ki we obtain the representation matrix Kλ given in Table 10. The known identities of degree n for RALT[X]b are represented by the nonzero rows of the row canonical form RCF(Kλ) of Kλ.
We denote by all the rank of the lower right block of RCF(Mλ) and by known the rank of RCF(Kλ). The nonzero rows that appear in the lower right block of RCF(Mλ) but not in RCF(Kλ) are the irreducible new identities for RALT[X]b.
The number of new identities is new = all - known.
Partition λ dλ known all new 4 1 5 5 31 3 12 12 22 2 7 8 1 211 3 9 10 1 1111 1 3 3 Table 11. Degree 4: ranks of matrix representations For n = 4 and n = 5 the ranks are given in Tables 11 and 12. There are 2 new identities in degree 4 and 3 in degree 5. For n = 4 the matrix RCF(Kλ) are REPRESENTATIONS AND IDENTITIES 23 Partition λ dλ known all new 5 1 13 13 41 4 47 47 32 5 55 56 1 311 6 64 65 1 221 5 51 52 1 2111 4 39 39 11111 1 9 9 Table 12. Degree 5: ranks of matrix representations obtained using (12) - (14). Since we know that (15) is an identity of RALT[X]b, we append the matrix φλ((15)) to the bottom of Kλ. Since the row canonical form of this new matrix coincides with the lower right block of RCF(Mλ), we have that the new identities of degree 4 of RALT[X]b are consequences of (15). For n = 5 we obtain RCF(Kλ) using (12) - (15), and by a similar procedure we verify that the new identities of degree 5 of RALT[X]b are consequences of (16). For n = 6, 7, known = all for all partitions and there are no new identities. Therefore we have: Theorem 4.5. (Hentzel and Peresi 2012) (i) A degree 4 element in BT[X] is an identity of RALT[X]b if and only if it is a consequence of identities (12)-(15).
(ii) A degree 5, 6 or 7 element in BT[X] is an identity of RALT[X]b if and only if it is a consequence of identities (12)-(16).
As a consequence of Theorem 4.5 we obtain that there are no special identity of degree < 8. For n = 8 the ranks are given in Table 13. Therefore we have: Theorem 4.6. (Hentzel and Peresi 2012) There are 13 irreducible special identities of degree 8. They are distributed as follows: Partition λ 62 53 521 44 431 3311 new 1 4 1 2 4 1 4.4. Special identities in partition 62. Given a partition λ = λ1λ2 . . . λk of n, we give a general method to construct special identities of degree n ≥8. We then apply this method to n = 8 and λ = 62. The method is based on the algorithm de-scribed by Hentzel and Jacobs in 1991. The computer implementation Albert of this algorithm is given by Jacobs .
The algebra Aλ[x1, . . . , xk]. The free right alternative algebra RALT[x1, . . . , xk] is the quotient algebra F[x1, . . . , xk]/I, where I is the T-ideal of F[x1, . . . , xk] gener-ated by the right alternative law (a, b, b) = 0. Let L be the subspace of F[x1, . . . , xk] spanned by the words w, where for some i in {1, . . . , k}, the degree of xi in w is > λi. Then L is an ideal of F[x1, . . . , xk]. The algebra Aλ[x1, . . . , xk] is the quotient algebra F[x1, . . . , xk]/(I + L). We denote by ψ : F[x1, . . . , xk] →Aλ[x1, . . . , xk] the homomorphism extending the map {x1, . . . , xk} →Aλ[x1, . . . , xk] that sends xi to xi+(I+L) for i = 1, . . . , k. We denote by r(n, λ) the dimension of Aλ[x1, . . . , xk]. A basis and multiplication table for Aλ[x1, . . . , xk] can be constructed using Albert.
24 LUIZ A. PERESI Partition λ dλ known all new 8 1 354 354 71 7 2457 2457 62 20 6959 6960 1 611 21 7288 7288 53 28 9665 9669 4 521 64 22000 22001 1 5111 35 11981 11981 44 14 4815 4817 2 431 70 23896 23900 4 422 56 19052 19052 4211 90 30503 30503 41111 35 11780 11780 332 42 14209 14209 3311 56 18884 18885 1 3221 70 23488 23488 32111 64 21344 21344 311111 21 6921 6921 2222 14 4653 4653 22211 28 9251 9251 221111 20 6549 6549 2111111 7 2250 2250 11111111 1 313 313 Table 13. Degree 8: ranks of matrix representations λ new b(n, λ) s(n) r(n, λ) t(n) null(Nλ) fewest 62 1 297 142 659 309 1 81 53 4 740 389 1510 766 5 187 521 1 2519 1333 5166 2677 7 409 44 2 984 530 1966 1020 7 186 431 4 4712 2605 9298 5003 17 359 3311 1 21996 12552 42730 23820 45 502 Table 14. Degree 8: data from creating the special identities The algebra Bλ[x1, . . . , xk]. The free Bol algebra Bol[x1, . . . , xk] is the quo-tient algebra BT[x1, . . . , xk]/I, where I is the T-ideal of BT[x1, . . . , xk] generated by (12)-(16).
Let L be the subspace of BT[x1, . . . , xk] spanned by the words w, where for some i in {1, . . . , k}, the degree of xi in w is > λi.
Then L is an ideal of BT[x1, . . . , xk].
The algebra Bλ[x1, . . . , xk] is the quotient algebra BT[x1, . . . , xk]/(I + L). We denote by b(n, λ) the dimension of Bλ[x1, . . . , xk]. A basis and multiplication table for Bλ[x1, . . . , xk] can be constructed using a method similar to the one used in Albert (see , p. 2327).
Using the operations [a, b] := ab −ba and {a, b, c} := ⟨b, c, a⟩, we expand the degree n basis elements of Bλ[x1, . . . , xk] in terms of the degree n basis elements REPRESENTATIONS AND IDENTITIES 25 of Aλ[x1, . . . , xk]. This gives us a matrix Nλ with s(λ) rows and t(λ) columns.
The (i, j) entry of Nλ is the j-th coordinate of the expansion of the i-th degree n basis element of Bλ[x1, . . . , xk]. Any dependence relation between these expanded basis elements becomes a special identity, since it is not an identity in the free Bol algebra because it is a linear combination of basis elements of Bλ[x1, . . . , xk]. It is an identity in any algebra Ab, where A is a right alternative algebra, because it expands to zero in the free right alternative algebra RALT[x1, . . . , xk]. We compute the left null space of Nλ, and denote its dimension by null(Nλ). Also, we use lattice basis reduction to attempt to find a special identity with fewest terms. The number of terms is not necessarily the smallest. For n = 8 the data is given in Table 14.
For n = 8 and λ = 62 we obtain a special identity with 81 terms, each term having 6 a’s and 2 b’s. Using the identity (14), we can express this identity using only 32 terms. This simpler form is called S(a, b) and it is given below.
We know that S(a, b) is not zero in the free Bol algebra Bol[a, b]. We can also verify this by giving an example of a Bol algebra where S(a, b) does not hold. We consider the Bol algebra B2 over a field of characteristic not 2 (§4.1, Example 4.2) and verify that S(e2, e1) = −2e2 ̸= 0.
We know that S(a, b) is an identity of RALT[a, b]b and it follows that S(a, b) is an identity in Ab for any right alternative algebra A. We can also verify that S(a, b) is an identity of RALT[a, b]b as follows. We construct the algebra A62[a, b] of dimension 659 and the homomorphism ψ : F[a, b] →A62[a, b]. Using the opera-tions [a, b] := ab −ba and {a, b, c} := ⟨b, c, a⟩, the identity S(a, b) becomes T(a, b).
Expanding the commutators and Jordan associators, and simplifying, we obtain that ψ(T(a, b)) = 0. Then, by Hentzel and Jacobs (Corollary of Theorem), we have that T(a, b) is an identity of RALT[a, b]. Therefore, S(a, b) is an identity RALT[a, b]b.
Finally, using φ62 : FS8 →M20(F), we verify that S(a, b) is a special identity, and that all special identities in partition 62 are consequences of S(a, b). We construct the matrix K62 representing the 1885 multilinear identities that are a generating set for the lifted identities of (12)-(16) to degree 8. The rank of the row canonical form RCF(K62) is 6959. Denote by LS(a, b) the linearized form of S(a, b). The binary-ternary polynomial LS(a, b) is a linear combination of 46080 multilinear binary-ternary monomials. We append to the bottom of the matrix K62 the 20×20 blocks that come from the representation matrix φ62(LS(a, b)). We calculate the row canonical form of this new matrix; the rank increases by 1 and is now 6960.
Therefore, S(a, b) is not an identity of the free Bol algebra. Also the row canonical form of this new matrix is the same matrix as the lower right block of RCF(M62).
Therefore S(a, b) is a special identity and all the special identities for partition 62 are consequences of S(a, b).
26 LUIZ A. PERESI Theorem 4.7. (Hentzel and Peresi 2012) The binary-ternary polynomial S(a, b) is a special identity. All special identities in partition six-two are conse-quences of the special identity S(a, b), where S(a, b) := 2{[{a, b, a}, a], a, +[{ −[{[[b, a], a], a, [[b, a], a]}, a] −2[{[{a, b, a}, a], a, [b, a]}, a] + [{[{a, b, a}, a], [b, a], a}, a] −2[{[[[b, a], a], a], a, [b, a]}, a] + [{[[[b, a], a], a], [b, a], a}, a] +[{{{a, b, a}, a, a}, a, b}, a] −3[{{{a, b, a}, a, b}, a, a}, a] +2[{{{a, b, b}, a, a}, a, a}, a] −[{{[[b, a], a], a, a}, a, a}, b] −8[{{[[b, a], a], a, a}, b, a}, a] + 5[{{[[b, a], a], a, a}, a, b}, a] −[{{[[b, a], a], b, a}, a, a}, a] + 5[{{[[b, a], a], a, b}, a, a}, a].
Acknowledgements The author thanks CIMAT (Centro de Investigaci´ on en Matem´ aticas, Guana-juato, Mexico) for financial support and hospitality during the Research School which was sponsored by CIMPA (Centre International de Math´ ematiques Pures et Appliqu´ ees, France). The author thanks the other members of the Scientific Committee: Murray R. Bremner (University of Saskatchewan, Canada), Ra´ ul Fe-lipe (CIMAT), and Juana S´ anchez-Ortega and Mercedes Siles Molina (University of M´ alaga, Spain).
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Department of Mathematics, University of S˜ ao Paulo, Brazil E-mail address: peresi@ime.usp.br |
5151 | https://www.math.purdue.edu/files/academic/courses/2013fall/MA22000/MA220lesson19notes.pdf | 1 MA 22000 Notes, Lesson 19 Textbook (calculus part) Section 2.4 Exponential Functions In an exponential function, the variable is in the exponent and the base is a positive constant (other than the number 1). Exponential Function: An exponential function with base a is defined as ( ) , where 0 and 1 x f x a a a . Note: If a (the base) in the above definition was 1, the function would be constant; a horizontal line, y = 1. Exponential functions often describe what is called exponential growth or exponential decay in real life examples. Example of an exponential function: Many real life situations model exponential functions. One example models the average amount spent(to the nearest dollar) by a person at a shopping mall after x hours and is the function, ( ) 42.2(1.56)x f x . The base of this function is 1.56. Notice there is also a ‘constant’ (42.2) multiplied by the power. Be sure to follow the order of operations; find the exponent power first, then multiply that answer by the 42.2. Suppose you wanted to find the amount spent in a mall after browsing for 3 hours. Let x = 3. 3 (3) 42.2(1.56) 42.2(3.796416) 160.2087552 f To the nearest dollar, a person on average would spend $160. Graph of an exponential function. Graphing Exponential Functions: To graph an exponential function, make a table of ordered pairs as you have for other types of graphs. Notice: If x = 0 for x b , the value is 1 (zero power is 1). For a basic exponential function, the y-intercept is 1. Also, notice that y values will always be positive, so the graph always lies above the x-axis. 2 Graph each exponential function. x y a 2 3 ) x x f b 3 1 ) ( ) What do you notice about the graphs above? Characteristics of Exponential Functions of the form x f(x) = a (basic exponential function) 1. The domain of the function is all real numbers ( , ) and the range is all positive real numbers (0, ) (graph always lies above the x-axis). 2. Such a graph will always pass through the point (0, 1) and the y-intercept is 1. There will be no x-intercept. 3. If the base b is greater than 1 ( 1) b , the graph increases left to right and is an increasing function. The greater the value of b, the steeper the increase (exponential growth). 4. If the base is between 0 and 1 (0 1) x , the graph decreases left to right and is a decreasing function (exponential decay). The smaller the value of b, the steeper the decrease. 5. The graph represents a 1-1 function and therefore will have an inverse. 6. The graph approaches but does not touch the x-axis. The x-axis is known as an asymptote. Solving exponential equations: There are a couple of ways to solve equations with the variable in an exponent. The first way is to rewrite both sides of the equation so the bases are the same. 1 2 1 2 4 4 8 x y -2 -1 0 1 2 x y -1 -2 0 1 2 3 Ex 1: Solve this equation: 3 2 25 125 x x Ex 2: Solve: 3 1 5 9 32 16 x x Compound Interest: One of the best examples in real life where an exponential function is used is in the banking business, compound interest. You know the simple interest formula I Prt . However, most banks periodically determine interest and add in the account. The amount in an account with an initial amount of P dollars invested at an annual interest rate of r (as a decimal), compounded m times per year for t years, returns a compound amount given by this formula. 1 mt r A P m Ex 2: Find the amount and the interest earned on $5800 at 4.3% interest compounded semiannually for 6 years. You will need to know how to use the ‘power’ key on your 1-line scientific calculator when you use this formula. 4 Definition of the number e: Letting P equal $1 and letting r be 100% and t equal 1 year. 1 1 becomes 1 mt m r A P m m Let the value of m become extremely large. Then 1 1 m m becomes closer and closer to a number we call e, whose approximate value is 2.718281828. (To find the value of e on a TI-30XA calculator use these steps: Enter the number 1, press the 2nd key then the LN key (notice that x e is above the LN key, so we are finding 1 e . You should get the approximation given above. As the amount of money in an account is compounded continuously, rather than periodically as with the compound interest formula earlier; the formula uses this number e. Continuous Compounding Interest: rt A Pe The steps to convert the regular compound interest formula to the formula above are shown on pages 82-83 in the textbook. It is a difficult process, so I will not demonstrate it in class. You can examine it on these pages, if you are interested. Ex 3: Suppose $800 is invested at 4 ½ % interest for 5 years. Find the accumulated amount in the account and the interest earned if… (a) the money is compounded quarterly. (b) the money is compounded continuously. 5 Ex 4: Suppose that a certain type of bacteria grows rapidly in a warm spot. If 500 bacteria of these bacteria are placed in a dish in a warm location, and the number present after x hours is given by the model 2 ( ) 500 3 x P x . Find the number of bacteria after (a) 1 hour, (b) 3 hours. |
5152 | https://dspace.mit.edu/bitstream/handle/1721.1/99978/Demaine_Non-crossing%20matches.pdf?sequence=1&isAllowed=y | MIT Open Access Articles Non-crossing matchings of points with geometric objects The MIT Faculty has made this article openly available. Please share how this access benefits you. Your story matters.
Citation: Aloupis, Greg, Jean Cardinal, Sebastien Collette, Erik D. Demaine, Martin L. Demaine, Muriel Dulieu, Ruy Fabila-Monroy, et al. “Non-Crossing Matchings of Points with Geometric Objects.” Computational Geometry 46, no. 1 (January 2013): 78–92.
As Published: Publisher: Elsevier Persistent URL: Version: Author's final manuscript: final author's manuscript post peer review, without publisher's formatting or copy editing Terms of use: Creative Commons Attribution-Noncommercial-NoDerivatives Non-crossing Matchings of Points with Geometric Objects1 Greg Aloupisa, Jean Cardinala, S´ ebastien Collettea,2, Erik D. Demaineb, Martin L. Demaineb, Muriel Dulieuc, Ruy Fabila-Monroyd, Vi Harte, Ferran Hurtadof, Stefan Langermana,3, Maria Saumellf, Carlos Searaf, Perouz Taslakiana aUniversit´ e Libre de Bruxelles, CP212, Bld. du Triomphe, 1050 Brussels, Belgium.
bMIT Computer Science and Artificial Intelligence Laboratory, 32 Vassar St., Cambridge, MA 02139, USA.
cPolytechnic Institute of NYU, USA.
dDepartamento de Matem´ aticas, CINVESTAV, M´ exico DF, M´ exico eStony Brook University, Stony Brook, NY 11794, USA.
fUniversitat Polit` ecnica de Catalunya, Jordi Girona 1–3, E-08034 Barcelona, Spain.
Abstract Given an ordered set of points and an ordered set of geometric objects in the plane, we are interested in finding a non-crossing matching between point-object pairs. In this paper, we address the algorithmic problem of deter-mining whether a non-crossing matching exists between a given point-object pair. We show that when the objects we match the points to are finite point sets, the problem is NP-complete in general, and polynomial when the ob-Email addresses: galoupis@ulb.ac.be (Greg Aloupis), jcardin@ulb.ac.be (Jean Cardinal), secollet@ulb.ac.be (S´ ebastien Collette), edemaine@mit.edu (Erik D.
Demaine), mdemaine@mit.edu (Martin L. Demaine), mdulieu@gmail.com (Muriel Dulieu), ruyfabila@math.cinvestav.edu.mx (Ruy Fabila-Monroy), vi@vihart.com (Vi Hart), ferran.hurtado@upc.edu (Ferran Hurtado), slanger@ulb.ac.be (Stefan Langerman), maria.saumell@upc.edu (Maria Saumell), carlos.seara@upc.edu (Carlos Seara), perouz.taslakian@ulb.ac.be (Perouz Taslakian) 1This work was partially supported by projects MTM2009-07242 and Gen. Cat. DGR 2009SGR1040; the Communaut´ e fran¸ caise de Belgique - ARC; and the ESF EUROCORES programme EuroGIGA, CRP ComPoSe: Fonds National de la Recherche Scientifique (F.R.S.-FNRS) - EUROGIGA NR 13604, for Belgium, and MICINN Project EUI-EURC-2011-4306, for Spain.
2Charg´ e de recherches du F.R.S.-FNRS.
3Maˆ ıtre de recherches du F.R.S.-FNRS.
Preprint submitted to Elsevier October 14, 2011 jects are on a line or when their size is at most 2. When the objects are line segments, we show that the problem is NP-complete in general, and polyno-mial when the segments form a convex polygon or are all on a line. Finally, for objects that are straight lines, we show that the problem of finding a min-max non-crossing matching is NP-complete.
1. Introduction Finding a matching between pairs of plane objects that connects these objects by a set of non-crossing line segments is a natural problem that has been frequently studied in computational geometry. It is well known, for instance, that given two sets of n points in the plane, say n red points and n blue points, there always exists a non-crossing perfect matching between red and blue points. In particular, it is not difficult to show that the minimum Euclidean length matching, if it exists, is non-crossing. Kaneko and Kano survey a number of related results. Algorithms for finding minimum sum and minimum bottleneck distance red-blue matchings are given in [18, 32].
In this paper, we investigate related questions for general plane objects instead of points. Again, matchings are represented by line segments, but here the endpoints can be placed anywhere inside the corresponding matched objects. Note that as a consequence of the aforementioned result on points, there always exists a non-crossing matching between two sets of objects. Here we consider the problem where we are given object pairs (i.e. a point and the geometric object it must be matched to) and need to find a set of non-crossing matching edges, if one exists. This can be seen as a 1-regular graph drawing problem with constraints on the location of vertices.
Related work. Problems on matchings have an important role in combina-torial graph theory, both for theoretical and applied aspects; hence a lot of research is devoted to the study of these problems (for example, see ).
Suppose we are given an embedding of a graph in the Euclidean plane, where the vertices are points in the plane, edges are rectilinear line segments, and weights on these edges represent the Euclidean distance between the vertices they connect. Elementary geometry tells us that the sum of any pair of op-posite sides of a convex quadrilateral is strictly smaller than the sum of the diagonals. Remarkably, this implies that the minimum weight matching in any straight line embedding of the complete graphs K2n and Kn,n consists of pairwise non-crossing segments. These geometric graph problems can be 2 p1 p2 p3 t2 t1 t3 Figure 1: An example of non-crossing matching for a set P={p1, p2, p3} of points and a set T={t1, t2, t3} of plane objects.
solved using generic algorithms for weighted graphs. However, in the planar case just mentioned, Vaidya proved that it is possible to obtain special-ized algorithms with better running times (the title of his paper is especially suggestive: Geometry helps in matching).
In particular, in the run-ning time of the generic algorithm for the bipartite case was reduced from O(n3) to O(n2.5 log n). This was later improved to O(n2+ε) by Agarwal et al. . Similar results have been obtained for other matching variations, such as bottleneck matching or uniform matching, in the work of Efrat, Itai and Katz . The authors consider matchings as an approach for the problem of matching a point set A with a point set B, where A must be moved in some way to coincide as much as possible with B or one of its subsets. This is a fundamental problem in pattern recognition [7, 10, 11, 13, 14, 15, 23, 24, 25].
Another matching variation is C-matching as described by ´ Abrego et al. .
Here the authors consider the problem of matching a given set of points with a set of geometric objects such that every geometric object contains exactly two points. The objects they consider are circles and isothetic squares, and show the existence and properties of such matchings. Bereg et al. consider C-matchings for axis-aligned squares and rectangles.
The non-crossing requirement in our problems is quite natural in geo-metric scenarios (see for example [3, 4, 31]), and the family of geometric problems that we consider has several applications; these applications in-clude geometric shape matching [6, 16, 21, 22] (see also the references we give for geometric pattern recognition), colour-based image retrieval , and computational biology [17, 20].
3 Our results. Throughout the paper, we let P := {p1, p2, . . . , pn} be a set of points in the plane and T := {t1, t2, . . . , tn} be a set of plane objects. A matching for a pair (P, T) consists of a set of line segments, called edges, of the form {p1m1, p2m2, . . . , pnmn}, where mi ∈ti. A matching is said to be non-crossing if no pair of matching edges properly cross. This is illustrated in Figure 1.
We consider the problem of deciding whether a non-crossing matching exists for a given pair (P, T). In cases where a non-crossing matching always exists, we consider the problem of finding the matching that minimizes either the length of the longest edge, or the sum of the lengths of all the edges.
In Section 2, we study the case where the objects ti are finite point sets.
We prove that the decision problem is NP-complete in general, but becomes polynomial when every ti has size at most two, or when all the ti are on a line. In Section 3 we consider T to be a set of line segments and prove that the (P, T) matching problem is NP-complete. We also consider special cases, such as the case when the line segments form a convex polygon surrounding all points in P (Section 4), or the case when segments belong to a single line (Section 5). We show that these special cases have polynomial solutions.
Finally, in Section 6, we consider the problem of matching points with lines.
In this variation, a non-crossing matching always exists; but we show that the optimization problems are NP-hard.
2. Matching points with finite point sets We first prove that if the objects ti are pairs of points, then we can decide whether there exists a non-crossing matching in polynomial time. On the other hand, if the sets ti may contain three points or more, the problem becomes NP-complete. This situation is similar to that of the k-satisfiability problem (k-SAT). In k-SAT we are given a boolean formula f of the form C1 ∧C2 ∧· · ·∧Cm (where each Ci is an OR clause of k variables), and we are required to find a truth assignment of its variables that satisfy the formula.
It is well-known that 2-SAT has a polynomial-time solution whereas k-SAT is NP-complete for k ≥3. The 2-SAT problem can be solved in polynomial time by exploiting the fact that, if in a clause a variable is set to false, it forces the other variable to be set to true.
4 Theorem 1. Given an ordered set4 P of points and an ordered set T of pairs of points, there is an algorithm that decides in O(n2) time whether (P, T) has a non-crossing matching.
Proof. We will prove the theorem by showing that the given matching prob-lem reduces to 2-SAT, which is known to have an O(n2) running time. As-sume that the elements of each ti are labeled arbitrarily “Ti” and “Fi” (thus ti = {Ti, Fi}). We think of each pi as a boolean variable, so that if we match pi with Ti then pi is set to “true”, and if pi is matched with Fi, it is set to “false”. Let Xi equal to Ti or Fi, and Yj equal to Tj or Fj. In O(n2) time, we construct a 2-SAT instance having variables x0, x1, . . . , xn−1 as follows: Consider the segments pi, Xi for all i = 0, 1, . . . , n −1.
For each pair of intersecting segments pi, Xi and pj, Yj, we construct the 2-SAT clause • (xi, xj) if Xi = Fi and Yj = Fj, • (xi, ¬xj) if Xi = Fi and Yj = Tj, • (¬xi, xj) if Xi = Ti and Yj = Fj, or • (¬xi, ¬xj) if Xi = Ti and Yj = Tj.
With this construction is it easy to see that if there is a solution for (P, T) where the two vertices pi and pj have a valid non-crossing perfect matching pi, Xi and pj, Yj, then the corresponding 2-SAT clause has a valid truth assignment if we set xi to Xi and xj to Xj. Conversely, if there exists a truth assignment that sets a 2-SAT clause (xi, xj) to “true” then there exists a matching for pi and pj. Therefore, the matching instance (P, T) has a non-crossing perfect matching if and only if the corresponding 2-SAT instance has a valid truth assignment. Since the 2-SAT instance is constructed in O(n2) time and solving 2-SAT is known to be possible in O(n2) time, the overall complexity of the matching algorithm is O(n2).
2.1. Matching points with triples Theorem 2. Given an ordered set P of points and an ordered set T of triples of points, it is NP-complete to decide whether (P, T) has a non-crossing matching. The problem remains NP-complete even if each triple of points is horizontally collinear.
4Here, and throughout the rest of the paper, by an ordered set we mean a totally ordered set.
5 Proof. First we argue that the problem is in NP. Our input is a set of point-triple pairs. A matching can be specified combinatorially by listing which point in each triple ti gets matched with the corresponding point pi. In time polynomial in the length of the input, we can check whether such a matching is non-crossing. Hence the problem is in NP.
It remains to show that the problem is NP-hard. We reduce from the planar 3-SAT problem, which is a version of 3-SAT whose implication graph (the bipartite graph having the variables on one side, the clauses on the other, and an edge between a variable x and a clause C if and only if x appears in C) is planar. Planar 3-SAT is known to be NP-hard . Given an instance SAT of the planar 3-SAT, we will construct an instance (P, T) of the problem of matching a set of points P to a set of triples T such that SAT has a valid truth assignment if and only if (P, T) has a non-crossing matching between point-triple pairs. Every boolean variable in SAT is represented in (P, T) by points (or a triple in which two points are identical). Thus every point vi can be matched in exactly two ways (see Figure 2). To each variable, we associate a wire gadget that is composed of a set of pairs (vi, qi) (see Figure 3). These pairs are chosen so that once the edge for one of the points is selected, all the others are determined (given that we require a non-crossing matching).
Hence in a non-crossing matching, the wire can only be in one of two distinct states, corresponding to the value of the variable. Such a wire can be split using the gadget shown in Figure 4.
True False vi vi Figure 2: Variable gadget.
Finally, we associate a pair (pj, tj), pj ∈P, tj ∈T to the j-th clause of the given 3-SAT formula, where tj is a triple of points. The three possible edges connecting pj to tj correspond to the choice of the literal that will satisfy the clause. The three line segments between pj and the three points of tj interfere with the wires corresponding to the three variables used in the clause. Using the layout of Figure 5, a matching edge for the clause crosses an edge of the wire if and only if the value of the literal encoded in the edge is not compatible with the value of the variable encoded in the wire. In other words, pj connects to point a of tj (representing some variable vi of the j-th 6 (a) Wire gadget (b) True signal (c) False signal vi vi vi Figure 3: Wire gadget.
Figure 4: Splitter gadget.
7 clause of the 3-SAT formula) if and only if matching the variable gadget vi as either true or false sets the literal representing a to true.
a b c a ∨b ∨c a b c a ∨b ∨c Figure 5: Clause gadget.
Using standard layout techniques for planar graphs (see ), we can represent the variable-clause incidence graph of the given 3-SAT formula using the wire and clause constructions above. This layout guarantees that there exists a satisfying assignment for the 3-SAT instance if and only if there exists a non-crossing matching for (P, T). If the 3-SAT instance has a valid truth assignment, then every clause has at least one literal set to “true”. In the constructed matching instance, this is equivalent to connecting every pi to at least one of the points in ti by a segment that does not cross any other.
On the other hand, assume that (P, T) has a valid non-crossing matching.
Then every variable gadget has a non-crossing matching that connects a point vi in either of two ways, “true” or “false”; moreover, this matching ensures that in every clause gadget, a vertex pj has a non-crossing matching to at least one of the three points of tj. If we now assign the values of the variable gadgets in (P, T) to the variables of the 3-SAT instance SAT, then every clause in SAT will have at least one literal set to “true”. To conclude the proof, we note that the number of points created in (P, T) for every variable and clause in SAT is a polynomial function of the input to the problem; hence, our reduction is polynomial in the size of the input to the 3-SAT instance.
Finally, observe that our wire and clause layout may be constructed such 8 that the points in a triple are collinear.
Thus the problem remains NP-complete even in this restricted version.
2.2. Matching points with k-tuples Theorem 3. Given an ordered set P of points and an ordered set T of k-tuples (where each ti is a set {ti1, ti2, . . . , tik} of k points), if every edge [pitij] crosses at most c < 0.183k other edges of the form [pi′ti′j′], then there exists a non-crossing matching between P and T.
Proof. We apply the probabilistic method , and match every point pi with tij, where j is chosen randomly in {1, . . . , k}. We need to show that there is a positive probability that the resulting matching is non-crossing. Let M denote the random matching.
We define a bad event as two edges of M of the form [pitij] and [pi′ti′j′] that cross. A bad event has probability either equal to 0 (if the edges are not crossing) or to exactly q := 1/k2. Two bad events are dependent whenever the two pairs of points of P involved intersect. Hence every bad event depends on at most d := 2ck other bad events (since there are k possible edges for each of two points, and every such edge intersects at most c others). By Lov´ asz’ Local Lemma , if eq(d + 1) ≤1 (where e is Euler’s number), then there is a nonzero probability that no bad event occurs. This means that a non-crossing matching exists. This yields e 1 k2(2ck + 1) ≤ 1 (1) c ≤ k 2e −1 2k ≃0.183k (2) Note that our proof does not use geometry, so it is likely that the constant 0.183 can be improved. The proof can also be made constructive using a recent result from Moser .
2.3. Matching points with k-tuples on a line Theorem 4. Given an ordered set P of points and an ordered set T of k-tuples of points on a line, we can decide in O(k3n2 + k2n3) time whether (P, T) has a non-crossing matching.
9 Proof. Without loss of generality, assume all the tuples are on a horizontal line L. Assume also that all points are on one side of L; otherwise we may consider each problem separately as the matching edges on each side of L do not interact. We now show how to build a dynamic programming table that solves the problem.
A p L Figure 6: Definition of a sub-problem.
In any solution to the problem, if a matching edge e is part of the solu-tion, then there is no matching edge that intersects e. Therefore, we can con-sider the regions on each side of e (sub-problems) separately and determine whether they in turn have a valid solution. To achieve this, we will consider the points of P top-to-bottom – the points with largest y-coordinate first, – and based on possible matching edges, split the problem into independent sub-problems. A sub-problem (P ′, T ′) is defined as follows (see Figure 6): given a trapezoid A with one edge adjacent to L and an edge parallel to L, we want to decide if it is possible to find a non-crossing matching completely contained in the region A for all the points contained in A, i.e., we want to solve the problem with P ′ = P ∩A and T ′ containing the subsets of the tuples of T contained in A. If A does not contain at least one point of P (sub-problem of size 0), it is trivially true that there is a non-crossing match-ing. Otherwise, to solve the sub-problem we consider the topmost point p in A. It has at most k possible matching edges. If it has no possible matching edge, i.e., if all points that p could be matched to in T are out of A, then there is no valid matching.
Each of the possible matching edges for p defines two new independent sub-problems (see Figure 7) in the trapezoids A1 and A2, whose sizes are strictly smaller than that of the original problem, as there is one less point to match. Each of the trapezoids A1 and A2 is defined by a possible matching edge of p, an edge bounding A, the line L, and a line through p parallel to L. Note that as p is the topmost point of A, then the region A \ (A1 ∪A2) contains no points of P; this implies that the union of the regions of the 10 A1 p A2 A1 p A2 A1 p A2 Figure 7: The three pairs of sub-problems to consider to decide if p can be matched.
sub-problems of A will contain all the points in A, and hence no point of P will be ignored in the process.
To decide whether a matching exists for the original sets P and T, we solve the sub-problem defined by the bounding box of both P and T. Notice that all the sub-problems correspond to trapezoids defined by a pair of possible matching edges or by the edges of the bounding box.
The dynamic programming table has kn + 2 rows and kn + 2 columns, each of which corresponds to a possible matching edge or one of the left and right edges of the bounding box; the cells correspond to sub-problems (a pair of non-adjacent edges defines a trapezoid), and we fill them with true or false values depending on whether or not a matching exists for the considered sub-problem. Filling a cell of the table corresponds to first finding the topmost point within the sub-problem in linear time, and then solving at most k pairs of sub-problems, which implies at most 2k lookups in the table for each of the O(k2n2) cells. Therefore, the total time and space required to solve the problem is O(k2n2(k + n)) = O(k3n2 + k2n3).
Corollary 1. Given an ordered set P of points and an ordered set T of triples of points on a line, we can decide in O(n3) time whether (P, T) has a non-crossing matching.
This corollary shows that the additional restriction of having points on a line greatly simplifies the problem, because the problem is NP-hard in the general case, but is polynomial for points on a line.
3. Matching points with line segments: general case In this section we show that deciding the existence of a non-crossing matching between a set of points and a set of line segments is NP-complete, even if the segments are all horizontal.
11 Theorem 5. Given an ordered set P of points and an ordered set T of line segments, it is NP-complete to decide whether (P, T) has a non-crossing matching. The problem remains NP-complete even if all line segments in T are horizontal.
Proof. First we argue that the problem is in NP. It suffices to show that only a polynomial number of points along segments of T need to be considered for a non-crossing matching. We construct the arrangement of lines between all pairs of points among the union of P and the endpoints of all segments in T. This arrangement divides each segment of T into subsegments, with the property that all points in the relative interior of a subsegment are equivalent matching solutions. Thus we can choose the midpoint and the endpoints of each subsegment as the canonical points representing possible choices for a non-crossing matching of (P, T). Any matching can be rounded to use only points in P and canonical points of subsegments in T, without adding any additional crossings. Therefore a matching can be represented as a combina-torial object on a polynomial number of points. Given such a representation, we can test in polynomial time whether the matching is non-crossing. Hence the problem is in NP.
It remains to show that the problem is NP-hard. We reduce from the non-crossing matching problem for an ordered set P ′ of points and an ordered set T ′ of horizontally collinear triples of points, which is NP-hard by Theorem 2.
For each point p ∈P ′ and corresponding triple t ∈T ′, we place three points p1, p2, p3 in P and three corresponding triples of segments t1, t2, t3 in T; refer to Figure 8.
p2 p3 t2 t3 a b c p1 t1 Figure 8: Matching points with segments is at least as hard as matching points with triples.
Suppose t = (a, b, c) with a, b, and c appearing left to right along a 12 horizontal line. Let p1 = p and t1 be the segment from a to c. Next, we choose a small subsegment of t1 containing a, and similarly we choose small subsegments containing b and c (“small” means that the subsegments do not cross any lines of the arrangement described above). Connecting the endpoints of these subsegments to p gives us three narrow triangles.
We place p2 on the right edge of the triangle containing a; we place two short horizontal segments t2 and t3 both having their left endpoint on the left edge of the triangle containing b and right endpoint on the right edge of the triangle containing b; and we place p3 on the left edge of the triangle containing c.
Any matching edges connecting p2 to t2 and p3 to t3 block the ranges between the narrow triangles. This forces the matching edge connecting p1 to t1 to lie in one of the three narrow triangles, effectively matching p with either a, b, or c. Thus (P, T) has a non-crossing matching if and only if (P ′, T ′) has a non-crossing matching. Note that the proof holds when the segments ti are horizontal.
4. Matching points with an enclosing convex polygon In this special case of matching points with line segments, we assume the segments are the edges of a convex polygon and the points to be matched are inside the polygon.
We first describe some geometric properties of the input of this problem.
We then describe an algorithm that finds a non-crossing matching (if one ex-ists) between a given set of point-segment pairs where the line segments form a convex polygon enclosing the points. Our algorithm runs in O(n log2 n) time and allows a minimum-length and minimum max-edge-length matching to be extracted easily.
4.1. Structural properties Let Do = {∆o 1, ∆o 2, . . . , ∆o n} be a set of triangles where each ∆o i is the triangle with apex pi and base ti. Any valid matching edge ei must lie inside ∆o i. Depending on the positions of other triangles in Do, some candidate positions for ei can be identified as invalid because they would always cross other matching edges. By identifying such cases, triangle ∆o i can be reduced to a smaller triangle ∆i. At any time, the reduced triangle ∆i has apex pi but its opposite base is a subsegment of ti. Initially, ∆i = ∆o i.
13 There are four ways in which two triangles ∆i and ∆j interact.
The second case leads to a reduction rule. We describe the four cases below (see Figure 9): 1. ∆i, ∆j are disjoint. In this case there will never be a direct interaction between the two.
2. pj is in ∆i, but pi is not in ∆j. In this case ∆i should be reduced so that the two triangles become tangent (so that pj is no longer in ∆i).
3. pi is in ∆j and pj is in ∆i. We call ∆i and ∆j inverted triangles, and cannot immediately make a reduction.
4. Both edges incident to each of pi and pj pairwise intersect. Then no non-crossing matching exists.
Note that in case (2) there is no choice but to reduce. The matching edge ej that is finally chosen will block any candidate ei that is outside the newly reduced ∆i. In case (3) there are two combinatorially valid placements for ei, ej, with respect to the positions of pi, pj. There is no reason to choose arbitrarily before verifying that neither triangle will be reduced further.
∆i ∆o i pi pi pi pi Figure 9: Left: ∆o i is reduced to ∆i (case 2). Middle: inverted triangles – no immediate reduction is possible (case 3). Right: no solution exists (case 4).
4.2. Properties of a reduced set of triangles Here we describe certain properties that must hold after we exhaustively apply our reduction rule to a set of triangles.
Let two (three) pairwise inverted triangles be called an inverted pair (triple).
An inverted triple is shown in Figure 10.
Consider an inverted triple ∆0, ∆1, ∆2. The clockwise radial ordering of the triangles ∆0, ∆1, ∆2 with respect to a point p is the circular ordering by angle around p of the 14 bases of these triangles that are visible to p. Note that since the bases of ∆0, ∆1, ∆2 do not cross (input segments are non-intersecting), then every triangle base appears exactly once in this ordering.
p1 p3 p2 Figure 10: An inverted triple (∆1, ∆2, ∆3).
Lemma 1. Let (∆1, ∆2, ∆3) be an inverted triple, and let pi be the apex of ∆i for i = 1, 2, 3. Then the clockwise order of p1, p2 and p3 along their convex hull is identical to the clockwise radial order of ∆1, ∆2, ∆3 from any of the points p1, p2, or p3.
Proof. Let c be the barycenter of p1, p2 and p3, and consider the oriented line ℓthrough c rotating clockwise. Note that c is in the intersection of ∆1, ∆2 and ∆3. When ℓis incident to pi with pic in the positive orientation of ℓ, the positive halfline of ℓfrom pi intersects the base of ∆i (because c ∈∆i). Thus, this halfline visits the points pi in the same clockwise order as the bases of the triangles ∆i.
Lemma 2. Let (∆1, ∆2) and (∆1, ∆3) be two inverted pairs. If a solution exists, then applying the reduction rules to ∆1, ∆2, ∆3 will result in either (∆1, ∆2, ∆3) forming an inverted triple or becoming disjoint.
Proof. If ∆3 is also inverted with ∆2, then we have an inverted triple. Oth-erwise, note that ∆2 and ∆3 cannot be disjoint, since they both contain the apex of ∆1. Assuming case (4) does not apply to ∆2 and ∆3 (in which case no solution would exist), we are left with case (2). Assume without loss of generality that ∆2 contains p3 and ∆3 does not contain p2 (see Figure 11).
Then ∆2 is reduced by ∆3, which implies that it is no longer inverted with ∆1.
Thus ∆1 gets reduced, and then so does ∆3. All triangles end up disjoint.
Lemma 3. Let (∆1, ∆2, ∆3) be an inverted triple. If ∆1 also has the in-verted property with some triangle ∆4, then if we apply the reduction rule to 15 p1 p3 p2 ∆1 ∆2 ∆3 Figure 11: Interaction of a new triangle with an inverted pair. The grey triangles are the three disjoint triangles after applying reductions.
∆1, ∆2, ∆3, ∆4, either all four triangles become disjoint or no non-crossing matching exists.
Proof. Assuming a non-crossing matching exists, then for all i, j ∈{1, 2, 3} and by Lemma 2, either (∆i, ∆j, ∆4) becomes an inverted triple, or ∆i, ∆j, and ∆4 become disjoint after applying the reduction rule. The latter case implies, again by the same lemma, that all four triangles would be disjoint.
Otherwise, in the former case, every triple from ∆1, ∆2, ∆3, ∆4 is inverted, and so by Lemma 1, every triple in p1, p2, p3, p4 has the same clockwise ori-entation as the corresponding bases. This implies that p1, p2, p3, p4 form a convex quadrilateral Q. The angle of ∆i at pi is larger than the interior angle of Q at pi since ∆i contains the 3 other points. Therefore, the sum of the angles of ∆i at pi is at least 2π. Let c be the barycenter of p1, p2, p3, p4. The angle from c to the base of ∆i is strictly larger than the angle of ∆i at pi.
The sum of the angles from c to the bases of ∆i is at most 2π because these wedges from c do not overlap. Therefore, the sum of the angles of ∆i at pi is strictly less than 2π, a contradiction.
Let a unit be a (possibly reduced) triangle, an inverted pair, or an inverted triple. Any time a triangle intersects a unit, the unit will be unaffected, or reduced according to Lemmas 3 and 2, or be “upgraded” to an inverted pair or triple (if it was a triangle or inverted pair, respectively). Two units are said to be disjoint if their interiors do not overlap. This establishes that units are the only possible structures that can remain after applying all possible deterministic reductions, if the decision problem has a positive answer. There can be an arbitrary number of any types of units in the final configuration.
16 Lemma 4. When a triangle ∆p is added to a set of n disjoint units not already containing ∆p, after possibly applying reductions we obtain a new set of disjoint units.
This new set is either disjoint to ∆p, or ∆p joins exactly one member of the set and becomes disjoint to all others. Furthermore, adding ∆p to the existing disjoint units and applying (possible) reductions takes O(n) time in the worst case.
Proof. Any triangle in a unit will be unaffected or reduced by interacting with ∆p, so the new set will end up disjoint. Since units are disjoint, p can be inside at most one unit u. This means that for all triangles not belonging to u, the interaction of ∆p will lead to case (4), or cause no change, or cause a reduction of ∆p. Furthermore, ∆p will become disjoint to all such triangles.
Now consider the interaction between ∆p and u. If u is a triangle, we either get a reduction of u or we obtain a new inverted pair.
If u is an inverted pair, by Lemma 2 we either obtain case (4), or an inverted triple, or three disjoint triangles, or ∆p reduces u (case (2)) without destroying the inverted pair of u. If u is an inverted triple, by Lemma 3 we either obtain case (4), or we get four disjoint triangles, or ∆p reduces u (case (2)) without destroying the inverted triple of u.
In all cases, ∆p either becomes part of an inverted unit or is left disjoint to all triangles. Since we only compare ∆p to every triangle in the set, this procedure takes linear time.
4.3. Algorithm Theorem 6. Given an ordered set P of points inside a convex polygon having an ordered set T of line segments as edges, deciding whether (P, T) admits a non-crossing matching can be done in O(n log2 n) time.
Proof. We provide an algorithm where we employ a divide-and-conquer tech-nique. Suppose that we have solved the problem separately on two consec-utive convex chains (we can transform a chain into a polygon by adding 3 fake edges and points; thus, solving the problem on a chain is equivalent to solving the polygonal version).
We claim that we can merge the two solutions in O(n log n) time. Each solution is a set of disjoint triangles and inverted pairs or triples. Refer to Figure 12.
17 Figure 12: Merging two solved sub-problems. In the left diagram, the grey regions in the left (black) sub-problem cannot contain points from the right (blue) sub-problem if there is a valid solution. In the right diagram, we see the type of event that we must check for after some initial reductions.
Let A and B be two solved sub-problems of size k.
We construct a standard point-location data structure5 on each in O(k) time . Now, for every point pi in B, we locate pi in A to determine if it is inside a unit in A.
Note that pi can be in at most one unit. If it is, we determine if ∆i reduces this unit by case (2). Likewise, for every point pj in A, we locate pj in B to determine if it is inside a unit in B and apply the appropriate reductions.
Note that if at some moment ∆i (belonging to B) gets reduced, this will not affect its corresponding unit in A; the same holds for all ∆j in A that get reduced.
Of course, it is possible that ∆i will be inverted with a triangle in A.
In this case we simply determine if there are reductions and, if applicable, we merge the two units.
Therefore a constant number of reductions are applied per point, which means we spend O(log k) time per point for the point-location step.
The only unresolved issue is to detect if case (4) will occur between tri-angles of A and B (see the right diagram in Figure 12). For this we can use the Bentley-Ottmann line segment intersection algorithm and stop as soon as a bad intersection is found . Given that all triangles have been reduced 5To construct their point-location data structure, Kirkpatrick et al. triangulate each subdivision in O(k log k) time, and hence their algorithm requires O(k log k) time in the worst case. Using Chazelle’s linear-time triangulation algorithm , we can reduce this running time to O(n).
18 and merged into units, essentially we are verifying that no segments inter-sect. For k segments, such queries take O(k log k + h log k) time, where h is the number of intersections reported. As we stop as soon as we report an intersection, h = 1 and hence the total time is O(log k) per point. Therefore, our merge procedure takes O(k log k) time. By a simple recurrence analysis, we determine that the entire algorithm takes O(n log2 n) time.
The algorithm described in the proof of Theorem 6 either decides that no solution exists, or otherwise produces a final set of reduced triangles that represents all valid solutions to the problem. In the latter case, every resulting unit is disjoint and thus independent of all others. So in each triangle we can easily pick the shortest joining segment, and in each inverted pair/triple, we try out the two possible choices and take the best matching. Therefore, after the algorithm finds a solution, the min-max and min-sum optimization problems can be solved in linear time.
5. Matching points with segments on a line As another special case of matching points to line segments, we now consider the case when the input line segments belong to one single line L.
Throughout this section we will assume, without loss of generality, that L is horizontal. As no matching edge will cross over L, our problem is split into two disjoint sub-problems, and we focus on points above L.
We consider two cases, depending on whether the segments are disjoint or not.
5.1. Matching points with disjoint segments on a line Theorem 7. Given an ordered set P of points above a horizontal line L and an ordered set T of disjoint line segments belonging to L sorted in or-der of smallest x-coordinate, deciding whether (P, T) admits a non-crossing matching can be done in linear time. In the affirmative, the matching that minimizes either the sum of the lengths of the edges or the maximum edge length can be found within the same time bound.
Proof. We denote by [ai, bi] the interval corresponding to segment ti, for i = 1, ..., n. Since the intervals are given in sorted order, we have a1 ≤b1 < a2 ≤b2 < · · · < an ≤bn.
19 If (P, T) admits some non-crossing matching {p1m1, p2m2, . . . , pnmn}, where ai ≤mi ≤bi for all i = 1, 2, . . . , n, we can always slide the point mi in-side ti to a position mL i as far to the left as possible (see Figure 13). This gives the unique leftmost non-crossing matching for (P, T), {p1mL 1 , p2mL 2 , . . . , pnmL n}.
Notice that either mL i = ai, or pi and mL i are collinear with some pj with j < i.
a a p b b b b ba a a a a p p p p p b b b b ba a a p p p p Figure 13: Leftmost non-crossing matching (right) obtained from an initial non-crossing matching (left).
Next we describe an algorithm for finding the leftmost non-crossing match-ing, if it exists. The algorithm considers points in a sequential greedy fashion, in the left-to-right order of the corresponding segments.
For p1, the leftmost matching is simply given by the segment p1a1. We then consider the rays from the endpoints of this segment in the direction of the negative semiaxis of abscissae; their points at infinity can be symbolically described as q0 = (−∞, 0) and q1 = (−∞, y(p1)).
The forbidden region is the (unbounded) region enclosed by an alternating sequence of horizontal line segments and subsegments of matched edges (see Figure 14). This region is updated at every step of the algorithm. Initially, it is described clockwise by its vertices, namely q1p1a1q0. Observe that if p2 is inside the forbidden region, then a non-crossing matching (P, T) would be impossible. If p2 is outside the forbidden region, a matching is possible if and only if there is some point m2 in the interval a2b2 such that the segment p2m2 does not cross the forbidden region. In the affirmative, we slide m2 to its leftmost possible position, and shoot a ray from p2 in the direction of the negative semiaxis of abscissae, which may go to infinity, or stop by hitting the segment p1a1. The forbidden region is updated in each case, and is always defined by alternating horizontal edges with portions of segments from the matching. See Figure 14.
Assume that, in a generic step, we have obtained the leftmost matching {p1mL 1 , p2mL 2 , . . . , pj−1mL j−1} and we are processing pj. Let qi1pi1qi2pi2...qikpik mL ikq0 be the current forbidden region (refer to Figure 14). Observe that if 20 forbidden region p i p i p k i pj b j m j m k i q i q q i q k i L Figure 14: Forbidden region and incremental step.
there is some mj ∈[aj, bj] such that the segment pjmj can be added to the edges found so far, getting a non-crossing matching, the segment pjbj is also valid. We show next how to check the validity of pjbj.
We first check the y coordinates of the points mik, pik, pik−1..., which form an increasing sequence, until we find that y(pit) ≥y(pj) ≥y(pit+1) (the case in which y(pj) is a maximum is completely analogous). Then, we check whether the segment pjbj crosses the segments mikpik, qik−1pik−1, ..., qit−1pit−1.
In the affirmative, the algorithm is over, as no crossing-free matching is possible. Otherwise, the segment pjbj is valid. We slide the point matched with pj as much to the left as possible (Figure 15), which can be done by finding the angularly closest point among pit+1, pit+2, ..., pik, aj.
pj pj bj bj mj aj aj L Figure 15: Moving the new edge to the leftmost position.
If we shoot a ray from pj in the direction of the negative semiaxis of abscissae, we hit the boundary of the forbidden region in a point qj, possibly at infinity, and the forbidden region is updated to be qi1pi1qi2pi2...pitqjpjmL j q0.
The cost of the step for pj is proportional to the size of the forbidden polygonal region that disappears, and that will never be processed again.
Therefore, the amortized cost of one step is constant and the global cost 21 of the algorithm is O(n).
At the end we obtain the leftmost matching {p1mL 1 , p2mL 2 , . . . , pnmL n}, unless no matching is possible.
If (P, T) admits a non-crossing matching, with a symmetric algorithm we can obtain the rightmost matching {p1mR 1 , p2mR 2 , . . . , pnmR n}. Then any points mi in the intervals [mL i , mR i ] provide a non-crossing matching {p1m1, p2m2, . . . , pnmn}. In particular, in each interval [mL i , mR i ] we can pick the matching point mi which is closest to pi, and hence obtain the matching that minimizes the sum of the lengths of the edges or the maximum edge length in additional O(n) time.
Observation. If the input disjoint segments t1, ..., tn are not given in sorted order along the line, then, we can always sort them in O(n log n) time as a preprocessing step. An Ω(n log n) lower bound holds for this problem of matching points with disjoint unsorted segments on a line, by reduction from the problem of integer uniqueness, which is known to have an Ω(n log n) lower bound in the algebraic decision tree model of computation.
Let x1, ..., xn be a set of given integers. We associate to them 2n points and 2n segments defining pi = (xi, 2i), p′ i = (xi, 2i+1), ti = [xi−2/5, xi−1/5], t′ i = [xi −4/5, xi −3/5], for i = 1, ..., n. Points pi and p′ i are to be matched with the segments ti and t′ i, respectively, for i = 1, ..., n.
If a number xi is unique, then the matching is possible (Figure 16, left).
However, if two values are equal, xi = xj, then a crossing is unavoidable (Figure 16, right). Therefore, a non-crossing matching exists if and only if the numbers x1, ..., xn are all different, which proves the claim.
p′ i = (xi, 2i + 1) pi = (xi, 2i) xi ti t′ i xi−1 p′ i = (xi, 2i + 1) pi = (xi, 2i) xi ti t′ i xi−1 p′ j = (xj, 2j + 1) pj = (xj, 2j) Figure 16: There is a crossing when some integer is repeated.
5.2. Matching points with arbitrary segments on a line In this section, we show that when the given segments are confined to a line and possibly intersect, we can determine the existence of a non-crossing matching in polynomial time. The proof first discretizes the problem, and 22 then uses the same approach as in the proof of Theorem 4 for k-tuples with k = O(n2).
Theorem 8. Given an ordered set P of points above a horizontal line L and an ordered set T of line segments belonging to L, deciding whether (P, T) admits a non-crossing matching can be done in O(n8) time.
Proof. We solve the problem by discretizing it: we transform it into matching the set of points P with O(n2)-tuples, corresponding to all combinatorially distinct matchings for each point.
Consider all lines through every pair of points in P. These lines intersect the horizontal line L. Let I be the set of all these intersection points and of all the endpoints of segments in T. I has size O(n2), as there are 2n endpoints and at most n 2 intersections.
I splits L into O(n2) regions. If any subset S of the points in P are matched with an edge incident to one region r of L, we can pick an arbitrary point x inside r and match all the points of S with an edge incident to x, still preserving the existence of a non-crossing matching. In other words, for each point, there are only O(n2) combinatorially different matching edges, and we can thus apply our algorithm of Theorem 4 for matching with k-tuples.
Therefore, the complexity of finding a matching is O((n2)3n2 + (n2)2n3) = O(n8).
6. Matching points with lines In the case where points are matched with lines, it is easy to see that a non-crossing matching always exists: choose an arbitrary direction, not parallel to any line, and project each point on its corresponding line in that direction. Here we show that the optimization problem of minimizing the maximum length over all matching edges in NP-complete. We consider the decision version of the min-max problem.
Theorem 9. Given an ordered set P of points, an ordered set T of lines, and a number y, deciding whether there exists a non-crossing matching of (P, T) whose longest edge has length at most y is NP-complete.
Proof. We argue that the problem is in NP. We will first show that only a polynomial number of points along lines of T need to be considered for a non-crossing matching.
We then show how, given a solution based on 23 these canonical points, we can determine in polynomial time whether the length of each matched edge is at most y. We construct the arrangement of lines between all pairs of points among the union of P and the points of intersection of the lines in T. We then place a bounding box enclosing the union of P and the points of intersection of the lines in T. Together with the bounding box, this arrangement divides each line of T into subsegments, with the property that all points in the relative interior of a subsegment are equivalent non-crossing matching solutions. We can now choose the midpoint and the endpoints of each subsegment, as the canonical points representing possible choices for a non-crossing matching of (P, T). Any matching can be rounded to use only points in P and canonical points of subsegments in T, without adding any additional crossings. Therefore a matching can be represented as a combinatorial object on a polynomial number of points.
Given a solution with such a representation, we can test in polynomial time whether the matching is non-crossing. Now, we still need to check whether the length of every matched edge is at most y. Let mi be the point on ti such that the distance between pi and the subsegment to which pi is matched is the shortest. For every pi, we can check in polynomial time whether the distance between pi and mi is at most y. Note that matching pi to mi will not introduce any crossings: suppose two points pi and pj are matched to the same canonical point in the given solution, and suppose matching pi to mi and pj to mj will cause the segments pimi and pjmj to intersect. Then this would imply that segment pimj is shorter than pimi, contradicting the fact that pimi is the shortest segment. Therefore, in polynomial time it is possible to check whether a given solution is non-crossing with the distance of every matching edge equal to at most y. Hence the problem is in NP.
We reduce from the problem of deciding the existence of a non-crossing matching between a set of points and a set of segments. In Section 3 we proved that this problem is NP-complete. Given an instance (Q, S) of the point-to-segment matching problem, we construct an instance (P, T) of our min-max problem as follows. For each pair (qi, si) in (Q, S), we include the point qi in P and the line ti supporting the segment si in T. We then include a number of pairs (x, ℓ), x ∈P, ℓ∈T, such that the edge matching qi with ti is forced to have its endpoint within the boundaries of si in order not to create a long edge between a pair (x, ℓ). Thus any non-crossing matching of (P, T) with maximum edge length y, when restricted to the pairs (qi, ti), will also be a non-crossing matching of (Q, S). The gadget is illustrated in Figure 17.
24 si a qi x ℓ ε a′ δ si a qi Figure 17: Illustration of the reduction for the point-to-lines problem.
Let a be one endpoint of the segment si. We include a point x on the segment qia, at an arbitrarily small distance δ from qi. The corresponding line ℓis parallel to qia, at a distance ε from x. Note that ℓis positioned so that it does not intersect si. Then, for ε sufficiently small, an edge connecting qi to ti on the right of a will force the edge between x and ℓto be long. More precisely, let a′ be the point on ti such that the angle ∠aqia′ equals θ for some small positive value θ. Then if qi is matched with ti at any point p ∈aa′, the length of the edge matching x with ℓcan be made arbitrarily close to ε/ sin θ.
If we fix the value of θ, we can reproduce the gadget at regular angular intervals around qi, covering the whole range of possible edge angles with a constant number of pairs (x, ℓ) (see Figure 17). The same construction is used for the other endpoint of si.
Let y = maxi{d(qi, si) : qi ∈Q, si ∈S}. We choose ε and θ such that ε/ sin θ > y > ε. If there exists a non-crossing matching for (Q, S), then qi can be matched to ti within the boundaries of si, and every x can be matched to the corresponding ℓusing an edge orthogonal to ℓ, of length ε. Hence every edge has length at most y. On the other hand, if no non-crossing matching exists for (Q, S), then a point qi ∈P needs to be matched with ti outside of si, and one (x, ℓ) gadget is triggered, creating an edge of length ε/ sin θ > y.
Note that we simultaneously require that ε be a constant, and ε/ sin θ > y, hence that θ < arcsin(ε/y). So the value maxi{d(qi, si) : qi ∈Q, si ∈S} must be bounded by a constant. Also, the gadget pairs (x, ℓ) should not interfere with other edges of the matching. Since δ can be made arbitrarily small, we require the existence of a ball of radius strictly greater than ε, around every point of Q, that is never intersected by any edge in a non-crossing matching.
25 These two conditions (that the largest distance to a segment is bounded, and that there exists an empty ball of constant radius around each point) are satisfied by the hard instances constructed in the reduction of Theorem 5.
This concludes the proof.
We observe that if the lines have only a bounded number k of distinct directions, then there is a simple approximation algorithm for the min-max matching problem. Consider the set of directions (that is, angles with respect to the horizontal axis) of the lines, and find the largest absolute difference between two consecutive angles. Let α and β be the two consecutive angles maximizing the difference γ = |α −β|. We have that γ ≥π/k. If we project the points pi on their respective lines at an angle (α + β)/2, then the length of the matching edge between pi and ti is at most 1/ sin(γ/2) times the distance between pi and ti. Thus this directly yields an approximation factor of 1/ sin(γ/2).
7. Concluding remarks Non-crossing matchings of points with geometric objects is part of a more general class of problems where non-crossing matchings between geometric sets are considered. In this latter class, the first interesting set of problems is that of matching points with sets of points/segments/lines, as the NP-hardness results for these problems apply to problems of finding non-crossing matchings between sets and sets of many different classes.
One example would be all classes of objects that include all segments, such as convex sets.
However, it is still unclear whether our results imply anything about the general problem of finding non-crossing matchings between sets of geomet-ric objects. It is not clear for example that our hardness result for finding a non-crossing matching between points and segments (Theorem 5) has any implication on the problem of finding non-crossing matchings between points and either orthogonal polygons with a fixed number of edges or fat convex objects. Could the algorithm for finding a non-crossing matching for seg-ments in convex position (Theorem 6) be extended to also work for sets of segments that have the same radial ordering about every point pi? Also, for matching points with lines, is the problem still NP-complete when the lines have a bounded number of directions? And is the condition of having lines with a bounded number of directions necessary for having an approxi-mation algorithm? These and many similar questions raise interesting open 26 problems in the study of non-crossing matchings between sets of geometric objects.
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30 |
5153 | https://www.wiley.com/en-us/Quantum+Mechanics,+Volume+2:+Angular+Momentum,+Spin,+and+Approximation+Methods,+2nd+Edition-p-9783527345540 | Quantum Mechanics, Volume 2: Angular Momentum, Spin, and Approximation Methods, 2nd Edition
Claude Cohen-Tannoudji, Bernard Diu, Franck Laloë
ISBN: 978-3-527-34554-0
December 2019
688 pages
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## Quantum Mechanics, Volume 2: Angular Momentum, Spin, and Approximation Methods, 2nd Edition
Claude Cohen-Tannoudji,Bernard Diu,Franck Laloë
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June 2020
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This new edition of the unrivalled textbook introduces concepts such as the quantum theory of scattering by a potential, special and general cases of adding angular momenta, time-independent and time-dependent perturbation theory, and systems of identical particles. The entire book has been revised to take into account new developments in quantum mechanics curricula.The textbook retains its typical style also in the new edition: it explains the fundamental concepts in chapters which are elaborated in accompanying complements that provide more detailed discussions, examples and applications. The quantum mechanics classic in a new edition: written by 1997 Nobel laureate Claude Cohen-Tannoudji and his colleagues Bernard Diu and Franck Laloë As easily comprehensible as possible: all steps of the physical background and its mathematical representation are spelled out explicitly Comprehensive: in addition to the fundamentals themselves, the book contains more than 170 worked examples plus exercisesClaude Cohen-Tannoudji was a researcher at the Kastler-Brossel laboratory of the Ecole Normale Supérieure in Paris where he also studied and received his PhD in 1962. In 1973 he became Professor of atomic and molecular physics at the Collège des France. His main research interests were optical pumping, quantum optics and atom-photon interactions. In 1997, Claude Cohen-Tannoudji, together with Steven Chu and William D. Phillips, was awarded the Nobel Prize in Physics for his research on laser cooling and trapping of neutral atoms.Bernard Diu was Professor at the Denis Diderot University (Paris VII). He was engaged in research at the Laboratory of Theoretical Physics and High Energy where his focus was on strong interactions physics and statistical mechanics.Franck Laloë was a researcher at the Kastler-Brossel laboratory of the Ecole Normale Supérieure in Paris. His first assignment was with the University of Paris VI before he was appointed to the CNRS, the French National Research Center. His research was focused on optical pumping, statistical mechanics of quantum gases, musical acoustics and the foundations of quantum mechanics.
About the Author
Claude Cohen-Tannoudji is a researcher at the Kastler-Brossel laboratory of the Ecole Normale Supérieure in Paris where he also studied and received his PhD in 1962. In 1973 he became Professor of atomic and molecular physics at the Collège des France. His main research interests were optical pumping, quantum optics and atom-photon interactions. In 1997, Claude Cohen-Tannoudji, together with Steven Chu and William D. Phillips, was awarded the Nobel Prize in Physics for his research on laser cooling and trapping of neutral atoms.
Bernard Diu was Professor at the Denis Diderot University (Paris VII). He was engaged in research at the Laboratory of Theoretical Physics and High Energy where his focus was on strong interactions physics and statistical mechanics.
Franck Laloë is a researcher at the Kastler-Brossel laboratory of the Ecole Normale Supérieure in Paris. His first assignment was with the University of Paris VI before he was appointed to the CNRS, the French National Research Center. His research was focused on optical pumping, statistical mechanics of quantum gases, musical acoustics and the foundations of quantum mechanics.
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Recurso Economico
Los recursos económicos son medios materiales o inmateriales que permiten satisfacer necesidades en la producción o comercio de una empresa. Son necesarios para operaciones económicas, comer…
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RECURSO ECONOMICO
Los recursos económicos son los medios materiales o inmateriales que permiten satisfacer ciertas necesidades dentro del proceso productivo o la actividad comercial de una empresa. Recursos económicos Estos recursos, por lo tanto, son nece sarios para el desarrollo de las operaciones económicas, comerciales o industriales. Acceder a un rec urso económico implica una inversión de dinero: lo importante para que la empresa sea re ntable es que dicha inversión pueda ser recuperada con la utilización o la explotación del recurso. Por ejemplo: un campo es un recurso económico que permite el desarrollo de la agricultura. Dic ho recurso puede volverse inviable desde el punto de vista e conómico si se encuentra en medio de la montaña o en alguna zona geográfica que re quiera de demasiado dinero para su explotación. En estos momentos se habla mucho de este tipo de recursos, en concreto de su escasez. Y es que debido a la crisis económica es frecuente exponer que muchas familias en el mundo se encuentran ahora sin recursos económicos para poder sobrevivir.
RECURSO LIBRE
Un recurso es una fuente o suministro del cual se produce un beneficio.1 Normalmente, los recursos son materiales u otros activos que son t ransformados para producir un beneficio y en el proceso pueden ser consumidos o no estar más disponibles. Desde una perspectiva humana, un recurso natural es cualquier elemento obtenido del medio ambiente para satisfacer las necesidades y los deseos humanos.2 Desde un punto de vista e cológico o biológico más amplio, un recurso satisface las necesidades de un organismo vivo.3 El concepto de recurso se ha aplicado en diversos ámbitos, en particular, con respecto a la economía, biología, ciencias de la computación, manejo del recurso tierra y recursos humanos. Además, está relacionado con los conceptos de competencia, sostenibilidad y conservación ambiental. Los recursos tienen tres caracte rísticas principales: utilidad, 2) disponibilid ad limitada y 3) potencial de agotamiento o consumo. Los recursos han sido c ategorizados como bióticos y abióticos, renovables y no renovables, potenciales y reales, además de clasificaciones más complejas.
ECONOMIA
Se llama actividad económica a cualquier actividad laboral donde se generan e intercambian productos, bienes o servicios para cubrir las necesidades de las poblaciones. La actividad económica permite la generación de riqueza dentro de una comunidad (ciudad, región y país) mediante la extracción, transformación y distribución de los recursos naturales o bien de algún tipo de servicio y recurso.1 En un se ntido amplio, la economía se refiere a la or ganización del uso de recursos escasos (limitado o infinito) cuando se implementan para satisfacer las necesidades individuales o colectivas, por lo que es un sistema de int eracciones que garantiza ese tipo de organización, también conocido como el sistema económico.23
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Las actividades económicas abarcan tres fases: producción, distribución y consumo.4 Como la producción depende del consumo, la economía también analiza el comportamiento de los consumidores con respecto a los productos.
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5156 | https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_7C_-_General_Physics/11%3A_Electromagnetism/11.9%3A_Magnetic_Induction | Skip to main content
11.9: Magnetic Induction
Last updated
: Feb 15, 2023
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11.8: Magnetic Force
Back Matter
Page ID
: 80102
Dina Zhabinskaya
UC Davis
( \newcommand{\kernel}{\mathrm{null}\,})
What Is Induction?
We have seen how Ørsted was able to demonstrate that electric currents can produce magnetic fields. The English physicist Michael Faraday, a brilliant experimentalist, was the first to demonstrate the converse effect just a few years later in 1831: magnetic fields can be used to induce electric currents. This is now called the principle of magnetic induction. It is interesting to note that Faraday had little formal schooling, so mathematics was by no means his strength. Nevertheless, he was one of the most influential scientists not just of his time, but his contributions continue to find applications to this day.
For example, when he demonstrated that magnetic fields could be used to produce a current in a wire loop, politicians at the time were not impressed as they failed to see the use of it. It turns out that this was the critical step in creating generators and power plants, which make electricity available without flying a kite in a storm or carrying large arrays of batteries. The alternating-current (AC) circuits that power all the electrical grids of the world have as part of their components a generator that is based on magnetic induction.
More recently, highly fuel efficient vehicles such as the gas-electric hybrid cars employ a technology called regenerative braking. This uses a device that can give power to the wheels of the car by means of an electric battery, and can recharge the battery during braking by running the circuit in “reverse”, transforming the kinetic energy of the car’s motion into electric potential energy stored in the battery and saving fuel as a result. This recent application has large implications for the world’s economy and is of global environmental impact, and at its core lies Faraday’s principle of induction: that we can transform magnetic fields to electric currents.
Magnetic Flux
Before we tackle the principle of magnetic induction, we first need to define a quantity which is crucial to understand it quantitatively: the concept of magnetic flux.
Let us discuss first the idea of flux in general using a familiar example: rain falling on the windshield of a car. Suppose that we want to quantitatively determine the amount of rain that hits the windshield of the car. For simplicity, first assume that the rain is falling vertically down, and that the shape of the windshield is a rectangle. Let us further simplify by assuming you are in a parked car, i.e. it is not moving. If we want to find how much rain hits the windshield, we mainly need to consider these three variables:
The amount of rain
The size of the windshield
The orientation of the windshield relative to the rain
Let’s discuss each in turn. If it is raining hard, there will be a lot more raindrops hitting the windshield than if it is raining lightly. Likewise, If the size of the windshield is large, more raindrops will hit it than if it were small. The orientation between the rain and the windshield will also determine how much rain hits the windshield: if the windshield were arranged vertically, there would be no rain hitting the windshield (in the idealization that the windshield is infinitely thin). Conversely, the most amount of rain will hit if it's arranged perpendicular to the rain, or horizontally (like the sunroof on top of the car). In general, the flux of some quantity through a surface area is a measure of how much of that quantity passes through the area.
This idea of calculating the amount of rain hitting a surface is analogous the concept of magnetic flux, Φ. We can consider the flux of the magnetic field passing through an arbitrarily-chosen surface. In our example, the rain falling vertically is our vector field, and the windshield is our arbitrary area. More rain means that the magnitude, |B⃗ |, of the vector field increases. A larger windshield means a greater surface area. The orientation is measured by the angle between the direction of the vector field and the normal vector, n^, to the surface area. The normal vector is a unit vector which is perpendicular to the surface through which flux is being calculated. For the particular case of the magnetic field vector B⃗ , we define the magnetic flux Φ through an area A as:
Φ=B⃗ ⋅An^=|B⃗ ||A|cosθ(11.9.1)
where the angle θ is the angle between the magnetic field vector, B⃗ and the vector normal, n^ to the surface area A, as depicted in the figure below, for some arbitrary alignment of the magnetic field with a surface represented by a plane.
Figure 11.9.1: Magnetic Flux
From our rain example, you can see that when the rain is falling vertically down, and the windshield surface is horizontal, the normal vector to the area will be vertical. Hence, the rain and the normal vector are parallel to each other, and the angle in Equation 11.9.1 will be θ=0∘. Since cosθ=1, this orientation maximizes flux. If the windshield surface is vertical, the vector normal to the surface area will now be horizontal, the angle will be θ=90°, resulting in cosθ=0 and the flux will vanish (no rain hits the vertical windshield). Note that there is freedom to choose the normal vector on either side of the surface. This will have no physical effect, but will simply change the value of the angle by 180°. In other words, the flux as we defined it will change sign, it can be either positive or negative. We will see shortly that changes in flux are what is important physically rather than the actual value of the flux. So any of the choices we make for convention will lead to identical changes in flux, resolving any ambiguity.
To summarize, the variables of interest when calculating the magnetic flux through an area will be:
The magnitude of the magnetic field, |B⃗ |.
The size of the surface area, A.
The angle between the magnetic field vector, B⃗ , and the vector normal, n⃗ , to the area.
Faraday's Law of Induction
Now that we have defined the magnetic flux, Φ, we can describe Faraday’s observations quantitatively. He sought out to describe a connection between the current in a wire in the presence of the field. For a current to flow, we must have a closed wire loop (a circuit). But unlike in a circuit the loop is not connected to any power source, such as a battery. The area we will consider for magnetic flux will be the area enclosed by our wire loop. Note that the wire can be looped in a circular, square, or arbitrarily complicated shape.
Considering the magnetic flux through a wire loop, Faraday asked what happened if you placed a magnet close to the loop and let it sit there. Would a current start to flow in the presence of the magnet? He carried out the experiment and found that there was no current in the loop in this case. However, if you move the magnet away, then for a brief instant a current appears. If you move it back toward the loop, then a current appears again.
What Faraday found is there is an induced current (and therefore induced voltage) only when the magnetic flux changes over time. We say that the current is induced because it's not created by a battery, or some connected voltage source like in a "standard" circuit. The current is induced in the wire by the changing magnetic field. He called the induced voltage the induced electromotive force, or induced EMF for short, denoted by E. We therefore refer to his findings as Faraday’s Law of Magnetic Induction. Specifically what he found was that:
The induced voltage E is proportional to the rate of change of the flux with time, ΔΦΔt.
If you add loops to the wire coil, each loop will contribute equally to E: if you have N coils, the induced voltage will be N times as strong.
We can now summarize these findings that embody Faraday's Law quantitatively:
E=−NΔΦΔt(11.9.2)
The induced current is the given by:
I=ER(11.9.3)
where R is the resistance in the wire loop. Equation 11.9.2 tells us that we need to have a changing magnetic flux to produce an induced voltage. If the magnetic flux does not change with time, then there will be no current. Based on Equation 11.9.1 there are three way that the flux can be changed: changing the strength of the field, |B⃗ |, changing the area, A, (this can be accomplished by part of a loop that is free to move), or by changing θ, a rotation of the field relative to the loop or the loop relative to the field.
Furthermore, the fasterthe flux changes, the larger the induced voltage. You can picture this last statement in the following way. If you are inducing current by moving a magnet close to a wire, the current will be larger if you move the magnet quickly than if you move it slowly. The magnitude of the rate of change is proportional to the voltage, the faster the magnetic field changes, the greater the induced current and induced voltage. The negative sign that appears in front of the equation will be explained shortly. Note that Faraday’s law focuses only on the effect of a changing magnetic field on a wire. For simplicity, we discussed using a permanent magnet as the source of our field. However, we could also use the magnetic field produced by current in anotherwire. In fact, this is how Faraday studied induced current and induced voltages.
Lenz's Law: The Direction of Induced Current
Equation 11.9.2 for the EMF, E, derived from Faraday’s Law, also has a sign. What is its significance? The sign gives the direction of the induced current in the loop. So far we have not discussed how we are to choose between the two possibilities for the current's direction. Experimentally, if we change a magnetic flux to induce a current, the current will flow to produce a new flux that opposes our change. This is known as Lenz’s Law. The best way to understand Lenz's Law is by looking at a few examples.
Consider a circular loop of wire in the figure below. Initially, there is no magnetic field in the region of the wire. At t=0 we create a changing magnetic field by moving a bar magnet toward the loop as shown. We make the magnitude of the field increase linearly with time. From Faraday’s Law, we know that there will be an induced voltage, because the flux through the loop is increasing as the field strength increases.
Figure 11.9.2: Changing Flux Example
The area enclosed by the circuit is constant. The angle between the normal to the area of the circle loop and the field is constant, θ=0 and not changing. The magnitude of the magnetic field, |B⃗ |, field increases with time. Therefore, the changing magnetic field is the only contribution to the change in flux:
ΔΦΔt=−AΔBΔt(11.9.4)
We can turn the derivative in Equation 11.9.1 into just changes, "Δs", since we are assuming that the field is changing linearly with time. Since the magnitude increases linearly with time, the rate of change the magnetic field magnitude versus time is a constant: therefore the induced voltage will be constant.
The question we want tot answer is: will the current flow clockwise or counterclockwise? Now we use Lenz’s Law. To consider flux, let's choose the normal to be parallel to the magnetic field. In this example, since the magnetic field points from the north side of the magnet, this means that the normal vector will point to the left (as shown in the figure). This means the flux is positive and it increases with time. Based on Lenz's Law the induced current should oppose this. Thus, it will produce a negative magnetic flux through the loop that tries to cancel the increase in the positive flux. Since a positive flux means the field points to the left, then the magnetic field produced by the induced current, which we call the induced field, B⃗ ind, will point to the right. Now can use the current right-hand-rule to established the direction of current based on the magnetic field, B⃗ ind, it produces. When you align your thumb along the tangent to the loop, and curl your fingers towards the induced magnetic field, your thumb will indicate the direction of the induced current. Since the induced field points to the right, we see that the induced current for this case is flowing in a counterclockwise direction as viewed from the right.
Below is another example of a conducting loop moving to the right into a uniform field which points into the page. Initially, there is no field through the loop. As it enters the field there is a flux into the page. To oppose this change the induced field will be out of the page. According to the right-hand-rule, this implies that the induced current will be counterclockwise as shown.
Figure 11.9.3: Conducting Loop Enters Uniform Magnetic Field
Next, as the loop continues to move to the right within the field, there is no longer a change in flux, this results in zero induced field and thus no current will flow.
Figure 11.9.4: Conducting Loop Stays Within the Uniform Magnetic Field
Finally, as the loop starts to exit the magnetic field, the flux that points into the page starts to decrease. To oppose this change, the induced flux and thus field will point into the page. With the help of the right-hand-rule we find that this must result in a clockwise induced current.
Figure 11.9.5: Conducting Loop Exits the Uniform Magnetic Field
We can explore directly how the changing flux is related to current by making a plot of flux as a function of time, as depicted in the figure below. Flux will always be into the page for this scenario since the magnetic field points into the page. If we will define the normal vector to point out of the page, the flux will be negative since the angle between flux and the normal vector is 180o in this example. We also assume that flux changes linearly as the loop moves to the right at a constant speed. The flux starts at zero and its magnitude increases as the loop moves into the field. Since the flux is negative, this means that it becomes more negative. The constant flux region in the plot represent the part of the motion when the loop is fully immersed in the field, thus, the amount of flux does not change as it moves through it. As it starts exiting the field, the magnitude of flux starts to decrease or become less negative until it reaches zero when it fully exists the field.
Figure 11.9.6: Flux and Current as a Function of Time
The plot also shows the induced current as a function of time. Equation 11.9.2 tells us that the induced current is the negative of the derivative of flux as a function of time. Initially, since the slope of the flux is constant and negative, this results in a constant positive current. This means that by defining flux into the page as negative, the change in flux will be also negative, so the induced field that opposes this change has to be positive. This leads to a counterclockwise induced current which by convention is then defined as positive. In the region where the flux is constant, the slope is zero, resulting in zero induced current. When the flux starts to decrease (become less negative), the slope is positive, resulting in a negative (clockwise) current.
Applications
We started discussing Faraday’s law by considering moving a magnet near a loop of wire. We have found that this produces an induced current in the wire. This phenomenon has found many familiar applications in the modern world:
Seismograph: One way to exploit Faraday’s Law is to attach a magnet to anything that moves and place it near a loop of wire; any movement or oscillation in the object can be detected as an induced current in the wire loop. In this way we can translate physical movements and oscillations into electrical impulses. In all devices of this kind, the movement or oscillation is measured between the position of a coil relative to a magnet, whose movement causes the current in the coil to vary, generating an electrical signal. For example, as the vibrations produced by an earthquake pass through a seismograph, a magnet's vibrations produce a current that can be amplified to drive a plotting pen. This is how the seismograph operates.
Guitar Pickup: Les Paul, a pioneer musician of pop-jazz guitar, applied Faraday’s Law to the making of musical instruments and invented the first electric guitar. The “pickup” of an electric guitar consists of a permanent magnet with a coil of wire wrapped around it several times. The permanent magnet is placed very close to the metal guitar strings. The magnetic field of the permanent magnet causes a part of the metal string of the guitar to become magnetized. When one plucks the string, it vibrates, creating a changing magnetic flux through the coil of wire surrounding the permanent magnet. The coil “picks up” the vibrations that generate an induced current and sends the signal to an amplifier, to the pleasure of rock fans everywhere.
Electric Generator: An electric generator is used to efficiently convert mechanical energy to electrical energy. The mechanical energy can be provided by any number of means, such as falling water (like in a hydroelectric generator), expanding steam (as in coal, oil, and nuclear power plants), or wind (as in wind turbine generators). In all cases, the principle is the same, the mechanical energy is used to move a conducting wire coil inside a magnetic field (usually by rotating the wire). In this case, the area of the coil is the constant, the magnitude of the field is constant, so the angle term in the equation for Faraday's law is responsible for the changing flux. This is caused by the change in the relative orientation between the magnetic field and the normal to the area of the coil. Consider the simple scenario where we rotate the coil with constant angular speed ω. The rotation angle is given by θ=ωt, and the flux will be proportional to cosωt. Using calculus, the time rate of change of the flux will then be proportional to ωsinωt. This means the induced current will oscillate sinusoidally. In other words, the current in the coil alternates in direction, flowing in one direction for half the cycle and flowing the other direction for the other half. This kind of generator is referred to as an alternating current generator, or simply an AC generator. The standard plugs you use to power all of your electrical appliances are all powered by an electric generator of this form.
Electric Motor: Electric motors work in basically the reverse principle that operates electric generators: an alternating electric current causes an electromagnetic cylinder to periodically switch poles, which interacts with the field of an inlaid magnet to turn it. Some motors use electromagnets for both components, but the principle is the same. The stationary magnetic piece is called the stator and the magnetic piece that rotates is called that rotor
Hybrid Cars: Regenerative breaking discussed above.
Example 11.9.1
Below are 4 cases: the left panels show the initial and the right panels show the final configurations. The arrows indicate the direction and magnitude of the external magnetic field.
Fill in the table below for each case
Solution
Example 11.9.2
Consider a magnetic field is created by a large solenoid magnet. The solenoid is 1.5 meters long, has 5000 turns, a resistance of 4Ω, and a one-meter radius. A coil located inside the solenoid has a single loop (as depicted below), a resistance of 0.6Ω, and a 0.4 m radius. The solenoid initially has a current which produces a 0.2 T magnetic field. The solenoid’s current is then reduced linearly to zero in 0.1 seconds as shown in the left plot below.
a) Calculate Imax which is marked on the plot below.
b) Make a graph of the current in the coil on the right plot below. Make sure to indicate numerical values and explain your choice of sign for the current.
Solution
: a) The magnetic field for a solenoid is B=μoINL. The maximum current with be when the magnetic field is at 0.2 T, since the current is reduced with time:
Imax=BLμoN=0.2T×1.5m4×10−7NA2×5000=47.7 A
b) The magnetic field is upward using the current RHR, which is reduced to zero. Thus, the induced magnetic field will also be upward to oppose the change of flux decreasing in the upward direction. Using the same RHR again, this results in an induced current which is counterclockwise as viewed from top. In other words, the direction of the current in the coil is the same as the direction of the current in the solenoid.
Now that we have established the direction of current, we just need to worry about the magnitude of the induced emf and current. The magnitude of induced emf for the wire which only contains one loop is:
|E|=dΦdt=AΔBΔt=πr2dBdt=(0.42π)m20.2T0.1s=1.68A
Solving for induced current:
Icoil=|E|R=1.0V0.6Ω=1.68A
11.8: Magnetic Force
Back Matter |
5157 | https://math.stackexchange.com/questions/2540629/how-to-find-critical-points-when-you-get-constant-value | calculus - How to find critical points when you get constant value - Mathematics Stack Exchange
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How to find critical points when you get constant value
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To find these critical points you must first take the derivative of the function. Second, set that derivative equal to 0 and solve for x. Each x value you find is known as a critical number.
But what happens if you take derivative and you get a constant value like -1?
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edited Nov 28, 2017 at 4:24
Gerry Myerson
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asked Nov 28, 2017 at 3:57
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4 It means that your function is linear and its graph is a not horizontal infinite straight line, so it does not have any critical point.Koto –Koto 2017-11-28 04:00:22 +00:00 Commented Nov 28, 2017 at 4:00
1 i can accept this answer if you'd like mathguy –mathguy 2017-11-28 04:02:14 +00:00 Commented Nov 28, 2017 at 4:02
Let's wait for more creative people to show up. =D Koto –Koto 2017-11-28 04:02:57 +00:00 Commented Nov 28, 2017 at 4:02
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The definition of the derivative of a function is the rate of change of a function, say f(x)f(x). It defines the gradient of the tangent at a specific point x x. For a parabola, we visualise that the gradient is always changing, so the derivative of a parabola would be f′(x)=a x+b f′(x)=a x+b.
If the derivative of a function is a constant value, say, 2 2, then this implies that the function's rate of change is 2 2 through the whole function, for x∈R x∈R. The only graph where this occurs is a linear graph, which is in the form f(x)=a x+b f(x)=a x+b. This also happens to be the derivative of a parabola. Obviously, a line that extends infinitely in either direction does not have a critical point.
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answered Nov 28, 2017 at 4:46
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5158 | https://www.youtube.com/watch?v=PZ-3d7u_TU0 | Proof: Recursive Identity for Binomial Coefficients | Combinatorics
Wrath of Math
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The binomial coefficient n choose k is equal to n-1 choose k + n-1 choose k-1, and we'll be proving this recursive formula for a binomial coefficient in today's combinatorics lesson!
This is the identity implicitly being used when we typically construct Pascal's triangle, and we can easily see why it's true by cleverly splitting the number of ways we can select k objects from n objects into two separate counts! Full details in the lesson!
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Transcript:
Introduction howdy folks in today's wrath of math lesson we'll be proving this neat recursive identity for binomial coefficients the number of ways we can choose k objects from a collection of n objects is equal to the number of ways we can choose k objects from a collection of n minus 1 objects plus a number of ways we can choose k minus 1 objects from a collection of n minus 1 objects so this gives us a way to Restrictions calculate a binomial coefficient with some particular n value as a sum of binomial coefficients with the previous n value n minus 1. now we might want to specify our typical restrictions on n and k so we might say that n has to be greater than or equal to 0 and k can range from 0 to n inclusive in the case that k equals n we can still say that this identity holds because we'll have n choose n which is one is equal to n minus one choose n there's zero ways to select n objects from n minus one objects so that would be zero and then this would be n minus one choose n minus 1 which is 1 so we'd have 1 equals 1 everything would be dandy now if you have written out pascal's triangle before whether or not you know it you are very familiar with this equality even though you know you might not recognize it because if we start to write out pascal's triangle this is we're using this to write out these numbers it's more apparent if we go down a few rows so let's go down a few rows into the triangle and how do we calculate the next row well we first we add up one and nothing which is one then we add up these two numbers they add to four then we add up these two numbers they add to six then we add up these two numbers they add to four one and nothing is just one so look at this just for example four choose two is six and we found it by adding three choose two plus three choose two minus 1 which is 3 choose 1. so that's how we calculate the binomial coefficients when we're writing out pascal's triangle like this we use this identity whether we know it or not so pretty cool so we might as well go ahead and prove it since you clicked on the video and that's what i promised Proof let's prove it can be proven pretty easily by separating our total count our total number of n choose k separating it into two separate counts so let's consider just a generic object a generic set of n objects excuse me so a1 a2 and so on up through some nth object a n so we're considering the number of ways we can select k elements from this n element set now if we're going to get the identity we want to pop out we're going to have to find some way to count the number of ways of selecting k objects from n objects but somehow reduce our total number of options to n minus 1. so how could we do that we could do it by separating the total count based on the number of collections that do and don't include a particular element so we can basically fix an element to either be included or not be included and that's going to turn out to work just fine so what exactly do i mean Solution well again we're considering the number of k element subsets that we can make from this n element set let's separate those collections of k objects or k elements let's separate them into the collections that don't include a n and the collections that do include a n a n is just an arbitrary choice we could just as well separate them using any other element of the set but let's just fix a n doesn't matter which we choose we'll fix a n so let's first consider the collections of k elements from this n element set that don't include a n okay so how many ways can we choose k elements from this set if we're not gonna include a n well we have a total of n minus one options right because the original set has n options but we're not going to pick a n so we have n minus 1 other elements that we could choose and then we just have to choose k of them so the total number of ways we could do that is n minus 1 choose k pretty straight forward so now we have counted all of the collections of k objects from this n element set that don't include a n now we just have to count up the ones that do include a n and notice of course we can see that we're on the right track because this is a term in our sum that we're going for that h is really far away from the rest of the word so now we're counting the collections that do include a n so we fixed a n it's already going to be in these these collections that we're counting now so how many other objects can we choose from well we've already chosen a n so there are again n minus one other objects to choose from and we want to choose a total of k objects because remember that we're counting the number of ways we can pick k objects from a collection of n objects we want to pick a total of k objects since we've already decided we're going to include a n in these collections we have k minus 1 choices left to make among the other n minus 1 objects in the set so we've just counted all of the collections of size k from a set of n elements that do include this element a n that we have fixed and that's it we've separated this total count into the groups or the collections that do do not include a n and the ones that do include a m so to get the total we just add them together and of course that gives us precisely what we want it's n minus one choose k the collections of size k that don't include a n plus n minus 1 choose k minus 1 the collections of size k that do include a n and again we've got that k minus 1 there because we have fixed this count to count the collections that do include a n so we only have k minus one choices left to make we've already determined we're going to include a n in those collections and so that's just another way that we could count the number of ways of choosing k objects from a collection of n objects and that proves the equality and choose k is equal to n minus 1 choose k plus n minus 1 choose k minus Outro 1. pretty neat a really simple strategy of just separating out the collections you know that we're trying to count separating them into the collections that don't include some arbitrary fixed element of our set and the collections that do include this element and then boom we get the equality pretty sweet and we can actually use this to help us prove the binomial theorem if we want to do that so hope this video helped you understand the proof let me know in the comments if you have any questions need anything clarified or of any other video requests thank you very much for watching if you appreciate these lessons on wrath of math i'd really appreciate it if you would make a small donation on paypal or small monthly pledge on patreon leave links to those in the description i'll see you next time and be sure to subscribe for the swankiest math lessons on the internet [Music] you |
5159 | https://prepp.in/question/for-orifice-meters-with-sharp-edges-the-coefficien-663277ed0368feeaa554a9ab | For orifice meters with sharp edges, the coefficient of velocity is ______
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Fluid Mechanics
Orificemeter
for orifice meters with sharp edges the coefficien
Question
Fluid Mechanics
For orifice meters with sharp edges, the coefficient of velocity is ______
0.98
0.1
0.4
1.9
Solution
The correct answer is
0.98
Orifice Meter Coefficient of Velocity Explained
An orifice meter is a device used to measure the flow rate of fluids (like liquids or gases) through a pipe. It operates by inserting a thin plate with a precisely sized hole (the orifice) into the pipe. As the fluid passes through this orifice, its velocity increases, and its pressure decreases, creating a measurable pressure difference.
The coefficient of velocity (C v C_v C v) is a factor used in fluid mechanics. It represents the ratio between the actual velocity of the fluid jet at its narrowest point (the vena contracta) and the theoretical velocity calculated based on energy conservation principles (like Bernoulli's equation). It essentially quantizes how much friction or other energy losses affect the fluid's speed.
The relationship is given by: C v=Actual Velocity Theoretical Velocity C_v = \frac{\text{Actual Velocity}}{\text{Theoretical Velocity}}C v=Theoretical Velocity Actual Velocity
Coefficient of Velocity for Sharp Edges
In many standard orifice meters, the edges of the hole are kept very sharp and square. When the fluid flows through this sharp-edged orifice, the flow stream separates from the edges. This causes the jet to narrow down to a minimum cross-sectional area some distance downstream of the orifice plate, a point known as the vena contracta.
The value of the coefficient of velocity for these sharp-edged orifices is influenced by factors like fluid viscosity and turbulence. Typically, experimental data shows that the actual velocity is very close to the theoretical velocity. For a sharp-edged orifice, the C v C_v C v value is approximately 0.98.
Understanding Orifice Meter Coefficients
The coefficient of velocity (C v C_v C v) is related to other coefficients used for orifice meters:
Coefficient of Contraction (C c C_c C c): This measures how much the fluid stream contracts. For sharp-edged orifices, C c C_c C c is usually around 0.61 to 0.65.
Coefficient of Discharge (C d C_d C d): This is the overall factor relating the actual flow rate to the theoretical flow rate. It's calculated as C d=C c×C v C_d = C_c \times C_v C d=C c×C v. For sharp-edged orifices, C d C_d C d is typically around 0.60 to 0.62.
The high value of C v C_v C v (around 0.98) for sharp-edged orifices indicates that the velocity loss itself is relatively small. The main reason the coefficient of discharge (C d C_d C d) is significantly less than 1 is the flow contraction effect captured by C c C_c C c.
Conclusion on Orifice Meter Velocity Coefficient
Based on established principles in fluid mechanics and empirical data, the coefficient of velocity for an orifice meter with sharp edges is generally accepted to be approximately 0.98.
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Important Questions from Orificemeter
The actual velocity at vena contracta for flow through an orifice from a reservoir of height H =
(A) 2 g H\sqrt {2gH} 2 g H
(B)C v 2 g H{C_v}\sqrt {2gH} C v2 g H
(C)2 g H C v\frac{{\sqrt {2gH} }}{{{C_v}}}C v2 g H
(D)C d 2 g H{C_d}\sqrt {2gH} C d2 g H Fluid Mechanics View Answer
2. The relation between the co-efficient of discharge (C d), co-efficient of contraction (C c), and co-efficient of velocity (C v) for an orifice-meter is given by equation: Fluid Mechanics View Answer
3. Orifice meter is used to measure: Fluid Mechanics View Answer
4. The head of water over Centre of orifice of diameter 20 mm is 3 m. The actual discharge through orifice is 1.45 l.p.s. Calculate C d of orifice. Fluid Mechanics View Answer
5. A fluid jet is discharging from a 100 mm nozzle and the vena contracta formed has a diameter of 90 mm. if the coefficient of velocity is 0.95, then the coefficient of discharge for the nozzle is Fluid Mechanics View Answer
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5160 | https://research.vu.nl/en/publications/work-together-or-fight-together-modeling-adaptive-cooperative-and | Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making - Vrije Universiteit Amsterdam
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Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making
Laila van Ments, Jan Treur
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Artificial Intelligence
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Research output: Chapter in Book / Report / Conference proceeding › Chapter › Academic › peer-review
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Abstract
In this chapter, joint decision making processes are studied and the role of cognitive metaphors as mental models in them. A second-order self-modeling network model is introduced based on mechanisms known from cognitive and social neuroscience and cognitive metaphor and mental model literature. The cognitive metaphors were modeled as specific forms of mental models providing a form of modulation within the joint decision making process. The model addresses not only the use of these mental models in the decision making, but also their Hebbian learning and the control over the learning. The obtained self-modeling network model was applied to two types of metaphors that affect joint decision making in different manners: a cooperative metaphor and a competitive metaphor. By a number of scenarios it was shown how the obtained self-modeling network model can be used to simulate and analyze joint decision processes and how they are influenced by such cognitive metaphors.
| Original language | English |
| Title of host publication | Mental Models and their Dynamics, Adaptation and Control |
| Subtitle of host publication | A Self-Modeling Network Modeling Approach |
| Editors | Jan Treur, Laila Van Ments |
| Publisher | Springer Nature Switzerland AG |
| Chapter | 10 |
| Pages | 255-284 |
| Number of pages | 30 |
| ISBN (Electronic) | 9783030858216 |
| ISBN (Print) | 9783030858209 |
| DOIs | |
| Publication status | Published - 2022 |
Publication series
| Name | Studies in Systems, Decision and Control |
| Volume | 394 |
| ISSN (Print) | 2198-4182 |
| ISSN (Electronic) | 2198-4190 |
Bibliographical note
Publisher Copyright:
© 2022, The Author(s), under exclusive license to Springer Nature Switzerland AG.
Keywords
Joint decision making
Mental model
Metaphor
Second-order adaptive
Self-modeling network model
Access to Document
10.1007/978-3-030-85821-6_10
Ments-Treur2022_Chapter10_WorkTogetherOrFightTogetherFinal published version, 885 KB Licence: Article 25fa Dutch Copyright Act
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van Ments, L., & Treur, J. (2022). Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making. In J. Treur, & L. Van Ments (Eds.), Mental Models and their Dynamics, Adaptation and Control: A Self-Modeling Network Modeling Approach(pp. 255-284). (Studies in Systems, Decision and Control; Vol. 394). Springer Nature Switzerland AG.
van Ments, Laila ; Treur, Jan. / Work Together or Fight Together : Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making. Mental Models and their Dynamics, Adaptation and Control: A Self-Modeling Network Modeling Approach. editor / Jan Treur ; Laila Van Ments. Springer Nature Switzerland AG, 2022. pp. 255-284 (Studies in Systems, Decision and Control).
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abstract = "In this chapter, joint decision making processes are studied and the role of cognitive metaphors as mental models in them. A second-order self-modeling network model is introduced based on mechanisms known from cognitive and social neuroscience and cognitive metaphor and mental model literature. The cognitive metaphors were modeled as specific forms of mental models providing a form of modulation within the joint decision making process. The model addresses not only the use of these mental models in the decision making, but also their Hebbian learning and the control over the learning. The obtained self-modeling network model was applied to two types of metaphors that affect joint decision making in different manners: a cooperative metaphor and a competitive metaphor. By a number of scenarios it was shown how the obtained self-modeling network model can be used to simulate and analyze joint decision processes and how they are influenced by such cognitive metaphors.",
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van Ments, L& Treur, J 2022, Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making. in J Treur & L Van Ments (eds), Mental Models and their Dynamics, Adaptation and Control: A Self-Modeling Network Modeling Approach. Studies in Systems, Decision and Control, vol. 394, Springer Nature Switzerland AG, pp. 255-284.
Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making. / van Ments, Laila; Treur, Jan.
Mental Models and their Dynamics, Adaptation and Control: A Self-Modeling Network Modeling Approach. ed. / Jan Treur; Laila Van Ments. Springer Nature Switzerland AG, 2022. p. 255-284 (Studies in Systems, Decision and Control; Vol. 394).
Research output: Chapter in Book / Report / Conference proceeding › Chapter › Academic › peer-review
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AU - Treur, Jan
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PY - 2022
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N2 - In this chapter, joint decision making processes are studied and the role of cognitive metaphors as mental models in them. A second-order self-modeling network model is introduced based on mechanisms known from cognitive and social neuroscience and cognitive metaphor and mental model literature. The cognitive metaphors were modeled as specific forms of mental models providing a form of modulation within the joint decision making process. The model addresses not only the use of these mental models in the decision making, but also their Hebbian learning and the control over the learning. The obtained self-modeling network model was applied to two types of metaphors that affect joint decision making in different manners: a cooperative metaphor and a competitive metaphor. By a number of scenarios it was shown how the obtained self-modeling network model can be used to simulate and analyze joint decision processes and how they are influenced by such cognitive metaphors.
AB - In this chapter, joint decision making processes are studied and the role of cognitive metaphors as mental models in them. A second-order self-modeling network model is introduced based on mechanisms known from cognitive and social neuroscience and cognitive metaphor and mental model literature. The cognitive metaphors were modeled as specific forms of mental models providing a form of modulation within the joint decision making process. The model addresses not only the use of these mental models in the decision making, but also their Hebbian learning and the control over the learning. The obtained self-modeling network model was applied to two types of metaphors that affect joint decision making in different manners: a cooperative metaphor and a competitive metaphor. By a number of scenarios it was shown how the obtained self-modeling network model can be used to simulate and analyze joint decision processes and how they are influenced by such cognitive metaphors.
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KW - Mental model
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van Ments L, Treur J. Work Together or Fight Together: Modeling Adaptive Cooperative and Competitive Metaphors as Mental Models for Joint Decision Making. In Treur J, Van Ments L, editors, Mental Models and their Dynamics, Adaptation and Control: A Self-Modeling Network Modeling Approach. Springer Nature Switzerland AG. 2022. p. 255-284. (Studies in Systems, Decision and Control). Epub 2022 Jan 26. doi: 10.1007/978-3-030-85821-6_10
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5161 | https://www.mayoclinic.org/diseases-conditions/asthma/in-depth/asthma-medications/art-20045557 | Asthma medications: Know your options - Mayo Clinic
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Asthma medications: Know your options
Confused about your asthma medications? Here's what you need to know to sort out the main classes and numerous subtypes of asthma drugs.By Mayo Clinic Staff
The types and doses of asthma medications you need depend on your age, your symptoms, the severity of your asthma and medication side effects.
Because your asthma can change over time, work closely with your health care provider to track your symptoms and adjust your asthma medications, if needed.
Types of asthma medications
| Category | Purpose | Types |
---
| Long-term asthma control medications | Taken regularly to control chronic symptoms and prevent asthma attacks — the most important type of treatment for most people with asthma | Inhaled corticosteroids Leukotriene modifiers Long-acting beta agonists (LABAs) Long-acting muscarinic antagonists (LAMAs) Combination inhalers Theophylline |
| Quick-relief medications (rescue medications) | Taken as needed for rapid, short-term relief of symptoms — used to prevent or treat an asthma attack | Short-acting beta agonists such as albuterol Ipratropium (Atrovent HFA) Oral corticosteroids (for severe asthma attacks) |
| Medications for allergy-induced asthma | Taken regularly or as needed to reduce your body's sensitivity to a particular allergy-causing substance (allergen) | Allergy shots (immunotherapy) Under-the-tongue (sublingual) immunotherapy tablets Allergy medications |
| Biologics | Taken with control medications to stop underlying biological responses that cause inflammation in the lungs — used to better manage severe asthma symptoms | Benralizumab (Fasenra) Dupilumab (Dupixent) Mepolizumab (Nucala) Omalizumab (Xolair) Reslizumab (Cinqair) Tezepelumab-ekko (Tezspire) |
Long-term control medications
Many people with asthma need to take long-term control medications daily, even when they don't have symptoms. There are several types of long-term control medications, including the following.
Inhaled corticosteroids
These anti-inflammatory drugs are the most effective and commonly used long-term control medications for asthma. They reduce swelling and tightening in your airways. You may need to use these medications for several months before you get their maximum benefit.
Inhaled corticosteroids include:
Fluticasone (Flovent HFA, Arnuity Ellipta, others)
Budesonide (Pulmicort Flexhaler)
Mometasone (Asmanex Twisthaler)
Beclomethasone (Qvar RediHaler)
Ciclesonide (Alvesco)
Regular use of inhaled corticosteroids helps keep asthma attacks and other problems linked to poorly controlled asthma in check. In children, long-term use of inhaled corticosteroids can delay growth slightly, but the benefits of using these medications to maintain good asthma control generally outweigh the risks.
Inhaled corticosteroids don't generally cause serious side effects. When side effects occur, they can include mouth and throat irritation and oral yeast infections. If you're using a metered dose inhaler, use a spacer and rinse your mouth with water after each use to reduce the amount of drug remaining in your mouth.
Leukotriene modifiers
These medications block the effects of leukotrienes, immune system chemicals that cause asthma symptoms. Leukotriene modifiers can help prevent symptoms for up to 24 hours. Examples include:
Montelukast (Singulair)
Zafirlukast (Accolate)
Zileuton (Zyflo)
In rare cases, montelukast is linked to psychological reactions, such as agitation, aggression, hallucinations, depression and suicidal thinking. See your health care provider right away if you have any unusual reaction.
Long-acting beta agonists (LABAs)
These bronchodilator (brong-koh-DIE-lay-tur) medications open airways and reduce swelling for at least 12 hours. They're used on a regular schedule to control moderate to severe asthma and to prevent nighttime symptoms. Although they're effective, they've been linked to severe asthma attacks. For this reason, LABAs are taken only in combination with an inhaled corticosteroid.
The most commonly used LABAs for asthma are:
Salmeterol (Serevent)
Formoterol
Long-acting muscarinic antagonists (LAMAs)
These medications are also bronchodilators. If an LABA cannot be used, a LAMA along with an inhaled corticosteroid can be an option. A LAMA can also be added to a LABA and inhaled corticosteroid if better control of your asthma is needed.
The LAMA tiotropium (Spiriva Respimat) may be added to the treatment plan for severe asthma.
Combination inhalers:
Corticosteroids and long-acting beta agonists
Some inhaled asthma medication combinations contain both a corticosteroid and a bronchodilator:
Fluticasone and salmeterol (Advair Diskus, AirDuo Digihaler, others)
Budesonide and formoterol (Symbicort)
Mometasone and formoterol (Dulera)
Fluticasone and vilanterol (Breo Ellipta)
Quick-relief medications
These asthma medications — sometimes called short-acting beta agonists (SABA) — open the lungs by relaxing airway muscles. Often called rescue medications, they can ease worsening symptoms or stop an asthma attack in progress. They begin working within minutes and are effective for 4 to 6 hours. They're not for daily use.
Some people use a quick-relief inhaler before exercise to help prevent shortness of breath and other asthma symptoms. Possible side effects include jitteriness and a rapid heartbeat (palpitations).
Quick-relief medications include:
Albuterol (ProAir HFA, Ventolin HFA, others)
Levalbuterol (Xopenex HFA)
If your symptoms are minor and infrequent or if you have exercise-induced asthma, you might manage your symptoms with one of these medications alone. However, most people with persistent asthma also need an inhaled corticosteroid or other long-term control medication.
If you need to use your inhaler more often than your health care provider recommends, your asthma is not under control — and you may be increasing your risk of a serious asthma attack. Talk with your provider about adjusting your therapy.
Ipratropium
Ipratropium (Atrovent HFA) is a short-acting bronchodilator that's usually prescribed for emphysema or chronic bronchitis but is sometimes used to treat asthma attacks. It may be used either with or instead of a SABA.
A combination of ipratropium and albuterol (Combivent) may sometimes be used to treat asthma.
Theophylline
You take this bronchodilator daily in pill form to treat mild asthma. Theophylline (Theo-24) relaxes the airways and decreases the lungs' response to irritants. Although theophylline isn't used much anymore, it can be helpful for nighttime asthma symptoms. You might need regular blood tests to make sure you're getting the correct dose.
Oral corticosteroids for serious asthma attacks
These medications may be taken — usually for a limited time — to treat severe asthma attacks. Examples include:
Prednisone
Methylprednisolone
Oral corticosteroids can cause bothersome short-term side effects and more-serious side effects if they're taken for a long period. Long-term side effects can include cataracts, thinning bones (osteoporosis), muscle weakness, decreased resistance to infection, high blood pressure and reduced growth in children.
Medications for asthma triggered by allergies
Medications that focus on treating allergy triggers (allergens) that can cause asthma include:
Allergy shots. Allergy shots (immunotherapy) may be an option if you have allergic asthma that can't be controlled by avoiding triggers. You begin with skin tests to determine which allergens trigger your asthma symptoms. Then you'll get a series of injections containing small doses of those allergens.
You generally receive injections once a week for a few months, and then once a month for 3 to 5 years. In some cases, immunotherapy can be done more quickly. Over time, you should lose your sensitivity to the allergens.
Under-the-tongue (sublingual) immunotherapy tablets. Rather than getting shots, you take tiny amounts of allergen in pill form that dissolves in your mouth. Pills are usually taken daily. Sublingual immunotherapy tablets don't work for all allergens but can be helpful for grass and ragweed pollens and dust mites.
Allergy medications. These include oral and nasal spray antihistamines and decongestants, as well as corticosteroid and cromolyn nasal sprays. Allergy medications are available in nonprescription and prescription form. They can help with allergic symptoms such as a runny nose, itchy eyes, congestion, sneezing and sinus pressure, but aren't substitutes for asthma medications.
Nasal corticosteroid sprays help reduce inflammation and are a safe, long-term treatment for most people. Because it has few, if any, side effects, cromolyn is safe to use over long periods of time. Decongestant nasal spray should not be used for more than 2 to 3 days at a time because it can worsen symptoms when used continuously (rebound swelling).
Biologics
Your health care provider may recommend treatment with biologics if you have severe asthma with symptoms that are not easily managed by control medications. Biologics reduce inflammation by targeting immune system antibodies or specific cells. Taken together with other asthma medications, biologics help people with more-severe forms of asthma gain greater symptom control.
Your provider will typically do bloodwork to decide which biologic will work best for you. Some biologics are given as a shot (injection) and some are given by a vein (intravenously).
Biologic for allergic asthma
Omalizumab (Xolair) is sometimes used to treat asthma triggered by airborne allergens. If you have allergies, your immune system identifies a harmless airborne substance — such as pollen, dust mites and pet dander — as being harmful. This substance is called an allergen.
Your immune system is how your body protects itself, so it produces immunoglobulin E (IgE) antibodies to protect against this allergen. These antibodies signal your immune system to release chemicals into your bloodstream, causing a reaction that leads to allergy and asthma symptoms. Omalizumab blocks the action of these antibodies, reducing the immune system reaction.
Omalizumab is given as an injection every 2 to 4 weeks. It isn't generally recommended for children under 12 years of age. In rare cases, this medication has triggered a life-threatening allergic reaction (anaphylaxis). In addition, the U.S. Food and Drug Administration (FDA) has issued a warning about a slightly increased risk of heart and brain blood vessel problems while taking this drug.
Anyone who gets an injection of this drug should be monitored closely by health care professionals in case of a severe reaction.
Biologics for eosinophilic asthma
A newer class of biologic drugs has been developed to target specific substances produced by certain immune system cells. For some people, certain white blood cells, called eosinophils, build up within body tissues. Eosinophils release substances, called cytokines, which cause inflammation. These biological drugs target eosinophils and cytokines, reducing their numbers within the body and lowering inflammation.
These medications include:
Benralizumab (Fasenra)
Dupilumab (Dupixent)
Mepolizumab (Nucala)
Reslizumab (Cinqair)
Biologic for either eosinophilic asthma or allergic asthma
Tezepelumab-ekko (Tezspire) is a recently FDA-approved biologic for people with severe asthma. This medication can be used to treat eosinophilic asthma or allergic asthma by reducing certain kinds of cytokines that cause inflammation.
Making the most of your asthma medications
Tracking symptoms and side effects and adjusting your treatment based on changes is key to keeping your asthma symptoms under control. With your health care provider, create a written detailed plan for taking long-term control medications and for managing an asthma attack. Then follow your plan.
Know when to adjust your medications, when to see your provider and how to recognize an asthma emergency. If your provider has prescribed a peak flow meter to measure how well your lungs are working, use it according to your plan. Even if you feel well, take your medications as prescribed and track your symptoms until you talk to your provider.
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Show references
Fanta CH. Treatment of intermittent and mild persistent asthma in adolescents and adults. Accessed May 12, 2022.
Peters S, et al. Treatment of moderate persistent asthma in adolescents and adults. Accessed May 12, 2022.
Asthma medications. American College of Allergy, Asthma & Immunology. Accessed May 10, 2022.
Drug guide. American Academy of Allergy, Asthma & Immunology. Accessed May 12, 2022.
Asthma treatment. American College of Allergy, Asthma & Immunology. Accessed May 12, 2022.
Agache I, et al. Efficacy and safety of treatment with biologicals (benralizumab, dupilumab, mepolizumab, omalizumab and reslizumab) for severe eosinophilic asthma. A systematic review for the EAACI guidelines — Recommendations on the use of biologicals in severe asthma. Allergy. 2020; doi:10.1111/all.14221.
Allergy shots (immunotherapy). American Academy of Allergy, Asthma & Immunology. Accessed May 12, 2022.
SLIT treatment for allergic rhinitis: Nothing to sneeze about. American Academy of Allergy, Asthma & Immunology. Accessed May 12, 2022.
Huang J, et al. New treatments for asthma. Immunology and Allergy Clinics of North America. 2021; doi:10.1016/j.iac.2021.07.013.
Cloutier MM, et al. 2020 focused updates to the Asthma Management Guidelines: A report from the National Asthma Education and Prevention Program Coordinating Committee Expert Panel Working Group. Journal of Allergy and Clinical Immunology. 2020; doi:10.1016/j.jaci.2020.10.003.
AskMayoExpert. Asthma pharmacotherapy options. Mayo Clinic; 2021.
Saag KG, et al. Major side effects of systemic glucocorticoids. Accessed May 13, 2022.
FDA drug safety communication: FDA approves label changes for asthma drug Xolair (omalizumab) including describing slightly higher risk of heart and brain adverse events. U.S. Food and Drug Administration. Accessed May 13, 2022.
Tezspire (prescribing information). AstraZeneca; 2021. Accessed May 13, 2021.
Muiser S, et al. Understanding the role of long-acting muscarinic antagonists in asthma treatment. Annals of Allergy, Asthma & Immunology. 2022; doi:10.1016/j.anai.2021.12.020.
FDA requires Boxed Warning about serious mental health side effects for asthma and allergy drug montelukast (Singulair); Advises restricting use for allergic rhinitis. U.S. Food and Drug Administration. Accessed May 13, 2022.
Medical review (expert opinion). Mayo Clinic. June 29, 2022.
Long-acting muscarinic antagonists. National Heart, Lung, and Blood Institute. Accessed June 1, 2022.
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5162 | https://www.mathsisfun.com/fractions_multiplication.html | Introduction to Fractions
Simplifying Fractions
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Multiplying Fractions
Multiply the tops, multiply the bottoms.
There are 3 simple steps to multiply fractions
Multiply the top numbers (the numerators).
Multiply the bottom numbers (the denominators).
Simplify the fraction if needed.
Example: 1 2 × 2 5
Step 1. Multiply the top numbers:
1
2
×
2
5
=
1 × 2
=
2
Step 2. Multiply the bottom numbers:
1
2
×
2
5
=
1 × 2
2 × 5
=
2
10
Step 3. Simplify the fraction:
2
10
=
1
5
With Pizza
Here you can see it with pizza ...
Do you see that half of two-fifths is two-tenths?
Do you also see that two-tenths is simpler as one-fifth?
With Pen and Paper
And here is how to do it with a pen and paper (press the play button):
Another Example: 1 3 × 9 16
Step 1. Multiply the top numbers:
1
3
×
9
16
=
1 × 9
=
9
Step 2. Multiply the bottom numbers:
1
3
×
9
16
=
1 × 9
3 × 16
=
9
48
Step 3. Simplify the fraction:
9
48
=
3
16
(This time we simplified by dividing both top and bottom by 3)
The Rhyme
♫ "Multiplying fractions: no big problem,
Top times top over bottom times bottom. "And don't forget to simplify,
Before it's time to say goodbye" ♫
Fractions and Whole Numbers
What about multiplying fractions and whole numbers?
Make the whole number a fraction, by putting it over 1.
Example: 5 is also
5
1
Then continue as before.
Example: 2 3 × 5
Make 5 into
5
1
:
2
3
×
5
1
Now just go ahead as normal.
Multiply tops and bottoms:
2
3
×
5
1
=
2 × 5
3 × 1
=
10
3
The fraction is already as simple as it can be.
Answer =
10
3
Or you can just think of the whole number as being a "top" number:
Example: 3 × 2 9
Multiply tops and bottoms:
3
×
2
9
=
3 × 2
9
=
6
9
6
9
=
2
3
Mixed Fractions
You can also read how to multiply mixed fractions
Copyright © 2023 Rod Pierce |
5163 | https://www.achievementnetwork.org/resource-center/new-new-math | Developing conceptual understanding alongside procedural skill
by Holly Rola
Learning to multiply decimals is nothing new to 5th grade math classrooms, but with Common Core instruction, one key shift makes a big difference. That shift? Developing conceptual understanding.
Here’s how I learned to multiply 2.1 and 4.6 in the pre-Common Core era:
I learned to leave a space in the one’s place when I multiplied by the 4 and to move the decimal two places to the left in the final product, because there are two numbers to the right of the decimal places in the factors. But, why? Well, I wasn’t really sure. I just followed the steps.
What is rigor in math?
With strong Common Core instruction, students are supported to develop conceptual understanding alongside the procedural skills and fluencies: they understand why the algorithm works. (Remember: conceptual understanding, procedural skills and fluency, and application are the three aspects of rigor, a key shift called for by the Common Core.) By balancing conceptual understanding with procedural skills when multiplying decimals, students can perform the operation fluently but also understand what they’re doing. And they’re developing number sense.
If you’re still unconvinced (or just want know why some Common Core math problems look so different), spend 3 minutes watching this video.
Building procedural skills from conceptual understanding
Ms. Caponigro is a 5th grade math teacher at McKinley Elementary School in Massachusetts. Watch how Ms. Caponigro helps her students build procedural skill from conceptual understanding:
Later in the lesson, Ms. Caponigro continues to help students build on the knowledge of place value:
Does this approach work?
On their ANet assessment, Ms. Caponigro’s 5th graders outperformed the network average by 19 points on Standard 5.NBT.7, which requires students to add, subtract, multiply and divide decimals.
In fact, one of the 5.NBT.7 questions on the ANet assessment asked students to solve 2.1 x 4.6.
Ms. Caponigro’s students, again, outperformed. Her students averaged 12 points higher than students across the country.
What is conceptual understanding in math?
Cheers to Ms. Caponigro and her students! Their hard work to develop conceptual understanding alongside procedural skills is paying off.
Want to learn more about teaching conceptual understanding alongside procedural skill?
Check out EngageNY’s Math Studio Talk: Instruction for 5.NBT to see how students can build on their knowledge of the base ten place value sytem and use concrete models to understand mathematical operations. (Note: It’s a lengthy video. It’s worth the invesment, but if you want to skip ahead to instruction aligned to standard 5.NBT.7, it begins at minute 14:30.)
Visit LearnZillion.com and explore the instructional videos and lesson plans, including this 5.NBT.7 lesson plan to apply understanding of decimal multiplication using an area model. The lesson supports students to decompose decimals numbers to better understand and solve problems that require decimal multiplication, as shown in the lesson excerpt below.
If you’d like to see more 5.NBT.7 instruction in action, watch this classroom video from the Teaching Channel, where Ms. Poirer helps her students in Phoeniz, AZ add tenths and hundreths.
And, by the way, great Common Core math instruction will help students develop number sense across all grade levels and domains, not just when multiplying decimals.
Holly is the Analysis and Reporting Manager at ANet.
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5164 | https://www.youtube.com/watch?v=czyhB6_7CNI | How to Prove that the Derivative of a Constant is Zero(0) - Calculus
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Transcript:
hi guys today we'll be proving uh that the derivative of a constant gives us a zero so we can use two methods to prove this the first one of course is the power rule then we're also going to try um using the first principle so using the power rule given our function f of x is equals to a constant c how can we find the derivative of this how can we show that the derivative is zero first of all we know that this c has x to the power of 0 because you know that any number or any variable or number is the power 0 is 1. so this x to the power 0. then using the power rule to find our divided x we're going to say the power times the coefficient which is zero so c times zero is zero x to the power we know that to the power of x we have to subtract one so that would be zero minus one now divided x will be equals to since zero is multiplying this whole function then dividing x is just equals to zero so now this is the proof using the power rule now i want to try using um [Music] the first principles so again divided f of x is equal to c so what is our f of x plus h it will be c as well because we don't have x to replace x and to replace um x with x plus h so the formula for the first principle is dy dx is equals to the limit as h approaches 0 then we have the function f of x plus h minus f of x all this is divided by h like that so let's do our replacements divided x is equals to the limit as h approaches zero what we have was f of x plus h is simply c minus what we have as f of x is also c divided by h like that then um drag x is equals to the limit as h approaches zero so c minus c that is zero over h do not replace this edge with a zero don't apply the limit here because zero over zero gives you something that is undefined so here's what we do is we say divide x is equals to the limit as h approaches zero so zero over h gives us a 0 like that then from there we're going to say if we apply the limit by replacing h with uh zero of which now we don't have so we can simply confirm that divided x is equals to zero so the core mistake that you can make is right here the common mistake you can make is by saying divide x is equal to the limit as h approaches zero then you replace this edge with a zero like this this one so you just have to give it a zero first then you apply the limit of which is h then just include that the derivative is equals to zero okay thank you very much you |
5165 | https://people.maths.ox.ac.uk/scott/Papers/graphint.pdf | Intersections of graphs B´ ela Bollob´ as ∗†‡ Alex Scott § January 20, 2010 Abstract Let G and H be two graphs of order n. If we place copies of G and H on a common vertex set, how much or little can they be made to overlap? The aim of this paper is to provide some answers to this question, and to pose a number of related problems. Along the way, we solve a conjecture of Erd˝ os, Goldberg, Pach and Spencer.
1 Introduction Let G and H be two graphs of order n, with e(G) = p n 2 and e(H) = q n 2 . If we place G and H at random onto the same vertex set, then we expect them to overlap in pq n 2 edges. How much or little can we make them overlap? Let us write disc+(G, H) for the largest amount by which we can exceed pq n 2 , and disc−(G, H) for the largest amount less than pq n 2 that we can achieve (more formal definitions are given below). We shall refer to these quantities as the positive and negative discrepancy of G with respect to H, and write disc(G, H) = max{disc+(G, H), disc−(G, H)}.
∗Department of Pure Mathematics and Mathematical Statistics, University of Cam-bridge, Cambridge CB3 0WB, UK and †Department of Mathematical Sciences, University of Memphis, Memphis TN 38152, USA ‡Research supported in part by NSF grants CNS-0721983, CCF-0728928, DMS-0906634 and CCR-0225610, and ARO grant W911NF-06-1-0076 §Mathematical Institute, 24-29 St Giles’, Oxford, OX1 3LB, UK; email: scott@maths.ox.ac.uk.
1 The main aim of this paper is to prove a result (Theorem 1) of form disc+(G, H)disc−(G, H) ≥c(p, q)n3.
In particular, this implies immediately that disc(G, H) ≥c′(p, q)n3/2.
By taking H to belong to specific families of graphs, we obtain results on the distribution of edges in G. For instance, taking H to be a complete graph of order n/2 together with n/2 isolated vertices, we obtain results on the discrepancy of G (see [10, 3], and the next two subsections). Taking G = Kn/2,n/2 gives a bound on the bipartite discrepancy of G and (Corollary 3) proves a conjecture of Erd˝ os, Goldberg, Pach and Spencer.
The rest of the paper is organized as follows. In Section 1.1, we discuss the discrepancy of a single graph, while in Section 1.2, we talk about the discrepancy of one graph with respect to another; we finish Section 1 with a discussion of our notation and conventions. The proof of Theorem 1 depends on results from the following four sections. In Section 2, we give a lower bound on disc+(G, H)disc−(G, H) in terms of another parameter ∆abs of the pair (G, H). In order to bound ∆abs, we look at certain subgraphs of G and H that we shall refer to as good 4-cycles. We discuss good 4-cycles in Section 3, and in Section 4 we give a lower bound on ∆abs in terms of the number of good 4-cycles in G and the number of good 4-cycles in H (thus bounding a parameter of G and H jointly in terms of simple parameters of G and H separately). In Section 5, we put these results together and prove our main result, namely a lower bound on disc+(G, H)disc−(G, H). In the final section we present a substantial number of conjectures and open questions.
We remark that there are various other notions of graph discrepancy. For general accounts of discrepancy, see S´ os , Beck and S´ os , Matouˇ sek and Chazelle ; for ideas related to negative discrepancy, see Erd˝ os, Faudree, Rousseau and Schelp , Krivelevich and Keevash and Sudakov .
1.1 The discrepancy of a graph For a graph G with n vertices and e(G) = p n 2 , we define the discrepancy1 of G to be disc(G) = max S⊂V (G) |e(S) −p |S| 2 |.
1The definition of discrepancy in , which we followed in , was maxS⊂V (G) |e(S) − 1 2 |S| 2 |. Here we follow , as it seems more natural for graphs with density p ̸= 1/2.
2 Thus the discrepancy is the maximum difference between the number of edges in an induced subgraph and the average number of edges in subgraphs of that order. Erd˝ os and Spencer proved that every graph G of order n has a set S ⊂V (G) with |e(S) −1 2 |S| 2 | ≥cn3/2; more generally, they showed that if H is a k-uniform hypergraph of order n then there is S ⊂V (H) with |e(S) −1 2 |S| k | ≥cn(k+1)/2. Thus every graph of density 1/2 has discrepancy at least cn3/2. Erd˝ os, Goldberg, Pach and Spencer extended this to graphs of arbitrary density, showing that if e(G) = p n 2 , where 2/(n −1) < p < 1 −2/(n −1), then disc(G) ≥c√rn3/2, where r = min{p, 1 −p} and c is an absolute constant.
A subset S of vertices with large discrepancy can have either more or fewer edges than p |S| 2 . Let us define the positive discrepancy of a graph G with n vertices and p n 2 edges by disc+(G) = max S⊂V (G) e(S) −p |S| 2 and the negative discrepancy by disc−(G) = max S⊂V (G) p |S| 2 −e(S) , so disc(G) = max{disc−(G), disc+(G)}. Note that disc+ and disc−are both nonnegative, since (for any k) choosing S uniformly at random from all k-sets we have E e(S) = p |S| 2 .
By considering random graphs G ∈G(n, 1/2), it can be seen that disc(G) can be as small as O(n3/2). However, it is possible to have smaller one-sided discrepancies: Kn/2,n/2 and its complement 2Kn/2 each have discrepancy O(n) on one side, although we pay by having discrepancy Ω(n2) on the other side (see and for further discussion, and extremal results on one-sided discrepancy). The trade-offwas quantified in , where it was shown that for every graph G with n vertices and p n 2 edges, with p(1 −p) ≥1/n, disc+(G)disc−(G) ≥cp(1 −p)n3.
(1) A similar result holds for k-uniform hypergraphs (see and section 6.1 below). Note that this extends the results of Erd˝ os and Spencer and of Erd˝ os, Goldberg, Pach and Spencer mentioned above.
3 1.2 Discrepancy of G with respect to H In this paper, we consider the discrepancy of a pair of graphs G, H. Given graphs G and H of order n, with e(G) = p n 2 and e(H) = q n 2 , a random placement of both graphs onto the same vertex set has an overlap of expected size pq n 2 . We define the positive discrepancy of G with respect to H by disc+(G, H) = max G′∼ =G |E(G′) ∩E(H)| −pq n 2 (2) and the negative discrepancy of G with respect to H by disc−(G, H) = pq n 2 −min G′∼ =G |E(G′) ∩E(H)|, (3) where in both cases the maximum/minimum is taken over graphs G′ isomor-phic to G and with the same vertex set as H. Clearly both discrepancies are nonnegative, and are symmetric in G and H. We define disc(G, H) = max{disc+(G, H), disc−(G, H)}.
(4) The discrepancy measures the maximum and minimum overlap of edges that we can get by defining G and H on a common vertex set. If G and H are graphs with mG and mH edges respectively, then the maximum possible value of disc+(G, H) is attained when one of the graphs is isomorphic to a subgraph of the other, while the maximum possible value of disc−(G, H) is attained when the two graphs have an edge-disjoint packing into the complete graph or can cover its edges (equivalently, one of G and H is a subgraph of the other).
By restricting H to particular families we can pick out various parameters of G. For instance, if H is a clique of order αn together with (1−α)n isolated vertices and G has p n 2 edges, then pq n 2 = p αn 2 and so disc(G, H) is related to the discrepancy of G, as disc+(G, H) = max S∈V (G)(αn) eG(S) −p αn 2 ≤disc+(G) and disc−(G, H) = max S∈V (G)(αn) p αn 2 −eG(S) ≤disc−(G).
4 Indeed, disc(G) = max i disc(G, Ki ∪En−i), where Ki ∪En−i is the graph obtained by adding n −i isolated vertices to Ki.
The main aim of this paper is to prove a lower bound in the product form (1) for the discrepancy of pairs of graphs. We shall show the following.
Theorem 1. Let G and H be graphs of order n, and suppose that e(G) = p n 2 and e(H) = q n 2 , where 16/n ≤p, q ≤1 −16/n. Then disc+(G, H)disc−(G, H) ≥p4(1 −p)4q4(1 −q)4n3/1020.
By taking H to be a complete graph of order n/2 together with n/2 isolated vertices and G to be a random graph we see that (as in (1)) the n3 bound is sharp, although the dependence on p and q can probably be improved (see Section 6.2 below for further discussion). Some upper and lower bound on p and q is necessary, however: if we take H = K1,n−1 and G to be any regular graph of order n, then disc+(G, H) = disc−(G, H) = disc(G, H) = 0.
The following corollary of Theorem 1 is immediate.
Corollary 2. Let G and H be graphs of order n, and suppose that e(G) = p n 2 and e(H) = q n 2 , where 16/n ≤p, q ≤1 −16/n. Then disc(G, H) ≥p2(1 −p)2q2(1 −q)2n3/2/1010.
If H = K⌈n/2⌉,⌊n/2⌋then disc(G, H) is the bipartite discrepancy bdis(G) defined by Erd˝ os, Goldberg, Pach and Spencer , who conjectured that if 1 2 n 2 ≤e(G) ≤(1 −ǫ) n 2 then the bipartite discrepancy of G bdisc(G) ≥δn3/2 (5) for some δ = δ(ǫ). The conjecture of Erd˝ os, Goldberg, Pach and Spencer follows as an immediate consequence of Theorem 1 Corollary 3. Let G be a graph of order n. Then suppose that e(G) = p n 2 where 16/n ≤p ≤1 −16/n. Then bdis(G) ≥p2(1 −p)2n3/2/1012.
5 1.3 Definitions and notation Throughout the rest of the paper, we will assume that G and H are graphs of order n with vertex set V = [n]. We define an action of the symmetric group Sn on G by π(G) ≡Gπ = (V, Eπ), (6) where Eπ = {π(i)π(j) : ij ∈E(G)}.
(7) Consider the vector space R(n 2), where standard coordinate directions are indexed by elements of V (2) = n (in any canonical manner). We identify G with the vector g ∈R(n 2) with coordinates g(ij) = 1G(ij) −e(G)/ n 2 , (8) where 1G(ij) = 1 if ij ∈E(G) and 0 otherwise. Similarly, we identify Gπ with the corresponding vector gπ. Writing 1 = (1, 1, . . . , 1) = 1Kn, we get ⟨gπ, 1⟩= ⟨g, 1⟩= X 1≤i 0} and get ⟨gπI, φ⟩= ⟨gπ, φ⟩+ ∆+ π ≈∆/2, which would contradict (18). But the transpositions do not behave independently: for instance, comparing the effects of performing ti and tj separately with the effect of performing them together shows that the edges between {vi, wi} and {vj, wj} can make differ-ent contributions in the two cases. In order to satisfy (18), the interactions between transpositions must therefore tend to decrease the inner product.
In fact, we will show that performing the set of all transpositions {ti : i ∈I} simultaneously must decrease the inner product enough to satisfy (19).
Of course, in order to make such an argument work, we must find a suit-able permutation π. However, we cannot expect to find π such that ⟨gπ, φ⟩ and ∆+ π are both close to their expected values simultaneously. Instead, we will work with a linear combination of the two quantities. (A similar ap-proach, in a rather simpler setting, is used in .) We now proceed with the argument.
9 Let π ∈Sn be chosen uniformly at random. Then, for 1 ≤i ≤⌊n/2⌋, E∆i π = E(⟨gπi, φ⟩−⟨gπ, φ⟩) = 0, as πi is also uniformly distributed over Sn. Summing over i, we get E(∆+ π −∆− π ) = 0.
It follows from (15) and (16) that E∆+ π = E∆− π = ∆/2.
Since E⟨gπ, φ⟩= 0, we have E((α + 1)⟨gπ, φ⟩+ ∆+ π ) = ∆ 2 , and so we can choose π such that (α + 1)⟨gπ, φ⟩+ ∆+ π ≥∆ 2 .
(20) Let I = {i : ∆i π > 0} and V + = [ i∈I {vi, wi}.
Thus V + is the union of pairs of vertices whose reversal on their own increases the inner product with φ.
We decompose ∆i π by expanding the inner products on the right of (13) and regrouping. Since ∆i π = P e(gπi(e) −gπ(e))φ(e) and gπi(e) = gπ(e) unless e contains exactly one vertex from {π(vi), π(wi)}, we have ∆i π = X x̸∈{vi,wi} y∈{vi,wi} (gπi(xy) −gπ(xy))φ(xy) = X x̸∈{vi,wi} y∈{vi,wi} (gπ(xti(y)) −gπ(xy))φ(xy) = X x̸∈{vi,wi} δi(x), 10 where δi(x) = (gπ(wix) −gπ(vix))φ(vix) + (gπ(vix) −gπ(wix))φ(wix) = (gπ(wix) −gπ(vix))(φ(vix) −φ(wix)) (21) is the contribution to ∆i π of the pairs {x, vi} and {x, wi}. We write ∆i π = ∆i in + ∆i out, where ∆i in = X x∈V +{vi,wi} δi(x) and ∆i out = X x∈V \V + δi(x).
We define the sum over I of these contributions to be ∆in = X i∈I ∆i in and ∆out = X i∈I ∆i out.
So ∆+ π = ∆in + ∆out.
(22) Informally, ∆in measures the contribution to ∆+ π from edges inside V + and ∆out measures the contribution from edges between V + and V \ V + (note that edges outside V + do not contribute anything).
We shall prove (19) by flipping a random subset of pairs from V +, so we shall also have to consider the interactions when we perform more than one transposition. Let ∆all = ⟨gπI, φ⟩V + −⟨gπ, φ⟩V +, where the inner product uses the notation defined at (12). Thus ∆all mea-sures the effects (inside V +) of applying all the transpositions {ti : i ∈I} simultaneously, whereas ∆in measures the effect (inside V +) of applying them individually. We have ∆all = ⟨gπI, φ⟩−⟨gπ, φ⟩−∆out, (23) since πI and π agree on vertices outside V +, and (a short calculation shows that) ∆out is the sum of gπI(e)φ(e) −gπ(e)φ(e) over edges between V + and its complement.
11 After all this notation, we can now complete the argument. Let J ⊂I be a random subset, where each i ∈I is chosen independently with probability 1/α. Note that if vw is an edge outside V + then πJ and π agree on vw; if vw is an edge from V + to V \ V + then it is shifted with probability 1/α; if vw is inside V + (and is not of form viwi) then it has each end shifted on its own with probability 1 α(1 −1 α) and both ends shifted (leading to an interaction between transpositions) with probability 1/α2. Grouping terms together, we get E⟨gπJ, φ⟩ = ⟨gπ, φ⟩+ 1 α X i∈I ∆i out + 1 α 1 −1 α X i∈I ∆i in + 1 α2∆all = ⟨gπ, φ⟩+ 1 α∆out + 1 α 1 −1 α ∆in + 1 α2∆all = ⟨gπ, φ⟩+ 1 α∆+ π + 1 α2(∆all −∆in).
(24) Since, by (18), ⟨gπJ, φ⟩≤∆/5α for all π, we have E⟨gπJ, φ⟩≤∆/5α and so, by (24), ∆all ≤α∆ 5 −α2⟨gπ, φ⟩−α∆+ π + ∆in.
(25) It follows from (23), (25) and (22) that ⟨gπI, φ⟩= ⟨gπ, φ⟩+ ∆out + ∆all ≤α∆ 5 + (1 −α2)⟨gπ, φ⟩−α∆+ π + ∆in + ∆out = α∆ 5 + (1 −α2)⟨gπ, φ⟩−(α −1)∆+ π .
(26) By (20) we have (α2 −1)⟨gπ, φ⟩+ (α −1)∆+ π ≥(α −1)∆ 2 and so by (26) ⟨gπI, φ⟩≤α∆ 5 −(α −1)∆ 2 ≤−α∆ 20 , since α ≥2.
In order to apply Theorem 4, we need some lower bound on ∆abs(G, φ), when φ corresponds to a graph H. This is our next task.
12 3 Good 4-cycles In order to give a bound on ∆abs(G, H), we will need a lemma about graphs.
Given a graph G with vertex set V , and a 4-cycle C in the complete graph KV , we say that C is good (in G) if either |E(G) ∩E(C)| is odd, or E(G) ∩E(C) is a pair of vertex-disjoint edges. (So the bad cases are when E(C) ⊂E(G) or E(C) ∩E(G) = ∅or E(G) ∩E(C) is a path of length 2.) Good 4-cycles Lemma 5. Suppose that G has n vertices and p n 2 edges, where min{p, 1 − p} ≥16/n. Then the probability that a 4-cycle chosen uniformly at random is good is at least p2(1 −p)2/5040.
Proof. Note that n ≥32 by the condition on p and 1 −p.
We may as-sume p ≤1/2, since goodness is invariant under taking complements. We first consider a random 8-tuple X = {x1, . . . , x8}. Then with probability at least p2(1 −p)2/24 we have x1x2, x3x4 ∈E(G) and x5x6, x7x8 ̸∈E(G) (since x1x2 ∈E(G) with probability p, and e(G −x1 −x2) ≥e(G) −2n + 3, so P(x3x4 ∈E(G)|x1x2 ∈E(G) ≥p −5/n), and so on). If X does not con-tain a good C4 then considering the 4-cycles x1x2x3x4 and x1x2x4x3 shows that {x1, x2, x3, x4} induces a complete graph; similarly, considering x5x6x7x8 and x5x6x8x7 shows that {x5, x6, x7, x8} is an independent set.
We can now find two vertex-disjoint edges or two vertex-disjoint non-edges between {x1, x2, x3, x4} and {x5, x6, x7, x8}, say x1x5 and x2x6. Then x1x5x6x2 is a good C4. Thus X contains a good C4 with probability at least p2(1−p)2/24.
Since each good C4 is contained in n−4 4 8-sets, the number of good C4s must be at least p2(1 −p)2 24 · n 8 n−4 4 = p2(1 −p)2 8!
n(n −1)(n −2)(n −3).
13 There are 3 n 4 copies of C4 in total, so a random C4 is good with probability at least p2(1 −p)2 8!
· 4!
3 = p2(1 −p)2/5040.
The constants in Lemma 5 could probably be substantially improved.
However, note that a lower bound on p of form Ω(1/n) is necessary, as the star K1,n has p ∼2/n and no good cycles.
4 Inner products and good cycles Let G and H be graphs, and g, h the associated vectors defined by (8). We are interested in the inner product ⟨gπ, h⟩for various choices of the permutation π. In particular, in order to bound ∆abs in the next section, we shall want a lower bound on expressions of form E |⟨gtπ, h⟩−⟨gπ, h⟩|, (27) where t is a given transposition and π is chosen uniformly at random. We shall approach (27) by considering the effect of replacing h by hσ for various permutations σ, and decomposing ⟨gtπ, h⟩−⟨gπ, h⟩into a sum of simpler quantities.
Given a 4-cycle C = uvwx, we define ⟨g, h⟩C = ⟨g, h⟩E(C), so ⟨g, h⟩C = g(uv)h(uv) + g(vw)h(vw) + g(wx)h(wx) + g(xu)h(xu).
The following lemma is crucial.
Lemma 6. If the 4-cycle C = vivjwiwj is good in both G and H, and ti and tj are the transpositions (viwi) and (vjwj) respectively, then |(⟨gti, h⟩C −⟨g, h⟩C) −(⟨gti, htj⟩C −⟨g, htj⟩C)| ≥1.
(28) Proof. This is a straightforward case analysis.
For instance, consider the example in the diagram.
14 wi vi wj vj H wi vi wj vj Htj wi vi wj vj G wi vi wj vj Gti Then ⟨g, h⟩C, ⟨gti, h⟩C and ⟨g, htj⟩C are all equal to (1 −p)(−q) + (1 −q)(−p) + 2(−p)(−q) = 4pq −p −q, and ⟨gti, htj⟩C = (1 −p)(1 −q) + 3(−p)(−q) = 1 −p −q + 4pq, so ⟨gti, h⟩C −⟨g, h⟩C = 0, while ⟨gti, htj⟩C −⟨g, htj⟩C = (1 −p −q + 4pq) −(4pq −p −q) = 1.
The other cases are similar.
We can now give a lower bound on (27) in terms of the numbers of good cycles in G and H.
Lemma 7. Let G and H be graphs with vertex set V with |V | = n ≥15.
Suppose that the pair {v, w} belongs to rh · n−2 2 good 4-cycles vxwy in H and that G contains rg · 3 n 4 good 4-cycles. Let τ be the transposition (vw), and suppose that π ∈Sn is chosen uniformly at random. Then E|⟨gτπ, h⟩−⟨gπ, h⟩| ≥rgrh √n/10.
15 Proof. A 4-cycle vxwy containing v and w at distance 2, and with {x, y} ⊂ V \ {v, w} chosen uniformly at random, is good in H with probability rh.
Thus there is a sequence x1, y1, . . . , x⌊(n−2)/2⌋, y⌊(n−2)/2⌋of distinct vertices in V \ {v, w} such that at least rh⌊(n −2)/2⌋≥rh(n −3)/2 of the 4-cycles vxiwyi are good (consider a random choice).
We generate our random permutation π in two steps: let ρ ∈Sn be chosen uniformly at random, and let σ be the product of a random subset of the transpositions {(xiyi) : 1 ≤i ≤⌊(n −2)/2⌋}, taking each transposition independently with probability 1/2. Thus σ = (x1y1)ǫ1 · · · (x⌊n/2⌋, y⌊(n−2)/2⌋)ǫ⌊(n−2)/2⌋, where ǫi ∈{0, 1} for each i. We let π = σρ, so that π has uniform distribution in Sn as required.
For any g′, and any γ ∈Sn, we have ⟨g′ γ, hγ⟩= ⟨g′, h⟩. Since σ = σ−1, we have ⟨g′ σ, h⟩= ⟨g′, h′ σ−1⟩= ⟨g′, h′ σ⟩. Furthermore, σ and τ commute, so ⟨gτπ, h⟩−⟨gπ, h⟩= ⟨gτσρ, h⟩−⟨gσρ, h⟩= ⟨gτρ, hσ⟩−⟨gρ, hσ⟩.
(29) (Note that gτσρ = gστρ = (gτρ)σ.) Since τ is fixed, and π, ρ are both uniformly distributed, E|⟨gτπ, h⟩−⟨gπ, h⟩| = E|⟨gτρ, h⟩−⟨gρ, h⟩|.
Thus the triangle inequality gives E|⟨gτπ, h⟩−⟨gπ, h⟩| = 1 2E (|⟨gτπ, h⟩−⟨gπ, h⟩| + E|⟨gτρ, h⟩−⟨gρ, h⟩|) ≥ 1 2E|(⟨gτπ, h⟩−⟨gπ, h⟩) −(⟨gτρ, h⟩−⟨gρ, h⟩)| = 1 2E|⟨gτρ, hσ⟩−⟨gρ, hσ⟩+ ⟨gρ, h⟩−⟨gτρ, h⟩|, where the last line follows from (29) (expectations are over both σ and ρ).
But ⟨gτρ, hσ⟩−⟨gρ, hσ⟩+ ⟨gρ, h⟩−⟨gτρ, h⟩can be decomposed as ⌊(n−2)/2⌋ X i=1 (⟨gτρ, hσ⟩Ci −⟨gρ, hσ⟩Ci +⟨gρ, h⟩Ci −⟨gτρ, h⟩Ci) = ⌊(n−2)/2⌋ X i=1 ǫiλi, (30) where Ci = vxiwyi, the λi depend only on ρ and, by Lemma 6, |λi| ≥1 whenever Ci is good in both Gρ and H.
(Note that only edges between 16 {v, w} and {x1, y1, . . . , x⌊(n−2)/2⌋, y⌊(n−2)/2⌋} contribute to the sum: the rest are cancelled out.) Let d = d(ρ) (≤⌊n/2⌋) be the number of Ci that are good in both G and H. Each ǫi independently in (30) is 0 or 1 with probability 1/2. Using basic facts about random sums (see, for instance, ) we get that, conditioning on a fixed ρ, Eσ|⟨gτπ, h⟩−⟨gπ, h⟩| ≥ 1 2Eσ| ⌊(n−2)/2⌋ X i=1 ǫiλi| ≥ 1 2 r d(ρ) 8 ≥ d(ρ) 4√n.
But then, by the tower law for expectation, Eπ|⟨gτπ, h⟩−⟨gπ, h⟩|Eρ d(ρ) 4√n.
Since at least rh(n −3)/2 of the Ci are good in H, by our initial choice of x1, y1, . . ., and a (random) 4-cycle is good in G with probability rg, the expected number of 4-cycles Ci that are good in both G and H is at least rgrh(n −3)/2. It follows that Eπ|⟨gτπ, h⟩−⟨gπ, h⟩| ≥rgrh(n −3)/8√n ≥rgrh √n/10.
5 The main result In this short section, we put together the results we have proved in previous sections to give our main result.
Given graphs G and H, we know from Theorem 4 that we can bound disc+(G, H)disc−(G, H) in terms of the quantity ∆abs(G, H) (or, more pre-cisely, ∆abs(G, h)) defined in (16). Because we defined this quantity by using a restricted set of permutations, and then took an expectation over permuta-tions applied to G, we have ∆abs(Gπ, H) = ∆abs(G, H) for any permutation π. However, ∆abs(G, Hπ) may be different from ∆abs(G, H). The idea of 17 the proof is therefore to show that there is a placement Hρ of H for which ∆abs(G, Hρ) is large and then apply Theorem 4.
We will use the following corollary of Lemma 7.
Corollary 8. Suppose G and H are graphs of order n ≥15, and that a random 4-cycle in G is good with probability pg and a random 4-cycle in H is good with probability ph. let ρ ∈Sn be chosen uniformly at random. Then E∆abs(G, Hρ) ≥pgphn3/2/22.
Proof. Let us begin by fixing a permutation ρ and consider the graph Hρ, with associated vector hρ ∈R(n 2). Let v1, w1, . . . , v⌊n/2⌋, w⌊n/2⌋be the first 2⌊n/2⌋vertices of V (G) (recall that the vertex set has a canonical ordering), and define r1, . . . , r⌊(n−2)/2⌋so that ri n−2 2 is the number of good 4-cycles of form vixwiy in H. Set R = P⌊n/2⌋ i=1 ri. Note that R and the quantities ri depend on ρ.
Writing ti = (viwi), Lemma 7 implies that Eπ (|⟨gtiπ, hρ⟩−⟨gπ, hρ⟩|) ≥pgri √n/10, and so (using the notation defined at (15)), ∆abs(G, Hρ) = E(∆abs π ) ≥ ⌊n/2⌋ X i=1 pgri √n/10 = Rpg √n/10.
Now, choosing ρ uniformly at random, ER = ⌊n/2⌋Er1 = ⌊n/2⌋ph, so E∆abs(G, Hρ) ≥ jn 2 k pgph √n/10 ≥ pgphn3/2/22.
Lemma 5 allows us to convert this into an estimate in terms of e(G) and e(H).
Theorem 9. Let G and H be graphs with a common vertex set of order n, with e(G) = p n 2 and e(H) = q n 2 , where 16/n ≤p, q ≤1 −16/n. There is a permutation ρ such that ∆abs(G, Hρ) ≥p2(1 −p)2q2(1 −q)2n3/2/109.
(31) 18 Proof. Lemma 5 implies that pg ≥p2(1−p)2/5040 and ph ≥q2(1−q)2/5040.
We are done by Corollary 8.
We can now complete the proof of our main result.
Proof of Theorem 1. This follows immediately from (17) and (31).
6 Discussion and open problems A number of open questions remain, and we gather a selection of these to-gether under various headings.
6.1 Extensions to hypergraphs The definition of discrepancy extends naturally to r-uniform hypergraphs. If H is k-uniform, with n vertices and p n k edges, we define disc+(H) = max S⊂V (H) e(S) −p |S| k , disc−(H) = max S⊂V (H) p |S| k −e(S) and disc(H) = max{disc−(H), disc+(H)}.
It was shown in that if p(1 −p) ≥1/n then disc+(H)disc−(H) ≥ckp(1 −p)nk+1.
Similarly, we can define disc+(G, H), disc−(G, H) and disc(G, H) for k-uniform hypergraphs G and H (just replace n 2 by n k in (2) and (3); (4) is unchanged).
An obvious question is whether Theorem 1 has an extension to hyper-graphs. We conjecture that it does.
Conjecture 10. Let G and H be k-uniform hypergraphs with vertex set [n], and suppose e(G) = p n k and e(H) = q n k , where 1/n ≤p, q ≤1 −1/n.
Then disc+(G, H)disc−(G, H) ≥c(p, q)nk+1.
19 It seems likely that it should be possible to take c(p, q) to be a polynomial in p and q.
More generally, it may be possible to let r grow with n. As a first step in this direction, we raise the following question.
Problem 11. Consider the family X of (n/2)-uniform hypergraphs on n vertices, with 1 2 n n/2 edges. What is min G∈X disc(G)?
Of course, there are many variants of this problem; and it would also be desirable to answer the product form of the question, that is to determine min G∈X disc+(G)disc−(G).
More ambitiously, it would be very interesting to know the values of min G,H∈X disc+(G, H)disc−(G, H).
(32) 6.2 Sharpness of constants and random graphs How good is the constant p4(1 −p)4q4(1 −q)4 in Theorem 1, as a function of p and q? For single graphs with density p, we know that disc+(G)disc−(G) ≥ cp(1−p)n3, which is much larger when p is small. A natural conjecture would be that when we have two graphs with densities p and q we should have a constant of form cp(1−p)q(1−q). However, this is not correct. For p < 1/2, consider the graphs G = Kpn,(1−p)n and H = Kn/2,n/2, so e(G) = p(1 −p)n2 and e(H) = n2/4. The maximum overlap between G and H is obtained by putting the smaller vertex class from G on one side of H, while the minimum overlap is obtained by splitting it equally between the two sides. Thus max π |E(Gπ ∩E(H)| = pn · n 2 and min π |E(Gπ ∩E(H)| = 2pn 2 n 2 −pn 2 = p(1 −p)n2/2.
Since E|E(Gπ) ∩E(H)| = p(1 −p)n2 · n2 4 n 2 = 1 2p(1 −p)n3/(n −1) = p(1 −p)n2/2 + p(1 −p)n/2 + O(1), 20 we have disc+(G, H) ∼p2n2/2 and disc−(G, H) ∼p(1 −p)n/2, which gives disc+(G, H)disc−(G, H) ∼p3(1 −p)n3/4.
So with p ≤1/2, noting that (1 −p)3 ≥1/8, we obtain an upper bound of form cp3(1 −p)3.
A different upper bound comes from considering random graphs. Fixing p and q, let G ∈G(n, p) be a random graph and let H be a clique of order √qn (plus some isolated vertices), so e(H) ∼q n 2 . We shall use a version of Chernoff’s inequality (see Janson, Luczak and Rucinski ): if X ∼ B(N, p) and t > 0, then P(|X −Np| ≥t) ≤2e−t2/(2Np+2t/3). For a random permutation π, and N = √qn 2 ∼qn2/2, |E(Gπ) ∩E(H)| has distribution B(N, p). Since E |E(Gπ) ∩E(H)| = pN ∼pqn2/2, and there are n √qn = O n−1/2 √q √q(1 −√q)1−√q−n distinct ways for Gπ to overlap with H, we have P(disc(G, H) > λ) < 1/n provided n √qn P(|B(N, p) −Np| > λ) < 1 n.
(33) The left hand side of (33) is at most cn−1/2 √q √q(1 −√q)1−√q−n e−λ2/(2Np+2λ/3), which, assuming λ = o(n2) and q < 1/2, is smaller than 1/n provided λ2 > (2Np + 2λ/3) · n(−√q log(√q) −(1 −√q) log(1 −√q)) + O(log n) = Θ(pqn2 · n · √q log(1/√q)).
Since e(G) = p n 2 with probability at least 1/n (provided p n 2 is an inte-ger), we deduce that (for q ≤1/2) there is some G with p n 2 edges and disc+(G, H)disc−(G, H) < c′pq3/2 log(1/q)n3.
We can also try working with both G and H random graphs. However, there is an additional problem. When we want to bound disc(G) for a single 21 random graph G, and H is a clique plus isolated vertices, we only need to consider O(2n) ways of superimposing the two graphs (as H has a large automorphism group). But when considering a pair G, H of random graphs, there are n! ways of placing the two graphs together, leading to a bound of form c(p)n3/2√log n rather than c(p)n3/2 (we have to avoid n! bad events rather than 2n, so we want the probability of each bad event to be at most 1/n! rather than at most 2−n; the bound then follows by considering the tail of the binomial distribution). It would be interesting to know the correct value.
Problem 12. Let G and H be random graphs chosen independently from G(n, p). What is E disc(G, H)?
From the remarks above, it is clear that c(p)n3/2 < E disc(G, H) < c′(p)n3/2√log n. However, it is not obvious where in this range the correct value lies, and it is equally unclear when G and H are random graphs with different densities.
What about E disc+(G, H)disc−(G, H)? It turns out that this is closely related to disc(G, H).
Indeed, consider the edge-exposure martingale ob-tained by revealing the edges of G and H one at a time, and taking the con-ditional expectation of disc+(G, H). There are 2 n 2 edges, and differences are bounded by 1, so by the Azuma-Hoeffding inequality the probability we are more than λ from expectation is at most 2 exp(−λ2/2n2). It follows that the discrepancy is concentrated close to some value. If ω(n) →∞as n →∞ then, with high probability, |disc+(G, H)−E disc+(G, H)| ≤nω(n) and, with probability 1−O(1/n5), |disc+(G, H)−E disc+(G, H)| ≤n√10 log n. Apply-ing the same argument to disc−(G, H) and noting that disc(G, H) = O(n2), we see that E disc+(G, H)disc−(G, H) = (1 + o(1))(E disc(G, H))2. However, this does not help with finding the value of E disc(G, H).
It would also be interesting to look at graphs (and hypergraphs) with low discrepancy. Graphs G with density 1/2 and disc(G) = o(n2) are known to be quasi-random, and therefore share certain properties with random graphs (see Thomason , Chung, Graham and Wilson and Chung and Graham [6, 7]; for related results see Mubayi and R¨ odl ). What can we say about graphs for which disc+(G)disc−(G) is small? For instance, we know from that if G has density 1/2 then disc+(G)disc−(G) ≥cn3. This bound is achieved by random graphs in G(n, 1/2), as well as the complete bipartite 22 graph Kn/2,n/2 and its complement. Must graphs that are close to extremal look like these examples?
Problem 13. Let G be a graph with density 1/2 and disc+(G)disc−(G) = O(n3). Must G be close to one of Kn/2,n/2, 2Kn/2 or a quasi-random graph?
Of course, as stated the problem is a little imprecise, as a suitable mea-sure of “closeness” needs to be specified.
However, it seems likely that there should be some sort of “stability theory”, saying that graphs that are “close” to extremal should also be “close” to graphs of one of a few different types. We could also demand a very small one-sided discrepancy: what do graphs G with disc(G) = O(n3/2) look like? Is there a stability theory for these? Perhaps there is a small family of examples that, together with graphs satisfying strong quasi-random properties, essentially character-ize graphs with disc+(G)disc−(G) = O(n3). Or there may be a decomposi-tion of G into a small number of such graphs. The weaker condition that disc+(G)disc−(G) = o(n4) is also interesting in this context. Does this imply that the graph is in some sense quasi-random? The focus here is on graphs with small one-sided discrepancy: do graphs G with disc+(G) = o(n2) have nice properties like quasirandomness? These questions, and the ones above, also arise for graphs (and hypergraphs) with densities p ̸= 1/2. Of course, it could also be very interesting to understand the structure of pairs of graphs G and H for which disc(G, H) or disc+(G, H) is small.
In a similar vein, it would be interesting to understand the relationship between disc(G) and the eigenvalues of G. For instance, what bounds can be given for disc(G) in terms of the density and spectral gap of G?
6.3 Other norms Given graphs G and H, we can consider the vector vG,H = (⟨gπ, h⟩)π∈Sn ∈Rn!.
Then disc+(G, H) is the maximum entry of vG,H and disc−(G, H) is the minimum entry, while disc(G, H) = ||vG,H||∞.
It would be interesting to look at other norms. For instance, ||vG,H||1 = n! E|⟨gπ, h⟩|, 23 where π ∈Sn is chosen with the uniform distribution. How large is ||vG,H||1?
For random graphs it can be calculated precisely, but the extremal question seems to be less trivial.
Problem 14. Let G and H be graphs with e(G) = p n 2 and e(H) = q n 2 , and let π ∈Sn be chosen uniformly at random. What is the minimum possible value of E |E(Gπ) ∩E(H)| −pq n 2 ?
Similarly, what happens if we consider the l2 norm?
Another interesting problem is to find a fractional version of Theorem 1.
Suppose that G is a real-valued edge-weighting of Kn with total weight zero and ||g||1 = p n 2 . For a graph H with q n 2 edges, what can we say about disc(G, H)? What if H is also edge-weighted? Note that Theorem 4 goes through in this context; and perhaps there is a fractional version of Theorem 9. Of course, the same problem arises for pairs of hypergraphs (results for disc(H), where H is an edge-weighted hypergraph can be found in ).
6.4 Group actions and set systems Finally, let us note that these problems can be raised in the more general context of a group G acting transitively on a set X. Given A, B ⊂X, we define disc+ G(A, B) = max g∈G |g(A) ∩B| −|A| · |B| |X| and disc− G(A, B) = max g∈G |A| · |B| |X| −|g(A) ∩B| .
There are many interesting examples of this, and we would expect bounds in general to involve both the size of G and X and the number of automorphisms of A and B.
The problem can also be stated in the infinite context, with an appropriate probability measures on G and on X. For instance, given subsets A, B ⊂Sn, taking normalized Lebesgue measure on Sn and letting G be the group of isometries of Sn with normalized Haar measure, what can we say about disc+(A, B) and disc−(A, B)? If λ(A) = 1/2 or λB = 1/2, then we could have zero discrepancy: for instance, if one set is a half-sphere and the other 24 set is symmetric in the origin. Similarly, in S1 we can construct pairs of sets with λ(A) = r/s, any λ(B) and zero discrepancy by taking A to be an interval and demanding that B has a rotational symmetry of order s. More generally, we can generate pairs of sets in Sn with arbitrarily low discrepancy by taking A to be a spherical cap (or any reasonably nice set) and B to be a subset of spherical cap obtained by dividing it into sets of small diameter and throwing away half of each set. These examples suggest that the symmetry group of Sn is too small (or that the class of sets is too large) for us to generate atypical overlaps. However, it may be possible to obtain more interesting results if we restrict the sets A and B to be sufficiently “nice”.
Returning to the finite context, we consider actions of Sn. There appears to be a trade-offbetween discrepancy and the size of the set on which Sn acts: when Sn acts on the singletons in [n], we get the maximum possible discrepancy, as any two sets can be shifted to overlap as much or as little as possible. On the other hand, if we allow Sn to act on itself, we get very small discrepancy: let H be a subgroup of Sn and choose A to be a union of left cosets of H and B to be a set of left coset representatives. Then |π(A)∩B| takes the same value for every π ∈Sn. There is a similar problem when we allow Sn to act on the cube {0, 1}n by permuting coordinates. As in the case of Sn acting on itself, it is easy to construct pairs of sets with small discrepancy: for instance, let A be a face of the cube and let B be closed under pointwise complementation. However, the question becomes more challenging if we enlarge the group to the full isometry group Tn of the cube {0, 1}n, which has 2nn! elements.
Problem 15. For p ∈[0, 1], let Qp,n be the collection of subsets of {0, 1}n with p2n elements. What is min A∈Qp,n,B∈Qq,n discTn(A, B)?
Interesting questions arise if we consider the action of Sn on {0, 1}n by permuting coordinates, and restrict our sets to be upsets. By the Harris-Kleitman inequality , if A and B are upsets then |A ∩B| ≥|A| · |B| 2n .
How much better than this can we do if we can permute coordinates?
Problem 16. What is the minimum of maxπ∈Sn |π(A) ∩B| over upsets A, B ⊂{0, 1}n with |A| = p2n and |B| = q2n?
25 Could the pair (A, B) with A = {x ∈{0, 1}n : xi = 1 for i ≤k} and B = {x ∈{0, 1}n : P xi ≥l} be extremal for this problem?
The same question arises for the discrepancy.
Problem 17. What is the minimum of discSn(A, B) over upsets A, B ⊂ {0, 1}n with |A| = p2n and |B| = q2n?
In all the questions above, it is also natural to ask about bounds on the product form (32).
References J. Beck and V. T. S´ os, Discrepancy theory, in Handbook of Combina-torics, Vol. 2, 1405–1446, Elsevier, Amsterdam, 1995 B. Bollob´ as, Random Graphs, Second Edition, Cambridge Studies in Advanced Mathematics, Cambridge University Press, 2001, xviii+498pp B. Bollob´ as and A.D. Scott, Discrepancy in graphs and hypergraphs, in More sets, graphs and numbers, Ervin Gyori, Gyula O.H. Katona and Laszlo Lov´ asz, eds, pp. 33–56, Bolyai Soc. Math. Stud. 15, Springer, Berlin, 2006.
B. Chazelle, The discrepancy method, Cambridge University Press, Cam-bridge, 2000, xviii+463 pp.
F. R. K. Chung, R. L. Graham and R.M. Wilson, Quasi-random graphs, Combinatorica 9 (1989), 345–362 F. R. K. Chung and R. L. Graham, On graphs not containing prescribed induced subgraphs, in A Tribute to Paul Erd˝ os, A. Baker, B. Bollob´ as and A. Hajnal, eds, pp. 111–120, Cambridge Univ. Press, Cambridge, 1990.
F. R. K. Chung and R. L. Graham, Quasi-random set systems, J. Amer.Math. Soc. 4 (1991), 151–196 P. Erd˝ os, R. Faudree, C. Rousseau and R. Schelp, A local density con-dition for triangles, Discrete Math. 127 (1994), 153–161 26 P. Erd˝ os, M. Goldberg, J. Pach and J. Spencer, Cutting a graph into two dissimilar halves, J. Graph Theory 12 (1988), 121–131 P. Erd˝ os and J. Spencer, Imbalances in k-colorations, Networks 1 (1971/2), 379–385 T. E. Harris, A lower bound for the critical probability in a certain percolation process, Proc. Cambridge Philos. Soc. 56 (1960), 13–20 S. Janson, T. Luczak and A. Rucinski, Random graphs, Wiley Inter-science, New York, 2000. xii+333 pp.
P. Keevash and B. Sudakov, Local density in graphs with forbidden subgraphs, Combinatorics, Probability and Computing 12 (2003), 139– 153 M. Krivelevich, On the edge distribution in triangle-free graphs, J. Com-binatorial Theory, Ser. B 63 (1995), 245–260 J. Matouˇ sek, Geometric discrepancy, Algorithms and Combinatorics 18, Springer-Verlag, Berlin, 1999, xii+288 pp.
D. Mubayi and V. R¨ odl, Uniform edge distribution in hypergraphs is hereditary, Electronic Journal of Combinatorics 11 (2004), Paper R55 V. T. S´ os, Irregularities of partitions: Ramsey theory, uniform distri-bution, in Surveys in Combinatorics (Southampton, 1983), 201–246, London Math. Soc. Lecture Note Ser., 82, Cambridge Univ. Press, Cambridge-New York, 1983 A. Thomason, Pseudo-random graphs, in Random Graphs ’85, Ann. Dis-crete Maths 33 (1987), 307–331 27 |
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上海虹桥火车站站台&列车候车室分布
2014-09-12 11:10:00
上海虹桥站共有30个站台，出口处统一在地下一层，分为南、北两向。
上海虹桥站，于2010年7月1日启用，是一座高度现代化的中国铁路客运车站，为亚洲最大铁路枢纽之一。该站是沪杭高铁、沪宁高铁与京沪高铁的交汇点，位于上海市偏西部，东邻上海虹桥国际机场T2航站楼，是上海虹桥综合交通枢纽的重要组成部分。
上海虹桥站总占地面积超过130万平方米，位于上海虹桥机场西侧，站房总建筑面积约23万平方米，其中铁路站房约10万平方米，雨棚面积约11万平方米，立体共分5层。北端引接京沪高铁、京沪铁路、沪宁城际;南端与沪昆铁路、沪杭甬客专、沪杭城际接轨。与运营中的上海站、上海南站实现明确的功能划分，上海虹桥站全面开行CRH动车组列车，而上海站则主要为动车、普速列车，上海南站主要为普速列车。上海虹桥站的车场总规模为16台30线，其中高速铁路场10台19线，城际普速场6台11线。
1、站台：
上海虹桥站共有30个站台，出口处统一在地下一层，分为南、北两向。站台两端有两条出站地下通道，乘车旅客下车后无论从哪条通道，都可以到达出口处，通道上方有标识指明“到达南”和“到达北”。
接站的市民也不用担心会与亲友错过，因为每座站台都会对应一个出站区域，具体是：
1—7号站台对应到达南(北)1—4号口;
8—19号站台对应到达南(北)5—10号口;
20—30号站台对应到达南(北)11—15号口
接站口的玻璃门上都有标识，双方只要约定是从南或北地道出来就可以了。
上海虹桥火车站站台平面图（黄色区域为各 站台所在）：
上海虹桥 火车站站台效果 图：
上海虹桥 火车站站台实景 图：
2、候车大厅：
上海虹桥火车站现有1个候车大厅，位于2层出发层，此候车大厅最多可同时容纳1万人候车。
上海虹桥 火车站候车大厅 图：
上海虹桥火车站候车实景 图：
3、售票处：
虹桥火车站非常大，售票处很多。亲们只要看虹桥火车站内的导航标志就能找到，有人工售票和自助售票哟
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5167 | https://www.escardio.org/Journals/E-Journal-of-Cardiology-Practice/Volume-15/Post-cardiac-injury-syndrome-aetiology-diagnosis-and-treatment | Post-cardiac injury syndrome: aetiology, diagnosis, and treatment
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E-Journal of Cardiology Practice - Volume 15
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Post-cardiac injury syndrome: aetiology, diagnosis, and treatment
Vol. 15,
N° 21
-
31 Oct 2017
Dr. Tom Sasse
Prof. Urs Eriksson ,
FESC FHFA
Post-cardiac injury syndrome (PCIS) represents a leading cause of pericarditis. Any invasive cardiac procedure may lead to myocardial, epicardial and pericardial trauma. Tissue damage results in accumulation of debris and blood in the pericardium. This results in immune responses, which maintain inflammation in susceptible individuals. Clinical presentation is variable, and recurrent episodes enhance morbidity after surgery. Colchicine prevents PCIS at the cost of adverse events. Mainstays of therapy are aspirin and ibuprofen. Steroids are used in severe cases. Colchicine and immunotherapy are effective in idiopathic pericarditis, but their benefit in treatment of PCIS is less evident.
Pericardial Disease
Definition
Post-cardiac injury syndrome (PCIS) refers to an aetiologic heterogenous group of autoimmune-mediated conditions of pericardial, epicardial, and myocardial inflammation. It is important to realise that inflammation is not confined to the pericardium and epicardium but also affects the myocardium to some extent. Nevertheless, symptoms of epicardial and pericardial inflammation, such as pain, effusions, and fever, dominate the clinical picture. Signs of ventricular dysfunction such as heart failure, ventricular arrhythmias or severe conduction abnormalities are very rare. If present, they rather reflect worsening of pre-existing heart failure by a second systemic inflammatory hit. By definition, post-cardiac injury syndromes include: post-myocardial infarction pericarditis triggered by ischaemic myocardial necrosis, post-pericardiotomy syndrome (PPS) after surgical trauma and, finally, post-traumatic pericarditis due to either iatrogenic or accidental injury (Figure 1).
Figure 1. Post-cardiac injury syndrome and its subentities.
Epidemiology and clinical picture
A prospective cohort study of 1,162 cases identified post-cardiac injury as the leading cause of pericarditis, accounting for 21% of patients .
Post-myocardial infarction pericarditis can be differentiated into early infarct-associated pericarditis and late post-myocardial infarction pericarditis, also known as Dressler’s syndrome. The former occurs within five days after myocardial infarction. With an incidence of <5% it has become quite rare in developed countries. A major risk factor is delayed angioplasty. The risk of recurrence however is very low. Dressler’s syndrome, on the other hand, occurs two to eight weeks after infarction and is also very rare nowadays in clinical settings. Prior to the reperfusion era it had been reported with an incidence of 5%. Newer studies have reported the de facto disappearance of Dressler’s syndrome, a fact that is not yet fully understood. On the one hand, this observation might reflect under-reporting of subclinical cases. However, diminution of infarct size by means of early reperfusion and immune-modulatory effects of standard-of-care drugs, such as ACE inhibitors, some beta-blockers, statins, and aspirin, mainly explain these observations. Importantly, and in marked contrast to early infarct-associated pericarditis, Dressler’s syndrome implies a relevant risk of recurrence. Nevertheless, complications such as late cardiac tamponade and constrictive pericarditis are rare for early infarct-associated pericarditis as well as for Dressler’s syndrome [3,4].
Post-pericardiotomy syndrome (PPS) has been found to be a relatively frequent complication of cardiac surgery with an incidence of 10-40% in patients undergoing cardiac operation. This wide range of incidence largely reflects inhomogeneous diagnostic criteria. PPS classically occurs days to several weeks after cardiac surgery [5–7]. It is of importance to distinguish PPS from uncomplicated postoperative effusion, which affects up to 85% of patients undergoing cardiac surgery. Uncomplicated postoperative effusions lack signs of systemic inflammation .
Last, but not least, post-traumatic pericarditis spans entities ranging from accidental blunt or sharp cardiac trauma to a wide variety of iatrogenic trauma. Accidental trauma-associated pericarditis is either rare or – more likely – under-reported. Accordingly, there are no systematic and valid data available in this context. Pericarditis due to iatrogenic trauma, on the other hand, has an incidence of 0.5% to 5% depending on the interventional trigger. Percutaneous coronary interventions, pacemaker lead insertions, radiofrequency ablation and even Swan-Ganz catheterisation can predispose to myocardial injury of various degrees and may culminate in post-traumatic pericarditis [9-11].
Diagnostic criteria
A major drawback in the study, understanding and treatment of PCIS is the lack of reliable and generally accepted diagnostic criteria defining the disease entity with appropriate specificity. This is largely due to its clinical and aetiological heterogeneity and the absence of a single consistent pathognomonic finding. Patients suffering from PCIS frequently seek medical assistance because of chest pain (>80%), low grade fever (50-60%) and dyspnoea (50-60%). Clinical evaluation usually reveals mild to moderate effusions both in the pericardium (>80%) and in the pleural space (>60%), and sometimes pericardial friction rubs (30-60%) on auscultation. Laboratory analysis shows systemic inflammation with elevation of C-reactive protein (CRP) (74%), and elevated blood leucocytes. ECG can provide evidence of typical alterations associated with pericarditis, such as widespread concave ST-segment elevations and PR depressions in multiple leads (except for aVR and sometimes V1, where ST-segment depressions and PR elevations are predominant). Identification and interpretation of the aforementioned findings is, however, often complicated by the underlying pathological condition which triggered PCIS. A diagnosis of PCIS may be considered in patients with chest pain, fever, pericardial or pleural effusions, and a re-emerging systemic inflammatory response syndrome after an appropriate latency period following prior myopericardial or pleural injury. Importantly, alternative causes for inflammation and effusion need to be ruled out before a diagnosis of PCIS can be established . Diagnostic criteria for PPS validated in three large clinical trials encompass five features, two of which have to be met in order to diagnose PPS: fever without alternative causes, pleuritic chest pain, friction rub and evidence of new or worsening pleural or pericardial effusion [5-7, 12].
Aetiology
Damage to pericardial or pleural mesothelial cells in combination with blood entering the pericardial space represents the initial trigger of PCIS. There is evidence for a correlation between elevated antimyocardial antibody titers and the incidence of PCIS after cardiac surgery. Patients undergoing open heart surgery in a small prospective study were tested for baseline and follow-up anti-myosin antibody titers. PPS occurred only in patients with detectable anti-myosin antibody titers and positively correlated with its severity. Based on these observations and insights from animal models, we therefore hypothesise that systemic release of cardiac antigens, together with non-specific, injury-mediated activation of the innate immune system initiate adaptive self-antigen specific responses, as evidenced by an increase of antimyocardial antibody titers and expansion of heart-specific CD4 T cells, which in turn promote recruitment of myelo-derived inflammatory cells to the heart. It is important to realise in this context that the epicardium, pericardium and pleura are much more susceptible to the accumulation of inflammatory cells than the cardiac interstitium, even in the presence of cardiomyocyte specific auto-immune responses. Taken together, the latency period between cardiac injury and PCIS, as well as the generally favourable response to anti-inflammatory drugs, both strongly support the hypothesis of an autoimmune-mediated pathogenesis [1,13,14].
Prevention
Prevention is not an issue in post-myocardial infarction pericarditis and post-cardiac injury pericarditis due to its low incidence. PPS, however, markedly contributes to morbidity in patients after cardiac surgery. Accordingly, most preventive studies have been conducted in the setting of PPS. Meta-analytic pooling showed that perioperative colchicine decreased the risk of PPS (OR 0.38, 0.22-0.65). Administration of methylprednisolone (OR 1.13, 0.57-2.25) and aspirin (OR 1.00, 0.16-6.11), however, showed no benefit [15,16].
The efficacy of colchicine as a potent preventive drug against PPS was confirmed in the randomised, double-blind, multicentre COPPS trial. Patients were randomised on the third postoperative day to receive either placebo or colchicine for one month. An initial loading dose of 1.0 mg twice daily was administered on the first day followed by a maintenance dose of 0.5 mg twice daily with a weight-adjusted dose reduction of 50% for patients weighing below 70 kg. The study showed superiority of colchicine vs. placebo with a number needed to treat (NNT) of 8 (95% CI: 27.3-75.6). Subsequently, the COPPS-2 trial was conducted to gain additional information on efficacy and safety. There was a slight change in the regimen in terms of already starting colchicine administration two to three days before surgery without a loading dose. The COPPS-2 trial strengthened the findings of its predecessor with a NNT of 10. However, an increase of adverse gastrointestinal effects was documented, with a number needed to harm (NNH) of 12. Observed adverse events primarily consisted of gastrointestinal intolerance without serious complications [6,7].
Moreover, a retrospective, single-centre study suggested that in-hospital administration of diclofenac after cardiac surgery protects from PPS (OR 0.34, 0.18-0.65) .
Treatment
Mainstays of treatment are anti-inflammatory NSAID and the adjunct use of colchicine, albeit that there are no randomised controlled trials addressing specific treatment approaches for PCIS. Treatment decisions must be taken with care given the potential side effects and risks associated with NSAID and colchicine in cardiac and post-surgery patients. Simple postoperative pericardial effusions without evidence for systemic inflammation do not warrant treatment .
Aspirin is the first choice for patients already under antiplatelet therapy. Dosages vary between 750 mg and 1,000 mg every six to eight hours. Known coronary artery disease strongly supports the use of aspirin in such patients. A treatment period of one to two weeks before tapering is usually sufficient. Recurrent symptoms demand prolonged treatment over two to four weeks. Alternatively, ibuprofen 600 mg every eight hours may be prescribed over one to two weeks in case of initial diagnosis or over two to four weeks in case of recurrence. However, it is important to be aware that ibuprofen interferes with the antiplatelet aggregation effects of aspirin in patients after coronary bypass grafting or PCI. Treatment with NSAID is continued until symptoms resolve and CRP normalises followed by subsequent tapering. Careful monitoring of renal and cardiac function is essential. We strongly suggest administration of proton pump inhibitors for adverse gastrointestinal effects prevention. NSAID should be avoided or at least closely monitored in patients with impaired kidney function, volume depletion or known heart failure.
Colchicine has been proven effective for the reduction of recurrences in pericarditis. However, there are insufficient data on its use in PCIS treatment. One retrospective, single-centre study showed a reduction of procedural intervention if PPS was treated with colchicine in combination with an anti-inflammatory drug as compared to anti-inflammatory drugs only (OR 0.43, 0.95-0.99). This study, however, was not designed to evaluate specifically the efficacy of colchicine in PPS or PCIS . Nevertheless, the 2015 European Society of Cardiology guidelines endorsed administration of colchicine in PCIS. Dosing is the same as in PPS prevention with 0.5 mg twice a day and a dose reduction of 50% for patients weighing below 70 kg. Patients should be monitored for creatinine elevation and blood panel changes at two- to three-week intervals. Moreover, we instruct all our patients to be aware of diarrhoea and nausea heralding colchicine overdose.
Glucocorticoid administration is required for a minority of patients with contraindications for aspirin and NSAIDs or in case of refractory symptoms. Data for PCIS are lacking, but best clinical practice suggests similar dosing to that administered in pericarditis, e.g., 0.25 to 0.5 mg/kg/d. If steroids are used, it is of the upmost importance to taper dosages very slowly to avoid recurrences. In patients treated with steroids and NSAID/colchicine, steroids must be tapered before colchicine is stopped.
Recurrence of PCIS has been reported to be lower than recurrences of idiopathic pericarditis, ranging between 10% and 15% . In contrast to idiopathic pericarditis, however, the risk of pericardial constriction is substantial for PCIS, approximating 2.8% over 72 months. Therefore, affected patients need long-term follow-up . Our institutional policy recommends clinical follow-up examination and echocardiography one and three years after PCIS.
In our opinion, the best standard of care includes: colchicine for prevention of PPS, aspirin in combination with colchicine for initial treatment, and glucocorticoids for recurrence or contraindications to aspirin. Proton pump inhibitors are required in combination with aspirin (Table 1).
Table 1. Treatment recommendations.
| | |
--- |
| PPS prevention | Colchicine 0.5 mg b.i.d., 3 days before and 1 month after surgery |
| PCIS treatment | Aspirin 750 mg t.i.d. for 2 weeks + Colchicine 0.5 mg b.i.d. for 6 months + Proton pump inhibitor |
| PCIS recurrence | Prednisolone 0.5 mg/kg o.d. for 1 month, subsequent very slow tapering + Aspirin 750 mg t.i.d. for 2 weeks + Colchicine 0.5 mg b.i.d. for 6 months + Proton pump inhibitor Do not stop colchicine before steroids! |
| | |
--- |
| PPS prevention | <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function |
| PCIS treatment | Contraindications to NSAID: no aspirin; use prednisolone instead: 0.5 mg/kg o.d. for 1 month, subsequent slow tapering Adverse effects of aspirin: consider reduction to 500 mg t.i.d. <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function |
| PCIS recurrence | Contraindications to NSAID: no aspirin; consider azathioprine 2 mg/kg o.d. together with steroids Adverse effects of aspirin: consider reduction to 500 mg t.i.d. <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function |
| PPS prevention | PCIS treatment | PCIS recurrence |
---
| Colchicine 0.5 mg b.i.d., 3 days before and 1 month after surgery | Aspirin 750 mg t.i.d. for 2 weeks + Colchicine 0.5 mg b.i.d. for 6 months + Proton pump inhibitor | Prednisolone 0.5 mg/kg o.d. for 1 month, subsequent very slow tapering + Aspirin 750 mg t.i.d. for 2 weeks + Colchicine 0.5 mg b.i.d. for 6 months + Proton pump inhibitor Do not stop colchicine before steroids! |
| <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function | Contraindications to NSAID: no aspirin; use prednisolone instead: 0.5 mg/kg o.d. for 1 month, subsequent slow tapering Adverse effects of aspirin: consider reduction to 500 mg t.i.d. <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function | Contraindications to NSAID: no aspirin; consider azathioprine 2 mg/kg o.d. together with steroids Adverse effects of aspirin: consider reduction to 500 mg t.i.d. <70 kg: reduce colchicine to 0.5 mg o.d. GFR <30 ml/min or unstable kidney function: no colchicine GFR 30-60 ml/min: adjust colchicine to kidney function |
References
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Imazio M, Negro A, Belli R, Beqaraj F, Forno D, Giammaria M, Trinchero R, Adler Y, Spodick D. Frequency and prognostic significance of pericarditis following acute myocardial infarction treated by primary percutaneous coronary intervention. Am J Cardiol. 2009 Jun 1;103(11):1525–9.
Shahar A, Hod H, Barabash GM, Kaplinsky E, Motro M. Disappearance of a syndrome: Dressler's syndrome in the era of thrombolysis. Cardiology. 1994;85(3-4):255-8.
Finkelstein Y, Shemesh J, Mahlab K, Abramov D, Bar-El Y, Sagie A, Sharoni E, Sahar G, Smolinsky AK, Schechter T, Vidne BA, Adler Y. Colchicine for the prevention of postpericardiotomy syndrome. Herz. 2002 Dec;27(8):791-4.
Imazio M, Trinchero R, Brucato A, Rovere ME, Gandino A, Cemin R, Ferrua S, Maestroni S, Zingarelli E, Barosi A, Simon C, Sansone F, Patrini D, Vitali E, Ferrazzi P, Spodick DH, Adler Y; COPPS Investigators. COlchicine for the Prevention of the Post-pericardiotomy Syndrome (COPPS): a multicentre, randomized, double-blind, placebo-controlled trial. Eur Heart J. 2010 Nov;31(22):2749-54.
Imazio M, Brucato A, Ferrazzi P, Pullara A, Adler Y, Barosi A, Caforio AL, Cemin R, Chirillo F, Comoglio C, Cugola D, Cumetti D, Dyrda O, Ferrua S, Finkelstein Y, Flocco R, Gandino A, Hoit B, Innocente F, Maestroni S, Musumeci F, Oh J, Pergolini A, Polizzi V, Ristic A, Simon C, Spodick DH, Tarzia V, Trimboli S, Valenti A, Belli R, Gaita F; COPPS-2 Investigators. Colchicine for Prevention of postpericardiotomy syndrome and postoperative atrial fibrillation: the COPPS-2 randomized clinical trial. JAMA. 2014 Sep 10;312(10):1016-23.
Weitzman LB, Tinker WP, Kronzon I, Cohen ML, Glassman E, Spencer FC. The incidence and natural history of pericardial elfusion after cardiac surgery - an echocardiographic study. Circulation. 1984 Mar;69(3):506-11.
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Nomura Y, Yoshinaga M, Haraguchi T, Oku S, Noda T, Miyata K, Umebayashi Y, Taira A. Relationship between the degree of injury at operation and the change in antimyosin antibody titer in the postpericardiotomy syndrome. Pediatr Cardiol. 1994 May-Jun;15(3):116-20.
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Bunge JJ, van Osch D, Dieleman JM, Jacob KA, Kluin J, van Dijk D, Nathoe HM; Dexamethasone for Cardiac Surgery (DECS) Study Group. Dexamethasone for the prevention of postpericardiotomy syndrome: A DExamethasone for Cardiac Surgery substudy. Am Heart J. 2014 Jul;168(1):126-31.
Sevuk U, Baysal E, Altindag R, Yaylak B, Adiyaman MS, Ay N, Alp V, Beyazıt U. Role of diclofenac in the prevention of postpericardiotomy syndrome after cardiac surgery. Vasc Health Risk Manag. 2015 Jun 30;11:373-8.
Meurin P, Lelay-Kubas S, Pierre B, Pereira H, Pavy B, Iliou MC, Bussière JL, Weber H, Beugin JP, Farrokhi T, Bellemain-Appaix A, Briota L, Tabet JY; French Society of Cardiology. Colchicine for postoperative pericardial effusion: a multicentre, double-blind, randomised controlled trial. Heart. 2015 Nov;101(21):1711-6.
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Imazio M, Brucato A, Maestroni S, Cumetti D, Belli R, Trinchero R, Adler Y. Risk of constrictive pericarditis after acute pericarditis..
Notes to editor
Authors:
Tom Sasse1, MD; Urs Eriksson1,2, MD, MSc-ETH, FESC, FHFA
Department of Medicine, GZO Regional Health Center, Wetzikon, Switzerland;
2.Center for Molecular Cardiology, University of Zurich, Schlieren, Switzerland
Address for correspondence:
Prof. Urs Eriksson, MD, MSc, FESC, FHFA
GZO- Regional Health Center
Spitalstrasse 66
CH-8620 Wetzikon
Switzerland
E-mail: urs.eriksson@uzh.ch
Author disclosures:
The authors report no conflicts of interest with regard to this paper.
The content of this article reflects the personal opinion of the author/s and is not necessarily the official position of the European Society of Cardiology.
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5168 | https://www.montgomeryschoolsmd.org/siteassets/schools/middle-schools/a-f/bannekerms/uploadedfiles/departments/math/unit20120topic20220flexbook.pdf | Unit 1 Topic 2: Linear Inequalities in One Variable.
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Printed: August 13, 2014 AUTHOR Algebra 1 Summit iii Contents www.ck12.org Contents 1 Unit 1 Topic 2: Linear Inequalities in One Varaible 1 iv www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible CONCEPT 1 Unit 1 Topic 2: Linear Inequalities in One Varaible Dear Parents & Guardians, Unit 1 Topic 2 focuses on writing and solving inequalities. Students develop fluency and mastery writing, interpret-ing, and translating between various forms of linear equations and inequalities in one variable, and using them to solve problems. This topic includes: • Writing inequalities to represent a situation • Solving inequalities in one variable • Justifying solution methods Like equations, inequalities show a relationship between two expressions. We solve and graph inequalities in a similar way to equations but an inequality is usually a set of values instead of just one value.
When writing inequalities we use the following symbols: > is greater than ≥i s greater than or equal to < is less than ≤is less than or equal to TABLE 1.1: Properties of Inequalities Property Rule Example Addition Property If x < y, then x+z < y+z if 3 < 6, then 3+5 < 6+5 8 < 11 is still true Subtraction Property If x < y, then x−z < y−z If 8 < 12, then 8−5 < 12−5 3 < 7 is still true Multiplication Property If x < y, and z > 0 then x∗z < y∗z If 12 < 15, and z = 5 then 12 ∗5 < 15∗5 60 < 75 is still true Multiplication Property with a neg-ative number If x < y, and z < 0 then x∗z > y∗z If 5 < 10, and z = −3 then 5∗−3 > 10∗−3 since 5 and 10 were multiplied by a negative, for the expression to remain true the sign must flip -15 > -30 is true that the sign flipped Division Property If x < y, and z > 0 then x÷z < y÷z If 4 < 8, and z = 2 then 4÷2 < 8÷2 2 < 4 stays true Division Property with a negative number If x < y, and z < 0 then x÷z > y÷z If 6 < 12, and z = −3 then 6÷−3 > 12÷−3 since 6, and 12 were divided by a negative, for the equation to remain true the sign must flip -2 > -4 is true that the sign flipped 1 www.ck12.org TABLE 1.1: (continued) Transitive Property If x > y and y > z, then x > z If 9 > 6 and 6 > 2, then 9 > 2 2 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible Write and Graph Inequalities in One Variable on a Number Line Let’s start with the simple inequality x > 3.
We read this inequality as "x is greater than 3". The solution is the set of all real numbers that are greater than three.
We often represent the solution set of an inequality with a number line graph. The reason the dot on three is open is that the inequality reads x is greater than three, therefore three is not included in the solution set.
Consider another simple inequality: x ≤4.
We read this inequality as "x is less than or equal to 4". The solution is the set of all real numbers that are equal to four or less than four. We can graph this solution set on a number line. The reason we used a closed dot on x is less than or equal to four, is that four is part of the solution set.
Examples Graph the following inequalities on a number line.
a) x < −3 b) x ≥6 c) x > 0 d) x ≤8 Solutions: a) The inequality x < −3 represents all numbers that are less than -3. The number -3 is not included in the solution, so it is represented by an open circle on the graph.
License: CC BY-NC 3.0 [Figure3] b) The inequality x ≥6 represents all numbers that are greater than or equal to 6. The number 6 is included in the solution, so it is represented by a closed circle on the graph.
License: CC BY-NC 3.0 [Figure4] 3 www.ck12.org c) The inequality x > 0 represents all numbers that are greater than 0. The number 0 is not included in the solution, so it is represented by an open circle on the graph.
License: CC BY-NC 3.0 [Figure5] d) The inequality x ≤8 represents all numbers that are less than or equal to 8. The number 8 is included in the solution, so it is represented by a closed circle on the graph.
FIGURE 1.1 License: CC BY-NC 3.0 [Figure6] Example Write the inequality that is represented by each graph.
a) FIGURE 1.2 License: CC BY-NC 3.0 [Figure7] b) FIGURE 1.3 License: CC BY-NC 3.0 [Figure8] c) FIGURE 1.4 License: CC BY-NC 3.0 [Figure9] d) FIGURE 1.5 License: CC BY-NC 3.0 [Figure10] 4 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible Solutions: a) x ≤−12 b) x > 540 (Since the open circle is not on an exact number, give a reasonable estimate) c) x < 6.5 d) x ≥85 5 www.ck12.org Inequalities appear regularly in real life.
Here are some common words or phrases that can be represented by inequalities. Here are some examples that are commonly used in real-world applications.
TABLE 1.2: > > < < more than minimum smaller than maximum greater than at least less than at most larger than no less than below no more than above greater than or equal to up to can not exceed exceeds fewer than less than or equal to Example Write each statement as an inequality and graph it on the number line.
a) You must be at least 48 inches tall to ride the Thunderbolt Roller coaster.
b) The speed limit on the interstate is 65 miles per hour or less.
Solutions: [Figure11] a) The words "at least" imply that the value of 48 inches is included in the solution set, so the inequality is written ash ≥48, where h is the height.
In the context of this problem, it is not realisitic that someone is 100 inches tall. This would be considered a non-viable solution. Viable solutions include 48 inches, 49 inches, etc.
License: CC BY-NC 3.0 [Figure12] b) Speed limit means the highest allowable speed, so the inequality is written as s ≤65, where s is the speed limit.
Any value below zero would not be realistic in this situation and would be considered a non-viable solution. Viable solutions include any speed between zero and 65.
License: CC BY-NC 3.0 [Figure14] Solving Inequalities Using Multiplication and Division We can also solve inequalities by multiplying or dividing both sides by a constant. For example, to solve the inequality 5x < 3, we would divide both sides by 5 to get x < 3 5.
However, something different happens when we multiply or divide by a negative number. We know, for example, that 5 is greater than 3. But if we multiply both sides of the inequality 5 > 3 by -2, we get −10 > −6. And we know thats not true; -10 is less than -6.
6 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible This happens whenever we multiply or divide an inequality by a negative number. We have to flip the sign around to make the inequality true. For example, to multiply 2 < 4 by -3, first we multiply the 2 and the 4 each by -3, and then we change the < sign to a > sign, so we end up with −6 > −12.
This is the same principle applies when solving an inequality in one variable.
7 www.ck12.org Example Solve each inequality.
a) 4x < 24 b) −5x ≤21 c) x 25 < 3 2 d) x −7 ≥9 Solutions: a) TABLE 1.3: 4x < 24 Original Problem 4x 4 < 24 4 Divide both sides by 4 Division Property of Inequalities (Dividing by a positive number does not affect the inequality sign) x < 6 Simplify b) TABLE 1.4: −5x ≤21 Original problem −5x −5 ≥21 −5 Divide both sides by -5 and f lip the inequality sign Division Property of Inequalities (Dividing by a negative number af-fects the inequality sign) x ≥−21 5 Simplify c) TABLE 1.5: x 25 < 3 2 Original problem 25· x 25 < 3 2 ·25 Multiply both sides by 25 Multiplication Property of In-equalities (multiplying by a positive number does not affect the inequality sign) x < 75 2 or x < 37.5 Simplify 8 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible d) TABLE 1.6: x −7 ≥9 Original problem: −7· x −7 ≤9·(−7) Multiply both sides by -7 and flip the inequality sign.
Multiplication Property of In-equalities (multiplying by a positive number affects the inequality sign) x ≤−63 Simplify Solving Multi-Step Inequalities In the last two sections, we studied inequalities which required only one step to obtain the solution. However, solving most inequalities requires several steps. As with solving equations, we must use the order of operations to find the correct solution. In addition, remember that when we multiply or divide the inequality by a negative number, the direction of the inequality sign changes.
The general procedure for solving multi-step inequalities is almost exactly like the procedure for solving multi-step equations: Example Solve each of the following inequalities and graph the solution set.
a) 9x 5 −7 ≥−3x+12 b) −25x+12 ≤−10x−12 Solution a) TABLE 1.7: 9x 5 −7 ≥−3x+12 Original problem 9x 5 +3x−7 ≥−3x+3x+12 Add 3x to both sides Addition Property of Inequalities 24x 5 −7 ≥12 Simplify 24x 5 −7+7 ≥12+7 Add 7 to both sides Addition Property of Inequalities 24x 5 ≥19 Simplify 5· 24x 5 ≥5·19 Multiply 5 to both sides Multiplication Property of In-equalities 24x ≥95 Simplify 24x 24 ≥95 24 Divide both sides by 24 Division Property of Inequalities x ≥95 24 Simplify x ≥323 24 or 3.96 Solution Graph: 9 www.ck12.org License: CC BY-NC 3.0 [Figure19] 10 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible b) TABLE 1.8: −25x+12 ≤−10x−12 Original problem −25x+10x+12 ≤−10x+10x−12 Add 10x to both sides Addition Property of Inequalities −15x+12 ≤−12 Simplify −15x+12−12 ≤−12−12 Subtract 12 from both sides Subtraction Property of Inequal-ities −15x ≤−24 Simplify −15x −15 ≥−24 −15 Divide both sides by -15 and flip the inequality sign Division Property of Inequalities x ≥8 5 Simplify x ≥13 5 or 1.6 solution Graph: License: CC BY-NC 3.0 [Figure20] 11 www.ck12.org Writing Inequalities to Represent a Situation Instructional video: Example Dita has a budget of $350 to spend on a rental car for an upcoming trip, but she wants to spend as little of that money as possible. If the trip will last five days, what daily rental rates should she be willing to consider? Let r represent the daily rental rate.
Inequality to represent the situation: 5r ≤350 5 is the number of days r represents the daily rental rates ≤less than or equal to. She can’t spend more than her budget allows but she can spend less than or equal to the amount.
$350 is her maximum budget for the rental car Solution: r ≤70 Dita can consider any rental rate up to $70 per day to stay within her budget.
12 www.ck12.org Concept 1.
Unit 1 Topic 2: Linear Inequalities in One Varaible Example Maria is two years more than three times her younger brother’s age. If the sum of their ages is at most 18, then find: (Let y represent the age of her younger brother) 1. the greatest age that Maria’s brother could be.
2. an inequality that represents all possible values of Maria’s age.
What do we know?
Maria = 2 + 3y (two years more than three times her younger brother’s age) Brother = y (her brother’s age is an unknown value) The sum of their ages (Maria and her brother) is 2+3y + y Combine like terms to get the inequality 4y + 2 ≤18 Maria is two years more than three times her younger brother’s age. If the sum of their ages is at most 18, then find: 1. the greatest age that Maria’s brother could be. y ≤4 Maria’s brother could be no more than 4 years old, by solving the above inequality for y, Maria’s brother could be at most 4 years old.
2. An inequality that represents all possible values of Maria’s age. 3y + 2 ≤14 assume her brother is his maximum age, she can be up to 14 years old Example Kevin is saving money for college. He will graduate high school in three years. He currently has $5,000 saved. How much money does he need to save each month if he wants to save at least $7,000? Create and solve an inequality to determine the answer.
Solution: 5,000 + 36m ≤7,000 5,000 is the amount he has already saved + add because he saving more money to add to his current savings 36 is the number of months in 3 years (3 years x 12 months) m represents the monthly savings Solving Inequalities MEDIA Click image to the left for more content.
13 |
5169 | https://app.oncoursesystems.com/curriculum/11730/14382773/unit/14382922 | Unit 2 - Decimals
Content Area: Mathematics
Course(s): Mathematics
Time Period: Week 7
Length: 3 Weeks
Status: Published
Unit Overview
Students will begin this unit by completing operations with multi-digit numbers, with special attention on dividing multi-digit numbers. This will prepare them for operations with decimals with multiple digits. The purpose of this unit is to have students add, subtract, multiply, and divide decimals. Students will use estimation as well as the actual procedures to complete these operations. Along with operations on decimals, students will use their knowledge of place values to round, compare, and order decimals as a review of the previous year. Students will then apply their skills on operating with decimals in real world problems, such as using decimal multiplication to find the percent of a number.
Standards
MA.6.RP.A.3c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
MA.5.NBT.A.3 Read, write, and compare decimals to thousandths.
MA.5.NBT.A.4 Use place value understanding to round decimals to any place.
MA.5.NBT.B Perform operations with multi-digit whole numbers and with decimals to hundredths.
MA.6.NS.B Compute fluently with multi-digit numbers and find common factors and multiples.
MA.6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm.
MA.6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
Essential Questions
• How can estimating before computing with decimals help?
• How do operations with decimals compare with operations with whole numbers?
• How do place values play a role in each different operation of decimals?
• How can you use decimals in real life?
Application of Knowledge: Students will know that...
• multiplication of decimals can be used to find the percent of a number. .
• the divisor always has to be a whole number. .• to get an exact answer when dividing, zeros can be added to the dividend until there is no remainder left.
.
•to place the decimal point when multiplying decimals, find the sum of the number of decimal places in each factor.
.
•when adding or subtracting decimals, the decimal point must be lined up so that the place values will be in line.
.
•when adding or subtracting decimals, zeros are added as needed to ensure the numbers have the same number of decimal places.
.
Application of Skills: Students will be able to...
•add and subtract decimals. .
•divide decimals. .
•divide multi-digit numbers. .
•find the percent of a number by using multiplication of decimals. .
•multiply decimals. .
Assessments
• Do Now: These daily assessments will include a few questions to check for prior knowledge and to determine mastery of particular topics. Remediation can also be done through this activity on an as needed basis.
• Exit Tickets and Quick Checks: These will be used to measure student understanding of the lesson and assist in determining whether remediation is needed for the topic or if there were any common misconceptions amongst the students.
• Communicator Practice: During guided practice, this will be used as a quick whole-class assessment tool to check for complete comprehension.
• IXL Practice: This online tool will be used to formatively assess students during independent practice. This will provide students with practice and immediate self-check.
• Homework and Classwork: These will be used to formatively assess students. Some examples of activities that can be used in class as assessments are listed in suggested activities (Shopping List, Multiplying with Decimals Showdown, Decimal Operation Scavenger Hunt, and Decimals Board Game)
• Marzano learning goals self-assessment: Students will complete tiered questions to determine their own proficiency in the topic on a scale of 0 to 4
• Informal Observations: Walking around the room, listening to productive conversations, and checking in on students will help to formatively assess their learning.
• Mid-Chapter Quiz: This will be used to formatively assess students halfway through the chapter.
• Chapter Test: This will be used to summatively assess students at the end of the chapter.
• Information from this unit will be included on a locally developed, mid-year or end of year benchmark assessment that may take the form of a test, performance based project, or other summative assessment. Suggested Activities
• Grade 6 Digits Topic 7 Launches
• Inquiry labs using hundreds blocks to show operations with decimals
• Student centered Smart Board lessons using hundreds blocks that students can drag and drop to perform operations with decimals
• Review games using communicators
• Marzano learning goals self-assessment: Students will complete tiered questions to determine their own proficiency in the topic on a scale of 0 to 4
• Shopping List: Students go around the room to different stores and buy items of decimal value. This will enforce their skills in adding and multiplying decimals
• Multiplying with Decimals Showdown: Students will work in groups. They will receive task cards that they must solve individually and then "showdown" when each student has a possible solution. Students will talk about their mistakes if there are any.
• Dividing Decimals Board Game: This is a board game focusing solely on dividing decimals with problems where the divisor is a whole number or decimal.
• Decimal Operation Scavenger Hunt: Students will walk around the room answering skill and problem-solving questions on decimal operations. Once they have the answer, they will use this to find and complete the next problem.
• Decimals Board Game: Students will play in a group and complete the problem presented to them when they land on a square on the board game.
Activities to Differentiate Instruction
Differentiation for special education:
• General modifications may include:
o Modifications & accommodations as listed in the student’s IEP
o Assign a peer to help keep student on task
o Modified or reduced assignments
o Reduce length of assignment for different mode of delivery
o Increase one-to-one time
o Working contract between you and student at risk
o Prioritize tasks
o Think in concrete terms and provide hands-on-tasks
o Position student near helping peer or have quick access to teacher
o Anticipate where needs will be
o Break tests down in smaller increments
• Content specific modifications may include:
o Use hundreds blocks to help student visualize operations with decimals through manipulatives
o Use grid lines to help student line up places and organize work when computing
o Provide guided notes and step-by-step instructions for operations with decimals
o Provide completed worked out examples on classwork and homework that students may use as a guide
Differentiation for ELL's:
• General modifications may include:
o Strategy groups o Teacher conferences
o Graphic organizers
o Modification plan
o Collaboration with ELL Teacher
• Content specific vocabulary important for ELL students to understand include:
o Place values (tenth, hundredth, thousandth, etc.), annex, estimate, round, sum, difference, product, quotient, multi-digit numbers
Differentiation to extend learning for gifted students may include:
• Complete operations of decimals with more than two numbers.
• Include more digits in decimals
• Identify and analyze errors in worked out problems
Technology Integration
• iPads or Chromebooks as appropriate to the activity.
• Online learning components including use of the Digits digital textbook, Buzzmath, KhanAcademy, and other resources.
• Teacher integration of the SMART board to facilitate active student engagement throughout the course of the lesson.
• Software or online programs that teachers may use to create students materials or generate problems such as Kuta software.
• Additional practice provided through the use of IXL
Integrated/Cross-Disciplinary Instruction
• ELA: Using grammatically correct sentences, descriptive words, and transitions when analyzing and comparing decimals
• Science: Scientific notation or converting in the metric system
• Economics: Finding the percent of a number using decimal multiplication
Resources
• Digits student access and support: www.MyMathUniverse.com
• Digits teacher materials and support: www.pearsonrealize.com
• IXL: www.ixl.com
• SMART Exchange: www.exchange.smartteach.com
• SMART Board Lessons
• Pizzazz worksheets (self-correcting)
• Kuta software generated worksheets • Khanacademy: www.khanacademy.org
• Buzzmath: www.buzzmath.com
• NCTM Illuminations: www.illuminations.nctm.org
• New Jersey Center for Teaching and Learning: www.njctl.org
21st Century Skills
CRP.K-12.CRP4 Communicate clearly and effectively and with reason.
CRP.K-12.CRP8 Utilize critical thinking to make sense of problems and persevere in solving them.
CRP.K-12.CRP12 Work productively in teams while using cultural global competence. |
5170 | https://photonics101.com/fs-charge-configurations/the-electric-field-of-a-point-charge.html | The Electric Field of a Point Charge
Electrodynamics
Math Basics
Blog
The Electric Field of a Point Charge
Published in Electrodynamics / Electrostatics / Charges and Electric Fields
Tags: Gauss's law
If we would make a poll about the most fundamental question of electrostatics, the field of a point charge is very likely the winner. You may already know the answer but are you able to derive it directly from Maxwell's equations?
Problem Statement
Consider a point chargeq in vacuum, e.g. an electron or proton. What is the resulting electric field E(r)?
Can you relate Coulomb's law to this force?
Background: From Forces to Fields
The electric field of a point charge is the most fundamental concept in electromagnetism. Historically, the force between two charged objects was found to scale with the product of the object's charges their inverse squared distance. This force is now known as Coulomb's law and was found in the second half of the 18th century.
The field concept is however a little different. Given a charge, one can calculate the force acting on a test object with unit charge. This force, in appropriate units, is then called the electric field. You can use the results from this problem to find the electric field of an arbitrary charge distribution using the principle of superposition see the electric field of two point charges.
Hints
Which one of Maxwell's equations relates the electric field to charges?
What is the corresponding integral form?
Now, let's jump into the mathematical solution of the electric field for a point charge. It will not be complicated but very insightful!
First of all we will derive the field of a point charge from Gauss's law when the charge is centered at the origin. Then we can generalize our result to an arbitrary position. Finally, let us relate the field to Coulomb's force.
Calculating the Field from First Principles
Without loss of generality we can place the point charge at the origin r=0. The only Maxwell equation we have to use is Gauss's law
∇⋅E(r)=ρ(r)ϵ 0
relating the charge density ρ(r) to the electric field E(r). We obtain that after an integration over a volume V containing the charge q we are left with
∫V∇⋅E(r)d V=1 ϵ 0∫V ρ(r)d V=q ϵ 0.
At this point we can use the divergence theorem
∫V∇⋅F(r)d V and
=∮∂V F(r)⋅n d S
with the normal vector of the surface n. This transforms Gauss' law into a surface integral:
∫S E(r)⋅n d S=∫S E(r)⋅e r r 2 sin(θ)d θ d φ=4 π r 2 E r(r,θ,φ)=q ϵ 0.
Here, E r and e r indicate the radial component of the electric field an the unit vector in radial direction, respectively.
The charge is situated in the center of the coordinate system. So, there cannot be any other electric field component than the radial one. We find that
E(r,θ,φ)=1 4 π ϵ 0 q r 2 e r.
A Charge in an Arbitrary Position
If we want to know the field of a point charge situated at an arbitrary positionr q, the use of a spherical coordinate system is possible but not very clever. We have to transform the field into a cartesian form. The radial unit vector is given by e r=r/|r|, so we find the electric field as E(r)=q 4 π ϵ 0 r|r|3.Now we have to consider the new position of the charge. This can be realized by a shift using r q:E(r)=q 4 π ϵ 0 r−r q|r−r q|3.
Now that we know the electric field for one point charge in a rather general form, we can calculate the resulting field for general charge configurations. This can be done using the principle of superposition. One example is the electric field of two point charges.
The Coulomb Force
From a general point of view, a charge can be seen as a coupling constant of an object to a given field. The electrostatic force acting on that object q′ is then simply given by
F(r)=q′E(r).
For the field of a point charge which we have just calculated, we find that the force is just Coulomb's law.
Congratulations, you have just found the solution to the most basic charge configuration!
Way to go, there are a lot of topics to cover in the electrodynamics course!
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5171 | https://space.stackexchange.com/questions/48971/why-is-specific-impulse-equivalent-to-effective-exhaust-velocity | Skip to main content
Space Exploration
Why is specific impulse equivalent to effective exhaust velocity?
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I am very comfortable with specific impulse in seconds. I know how to calculate it, I know what it is, I know what happens to a rocket when you double in or cut it in half. But it feels like the strangest coincidence that specific impulse in units of velocity is equivalent to exhaust velocity.
Here's what I mean:
If we use the interpretation of specific impulse as the thrust per unit of fuel mass flow, we've got units of
Nkgs
Split up our Newtons using Newton's Second Law (F=ma)
kg⋅ms2kgs=>kg⋅mskg=>ms
Ok, great, like magic we have meters per second. Why? The math said so. But somehow this is supposed to be the exhaust velocity? The m in m/s came from the thrust of the rocket in the math, not from any property of the exhaust itself. The exhaust kg and s just cancel out some components of thrust.
Is there a good intuition to explain why this is equivalent to exhaust velocity, beyond "because the math says so"?
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asked Dec 5, 2020 at 21:55
TheEnvironmentalistTheEnvironmentalist
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I find it interesting that you are "comfortable with specific impulse in seconds". I've always considered it a historical relic, an arbitrary conversion factor thrown in to simplify hand calculations that I've always found to confuse matters. It's immediately obvious what a rocket with an exhaust velocity of 3 km/s implies, how it relates with mass flow rate to thrust and so on. 306 seconds, though? It's a duration, but not one that seems very interesting except for being related to exhaust velocity by g0.
– Christopher James Huff
Commented
Dec 5, 2020 at 22:45
@ChristopherJamesHuff oddly the very early writers like Willy Ley only wrote about exhaust velocity when discussing efficiency; specific impulse didn't start to be the standard measure until later. "In addition to using the exhaust velocity as a yardstick...performance...can also be expressed by the specific impulse" - the only time the term appears in Ley, 1954
– Organic Marble
Commented
Dec 5, 2020 at 23:00
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"The m in m/s came from the thrust of the rocket in the math, not from any property of the exhaust itself." The two critical properties of the rocket exhaust are its mass and its velocity; the thrust is the product of mass flow rate and velocity.
– Russell Borogove
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Dec 5, 2020 at 23:05
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Rockets produce thrust by ejecting reaction mass at some velocity. The fundamental quantities involved are mass flow rate and exhaust velocity, thrust is the consequence of these.
It's no coincidence that specific impulse in units of velocity equals exhaust velocity, that's what specific impulse is. The exhaust velocity gives you the impulse per unit of reaction mass simply because p=m∗v.
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answered Dec 5, 2020 at 22:31
Christopher James HuffChristopher James Huff
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That parameter is known as the Effective Exhaust Velocity because...
Sutton, 4th edition:
When [the exit plane pressure is equal to the ambient pressure], the effective exhaust velocity is equal to the average actual exhaust velocity of the propellant gases...the effective exhaust velocity is usually close in value to the actual exhaust velocity.
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edited Dec 5, 2020 at 22:58
answered Dec 5, 2020 at 22:06
Organic MarbleOrganic Marble
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When Specific Impulse (Isp) is expressed in units of time, the rate of fuel usage is not expressed in units of mass per second.
It's in units of weight (under one Standard Gravity, g0) per second.
As such, the unit conversion comes from:
NNs=s
You must multiply the time-valued Specific Impulse by g0 or 9.80665m/s2, to get the Effective Exhaust Velocity(ve). As such, the actual unit conversion looks like this when going from Isp to ve:
NNs˙ms2=ms
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edited Dec 5, 2020 at 22:47
answered Dec 5, 2020 at 22:34
notovnynotovny
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Given:
You're in a boat on a calm, windless lake and you're several meters from the dock.
It's essential that you get to the dock while staying in the boat, but you've lost your oars1.
There are several massive objects in the boat that you can throw away from the dock in order to accelerate towards it.
Question:
With all that given, what could be more fundamental and critically important than the velocity with which you can throw those objects?
Answer: No, velocity is the most fundamental and critically important here.
Forget all those equations; velocity is what's central to the problem. You know how much fuel (or objects) you have, you know the spacecraft (boat) will get lighter after you throw some of them, you want the most force per unit mass thrown, and starting from velocity you can calculate all the other stuff using math.
1or after you pushed off from the dock you realize that you've forgotten to bring the "keys to the oar locks" (humor).
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answered Dec 5, 2020 at 23:01
uhohuhoh
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Of course, specific impulse is "a measure of how effectively a rocket uses propellant", but that's quite a vague statement. So is there a physical interpretation of the value of specific impulse? What does it mean if we have a rocket with a specific impulse of 4 km/s? Where do we end up with the 4 km/s?
It turns out, there is. It's an answer to the question:
```
What speed could we give to a rocket with constant mass, when expelling a mass equal to itself?
```
So if we had a rocket of 1000 kg, that had a fuel tank of 1000 kg, where mass would magically reappear so that its total mass remained 1000 kg even when expelling fuel, and this rocket had engines with a specific impulse of 4 km/s, we can now conclude that this rocket could reach a maximum speed of 4 km/s.
Now such a rocket is impossible. But we could approximate it by starting with a 1000 kg rocket, and asking what speed it could attain by expelling 1 kg of fuel (where rocket mass is almost constant). Well, that's 1/1000th the fuel, so it can attain 1/1000th of the speed.
And then we can apply conversation of momentum. We expel a mass of 1/1000th the mass of the rocket, which gives our rocket a momentum of mr⋅(Isp/1000). And the counteracting momentum of the exhaust is (mr/1000)⋅vexh
So there we have it, Isp must be equal to vexh.
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answered Dec 6, 2020 at 12:42
Emil BodeEmil Bode
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5172 | https://www.onemathematicalcat.org/Math/Precalculus_obj/graphInverse.htm | The Graph of an Inverse Function
(Click for cat book)
Index card: 24ab
One-to-one functions and inverse functions are explored in these earlier sections:
Using a Function Box ‘Backwards’
One-to-One Functions
Undoing a One-to-One Function; Inverse Functions
Properties of Inverse Functions
Finding Inverse Functions (when there's only one $x$ in the formula)
Finding Inverse Functions (switch input/output names method)
If a point $\,(a,b)\,$ is on the graph of a one-to-one function $\,f\,,$ then the point $\,(b,a)\,$ is on the graph of its inverse $\,f^{-1}\,.$ This is because the input/output roles for a function and its inverse are switched—an input to one is an output from the other.
As discussed below, the point $\,(b,a)\,$ is found by reflecting the point $\,(a,b)\,$ about the line $\,y = x\,,$ when graphed in a coordinate system where ‘$\,1\,$’ on the $x$-axis is the same as ‘$\,1\,$’ on the $y$-axis.
From this observation, we will see below that the graph of an inverse function is very easy to obtain if we have the graph of the original function—just reflect the original graph about the line $\,y = x\,.$
Finding a Point $\,(b,a)\,$ from a Given Point $\,(a,b)\,$
Start with a coordinate system where ‘$\,1\,$’ on the $x$-axis is the same as ‘$\,1\,$’ on the $y$-axis. This forces distances measured on either axis to have exactly the same length.
Also, the line $\,y = x\,$ (which has slope $\,1\,$) will be at a $\,45^{\circ}\,$ angle in such a coordinate system.
Plot a point $\,(a,b)\,.$ This is the big green point in the diagram above.
Locate the number $\,b\,$ on the $x$-axis, as follows (use the blue arrows):
Go from the green point, horizontally, to the line (horizontal blue arrow).
Points on the line $\,y = x\,$ have the same $\,x\,$ and $\,y\,$ values.
Since we know this point has $y$-value equal to $\,b\,,$ its $x$-value must also be $\,b\,.$
Mark $\,b\,$ on the $x$-axis (vertical blue arrow).
Locate the number $\,a\,$ on the $y$-axis, as follows (use the purple arrows):
Go from the green point, vertically, to the line (vertical purple arrow).
Points on the line $\,y = x\,$ have the same $\,x\,$ and $\,y\,$ values.
Since we know this point has $x$-value equal to $\,a\,,$ its $y$-value must also be $\,a\,.$
Mark $\,a\,$ on the $y$-axis (horizontal purple arrow).
Observe that the green shaded region is a square: all sides have length $\,a - b\,,$ and all angles are $\,90^{\circ}\,.$ The desired point $\,(b,a)\,$ (large black point) has been located as the corner of the square opposite the original green point.
When you fold a square along a diagonal, the opposite corners meet. Thus, the green and black dots will coincide after folding.
Based on these results, simplify the process: to get from the green point to the black point, just fold along the line $\,y = x\,.$ That is, to get from a point $\,(a,b)\,$ to a point $\,(b,a)\,,$ just fold along the line $\,y = x\,.$
Notice that the line $\,y = x\,$ is the perpendicular bisector of the segment connecting the original point $\,(a,b)\,$ and the reflected point $\,(b,a)\,.$
The Graph of an Inverse Function
By definition, the graph of a function is the set of all its input/output pairs, as the inputs vary over the domain.
Thus: $$ \begin{align} &\cssId{s40}{\text{the graph of } f^{-1}}\cr &\qquad \cssId{s41}{= {(x,f^{-1}(x))\ |\ x\in\text{dom}(f^{-1})}}\ \end{align} $$
Since $\,x\,$ is a dummy variable, this set can also be written using the dummy variable $\,y\,$: $$ \begin{align} &\cssId{s44}{\text{the graph of } f^{-1}}\cr &\qquad \cssId{s45}{= {(y,f^{-1}(y))\ |\ y\in\text{dom}(f^{-1})}} \end{align} $$
Now, we have:
$$ \begin{align} &\cssId{s47}{\text{the graph of $\,f^{-1}\,$}}\cr\cr &\quad \cssId{s48}{= {(y,f^{-1}(y))\ |\ y\in\text{dom}(f^{-1})}}\cr &\qquad\ \ \cssId{s49}{\text{(definition of the graph of a function)}}\cr\cr &\quad \cssId{s50}{= {(y,f^{-1}(y))\ |\ y\in\text{ran}(f)}}\cr &\qquad\ \ \cssId{s51}{\text{(since $\text{ran}(f) = \text{dom}(f^{-1})$)}}\cr\cr &\quad \cssId{s52}{= {(f(x),x)\ |\ x\in\text{dom}(f)}} \end{align} $$
For the last step:
$y\in\text{ran}(f)\,$if and only ifthere exists $\,x\in\text{dom}(f)\,$ such that $\,f(x) = y$
And: $$ \begin{gather} f(x) = y\cr \text{if and only if}\cr x = f^{-1}(y) \end{gather} $$
Thus, the graph of $\,f^{-1}\,$ is precisely the set of points that make up the graph of $\,f\,,$ but with the coordinates switched!
We've already seen that ‘switching coordinates’ is accomplished by reflecting about the line $\,y = x\,$ in a coordinate system where the scales on the horizontal and vertical axes are identical. Thus, we have:
GRAPHING AN INVERSE FUNCTION
Let $\,f\,$ be a one-to-one function, so that it has an inverse $\,f^{-1}\,.$
The graph of $\,f^{-1}\,$ is found as follows:
Create a coordinate system where ‘$1$’ on the $x$-axis is the same as ‘$1$’ on the $y$-axis.
Graph $\,f\,$ in this coordinate system.
Reflect (‘mirror’) the graph of $\,f\,$ about the line $\,y = x\,.$
The reflected graph is the graph of $\,f^{-1}\,.$
Several different one-to-one functions and their reflections in the line $\,y = x\,$ are shown below:
Concept Practice |
5173 | https://www.studocu.com/en-us/messages/question/8085233/read-the-chemical-equationn2-3h2-2nh3using-the-volume-ratio-determine-how | [Solved] Read the chemical equation N2 3H2 2NH3 Using the volume ratio - Organic Chemistry I (CHEM 231) - Studocu
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Read the chemical equation. N2 + 3H2 → 2NH3 Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2, if all measurements are taken at the same temperature and pressure?Organic Chemistry I (CHEM 231)
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Capital University, Columbus Ohio
Organic Chemistry I
Question
Read the chemical equation N2 3H2 2NH3 Using the volume ratio
Capital University, Columbus Ohio
Organic Chemistry I
Question
### jay
1 year ago
Read the chemical equation. N2 + 3H2 → 2NH3 Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2, if all measurements are taken at the same temperature and pressure?
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Answer Created with AI
1 year ago
Answer
The chemical equation you provided is:
js
N2 + 3H2 → 2NH3
This equation tells us that 3 volumes of hydrogen gas (H2) react with nitrogen gas (N2) to produce 2 volumes of ammonia gas (NH3). This is a volume ratio of 3:2 for H2:NH3.
Given that 4.2 liters of H2 is reacted, we can use the volume ratio to find the volume of NH3 produced.
To do this, we set up a proportion:
js
3 volumes of H2 produces 2 volumes of NH3
4.2 volumes of H2 produces x volumes of NH3
We can solve for x (the volume of NH3) by cross-multiplying and dividing:
js
(2 volumes NH3 / 3 volumes H2) 4.2 volumes H2 = x volumes NH3
This gives us:
js
x = (2/3) 4.2 = 2.8 liters of NH3
So, if 4.2 liters of H2 reacts with an excess of N2, 2.8 liters of NH3 will be produced, assuming all measurements are taken at the same temperature and pressure.
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5174 | https://www.ck12.org/section/ratios-proportions-and-percents-%3A%3Aof%3A%3A-grade-6-differentiated-instruction-%3A%3Aof%3A%3A-ck-12-middle-school-math-grade-6-teachers-edition/ | Elementary Math
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HomeMathematicsRatios, Proportions and Percents
Ratios, Proportions and Percents
Difficulty Level: Basic | Created by: CK-12
Last Modified: Aug 19, 2014
This eighth chapter Ratios, Proportions and Percents introduces students to ratios and rates, basic uses of proportions including understanding scale drawings, and percents. Also explored is the relationship among percents, decimals, and fractions. In this flexbook, titled Differentiated Instruction, you will be given teaching strategies to help reach a diverse community of learners. There will be suggestions for students with special needs as well as those with different learning styles. Alternative assessment options will also be suggested.
Lessons
The following lessons are part of this chapter.
Ratios
Rates
Solving Proportions
Proportions and Scale Drawings
Understanding Percent
Percents, Decimals and Fractions
Finding a Percent of a Number
Problem-Solving Strategy: Use a Proportion
Ratios
Alternative Teaching Strategies
One of the first ways that you can differentiate this lesson is to look at the three different ways of identifying ratios. We can write them as fractions, with a colon or by using the word “to”. The three different forms can be used together and students can practice interchanging the forms. It shouldn’t matter if a fraction is equivalent to a ratio in the form with a colon or not. This is an excellent way for students to be challenged during this lesson. For students with special needs, see the notes below.
To challenge students, look at these examples.
Directions: Ask students to write an equivalent fraction by simplifying. Use a different form of ratios for each answer.
15 to 45
69 to 13
Alternative Assessment Options
Assessment can be done through observation. Here are some questions to think about.
Is the student able to understand how to simplify?
Can the student simplify the ratio in all three forms?
Can the students use the three forms interchangeably?
Does the student need more practice with simplifying?
Classroom Modifications for Diverse Learners
A modification for students with special needs is to only work with one form of the ratio at a time. In other words, students would be simplifying ratios only in fraction form first. Fraction form is the simplest way for students to first work with simplifying ratios because it is the form that they are already familiar with. This is an excellent place to start. Begin by having the students simplify ratios in fraction form. Here are a few examples.
Then have them move on to simplifying ratios written using a colon. You can begin by using the same three examples above so that the students can make the connection of them being the same.
6:10
4:12
6:9
Finally, move to using the word “to”. Again, you can use the same three examples. By now students will see how to work with them and be able to interchange the forms of the ratios when simplifying or finding equivalent ratios.
Additional Resources
– This is a resource page for teachers on how to engage middle school students when teaching ratios.
Pythagoras and the Ratios: A Math Adventure by Julie Ellis is a book that could be used to differentiate ratios through a geometry integration.
Rates
Alternative Teaching Strategies
A fun way to differentiate working with rates is to use actual advertisements from a newspaper or from a supermarket advertisement. Many times, these advertisements will have notes about how much per pound, etc. For example, in thinking about Thanksgiving, you can say that turkeys are .89 a pound. This is the rate. It is being compared to 1.
.89 to 1 pound of turkey
A possible problem would be to figure out the cost of a 10 or 12 pound turkey.
To work with the students, you can begin with an example like this one. Then have them work in small groups to write a problem that uses the same rate as in the example, but a different number of pounds. Working in this way helps to engage many different learning styles including auditory, visual, kinesthetic, interpersonal and intrapersonal. Allow time for students to share their work at the end of the lesson.
This will also be a good warm – up for the enrichment activity presented in another flexbook.
Alternative Assessment Options
When assessing student work, you can use the following questions as a guide.
Did the students understand that they needed to multiply?
What did the students multiply to figure out the price?
Did they do this correctly?
Is the price accurate?
Are the students able to explain their thinking to their classmates?
Did everyone in the group participate in the work?
Classroom Modifications for Diverse Learners
Provide special needs students with a format for working on the problem. Here is a suggested set of ratios.
Then students can see that they need to figure out how much a 12 pound turkey costs if it is .89 per pound. Many will know that they need to multiply. If the proportion is confusing, then allow students the chance to work on the problem intuitively.
Additional Resources
– This is a resource page for teachers about teaching ratios and fractions as rates.
– This website has worksheets designed for graders on rates and ratios. These could be used as a challenge for students needing some additional skill work.
– This website has a guided tutorial and several worksheets on rates and unit rates. One is a warm – up and then another can be used for additional practice for the students.
Solving Proportions
Alternative Teaching Strategies
There are two different methods taught in this lesson for identifying and solving proportions. The first is to help students to simplify the ratios to see if they are equivalent. The second is to use cross–products. To help the students to remember these two ways, make a list of them on the board. You want to keep the list visual while the students are working through the lesson. When things are left in front of the students, you can help them to use their reasoning skills to figure out that if they can’t use simplifying, then they should use a cross–product.
Word problems are a great way to challenge advanced students. You can provide them with a word problem like the one in the text and ask them to solve it using a proportion. Be very specific about the proportion as the method of solving it. Then you can ask them to illustrate how they solved the problem and their solution. Illustrating word problems is excellent for all students. The students really enjoy drawing it out and many will be very creative in their methods.
Alternative Assessment Options
When assessing how a student has solved a word problem using an illustration, you can look to see if they could show which values were being compared in the proportion. Look at this example from the text.
This is a proportion from the text. The students would be illustrating this proportion and it’s solution. An acceptable drawing would be one where one tent is drawn with an arrow or equals sign with 25 campers drawn. Students could also draw one big tent with 25 people drawn inside it. Wow! A big tent! Then the student would draw the correct number of tents for 75 people (3 tents) and show how this ratio is equal to the other ratio.
Classroom Modifications for Diverse Learners
To modify this lesson, begin by having special needs students work on simplifying problems first. Once they have mastered determining proportions through simplifying, you can move on to cross–products. For students with where English is a second language or who have a language based learning disability, keep the word problems for last. Then you can allow them the chance to work with a partner or in a small group with you as the leader.
Additional Resources
– This website has interactive word problems that require students to solve proportions.
– This website has many internet links for parents and teachers to work with students on ratios and proportions.
Proportions and Scale Drawings
Alternative Teaching Strategies
When teaching this lesson, you can differentiate the instruction by first teaching the content and then using the problem from the introduction as the differentiated activity. After teaching the students how to identify proportions and work with scale measurements, ask the students to create the actual poster. They have been given a scale and the problem. Students can use their own creativity to make the poster. This is an excellent opportunity to connect the content to a real world activity. Students with creative ability can be leaders in this activity, and students who excel in mathematics can help their groups with the measurement. You can use the problem solution at the end of the lesson so that students can check their work when finished.
Alternative Assessment Options
Assessment can be measured a couple of different ways. Here are some questions to think about as you assess student work.
First, did the students write the proportion accurately?
Is the measurements accurate?
Is the art work creative and shows a best effort?
Can the students explain how they arrived at the answer that they did?
Do the measurements match the solution in the text?
Classroom Modifications for Diverse Learners
One way to modify this lesson for special needs students is to show a set – up proportion that students will need to solve. Remind the students that they can solve a proportion through simplifying or through cross products. Here is an example of a proportion for the problem in the introduction.
This example gives the students the dimension for width. This is one way to modify the assignment. The final step is for the students to complete. They will need to figure out the measurement of the length.
The correct answer is 4 feet because which means the measure is 48 inches or 4 feet.
Additional Resources
– This website has a wonderful tutorial to scale and map measurements.
– This website has an article for teachers about an educator’s experience teaching ratios and proportions to students.
Understanding Percent
Alternative Teaching Strategies
In this lesson, students are going to make the connection between a fraction, a decimal and a percent. Fractions, decimals and percents are all part of a whole. Because of this, we can use them interchangeably. There are two goals in working with this content. First, we need the students to understand these parts conceptually and then procedurally understand how to work with them.
To create an interactive game with this content, you can use the following list to create index cards. Each card has either a fraction, decimal or percent or written words describing one of the equal ratios. Then the students work in pairs to identify all of the other forms of the equivalent ratios.
Alternative Assessment Options
Student work can be collected an evaluated for accuracy. Students should have four columns to show the words, the fraction, the decimal and the percent that they have been given.
Classroom Modifications for Diverse Learners
A modification for diverse learners is to actually receive a paper with four columns created on it. Each column is labeled with the following headings: words, fraction, decimal, percent. Then the students only have to fill in the missing values on the chart.
Additional Resources
Here are possible choices for the cards:
14 out of 100
12%
34%
13 out of 100
50 out of 100
60 out of 100
.40
.55
.60
.80
90%
Students use these given values and then work together to write the missing ratios.
Percents, Decimals and Fractions
Alternative Teaching Strategies
If you did the activity in the last lesson, then many of your students have already had practice working with converting basic fractions, decimals and percentages. The tricky part is working with ones that aren’t basic such as one – third. This is where students are introduced to terminating and repeating decimals.
To help students to understand these concepts, you can make this visual by writing examples on the board. You can also challenge the students to come up with as many repeating decimals as they can. This is a fun way to engage them in working together to figure it out. The students pair off or work in small groups of three. The students work together and create a list of repeating decimals.
Here are some possible options:
has repeating parts
Alternative Assessment Options
You can assess student understanding through observations. Here are some questions to think about.
Do the students understand how to convert a fraction to a decimal through division?
Are the decimal points in the correct place?
Do the students understand the difference between terminating and repeating decimals?
Classroom Modifications for Diverse Learners
A modification for diverse learners is to have them simply work on converting fractions to decimals through division. Many special needs students won’t understand how to test fractions to see if they turn into repeating or terminating decimals. Therefore, you can give them a list of fractions and have them divide them to turn them into decimals. Only when a repeating decimal comes up do you introduce the term. Then it has a real connection for the students. Let them discover first and then explain the different terminology.
Additional Resources
– This is a great tutorial on converting fractions to different types of decimals. It also has some fun suggestions for teachers on how to make this an engaging activity for students.
Finding a Percent of a Number
Alternative Teaching Strategies
Alternative Assessment Options
When assessing student understanding, you will be able to determine whether or not students understand percents as a concept by checking for accuracy in this lesson. Up until now, there have been specific procedures for working with percentages. This lesson approaches percentages from a conceptual point of view. If students understand the concepts involved in this lesson, they will be able to complete the work accurately. If their understanding is strictly procedural, as in knowing “what to do” they will struggle with the lesson.
Classroom Modifications for Diverse Learners
When working with special needs students, you can provide them with the following steps.
Change the percent to a decimal over 100.
Then use the number as the second part of the proportion.
Simplify or use cross–products.
Now solve = 13
This is a number not a percent. We found 50% of the number 26 which is 13. It is a common error for students to want to label everything with a % sign. Be on watch for this it shows that a student does not completely understand the concept being taught.
Additional Resources
– This website has examples that are color coded and use illustrations. A nice website for a visual learner or for someone who needs an additional tutorial.
– Here is a website of resources for teachers for teaching fractions, decimals and percents. Many have links to PDF downloads.
Problem-Solving Strategy: Use a Proportion
Alternative Teaching Strategies
The best way to differentiate this lesson is to have the students work with the information in the problem along with their own height to use a proportion to solve the problem. In the introduction, Tim’s height is being used, but this problem would be more relevant to the students if they were using their own heights. You want the student’s to figure out how to use the proportion to solve the problem. Therefore, don’t give them too much information. Read the problem. Tell them to use their own height and then let them work. Allow students the chance to work with a partner so that they have someone to talk things over with. Ask students to illustrate their solution.
Alternative Assessment Options
Student work can be assessed by thinking through the following questions.
Did the students accurate write a proportion that compares height to distance and height to distance?
Does the illustration accurately show this proportion as well?
Is the work creative and does it show best effort?
Is the answer accurate given the students height?
Classroom Modifications for Diverse Learners
A modification for special needs students would be to show them how to set up the proportion.
Remind students to keep in mind that this jump length is 20 times the height.
Additional Resources
– This is a great resource site for teachers and is all about problem solving and proportions with middle school students. It is a wonderful site for teaching strategies and contains some wonderful problems to challenge those students who are in need of an additional challenge.
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Description
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Difficulty Level
Basic
Tags
variable expressions, patterns, measurement, equations, decimals, fractions, proportions, ratios, percents, functions, statistics, probability, geometry, multiplication, addition, subtraction, operations with numbers, CK.MAT.ENG.TE.1.Math-Grade-6.3, (15 more)
Subjects
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Grades
6
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Language
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5175 | https://www.drmaryling.com.au/microdochetomy | (02) 4321 0302
URGENT APPOINTMENT
(02) 4321 0302
MICRODOCHECTOMY & TOTAL DUCT EXCISION
Nipple discharge, even in the absence of clinical or imaging findings, may require surgery to remove the ducts behind the nipple to stop the discharge and to provide tissue to the pathologist to identify the cause of the discharge.
Microdochectomy is a targeted surgical procedure where a single duct behind the nipple is removed. It is usually recommended for discharge from a single duct in the nipple. It is suitable for younger women who wish to preserve the ability to breastfeed after surgery.
Total duct excision is a surgical procedure where all the ducts behind the nipple are removed. It is usually recommended for discharge from multiple ducts in the nipple or persistent discharge after microdochectomy.
How are microdochectomy and total duct excision performed?
Microdochectomy and total duct excision are performed under general anaesthesia. A single dose of intravenous antibiotics is given at the start of surgery.
The incision is placed along the edge of the brown area around your nipple, known as periareolar incision.
Microdochectomy involves the removal of a single duct. The specific duct responsible for the discharge is identified and isolated with a fine probe. The duct is removed without damaging the remaining ducts.
Total duct excision involves removal of all the major ducts from behind the nipple.
The incision is infiltrated with local anaesthetic and closed with dissolvable sutures and skin glue, and covered with a waterproof dressing.
The procedure takes around 60 minutes.
What is the expected recovery after microdochectomy and total duct excision?
Hospital Stay. Most people go home on the day of surgery. It is recommended that you be accompanied home by a carer who will stay with you (or very close by) for the first 24 to 48 hours.
Wound Care. Your wound is closed with dissolvable sutures and skin glue and covered with a waterproof dressing. Remove your dressings in 2 weeks - your wounds should be healed by then and you do not need further dressings. You can massage your scar at 3 weeks using a plain moisturiser with firm circular motions for at least 10 minutes twice a day.
Bras. You may feel more comfortable wearing a supportive wire-free bra for the first few days.
Driving. You should not drive for at least 24 hours after general anaesthesia.
Physical Activity. You are encouraged to do as much walking as is comfortable. Avoid lifting (over 5 kg), pushing or pulling for 2 weeks - this includes lifting children and housework such as vacuuming or hanging out the washing. Avoid exercises that create a lot of 'breast bounce' for 2 weeks - such as jogging or cardio sessions.
What are the risks of microdochectomy and total duct excision?
All surgery has risks despite the highest standard of practice. Some of the possible risks associated with microdochectomy and total duct excision include:
Bleeding
Infection
Unacceptable scarring, including hypertrophic or keloid scars
Changes in nipple sensation (temporary or permanent)
Loss of nipple skin
Breastfeeding is possible after microdochectomy but not after total duct excision
Recurrence of symptoms
What is the cost of surgery?
For patients with private health insurance, there is no out of pocket fee for the surgeon (as long as you are covered for item numbers 31553 - microdochectomy or 31556 - total duct excision). There may be additional costs related to anaesthetist’s fee, hospital excess and pathology services.
For patients without private health insurance, please contact us for a quote.
Dr Mary Ling performs breast surgery at Gosford Private Hospital and Brisbane Waters Private Hospital on the Central Coast. Dr Ling welcomes patients from Newcastle, Lake Macquarie, the Hunter Region and Northern NSW.
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5176 | https://brainly.com/question/40279856 | [FREE] Question 25 of 25 To solve x - 8 = 16, Cassie added 8 to both sides of the equation. Which property - brainly.com
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Question 25 of 25
To solve x−8=16, Cassie added 8 to both sides of the equation. Which property justifies this step?
A. Addition property of equality
B. Subtraction property of equality
C. Distributive property
D. Commutative property of addition
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The property that justifies adding 8 to both sides of the equation is the Addition property of equality.
Explanation
The property that justifies Cassie's step of adding 8 to both sides of the equation x-8 = 16 is the Addition property of equality. This property states that if you add the same number to both sides of an equation, the sides remain equal. By adding 8 to both sides, Cassie is maintaining the equality of the equation.
Learn more about Addition property of equality here:
brainly.com/question/10617252
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Cassie used the Addition Property of Equality when she added 8 to both sides of the equation x−8=16 to maintain equality and isolate x. This property allows equal operations on both sides of the equation without changing their equality.
Explanation
To solve the equation x−8=16, Cassie added 8 to both sides. This step can be justified by the Addition Property of Equality.
Understanding the Addition Property of Equality:
The Addition Property of Equality states that if you add the same number to both sides of an equation, the two sides remain equal.
In this scenario, Cassie is trying to isolate x. So she adds 8 to both sides of the equation to eliminate the −8 on the left side.
Here’s the step-by-step breakdown:
Start with the original equation: x−8=16.
Apply the Addition Property by adding 8 to both sides: x−8+8=16+8
Simplify both sides: x=24
By using the Addition Property of Equality, Cassie successfully solved for x.
Examples & Evidence
An example of the Addition Property of Equality is solving 3+x=10. To isolate x, you can subtract 3 from both sides: 3+x−3=10−3 leading to x=7. This process shows how adding or subtracting the same number on both sides keeps the equation balanced.
The Addition Property of Equality is a fundamental concept in algebra that is widely accepted and taught in mathematics. It is used extensively in solving equations and can be found in various mathematics texts and resources.
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### Free Mathematics solutions and answers
Community Answer 43 To solve x-8=16 cassie added 8 to both side of the equation which property justifies this step? A. Subtraction property of equality B. Addition Property of Equality C. Distributive property D. Commutative property of addition
Community Answer 40 Consider the equation and its solution. 8(x-2)=64 8x-16=64 8x=80 x=10 Which property is used in the last step to find that x=10? distributive property addition property of equality subtraction property of equality division property of equality
Community Answer 5.0 8 ‼️‼️HURRY‼️‼️WILL GIVE BRAINLIEST‼️‼️ Consider the equation and its solution. 8(x-2) = 64 8x-16 = 64 x 8x = 80 X= 10 Which property is used in the last step to find that X= 10? O distributive property O addition property of equality O subtraction property of equality O division property of equality
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
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5177 | https://www.crystallography.net/cod//1000041.html | Crystallography Open Database: Information card for entry 1000041
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Information card for entry 1000041
1000040<<1000041>>1000042
Preview
Jmol._Canvas2D (Jmol) "jmolApplet0"[x]
| Coordinates | 1000041.cif |
| Original paper (by DOI) | HTML |
| External links | AMCSD |
[x] Structure parameters
| Chemical name | Sodium chloride |
| Formula | Cl Na |
| Calculated formula | Cl Na |
| SMILES | [Na+].[Cl-] |
| Title of publication | Accuracy of an automatic diffractometer. measurement of the sodium chloride structure factors |
| Authors of publication | Abrahams, S C; Bernstein, J L |
| Journal of publication | Acta Crystallographica (1,1948-23,1967) |
| Year of publication | 1965 |
| Journal volume | 18 |
| Pages of publication | 926 - 932 |
| a | 5.62 Å |
| b | 5.62 Å |
| c | 5.62 Å |
| α | 90° |
| β | 90° |
| γ | 90° |
| Cell volume | 177.5 Å 3 |
| Number of distinct elements | 2 |
| Space group number | 225 |
| Hermann-Mauguin space group symbol | F m -3 m |
| Hall space group symbol | -F 4 2 3 |
| Residual factor for all reflections | 0.022 |
| Has coordinates | Yes |
| Has disorder | No |
| Has F obs | No |
[x] Version history
| Revision | Date | Message | Files |
--- --- |
| 130149 (current) | 2015-01-27 | cod/ (saulius@kolibris) Deriving Hall space group symbols for 12003 CIFs using the 'cif_filter --estimate-spacegroup' command. | 1000041.cif |
| 120071 | 2014-07-11 | Adding DOIs to range 1 structures. | 1000041.cif |
| 35911 | 2012-02-28 | cif/ Reorganising ranges 1, 3, 5, 6 and 8 into a prefix-directory tree. | 1000041.cif |
| 966 | 2010-01-30 | cif/ Adding cod_database_code tags to all COD entries. Adding tag description to the cif_cod.dic dictionary so that COD entries can be validated. | 1000041.cif |
| 900 | 2009-11-21 | cif/ Renaming data blocks of all COD CIFs so that they have COD number (taken from the COD file name) as their datablock number: find ? -name .cif \ | xargs \ perl \ -MFile::Basename \ -i \ -lpe 's/^data([^\s])/"data_".basename($ARGV, ".cif")/e' After this change, 'cif2cod' will give correct keys in the generated table. Also, data block names are guaranteed to be distinct, so any subset of COD CIFs can be concatenated into one file without introducing ambiguity. | 1000041.cif |
| 177 | 2008-02-25 | 14 files corrected. | 1000041.cif |
| 20 | 2008-01-26 | Removing white space at the end of all lines from all COD CIF files: in the cod/cif/ directory: ff .cif | xargs perl -i -pe 's/\s+\n$/\n/' | 1000041.cif |
| 19 | 2008-01-26 | Changing all end-of-line codes from DOS style to UNIX style in all COD CIF files, for more convenient processing: In the cod/cif/ directory: ff .cif | xargs perl -i -pe 's/\r\n/\n/' | 1000041.cif |
| 14 | 2008-01-14 | Appending a header with SVN keywords, COD URL and copying policy to all COD CIF files. | 1000041.cif |
| 13 | 2008-01-13 | Setting the svn:keywords property to "Author Date Revision URL Id" for all CIF files in the cod/cif directory. | 1000041.cif |
| 10 | 2008-01-04 | BUG FIX: renaming all CIF files that had extension with all capital letters (.CIF) into files written in all lowercase (.cif). On Linux, where filenames are strictly cae-sensitive, the Web site engine would not find files that had .CIF extension (written in all caps). | 1000041.cif |
| 1 | 2007-11-30 | Adding all .cif files from the COD, downloaded in a Cod.zip file on 2007.02.07 (and the same Cod.zip is downloadable today, on 2007.11.29) | 1000041.cif |
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All data in the COD and the database itself are dedicated to the public domain and licensed under the CC0 License . Users of the data should acknowledge the original authors of the structural data. |
5178 | https://www.aimsciences.org/article/doi/10.3934/dcds.2015.35.1843 | | |
| `x^2+y_1+z_12^34` |
Discrete and Continuous Dynamical Systems
A variational approach to reaction-diffusion equations with forced speed in dimension 1
Sorbonne Universités, UPMC Univ Paris 06, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris
Abstract
| |
| Citation: shu |
| |
| Citation: shu |
References
| | |
--- |
| | H. Berestycki, L. Desvillettes and O. Diekmann, Can climate change lead to gap formation?, in preparation. |
| | H. Berestycki, O. Diekmann, C. J. Nagelkerke and P. A. Zegeling, Can a species keep pace with a shifting climate?, Bull. Math. Biol., 71 (2009), 399-429.doi: 10.1007/s11538-008-9367-5. |
| | H. Berestycki and L. Rossi, Generalizations and properties of the principal eigenvalue of elliptic operators in unbounded domains, in preparation, URL doi: 10.1002/cpa.21536. |
| | H. Berestycki and L. Rossi, Reaction-diffusion equations for population dynamics with forced speed. I. The case of the whole space, Discrete Contin. Dyn. Syst., 21 (2008), 41-67.doi: 10.3934/dcds.2008.21.41. |
| | H. Berestycki and L. Rossi, Reaction-diffusion equations for population dynamics with forced speed. II. Cylindrical-type domains, Discrete Contin. Dyn. Syst., 25 (2009), 19-61.doi: 10.3934/dcds.2009.25.19. |
| | K. J. Brown and H. Budin, On the existence of positive solutions for a class of semilinear elliptic boundary value problems, SIAM J. Math. Anal., 10 (1979), 875-883.doi: 10.1137/0510082. |
| | Y. Du and H. Matano, Convergence and sharp thresholds for propagation in nonlinear diffusion problems, J. Eur. Math. Soc. (JEMS), 12 (2010), 279-312.doi: 10.4171/JEMS/198. |
| | L. C. Evans, Partial Differential Equations, vol. 19 of Graduate Studies in Mathematics, 2nd edition, American Mathematical Society, Providence, RI, 2010. |
| | P. C. Fife and J. B. McLeod, The approach of solutions of nonlinear diffusion equations to travelling front solutions, Arch. Ration. Mech. Anal., 65 (1977), 335-361. |
| | R. Fisher, The advance of advantageous genes, Annals of Eugenics, 7 (1937), 355-369.doi: 10.1111/j.1469-1809.1937.tb02153.x. |
| | T. Gallay and R. Joly, Global stability of travelling fronts for a damped wave equation with bistable nonlinearity, Ann. Sci. Éc. Norm. Supér. (4), 42 (2009), 103-140. |
| | T. Gallay and E. Risler, A variational proof of global stability for bistable travelling waves, Differential Integral Equations, 20 (2007), 901-926. |
| | J. K. Hale and G. Raugel, Convergence in gradient-like systems with applications to PDE, Z. Angew. Math. Phys., 43 (1992), 63-124.doi: 10.1007/BF00944741. |
| | S. Heinze, A variational approach to traveling waves, Technical Report 85, Max Planck Institute for Mathematical Sciences, Leipzig. |
| | A. Kolmogorov, I. Petrovskii and N. Piskunov, Etude de l'équation de la diffusion avec croissance de la quantité de la matière et son application à un problème biologique, Bull. Univ. Etat Mosc. Sér. Int. A, 1 (1937), 1-26. |
| | M. Lucia, C. B. Muratov and M. Novaga, Linear vs. nonlinear selection for the propagation speed of the solutions of scalar reaction-diffusion equations invading an unstable equilibrium, Comm. Pure Appl. Math., 57 (2004), 616-636.doi: 10.1002/cpa.20014. |
| | M. Lucia, C. B. Muratov and M. Novaga, Existence of traveling waves of invasion for Ginzburg-Landau-type problems in infinite cylinders, Arch. Ration. Mech. Anal., 188 (2008), 475-508.doi: 10.1007/s00205-007-0097-x. |
| | H. Matano, Convergence of solutions of one-dimensional semilinear parabolic equations, J. Math. Kyoto Univ., 18 (1978), 221-227. |
| | C. B. Muratov, A global variational structure and propagation of disturbances in reaction-diffusion systems of gradient type, Discrete Contin. Dyn. Syst. Ser. B, 4 (2004), 867-892.doi: 10.3934/dcdsb.2004.4.867. |
| | A. B. Potapov and M. A. Lewis, Climate and competition: The effect of moving range boundaries on habitat invasibility, Bull. Math. Biol., 66 (2004), 975-1008.doi: 10.1016/j.bulm.2003.10.010. |
| | E. Risler, Global convergence toward traveling fronts in nonlinear parabolic systems with a gradient structure, Ann. I. H. Poincaré, 25 (2008), 381-424.doi: 10.1016/j.anihpc.2006.12.005. |
| | L. Roques, A. Roques, H. Berestycki and A. Kretzschmar, A population facing climate change: Joint influences of allee effects and environmental boundary geometry, Population Ecology, 50 (2008), 215-225.doi: 10.1007/s10144-007-0073-1. |
| | H. H. Vo, Traveling fronts for equations with forced speed in mixed environments, in preparation. |
| | T. Zelenyak, Stabilization of solutions of boundary value problems for a second order parabolic equation with one space variable, Differentsial'nye Uravneniya, 4 (1968), 34-45. |
| | Y. Zhou and M. Kot, Discrete-time growth-dispersal models with shifting species ranges, Theor Ecol, 4 (2011), 13-25.doi: 10.1007/s12080-010-0071-3. |
H. Berestycki, L. Desvillettes and O. Diekmann, Can climate change lead to gap formation?, in preparation.
H. Berestycki, O. Diekmann, C. J. Nagelkerke and P. A. Zegeling, Can a species keep pace with a shifting climate?, Bull. Math. Biol., 71 (2009), 399-429.doi: 10.1007/s11538-008-9367-5.
H. Berestycki and L. Rossi, Generalizations and properties of the principal eigenvalue of elliptic operators in unbounded domains, in preparation, URL doi: 10.1002/cpa.21536.
H. Berestycki and L. Rossi, Reaction-diffusion equations for population dynamics with forced speed. I. The case of the whole space, Discrete Contin. Dyn. Syst., 21 (2008), 41-67.doi: 10.3934/dcds.2008.21.41.
H. Berestycki and L. Rossi, Reaction-diffusion equations for population dynamics with forced speed. II. Cylindrical-type domains, Discrete Contin. Dyn. Syst., 25 (2009), 19-61.doi: 10.3934/dcds.2009.25.19.
K. J. Brown and H. Budin, On the existence of positive solutions for a class of semilinear elliptic boundary value problems, SIAM J. Math. Anal., 10 (1979), 875-883.doi: 10.1137/0510082.
Y. Du and H. Matano, Convergence and sharp thresholds for propagation in nonlinear diffusion problems, J. Eur. Math. Soc. (JEMS), 12 (2010), 279-312.doi: 10.4171/JEMS/198.
L. C. Evans, Partial Differential Equations, vol. 19 of Graduate Studies in Mathematics, 2nd edition, American Mathematical Society, Providence, RI, 2010.
P. C. Fife and J. B. McLeod, The approach of solutions of nonlinear diffusion equations to travelling front solutions, Arch. Ration. Mech. Anal., 65 (1977), 335-361.
R. Fisher, The advance of advantageous genes, Annals of Eugenics, 7 (1937), 355-369.doi: 10.1111/j.1469-1809.1937.tb02153.x.
T. Gallay and R. Joly, Global stability of travelling fronts for a damped wave equation with bistable nonlinearity, Ann. Sci. Éc. Norm. Supér. (4), 42 (2009), 103-140.
T. Gallay and E. Risler, A variational proof of global stability for bistable travelling waves, Differential Integral Equations, 20 (2007), 901-926.
J. K. Hale and G. Raugel, Convergence in gradient-like systems with applications to PDE, Z. Angew. Math. Phys., 43 (1992), 63-124.doi: 10.1007/BF00944741.
S. Heinze, A variational approach to traveling waves, Technical Report 85, Max Planck Institute for Mathematical Sciences, Leipzig.
A. Kolmogorov, I. Petrovskii and N. Piskunov, Etude de l'équation de la diffusion avec croissance de la quantité de la matière et son application à un problème biologique, Bull. Univ. Etat Mosc. Sér. Int. A, 1 (1937), 1-26.
M. Lucia, C. B. Muratov and M. Novaga, Linear vs. nonlinear selection for the propagation speed of the solutions of scalar reaction-diffusion equations invading an unstable equilibrium, Comm. Pure Appl. Math., 57 (2004), 616-636.doi: 10.1002/cpa.20014.
M. Lucia, C. B. Muratov and M. Novaga, Existence of traveling waves of invasion for Ginzburg-Landau-type problems in infinite cylinders, Arch. Ration. Mech. Anal., 188 (2008), 475-508.doi: 10.1007/s00205-007-0097-x.
H. Matano, Convergence of solutions of one-dimensional semilinear parabolic equations, J. Math. Kyoto Univ., 18 (1978), 221-227.
C. B. Muratov, A global variational structure and propagation of disturbances in reaction-diffusion systems of gradient type, Discrete Contin. Dyn. Syst. Ser. B, 4 (2004), 867-892.doi: 10.3934/dcdsb.2004.4.867.
A. B. Potapov and M. A. Lewis, Climate and competition: The effect of moving range boundaries on habitat invasibility, Bull. Math. Biol., 66 (2004), 975-1008.doi: 10.1016/j.bulm.2003.10.010.
E. Risler, Global convergence toward traveling fronts in nonlinear parabolic systems with a gradient structure, Ann. I. H. Poincaré, 25 (2008), 381-424.doi: 10.1016/j.anihpc.2006.12.005.
L. Roques, A. Roques, H. Berestycki and A. Kretzschmar, A population facing climate change: Joint influences of allee effects and environmental boundary geometry, Population Ecology, 50 (2008), 215-225.doi: 10.1007/s10144-007-0073-1.
H. H. Vo, Traveling fronts for equations with forced speed in mixed environments, in preparation.
T. Zelenyak, Stabilization of solutions of boundary value problems for a second order parabolic equation with one space variable, Differentsial'nye Uravneniya, 4 (1968), 34-45.
Y. Zhou and M. Kot, Discrete-time growth-dispersal models with shifting species ranges, Theor Ecol, 4 (2011), 13-25.doi: 10.1007/s12080-010-0071-3.
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How to Calculate Profit Percentage: A Comprehensive Guide for Retailers
Comprehensive Guide for Retailers
Calculating profit percentage is a fundamental skill for anyone running a business. Whether you're analyzing performance, setting prices, or making investment decisions, understanding your profit margins gives you clarity about your financial health. For businesses managing multiple fulfillment sources and complex supply chains, this becomes even more critical.
What Is Profit Percentage?
Profit percentage (or profit margin) measures how much of your revenue actually becomes profit after covering all costs. It's expressed as a percentage rather than a dollar amount, making it easier to compare performance across different products, time periods, or businesses.
Put simply, profit percentage answers: "For every dollar of sales, how many cents do I keep as profit?"
Why Profit Percentage Matters
Before diving into calculations, let's talk about why this number matters. It helps you make pricing decisions based on facts, not guesses. It shows which products or services are most profitable. It helps you track improvement (or decline) over time. It makes comparing your business to industry benchmarks possible. And it gives investors and lenders a clear picture of your business health.
The Basic Formula for Profit Percentage
The basic formula is straightforward:
Profit Percentage = (Net Profit ÷ Revenue) × 100
Let's break this down: Revenue is the total amount of money from sales before any expenses. Net Profit is what's left after subtracting all costs and expenses from revenue. The multiplication by 100 converts the decimal to a percentage.
For example, if your business made $100,000 in sales and your net profit was $25,000:
Profit Percentage = ($25,000 ÷ $100,000) × 100 = 25%
This means you keep 25 cents of every dollar sold as profit.
Different Types of Profit Percentages
There are actually several types of profit percentages, each serving a different purpose.
Gross Profit Percentage
This measures profitability after accounting for direct costs of goods sold (COGS), but before operating expenses.
Gross Profit Percentage = ((Revenue - COGS) ÷ Revenue) × 100
Gross profit percentage helps you understand how profitable your core product or service is before overhead.
Operating Profit Percentage
This includes both COGS and operating expenses but excludes taxes and interest.
Operating Profit Percentage = ((Revenue - COGS - Operating Expenses) ÷ Revenue) × 100
Operating profit percentage shows how efficiently you're running day-to-day operations.
Net Profit Percentage
This is the "bottom line" percentage after all expenses, including taxes and interest.
Net Profit Percentage = ((Revenue - All Expenses) ÷ Revenue) × 100
Net profit percentage reveals your true profitability after everything is paid.
Beyond Basic Profit Calculations for Multi-Channel Retailers
Standard profit formulas don't account for the complexity in multi-channel retail environments. If you're running a marketplace with third-party vendors, managing dropship relationships, or balancing direct and distributed fulfillment, your approach needs to be more sophisticated.
The Multi-Source Reality
When you're fulfilling orders through a mix of internal warehouses, multiple 3PL partners, dropship vendors, print-on-demand suppliers, and marketplace models like FBA, each channel has its own cost structure, fee arrangements, and operational implications that directly impact profitability.
Channel-Specific Profit Percentage Formulas
For Dropship/Marketplace Operations
Channel Profit % = ((Retail Price - Vendor Cost - Commission - Platform Fees - Payment Processing - Channel-Specific Shipping Subsidies) ÷ Retail Price) × 100
This captures the true margin after accounting for the unique costs in vendor-fulfilled models, including vendor/supplier wholesale costs, platform commissions (typically 10-30%), listing and transaction fees, and channel-specific shipping policies and customer expectations.
For Internal + 3PL Mixed Fulfillment
Blended Profit % = ((Revenue - COGS - (Internal Fulfillment Costs × Internal %) - (3PL Costs × 3PL %) - (FBA Fees × FBA %)) ÷ Revenue) × 100
This weighted approach accounts for variable fulfillment costs across locations, inventory carrying costs by location, labor efficiencies and inefficiencies, and volume-based fee structures.
For B2B Suppliers with Multiple Channels
B2B Profit % = ((B2B Revenue - COGS - Volume Discounts - Service Level Costs - Channel Acquisition Costs - Sales Commissions) ÷ B2B Revenue) × 100
This recognizes the unique aspects of B2B operations like tiered pricing structures, service level agreements costs, sales team commissions, and customer acquisition costs that vary dramatically by channel.
Step-by-Step Calculation Example
Let's walk through a complete example for a business selling custom electronics. Imagine you run a mid-sized electronics business with $150,000 in monthly revenue. Your COGS is $60,000, operating expenses total $45,000, and you have $5,000 in interest expenses plus $10,000 in taxes.
Starting with gross profit, subtract your COGS from revenue: $150,000 - $60,000 = $90,000. This gives you a gross profit of $90,000. To find your gross profit percentage, divide this by your revenue and multiply by 100: ($90,000 ÷ $150,000) × 100 = 60%. This means after covering the direct costs of your products, you keep 60% of your revenue.
Next, calculate your operating profit by subtracting operating expenses from gross profit: $90,000 - $45,000 = $45,000. Your operating profit percentage is ($45,000 ÷ $150,000) × 100 = 30%. At this stage, 30% of your revenue remains after covering both product costs and day-to-day operations.
Finally, determine your net profit by subtracting interest and taxes from operating profit: $45,000 - $5,000 - $10,000 = $30,000. Your net profit percentage is ($30,000 ÷ $150,000) × 100 = 20%. This bottom-line figure shows that after all expenses, you keep 20 cents of every dollar in sales.
Case Study: Multi-Source Brand Profit Analysis
Now let's explore a more complex scenario with a specialty retailer using three different fulfillment methods. This business generates $2.5 million in monthly revenue, with 40% fulfilled from their owned warehouse, 35% dropshipped from vendors, and 25% handled by an external 3PL.
For the owned warehouse channel, costs include $400,000 in COGS, $120,000 in labor, $75,000 in warehouse overhead, $35,000 in packaging, and $150,000 in shipping. Total channel costs reach $780,000 against $1 million in revenue (40% of total), yielding a channel profit of $220,000 or 22%.
The dropship channel involves $525,000 in vendor product costs, $8,000 in integration fees, $26,250 in platform transaction fees, and $18,000 in returns processing. With channel costs of $577,250 against $875,000 in revenue (35% of total), this channel generates $297,750 in profit, for an impressive 34% profit margin.
The 3PL channel includes $250,000 in COGS, $62,500 in 3PL fees, $15,000 in storage fees, $43,750 in pick/pack fees, and $93,750 in shipping. Total costs reach $465,000 against $625,000 in revenue (25% of total), creating $160,000 in profit at a 25.6% margin.
In addition to these channel-specific expenses, the business has shared costs: $250,000 in marketing, $85,000 in customer service, $105,000 in admin/overhead, and $45,000 in software/tech. These shared costs total $485,000.
After accounting for all channel-specific and shared costs, the total profit is $192,750 ($220,000 + $297,750 + $160,000 - $485,000), resulting in an overall profit percentage of just 7.7%.
This analysis reveals something crucial: while the dropship channel appears most profitable at 34% margin, the business's overall profit is significantly diluted by shared costs and lower-margin channels. This highlights why channel-specific analysis is essential for strategic decision-making in multi-source retail operations.
Interpreting Your Profit Percentages
Now that you've calculated these percentages, what do they tell you? A gross profit percentage of 60% means that after paying for direct costs, you have 60% of revenue left to cover operating expenses and generate profit. An operating profit percentage of 30% means that after core business expenses, you're converting 30% of sales into profit before interest and taxes. A net profit percentage of 20% means your business keeps 20 cents of every dollar sold after all expenses.
These numbers vary widely by industry. A grocery store might have a net profit percentage of 2-3%, while a software company might have 15-30%.
Advanced Profit Percentage Considerations
For Mixed Fulfillment Retailers
When managing multiple fulfillment sources, you need to account for hidden costs that basic formulas miss. Inventory balancing costs come from redistributing inventory between fulfillment locations. Split-order impacts occur when a single order requires multiple shipments from different sources. Channel conflict costs arise from price matching and competitive adjustments across marketplaces. Return routing complexity adds expenses when managing returns to different fulfillment sources.
An actionable insight: Calculate your "split order percentage" and the average additional cost per split order. For many retailers, each split order reduces profit by 15-25% compared to single-source fulfillment.
For Marketplace Retailers
Commission structure analysis becomes critical when working with marketplace models. Tiered commission impact shows how volume thresholds affect effective commission rates. Category-specific profitability identifies which product categories deliver higher retained profit. Vendor performance impact correlates vendor reliability metrics with true profitability.
An actionable insight: Build a vendor scorecard that incorporates both direct costs (commission rates, wholesale costs) and indirect costs (order accuracy, fulfillment speed, return rates) to identify your truly profitable vendors.
For B2B Suppliers
Customer acquisition costs (CAC) vary dramatically by channel. Direct sales CAC typically runs $800-1,500 per new account. Distribution partner CAC is usually $300-600 per new account. Self-service eCommerce CAC is often $150-400 per new account.
A more accurate B2B profit calculation includes CAC amortization:
True B2B Profit % = ((Revenue - COGS - Operating Expenses - (New Accounts × CAC ÷ Average Customer Lifespan)) ÷ Revenue) × 100
An actionable insight: Track customer acquisition payback period by channel. Most B2B suppliers find their direct sales channel has a 12-18 month payback period, while their eCommerce channel pays back in 3-6 months.
Common Pitfalls to Avoid
When calculating profit percentages, watch out for these common mistakes. Using the wrong revenue figure can skew results, so make sure you're using gross revenue (before returns) for some calculations and net revenue for others. Forgetting hidden costs like depreciation, inventory shrinkage, and owner's salary often leads to overstated profit.
Comparing unlike periods gives misleading trends, so always compare equal time frames like month to month or quarter to quarter. Industry mismatch can lead to unrealistic expectations, so don't compare your software company's margins to a restaurant's. Ignoring channel-specific costs means missing the true profitability picture for each fulfillment method.
Using Technology to Improve Profit Percentage Accuracy
API-enabled real-time cost tracking can connect your order management system with your suppliers' systems to capture true per-order costs. This includes real-time shipping rates, volume-based supplier discounts, actual payment processing fees, and dynamic marketplace commissions.
Multi-source order routing optimization leverages algorithms to determine the most profitable fulfillment source for each order. This considers real-time inventory availability, distance to customer, current shipping carrier performance, fulfillment center capacity, and vendor SLA performance.
Attribution modeling for accurate channel profitability helps you understand which marketing channels drive the most profitable orders, how customer acquisition source correlates with lifetime profitability, and which products serve as profit leaders versus loss leaders across channels. This includes evaluating the true ROI of pay per click campaigns compared to organic channels, ensuring that budget allocation maximizes profitability.
Implementation Strategy for Profit Percentage Improvement
A 90-day profit enhancement plan might look like this:
During days 1-15, focus on baseline assessment. Audit your current profit calculation methodology. Identify missing cost factors specific to your fulfillment model. Establish channel-specific benchmarks.
For days 16-30, work on data integration. Connect all fulfillment sources to your central data repository. Standardize cost allocation methodology across channels. Implement uniform vendor performance metrics.
In days 31-60, pursue channel optimization. Adjust routing logic to prioritize higher-margin fulfillment paths. Renegotiate terms with lowest-performing vendors. Implement automated shipping carrier selection based on profit impact.
For days 61-90, refine your strategy. Develop SKU-level profitability analysis across fulfillment methods. Create dynamic pricing rules based on fulfillment source. Build reporting dashboards that highlight profit enhancement opportunities.
Tools to Help Calculate Profit Percentage
You don't need to do this math manually every time. Several solutions can streamline profit tracking and analysis:
Accounting software like QuickBooks or Xero provides basic profit calculation capabilities. Many retailers use spreadsheet templates with built-in formulas, though these require regular maintenance and manual data entry.
For multi-channel retailers, purpose-built platforms offer more sophisticated profit analytics. Order management systems with integrated profitability reporting can track margins across different fulfillment sources. Multi-channel retail platforms like Flxpoint include built-in profitability reports that reconcile revenue and expenses automatically.
Flxpoint's Profitability Report, for example, takes channel invoice data as revenue and source invoice data as expenses to calculate margins for specific orders. It allows you to analyze data over custom time periods and filter by metrics like total profit, most/least profitable products, and highest/lowest margin items. The platform features pre-built integrations with QuickBooks and other accounting systems, ensuring your financial data stays synchronized without manual reconciliation.
With these tools, you can stay on top of your margins without waiting to close your books at the end of each period, and quickly identify your best and worst performing products across all fulfillment channels, making more data-driven decisions about what and how you're selling.
Ready to gain deeper insights into your profit margins across all your sales channels? Request a demo of Flxpoint's Profitability Report to see how you can track and optimize your profit percentages in real-time.
Final Thoughts
Calculating profit percentage isn't just a financial exercise. It's about gaining insights that help you make better business decisions. By tracking these metrics regularly, you'll develop an intuitive sense for your business's financial health and spot opportunities for improvement.
For multi-channel retailers with complex fulfillment models, profit percentage isn't just a financial metric—it's the key to strategic decision-making. By understanding the true cost implications of each fulfillment channel and implementing sophisticated profit tracking, you can identify which aspects of your business deliver genuine value and which require optimization.
The most successful businesses don't just calculate these figures, they actively manage them. They understand which levers to pull to improve each type of profit percentage and make deliberate choices based on that knowledge.
Remember that the goal isn't simply to know your profit percentage, but to use that knowledge to continually refine your operational model, vendor relationships, and channel strategy. In today's complex retail landscape, this level of financial clarity isn't just helpful—it's essential for sustainable growth
Flxpoint – Powerful Dropship and Ecommerce Automation Platform
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | | Cross Section Practice (Polyhedra) MathBitsNotebook.com Topical Outline | JrMath Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | Directions: Read carefully and choose the best answer. | | | | | | | | | | --- --- --- --- | | | | | --- | | Given a right rectangular prism. If a cross section is made parallel to this base, which of the following statements regarding the area of the cross section and the area of the base of the prism is correct. | | | | | | | | | | | | | | | | | --- --- --- | | | | | | --- | 2. | A right square pyramid is sliced through its apex (vertex, top) perpendicular to its base, but parallel to a side of the base. The sides of the base have a measure of 12 inches. The perpendicular distance from the apex to the center of the base is 18 inches. What is the area of the cross section made by the slice, in square inches? | | | Choose: | | | | | | | | | | --- --- --- | | | | | | --- | 3. | Of the choices "square", "rectangle" and "triangle", which of the following are possible cross sections obtained when slicing a cube either parallel or perpendicular to its base? | | | Choose: | | | | | | | | | | --- --- --- | | | | | | --- | 4. | This right rectangular pyramid is sliced parallel to its base. Which of these two-dimensional shapes describes the cross section? | | | Choose: | | | | | | | | | | --- --- --- | | | | | | --- | 5. | A right triangular prism is sliced perpendicular to its base. Which of these two dimensional shapes best describes the cross section? | | | Choose: | | | | | | | | | --- --- | | | | --- | | 6. | Which of the following statements is FALSE regarding possible cross sections when slicing a right square pyramid? | | Choose: | | | | | | | | | | | --- --- --- | | | | | --- | 7. | Given a cube. A cross section is sliced diagonally through the center of the cube, as shown. Which of the following statements is true regarding the area of the cross section and the area of the cube's face? | | | Choose: | | | | | | | | | | | --- --- --- | | | | | | --- | 7. | A plane intersects a cube perpendicular to the base such that the cross sections are not parallel to a lateral side. Which of the following statements is NOT true regarding the two-dimensional cross sections that could be formed by the plane? | | | Choose: | | | | | | | | | | | | | --- --- --- --- | | | | | --- | 8. | Given a right triangular prism as shown at the right. A cross section is made parallel to the bases. Which of the following statements is true regarding the area of the cross section and the area of the base? | | | | | | Choose: | | | | | | | | | | --- --- | | | | --- | | 9. | A solid has a triangle as a cross section. Which of the following solids could NOT have a triangle as a cross section? Allow for all possible orientations (angles) of the cross sections. | | Choose: | | | | | | | | | | | --- --- --- | | | | | --- | | A cube and a right square pyramid are sitting on a flat surface. Which of the following statements are true? (Check all that apply, and hit SUBMIT!) | | | | | | | | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | JrMath Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of Use Contact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. | | |
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Interleave different length lists, elimating duplicates, and preserve order
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Asked 12 years, 8 months ago
Modified3 years, 9 months ago
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I have two lists, let's say:
python
keys1 = ['A', 'B', 'C', 'D', 'E', 'H', 'I']
keys2 = ['A', 'B', 'E', 'F', 'G', 'H', 'J', 'K']
How do I create a merged list without duplicates that preserve the order of both lists, inserting the missing elements where they belong? Like so:
python
merged = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K']
Note that the elements can be compared against equality but not ordered (they are complex strings). The elements can't be ordered by comparing them, but they have an order based on their occurrence in the original lists.
In case of contradiction (different order in both input lists), any output containing all elements is valid. Of course with bonus points if the solution shows 'common sense' in preserving most of the order.
Again (as some comments still argue about it), the lists normally don't contradict each other in terms of the order of the common elements. In case they do, the algorithm needs to handle that error gracefully.
I started with a version that iterates over the lists with .next() to advance just the list containing the unmatched elements, but .next() just doesn't know when to stop.
```python
merged = []
L = iter(keys1)
H = iter(keys2)
l = L.next()
h = H.next()
for i in range(max(len(keys1, keys2))):
if l == h:
if l not in merged:
merged.append(l)
l = L.next()
h = H.next()
elif l not in keys2:
if l not in merged:
merged.append(l)
l = L.next()
elif h not in keys1:
if h not in merged:
merged.append(h)
h = H.next()
else: # just in case the input is badly ordered
if l not in merged:
merged.append(l)
l = L.next()
if h not in merged:
merged.append(h)
h = H.next()
print merged
```
This obviously doesn't work, as .next() will cause an exception for the shortest list. Now I could update my code to catch that exception every time I call .next(). But the code already is quite un-pythonic and this would clearly burst the bubble.
Does anyone have a better idea of how to iterate over those lists to combine the elements?
Bonus points if I can do it for three lists in one go.
python
list
iterator
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edited Feb 2, 2019 at 15:04
double-beep
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asked Jan 9, 2013 at 16:05
Chaos_99Chaos_99
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3 I don't think that the list you want to compute is guaranteed to exist in general. What if keys1 = ['A', 'B', 'D']; keys2 = ['D', 'C', 'B']?Ryan C. Thompson –Ryan C. Thompson 2013-01-09 16:21:20 +00:00 Commented Jan 9, 2013 at 16:21
1 how should an algorithm solve this case: keys1 = ['A', '%', ''] and keys1 = ['A', '@', '?']tzelleke –tzelleke 2013-01-09 16:22:22 +00:00 Commented Jan 9, 2013 at 16:22
@RyanThompson There are solutions, namely ['A', 'B', 'D', 'C', 'B'] and ['A', 'D', 'C', 'B', 'D'], but how to choose which one to return? And is an element allowed to be repeated in the output sequence?Khaur –Khaur 2013-01-09 16:25:58 +00:00 Commented Jan 9, 2013 at 16:25
I guess that's the point. The question gives an example where the desired answer is made obvious by spacing and the use of alphabetic characters in order, but then says that the elements are unordered. So the example given doesn't fully specify the what the desired result is in the general case.Ryan C. Thompson –Ryan C. Thompson 2013-01-09 18:00:27 +00:00 Commented Jan 9, 2013 at 18:00
1 Thinking some more, I wonder if the OP isn't effectively asking for a solution to the shortest common superstring problem?Ryan C. Thompson –Ryan C. Thompson 2013-01-09 18:02:04 +00:00 Commented Jan 9, 2013 at 18:02
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What you need is basically what any merge utility does: It tries to merge two sequences, while keeping the relative order of each sequence. You can use Python's difflib module to diff the two sequences, and merge them:
```python
from difflib import SequenceMatcher
def merge_sequences(seq1,seq2):
sm=SequenceMatcher(a=seq1,b=seq2)
res = []
for (op, start1, end1, start2, end2) in sm.get_opcodes():
if op == 'equal' or op=='delete':
#This range appears in both sequences, or only in the first one.
res += seq1[start1:end1]
elif op == 'insert':
#This range appears in only the second sequence.
res += seq2[start2:end2]
elif op == 'replace':
#There are different ranges in each sequence - add both.
res += seq1[start1:end1]
res += seq2[start2:end2]
return res
```
Example:
```python
keys1 = ['A', 'B', 'C', 'D', 'E', 'H', 'I']
keys2 = ['A', 'B', 'E', 'F', 'G', 'H', 'J', 'K']
merge_sequences(keys1, keys2)
['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K']
```
Note that the answer you expect is not necessarily the only possible one. For example, if we change the order of sequences here, we get another answer which is just as valid:
```python
merge_sequences(keys2, keys1)
['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'J', 'K', 'I']
```
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edited Jan 9, 2013 at 17:00
answered Jan 9, 2013 at 16:48
interjayinterjay
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Stephane Rolland
Stephane RollandOver a year ago
+1 for the J in non-alphabetical order ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'J', 'K', 'I'], that's what I was trying to say in my comment.
2013-01-09T16:56:22.72Z+00:00
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Stephane Rolland
Stephane RollandOver a year ago
And by (sheer :-)) curiosity how does it does work with let's says: ['K','A','O','S'] and ['S','O','A','K','I','N','G'] :-)
2013-01-10T16:22:45.143Z+00:00
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interjay
interjayOver a year ago
@StephaneRolland: In that case you'll get some duplication, for example KAOSOAKING.
2013-01-10T21:08:45Z+00:00
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interjay
interjayOver a year ago
@StephaneRolland It can be useful for making file comparison tools, source control software, incremental patching, etc. I don't think I've used it myself.
2013-01-11T14:16:05.86Z+00:00
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aschmid00Over a year ago
this fails when trying to merge 2 lists with reversed items. merge_sequences([1,2,3], [3,2,1]) will end up in [1,2,3,2,1]. any suggestions?
2014-05-21T15:36:07.093Z+00:00
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I suspect that you may be asking for a solution to the shortest common supersequence problem, which I believe is NP-hard in the general case of an arbitrary number of input sequences. I'm not aware of any libraries for solving this problem, so you might have to implement one by hand. Probably the quickest way to get to working code would be to take interjay's answer using difflib and then use reduce to run it on an arbitrary number of lists (make sure to specify the empty list as the 3rd argument to reduce).
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answered Jan 9, 2013 at 18:23
Ryan C. ThompsonRyan C. Thompson
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Chaos_99
Chaos_99Over a year ago
Yes, the wiki definition seems about right for my problem. Thanks for pointing out the correct term and the note on using reduce for more then two input sequences.
2013-01-10T10:08:02.88Z+00:00
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I would use a Set (cf. python doc), that I'd fill with the elements of the two lists, one aafter the other.
And make a list from the Set when it's done.
Note that there is a contradiction/paradox in your question: you want to preserve order for elements that cannot be compared (only equality because "they are complex strings" as you said).
EDIT: the OP is right noticing that sets don't preserve order of insertion.
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edited Jan 21, 2013 at 9:29
answered Jan 9, 2013 at 16:07
Stephane RollandStephane Rolland
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Khaur
KhaurOver a year ago
From the example I guess the elements between matches should be inserted starting with the ones from the first list.
2013-01-09T16:19:35.88Z+00:00
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interjay
interjayOver a year ago
A set is unordered. He wants to preserve the order of the two lists here. And there is no contradiction: It's possible to preserve ordering in a list without comparing the list's elements.
2013-01-09T16:37:49.757Z+00:00
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Stephane Rolland
Stephane RollandOver a year ago
The ordering you refer about is the order of the list, not the order of the item's value ( since they are not orderly comparable). The question is biaised because it represents the two list with the implicit alphabetical order which is not applicable according to the OP (because these are complexe strings...) A + iB ? :-)
2013-01-09T16:46:55.543Z+00:00
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Chaos_99
Chaos_99Over a year ago
+1 for getting the difference between the order of elements and orderable elements. (-1 for the set)
2013-01-10T10:02:12.343Z+00:00
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Stephane Rolland
Stephane RollandOver a year ago
thx for pointing me out that sets would not preserve the order of insertion.
2013-01-10T16:41:13.357Z+00:00
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By using only lists, you can achieve this with few simple for loops and .copy():
```python
def mergeLists(list1, list2):
# Exit if list2 is empty
if not len(list2):
return list1
# Copy the content of list2 into merged list
merged = list2.copy()
# Create a list for storing temporary elements
elements = []
# Create a variable for storing previous element found in both lists
previous = None
# Loop through the elements of list1
for e in list1:
# Append the element to "elements" list if it's not in list2
if e not in merged:
elements.append(e)
# If it is in list2 (is a common element)
else:
# Loop through the stored elements
for x in elements:
# Insert all the stored elements after the previous common element
merged.insert(previous and merged.index(previous) + 1 or 0, x)
# Save new common element to previous
previous = e
# Empty temporary elements
del elements[:]
# If no more common elements were found but there are elements still stored
if len(elements)
# Insert them after the previous match
for e in elements:
merged.insert(previous and merged.index(previous) + 1 or 0, e)
# Return the merged list
return merged
In : keys1 = ["A", "B", "D", "F", "G", "H"]
In : keys2 = ["A", "C", "D", "E", "F", "H"]
In : mergeLists(keys1, keys2)
Out: ["A", "B", "C", "D", "E", "F", "G", "H"]
```
English is not my first language, and this one is pretty hard to explain, but if you care about the explanation, here's what it does:
There's a local list called elements which can store temporary elements.
There's a local variable called previous which stores the previous element that was in both lists.
When ever it finds an element that is NOT in list2 but is in list1, it will append that element to elements list and continue the loop.
Once it hits an element that is in both lists, it loops through the elements list, appending all elements after previous element to list2.
The new match is then stored into previous and elements is reset to [] and the loop continues.
Beginning of the lists and end of the lists are counted as a common element, if first or last element is not a common element in both lists.
This way it will always follow this format:
Previous common element
Elements from list1, between two common elements
Elements in list2, between two common elements
New common element
So for example:
python
l1 = ["A", "B", "C", "E"]
l2 = ["A", "D", "E"]
The revious common element A will be first in the merged list.
Elements from l1 between the previous common element A and the new common element E will be inserted right after A.
Elements from l2 between the previous common elmeent A and the new common elmeent E will be inserted right after the elements from l1.
The new common element E will be last element.
Back to step 1 if more common elements found.
["A", "B", "C", "D", "E"]
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edited Jan 10, 2013 at 17:00
answered Jan 9, 2013 at 18:55
user1632861 user1632861
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interjay
interjayOver a year ago
Neither of these solutions preserve the order of the lists. The second solution is close but can change the order of elements in the second list.
2013-01-10T09:41:02.003Z+00:00
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Chaos_99
Chaos_99Over a year ago
As interjay said: with sets, the order is not preserved. The second solutions does not interleave, but only append.
2013-01-10T10:09:46.363Z+00:00
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user1632861
user1632861Over a year ago
Fixed it, and removed the set option.
2013-01-10T17:00:40.173Z+00:00
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I recently had stumbled upon a similar issue while implementing a feature. I tried to clearly define the problem statement first. If I understand right, here is the problem statement
Problem Statement
Write a function merge_lists which will merge a list of lists with overlapping items, while preserving the order of items.
Constraints
If item A comes before item B in all the lists where they occur together, then item A must precede item B in the final list also
If item A and item B interchange order in different lists, ie in some lists A precedes B and in some others B precedes A, then the order of A and B in the final list should be the same as their order in the first list where they occur together. That is, if A precedes B in l1 and B precedes A in l2, then A should precede B in final list
If Item A and Item B do not occur together in any list, then their order must be decided by the position of the list in which each one occurs first. That is, if item A is in l1 and l3, item B is in l2 and l6, then the order in the final list must be A then B
Test case 1:
Input:
l1 = ["Type and Size", "Orientation", "Material", "Locations", "Front Print Type", "Back Print Type"]
l2 = ["Type and Size", "Material", "Locations", "Front Print Type", "Front Print Size", "Back Print Type", "Back Print Size"]
l3 = ["Orientation", "Material", "Locations", "Color", "Front Print Type"]
merge_lists([l1,l2,l3])
Output:
['Type and Size', 'Orientation', 'Material', 'Locations', 'Color', 'Front Print Type', 'Front Print Size', 'Back Print Type', 'Back Print Size']
Test case 2:
Input:
l1 = ["T", "V", "U", "B", "C", "I", "N"]
l2 = ["Y", "V", "U", "G", "B", "I"]
l3 = ["X", "T", "V", "M", "B", "C", "I"]
l4 = ["U", "P", "G"]
merge_lists([l1,l2,l3, l4])
Output:
['Y', 'X', 'T', 'V', 'U', 'M', 'P', 'G', 'B', 'C', 'I', 'N']
Test case 3:
Input:
l1 = ["T", "V", "U", "B", "C", "I", "N"]
l2 = ["Y", "U", "V", "G", "B", "I"]
l3 = ["X", "T", "V", "M", "I", "C", "B"]
l4 = ["U", "P", "G"]
merge_lists([l1,l2,l3, l4])
Output:
['Y', 'X', 'T', 'V', 'U', 'M', 'P', 'G', 'B', 'C', 'I', 'N']
Solution
I arrived at a reasonable solution which solved it correctly for all the data I had. (It might be wrong for some other data set. Will leave it for others to comment that). Here is the solution
```python
def remove_duplicates(l):
return list(set(l))
def flatten(list_of_lists):
return [item for sublist in list_of_lists for item in sublist]
def difference(list1, list2):
result = []
for item in list1:
if item not in list2:
result.append(item)
return result
def preceding_items_list(l, item):
if item not in l:
return []
return l[:l.index(item)]
def merge_lists(list_of_lists):
final_list = []
item_predecessors = {}
unique_items = remove_duplicates(flatten(list_of_lists))
item_priorities = {}
for item in unique_items:
preceding_items = remove_duplicates(flatten([preceding_items_list(l, item) for l in list_of_lists]))
for p_item in preceding_items:
if p_item in item_predecessors and item in item_predecessors[p_item]:
preceding_items.remove(p_item)
item_predecessors[item] = preceding_items
print "Item predecessors ", item_predecessors
items_to_be_checked = difference(unique_items, item_priorities.keys())
loop_ctr = -1
while len(items_to_be_checked) > 0:
loop_ctr += 1
print "Starting loop {0}".format(loop_ctr)
print "items to be checked ", items_to_be_checked
for item in items_to_be_checked:
predecessors = item_predecessors[item]
if len(predecessors) == 0:
item_priorities[item] = 0
else:
if all(pred in item_priorities for pred in predecessors):
item_priorities[item] = max([item_priorities[p] for p in predecessors]) + 1
print "item_priorities at end of loop ", item_priorities
items_to_be_checked = difference(unique_items, item_priorities.keys())
print "items to be checked at end of loop ", items_to_be_checked
print
final_list = sorted(unique_items, key=lambda item: item_priorities[item])
return final_list
```
I've also open sourced the code as a part of the library named toolspy. So you can just do this
```python
pip install toolspy
from toolspy import merge_lists
lls=
merge_lists(lls)
```
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answered Apr 4, 2018 at 12:48
suryasankarsuryasankar
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Here's a C# solution I came up with -- using an extension method -- for the case where the two lists might not contain the same type of elements, so it takes a compare method and a selector method (that returns an object of the target type given the source object). In this case, the first list ("me") is modified to contain the final result, but it could be modified to create a separate list.
```python
public static class ListExtensions
{
///
/// Merges two sorted lists containing potentially different types of objects, resulting in a single
/// sorted list of objects of type T with no duplicates.
///
public static void MergeOrderedList(this List me, IReadOnlyList other, Func compare = null, Func selectT = null)
{
if (other == null)
throw new ArgumentNullException(nameof(other));
if (compare == null)
{
if (typeof(TMe).GetInterfaces().Any(i => i == typeof(IComparable)))
{
compare = (a, b) => ((IComparable)a).CompareTo(b);
}
else
{
throw new ArgumentNullException(nameof(compare),
"A comparison method must be supplied if no default comparison exists.");
}
}
if (selectT == null)
if (typeof(TMe).IsAssignableFrom(typeof(TOther)))
{
selectT = o => (TMe)(o as object);
}
else
{
throw new ArgumentNullException(nameof(selectT),
$"A selection method must be supplied if the items in the other list cannot be assigned to the type of the items in \"{nameof(me)}\"");
}
if (me.Count == 0)
{
me.AddRange(other.Select(selectT));
return;
}
for (int o = 0, m = 0; o < other.Count; o++)
{
var currentOther = other[o];
while (compare(me[m], currentOther) < 0 && ++m < me.Count) {}
if (m == me.Count)
{
me.AddRange(other.Skip(o).Select(selectT));
break;
}
if (compare(me[m], currentOther) != 0)
me.Insert(m, selectT(currentOther));
}
}
}
```
Note: I did write unit tests for this, so it's solid.
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answered May 14, 2020 at 18:25
Tom BogleTom Bogle
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Assuming the problem is to keep the same relative order of as many common items as possible.
Consider the nodes of a graph represent the index pairs m, n having same value in the corresponding lists. E.g. [a, b, c] and [b, a, c] => [(0, 1), (1, 0), (2, 2)]
The relative order of two nodes m, n and m', n' can be satisfied if and only if (m < m' and n < n') or (m > m' and n > n'). In previous example (0, 1), (1, 0) do not satisfy this condition and therefore it is impossible to satisfy the relative order of a and b in both lists. Whereas (1, 0), (2, 2) satisfy this property and therefore it is possible to preserve the order of a and c.
Based on this condition, find edges between all node pairs O(n^2). To find the most optimal arrangement, find the largest maximal clique (which is NP-complete O(3^(n/3))) using Bron–Kerbosch algorithm. The resultant same value index pairs can be used as the anchors to generate the merged list.
If approximate ordering is acceptable for a polynomial solution, below method uses union-find (with path compression and rank optimization) to approximate the maximal clique and runs in O(n^2) time and takes O(n) space.
```python
from collections import defaultdict
def find(cur, d):
path = []
while d[cur] != cur:
path.append(cur)
cur = d[cur]
for n in path:
d[n] = cur
return cur
def f(o, n):
if o == n:
return o
first_list = list(reversed(o))
second_list = list(reversed(n))
first_list_dict = {v: i for i, v in enumerate(first_list)}
common = []
for i, v in enumerate(second_list):
if v in first_list_dict:
common.append((first_list_dict[v], i))
if not common:
o.extend(n)
return o
if len(common) == len(first_list):
return list({v: None for l in (o, n) for v in l})
if len(common) == len(second_list):
return o
d = {p:p for p in common}
rank = defaultdict(int)
for i, p1 in enumerate(common):
for p2 in common[i+1:]:
if (p1 - p2) (p1 - p2) > 0:
h1, h2 = find(p1, d), find(p2, d)
if rank[h1] > rank[h2]:
h1, h2 = h2, h1
elif rank[h1] == rank[h2]:
rank[h2] += 1
d[h1] = h2
freq = defaultdict(list)
for p in common:
freq[find(p, d)].append(p)
largest_clique = sorted(max(freq.values(), key=len))
res = []
seen = set()
l_idx1, l_idx2 = 0, 0
while largest_clique:
idx1, idx2 = largest_clique.pop()
new = first_list[l_idx1-1:idx1:-1]
new.extend(second_list[l_idx2-1:idx2:-1])
new.append(first_list[idx1])
res.extend(v for v in new if v not in seen)
seen.update(new)
l_idx1, l_idx2 = idx1, idx2
return res
for o, n in (
,
,
,
,
,
,
):
print(f"{str(o): <30}, {str(n): <30} => {str(f(o, n)): >40}")
```
Gives:
python
[0, 1, 2, 3, 4, 5] , [5, 0, 1, 3, 4] => [0, 1, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5] , [7, 3, 1, 2, 4, 5] => [0, 7, 3, 1, 2, 4, 5]
[0, 1, 2, 3, 4, 5] , [3, 4, 5, 0, 1, 2] => [0, 1, 2, 3, 4, 5]
['A', 'B', 'C', 'E'] , ['A', 'D', 'E'] => ['A', 'B', 'C', 'D', 'E']
['A', 'B', 'D', 'F', 'G', 'H'], ['A', 'C', 'D', 'E', 'F', 'H'] => ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H']
[0, 1, 2, 3, 4] , [5, 6, 7, 8] => [0, 1, 2, 3, 4, 5, 6, 7, 8]
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5182 | https://www.sciencedirect.com/topics/engineering/copper-loss | Copper Loss - an overview | ScienceDirect Topics
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Copper Loss
In subject area:Engineering
Copper losses refer to the energy lost due to the resistance in the wire of the primary and secondary windings, which is influenced by the current flowing through them. These losses can be mitigated by using wires with a larger cross-sectional area in coil manufacturing.
AI generated definition based on:Transformers and Motors, 1989
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You might find these chapters and articles relevant to this topic.
Chapter
Magnetic Materials for Inductors and Transformers
2017, Power Supplies for LED Driving (Second Edition)Steve Winder
12.6 Copper Losses
Copper loss is the term used to describe the energy dissipated by resistance in the wire used to wind a coil. In 99.9% of cases this wire will be made of copper, whose resistivity at 20°C is about 1.73×10–8 Ω m. However, coils often have to operate above room temperature and will be heated by the operating losses in any case. The wire resistance at any temperature can be estimated from Table 12.1, developed by Mullard (now Philips).
Table 12.1. Wire resistance versus temperature
| Temperature (°C) | Multiplying Factor |
--- |
| 20 | 1.000 |
| 40 | 1.079 |
| 60 | 1.157 |
| 80 | 1.236 |
| 100 | 1.314 |
Unfortunately, the resistance of wire also increases as the frequency of signals passing through it increases. The phenomenon of the “skin effect” is when the magnetic field caused by the current flow tends to force the electrons to flow down the outside of the wire. An alternating magnetic field produced by the current in the wire induces an electric field, strongest at the center of the wire, which repels the electrons and forces them to the outside surface of the wire. Thus changes in current produces a force that opposes those changes, which is inductance on a small scale.
The skin depth is given in Table 12.2.
Table 12.2. Skin depth versus frequency
| Frequency | Skin Depth (mm) |
--- |
| 50 Hz | 9.36 |
| 1 kHz | 2.09 |
| 100 kHz | 0.209 |
| 1 MHz | 0.0662 |
| 10 MHz | 0.0209 |
Fortunately, Terman has created a formula for a wire gauge (in millimeters) where the skin effect increases resistance by 10%, which is a nominal limit that allows reasonable losses:
D=200 f mm.
For example, suppose we are operating at 100 kHz, then D=0.63 mm. Using a larger-diameter wire than this does not give much benefit because the current will not be carried in the center of the wire. In fact, in an LED driver (or any PWM power supply) harmonics are present at many times the switching frequency. In the earlier case, a significant proportion of the signal will have a frequency of 300 kHz.
In some cases, it is necessary to suffer higher-copper losses that are desirable, to have a transformer of a reasonable size. The use of Litz wire may be justified (although it is expensive) if low-copper loss is essential at high-switching frequency.
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Book 2017, Power Supplies for LED Driving (Second Edition)Steve Winder
Chapter
Fundamental Concepts: Transformers
1989, Transformers and MotorsGeorge Patrick Shultz
Copper Losses
Copper losses are due to the resistance of the wire in the primary and secondary windings and the current flowing through them. These losses can be reduced by using wire with large cross-sectional area in the manufacturing of the coils.
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Book 1989, Transformers and MotorsGeorge Patrick Shultz
Chapter
Electric Machines
2018, Power ElectronicsJean Pollefliet
1.7.1 Copper losses
In the primary and secondary windings energy losses occur. If R P and R S are the respective resistances of the windings then the losses may be written as R P . I P 2 and R S . I S 2. The sum of both is the total energy loss. This is referred to as the copper losses of the transformer.
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Book 2018, Power ElectronicsJean Pollefliet
Chapter
Design of generators for offshore wind turbines
2016, Offshore Wind FarmsA. McDonald, J. Carroll
8.3.5 Generator losses
The major sources of losses in AC machines are copper losses and iron losses. There are also minor loss mechanisms due to bearing friction and windage (aerodynamic drag on the rotor surface).
Copper losses result from current flowing through the resistance of any windings (field or armature). This can be reduced by reducing resistance, either by increasing the cross-sectional area of conductors (generally increases the size of the machine), lowering the winding temperature (reduces the resistivity) or using materials which have lower resistivity. Copper is usually used as the conductor material and increasing current densities can be achieved – without significant additional losses – with improving cooling systems.
Iron loss is an important loss component in the components that conduct flux, which changes with time. They are sometimes described as ‘core losses’. The two loss mechanisms are hysteresis and eddy current losses. Both of these increase with increasing flux density in the teeth and back iron. Hysteresis losses are proportional to the electrical frequency and the eddy current losses are proportional to the square of the electrical frequency. The electrical steel used in the stator and rotor of induction machines is affected, whereas the rotors of synchronous machines are less affected as the main field is stationary relative to the rotor.
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Book 2016, Offshore Wind FarmsA. McDonald, J. Carroll
Chapter
Component Technology
2012, Valve Amplifiers (Fourth Edition)Morgan Jones
Copper Losses
Copper wire has resistance, and in a well-designed transformer the losses due to resistance are equal in primary and secondary, and will therefore be related by:
R s=R p N 2
where N is the primary to secondary turns ratio.
Having equalised copper losses between primary and secondaries, total copper losses may be traded against iron losses in a given transformer design so that two transformers may have different proportions of iron to copper to achieve equal power ratings.
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Book 2012, Valve Amplifiers (Fourth Edition)Morgan Jones
Review article
Thermal management challenges in hybrid-electric propulsion aircraft
2024, Progress in Aerospace SciencesMajid Asli, ... Panagiotis Laskaridis
3.1 Losses
Losses cause reduction in efficiency during the conversion from electrical to mechanical power (motor) or vice versa (generator). They manifest themselves either as a direct heat source or indirect heat source by causing increased energy consumption and reducing the efficiency of the machine.
In real machines, joule losses occur primarily. These are usually divided into copper losses, excitation losses and brush transition losses. Copper losses P C u refer to those losses caused by the flowing current I (direct current or root mean square value of alternating current) and the electrical resistance R in windings, see Eq. (1).
(1)P C u=I 2⋅R
Excitation losses P E x occur in electrical circuits of an excitation winding including the associated slip ring transmission and external variable resistors, expressed in Eq. (2). No rotational voltages are induced in excitation windings, however, they build up electric fields.
(2)P E x=U E⋅I E
Brush transition losses P B r occur in the part of the winding which is connected to the external supply line by means of a sliding contact and exists and carries current due to a voltage drop across these brushes. The voltage drop U B is not proportional to the current I, shown in Eq. (3).
(3)P B r=U B⋅I
Iron losses P I r o n occur everywhere in the magnetic circuit where time-varying inductions are present. They consist of hysteresis P H y s t and eddy current losses P E d d y, Eq. (4). Hysteresis losses are linear, eddy current losses are quadratically dependent on the frequency.
(4)P I r o n=P H y s t+P E d d y
Friction losses P F r i c can occur due to bearing friction, brush friction and air friction. Air friction losses include rotor air friction losses (windage) and the power requirements of any cooler attached to the machine shaft.
The last type of loss to be mentioned is additional losses P A d d. In contrast to the previous losses, these are difficult to determine in terms of their place of origin, since their cause is not directly related to the operating principle of the machine. For quantification, typically certain tests/experiments are run to determine the aforementioned losses. The measured total loss is reduced by these portions, so that the rest is defined as additional loss .
The speed and torque of electric motors can usually be controlled independently, resulting in the requested mechanical output power according to the flight mission. This allows heat generation due to losses to be influenced and addressed during the design process and in the control scheme. A typical loss distribution for a PMSM is shown in Fig. 3. It is clear that joule losses (copper losses in the windings) dominate in all cases. In operating conditions with high speeds (maximum speed condition), the mechanical losses increase significantly due to bearing and air friction. In phases in which high torque is required (peak load), the proportion of current-related losses increases significantly.
Fig. 3. Loss distribution of a PMSM for different working conditions based on .
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Journal2024, Progress in Aerospace SciencesMajid Asli, ... Panagiotis Laskaridis
Chapter
Electric Machines
2018, Power ElectronicsJean Pollefliet
1.7 Energy losses
1.7.1 Copper losses
In the primary and secondary windings energy losses occur. If R P and R S are the respective resistances of the windings then the losses may be written as R P . I P 2 and R S . I S 2. The sum of both is the total energy loss. This is referred to as the copper losses of the transformer.
1.7.2 Iron losses
In ferromagnetic materials, subjected to a varying magnetic field, hysteresis losses occur:
(16-15)P h=k h.f.B^n W/kg
whereby:
k h =material constant of the ferromagnetic material used in relation to hysteresis losses.
f =frequency (Hz)
B^ =amplitude of the magnetic induction (T=Wb/m 2)
n =empirical constant for the magnetic material (1<n<3).
Since the magnetic circuit of a transformer is constructed from metal plates, hysteresis losses occur. To limit these losses, it is desirable that the material constant be as small as possible.
A possibility in this case is an iron alloy using silicon (e.g. 3% silicon). If the core was made from solid iron, then considerable eddy currents would occur. These can be dramatically limited by making the magnetic circuit from plates which are insulated from each other and the surface of which is in the direction of the flux. As result of this the path of the eddy currents is limited.
The eddy current losses P w can be determined with a formula in the following form:
(16-16)P w=k w.δ 2.f 2.B^2 W/kg
Here in :
k w=material constant with respect to the eddy current losses
δ=plate thickness in mm.
By adding silicon the electrical resistance is also increased as a result of which the eddy current losses are reduced. According to the last formula, it is advantageous to have the plates as thin as possible. Typical plate thickness lies between 0.3 and 1 mm for 50 Hz operation. The plates can be 0.02 mm for high frequencies. For band wound cores thicknesses of 0.003 to 0.3 mm are possible.
In many applications it is possible that non-sinusoidal waveforms occur. The eddy current losses are proportional to the square of the form factor a=V R M S V A V so that:
(16-17)P w=K w 1.1 1 2⋅a 2⋅δ 2⋅f 2⋅B^2
1.11=form factor of sinusoidal voltage
a=form factor of the actual voltage.
Hysteresis and eddy current losses form the iron losses. They are sometimes called the constant losses of the transformer since they do not depend upon the load but only the magnetic induction. The magnetic induction only depends upon the applied voltage. The following table provides an idea of the iron losses with plates between 0.2 and 0.5 mm thick, and a frequency of 50 Hz with an induction of 1 Tesla.
| Material | Losses in W/kg |
--- |
| commercial iron | 5 … 10 |
| Si-Fe, warm rolled | 1 … 3 |
| Si-Fe, cold rolled and crystal orientated | 0.3 … 0.6 |
| 50% Ni-Fe | 0.2 |
| approximately 65% Ni-Fe | 0.06 |
Fig.16-11 shows the total iron losses at 50 Hz for toroidal band wound cores of 0.3 mm (data for cold rolled 3% Si-Fe cores). In fig.16-12, we see the influence of the frequency on the total iron losses for the same material. Such cores are used for power transformers, impulse transformers, welding transformers, line transformers, etc….
Fig.16-11. Iron losses as a function of induction
Fig.16-12. Iron losses with the frequency as a parameter
In order to limit the eddy current losses at higher frequencies, ferromagnetic cores are used. These ferrites consist of an alloy of iron oxide with modern materials such as manganese, nickel,… Oxides have a low electrical conductivity. Ferrites can be made with no losses into the MHz-range. The maximum self induction of ferrites (0.3 to 0.5 T) is less than that of Si-steel plates (1 to 1.5 T) since a large part of the volume is composed of oxygen atoms that are obviously nonmagnetic. Fig.16-13 shows the core losses for ferrite (Siemens) as a function of the inductance, while fig.16-14 shows the losses as a function of the frequency for the same material.
Fig.16-13. Core losses N27 (for R16 ring cores)
Fig. 16-14. Losses N27 (for R16 ring cores)
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Book 2018, Power ElectronicsJean Pollefliet
Chapter
Electric Machines, Design
2003, Encyclopedia of Physical Science and Technology (Third Edition)Enrico Levi
III.C Copper Losses
The Joule losses in the windings go by this name, because copper is the material most commonly used. The Joule loss per unit volume of the conductor is
(18)p Cu=J 2/γ=K 2/a 2 γ k dp 2,
and for the whole winding it is
P Cu=π D l co a J 2 γ=π D L(l co L)K 2 a γ k dp 2,
where
(19)l co∼(1+2 D pL)L∼2.27 p−0.16 L
denotes the average length of the conductors and γ is the electrical conductivity of the material calculated at two-thirds the maximum temperature allowed by the insulation class. Additional losses due to skin and proximity effects and harmonics may augment the dc loss by more than 50%. The copper losses amount to ∼35% of the total loss in the winding carrying the magnetizing current and ∼25% in the other winding.
A good starting guess for the copper loss in the conductor, which is critically stressed thermally, is
(20)P Cu=0.42 P i 0.75,
where the ideal power P i is given by the rated power P R divided by the product of the efficiency η R and power factor cos ϕ R.
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Reference work 2003, Encyclopedia of Physical Science and Technology (Third Edition)Enrico Levi
Review article
A review on methods of finding losses and cooling methods to increase efficiency of electric machines
2021, Ain Shams Engineering JournalEdison Gundabattini, ... Rosli Abu Bakar
2.1.4 Power losses
Fixed losses are losses in the active iron, and additional no load losses in other metal parts, Losses due to friction and windage loss in the machine. Load losses are copper losses in primary windings, losses in secondary windings and additional load losses. Stator losses,
a.
Stator Copper losses: losses are produced when the current passing through the stator windings, and generates the heat and consequently, the temperature of the motor rises. These losses are dependent on the square of stator current.
b.
Rotor bars losses: losses are produced when the current passing through the rotor bars, they are dependent on the square of rotor current.
c.
Iron losses or core losses: losses are generated in the conducting core laminations, due to hysteresis, eddy-current. These play an important role in design of the machine and in determination of its thermal rating.
d.
Mechanical losses contain air friction losses and bearing friction losses. Air friction losses result following the air circulation around the rotor during its operation. The air friction losses are divided in to two parts: the losses corresponding to the rotor seen as a rotating cylinder and the losses corresponding to the end surfaces of the rotor. Bearing friction losses are the results of the relative motion in bearings.
Copper losses are due to Joule losses, iron losses are due to eddy current and hysteresis effect. Compared to the conventional engine losses, they are quite low because the power is generally lower and the efficiency is much higher. Electric motors have very high efficiency, however their power losses produce a significant amount of heat. All the power losses become heat and increase the temperature of the component. The maximum temperature have an important effect on the de-rating of the power electronic components. In order to avoid premature de-rating, it is important to leave the temperature as low as possible . Core loss seen in Fig. 5 is much greater than other losses and is the main source of heat due to the high operating frequency. Therefore, it is of crucial importance to calculate the core loss accurately, taking operating temperature into account. The core eddy current loss and hysteresis loss are due to temperature-induced changes in core material resistivity . In order to design a high-efficiency motor, the iron loss generated in the motor should be reduced. The stator loss, armature winding loss, rotor loss, and axial structure loss are assigned as the heat sources in the electrical machine. The large rotor’s eddy current loss will cause the temperature of permanent magnet to increase considerably [29–32].
Fig. 5. Different type of losses acting as heat sources.
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Journal2021, Ain Shams Engineering JournalEdison Gundabattini, ... Rosli Abu Bakar
Chapter
Optimal network reconfiguration of active distribution network
2022, Active Electrical Distribution NetworkBaseem Khan, Ziad M. Ali
3.1 Power loss of distribution system
The impedance of the wires is a source of system losses. Copper losses , which are functions of the square of current flowing through the line, account for the majority of active power losses in the utility grid.
(1)P loss=I 2⁎R
where I represents the current and R represents the resistance of the conductor.
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Book 2022, Active Electrical Distribution NetworkBaseem Khan, Ziad M. Ali
Related terms:
Energy Engineering
Antenna
Amplifier
Power Supply
Permanent Magnet
Rotors
Chrominance
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Power Factor
Induction Motor
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5183 | https://pobchemteam.weebly.com/uploads/2/5/7/7/25772138/ch_12_prac_test_kinetics1.pdf | A.P. Chemistry Practice Test: Ch. 12, Kinetics MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) Consider the following reaction: 3A ¬ 2B The average rate of appearance of B is given by D[B]/Dt. Comparing the rate of appearance of B and the rate of disappearance of A, we get D[B]/Dt = _ x (-D[A]/Dt).
A) -2/3 B) -3/2 C) +2/3 D) +3/2 E) +1 2) A burning splint will burn more vigorously in pure oxygen than in air because A) nitrogen is a reactant in combustion and its low concentration in pure oxygen catalyzes the combustion.
B) oxygen is a reactant in combustion and the concentration of oxygen is higher in pure oxygen than it is in air.
C) oxygen is a product of combustion.
D) nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature.
E) oxygen is a catalyst for combustion.
3) Of the following, all are valid units for a reaction rate except _.
A) mol/L B) M/s C) mol/hr D) mol/L-hr E) g/s A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) ¬B(g). The following data are obtained for [A] as the reaction proceeds: Time (s) 1 10 20 30 40 Moles of A 0.124 0.110 0.088 0.073 0.054 4) The average rate of disappearance of A between 10 s and 20 s is _ mol/s.
A) 1.1 ˛ 10-3 B) 4.4 ˛ 10-3 C) 454 D) 9.90 ˛ 10-3 E) 2.2 ˛ 10-3 The peroxydisulfate ion (S2O82-) reacts with the iodide ion in aqueous solution via the reaction: S2O82- (aq) + 3I- ¬ 2SO4 (aq) + I3- (aq) An aqueous solution containing 0.050 M of S2O82- ion and 0.072 M of I- is prepared, and the progress of the reaction followed by measuring [I-]. The data obtained is given in the table below.
Time (s) 0 400 800 1200 1600 [I-] (M) 0.072 0.057 0.046 0.037 0.029 5) The concentration of S2O82- remaining at 800 s is __ M.
A) 0.015 B) 0.041 C) 0.076 D) 4.00 ˛10-3 E) 0.046 6) Of the units below, __ are appropriate for a first-order reaction rate constant.
A) L mol-1 s-1 B) s-1 C) M-1 s-1 D) mol/L E) M s-1 1 7) If the rate law for the reaction 2A + 3B ¬ products is first order in A and second order in B, then the rate law is rate = _.
A) k[A]2[B]3 B) k[A]2[B]2 C) k[A][B] D) k[A]2[B] E) k[A][B]2 8) The kinetics of the reaction below were studied and it was determined that the reaction rate increased by a factor of 9 when the concentration of B was tripled. The reaction is _ order in B.
A + B ¬ P A) zero B) first C) second D) third E) one-half The data in the table below were obtained for the reaction: 2 ClO2 (aq) + 2 OH- (aq) ¬ ClO3- (aq) + ClO2- (aq) + H2O (1) Experiment Number [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 9) What is the order of the reaction with respect to ClO2?
A) 4 B) 1 C) 0 D) 2 E) 3 10) What is the order of the reaction with respect to OH-?
A) 0 B) 1 C) 2 D) 3 E) 4 11) What is the overall order of the reaction?
A) 0 B) 1 C) 4 D) 3 E) 2 12) What is the magnitude of the rate constant for the reaction?
A) 115 B) 1.15 x 104 C) 713 D) 4.6 E) 230 13) For a first-order reaction, a plot of _ versus _ is linear.
A) ln [A]t, t B) 1 [A]t , t C) In [A]t, 1 t D) [A]t, t E) t, 1 [A]t 2 14) The rate law for a reaction is rate = k [A][B]2 Which one of the following statements is false?
A) If [B] is doubled, the reaction rate will increase by a factor of 4.
B) The reaction is second order in B.
C) The reaction is first order in A.
D) k is the reaction rate constant E) The reaction is second order overall.
15) The half-life of a first-order reaction _.
A) is constant B) is the time necessary for the reactant concentration to drop to half its original value C) can be calculated from the reaction rate constant D) does not depend on the initial reactant concentration E) All of the above are correct.
16) The reaction CH3-N”C ¬ CH3-C”N is a first-order reaction. At 230.3eC, k = 6.29 x 10-4 s-1. If [CH3-N”C] is 1.00 x 10-3 initially, [CH3-N”C] is _ after 1.000 x 103 s.
A) 4.27 x 10-3 B) 2.34 x 10-4 C) 5.33 x 10-4 D) 1.88 x 10-3 E) 1.00 x 10-6 3 17) Which one of the following graphs shows the correct relationship between concentration and time for a reaction that is second order in [A]?
A) B) C) D) E) 18) A first-order reaction has a rate constant of 0.33 min-1. It takes _ min for the reactant concentration to decrease from 0.13 M to 0.088 M.
A) 1.2 B) 1.4 C) 0.13 D) 0.85 E) 0.51 19) The rate constant for a second-order reaction is 0.13 M-1s-1. If the initial concentration of reactant is 0.26 mol/L, it takes _ s for the concentration to decrease to 0.13 mol/L.
A) 1.0 B) 4.4 x 10-3 C) 0.017 D) 0.50 E) 30 4 20) The graph shown below depicts the relationship between concentration and time for the following chemical reaction.
The slope of this line is equal to _.
A) -k B) -1/k C) k D) ln[A]o E) 1/k 21) The rate constant of a first-order process that has a half-life of 225 s is _ s-1.
A) 3.08 x 10-3 B) 12.5 C) 1.25 D) 4.44 x 10-3 E) 0.693 22) At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) ¬ CH33CN (g) The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9e C.
The rate constant for the reaction is _ s-1.
A) -5.2 ˛ 10-5 B) +1.9 ˛ 104 C) +6.2 D) -1.9 ˛ 104 E) +5.2 ˛ 10-5 23) The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2N2O5 (soln) ¬ 4NO2 (soln) + O2 (soln) The reaction is first order and has a rate constant of 4.82 ˛ 10-3 s-1 at 64eC. The rate law for the reaction is rate = ___.
A) k [N2O5]2 [NO2]4 [O2] B) k[N2O5]2 C) k[N2O5] D) 2k[N2O5] E) k [NO2]4 [O2] [N2O5]2 5 24) The rate of a reaction depends on _.
A) collision frequency B) collision orientation C) collision energy D) all of the above E) none of the above 25) Which energy difference in the energy profile below corresponds to the activation energy for the forward reaction?
A) x B) y C) x + y D) y - x E) x - y 26) In the energy profile of a reaction, the species that exists at the maximum on the curve is called the _.
A) product B) enthalpy of reaction C) atomic state D) activated complex E) activation energy 27) In the Arrhenius equation, k = Ae-Ea/RT _ is the frequency factor.
A) k B) A C) e D) R E) Ea 6 28) At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) ¬ CH3CN (g) The dependence of the rate constant on temperature is studied and the graph below is prepared from the results. The energy of activation of this reaction is _ kJ/mol.
A) 4.4 ˛ 10-7 B) 1.9 ˛ 104 C) 1.6 ˛ 105 D) 160 E) 4.4 ˛ 10-4 29) The mechanism for formation of the product X is: A + B ¬ C + D (slow) B + D ¬ X (fast) The intermediate reactant in the reaction is _.
A) A B) B C) C D) D E) X 30) For the elementary reaction NO3 + CO ¬ NO2 + CO2 the molecularity of the reaction is _, and the rate law is rate = _.
A) 4, k[NO3][CO][NO2][CO2] B) 2, k[NO3][CO]/[NO2][CO2] C) 4, k[NO2][CO2]/[NO3][CO] D) 2, k[NO2][CO2] E) 2, k[NO3][CO] 31) Of the following, ____ will lower the activation energy for a reaction.
A) increasing the concentrations of reactants B) adding a catalyst for the reaction C) raising the temperature of the reaction D) removing products as the reaction proceeds E) increasing the pressure 7 32) The rate law of the overall reaction A + B ¬ C is rate = k[A]2. Which of the following will not increase the rate of the reaction?
A) increasing the concentration of reactant A B) increasing the temperature of the reaction C) increasing the concentration of reactant B D) adding a catalyst for the reaction E) All of these will increase the rate.
33) A particular first-order reaction has a rate constant of 1.35 ˛ 102 s-1 at 25eC. What is the magnitude of k at 75eC if Ea = 85.6 kJ/mol?
A) 670 B) 3.47 ˛ 104 C) 3.85 ˛ 106 D) 1.93 ˛ 104 E) 1.36 ˛ 102 8 Answer Key Testname: CH_12_PRAC_TEST_KINETICS.TST MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) C ID: chem9b 14.1-1 2) B ID: chem9b 14.1-2 3) A ID: chem9b 14.1-3 4) E ID: chem9b 14.1-9 5) B ID: chem9b 14.1-19 6) B ID: chem9b 14.1-24 7) E ID: chem9b 14.1-26 8) C ID: chem9b 14.1-28 9) D ID: chem9b 14.1-37 10) B ID: chem9b 14.1-38 11) D ID: chem9b 14.1-39 12) E ID: chem9b 14.1-40 13) A ID: chem9b 14.1-41 14) E ID: chem9b 14.1-42 15) E ID: chem9b 14.1-45 16) C ID: chem9b 14.1-47 17) E ID: chem9b 14.1-49 18) A ID: chem9b 14.1-50 19) E ID: chem9b 14.1-52 20) A ID: chem9b 14.1-54 21) A ID: chem9b 14.1-60 1 Answer Key Testname: CH_12_PRAC_TEST_KINETICS.TST 22) B ID: chem9b 14.1-65 23) C ID: chem9b 14.1-67 24) D ID: chem9b 14.1-69 25) A ID: chem9b 14.1-70 26) D ID: chem9b 14.1-71 27) B ID: chem9b 14.1-72 28) D ID: chem9b 14.1-74 29) D ID: chem9b 14.1-75 30) E ID: chem9b 14.1-77 31) B ID: chem9b 14.1-80 32) C ID: chem9b 14.1-81 33) D ID: chem9b 14.2-12 2 |
5184 | https://math.stackexchange.com/questions/2020153/limit-of-the-sequence-a-n-sqrtn1n1-sqrtnn | Limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}$ - Mathematics Stack Exchange
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Limit of the sequence a n=(n+1)!−−−−−−√n+1−n!−−√n a n=(n+1)!n+1−n!n
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Find the limit of the sequence a n=(n+1)!−−−−−−−√n+1−n!−−√n.a n=(n+1)!n+1−n!n.
The limit is supposed to be e−1 e−1 and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is 1.1. Please tell me what's going wrong here.
Proof: We can write a n=n!−−√n(b n−1),a n=n!n(b n−1), where b n=(n+1)!√n+1 n!√n.b n=(n+1)!n+1 n!n. Hence
a n=n⋅n!−−√n n⋅b n−1 ln b n⋅ln b n=n!−−√n n⋅b n−1 ln b n⋅ln b n n.a n=n⋅n!n n⋅b n−1 lnb n⋅lnb n=n!n n⋅b n−1 lnb n⋅lnb n n.
But lim n→∞n!√n n=e−1,lim n→∞n!n n=e−1, so b n→1 b n→1 as n→∞.n→∞. On the other hand,
lim n→∞b n n=lim n→∞(n+1)!n!.1(n+1)!−−−−−−−√n+1=e.lim n→∞b n n=lim n→∞(n+1)!n!.1(n+1)!n+1=e.
Also
lim n→∞b n−1 ln b n=lim n→∞1 ln[1+(b n−1)]1/(b n−1)=1 ln e=1 lim n→∞b n−1 lnb n=lim n→∞1 ln[1+(b n−1)]1/(b n−1)=1 lne=1
And the author goes on to say that the limit is e−1.e−1. But doesn't it say that the limit is 1?1? I must have overlooked something so obvious.
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edited Feb 21 at 6:58
Riemann
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asked Nov 18, 2016 at 16:34
Bijesh K.SBijesh K.S
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lim n→∞a n=lim n→∞(n!−−√n n)lim n→∞(b n−1 log(b n))lim n→∞(log(b n n))=(e−1)(1)(log(e))=(e−1)(1)(1)=e−1 lim n→∞a n=lim n→∞(n!n n)lim n→∞(b n−1 log(b n))lim n→∞(log(b n n))=(e−1)(1)(log(e))=(e−1)(1)(1)=e−1
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edited Nov 18, 2016 at 16:48
answered Nov 18, 2016 at 16:41
Mark ViolaMark Viola
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OMG!!!!! Here I sit realizing how many hours I wasted. Thanks a lot.Bijesh K.S –Bijesh K.S 2016-11-18 16:45:43 +00:00 Commented Nov 18, 2016 at 16:45
4 You're welcome! My pleasure. -Mark Mark Viola –Mark Viola 2016-11-18 16:46:08 +00:00 Commented Nov 18, 2016 at 16:46
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5185 | https://metricviews.uk/2010/12/04/snow-calculations-made-simple-or-not/ | Snow calculations made simple, or not – Metric Views
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Snow calculations made simple, or not
One of our readers, John Frewen-Lord, has been housebound for four days by the recent heavy snow falls – about 60 cm deep in his area. This has prompted him to provide an illustration of the comparative simplicity of calculating snow loads in metric units.
John writes:
“Two winter calculations. Which is easier?
Like many parts of the UK in the beginning of December 2010, we in North East Lincolnshire have had a lot of snow. Up to now we’ve had around 60 cm. On the roofs of our cars it is about 50 cm deep. Like all good motorists, and having just spent three days digging one car out, we do not drive off with snow like this on the roof. Apart from it being dangerous to vehicles behind (and possibly illegal as well), it is a huge amount of extra weight to carry around! Carrying all that extra weight not only increases fuel consumption unnecessarily, but it is more likely to get you stuck in deep snow – a lighter vehicle will not bog down in snow as much as a heavier one.
So how much extra weight does this 50 cm of snow amount to? In metric it is a very easy calculation. The only things we need to know are that snow has about 1/10 th the density of liquid water, 1 litre (L) is 1000 th of a cubic metre, and that 1 L of water weighs 1 kg. Now from this it can be seen that 1 L = 1 m² x 1 mm deep.
Now I measured the pile of snow on the roof of my car, and it was approximately 2 m long x 1.5 m wide, or 3 m². Using the above information, 3 m² of snow x 50 cm deep is equivalent to 3 m² of water x 50 mm deep. Therefore the snow on the roof of my car will have an equivalent water volume of 3 x 50 = 150 L. Which will weigh 150 kg.
As the average adult person weighs 75 kg, that is equivalent to carrying an extra two people around. I think you will agree that those calculations are pretty simple, and could be easily done in your head. Could you do the same in imperial units? I very much doubt it.
Let’s say the snow on the roof of my car is 6ft 6 in long by 5 ft wide, and that it is 20 in deep. We need to convert the feet and inches to feet only, so the area becomes 6.5 x 5 = 32.5 ft². The 20 in of snow is equivalent to 32.5 ft² x 2 in deep of water. Where we do we go from here? The easiest way is to calculate the volume of water in cubic feet, by converting the 2 in to decimals of a foot. This then becomes 32.5 x 0.167 = 5.43 ft³. Now 1 ft³ = 6.22883288 imperial gallons (let’s round that to 6.23 gallons). Our water therefore has a volume of 5.43 x 6.23 = 33.8 gallons. How much does 1 gallon of water weigh? The imperial gallon (but not the US gallon) happens to weigh 10 lbs, so the weight of the snow on the roof of my car would be 33.8 x 10 = 338 lbs.
We’re not finished yet. That weight of 338 lbs, representing the weight of two people of 169 lbs each, needs to be converted to those stones that we British seem to love. There are 14 lbs in a stone, therefore 169 lbs = 12.07 stones = 12 stone 1 lb.
While you may not want to calculate the weight of snow on the roof of your car (though it is an eye-opener as to just how much it does weigh), this sort of calculation is surely typical of the kind that most of have to make continually in our lives. Whereas the metric calculations in this example involved just 4 calculations which were easily done in your head, and involved NO conversion factors, the imperial calculations involved no fewer than 9 calculations, of which 6 involved a conversion factor of one sort or another. And unless you have lots of time on your hands and like doing calculations manually with some awkward numbers, a calculator is necessary.
I am still amazed that we in the UK still want to torture ourselves with calculations like this.”
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Author Derek PollardPosted on 2010-12-04 2020-08-09Categories Education, General, TechnicalTags calculations, metric
8 thoughts on “Snow calculations made simple, or not”
John Steelesays: 2010-12-05 at 11:59 John,
I agree metric is easier, and with all your calculation methods. However, be aware that 10% of water density is a very crude approximation to snow density. It varies considerable with conditions. If the snow packs well into snowballs, it is appreciably heavier, 2X or more. You may find this article helpful. Apparently attempts to predict density from temperature and other conditions are only rough guides, and you really need to melt a core for snow-water-equivalent.
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2. dereksays: 2010-12-05 at 12:35 Many people deprecate the use of the decimetre (dm), but I found it useful here as 1 dm³ = 1 L. Hence, weight of snow = 20 x 15 x 5 x 0.1 = 150 kg. Just one step, compared with imperial’s nine.
John has chosen a topical example of the simplicity of metric calculations, but others can be found in every walk of life, at work and in the home. Time will show if those people, companies and countries which prefer expensive and inefficient practices can survive and prosper in global markets.
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3. philhsays: 2010-12-05 at 16:26 Just as a point of interest, I do know of an incident where someone was fined £60 and 3 penalty points for driving their car with snow on the roof.
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4. philhsays: 2010-12-06 at 09:26 I strongly agree with Derek about the usefulness of the decimetre.
We don’t have to limit ourselves to rectangular shapes either. For example the volume of a cylinder is quite close to 80% of the enclosing box shape (actually it’s pi/4 = 0.785 to 3 d.p.). So consider a tin can 70 mm in diameter and 100 mm tall, then converting to dm we have 0.7 x 0.7 x 1 x 0.8 = 0.392 or about 400 ml
Critics would point out that the convenient fluid mass relationship to the litre only applies to water. True, but at least the basic unit used for liquids does relate sensibly to linear units. Can’t say that for the pint versus the cubic inch or the gallon versus the cubic foot etc.
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5. Martin Vlietstrasays: 2010-12-06 at 12:15 John raised the interesting question – “Which is the easier to calculate”. The answer is self-evident, but was John asking the right question? In reality there is a mood of “Am I bovvered?” in the UK – fuelled by, amongst other things, the complex calculations needed in situations that could be simplified by using metric units.
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6. BrianACsays: 2011-06-14 at 18:03 This brings up another point, should we bother? Well, yes, we should. The point, houses, buying, doing them up and furnishing. Here in UK we have room sizes almost exclusively in ft and ins times the number of rooms as the single guide to spending hundreds of thousands of pounds on a house. Try asking the floor area of the house, try working it out! I have only bought one house outside UK but have rented plenty. Always the first thing you know is the area in sq. m.,the land in hectares, often a ground plan as well. No need to look at 100 homes to see that 95 of them are not what you want. Yet we seem to love the system here. Then buying materials, in France just about everything is sold buy the sq.m, just measure the room then buy the same amout of material and mostly the job is done. Replacing the boiler? already got the house size job done.
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7. Tiborsays: 2016-02-10 at 18:56 Everything is ok, I was looking at the statement you give, 1 / 10 density, only a little error; you forgot to divide 150 with 10, so that was only 15 kilos !!! It’s another question whether the density is constant; it could be a light snow but other times wetter like …I’m living on mountain…Kopaonik in Serbia, former Yugoslavia…
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8. Graham Tobinsays: 2018-03-02 at 18:01 @Tibor
I just shovelled snow off a trampoline and found this calculation.
It is 3.4 m 3.4 m and it was very heavy.
Regarding your correction. Please note the calc already has reduced by 10.
“3 m² of snow x 50 cm deep is equivalent to 3 m² of water x 50 mm deep”
3 m² of snow x 500 mm deep is equivalent to 3 m² of water x 50 mm deep
The 500mm to 50 mm here is the division by 10.
My snow was 20 cm deep and I definitely shoveled off 230 kg of snow rather than 23 kg.
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5186 | https://www.cuemath.com/ncert-solutions/the-ratio-of-exterior-angle-to-interior-angle-of-a-regular-polygon-is-1-4-find-the-number-of-sides-of-the-polygon/ | The ratio of exterior angle to interior angle of a regular polygon is 1:4. Find the number of sides of the polygon.
Solution:
Given, the ratio of exterior angle to interior angle of a regular polygon is 1 : 4
We have to find the number of sides of the polygon.
Let the exterior angle of a polygon be x.
Interior angle = 180° - exterior angle
So, interior angle = 180° - x
According to the question,
x/(180° - x) = 1/4
On cross multiplication,
4x = 180° - x
By transposing,
4x + x = 180°
5x = 180°
x = 180°/5
x = 36°
Number of sides of a polygon = 360°/exterior angle
= 360°/36°
= 10
Therefore, the number of sides of a polygon is 10.
✦ Try This: The ratio of exterior angle to interior angle of a regular polygon is 2 : 5. Find the number of sides of the polygon.
☛ Also Check: NCERT Solutions for Class 8 Maths
NCERT Exemplar Class 8 Maths Chapter 5 Solved Problem 32
The ratio of exterior angle to interior angle of a regular polygon is 1:4. Find the number of sides of the polygon.
Summary:
The ratio of exterior angle to interior angle of a regular polygon is 1:4. The number of sides of the polygon is 10.
☛ Related Questions: |
5187 | https://ojs.utlib.ee/index.php/ACUTM/article/download/ACUTM.2014.18.02/12091/18733 | ACTA ET COMMENTATIONES UNIVERSITATIS TARTUENSIS DE MATHEMATICA Volume 18, Number 1, June 2014 Available online at
Algebraic trigonometric values at rational multipliers of π
Pinthira Tangsupphathawat
Abstract. The problem of finding all algebraic values of α∈[−1,1] when arccos α, arcsin αand arctan αare rational multiples of πis solved. The values of such αof degree less than five are explicitly determined.
Introduction
Rational and higher degree algebraic values of the cosine function have been of much interest for quite some time, cf. [4, 3, 8, 10]. As early as 1933, Lehmer proved that if k/n, n > 2, is an irreducible fraction, then 2 cos(2 πk/n ) and 2 sin(2 πk/n ) are algebraic integers of degree ϕ(n)/2, where
ϕ(n) is the Euler’s totient function. Lehmer’s proof makes use of cyclotomic polynomials. As a consequence, we have [7, Theorem 6.16, pp. 308–309]: let
θ = rπ be a rational multiple of π. Then cos θ, sin θ and tan θ are irrational numbers apart from the cases where tan θ is undefined, and the exceptions cos θ = 0 , ±1/2, ±1; sin θ = 0 , ±1/2, ±1; tan θ = 0 , ±1.
Recently, Varona proved that if r ∈ Q ∩ [0 , 1] then arccos( √r) is a rational multiple of π if and only if r ∈ { 0, 1/4, 1/2, 3/4, 1}. His proof is elementary and is similar to the proof of [1, Theorem 4, p. 32]. In Section 2, we push further the result of Varona, using elementary trigonometric identities, to find all possible nonnegative rational and some quadratic (i.e., algebraic of degree 2) values of the cosine function at rational multiples of π. In Section 3, we adopt the approach of Lehmer in , i.e., using cyclotomic polynomials, to determine all other algebraic values. All algebraic values of degree less than 5 are explicitly worked out. In the last section, we consider the same problem for the sine and tangent functions.
Received December 2, 2013. 2010 Mathematics Subject Classification. 11J72, 33B10.
Key words and phrases. Trigonometric functions, algebraic values, multiples of π.
9
310 PINTHIRA TANGSUPPHATHAWAT
Nonnegative rational and some quadratic cosine values
From the well-known trigonometric identities 2 cos 2 θ = (2 cos θ)2 − 2 and 2 cos( n + 1) θ = (2 cos θ)(2 cos nθ ) − 2 cos( n − 1) θ (n ∈ N), using an idea from [7, Theorem 6.16, pp. 308–309], it follows that there exists a monic
fn(x) ∈ Z[x] of degree n such that 2 cos nθ = fn(2 cos θ). (2.1) Clearly,
f1(x) = x, f 2(x) = x2 − 2, f n+1 (x) = xf n(x) − fn−1(x) (n ∈ N). (2.2) The polynomials fn(x) are closely related to the Chebyshev polynomials of the first kind ([2, pp. 61–63]) defined by
T0(x) = 1 , T1(x) = x, Tn+1 (x) = 2 xT n(x) − Tn−1(x) ( n ≥ 1) ,
i.e., fn(2 x) = 2 Tn(x). Taking θ = 2 kπ/n , k ∈ N, in (2.1), we get
fn(2 cos(2 kπ/n )) − 2 = 0 . (2.3) Since fn ∈ Z[x] is monic of degree n, we have thus proved
Theorem 2.1. If k, n ∈ N, then 2 cos(2 kπ/n ) is an algebraic integer of degree ≤ n.
In , the author’s main idea is to take a “proof from THE BOOK” for the case 1
π
arccos(1 /√n) ( n ∈ N, n odd, n ≥ 3) and to do a simple variation of the proof to analyze the case 1
π
arccos( √r) ( r ∈ Q) that is much more general. We now proceed analogously using Theorem 2.1 to give a simple proof of the result in .
Corollary 2.2. Let r ∈ Q with 0 ≤ r ≤ 1. Then, the number 1
π
arccos( √r)
is rational if and only if r ∈ { 0, 1/4, 1/2, 3/4, 1}.Proof. Assume that 1
π
arccos( √r) = 2 k/n is a rational number. Theorem 2.1 shows then that 2 √r = 2 cos(2 kπ/n ) is an algebraic integer. Thus, 4r is also an algebraic integer. Since r ∈ Q ∩ [0 , 1], we deduce that 4 r is a rational integer in [0 , 4], which implies r ∈ { 0, 1/4, 1/2, 3/4, 1}. The converse is trivial.
An alternative simple derivation of Corollary 2.2 is to start from [7, The-orem 6.16] which asserts that for r ∈ Q, the values cos( rπ ) ∈ Q only for cos( rπ ) ∈ { 0, ±1, ±1/2}. By using cos 2(x) = (1 + cos(2 x)) /2, we get that cos 2(rπ ) ∈ Q if and only if cos(2 rπ ) ∈ { 0, ±1, ±1/2}. Consequently, cos 2(rπ )can only take the rational values (1 + 0) /2, (1 ± 1) /2 and (1 ± 1/2) /2, so the possible rational values for cos 2(rπ ) correspond to the cases cos( rπ ) ∈
{
0, ±1, ± 12 , ± 1
√2 , ±√32
}
.ALGEBRAIC TRIGONOMETRIC VALUES 11
Recall from [9, Theorem 6.6] that a quadratic integer in Q(√d), where d
is a square-free rational integer, takes one of the following two forms: 1. all numbers of the form a + b√d, where a and b are rational integers, and 2. if d ≡ 1(mod4), but not otherwise, all numbers of the form ( a+b√d)/2, where a and b are odd. This result is proved by analyzing solutions
a + b√dc (2.4) with a, d, c, d ∈ N ∪ { 0}, c, d 6 = 0, gcd( a, b, c ) = 1, of a quadratic equation
x2 + sx + t = 0 with s, t ∈ Z.Extending the proof of Corollary 2.2, we obtain the following generaliza-tion.
Theorem 2.3. Let
α = a + b√dc ∈ [0 , 1]
with a, b, c, d ∈ N ∪ { 0}, c, d 6 = 0 , gcd( a, b, c ) = 1 and d square free. Then, the number 1
π
arccos( α) is a rational number if and only if α takes one of the following values
0 = cos π
2
, 12 = cos π
3
, 1 = cos 0 ,
√22
= cos π
4
,
√32
= cos π
6
,
√5+1 4
= cos π
5
.
(2.5)
Proof. The sufficiency part is clear from (2.5). We proceed now to prove the necessity part. If 1
π
arccos( α) = 2 k/n is a rational number, then, by Theorem 2.1, 2 cos
( 2kπ n
)
= 2 α = 2
(
a + b√dc
)
∈ [0 , 2] is an algebraic integer. From the shapes of quadratic integers, given in (2.4), we must have c = 2 or 4, the latter possibility occuring only when d ≡ 1mod 4 and a, b odd. If c = 2, then 2 α = a + b√d ∈ [0 , 2]. Since a, b are nonnegative, the only possible values of 2 α are 0 , 1, 2, √2, √3. If c = 4 , d ≡ 1 mod 4 , a and b odd, then 2 α = ( a + b√d)/2 ∈ [0 , 2] implying that the only possible value of 2 α is (1 + √5) /2.
To deal with the case where a, b are integers, we need more information about cos(2 kπ/n ). We carry this out in the next section using Lehmer’s results in . 12 PINTHIRA TANGSUPPHATHAWAT
General algebraic cosine values
Let n ∈ N, n > 2 and ζn = e2πi/n . The nth cyclotomic polynomial (see, e.g., [6, pp. 33–36]) is defined by Φn(x) = ∏
1≤k≤n
gcd( k,n )=1
(x − ζkn). (3.1) Clearly, deg Φ n(x) = ϕ(n), where ϕ is the Euler totient function. It is well known that Φ n(x) ∈ Z[x]. Since the roots of Φ n(x) occur in reciprocal pairs, there is a monic polynomial ψn(x) ∈ Z[x] of degree ϕ(n)/2 such that
ψn(x + x−1) = x−ϕ(n)/2 Φn(x).
From [4, Theorem 1] and its proof, we have
Proposition 3.1. Let α ∈ [−1, 1] . If 1
π
arccos α = 2 k/n ∈ Q with k, n ∈
Z, n > 2 and gcd( k, n ) = 1 , then 2α = 2 cos(2 kπ/n ) is an algebraic integer of degree ϕ(n)/2 whose minimal polynomial is ψn(x).
Note in addition that the cases where n = 1 and 2 are trivial for cos(2 kπ ) = 1, cos( kπ ) = ( ±1) k. A special case of Theorem 3.1 is the following result, which is a slight extension of the main result in .
Corollary 3.2. The value of 2 cos( a◦ b′ c′′ ), where a, b and c are nonneg-ative integers, is an algebraic integer of degree ≤ 283352.Moreover, 2 cos( a◦ b′ c′′ ) is an algebraic integer of exact degree 283352 if and only if gcd( c, 30) = 1 .Proof. The result follows at once from Proposition 3.1 by noting that
a◦b′c′′ = (60 2a + 60 b + c)π
263453 .
Proposition 3.1 also enables us to answer the problem posed at the end of the last section, namely, to determine when 1
π
arccos
( a+b√dc
)
is rational for a, b being any integers, and much more.
Theorem 3.3. Let α ∈ [−1, 1] Assume that 1
π
arccos α = 2 k/n , k ∈ Z,
n ∈ N, gcd( k, n ) = 1 . Then
(i) the number 2α = 2 cos(2 πk/n ) is an algebraic integer of degree 1 if and only if n = 1 , 2, 3, 4, 6; in such cases, all the values taken by α
are
1 = cos 0 = −cos π, 0 = cos 2π
4= cos 6π
4,
12 = cos 2π
6= cos 10 π
6=−cos 2π
3=−cos 4π
3 ;ALGEBRAIC TRIGONOMETRIC VALUES 13
(ii) the number 2α = 2 cos(2 πk/n ) is an algebraic integer of degree 2 if and only if n = 5 , 8, 10 , 12 ; in such cases, all values of α are
√5 − 14 = cos 2π
5 = cos 8π
5 = − cos 6π
10 = − cos 14 π
10 ,
√5 + 1 4 = − cos 4π
5 = − cos 6π
5 = cos 2π
10 = cos 18 π
10 ,
√22 = cos 2π
8 = cos 14 π
8 = − cos 6π
8 = − cos 10 π
8 ,
√32 = cos 2π
12 = cos 22 π
12 = − cos 10 π
12 = − cos 14 π
12 ;
(iii) the number 2α = 2 cos(2 πk/n ) is an algebraic integer of degree 3 if and only if n = 7 , 9, 14 , 18 ; in such cases all values of α are
cos 2π
7 = cos 12 π
7 = − cos 10 π
14 = − cos 18 π
14 ,
cos 6π
14 = cos 22 π
14 = − cos 4π
7 = − cos 10 π
7 ,
cos 2π
14 = cos 26 π
14 = − cos 6π
7 = − cos 8π
7 ,
cos 2π
9 = cos 16 π
9 = − cos 14 π
18 = − cos 22 π
18 ,
cos 4π
9 = cos 14 π
9 = − cos 10 π
18 = − cos 26 π
18 ,
cos 2π
18 = cos 34 π
18 = − cos 8π
9 = − cos 10 π
9 ;
(iv) the number 2α = 2 cos(2 πk/n ) is an algebraic integer of degree 4 if and only if n = 15 , 16 , 20 , 24 , 30 ; in such cases all values of α are
18
(
1 + √5 +
√
6(5 − √5)
)
= cos 2π
15 = cos 28 π
15 = − cos 26 π
30 = − cos 34 π
30 ,
18
(
1 − √5 +
√
6(5 + √5)
)
= cos 4π
15 = cos 26 π
15 = − cos 22 π
30 = − cos 38 π
30 ,
18
(
1 + √5 −
√
6(5 − √5)
)
= cos 8π
15 = cos 22 π
15 = − cos 14 π
30 = − cos 46 π
30 ,
18
(
1 − √5 −
√
6(5 + √5)
)
= cos 14 π
15 = cos 16 π
15 = − cos 2π
30 = − cos 58 π
30 ,
√2 + √22 = cos 2π
16 = cos 30 π
16 = − cos 14 π
16 = − cos 18 π
16 ,
√2 − √22 = cos 6π
16 = cos 26 π
16 = − cos 10 π
16 = − cos 22 π
16 ,
√10 + 2 √54 = cos 2π
20 = cos 38 π
20 = − cos 18 π
20 = − cos 22 π
20 ,
√10 − 2√54 = cos 6π
20 = cos 34 π
20 = − cos 14 π
20 = − cos 26 π
20 ,
√3 + 1 2√2 = cos 2π
24 = cos 46 π
24 = − cos 22 π
24 = − cos 26 π
24 ,
√3 − 12√2 = cos 10 π
24 = cos 38 π
24 = − cos 14 π
24 = − cos 34 π
24 .
414 PINTHIRA TANGSUPPHATHAWAT
Proof. The following table gives the values of n for the first four values of ϕ(n)/2. The determination of these values can be found in the paper which gives an elementary approach to solve the equation ϕ(x) = k.
ϕ(n)/2n
13, 4, 6 25, 8, 10, 12 37, 9, 14, 18 415, 16, 20, 24, 30
Using the values of n in the preceding table, the corresponding minimal polynomials are shown in the next table.
n Φn(x) ψn(t) ( t = x + x−1)3 x2 + x + 1 t + 1 4 x2 + 1 t
5 x4 + x3 + x2 + x + 1 t2 + t − 16 x2 − x + 1 t − 17 x6 + x5 + x4 + x3 + x2 + x + 1 t3 + t2 − 2t − 18 x4 + 1 t2 − 29 x6 + x3 + 1 t3 − 3t + 1 10 x4 − x3 + x2 − x + 1 t2 − t − 112 x4 − x2 + 1 t2 − 314 x6 − x5 + x4 − x3 + x2 − x + 1 t3 − t2 − 2t + 1 15 x8 − x7 + x5 − x4 + x3 − x + 1 t4 − t3 − 4t2 + 4 t + 1 16 x8 + 1 t4 − 4t2 + 2 18 x6 − x3 + 1 t3 − 3t − 120 x8 − x6 + x4 − x2 + 1 t4 − 5t2 + 7 24 x8 − x4 + 1 t4 − 4t2 + 1 30 x8 + x7 − x5 − x4 − x3 + x + 1 t4 + t3 − 4t2 − 4t + 1
The stated values of α are computed by finding all the roots of ψn(t) using Mathematica.
Remark 3.4. The results in Proposition 3.1, Corollary 3.2, and Theo-rem 3.3 can be transformed to those of hyperbolic cosine function by noting that cosh( z) = cos( zi ).
Algebraic sine and tangent values
The following results are taken from [6, Theorems 3.9 and 3.11]. They were first considered by Lehmer ; however, the original Lehmer’s result for the sine function was incomplete.
Proposition 4.1. Let α ∈ [−1, 1] .
I. If 1
π
arcsin α = 2 k/n ∈ Q with k, n ∈ Z, n > 2, n 6 = 4 and gcd( k, n ) = 1, then 2α = 2 sin(2 kπ/n ) is an algebraic integer of degree ϕ(n),
ϕ(n)/4 or ϕ(n)/2 according as gcd( n, 8) < 4, gcd( n, 8) = 4 or
gcd( n, 8) > 4.ALGEBRAIC TRIGONOMETRIC VALUES 15
II. If 1
π
arctan α = 2 k/n ∈ Q with k ∈ Z, n > 2, n 6 = 4 and gcd( k, n ) = 1, then 2α = 2 tan(2 kπ/n ) is an algebraic integer of degree ϕ(n),
ϕ(n)/2 or ϕ(n)/4 according as gcd( n, 8) < 4, gcd( n, 8) = 4 or
gcd( n, 8) > 4.
The cases when n = 1 and n = 2 are trivial for 0 = sin(2 kπ ) = sin( kπ ) = tan(2 kπ ) = tan( kπ ), and the case when n = 4 is also trivial for sin( kπ/ 2) ∈{− 1, 1} and tan( kπ/ 2) = 0 or undefined. Using Proposition 4.1 and Theorem 3.3, all algebric values of degrees < 5of the sine and tangent functions are given in the following theorem, whose proof is omitted.
Theorem 4.2. Let α ∈ [−1, 1] .
A. Assume that 1
π
arcsin α = 2 k/n , k ∈ Z, n ∈ N, gcd( k, n ) = 1 . Then
(A1) the number 2α = 2 sin(2 πk/n ) is an algebraic integer of degree 1 if and only if n = 1 , 2, 4, 6, 12 ; in such cases, all the values taken by α
are
0 = sin 0 = sin π, −1 = sin 2π
4= sin 6π
4,
−12 = sin 14 π
12 = sin 22 π
12 ,12 = sin 10 π
12 = sin 26 π
12 ;
(A2) the number 2α = 2 sin(2 πk/n ) is an algebraic integer of degree 2 if and only if n = 3 , 6, 8, 20 ; in such cases, all the values taken by α
are √32 = sin 2π
3= sin 14 π
6=−sin 4π
3=−sin 10 π
6,
√22= sin 6π
8= sin 18 π
8=−sin 10 π
8=−sin 14 π
8,
√5−14= sin 18 π
20 = sin 42 π
20 =−sin 22 π
20 =−sin 38 π
20 ,
√5 + 1 4= sin 14 π
20 = sin 46 π
20 =−sin 26 π
20 =−cos 34 π
20 ;
(A3) the number 2α = 2 sin(2 πk/n ) is an algebraic integer of degree 3 if and only if n = 28 , 36 ; in such cases all the values taken by α are
cos 2π
7= sin 22 π
28 = sin 62 π
28 =−sin 30 π
28 =−sin 54 π
28 ,
cos 6π
7= sin 38 π
28 = sin 46 π
28 =−sin 18 π
28 =−sin 66 π
28 ,
cos 6π
14 = sin 26 π
28 = sin 58 π
28 =−sin 34 π
28 =−sin 50 π
28 ,
cos 2π
9= sin 26 π
36 = sin 82 π
36 =−sin 34 π
36 =−sin 74 π
36 ,
cos 8π
9= sin 50 π
36 = sin 58 π
36 =−sin 22 π
36 =−sin 86 π
36 ,
cos 10 π
18 = sin 38 π
36 = sin 70 π
36 =−sin 46 π
36 =−sin 62 π
36 ;16 PINTHIRA TANGSUPPHATHAWAT
(A4) the number 2α = 2 sin(2 πk/n ) is an algebraic integer of degree 4 if and only if n = 5 , 10 , 16 , 24 , 60 ; in such cases all values of α are
18 (1 + √5) + 14
√ 32 (5 − √5) = cos 2π
15 = sin 38 π
60 = sin 142 π
60 = − sin 82 π
60 = − sin 98 π
60 ,
18 (1 − √5) + 14
√ 32 (5 + √5) = cos 4π
15 = sin 46 π
60 = sin 134 π
60 = − sin 74 π
60 = − sin 106 π
60 ,
18 (1 + √5) − 14
√ 32 (5 − √5) = cos 8π
15 = sin 62 π
60 = sin 118 π
60 = − sin 58 π
60 = − sin 122 π
60 ,
18 (1 − √5) − 14
√ 32 (5 + √5) = cos 14 π
15 = sin 86 π
60 = sin 94 π
60 = − sin 34 π
60 = − sin 146 π
60 ,
√2 + √22 = cos 2π
16 = sin 10 π
16 = sin 38 π
16 = − sin 22 π
16 = − sin 26 π
16 ,
√2 − √22 = cos 6π
16 = sin 14 π
16 = sin 34 π
16 = − sin 18 π
16 = − sin 30 π
16 ,
√
58 +
√58 = cos 2π
20 = sin 6π
10 = sin 14 π
10 = − sin 14 π
10 = − sin 8π
5 ,
√
58 −√58 = cos 6π
20 = sin 4π
5 = sin 22 π
5 = − sin 6π
5 = − sin 18 π
10 ,
√3 + 1 2√2 = cos 2π
24 = sin 14 π
24 = sin 58 π
24 = − sin 34 π
24 = − sin 38 π
24 ,
√3 − 12√2 = cos 10 π
24 = sin 22 π
24 = sin 50 π
24 = − sin 26 π
24 = − sin 46 π
24 .
B. Assume that 1
π
arctan α = 2 k/n , k ∈ Z, n ∈ N, gcd( k, n ) = 1 . Then
(B1) the number 2α = 2 tan(2 πk/n ) is an algebraic integer of degree 1 if and only if n = 1 , 2, 8; in such cases, all the values taken by α are
0 = tan 0 = tan π,
1 = tan π
4 = tan 5π
4 = − tan 3π
4 = − tan 7π
4 ;ALGEBRAIC TRIGONOMETRIC VALUES 17
(B2) the number 2α = 2 tan(2 πk/n ) is an algebraic integer of degree 2 if and only if n = 3 , 6, 12 , 16 , 24 ; in such cases, all the values of α are
√3 = tan 4π
3 = tan 2π
6 = − tan 2π
3 = − tan 10 π
6 ,
1
√3 = tan 2π
12 = tan 14 π
12 = − tan 10 π
12 = − tan 22 π
12 ,
1 − √2 = tan 14 π
16 = tan 30 π
16 = − tan 2π
16 = − tan 18 π
16 ,
1 + √2 = tan 6π
16 = tan 22 π
16 = − tan 10 π
16 = − tan 26 π
16 ,
2 − √3 = tan 2π
24 = tan 26 π
24 = − tan 22 π
24 = − tan 46 π
24 ,
2 + √3 = tan 10 π
24 = tan 34 π
24 = − tan 14 π
24 = − tan 38 π
24 ;
(B3) the number 2α = 2 tan(2 πk/n ) is an algebraic integer of degree 4 if and only if n = 5 , 10 , 20 , 32 , 40 , 48 ; in such cases all the values taken by α are √
5 + 2 √5 = tan 2π
5 = tan 14 π
10 = − tan 6π
10 = − tan 8π
5 ,
√
5 − 2√5 = tan 6π
5 = tan 2π
10 = − tan 4π
5 = − tan 18 π
10 ,
√
5 − 2√55 = tan 2π
20 = tan 22 π
20 = − tan 18 π
20 = − tan 38 π
20 ,
√
5 + 2 √55 = tan 6π
20 = tan 26 π
20 = − tan 14 π
20 = − tan 34 π
20 ,
√
4 + 2 √2 − √2 − 1 = tan 2π
32 = tan 34 π
32 = − tan 30 π
32 = − tan 62 π
32 ,
√
4 − 2√2 − √2 + 1 = tan 6π
32 = tan 38 π
32 = − tan 26 π
32 = − tan 58 π
32 ,
√
4 − 2√2 + √2 − 1 = tan 10 π
32 = tan 42 π
32 = − tan 22 π
32 = − tan 54 π
32 ,
√
4 + 2 √2 + √2 + 1 = tan 14 π
32 = tan 46 π
32 = − tan 18 π
32 = − tan 50 π
32 .
Final remarks
As is easily checked from Proposition 4.1, II, there are no third degree algebraic values for the tangent function. 2. The result about all possible rational values of the sine, cosine and tangent functions derived in is properly contained in Theorem 3.3, (i), and Theorem 4.2, (A1), (B1).
Acknowledgements
The author wishes to thank the Thailand Research Fund for financial support, and also the anonymous referee for all valuable suggestions.
518 PINTHIRA TANGSUPPHATHAWAT
References
M. Aigner and G. M. Ziegler, Proofs from THE BOOK , Springer-Verlag, Berlin, 1998. E. W. Cheney, Introduction to Approximation Theory , Chelsea, New York, 1982. R. W. Hamming, The transcendental character of cos x, Amer. Math. Monthly 52
(1945), 336–337. D. H. Lehmer, A note on trigonometric algebraic numbers , Amer. Math. Monthly 40
(1933), 165–166. N. S. Mendelsohn, The equation ϕ(x) = k, Math. Mag. 49 (1976), 37–39. I. Niven, Irrational Numbers , The Carus Mathematical Monographs 11 , Wiley, New York, 1956. I. Niven, H. S. Zuckerman, and H. L. Montgomery, An Introduction to the Theory of Numbers , Wiley, New York, 1991. J. M. H. Olmsted, Rational values of trigonometric functions , Amer. Math. Monthly
52 (1945), 507–508. H. Pollard and H. G. Diamond, The Theory of Algebraic Numbers , The Carus Mathe-matical Monographs 9, The Mathematical Association of America, Washington, 1975. J. L. Varona, Rational values of the arccosine function , Cent. Eur. J. Math. 4 (2006), 319–322.
Department of Mathematics, Faculty of Science and Technology, Phranak-hon Rajabhat University, Bangkok 10220, Thailand
E-mail address : pinthira12@hotmail.com; pinthira@pnru.ac.th |
5188 | https://www.youtube.com/watch?v=-EnRggYVGlk | Magnetic Field using Biot-Savart law: Circular Loop and Long Wire
Physics Ninja
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Posted: 7 Nov 2022
Physics Ninja looks at the Biot-Savart law and uses the law to calculate the magnetic field produced by a current loop. In the second example i show how to evaluate the field produced by a long wire.
For the long wire i show how to solve using trigonometric substitution and also using integral tables.
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Transcript:
hi everybody physics ninja here today I'm going to look at the magnetic field produced by a steady state current and how do we use uh the bo Savard law in order to calculate this magnetic field so we're first going to review the equation which is written uh in this gray box and I'm going to apply it to a current Loop and I'm going to show you how to apply this equation I'm going to teach you what all the terms mean and then at the end we want to apply it to simpler geometries than the one I have depicted in this figure I'm going to apply it to a circular Loop and also to a very long but a straight wire so like with all my videos if you like it give it a thumbs up consider subscribing to my channel it's the best way to support what I do all right let's get started all right we want to look at the magnetic field produced by a steady current all right what do I mean by this steady current right here this is very very important is that the current is constant okay so we have a constant current I flowing in some wire and I have a wire here depicted here that's highlighted in yellow uh the wire is the blue section over here now let's have a look now I've defined a particular point of observation this point is the one that I've drawn here in the green okay and if I'm looking for the little bit of magnetic field okay the little bit of magnetic field is denoted by DB we'll see why it's D just means it's a little bit of field oh that little bit of magnetic field is given by this equation which is our bios of Ard law okay first of all there's a term in the bracket the bracket is simply mu 0 divided by four Pi okay a mu zero is simply the permeability of free space and it's given by this quantity right here 4 pi multiplied by 10 to the minus 7. it comes in these units of Newtons per amperes squared if I'm dealing in SI units I simply signifies the current flowing through that wire pretty straightforward that's measured in amps and what else now we have uh in the numerator up here there are two vectors the vector DL is like the red one I've drawn here this is one particular element of length along the direction of the current okay so it's simply a vector that has units of length so units of meters and the last one which is very important is a unit it's a unit Vector so it has a magnitude of one and the direction of that Vector now goes from the the element DL that I'm considering to the point of observation so I've drawn kind of a dotted line here that goes in that direction and the purple Vector represents the vector R hat for this case the last bit is what appears in the denominator it's an r squared term this is reminiscent of of the same r squared term that was in Newton's law of gravitation and in electrostatics it's simply the distance squared from the element to the point of observation all right this is the biosavard law and this is what we want to apply to our problems before we do that let's do a little dimensional analysis to make sure this makes sense so like I said um the term in the bracket here is Mu 0 divided by 4 Pi so the term in the bracket has to have the same units as the MU zero term and I gave you those units right those units were in Newtons per ampere squared uh the current we said had units of amps that's simply a and now this whole term here with the vectors uh divided by the distance squared the numerator has units of meters which only comes from the DL component the denominator has units of meters squared because it's r squared so if I put everything together at the end of the day I get Newtons per amps per meter okay this is what we call a Tesla if I group these units together this is what I refer to as one Tesla so one Tesla is the equivalent to one newton per ampere per meter okay so that is simply a definition so now let's go at let's go back and look at the properties of all of these terms okay I want to just before we use it to calculate the magnetic field from a loop and a current I want to make sure we understand some of the properties of this equation right here all right I want to highlight a couple things to know that are really important to know about the biosavard law the first thing is it is a 1 over r squared law right and that is very similar to Coulomb's law in Newton's law of gravitation right both of those had this r squared term in the denominator uh this Vector over here is the vector R hat okay our hat is simply a unit Vector okay it has a magnitude equal to one and I discussed the direction before okay super important properties now this multiplication involves two vectors the vector DL and the vector R hat this here is a cross product between two vectors so it's going to be important to know how do you evaluate the cross product between two vectors one property of cross products is that whatever I have here DL and R hat my resulting Vector which is DB is going to be perpendicular to both of those it has to be perpendicular to DL and it has to be perpendicular to the vector R hat so in this case here both of these vectors DL and R hat are into the plane the vector DB is actually pointing out of the plane but I'm going to illustrate that in just a minute we're going to look at a few examples okay so it's important to know how do you evaluate a cross product between two vectors so one way to evaluate the direction of a cross product we're only worried about the direction is something known as the right hand rule okay let me just highlight this term so um again the right hand rule I use it only to find the direction of the cross product between for example in this case A and B okay it's very useful also for magnetism as we're going to see so if I have have a cross product between two vectors A and B all you have to do is you use your right hand as illustrating this figure you place your fingers along the first Vector the first Vector that appears here is the vector a notice on this left hand side figure that the fingers are pointing along the a vector now what you do is see this Black Arrow here it means you curl your fingers toward the second Vector the second Vector here is the vector B so you curl your fingers as Illustrated here on this right hand side figure all right and what you get here is your thumb is pointing up in this figure this means that this Vector here has to point in the direction of the thumb okay so this is how you apply the right hand rule fingers along the first Vector curl toward the second Vector your thumb gives you the direction of this cross product pretty straightforward I do list this warning down here there are other versions of the right hand rule this is one example of the right hand rule and this is one we're going to use later on all right what if you actually wanted to evaluate well what is this resulting Vector this Vector C here which is the product between both of those vectors there are two ways to do that I'm just doing this to remind you how do you evaluate a cross product between two vectors one definition is the first one here so you take the magnitude of each vector and you simply multiply them together so these are simply numbers okay magnitudes Theta here sine of theta is the sine of the angle between both vectors so I've depicted here in the diagram here in the lower right hand side see notice I have vectors A and B the angle Theta is the angle between them that would be the angle that you would substitute in this cross product definition now again this here must give you a vector at the end of the day the If I multiply or I take a cross product between 2 I need a vector at the end of the day so here I'm simply writing that Vector as little n little N is a vector which is a unit vector and I've got so I've kind of written that over here maybe there's no need to rewrite it here so it has a magnitude of one in the direction of this unit Vector again it has to be perpendicular to A and B you can find this uh the direction of this Vector using our right hand rule that we just discussed in the previous slide okay so that will Define that Vector now the alternative way of finding the cross product between two vectors is you can write them in any coordinate system that you want but let's imagine there we know the components of a and b they're written in Cartesian coordinates so I know the ax a y and a z components similar for the vector B what you can do at the end is simply evaluate that cross product by substituting those components in to this final expression this is something you might have seen in your math class you end up getting this result by taking the determinant of this Matrix here where I've lined up both vectors as rows of this Matrix okay if you do that these are going to be the components of your final vectors so this here would be the CX component the Cy component and the CZ comp components so this is a mathematical approach that allows you to find the cross product between two vectors all right so our last section now is what we want to do is we started with this equation right this was the element of magnetic field produced by a tiny segment but imagine you would go back to my original Loop here what if I want to find the total magnetic field produced by each one of those segments that means you simply want to add them all up and the way you add everything up in physics is simply by integrating so what all I'm going to do is I'm going to integrate this expression of DB to get the total field and if I take out all the constants from the integral I am left with this equation down over here which I guess is the formal expression for the biosavard law so it simply means we have to integrate each one of these terms now this looks rather complicated but we're going to start off with some easy cases just to make sure we familiarize yourself with this expression and show you how to apply it now all right my first question is let's find the Magna uh the magnetic field at the center of the loop so here's my point right at the center of the loop this Loop here has a radius let's call the radius Define it here as Big R okay and here's my bio Savard law so now it looks really complicated because you have this integral and you also have this cross product but let's first look at the cross product okay the first let's examine this guy cross our hat okay so I've drawn a couple segments I've drawn one here and I've drawn another segment here they are different but let's focus on this first segment right here okay if I write down what this cross product is remember it is the magnitude of this guy multiplied by the magnitude of our hat multiplied by sine of the angle between both of those vectors multiplied by and hat okay so the magnitude of DL I just write it as DL the magnitude of our hat is simply one that's a definition now what is the angle between both of those vectors let me just kind of eliminate that green here and let's look at the angle so how would you define the angle here well again you have this Vector DL which goes along the loop and you have our hat which goes from The Edge toward the center you should be able to convince yourself that this angle is 90 degrees notice that if I look at the angle down over here it's also going to be 90 degrees and it's going to be the same everywhere along this circle okay although both of those vectors change orientation now the angle between them doesn't change and that's really important that means here I'm going to get sine of 90 degrees all right what about this last term the last term is this Vector over here for this I am going to use the right hand rule right the right hand rule says you take your right hand you place your fingers along the first vector and then you curl them toward the second Vector if you did this properly this Vector here should be in the direction of your thumb that means that this one here should be out of the page okay all right I would get the same answer if I did this one I'd place my right hand my fingers pointing down I would curl them toward the second vector if you do that your thumb should be pointing out of the page so you get the same direction everywhere you look for this circle so there are some simplifications here because look at sine of 90 sine of 90 is also one and the vector also hasn't changed it's always out of the page so this complicated look and term which involve the cross product of two vectors at the end simplifies to Simply a magnitude and then the vector is always out of the page I'm just going to call that direction z z hat okay just remind myself that it's out of page all right now we're going to go ahead and substitute it into our bio Savard law bio Savar says that the B total um as a vector has to be equal to now there's a whole bunch of constants you have mu zero I over 4 pi all right now we have this integral and I have a numerator term which I've simplified as DL Z hat I'm just going to stick it out here in the back that's it for the top what about the bottom the bottom is r squared what is r r was the distance from the element all the way to the point of observation well that is by definition the radius of the circle so all you have to do now is do the radius of the circle squared okay this is also a constant first Circle it's always the same distance every segment is a radius away from the point from the center in this case so what you can do at the end is simply take this term and remove it from the integral so that's what I'm going to do now instead of having it here I'm simply going to move that term to the outside just to simplify the expression now there's no more term in the denominator so I'm just going to get rid of this and let's write it a little bit more neatly just by moving the Z hat over all right now all we're left with for this problem is to examine what is this term here this is the integral of DL so now we have to think about it right all we're doing here is you're adding up all of the segments around this circle and you're just breaking them up in terms of a s right a whole series of infinitesimal distances right so you're adding up um lengths tiny ones around the circle so what happens if I add up all the lengths going around a circle by definition this here should be the perimeter of the circle okay so this whole integral of integrating over DL is nothing more than 2 pi multiplied by the radius that is the circumference of our Circle all right so now we continue so at the end we simply put everything together now so we have B total is Mu zero I over 4 pi r squared this whole integral now reduces to 2 pi r and then I still have the vector Z hat which is out of the page that I've mentioned before so now you simply have to look for common terms and cancel them out so I have a pie I can get rid of I have one R and I have an r squared I have a 2 and here I can get rid of one two and I'm still left with 2 at the denominator and now you simply group all the terms together you get to the final answer B total has to be equal to Mu 0 I over 2R and this is multiplied by Z hat that gives me the direction so that field looks like this expression here that's the field right at the center all right so if you're given a current for example imagine I'm given a current of 3 amps and imagine this radius R was say two centimeters I could calculate a magnitude you simply have to put the numbers in here so here you would do 4 pi times 10 to the minus 7. uh the current I would substitute for three here we do two now the only thing you have to be careful of is that this radius should be measured in meters if you want to get Teslas at the end all right and you put everything in the calculator you should get about 0.94 times 10 to the minus 4 Tesla that would be the magnitude of this field and again it is pointing out of the page if I'm looking for the total Vector of that magnetic field all right now let's now look at one other example of the infinite wire all right question two says find the magnetic field at this point P produced by this infinite current right here this infinite wire we have a current flowing to the right hand side like this and the point p is a distance a away from uh from this wire so how do we apply bio Savar to that problem again I'm going to first start by the DL cross our hat term again if I'm looking for the magnitude of this it's simply DL what else multiplied by one that's the magnitude of our hat multiplied by sine of the angle between them well we're going to call this angle here I'm going to call this say the angle Phi times sine of Phi and call it whatever you want Theta Phi doesn't matter and what else I'm left now with some vector and hat now the way you find n hat is using our right hand rule right hand rule says you take your right hand you put your fingers along the first vector and you curl them toward the second Vector again this n hat Vector should be out of the page that is the direction of the field produced by this tiny segment at this point what if you considered another segment well this would be the vector DL okay and the vector R hat would then go this way notice that in this case if I do DL cross with our hat using my right hand rule I still get a vector that is always in the same direction it doesn't matter which segment you choose you're always going to get the same direction over here for this n hat Factor so that's great now what's different from this one is that this Angle now is not always going to be 90 degrees right if this segment is far away from the center over here where the point is the angle could be quite small the middle one that I drew here the angle is actually 90 degrees so the angle does change depending on how far you are from from the point so you can't just simplify that term but we know that the direction has to be out of the page so now I know half I no longer have to worry about the direction of this guy um the next thing I'm going to do now is simply write DL I'm actually going to give it a direction now okay DL is actually a little segment in the X Direction so it makes sense simply to call it DX okay so now we're going to go back and substitute into our biosavard law so we're going to have that b total again I'm just going to write the magnitude down because we already know the direction mu 0 over 4 pi I the integral of so this top term now becomes this evaluation except I'm going to write DX and sine of Phi I'll still leave that as sine of Phi I'm not worried about the n hat because I'm only worried about the magnitude right now and divided by r squared all right so what is r squared let me just write it as r squared and we're going to look at our diagram R was the distance between the point which is here and any segment over here okay well again if I look at this figure right here this segment is a certain distance away from uh the Middle Point here so I'm just going to call this distance X I know that the vertical distance is a that was the distance between the point and the wire so the way that you would write down the distance r are in this case simply used Pythagorean it would simply be x squared plus a squared that is this distance R because this angle is 90 degrees so that's pretty straightforward that means if you have r squared has to be x squared plus a squared all right we're doing pretty good we could substitute this expression down here in the denominator let's go ahead and do that so B total mu zero multiplied by the current over 4 pi and again we're integrating DX I'm going to leave the sine of Phi here and here I'm going to divide by x squared plus a squared now I know we want to right away and again if we're integrating over DX think about where these elements go if this is an infinite wire an infinite wire basically means that X goes all the way this way and all the way that way right so this way we have X that goes toward minus infinity and here x goes toward Infinity so really my limits of integration here have to be like this and like this now the only thing that prevents us from evaluating this integral now is well we have two terms that depend on x i we talked about how this angle also depends on X because you can see when the angle is 90 degrees the segment is right underneath that point in that case the distance is short okay so if the angle Phi depends on X but let's look at our triangle right what if I look at what is the sine of the angle Phi sine of the angle Phi just use my trigonometry is the distance a divided by R right it is opposite over the hypotenuse and I know both of these right this is A and R I said it earlier R was given by square root of x squared plus a squared so now we have to Simply get rid of it okay so we're going to go back to our expression and before I do the final integral I am simply going to get rid of sine of Phi to have everything in terms of x so here I'm going to have DX in the numerator and sine of Phi I replace it with a over square root so you just have to be careful here because it looks like the same term you just have to be careful on how you're going to deal with this one this here has a power of one if I write this one in terms of exponent in terms of in terms of a bracket and an exponent this would be one half so over here I have to have three halves as my total exponent so now we're going to go on the next page and we're going to finish this off by looking how do you evaluate this integral right here there are many ways to do this I'm going to show you uh one way and discuss another method all right the first method we're going to use here is something called trig substitution let's call it trips trig Subs to find an angle Theta over here and it's kind of useful to use this angle Theta as an integration you'll see why in a minute but if I Define this as the angle Theta again the distance X goes from the center here all the way to wherever my segment is and you can have it on both sides it doesn't matter but one thing I could Define is a tangent of theta tangent of theta is the opposite which is X over The adjacent which is a well guess what if I bring the A on the other side I get a tangent of theta which equals to X so this is actually a really useful expression here now one thing you want to do is maybe you need to get rid of all the X terms so I can use this to get rid of the X in this term I also need to introduce an infinitesimal DX and the way you do that now is you simply differentiate if you differentiate this expression relative to Theta with respect to Theta a is simply a constant if you take the derivative of this it's secant squared theta okay and at the end you can write that DX is equal to a secant I'm just going to write it as 1 over cosine so if it's secant squared it's 1 over cosine squared and then you have D Theta like this so we're almost there the last thing I could do I can also write cosine and sine that's also pretty helpful but um I think we're pretty good to go right now so all you've got to do now is go back and substitute now what would happen to these limits well for that you have to look at this expression right what happens when X goes toward minus infinity minus infinity means you have a segment down here that's really really far away in that case uh the angle Theta has to go toward minus pi over two okay the other limit that I would have would be when X goes toward positive Infinity right when X goes toward positive Infinity this segment is all the way down on the right hand side the angle Theta then would go toward pi over 2 and you can get that from this expression here in the red box so now we have to go back and you simply substitute everything if you're not sure what to do with this denominator term another thing you could do is use my cosine of theta now right since Theta is this angle up here cosine of theta would simply be a over R and R was the square root of that term right in the denominator so you can write it as like this so now let's go back to this top expression and start substituting in our different values to write things in terms of the angle Theta and the infinitesimal D Theta to finish our trigonometric substitution all right so be total now it's just math at this point B total is Mu zero I over four Pilots be very careful now I'm just still going to just do one additional step and I apologize for this and again this term here is what um x squared plus a squared was r squared and then all of this is to the three halves okay so be a little bit careful here so if I simplify this I'm going to now just deal with those exponents this is our Cube down at the bottom okay all right now we do the trigonometric substitution what can I do with the DX term so first you can just rewrite the constants in the front I am going to be integrating now over the variable Theta so that it goes from minus pi over 2 to pi over 2. a I can bring it out in the front it is a constant all right DX is given by this term right here right DX is right here you simply substitute it into the integral it's a d Theta divided by cosine squared Theta what else I still have the r cube right there for the r Cube what I'm going to do is I'm going to use this expression right here because it's the easy right what would this mean this would be in that R cubed would be a over cosine Theta in each one of those terms is Cube so at the end you have to do something like this this here is a cube and divided by cosine cubed of theta now we simply have to cross out common terms you have cos squared you have cos cubed You're simply going to be left with COs in the numerator now look at all the A's I have a a a cube I can bring this one here out in the front so let me get rid of it here and write it out here all right this integral now is simplified quite a bit all right so you have mu 0 I over 4 Pi a integrating from minus pi over 2 to pi over 2. and here we're left with simply cos Theta D Theta and again these limits are only here because I'm going from minus infinity to Infinity so this integral is pretty straightforward okay the integral of cosine is simply sine so you do 4 Pi a this integral gives me sine Theta it's evaluated between both of those limits and now we substitute a mu zero I over 4 Pi a this becomes sine of pi over 2 the upper limit minus sine of minus pi over 2. well guess what each one of these is going to be this is one and this guy here since the angle is negative is minus one but this minus cancels out with that minus so at the end this whole bracket here ends up giving you the number two all right so if you have the two in the numerator you had a 4 in the denominator your final expression for the field produced by a infinite wire ends up being mu zero I divided by 2 pi multiplied by a and a was again the shortest distance from the point of observation to the wire so it's actually pretty lengthy to do this right so this is how one way you can do trigonometric substitution okay it's still not really easy right it still has a lot of steps to it there is another way you could do it it's actually starting directly from this first expression here that was only written in terms of X and it involves looking at integral tables I'll show you what I did in order to evaluate it using integral tables now okay all right I'm going to show you how to find this total magnetic field now by evaluating this integral here by using an integral table I used to have books full of integral tables nowadays it's more common just to simply pull up Google and type in integral tables I like to do like an image search and then I look for something that looks similar to what I've I'm looking for this is what I found I was looking for integrals of this form here and if I go down to number 29 that was something that looks similar to the case that I was looking at because I had the same exponent right I had the exponent three halves and then denominator like this so uh what I did was let's go back to this term and let's evaluate this term here again we're evaluating under these limits right from Infinity to minus infinity all right the other difference is this integral is in terms of U this is in terms of X but that's okay that's just any variable you can call this whatever you want I could just substitute this back to X I guess the only tricky part now is how do you evaluate this limit if you look at that denominator term you have to be a little bit careful right because imagine you had this just look at this this term over here plus x squared what if you took the limit when X goes toward minus infinity or plus infinity it doesn't matter right if you take the limit here well you can forget about a a is just the number right it was that distance from the point to the wire I mean if x is very very big in any direction forget about this right it's going to be negligible so at the end you're simply going to be left with this tends really toward the value X but you have an X in the numerator and now this here is approximately equal to X when evaluated under these limits so that means that this whole thing here tends to Ward really one over a squared for um for this guy okay so now let's evaluate that integral so you have to substitute both terms in there so you're going to get uh again infinite over a squared square root of a squared plus infinite squared and then minus now be a little bit careful here now it's minus infinity over a squared and then minus infinity squared right or if you again this is not the most mathematically rigorous but both of these terms tend toward the same so they cancel out again this negative sign is going to cancel with that one I'll just turn this into a positive and again this term is again similar to that one when the variable is very very big so that means that this integral here is one over a squared plus 1 over a squared which gives me 2 over a squared so if I go back now and I substitute look what you get mu zero I multiplied by a 4 pi and this entire integral now was 2 over a squared and I could take on my pen cancel out some of the common terms I'm left with 2 and lo and behold we are going to get mu zero I over 2 pi and a in the denominator that is the magnitude of our total magnetic field if I'm a distance a away okay this is how you could do it now using integral tables all right folks that's it for me hopefully you learned how to apply uh biosavard law to calculate at least a couple cases okay I have several other examples on my website so have a look at them just search them up I have a loop I have another rectangular Loop um also helmholtz coil where I kind of use this result also all right thanks for watching folks |
5189 | https://clinmedjournals.org/articles/ijwhw/international-journal-of-womens-health-and-wellness-ijwhw-4-084.php?jid=ijwhw | Vulvar Pruritus: Variability of Clinical Evaluation and Management
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Bedell SL, Lawson AA, Griffith WF, Werner CL (2018) Vulvar Pruritus: Variability of Clinical Evaluation and Management. Int J Womens Health Wellness 4:084. doi.org/10.23937/2474-1353/1510084
ORIGINAL ARTICLE | OPEN ACCESS DOI: 10.23937/2474-1353/1510084
Vulvar Pruritus: Variability of Clinical Evaluation and Management
Sarah L Bedell, Ashli A Lawson, William F Griffith and Claudia L Werner
Department of Obstetrics and Gynecology, University of Texas Southwestern Medical Center, USA
Abstract
Objective
We characterize the evaluation and initial management of patients with vulvar pruritus, including elements of history-taking, physical examination, laboratory testing, and treatments. We propose an algorithm for approaching this common clinical problem in a systematic way.
Methods
A retrospective chart review of patients with vulvar pruritus who presented to Gynecology or Vulvology Clinic at Parkland Health and Hospital System in 2012 informed this descriptive study.
Results
A total of 46 patients aged 19 to 70 years presented with a primary symptom of vulvar itching, of which 41% had additional irritative vulvar complaints. The average duration of symptoms was 24 months; 67% had used at least one self-treatment unsuccessfully prior to their visit. Only 13% of patients received a non-genital skin examination; 46% received a vaginal culture, and 20% underwent vulvar biopsy. The most common diagnoses were "nonspecific" (39%), and "dermatitis" (22%). The most common treatment prescribed was a topical steroid (78%). Vulvar skin care counseling was documented for 67% of patients. At 18 months, 46% were lost to follow up and of those who returned, 48% reported resolution of initial vulvar symptoms.
Conclusions
The assessment and treatment of vulvar pruritus is variable. An algorithm for the evaluation of this common clinical complaint is proposed.
Keywords
Vulvar itching, Vulvar irritation, Vulvar pruritis
List of Abbreviations and Acronyms
DNA: Deoxyribonucleic Acid; ICD-9: International Classification of Diseases, 9 th Revision; HRT: Hormone Replacement Therapy; PHHS: Parkland Health and Hospital System; VI: Vulvar Irritation; VP: Vulvar Pruritus, CG/CT: gonorrhea/Chlamydia
Introduction
Vulvar pruritus (VP), along with other symptoms of vulvar irritation (VI), is a common complaint for which women present to obstetrician-gynecologists and other primary care providers. In addition to itching, VI includes a sensation of irritation, chafing, pain or burning, for which women commonly self-treat with increased washing and over-the-counter hygiene or medicinal products. With a $3 billion feminine care product industry, women have become regular users of products that contain known vulvar irritants [1-5]. Paradoxically, products meant to treat VP/VI can cause vulvar dermatitis and worsen preexisting symptoms . While the exact prevalence of vulvar dermatitis is unknown, it has been shown to occur in up to 20 to 64% of patients with chronic vulvar symptoms [7,8]. It appears that both contact and irritant dermatitis play more significant roles than infection in the etiology of vulvar complaints [4,5,9,10].
Despite being common, vulvar symptoms can be challenging both diagnostically and therapeutically due to a lack of a standardized clinical approach. Additional obstacles in the management of VP/VI include the fact that no one specialty claims or prioritizes vulvar disease, and evaluation of these complaints is not always included in training programs [5,11]. While a small amount of expert opinion and recommendations are available, there are no widely practiced or standardized guidelines for clinicians. Additionally, there is little data regarding the relevance of personal hygiene habits and their impact on vulvar disease, and a targeted history to elicit these behaviors is not consistently obtained . Moreover, vulvar dermatoses can be difficult to recognize clinically as findings on exam may be subtle [5,13]. We suspected that both the assessment and management of patients with vulvar pruritus and related symptoms are inconsistent at our own institution, and this may be the case elsewhere. The lack of a systematic approach likely contributes to delayed diagnosis, multiple clinical visits, and patient or physician frustration. We characterize patients presenting with VP and other VI symptoms in a specialized clinic setting, describe provider approaches to patients with these complaints, and propose an algorithm (Appendix) to guide the evaluation and initial management based on these findings [14,15].
Methods
With Institutional Review Board approval from the University of Texas Southwestern Medical Center, we performed a retrospective chart review of all female patients 18 to 80 years of age who presented to Parkland Health and Hospital System (PHHS) from January to December, 2012 and whose charts were coded with ICD-9 698.1: pruritus of genital organs. Of these patients, we selected those who presented for an initial visit to the general Gynecology Clinic or the specialized Vulvology Clinic within the Gynecology Clinic. Charts were reviewed to confirm that the diagnostic code pertained to an actual vulvar complaint. We tabulated demographic and historic information including age, symptom duration, presence of additional irritative symptoms (irritation, burning, or pain), whether associated vaginal complaints (including discharge) were present, as well as the use of any self-treatments prior to the initial visit. We recorded whether the provider questioned the patient specifically regarding vulvar hygiene habits. Elements of physical examination were recorded including whether the patient received targeted vulvar, vaginal, non-genital skin, and oral mucosa exams. Any testing obtained was recorded including saline wet prep of vaginal secretions, vaginal fungal or aerobic culture, gonorrhea and Chlamydia DNA polymerase testing, and vulvar skin biopsy. We tabulated up to three consecutive treatments each patient received and whether the patient was counseled regarding vulvar skin care recommendations as part of management. The presumptive clinical diagnosis was recorded as well as whether the diagnosis changed by 18 months of surveillance. Charts were also reviewed to determine the number of additional visits the patient had for the same complaint by 18 months of follow up, and whether their symptoms were persistent, improved or resolved at their latest visit. Chart abstraction was performed by two of the authors (SLB and AAL) with reconciliation of any discrepancies by repeat, joint chart review. Analysis of the selected parameters was performed using simple descriptive statistics.
Results
During 2012, a total of 1,175 female patient visits at PHHS were coded as ICD-9 698.1: pruritus of genital organs; this included all outpatient and Emergency Department encounters. Of these patients, only 123 patients were seen in the Gynecology or Vulvology Clinics. Patients whose vulvar complaint was not the primary purpose of the visit, who had a specifically vaginal complaint, or who were not seen for an initial evaluation (i.e. surveillance visits only) were excluded (n = 77). Information was tabulated for the remaining 46 patients seen for initial evaluation of a vulvar complaint. The average age of these patients was 46 years with a range of 19 to 70 years. Patients were seen in the general Gynecology Clinic (22%) or a specialized Vulvology Clinic (78%) at PHHS. Symptom duration prior to presentation ranged from 1 to 120 months, with an average of 24 months. Concurrent additional complaints of vulvar burning, pain, irritation, or vaginal discharge were reported in 41% of patients while 30% also reported vaginal itching. Inquiry regarding vulvar hygiene habits such as methods of genital cleansing, soaps and/or laundry detergents used (names, amounts, frequency of use) was documented for 70% at initial patient encounters.
For 91% of patients, there was documentation of prior self-treatment; of these, 67% of patients used one, 25% used two, and one patient (4%) used three products before initial presentation. The most commonly used self-treatments were topical anti-fungals (36%), topical steroids (29%) or topical estrogens (14%) acquired previously (Table 1). Regarding physical examination, all patients received a vulvar examination by unaided or magnified inspection, 96% underwent a vaginal speculum examination and 13% underwent a documented non-genital skin examination. During initial evaluation, testing included a saline wet prep of vaginal secretions for 65% of patients, an aerobic vaginal culture for 46%, gonorrhea/Chlamydia testing for 46% and a vulvar biopsy for 20% of patients.
Table 1: Treatment(s) used by patients with complaints of vulvar itching prior to their initial visit. View Table 1
The most common initial clinical diagnoses were non-specific, including a restatement of patient symptoms (39%), irritant or contact dermatitis (22%), or specific vulvar dermatoses including lichen sclerosus, lichen planus or lichen simplex chronicus (17%) (Table 2). Of note, for 7% of patients, providers were unable to distinguish if a patient's symptoms were the result of a dermatitis or Candida. Patients received information regarding recommended vulvar care measures in the form of counseling and/or a printed handout at 67% of initial encounters. Approximately 87% were prescribed at least one treatment during their initial evaluation; of these, 33% received two treatments and 2% received three. The most commonly prescribed treatments were topical steroids (78%), topical anti-fungals (20%) and topical lidocaine (10%) (Table 3).
Table 2: Clinical diagnosis at the patient's initial visit. View Table 2
Table 3: Treatment(s) prescribed by providers at the patient's initial visit. View Table 3
Within 18 months, 54% of patients had encounters for vulvar symptom reassessment. Of these patients, 11% received a change in diagnosis. Of the 9 patients who underwent vulvar biopsy, 8 had a histologic diagnosis consistent with their clinical diagnosis; four showed lichen sclerosus, four demonstrated findings consistent with contact or irritant dermatitis, and one was consistent with a hidradenoma papilliferum, a benign apocrine tumor. Patients had an average of two return visits (range 0 to 8) for complaints of vulvar itching in the following 18 months. At 18 months, 20% of patients experienced recurrent or persistent symptoms, 32% had improvement, and 48% of patients reported resolution of their symptoms.
Discussion
Complaints of vulvar itching and/or irritation, particularly itching, are common and affect a wide age range; the duration of symptoms at presentation is highly variable. They often lead to multiple outpatient visits for evaluation and treatment. Over 1,100 visits in one year to our hospital system were related to such vulvar complaints. A detailed, focused history and knowledge of the common causes of vulvar irritation inform the proper diagnosis and management of these patients. In our experience, patients often assume that their symptoms are due to inadequate hygiene or infection, triggering escalating hygiene practices that in turn exacerbate irritation and lead to ineffective self-treatment. This study is consistent with previously published investigations demonstrating that patients frequently misdiagnose themselves, leading to the unindicated or ineffective use of nonprescription anti-fungals, anti-itch creams, and topical antibiotics . For example, more than a third of our patients self-treated with anti-fungals prior to presentation. Without a careful history, these potentially contributory behaviors can be overlooked and go unaddressed. Symptom duration is often prolonged, with an average time of 2 years reported by our patients. Our study confirms that the majority of vulvar itching and related complaints are due to noninfectious etiologies, and treatment with topical steroids are most often prescribed followed by antifungals. Aerobic cultures to identify fungal vulvovaginitis not detected by vaginal wet prep did not change the diagnosis in any case at initial encounter. Likewise, vulvar biopsies failed to alter the initial diagnosis. This implies that these two tests can be reserved for subsequent evaluation of refractory cases.
The strengths of our study include the careful and adjudicated data collection completed by two of the authors. The restriction of patients to those seen under one specific diagnostic code commonly used at our institution allowed focus on patients presenting specifically for the vulvar symptoms addressed by this study. Limitations include a relatively small number of study subjects. The inclusion of patients within only one diagnostic code for genital pruritus likely excluded a much larger number of patients coded for other specific vulvar complaints such as vulvar irritation, pain, and burning, or as having vaginal, vulvovaginal, or other coexistent gynecologic complaints. Indeed, approximately two-thirds of patients coded for vulvar pruritus were eliminated from the study due to miscoding of the primary complaint in the electronic medical record. Limiting this review to visits within the Department of Gynecology (including a specialty clinic for vulvar complaints) surely underestimates the number of visits to all types of providers for vulvar pruritus and irritation as well as the variation in evaluation and management that patients experience within this hospital system. Indeed, approximately ten times more visits coded as vulvar pruritis took place outside the Gynecology or Vulvology Clinics at PHHS. Treatments beyond the maximum of three per patient were not recorded during the study time period. Furthermore, a co-existing diagnosis of diabetes was not elicited. As such, the degree to which this affects patient symptoms and number of clinic visits was not studied, however this likely does contribute to persistent symptoms of vulvar pruritis, especially via infectious agents.
Conclusions
The clinical assessment and monitoring of patients with vulvar pruritus and additional irritative vulvar complaints is variable, even in a specialized clinical setting. The standardization of clinical evaluation and initial management of this common complaint is warranted in order to study best approaches to initial evaluation and management. A clinical algorithm is proposed in the Appendix that follows as a potential move toward this goal, encouraging a thoughtful, detailed, organized approach that will more efficiently and accurately aid the diagnosis, management, and study of vulvar pruritus and other irritative complaints (Figure 1).
Figure 1: Proposed algorithm for the evaluation and initial management of patients with irritative vulvar symptoms (Box 1 and Box 2). View Figure 1
Declaration of Interests
The authors have no conflicts of interest to declare. The authors have no financial support disclosures.
References
Schlosser BJ (2010) Contact dermatitis of the vulva. Dermatol Clin 28: 697-706.
Trager JD (2005) What's your diagnosis? Acute vulvar erythema, edema, and pruritus in a young woman. J Pediatr Adolesc Gynecol 18: 275-280.
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Margesson LJ (2006) Vulvar disease pearls. Dermatol Clin 24: 145-155.
Erekson EA, Martin DK, Brousseau EC, Yip SO, Fried TR (2014) Over-the-counter treatments and perineal hygiene in postmenopausal women. Menopause 21: 281-285.
Stewart EJC, Wojnarowska F, Marren PM (1995) Experience in a clinic dedicated to the study of vulvar disorders. Br J Obs Dermatol 133: 31-32.
Fischer G, Spurret B, Fischer A (1995) The chronically symptomatic vulva: Aetiology and management. Br J Obstet Gynaecol 102: 773-779.
Crone AM, Stewart EJ, Wojnarowska F, Powell SM (2000) Aetiological factors in vulvar dermatitis. J Eur Acad Dermatol Venereol 14: 181-186.
Nyirjesy P, Peyton C, Weitz MV, Mathew L, Culhane JF (2006) Causes of chronic vaginitis: Analysis of a prospective database of affected women. Obstet Gynecol 108: 1185-1191.
Edwards L (2010) Dermatologic causes of vaginitis: A clinical review. Dermatol Clin 28: 727-735.
Marin M-G, King R, Sfameni S, Dennerstein GJ (2000) Adverse behavioral and sexual factors in chronic vulvar disease. Am J Obstet Gynecol 183: 34-38.
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Burrows LJ, Shaw HA, Goldstein AT (2008) The vulvar dermatoses. J Sex Med 5: 276-283.
Farage MA (2005) Vulvar susceptibility to contact irritants and allergens: A review. Arch Gynecol Obstet 272: 167-172.
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Citation
Bedell SL, Lawson AA, Griffith WF, Werner CL (2018) Vulvar Pruritus: Variability of Clinical Evaluation and Management. Int J Womens Health Wellness 4:084. doi.org/10.23937/2474-1353/1510084
Volume 4 Issue 2
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Article Details
International Journal of Women's Health and Wellness
ISSN: 2474-1353
Int J Womens Health Wellness
Abbrevation: IJWHW
DOI: 10.23937/2474-1353/1510084
Pub Date: November 30, 2018
Article Type: Original Article
Pub Type: Open Access
Corresponding author
Claudia L Werner, MD, Department of Obstetrics and Gynecology, University of Texas Southwestern Medical Center, 5323 Harry Hines Boulevard, Dallas, TX, 75390-9032, USA, Tel: 214-648-3662, Fax: 214-648-9028.
Copyright
© 2018 Bedell SL, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
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5190 | https://en.wikipedia.org/wiki/Heat_transfer | Jump to content
Heat transfer
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From Wikipedia, the free encyclopedia
Transport of thermal energy in physical systems
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes. Engineers also consider the transfer of mass of differing chemical species (mass transfer in the form of advection), either cold or hot, to achieve heat transfer. While these mechanisms have distinct characteristics, they often occur simultaneously in the same system.
Heat conduction, also called diffusion, is the direct microscopic exchanges of kinetic energy of particles (such as molecules) or quasiparticles (such as lattice waves) through the boundary between two systems. When an object is at a different temperature from another body or its surroundings, heat flows so that the body and the surroundings reach the same temperature, at which point they are in thermal equilibrium. Such spontaneous heat transfer always occurs from a region of high temperature to another region of lower temperature, as described in the second law of thermodynamics.
Heat convection occurs when the bulk flow of a fluid (gas or liquid) carries its heat through the fluid. All convective processes also move heat partly by diffusion, as well. The flow of fluid may be forced by external processes, or sometimes (in gravitational fields) by buoyancy forces caused when thermal energy expands the fluid (for example in a fire plume), thus influencing its own transfer. The latter process is often called "natural convection". The former process is often called "forced convection." In this case, the fluid is forced to flow by use of a pump, fan, or other mechanical means.
Thermal radiation occurs through a vacuum or any transparent medium (solid or fluid or gas). It is the transfer of energy by means of photons or electromagnetic waves governed by the same laws.
Overview
[edit]
See also: Heat transfer physics
Heat transfer is the energy exchanged between materials (solid/liquid/gas) as a result of a temperature difference. The thermodynamic free energy is the amount of work that a thermodynamic system can perform. Enthalpy is a thermodynamic potential, designated by the letter "H", that is the sum of the internal energy of the system (U) plus the product of pressure (P) and volume (V). Joule is a unit to quantify energy, work, or the amount of heat.
Heat transfer is a process function (or path function), as opposed to functions of state; therefore, the amount of heat transferred in a thermodynamic process that changes the state of a system depends on how that process occurs, not only the net difference between the initial and final states of the process.
Thermodynamic and mechanical heat transfer is calculated with the heat transfer coefficient, the proportionality between the heat flux and the thermodynamic driving force for the flow of heat. Heat flux is a quantitative, vectorial representation of heat flow through a surface.
In engineering contexts, the term heat is taken as synonymous with thermal energy. This usage has its origin in the historical interpretation of heat as a fluid (caloric) that can be transferred by various causes, and that is also common in the language of laymen and everyday life.
The transport equations for thermal energy (Fourier's law), mechanical momentum (Newton's law for fluids), and mass transfer (Fick's laws of diffusion) are similar, and analogies among these three transport processes have been developed to facilitate the prediction of conversion from any one to the others.
Thermal engineering concerns the generation, use, conversion, storage, and exchange of heat transfer. As such, heat transfer is involved in almost every sector of the economy. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.
Mechanisms
[edit]
The fundamental modes of heat transfer are:
Advection
: Advection is the transport mechanism of a fluid from one location to another, and is dependent on motion and momentum of that fluid.
Conduction or diffusion
: The transfer of energy between objects that are in physical contact. Thermal conductivity is the property of a material to conduct heat and is evaluated primarily in terms of Fourier's law for heat conduction.
Convection
: The transfer of energy between an object and its environment, due to fluid motion. The average temperature is a reference for evaluating properties related to convective heat transfer.
Radiation
: The transfer of energy by the emission of electromagnetic radiation.
Advection
[edit]
By transferring matter, energy—including thermal energy—is moved by the physical transfer of a hot or cold object from one place to another. This can be as simple as placing hot water in a bottle and heating a bed, or the movement of an iceberg in changing ocean currents. A practical example is thermal hydraulics. This can be described by the formula:
where
is heat flux (W/m2),
is density (kg/m3),
is heat capacity at constant pressure (J/kg·K),
is the difference in temperature (K),
is velocity (m/s).
Conduction
[edit]
Main article: Thermal conduction
On a microscopic scale, heat conduction occurs as hot, rapidly moving or vibrating atoms and molecules interact with neighboring atoms and molecules, transferring some of their energy (heat) to these neighboring particles. In other words, heat is transferred by conduction when adjacent atoms vibrate against one another, or as electrons move from one atom to another. Conduction is the most significant means of heat transfer within a solid or between solid objects in thermal contact. Fluids—especially gases—are less conductive. Thermal contact conductance is the study of heat conduction between solid bodies in contact. The process of heat transfer from one place to another place without the movement of particles is called conduction, such as when placing a hand on a cold glass of water—heat is conducted from the warm skin to the cold glass, but if the hand is held a few inches from the glass, little conduction would occur since air is a poor conductor of heat. Steady-state conduction is an idealized model of conduction that happens when the temperature difference driving the conduction is constant so that after a time, the spatial distribution of temperatures in the conducting object does not change any further (see Fourier's law). In steady state conduction, the amount of heat entering a section is equal to amount of heat coming out, since the temperature change (a measure of heat energy) is zero. An example of steady state conduction is the heat flow through walls of a warm house on a cold day—inside the house is maintained at a high temperature and, outside, the temperature stays low, so the transfer of heat per unit time stays near a constant rate determined by the insulation in the wall and the spatial distribution of temperature in the walls will be approximately constant over time.
Transient conduction (see Heat equation) occurs when the temperature within an object changes as a function of time. Analysis of transient systems is more complex, and analytic solutions of the heat equation are only valid for idealized model systems. Practical applications are generally investigated using numerical methods, approximation techniques, or empirical study.
Convection
[edit]
Main article: Convective heat transfer
The flow of fluid may be forced by external processes, or sometimes (in gravitational fields) by buoyancy forces caused when thermal energy expands the fluid (for example in a fire plume), thus influencing its own transfer. The latter process is often called "natural convection". All convective processes also move heat partly by diffusion, as well. Another form of convection is forced convection. In this case, the fluid is forced to flow by using a pump, fan, or other mechanical means.
Convective heat transfer, or simply, convection, is the transfer of heat from one place to another by the movement of fluids, a process that is essentially the transfer of heat via mass transfer. The bulk motion of fluid enhances heat transfer in many physical situations, such as between a solid surface and the fluid. Convection is usually the dominant form of heat transfer in liquids and gases. Although sometimes discussed as a third method of heat transfer, convection is usually used to describe the combined effects of heat conduction within the fluid (diffusion) and heat transference by bulk fluid flow streaming. The process of transport by fluid streaming is known as advection, but pure advection is a term that is generally associated only with mass transport in fluids, such as advection of pebbles in a river. In the case of heat transfer in fluids, where transport by advection in a fluid is always also accompanied by transport via heat diffusion (also known as heat conduction) the process of heat convection is understood to refer to the sum of heat transport by advection and diffusion/conduction.
Free, or natural, convection occurs when bulk fluid motions (streams and currents) are caused by buoyancy forces that result from density variations due to variations of temperature in the fluid. Forced convection is a term used when the streams and currents in the fluid are induced by external means—such as fans, stirrers, and pumps—creating an artificially induced convection current.
Convection-cooling
[edit]
See also: Nusselt number
Convective cooling is sometimes described as Newton's law of cooling:
The rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.
However, by definition, the validity of Newton's law of cooling requires that the rate of heat loss from convection be a linear function of ("proportional to") the temperature difference that drives heat transfer, and in convective cooling this is sometimes not the case. In general, convection is not linearly dependent on temperature gradients, and in some cases is strongly nonlinear. In these cases, Newton's law does not apply.
Convection vs. conduction
[edit]
In a body of fluid that is heated from underneath its container, conduction, and convection can be considered to compete for dominance. If heat conduction is too great, fluid moving down by convection is heated by conduction so fast that its downward movement will be stopped due to its buoyancy, while fluid moving up by convection is cooled by conduction so fast that its driving buoyancy will diminish. On the other hand, if heat conduction is very low, a large temperature gradient may be formed and convection might be very strong.
The Rayleigh number () is the product of the Grashof () and Prandtl () numbers. It is a measure that determines the relative strength of conduction and convection.
where
g is the acceleration due to gravity,
ρ is the density with being the density difference between the lower and upper ends,
μ is the dynamic viscosity,
α is the Thermal diffusivity,
β is the volume thermal expansivity (sometimes denoted α elsewhere),
T is the temperature,
ν is the kinematic viscosity, and
L is characteristic length.
The Rayleigh number can be understood as the ratio between the rate of heat transfer by convection to the rate of heat transfer by conduction; or, equivalently, the ratio between the corresponding timescales (i.e. conduction timescale divided by convection timescale), up to a numerical factor. This can be seen as follows, where all calculations are up to numerical factors depending on the geometry of the system.
The buoyancy force driving the convection is roughly , so the corresponding pressure is roughly . In steady state, this is canceled by the shear stress due to viscosity, and therefore roughly equals , where V is the typical fluid velocity due to convection and the order of its timescale. The conduction timescale, on the other hand, is of the order of .
Convection occurs when the Rayleigh number is above 1,000–2,000.
Radiation
[edit]
Radiative heat transfer is the transfer of energy via thermal radiation, i.e., electromagnetic waves. It occurs across vacuum or any transparent medium (solid or fluid or gas). Thermal radiation is emitted by all objects at temperatures above absolute zero, due to random movements of atoms and molecules in matter. Since these atoms and molecules are composed of charged particles (protons and electrons), their movement results in the emission of electromagnetic radiation which carries away energy. Radiation is typically only important in engineering applications for very hot objects, or for objects with a large temperature difference.
When the objects and distances separating them are large in size and compared to the wavelength of thermal radiation, the rate of transfer of radiant energy is best described by the Stefan-Boltzmann equation. For an object in vacuum, the equation is:
For radiative transfer between two objects, the equation is as follows:
where
is the heat flux,
is the emissivity (unity for a black body),
is the Stefan–Boltzmann constant,
is the view factor between two surfaces a and b, and
and are the absolute temperatures (in kelvins or degrees Rankine) for the two objects.
The blackbody limit established by the Stefan-Boltzmann equation can be exceeded when the objects exchanging thermal radiation or the distances separating them are comparable in scale or smaller than the dominant thermal wavelength. The study of these cases is called near-field radiative heat transfer.
Radiation from the sun, or solar radiation, can be harvested for heat and power. Unlike conductive and convective forms of heat transfer, thermal radiation – arriving within a narrow-angle i.e. coming from a source much smaller than its distance – can be concentrated in a small spot by using reflecting mirrors, which is exploited in concentrating solar power generation or a burning glass. For example, the sunlight reflected from mirrors heats the PS10 solar power tower and during the day it can heat water to 285 °C (545 °F).
The reachable temperature at the target is limited by the temperature of the hot source of radiation. (T4-law lets the reverse flow of radiation back to the source rise.) The (on its surface) somewhat 4000 K hot sun allows to reach coarsely 3000 K (or 3000 °C, which is about 3273 K) at a small probe in the focus spot of a big concave, concentrating mirror of the Mont-Louis Solar Furnace in France.
Phase transition
[edit]
Phase transition or phase change, takes place in a thermodynamic system from one phase or state of matter to another one by heat transfer. Phase change examples are the melting of ice or the boiling of water.
The Mason equation explains the growth of a water droplet based on the effects of heat transport on evaporation and condensation.
Phase transitions involve the four fundamental states of matter:
Solid – Deposition, freezing, and solid-to-solid transformation.
Liquid – Condensation and melting / fusion.
Gas – Boiling / evaporation, recombination/ deionization, and sublimation.
Plasma – Ionization.
Boiling
[edit]
The boiling point of a substance is the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid evaporates resulting in an abrupt change in vapor volume.
In a closed system, saturation temperature and boiling point mean the same thing. The saturation temperature is the temperature for a corresponding saturation pressure at which a liquid boils into its vapor phase. The liquid can be said to be saturated with thermal energy. Any addition of thermal energy results in a phase transition.
At standard atmospheric pressure and low temperatures, no boiling occurs and the heat transfer rate is controlled by the usual single-phase mechanisms. As the surface temperature is increased, local boiling occurs and vapor bubbles nucleate, grow into the surrounding cooler fluid, and collapse. This is sub-cooled nucleate boiling, and is a very efficient heat transfer mechanism. At high bubble generation rates, the bubbles begin to interfere and the heat flux no longer increases rapidly with surface temperature (this is the departure from nucleate boiling, or DNB).
At similar standard atmospheric pressure and high temperatures, the hydrodynamically quieter regime of film boiling is reached. Heat fluxes across the stable vapor layers are low but rise slowly with temperature. Any contact between the fluid and the surface that may be seen probably leads to the extremely rapid nucleation of a fresh vapor layer ("spontaneous nucleation"). At higher temperatures still, a maximum in the heat flux is reached (the critical heat flux, or CHF).
The Leidenfrost Effect demonstrates how nucleate boiling slows heat transfer due to gas bubbles on the heater's surface. As mentioned, gas-phase thermal conductivity is much lower than liquid-phase thermal conductivity, so the outcome is a kind of "gas thermal barrier".
Condensation
[edit]
Condensation occurs when a vapor is cooled and changes its phase to a liquid. During condensation, the latent heat of vaporization must be released. The amount of heat is the same as that absorbed during vaporization at the same fluid pressure.
There are several types of condensation:
Homogeneous condensation, as during the formation of fog.
Condensation in direct contact with subcooled liquid.
Condensation on direct contact with a cooling wall of a heat exchanger: This is the most common mode used in industry:
Filmwise condensation is when a liquid film is formed on the subcooled surface, and usually occurs when the liquid wets the surface.
Dropwise condensation is when liquid drops are formed on the subcooled surface, and usually occurs when the liquid does not wet the surface.
Dropwise condensation is difficult to sustain reliably; therefore, industrial equipment is normally designed to operate in filmwise condensation mode.
Melting
[edit]
Melting is a thermal process that results in the phase transition of a substance from a solid to a liquid. The internal energy of a substance is increased, typically through heat or pressure, resulting in a rise of its temperature to the melting point, at which the ordering of ionic or molecular entities in the solid breaks down to a less ordered state and the solid liquefies. Molten substances generally have reduced viscosity with elevated temperature; an exception to this maxim is the element sulfur, whose viscosity increases to a point due to polymerization and then decreases with higher temperatures in its molten state.
Modeling approaches
[edit]
Heat transfer can be modeled in various ways.
Heat equation
[edit]
The heat equation is an important partial differential equation that describes the distribution of heat (or temperature variation) in a given region over time. In some cases, exact solutions of the equation are available; in other cases the equation must be solved numerically using computational methods such as DEM-based models for thermal/reacting particulate systems (as critically reviewed by Peng et al.).
Lumped system analysis
[edit]
Lumped system analysis often reduces the complexity of the equations to one first-order linear differential equation, in which case heating and cooling are described by a simple exponential solution, often referred to as Newton's law of cooling.
System analysis by the lumped capacitance model is a common approximation in transient conduction that may be used whenever heat conduction within an object is much faster than heat conduction across the boundary of the object. This is a method of approximation that reduces one aspect of the transient conduction system—that within the object—to an equivalent steady-state system. That is, the method assumes that the temperature within the object is completely uniform, although its value may change over time.
In this method, the ratio of the conductive heat resistance within the object to the convective heat transfer resistance across the object's boundary, known as the Biot number, is calculated. For small Biot numbers, the approximation of spatially uniform temperature within the object can be used: it can be presumed that heat transferred into the object has time to uniformly distribute itself, due to the lower resistance to doing so, as compared with the resistance to heat entering the object.
Climate models
[edit]
Climate models study the radiant heat transfer by using quantitative methods to simulate the interactions of the atmosphere, oceans, land surface, and ice.
Engineering
[edit]
Heat transfer has broad application to the functioning of numerous devices and systems. Heat-transfer principles may be used to preserve, increase, or decrease temperature in a wide variety of circumstances.[citation needed] Heat transfer methods are used in numerous disciplines, such as automotive engineering, thermal management of electronic devices and systems, climate control, insulation, materials processing, chemical engineering and power station engineering.
Insulation, radiance and resistance
[edit]
Thermal insulators are materials specifically designed to reduce the flow of heat by limiting conduction, convection, or both. Thermal resistance is a heat property and the measurement by which an object or material resists to heat flow (heat per time unit or thermal resistance) to temperature difference.
Radiance, or spectral radiance, is a measure of the quantity of radiation that passes through or is emitted. Radiant barriers are materials that reflect radiation, and therefore reduce the flow of heat from radiation sources. Good insulators are not necessarily good radiant barriers, and vice versa. Metal, for instance, is an excellent reflector and a poor insulator.
The effectiveness of a radiant barrier is indicated by its reflectivity, which is the fraction of radiation reflected. A material with a high reflectivity (at a given wavelength) has a low emissivity (at that same wavelength), and vice versa. At any specific wavelength, reflectivity=1 - emissivity. An ideal radiant barrier would have a reflectivity of 1, and would therefore reflect 100 percent of incoming radiation. Vacuum flasks, or Dewars, are silvered to approach this ideal. In the vacuum of space, satellites use multi-layer insulation, which consists of many layers of aluminized (shiny) Mylar to greatly reduce radiation heat transfer and control satellite temperature.[citation needed]
Devices
[edit]
A heat engine is a system that performs the conversion of a flow of thermal energy (heat) to mechanical energy to perform mechanical work. A thermocouple is a temperature-measuring device and a widely used type of temperature sensor for measurement and control, and can also be used to convert heat into electric power. A thermoelectric cooler is a solid-state electronic device that pumps (transfers) heat from one side of the device to the other when an electric current is passed through it. It is based on the Peltier effect. A thermal diode or thermal rectifier is a device that causes heat to flow preferentially in one direction.
Heat exchangers
[edit]
A heat exchanger is used for more efficient heat transfer or to dissipate heat. Heat exchangers are widely used in refrigeration, air conditioning, space heating, power generation, and chemical processing. One common example of a heat exchanger is a car's radiator, in which the hot coolant fluid is cooled by the flow of air over the radiator's surface.
Common types of heat exchanger flows include parallel flow, counter flow, and cross flow. In parallel flow, both fluids move in the same direction while transferring heat; in counter flow, the fluids move in opposite directions; and in cross flow, the fluids move at right angles to each other. Common types of heat exchangers include shell and tube, double pipe, extruded finned pipe, spiral fin pipe, u-tube, and stacked plate. Each type has certain advantages and disadvantages over other types.[further explanation needed]
A heat sink is a component that transfers heat generated within a solid material to a fluid medium, such as air or a liquid. Examples of heat sinks are the heat exchangers used in refrigeration and air conditioning systems or the radiator in a car. A heat pipe is another heat-transfer device that combines thermal conductivity and phase transition to efficiently transfer heat between two solid interfaces.
Applications
[edit]
Architecture
[edit]
Efficient energy use is the goal to reduce the amount of energy required in heating or cooling. In architecture, condensation and air currents can cause cosmetic or structural damage. An energy audit can help to assess the implementation of recommended corrective procedures. For instance, insulation improvements, air sealing of structural leaks, or the addition of energy-efficient windows and doors.
Smart meter is a device that records electric energy consumption in intervals.
Thermal transmittance is the rate of transfer of heat through a structure divided by the difference in temperature across the structure. It is expressed in watts per square meter per kelvin, or W/(m2K). Well-insulated parts of a building have a low thermal transmittance, whereas poorly-insulated parts of a building have a high thermal transmittance.
Thermostat is a device to monitor and control temperature.
Climate engineering
[edit]
See also: Anthropogenic heat
Climate engineering consists of carbon dioxide removal and solar radiation management. Since the amount of carbon dioxide determines the radiative balance of Earth's atmosphere, carbon dioxide removal techniques can be applied to reduce the radiative forcing. Solar radiation management is the attempt to absorb less solar radiation to offset the effects of greenhouse gases.
An alternative method is passive daytime radiative cooling, which enhances terrestrial heat flow to outer space through the infrared window (8–13 μm). Rather than merely blocking solar radiation, this method increases outgoing longwave infrared (LWIR) thermal radiation heat transfer with the extremely cold temperature of outer space (~2.7 K) to lower ambient temperatures while requiring zero energy input.
Greenhouse effect
[edit]
The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases and clouds, and is re-radiated in all directions, resulting in a reduction in the amount of thermal radiation reaching space relative to what would reach space in the absence of absorbing materials. This reduction in outgoing radiation leads to a rise in the temperature of the surface and troposphere until the rate of outgoing radiation again equals the rate at which heat arrives from the Sun.
Heat transfer in the human body
[edit]
See also: Wet-bulb temperature
The principles of heat transfer in engineering systems can be applied to the human body to determine how the body transfers heat. Heat is produced in the body by the continuous metabolism of nutrients which provides energy for the systems of the body. The human body must maintain a consistent internal temperature to maintain healthy bodily functions. Therefore, excess heat must be dissipated from the body to keep it from overheating. When a person engages in elevated levels of physical activity, the body requires additional fuel which increases the metabolic rate and the rate of heat production. The body must then use additional methods to remove the additional heat produced to keep the internal temperature at a healthy level.
Heat transfer by convection is driven by the movement of fluids over the surface of the body. This convective fluid can be either a liquid or a gas. For heat transfer from the outer surface of the body, the convection mechanism is dependent on the surface area of the body, the velocity of the air, and the temperature gradient between the surface of the skin and the ambient air. The normal temperature of the body is approximately 37 °C. Heat transfer occurs more readily when the temperature of the surroundings is significantly less than the normal body temperature. This concept explains why a person feels cold when not enough covering is worn when exposed to a cold environment. Clothing can be considered an insulator which provides thermal resistance to heat flow over the covered portion of the body. This thermal resistance causes the temperature on the surface of the clothing to be less than the temperature on the surface of the skin. This smaller temperature gradient between the surface temperature and the ambient temperature will cause a lower rate of heat transfer than if the skin were not covered.
To ensure that one portion of the body is not significantly hotter than another portion, heat must be distributed evenly through the bodily tissues. Blood flowing through blood vessels acts as a convective fluid and helps to prevent any buildup of excess heat inside the tissues of the body. This flow of blood through the vessels can be modeled as pipe flow in an engineering system. The heat carried by the blood is determined by the temperature of the surrounding tissue, the diameter of the blood vessel, the thickness of the fluid, the velocity of the flow, and the heat transfer coefficient of the blood. The velocity, blood vessel diameter, and fluid thickness can all be related to the Reynolds Number, a dimensionless number used in fluid mechanics to characterize the flow of fluids.
Latent heat loss, also known as evaporative heat loss, accounts for a large fraction of heat loss from the body. When the core temperature of the body increases, the body triggers sweat glands in the skin to bring additional moisture to the surface of the skin. The liquid is then transformed into vapor which removes heat from the surface of the body. The rate of evaporation heat loss is directly related to the vapor pressure at the skin surface and the amount of moisture present on the skin. Therefore, the maximum of heat transfer will occur when the skin is completely wet. The body continuously loses water by evaporation but the most significant amount of heat loss occurs during periods of increased physical activity.
Cooling techniques
[edit]
Evaporative cooling
[edit]
Evaporative cooling happens when water vapor is added to the surrounding air. The energy needed to evaporate the water is taken from the air in the form of sensible heat and converted into latent heat, while the air remains at a constant enthalpy. Latent heat describes the amount of heat that is needed to evaporate the liquid; this heat comes from the liquid itself and the surrounding gas and surfaces. The greater the difference between the two temperatures, the greater the evaporative cooling effect. When the temperatures are the same, no net evaporation of water in the air occurs; thus, there is no cooling effect.
Laser cooling
[edit]
In quantum physics, laser cooling is used to achieve temperatures of near absolute zero (−273.15 °C, −459.67 °F) of atomic and molecular samples to observe unique quantum effects that can only occur at this heat level.
Doppler cooling is the most common method of laser cooling.
Sympathetic cooling is a process in which particles of one type cool particles of another type. Typically, atomic ions that can be directly laser-cooled are used to cool nearby ions or atoms. This technique allows the cooling of ions and atoms that cannot be laser-cooled directly.
Magnetic cooling
[edit]
Main articles: Magnetic refrigeration and Magnetic evaporative cooling
Magnetic evaporative cooling is a process for lowering the temperature of a group of atoms, after pre-cooled by methods such as laser cooling. Magnetic refrigeration cools below 0.3K, by making use of the magnetocaloric effect.
Radiative cooling
[edit]
Radiative cooling is the process by which a body loses heat by radiation. Outgoing energy is an important effect in the Earth's energy budget. In the case of the Earth-atmosphere system, it refers to the process by which long-wave (infrared) radiation is emitted to balance the absorption of short-wave (visible) energy from the Sun. The thermosphere (top of atmosphere) cools to space primarily by infrared energy radiated by carbon dioxide (CO2) at 15 μm and by nitric oxide (NO) at 5.3 μm. Convective transport of heat and evaporative transport of latent heat both remove heat from the surface and redistribute it in the atmosphere.
Thermal energy storage
[edit]
Thermal energy storage includes technologies for collecting and storing energy for later use. It may be employed to balance energy demand between day and nighttime. The thermal reservoir may be maintained at a temperature above or below that of the ambient environment. Applications include space heating, domestic or process hot water systems, or generating electricity.[citation needed]
History
[edit]
Main article: Newton's law of cooling
Newton's law of cooling
[edit]
In 1701, Isaac Newton anonymously published an article in Philosophical Transactions noting (in modern terms) that the rate of temperature change of a body is proportional to the difference in temperatures (graduum caloris, "degrees of heat") between the body and its surroundings. The phrase "temperature change" was later replaced with "heat loss", and the relationship was named Newton's law of cooling. In general, the law is valid only if the temperature difference is small and the heat transfer mechanism remains the same.
Thermal conduction
[edit]
In heat conduction, the law is valid only if the thermal conductivity of the warmer body is independent of temperature. The thermal conductivity of most materials is only weakly dependent on temperature, so in general the law holds true.
Thermal convection
[edit]
In convective heat transfer, the law is valid for forced air or pumped fluid cooling, where the properties of the fluid do not vary strongly with temperature, but it is only approximately true for buoyancy-driven convection, where the velocity of the flow increases with temperature difference.
Thermal radiation
[edit]
In the case of heat transfer by thermal radiation, Newton's law of cooling holds only for very small temperature differences.
Thermal conductivity of different metals
[edit]
In a 1780 letter to Benjamin Franklin, Dutch-born British scientist Jan Ingenhousz relates an experiment which enabled him to rank seven different metals according to their thermal conductivities:
You remembre you gave me a wire of five metals all drawn thro the same hole Viz. one, of gould, one of silver, copper steel and iron. I supplyed here the two others Viz. the one of tin the other of lead. I fixed these seven wires into a wooden frame at an equal distance of one an other ... I dipt the seven wires into this melted wax as deep as the wooden frame ... By taking them out they were covred with a coat of wax ... When I found that this crust was there about of an equal thikness upon all the wires, I placed them all in a glased earthen vessel full of olive oil heated to some degrees under boiling, taking care that each wire was dipt just as far in the oil as the other ... Now, as they had been all dipt alike at the same time in the same oil, it must follow, that the wire, upon which the wax had been melted the highest, had been the best conductor of heat. ... Silver conducted heat far the best of all other metals, next to this was copper, then gold, tin, iron, steel, Lead.
Benjamin Thompson's experiments on heat transfer
[edit]
See also: Benjamin Thompson § Experiments on heat
During the years 1784 – 1798, the British physicist Benjamin Thompson (Count Rumford) lived in Bavaria, reorganizing the Bavarian army for the Prince-elector Charles Theodore among other official and charitable duties. The Elector gave Thompson access to the facilities of the Electoral Academy of Sciences in Mannheim. During his years in Mannheim and later in Munich, Thompson made a large number of discoveries and inventions related to heat.
Conductivity experiments
[edit]
"New Experiments upon Heat"
[edit]
In 1785 Thompson performed a series of thermal conductivity experiments, which he describes in great detail in the Philosophical Transactions article "New Experiments upon Heat" from 1786. The fact that good electrical conductors are often also good heat conductors and vice versa must have been well known at the time, for Thompson mentions it in passing. He intended to measure the relative conductivities of mercury, water, moist air, "common air" (dry air at normal atmospheric pressure), dry air of various rarefication, and a "Torricellian vacuum".
From the striking analogy between the electric fluid and heat respecting their conductors and non-conductors (having found that bodies, in general, which are conductors of the electric fluid, are likewise good conductors of heat, and, on the contrary, that electric bodies, or such as are bad conductors of the electric fluid, are likewise bad conductors of heat), I was led to imagine that the Torricellian vacuum, which is known to afford so ready a passage to the electric fluid, would also have afforded a ready passage to heat.
| Medium | Relative conductivity |
| Mercury | 1000 |
| Moist air | 330 |
| Water | 313 |
| Dry air (1 atm) | 80.41 |
| Dry air (1/4 atm) | 80.23 |
| Dry air (1/24 atm) | 78 |
| Torricellian vacuum | 55 |
For these experiments, Thompson employed a thermometer inside a large, closed glass tube. Under the circumstances described, heat may—unbeknownst to Thompson—have been transferred more by radiation than by conduction. These were his results.
After the experiments, Thompson was surprised to observe that a vacuum was a significantly poorer heat conductor than air "which of itself is reckoned among the worst", but only a very small difference between common air and rarefied air. He also noted the great difference between dry air and moist air, and the great benefit this affords.
I cannot help observing, with what infinite wisdom and goodness Divine Providence appears to have guarded us against the evil effects of excessive heat and cold in the atmosphere; for if it were possible for the air to be equally damp during the severe cold of the winter ... as it sometimes is in summer, its conducing power, and consequently its apparent coldness ... would become quite intolerable; but, happily for us, its power to hold water in solution is diminished, and with it its power to rob us of our animal heat.
Every body knows how very disagreeable a very moderate degree of cold is when the air is very damp; and from hence it appears, why the thermometer is not always a just measure of the apparent or sensible heat of the atmosphere. If colds ... are occasioned by our bodies being robbed of our animal heat, the reason is plain why those disorders prevail most during the cold autumnal rains, and upon the breaking up of the frost in the spring. It is likewise plain [why] ... inhabiting damp houses, is so very dangerous; and why the evening air is so pernicious in summer ... and why it is not so during the hard frosts of winter.
Temperature vs. sensible heat
[edit]
Thompson concluded with some comments on the important difference between temperature and sensible heat.
The ... sensation of hot or cold depends not intirely upon the temperature of the body exciting in us those sensations ... but upon the quantity of heat it is capable of communicating to us, or receiving from us ... and this depends in a great measure upon the conducing powers of the bodies in question. The sensation of hot is the entrance of heat into our bodies; that of cold is its exit ... This is another proof that the thermometer cannot be a just measure of sensible heat ... or rather, that the touch does not afford us a just indication of ... real temperatures.
Coining of the term "convection"
[edit]
In the 1830s, in The Bridgewater Treatises, the term convection is attested in a scientific sense. In treatise VIII by William Prout, in the book on chemistry, it says:
This motion of heat takes place in three ways, which a common fire-place very well illustrates. If, for instance, we place a thermometer directly before a fire, it soon begins to rise, indicating an increase of temperature. In this case the heat has made its way through the space between the fire and the thermometer, by the process termed radiation. If we place a second thermometer in contact with any part of the grate, and away from the direct influence of the fire, we shall find that this thermometer also denotes an increase of temperature; but here the heat must have travelled through the metal of the grate, by what is termed conduction. Lastly, a third thermometer placed in the chimney, away from the direct influence of the fire, will also indicate a considerable increase of temperature; in this case a portion of the air, passing through and near the fire, has become heated, and has carried up the chimney the temperature acquired from the fire. There is at present no single term in our language employed to denote this third mode of the propagation of heat; but we venture to propose for that purpose, the term convection, [in footnote: [Latin] Convectio, a carrying or conveying] which not only expresses the leading fact, but also accords very well with the two other terms.
Later, in the same treatise VIII, in the book on meteorology, the concept of convection is also applied to "the process by which heat is communicated through water".
See also
[edit]
Combined forced and natural convection
Heat capacity
Heat transfer enhancement
Heat transfer physics
Stefan–Boltzmann law
Thermal contact conductance
Thermal energy storage
Thermal physics
Thermal resistance
Citations
[edit]
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References
[edit]
Martin, Thomas (October 1951). "The Experimental Researches of Benjamin Thompson, Count Rumford". Bulletin of the British Society for the History of Science. 1 (6): 144–158. doi:10.1017/S0950563600000567. JSTOR 4024834.
Thompson, Benjamin (31 December 1786). "XIV. New experiments upon heat. By Colonel Sir Benjamin Thompson, Knt. F. R. S. In a letter to Sir Joseph Banks, Bart. P. R. S.". Philosophical Transactions of the Royal Society of London. 76: 273–304. doi:10.1098/rstl.1786.0014.
External links
[edit]
A Heat Transfer Textbook - (free download).
Thermal-FluidsPedia - An online thermal fluids encyclopedia.
Hyperphysics Article on Heat Transfer - Overview
Interseasonal Heat Transfer - a practical example of how heat transfer is used to heat buildings without burning fossil fuels.
Aspects of Heat Transfer, Cambridge University
Thermal-Fluids Central
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5191 | https://www.youtube.com/watch?v=WOQQQxJTQ2o | Chapter 10: Photosynthesis | Campbell Biology (Podcast Summary)
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Posted: 13 Mar 2025
Chapter 10 of Campbell Biology explains photosynthesis, the process by which plants, algae, and some prokaryotes convert light energy into chemical energy. Photosynthesis is essential for life on Earth, providing the oxygen we breathe and the organic molecules that form the base of the food chain.
The chapter details the two stages of photosynthesis—the light reactions (which convert solar energy into ATP and NADPH) and the Calvin cycle (which uses ATP and NADPH to fix carbon dioxide into organic molecules). It also explores adaptations such as C4 and CAM pathways, which allow plants to minimize water loss in hot and dry environments.
🔹 Key Concepts
🔹 Photosynthesis and Energy Flow
Autotrophs (producers) make their own organic molecules using light energy.
Heterotrophs (consumers) depend on autotrophs for food and oxygen.
The general equation for photosynthesis:
6 CO₂ + 6 H₂O + Light energy → C₆H₁₂O₆ + 6 O₂
Photosynthesis is a redox process:
Water is oxidized (loses electrons) → produces O₂.
Carbon dioxide is reduced (gains electrons) → forms glucose.
🔹 The Light Reactions (Occur in the Thylakoids)
✅ Convert solar energy into chemical energy (ATP & NADPH)
✅ Water is split, releasing O₂ as a byproduct.
✅ Key steps:
Photosystem II (PSII) absorbs light, exciting electrons from chlorophyll.
Electrons pass through the electron transport chain (ETC), pumping H⁺ into the thylakoid space.
ATP synthase uses this H⁺ gradient to generate ATP (photophosphorylation).
Photosystem I (PSI) absorbs light, re-energizing electrons to reduce NADP⁺ to NADPH.
🔹 The Calvin Cycle (Occurs in the Stroma)
✅ Uses ATP and NADPH from light reactions to fix CO₂ into sugar.
✅ Three Phases of the Calvin Cycle:
Carbon Fixation: CO₂ is attached to RuBP by the enzyme rubisco.
Reduction: ATP & NADPH are used to convert 3-PGA into G3P (a three-carbon sugar).
Regeneration of RuBP: ATP is used to regenerate RuBP, allowing the cycle to continue.
✅ For every three CO₂ molecules fixed, one G3P is produced.
🔹 Alternative Pathways of Carbon Fixation
C3 Plants: Use the Calvin cycle directly but are prone to photorespiration (wasting energy when O₂ levels are high).
C4 Plants (e.g., corn, sugarcane):
Use PEP carboxylase instead of rubisco, reducing photorespiration.
CO₂ is fixed into a 4-carbon compound and delivered to the Calvin cycle in bundle-sheath cells.
CAM Plants (e.g., cacti, pineapples):
Stomata open at night to fix CO₂ into organic acids, storing it for daytime use.
This reduces water loss in arid environments.
🔹 The Importance of Photosynthesis
Provides oxygen for aerobic life.
Forms the base of the food chain in ecosystems.
Converts solar energy into organic molecules for plant growth and energy storage.
📖 Glossary of Terms
🔹 Autotrophs – Organisms that produce their own food using inorganic molecules.
🔹 Calvin Cycle – The cycle that converts CO₂ into sugars using ATP & NADPH.
🔹 Chlorophyll – The green pigment in plants that captures light energy.
🔹 Electron Transport Chain (ETC) – A series of proteins that transfer electrons, creating an H⁺ gradient.
🔹 G3P (Glyceraldehyde-3-Phosphate) – The three-carbon sugar produced in the Calvin cycle.
🔹 NADPH – A high-energy electron carrier used in the Calvin cycle.
🔹 Photorespiration – A process where rubisco binds O₂ instead of CO₂, reducing photosynthetic efficiency.
🔹 Photophosphorylation – The process of generating ATP using light energy.
🔹 Photosystems – Protein-pigment complexes that absorb light for photosynthesis.
🔹 Rubisco (RuBP Carboxylase-Oxygenase) – The enzyme that fixes CO₂ in the Calvin cycle.
🔹 Stomata – Small pores in leaves that allow gas exchange.
🔹 Thylakoids – Membrane-bound sacs in chloroplasts where the light reactions occur.
Campbell Biology Chapter 10 summary, Photosynthesis explained, Light Reactions and Calvin Cycle, Electron Transport Chain in Photosynthesis, Photosystem I and II, ATP and NADPH, Carbon Fixation, C3 vs C4 vs CAM plants, AP Biology Photosynthesis, College Biology Notes 📘 Read full blog summaries for every chapter:
Transcript:
hey everyone welcome back ready for another deep dive today we're tackling something pretty fundamental to Life as We Know It photosynthesis yeah photosynthesis huh sounds a little dry maybe maybe but trust me it's anything but we're going Way Beyond the basic plants use sunlight to make food stuff right right we're going to really dig into the mechanics of how it all works like how do those green machines turn sunlight into well pretty much everything we depend on well that's what makes it's so fascinating I mean photosynthesis is really the foundation of almost every ecosystem on the planet yeah think about it it's this giant energy conversion system I mean it's turning sunlight into the food we eat and the oxygen we breathe you know so this chapter we're looking at it covers everything from the big picture role of photosynthesis like you're talking about down to the tiny details right absolutely yeah starts with the basics you've got your autotrofos those are the self-feeding organisms like uh plants and algae they use photosynthesis to grab that light energy and convert it into chemical energy no we're producers exactly and then you've got the heterotrophs like us we can't make our own food so we rely on those Auto trops you know they consumers yeah so it's kind of like plants are the chefs of the planet right yeah whipping up this Feast using sunlight and we're all just kind of showing up to enjoy the meal that's a great way to put it it really is and it's not just about the food either right I mean photosynthesis produces the oxygen we need to breathe yeah every breath it's pretty wild when you think about it like every breath we take is connected to this incredible process that's happening in you know plants all around us so that's the big picture but what about those tiny kitchens where all the magic actually happens the chloroplasts ah the chloroplasts yeah they've got this amazing internal architecture it's perfectly designed for capturing and converting that light energy you've got this double membrane enclosing this space called the stroma right okay and inside you've got these stacks of um flattened sacks called phids they look kind of like a stack of pancakes if you can imagine that okay I could picture that and embedded in those thid membranes that's where you find the star of the show chlorophyll chlorophyll right right it gives plants their green color it's that pigment that's grabbing all that sunlight but how does it actually work like at the molecular level well we got to think about light as a form of energy that travels in waves right and chlorophyll is specifically designed to capture certain wavelengths of light in the visible spectrum okay particularly um in the violet blue and red ranges Ah that's why plants look green to us they're absorbing all those other colors and reflecting back the green exactly and when chlorophyll absorbs a photon of light one of its electrons gets bumped up to a higher energy level it's like it gets this sudden burst of energy and jumps up a step okay so now we've got excited electrons what happens next well now if you just had Corell like floating around by itself those excited electrons they they would quickly lose their energy right fall back down releasing that energy as heat and a little bit of red light we call it fluorescence so that's why a spinach smoothie doesn't like make my kitchen glow in the dark right exactly but inside a chloroplast things get a lot more interesting the chlorophyll molecules are organized into these complexes we call them photosystems photosystems they sound a bit like something out of Star Wars yeah a little bit right but each photosystem it consists of a reaction Center complex surrounded by these light harvesting complexes and these light harvesting complexes act like antennas capturing photons of light and funneling that energy to the reaction Center okay so it's like a network of tiny solar panels all channeling energy to this Central Power Station Perfect Analogy and at the heart of that reaction Center is a special pair of chlorophyll molecules now we've got two types of photosystems photos system 2 and photos system 1 right each has its own special chlorophyll app pair p680 and p700 respectively okay so we got the light energy being captured exciting electrons in the special chlorophyll aairs but what happens to those energized electrons do they just like say excited that's where things get really interesting these excited electrons they go on this remarkable Journey it's a chain reaction of electron transfers that really drives this whole process of converting light energy into chemical energy lead the way we're ready to follow those electrons on their Adventure All right so we're back and those energized electrons they they're ready to start their Journey we call this linear electron flow and it's how light energy gets turned into the chemical energy that powers Life as we know it okay so it's like each of these electrons is carrying this Precious Cargo like a tiny little bundle of energy captured from the sunlight so where do they go first first up is photosystem 2 remember that special pair of chlorophyll molecules p680 yeah when light hits it and excites an electron in p680 that electron is really quickly snatched up by a molecule we call the primary electronic scepter okay so it's like a relay race right exactly p680 passes the energy baton to the primary electronic scepter but now p680 is missing an electron what happens then well this is where it gets really cool right to replace that missing electron p680 basically steals one from a water molecule well wait steals an electron from water how does that even work okay so there's this enzyme in photosystem 2 and this enzyme it can split water molecules and when it splits those water molecules what you get is electrons hydrogen ions protons and oxygen as a byproduct hold on are you saying that the oxygen we breathe like the stuff that keeps us alive that's basically just a leftover from plants splitting water yeah pretty much it's pretty amazing right that is mindblowing so those electrons from the water they go back and replenish p680 and that lets it keep capturing light energy and keep passing those energized electrons down the chain okay so the electrons on the Move where does it head to next from that primary electron acceptor the electron basically hops down this chain of molecules we call it the electron transport chain right like a series of steps exactly yeah and with each step down a little bit of energy is released and that reliefed energy that's what's used to pump protons across the Philo covid membrane right you got it from the stroma into the thilo covid space so we're building up this concentration of protons inside that thilo covid space yeah kind of like bumping air into a tire yeah Perfect Analogy and all those protons create what we call a proton gradient basically a difference in concentration on either side of that membrane oh okay I see where you're going with this yeah this is starting to sound familiar we talked about proton gradients back when we were discussing cellular respiration this is all about chemiosmosis isn't it you got it that proton gradient it's like potential energy like water stored behind a dam and just like a dam can release water to generate electricity that proton gradient can be used to power the production of ATP ATP the cell's energy currency right so we're literally seeing the energy from sunlight being transformed into this usable form that the cell can use exactly but while that first electron is busy helping create that proton gradient the story is not over yet photosystem huse is also absorbing light and exciting electrons so it's like we have two separate relay races happening side by side you got it so photosystem M has its own special chlorophyll aair it's p700 and it's got its own primary electron acceptor too okay the excited electron from p700 it gets passed down another electron transport chain but this time instead of creating a proton gradient the electron ultimately gets passed to a molecule called nadp plus laa reducing it to nadph nadph another energy carrier molecule right we've got ATP being produced through chemiosmosis and nadph being generated yeah and all this is happening thanks to the power of sunlight and these complex electron relay races that's the magic of the light reactions right there light energy has been transformed into chemical energy stored in ATP and that reducing power of nadph so the light reactions have set the stage and now we've got the fuel we need to actually start building those sugars exactly and those sugars are built in the next stage the Calvin cycle right the Sugar Factory we can't wait to see how it all comes together but before we jump into that the chapter mentioned another pathway called cyclic electron flow ah yeah good catch cyclic electron flow this one only involves photosystem one so no photosystem 2 no water splitting no oxygen production what's the point of this cyclic route then well in cyclic electron flow that excited electron from p700 it doesn't end up reducing nadp Plus or right side instead it Cycles back to that first electron transport chain and that helps contribute to that proton gradient and ultimately that means more ETP oh interesting so it's like a shortcut that focuses on boosting ATP production exactly and research suggests that cyclic electron flow it might be really important for protecting plants from damage especially under those really high light conditions Okay so we've got linear electron flow and cyclic electron flow both generating ATP but only one producing neph busy place these light reactions but now I think we're finally ready to enter the Sugar Factory itself the celvin cycle the celvin cycle this is where the carbon dioxide from the air is transformed into the building blocks of life now that sounds like some serious magic let's see how it all works yeah okay so we've been through the light reactions seen those electrons zipping around and how ATP and NPH are made but now let's get to the main course right the Calvin cycle it's like this is where those energy molecules get put to work building sugars from CO2 I mean it's happening right now in every leaf and every blade of grass out there it's pretty remarkable when you think about it the Calvin cycle is like this intricate metabolic machine you know it's just churning away inside the chloroplast stroma using all that ATP and nadph that was made in the light reactions to well convert that inorganic carbon dioxide into organic sugars so it's like the light reactions provide the ingredients and the Calvin cycle is like the chef following this recipe to make those sugar molecules I like that analogy and like any good recipe the Calvin cycle has these key steps so first up we have what we call carbon fixation where CO2 gets incorporated into an existing organic molecule so it's like grabbing that carbon from the air and attaching it to something that a plant can actually use exactly and the star of the step is an enzyme called rabisco you know it's actually the most abundant protein on Earth and it's probably one of the most important too because it catalyzes that first step in converting atmospheric CO2 into organic molecules rubisco the carbon capture got it but you mentioned something earlier called photorespiration and it sounded like rubisco can sometimes mess up and grab oxygen instead of CO2 yeah that's right rabisco evolved way back when the atmosphere had a lot less oxygen so back then this whole oxygen grabbing thing wasn't really a big deal but these days with all the oxygen we have photorespiration can kind of drag down a plant's efficiency so instead of fixing carbon sometimes Riso kind of throws a wrench in the works like wastes energy and releases CO2 doesn't that seem counterproductive I mean if it's so wasteful why hasn't Evolution gotten rid of it well it's not quite that simple I mean photorespiration does have a few benefits for example it can help protect plants from damage when there's a lot of light and some stud suggest it might even play a role in nitrogen metabolism but yeah it's definitely something that could be more efficient so even Nature's best enzyme has its quirks but the chapter mentions some plants that have evolved ways to actually minimize this photorespiration problem right especially the plants that are living in those really hot dry places that's right those are the C4 and cam plants you know like corn sugar cane cacti pineapples things like that they've come up with some really clever strategy IES to concentrate the CO2 in their cells which makes rabisco more efficient and cuts down on photorespiration okay let's break that down a bit first the C4 plants how do they manage to concentrate that CO2 well C4 plants they have this unique Leaf structure they have these special bundle sheath cells and those cells are where the Calvin cycle actually takes place and get this those cells are surrounded by mesophyll cells and in those mesophile cells the CO2 is fixed by a different enzyme it's called pep carboxylase and it has a much higher affinity for CO2 than rubisco does and it doesn't bind to oxygen ah so they've kind of bypassed rubisco for that initial carbon capture step that's pretty clever it is so this P carboxylase fixes the CO2 into a four carbon compound and then that gets transported to the bundle sheath cells and there that four carbon compound is broken down which releases the CO2 and concentrates it right near Risco and that lets the Calvin cycle run efficiently even when the plant stamata are partially closed to save water so it's like they've built this little CO2 pump delivering it right to where it's needed yeah it's a great way to think about it now seam implants they have a different approach they separate carbon fixation and the Calvin cycle in time instead of space Oh interesting so instead of different cells doing different jobs it's the same cells but working at different times exactly so cam plants they open their stamata at night when it's cooler and they don't lose as much water they take in the CO2 and they convert it into organic acids storing them in their Val vules and then during the day they close their stamata to conserve water and then they release that CO2 from the stored organic acids and that feeds into the Calvin cycle wow so it's like they're stockpiling CO2 overnight and then they use it during the day to make sugar when there's sunlight to power the process that's amazing it's pretty ingenious these adaptations let those C4 and cam plants thrive in those hot dry climates where other plants would really struggle it really shows how Evolution can come up with these really elegant solutions to those tough environmental challenges so we've covered the ins and outs of photosynthesis from the light reactions to the Calvin cycle and even those awesome adaptations like C4 and cam but what's the big takeaway here like why should we care so much about all of this well for me the most important takeaway is that photosynthesis it's not just this biological process it's the foundation of pretty much all life on Earth I mean it provides the food we eat the oxygen we breathe and it plays this huge role in regulating the climate of our entire planet it's easy to forget about that connection especially when you're focused on the details of chloroplasts and enzymes and all that it is but every bite of food every breath we take it's all thanks to the power of photosynthesis understanding how it works and how plants have adapted to all these different environments it's crucial if we want to tackle these Global challenges like climate change and food security I think next time I see a plan I'm going to look at it with a whole new level of respect I think we all should I mean these plants are quietly powering our planet and the more we understand them the better we can protect them in the ecosystems that we all depend on well that wraps up our Deep dive into photosynthesis we've gone from sunlight to Sugar explored the inner workings of chloroplasts and we even got into the ecological implications of different photosynthetic Pathways we hope you enjoyed the journey and learned something new about the wonders of the natural world thanks for joining us and we'll see you next time |
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Quantum Mechanical Model
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General Chemistry
Get ready to dive into the world of quantum physics and explore the quantum mechanical model of the atom. The Heisenberg Uncertainty Principle states that it is impossible to determine the position and momentum of an object with perfect accuracy. This principle highlights the issue with the Bohr model of the atom, which assumes electrons have definite positions and velocities. Instead, the quantum mechanical model describes electrons as behaving more like a cloud of charge, following the principles of quantum mechanics.
This model introduces four key quantum numbers. The principal quantum number (n) indicates the energy shell occupied by an electron, with larger values representing larger radii and higher energy levels. The azimuthal quantum number (l) determines the shape of the orbitals, while the magnetic quantum number (ml) specifies the particular orbital an electron occupies, such as s, p, d, and f orbitals. Lastly, the spin quantum number (ms) ensures that no two identical electrons can occupy the same place, as described by the Pauli Exclusion Principle.
Lesson Outline
Heisenberg Uncertainty Principle
Impossible to determine position and momentum of an object with perfect accuracy
Incompatibile with Bohr model of the atom
Quantum Mechanical Model: updated alternative to the Bohr model
Electrons behave like a cloud of charge
Electrons follow principles of quantum mechanics
Pauli Exclusion Principle
Electrons cannot share the same four quantum numbers
Two identical electrons cannot be in the same place
Quantum Numbers
Principal Quantum Number (n)
Indicates energy shell occupied by an electron
Relation between n value and energy level
Capacity of energy levels: maximum of 2n^2 electrons (where n is the principal quantum number of the energy level)
Azimuthal Quantum Number (l)
Range of possible l values: 0, 1, 2, 3...
Each value of l corresponds to one subshell, which implies a certain shape/type of orbital (0 = s, 1 = p, 2 = d, etc.)
Magnetic Quantum Number (ml)
Specifies the particular orbital an electron occupies
Range of ml values: integers from -l to +l, where l is the azimuthal quantum number for the orbital
Number of types for each orbital: (s = 1, p = 3, d = 5, f = 7)
Shapes of orbitals become more complex with each subshell
Spin Quantum Number (ms)
Refers to the intrinsic angular momentum of an electron
Can be one of two values: +1/2 and -1/2
Electrons with the same spin quantum number are considered parallel
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FAQs
How does the Quantum Mechanical Model differ from the Bohr Model in explaining atomic structure?
The Quantum Mechanical Model is a more advanced and accurate representation of atomic structure than the Bohr Model. In the Bohr Model, electrons are depicted as orbiting the nucleus in fixed circular paths, while the Quantum Mechanical Model describes the electron's location in terms of its probability distribution or electron cloud, based on its wave-like nature. Furthermore, the Quantum Mechanical Model incorporates various quantum numbers (principal, azimuthal, magnetic, and spin quantum numbers) to specify electron configuration within an atom, whereas the Bohr Model does not.
What is the Heisenberg Uncertainty Principle, and how does it relate to the Quantum Mechanical Model?
The Heisenberg Uncertainty Principle states that it is impossible to precisely determine both the position and momentum of a subatomic particle, such as an electron, simultaneously. This principle is fundamentally connected to the Quantum Mechanical Model, as it acknowledges the wave-particle duality of electrons, which means they behave both as particles and as waves. In turn, this leads to the concept of electron cloud or probability distribution within atoms to determine the probable region where the electron is located rather than a fixed orbit like in the Bohr Model.
What roles do the four quantum numbers play in defining electron configuration within an atom?
The four quantum numbers – principal (n), azimuthal (l), magnetic (ml), and spin (ms) – provide a way to specify the position, energy, and properties of electrons within an atom. The principal quantum number (n) designates the energy level or shell of an electron, determining its distance from the nucleus. The azimuthal quantum number (l) signifies the electron's orbital shape (s, p, d, or f) and its angular momentum. The magnetic quantum number (ml) indicates the orbital that an electron occupies. Lastly, the spin quantum number (ms) describes the intrinsic angular momentum of an electron, representing either "spin up" (+1/2) or "spin down" (-1/2).
What is the Pauli Exclusion Principle, and how does it impact electron configuration in the Quantum Mechanical Model?
The Pauli Exclusion Principle states that no two electrons in an atom can have the same set of quantum numbers (n, l, m, and s). This principle serves as a fundamental rule in determining electron configuration within an atom, as it ensures that each electron must occupy a unique position and state. In effect, this leads to the filling of electron orbitals in a specific order based on their energy levels, shaping the periodic properties of elements and influencing their chemical behavior.
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Why we count from 0 to 9? [duplicate]
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Why have we chosen our number system to be decimal (base 10)? (8 answers)
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In our daily lives, we generally use 10 10 digits to define the numbers, i.e., 0 0 to 9 9. Based on these ten digits, we define all the natural, rational, and irrational numbers, etc. This representation is known as the decimal representation.
I know that one could also define the number system in other ways. For example:
Unary: 1 11 111 1111 ...
Binary: 0 1 10 11 100 ...
Hexadecimal: 0 1 2 3 4 5 6 7 8 9 A B C D E F ...
Similarly, I can define my own number system, for example say: α α β β γ γ 9 9 α α α α β 2 α β 2 α...
But I am wondering, why do we use the decimal system to learn mathematics or say in our everyday lives. I know that a computer uses a binary number system since the most basic unit of storage is 0 0 or 1 1. However, I want to know, why humans use the decimal number system. Is there any logical advantage of using decimal representation over say Unary, Binary, or anything other? Thanks.
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asked Jan 29, 2021 at 16:46
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10 This would probably be a better fit for the history of math stackexchange.Noah Schweber –Noah Schweber 2021-01-29 16:48:23 +00:00 Commented Jan 29, 2021 at 16:48
2 The logical advantage is that it's the system everyone else uses.Misha Lavrov –Misha Lavrov 2021-01-29 16:49:40 +00:00 Commented Jan 29, 2021 at 16:49
6 This is generally attributed to the fact that we have ten fingers. And now that everyone uses base-10, the cost of switching to a more logical base system is just too great.Joe –Joe 2021-01-29 16:51:17 +00:00 Commented Jan 29, 2021 at 16:51
But also it's interesting to note that, the way human hands are designed, it seems like base six might have been a better choice for counting on fingers.littleO –littleO 2021-01-29 16:56:17 +00:00 Commented Jan 29, 2021 at 16:56
1 "Based on these ten digits, we define all the natural, rational, and irrational numbers" Not sure if this is just bad wording, but I would like to strongly note that none of the definitions for any of these sets of numbers relies on the decimal system. (It sounds as if you're thinking that in binary or hex, some numbers would be (ir)rational which are not in base 10 10. That's not the case.)Torsten Schoeneberg –Torsten Schoeneberg 2021-01-29 20:42:56 +00:00 Commented Jan 29, 2021 at 20:42
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We use the base 10 10 because we have 10 10 fingers, and we learn how to count with them.
Note that there were some people ( who used base 8 8 to count, and it was due to the fact that they would count between knucles instead of counting on fingers
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answered Jan 29, 2021 at 16:51
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1 Well I'm not convinced by this fingers (only) explanation, many old numerations systems where constructed on base 12 12 because we can count on our 12 12 phalanxes with the thumb. So the decimal system must have had other advantages to impose itself over the duodecimal one.zwim –zwim 2021-01-29 18:40:22 +00:00 Commented Jan 29, 2021 at 18:40
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5194 | https://www.gauthmath.com/solution/1835140668317745/fx-x2-4x- | Solved: f(x)=|-x^2-4x| [Math]
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f(x)=|-x^2-4x|
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Answer
The answer is $$f(x) = \begin{cases} x^{2} + 4x, & x < -4 \ -x^{2} - 4x, & -4 \leq x \leq 0 \ x^{2} + 4x, & x > 0 \end{cases}$$f(x)=⎩⎨⎧x 2+4 x,−x 2−4 x,x 2+4 x,x<−4−4≤x≤0 x>0
Explanation
Factor out -x from the expression inside the absolute value
$$f(x) = |-x(x+4)|$$f(x)=∣−x(x+4)∣
Apply the absolute value property $$|ab| = |a||b|$$∣ab∣=∣a∣∣b∣
$$f(x) = |-x||x+4|$$f(x)=∣−x∣∣x+4∣
Rewrite $$-x$$−x as $$|x|$$∣x∣
$$f(x) = |x||x+4|$$f(x)=∣x∣∣x+4∣
Consider the cases for the absolute value of x
Since $$|x|$$∣x∣ is always non-negative, we have two cases:
Case 1: $$x \geq 0$$x≥0, then $$|x| = x$$∣x∣=x, so $$f(x) = x(x+4) = x^{2} + 4x$$f(x)=x(x+4)=x 2+4 x
Case 2: $$x < 0$$x<0, then $$|x| = -x$$∣x∣=−x, so $$f(x) = -x(x+4) = -x^{2} - 4x$$f(x)=−x(x+4)=−x 2−4 x
Combine the cases into a piecewise function
$$f(x) = \begin{cases} x^{2} + 4x, & x \geq 0 \ -x^{2} - 4x, & x < 0 \end{cases}$$f(x)={x 2+4 x,−x 2−4 x,x≥0 x<0
Rewrite the function in a simpler form
Notice that $$f(x) = |-x^{2} - 4x| = |-(x^{2} + 4x)| = |x^{2} + 4x|$$f(x)=∣−x 2−4 x∣=∣−(x 2+4 x)∣=∣x 2+4 x∣
So, $$f(x) = |x^{2} + 4x|$$f(x)=∣x 2+4 x∣
Find the zeros of the quadratic expression inside the absolute value
$$x^{2} + 4x = 0$$x 2+4 x=0
$$x(x+4) = 0$$x(x+4)=0
$$x = 0$$x=0 or $$x = -4$$x=−4
Analyze the sign of $$x^{2} + 4x$$x 2+4 x in the intervals determined by the zeros
Interval 1: $$x < -4$$x<−4, e.g., $$x = -5$$x=−5, $$x^{2} + 4x = (-5)^{2} + 4(-5) = 25 - 20 = 5 > 0$$x 2+4 x=(−5)2+4(−5)=25−20=5>0
Interval 2: $$-4 < x < 0$$−4<x<0, e.g., $$x = -2$$x=−2, $$x^{2} + 4x = (-2)^{2} + 4(-2) = 4 - 8 = -4 < 0$$x 2+4 x=(−2)2+4(−2)=4−8=−4<0
Interval 3: $$x > 0$$x>0, e.g., $$x = 1$$x=1, $$x^{2} + 4x = (1)^{2} + 4(1) = 1 + 4 = 5 > 0$$x 2+4 x=(1)2+4(1)=1+4=5>0
Write the piecewise function based on the sign analysis
$$f(x) = \begin{cases} x^{2} + 4x, & x < -4 \ -(x^{2} + 4x), & -4 \leq x \leq 0 \ x^{2} + 4x, & x > 0 \end{cases}$$f(x)=⎩⎨⎧x 2+4 x,−(x 2+4 x),x 2+4 x,x<−4−4≤x≤0 x>0
Simplify the piecewise function
$$f(x) = \begin{cases} x^{2} + 4x, & x < -4 \ -x^{2} - 4x, & -4 \leq x \leq 0 \ x^{2} + 4x, & x > 0 \end{cases}$$f(x)=⎩⎨⎧x 2+4 x,−x 2−4 x,x 2+4 x,x<−4−4≤x≤0 x>0
Verify the function at the boundaries
At $$x = -4$$x=−4, $$f(x) = x^{2} + 4x = (-4)^{2} + 4(-4) = 16 - 16 = 0$$f(x)=x 2+4 x=(−4)2+4(−4)=16−16=0 and $$f(x) = -x^{2} - 4x = -(-4)^{2} - 4(-4) = -16 + 16 = 0$$f(x)=−x 2−4 x=−(−4)2−4(−4)=−16+16=0
At $$x = 0$$x=0, $$f(x) = -x^{2} - 4x = -(0)^{2} - 4(0) = 0$$f(x)=−x 2−4 x=−(0)2−4(0)=0 and $$f(x) = x^{2} + 4x = (0)^{2} + 4(0) = 0$$f(x)=x 2+4 x=(0)2+4(0)=0
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Failure of Fermat's Little Theorem for composite moduli, explained in terms of combinatorics?
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Notice that
3 4−3=81−3=78,3 4−3=81−3=78,
which is divisible by 2 2, but not by 4 4.
In your opinion, what "goes wrong" in this case?
This is a question meant to be answered via combinatorics, and to me, what I was able to note that there may be (i,j)(i,j) repetitions, such that 3∗3−3=6 3∗3−3=6, which 6 6 is not divisible by 4 4.
Can someone explain what goes wrong in terms of the combinatorics?
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edited Jan 24, 2012 at 2:41
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asked Jan 24, 2012 at 2:37
Buddy HollyBuddy Holly
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The combinatorial proof that p∣a p−a p∣a p−a when p p is prime is that a p−a a p−a counts the number of acyclic necklaces of length p p with a a colors. For any n n, the cyclic group Z/n Z Z/n Z acts by cyclic permutation on the set of necklaces of length n n with a a colors. When p p is prime, we can split up the set of necklaces as follows:
The necklaces all of whose colors are the same, of which there are a a, and
The acyclic necklaces, of which there are a p−a a p−a.
The action of Z/p Z Z/p Z on the first collection is trivial and the action on the second is free because Z/p Z Z/p Z has no nontrivial subgroups, so the action on the second collection splits up into a p−a p a p−a p orbits of size p p by the orbit-stabilizer theorem.
When n n is not prime, other orbit types may appear, since necklaces may have nontrivial cyclic structure such as R B Y R B Y R B Y R B Y, so there are orbits with nontrivial stabilizers. In this case the correct count of the number of acyclic necklaces (those on which Z/n Z Z/n Z acts freely) is given, not by a n−a a n−a, but by
∑d∣n μ(d)a n/d∑d∣n μ(d)a n/d
where μ μ is the Möbius function; the proof is a classical application of Möbius inversion and worth working out as an exercise. So the correct generalization to the non-prime case from the combinatorial point of view is that
n∣∑d∣n μ(d)a n/d.n∣∑d∣n μ(d)a n/d.
In the specific case a=3,n=4 a=3,n=4, the group Z/4 Z Z/4 Z has Z/2 Z Z/2 Z as its only nontrivial subgroup. Letting the colors be R,B,Y R,B,Y, the orbits with full stabilizer look like R R R R R R R R, the orbits with stabilizer Z/2 Z Z/2 Z look like R B R B R B R B, and the orbits with trivial stabilizer look like R B Y B R B Y B. There are 3 3 of the first kind and
μ(1)3 4+μ(2)3 2+μ(4)3 1 4=3 4−3 2 4=18 μ(1)3 4+μ(2)3 2+μ(4)3 1 4=3 4−3 2 4=18
of the third kind, so 3 3 of the second kind.
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edited Jan 24, 2012 at 3:33
answered Jan 24, 2012 at 2:43
Qiaochu YuanQiaochu Yuan
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Can you explain it without using the Mobius function? Our class has never studied that yet, and its not in the book. Thanks Buddy Holly –Buddy Holly 2012-01-24 02:46:15 +00:00 Commented Jan 24, 2012 at 2:46
Also, can you edit your answer to incorporate the Orbit Stabilizer Theorem? Thanks Buddy Holly –Buddy Holly 2012-01-24 02:51:01 +00:00 Commented Jan 24, 2012 at 2:51
7 @Buddy: I have never understood requests of this kind. The Mobius function naturally appears in the answer, so you are asking me not to explain part of the answer. You can just ignore that part of the answer and work out the orbit structure for specific choices of a a and n n if you want, but I don't see the point of editing out relevant information.Qiaochu Yuan –Qiaochu Yuan 2012-01-24 03:19:12 +00:00 Commented Jan 24, 2012 at 3:19
Hi Qiaochu, its just that we haven't learned it yet. Can you explain to me how the orbit structure for specific choices of a and n works for this problem? THanks Buddy Holly –Buddy Holly 2012-01-24 05:06:47 +00:00 Commented Jan 24, 2012 at 5:06
@Buddy: it would be a good idea to try to work this out for yourself. Try very small values of a a and n n first (e.g. try a=2 a=2 and n=2,3,4 n=2,3,4).Qiaochu Yuan –Qiaochu Yuan 2012-01-24 15:57:23 +00:00 Commented Jan 24, 2012 at 15:57
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The point of Qiaochu Yuan's answer is that firstly that in order to be surprised the a p−a a p−a is not divisible by p p when p p is composite, you should first have a (combinatorial in your case) reason to find it natural that it does hold when p p is a prime number. One such reason is that a p−a a p−a counts words (which I find more convenient to talk about than necklaces, but it is the same thing) of length p p over an a a-element alphabet that do not consist of p p identical letters. The reason this number is divisible by p p is that cyclic rotation partitions the set of such words into packets (orbits) of size p p. That this is so requires a bit of thought: if there were any non-trivial cyclic rotation (not necessarily by a single place) that transforms a word into itself, then by iterating that one can obtain the cyclic rotation by a single place, and any word invariant by that rotation clearly has all its letters equal. This amounts to the fact that any non-identity element of a cyclic group of p p element generates the entire group.
Now the second point of the Qiaochu Yuan's answer is to analyze the situation when p p is composite (I'll henceforth call the number n n to avoid confusion). Now it is no longer true that any non-identity element of a cyclic group of n n element generates the entire group; notably it can generate a subgroup of any order dividing n n. Consequently there are words that are invariant under some non-trivial cyclic rotation, but not under all of them, and in particular not under the rotation by a single place. The word RBYRBY is an example, invariant under rotation by 3 3 places, but not under rotation by one place. You may very well stop here; it is clear why the argument breaks down, and the numerical examples you gave show that there is no other argument that will make a n−a a n−a divisible by n n for all n n either.
If you insist on finding something that is counted by a number always divisible by n n, then you can still take all words that are not invariant under any non-trivial cyclic rotation, as these come in packets (orbits) of size n n. Call this number f(n)f(n), which will be useful later. Now rather than having f(n)=a n−a f(n)=a n−a, you need to exclude for every strict divisor d d of n n the words that come in packets of size d d. These are harder to count than words that come in singleton packets (as we saw there were a a of them), but can be done by the following reasoning. Such words necessarily consist of a sub-word s s of d d letters that is repeated d d times: w=s.s…s w=s.s…s. However s s cannot just be any word, it must not have any internal cyclic symmetry, since if it did it would belong to a shorter orbit.
But words without any cyclic symmetry is just what we were out to count, except that it was for length n n and now we are confronted with length d<n d<n. Hence the utility of a general function f f: it is defined implicitly (by induction) by the relation
f(m)=a m−∑d strictly divides m f(d)f(m)=a m−∑d strictly divides m f(d)
for all m≥1 m≥1. Note that in particular f(1)=a f(1)=a since 1 1 has no strict divisors. This equation can be formulated more simply by moving the summation to the other side, which also avoid having to mention "strict":
∑d∣m f(d)=a m∑d∣m f(d)=a m
for all m≥1 m≥1. The general technique the will solve this is called Möbius inversion. Since you haven't heard about that, here's its secret. Think of a vectors F(n)=(f(1),f(2),…,f(n))F(n)=(f(1),f(2),…,f(n)), the unknown we are after, and G(n)=(a,a 2,…,a n)G(n)=(a,a 2,…,a n) which tabulates the right-hand-sides of our last equation. This equation then says F(n)=D(n)⋅F(n)F(n)=D(n)⋅F(n), where D(n)D(n) is the lower triangular matrix with entries D(n)i,j=1 D(n)i,j=1 whenever j j divides i i, and D(n)i,j=0 D(n)i,j=0 otherwise. Actually the equation gives us the result of multiplying row m m of of the matrix D(n)D(n) and the vector G(n)G(n), so the full matrix equation takes care of "for all m m" (upto n n). Now all we need is the inverse matrix D(n)D(n), and it can (rather easily) be shown to be given by
D(n)−1 i,j={(−1)c 0 if i/j is integer and a product of c distinct primes otherwise D(n)i,j−1={(−1)c if i/j is integer and a product of c distinct primes 0 otherwise
which is usually written as
D(n)−1 i,j={μ(i/j)0 if j∣i otherwise D(n)i,j−1={μ(i/j)if j∣i 0 otherwise
where μ μ is the so-called (arithmetic) Möbius function. Now one can write F(n)=D(n)−1⋅G(n)F(n)=D(n)−1⋅G(n), giving
f(m)=∑d∣m μ(m/d)a d f(m)=∑d∣m μ(m/d)a d
which is the formula Qiaochu gave (with m=n m=n) for the number of entirely non-symmetric necklacesof length n n. (Actually he interchanged n/d n/d and d d, which can be seen to make no difference; personally I prefer to always put n/d n/d in the argument of μ μ.)
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edited Apr 15, 2013 at 6:36
answered Jan 24, 2012 at 10:59
Marc van LeeuwenMarc van Leeuwen
120k 8 8 gold badges 183 183 silver badges 373 373 bronze badges
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Thank you for this very thorough explanation, Marc! I confess that I assumed that the OP was already familiar with the combinatorial argument and that I was just resummarizing it to make sure we were on the same page.Qiaochu Yuan –Qiaochu Yuan 2012-01-24 15:56:18 +00:00 Commented Jan 24, 2012 at 15:56
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5196 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4762976/ | Oral Pemphigus Vulgaris: Case Report - PMC
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Oral Pemphigus Vulgaris: Case Report
Rai Arpita
Rai Arpita
1 Department of Oral Medicine and Radiology, Faculty of Dentistry, Jamia Millia Islamia, New Delhi, India
Find articles by Rai Arpita
1, Arora Monica
Arora Monica
2 Santosh Medical College, Ghaziabad, India
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2, Naikmasur Venkatesh
Naikmasur Venkatesh
3 SDM College of Dental Sciences, Dharwad, India
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3, Sattur Atul
Sattur Atul
3 SDM College of Dental Sciences, Dharwad, India
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3, Malhotra Varun
Malhotra Varun
2 Santosh Medical College, Ghaziabad, India
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2
Author information
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1 Department of Oral Medicine and Radiology, Faculty of Dentistry, Jamia Millia Islamia, New Delhi, India
2 Santosh Medical College, Ghaziabad, India
3 SDM College of Dental Sciences, Dharwad, India
✉
Corresponding Author: Rai Arpita, arpitadoc@gmail.com
Copyright © Jimma University, Research & Publications Office 2015
PMC Copyright notice
PMCID: PMC4762976 PMID: 26949302
Abstract
Background
Pemphigus is a potentially life threatening autoimmune disease that causes blisters and erosions of the skin and the mucous membrane. The epithelial lesions are a result of auto-antibodies that react with desmosomal glycoproteins that are present on the cell surface of the keratinocyte. The autoimmune reaction against these glycoproteins causes a loss of cell to cell adhesion, resulting in the formation of intraepithelial bullae. Eighty to ninety percent of patients with pemphigus vulgaris develop oral lesions and in 60% of cases oral lesions are the first sign. Timely recognition and therapy of oral lesion is critical as it may prevent skin involvement. If treatment is instituted during this time, the disease is easier to control and the chance for an early remission of the disorder is enhanced.
Case Details
This case report describes the case of a patient who complained of ulcers of the mouth and difficulty in swallowing since 20 days, who was diagnosed as having Pemphigus vulgaris. Due to early diagnosis, lower doses of medication for a shorter period of time could control the disease.
Conclusion
Dental professionals must be sufficiently familiar with the clinical manifestations of pemphigus vulgaris to ensure early diagnosis and treatment which in turn determines the prognosis and course of the disease.
Keywords: Pemphigus, oral lesions, mucous membrane, chronic oral ulcers, pemphigus vulgaris
Introduction
Pemphigus is a chronic inflammatory autoimmune bullous disease. There are 0.5 to 3.2 cases reported each year per 100,000 population, with the highest incidence in the 5 th and 6 th decade of life, with male to female ratio of 1:2. (1). Some rare cases have been reported in children and the elderly (2). The major variants of pemphigus are pemphigus vulgaris, pemphigus vegetans, pemphigus foliaceus, pemphigus erythematosus, paraneoplastic pemphigus (PNP) and drug related pemphigus. Pemphigus Vulgaris is the most common form of pemphigus, accounting for over 80% of cases (3). In the majority of patients, it affects the oral mucosa and is sometimes difficult to diagnose when only mucosal involvement is present.
Lesions may occur anywhere on the oral mucosa, but the buccal mucosa is the most commonly affected site followed by involvement of the palatal, lingual and labial mucosa. Gingiva is the least commonly affected site and desquamative gingivitis is the commonest manifestation of the disease when gingival is involved (3). In many patients, oral lesions are followed by the development of skin lesions (4,5). If oral pemphigus vulgaris can be recognized in its early stages, treatment may be initiated to prevent the progression of the disease to skin involvement. Diagnostic delays of greater than 6 months are common in patients with oral pemphigus vulgaris (6). The oral cavity may be the only site of involvement for a year or so, and this may lead to delayed diagnosis and inappropriate treatment of a potentially fatal disorder. This case report describes the case of a patient complaining of ulcers in the mouth and difficulty in swallowing since 20 days, who was diagnosed as having Pemphigus vulgaris.
Case Report
A 40-year-old male patient resident of Karwar, Karnataka, reported with the chief complaint of ulcers in the mouth and difficulty in swallowing solid food and liquids since 20 days. History revealed that the patient first noticed dysphagia for solid food which progressively increased in severity. At the time of presentation, the patient had dysphagia for liquid as well. The patient had noticed ulcers of the mouth which bled on brushing, and increased salivation in the morning was reported. The patient did not report of skin lesions or involvement of other mucosal sites. A review of medical and family history was noncontributory. The patient had poor oral hygiene with adverse habit of taking betel quid with tobacco 10 times daily and smoking 35 bidis per day since 10 years. He was also habit of alcohol consumption of 2 quarters per day from last 10 years.
On general examination, the patient was moderately built and signs of anemia were present. Submandibular lymph nodes were enlarged, palpable and tender bilaterally. Intra-oral examination revealed ulcerative lesions present on bilateral buccal mucosa along the line of occlusion extending from retrocommisural areas to the retromolar trigone posteriorly (Figure 1). Lesions extended superiorly from the line of occlusion and were irregular in shape covered by pseudomembrane with erythematous surrounding. On manipulation, bleeding was present. Similar lesions with irregular borders associated with flaacid bullae were present in the lower buccal vestibule in relation to molar region. Lesions were also present posteriorly on the lateral border of the tongue on left side. Erosive lesions were seen involving posterior hard plate, soft palate, faucial pillars and extending to the oropharynx (Figure 2). There were diffuse areas of erosions covered by pseudo membrane at some sites. Nikolysky's sign showed a positive reaction. Generalised teeth attrition and gingival inflammation with bleeding on probing were present.
Figure 1.
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Intra-oral photograph of the patient showing ulcerative lesions present on bilateral buccal mucosa along the line of occlusion; the lesions with irregular borders associated with flaacid bullae in lower buccal vestibule in relation to molar region seen
Figure 2.
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Intra-oral photograph of the patient showing erosive lesions involving posterior hard plate, soft palate, faucial pillars and extending to the oropharynx
The clinical presentation of chronic multiple oral ulcers, flaccid bullae and positive Nikolysky sign in this case led to provisional diagnosis of vesiculo-bullous lesion affecting the oral cavity. Differential diagnosis included pemphigus vulgaris, mucous membrane pemphigoid, bullous lichen planus, para neoplastic pemphigus, chronic ulcerative stomatitis, recurrent herpes lesions in immunocompromised patients and erythema multiforme.
Routine haematological and biochemical investigations were within normal limits except haemoglobin of 9.5mg%. Incisional biopsy was performed from peri lesional site of the right buccal mucosa. Histopathological examination revealed parakeratinized stratified squamous epithelium with intra-epithelial blister formation. Split area was covered by thick area of fibrinous exudate consisting of inflammatory cells. Occasional giant cells (Tzank cells) were seen in the split area. Chronic inflammatory cell infiltrate in the sub epithelial and perivascular region was evident (Figure 3). Based on the histopathological findings, a final diagnosis of pemphigus vulgaris was made.
Figure 3.
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Photomicrograph (X40, Hematoxylin and eosin stained) showing parakeratinized stratified squamous epithelium with intra-epithelial blister formation; occasional giant cells seen in the split area; chronic inflammatory cell infiltrate in the sub epithelial and perivascular region evident.
The treatment plan comprised oral prednisolone 60 mg/day for 4 days along with multi-vitamin and analgesic. Topical analgesic mouthwash and 0.1% Triamcinolone acetonide ointment were also prescribed to the patient. On the first follow-up, the patient had 50% reduction in symptoms with partial healing of lesions, erythema and inflammation in relation to ulcers had reduced (Figure 4). The dose of Wysolone was tapered to 40 mg/day for the next four days. On second follow-up, 20% reduction in symptom compared to the first follow-up was noticed. The number of lesions reduced. Biopsy site healed. Suture removal was done and prednisolone dose was tapered to 20mg/day for 4 days. On the third follow-up, 90% reduction in symptom was noticed; lesions healed almost completely (Figure 5).
Figure 4.
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Intra-oral photograph of the patient showing reduction in the size of the lesions and reduction in the erythema and inflammation associated with it
Figure 5.
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Intra-oral photograph of the patient showing almost complete healing of ulcerative lesions.
Pemphigus vulgaris is a rare cause of chronic ulceration of the oral mucosa. The mouth may be the only site of involvement for a year or so. In this case, early diagnosis and lower doses of medication for shorter period of time was could control the disease.
Discussion
Pemphigus is defined as a group of life threatening blistering disorder of skin and mucous membrane characterized by acantholysis (loss of keratinocyte to keratonicyte adhesion). The process of acantholysis is induced by circulatory autoantibodies to intercellular adhesion molecules (6). In most cases (70–90%), the first sign of the disease appears on the oral mucosa. While the lesions can be located anywhere within the oral cavity, they are most commonly found in areas subjected to frictional trauma such as the cheek mucosa, pharynx, larynx, esophagus, genital mucosa as well as the skin where intact blisters are commonly seen (7).
The major variants of pemphigus are pemphigus vulgaris, pemphigus vegetans, pemphigus foliaceus, pemphigus erythematous, paraneoplastic pemphigus and drug related pemphigus. Each form of this disease has antibodies directed against different target cell surface antigen, resulting in a lesion forming in different layers of the epithelium (8). Pemphigus vulgaris is the most common form of pemphigus, accounting for over 80% of cases (5).
The underlying mechanism responsible for causing the intraepithelial lesions of pemphigus vulgaris is the binding of Ig G autoantibodies to desmoglein 3, a transmembrane glycoprotein adhesion molecule present on desmosome. The binding of Pemphigus Vulgaris antibody activates protease, whereas more recent evidence support the theory that the Pemphigus vulgaris antibodies directly block the adhesion function of the desmoglein (9, 10, and 11 The separation of cells called acantholysis takes place in the lower layers of stratum spinosum, which results in the formation of suprabasilar bulla. The bulla increasingly involves a larger area of epithelim, resulting in loss of large area of skin and mucosa.
The classical lesion of pemphigus is a thin walled bulla arising on otherwise normal skin or mucosa. A characteristic sign of the disease may be obtained by the application of pressure to intact bullae. In a patient with Pemphigus Vulgaris, bullae enlarges by extension to an apparently normal surface. Another characteristic sign of the disease is the pressure to apparently normal area resulting in the formation of a new lesion. This phenomenon, called Nikolysky sign, results from the upper layer of the skin pulling away from the basal layer. Nikolysky sign is also found to be positive in toxic epidermal necrolysis, scalded skin syndrome (both of which are acute conditions) and mucous membrane pemphigoid (12).
If proper history is taken, the clinician should be able to distinguish the lesions of pemphigus from those caused by acute viral infections like herpes and erythema multiforme. Immunocomprised patients present with recurrent herpetic simplex infections in the form of atypical ulcers, which may last several weeks or months if undiagnosed and untreated. Moreover, the presence of Tzank cells may complicate the diagnosis. Since in the present case, the patient did not give history of immunocompromise like chemoptherapy, organ transplant or acquired immune deficiency syndrome, recurrent herpes simplex infection could be safely ruled out.
Differential diagnosis of Pemphigus Vulgaris can be done from other similar conditions by biopsy and direct immunoflorescence. Biopsies are best done on intact vesicles and bullae less than 24 hours old. The biopsy specimen should be taken from the advancing edge of the lesion, where the area of characteristic suprabasilar acantholysis can be observed by the pathologist. Supra basilar split seen in Pemphigus Vulgaris helps distinguish this condition from sub-epithelial blistering diseases such as mucous membrane pemphigoid, bullous lichen planus and chronic ulcerative stomatitis. Indirect immunoflorescence is helpful in further distinguishing pemphigus from pemphigoid and other chronic oral lesions and is useful in following the progress of patient for pemphigus. The diagnosis is confirmed by the characteristic deposition of IgG and other C3 antibodies that bind to cell surface of perilesional skin or mucosa (13,14). Indirect immunoflorescence is less sensitive than direct immunofluorescence, but may be helpful if biopsy is difficult. ELISA has been developed that can detect desmoglein 1 and 3 in serum sample of patients with Pemphigus Vulgaris (14). The presence of anemia and submandibular gland lymphadenopathy along with painful stomatitis and acantholysis on histopathological investigations in the present case led to the differential diagnosis of paraneoplastic pemphigus. Absence of underlying lymphoproliferative disorder, the reduced severity of the lesions and absence of inflammation at the dermal-epidermal junction and keratinocyte necrosis in addition to characteristic acantholysis helped rule out this condition. Also, direct immunoflorescence of para neoplastic pemphigus shows deposition of IgG and complement along the basement membrane as well as on the keratinocyte surface in an intercellular location (15).
An important aspect of patient management is early diagnosis when lower doses of medication can be used for shorter periods of time to control the disease. Dental professionals must be sufficiently familiar with clinical manifestation of pemphigus vulgaris to ensure early diagnosis and treatment, since this in turn determines the prognosis and course of the disease. Institution of early treatment could prevent serious involvement of other mucosa and cutaneous sites and fatal complications. Pemphigus Vulgaris is generally managed with local and systemic corticosteroid therapy. Treatment is administered in 2 phases: a loading phase, to control the disease, and a maintenance phase, which is further divided into consolidation and treatment tapering. Local treatment consists of a paste, an ointment or a mouthwash administered alone or in conjunction with systemic treatment. Intralesional injections of corticosteroids have been used for the management of persistent lesions (16). In cases of extensive oral lesions or involvement of other mucosa and skin, systemic corticosteroid therapy is initiated immediately. The initial of prednisone 0.5–2 mg/kg is recommended (17). Depending on the response, the dose is gradually decreased to the minimum therapeutic dose, taken once a day in the morning to minimize side effects. When steroids are used for longer periods of time, adjuvants such as Azathioprine or Cyclophosphamide are added to the regimen to reduce the complications of long term corticosteroid therapy. Before the advent of corticosteroid therapy, pemphigus was fatal, with a mortality rate of up to 75% in the first year. It is still a serious disorder, but the 5% to 10% mortality rate is now primarily due to the side effects of therapy (18).
References
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The Graph Isomorphism Problem
Exploring the theoretical and practical aspects of the graph isomorphism problem.
By Martin Grohe and Pascal Schweitzer
Posted Nov 1 2020
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Introduction
Key Insights
Combinatorial Algorithms
Group theoretic algorithms
Babai's Quasipolynomial-Time Algorithm
Practical Graph Isomorphism
Applications
Concluding Remarks
References
Authors
Deciding whether two graphs are structurally identical, or isomorphic, is a classical algorithmic problem that has been studied since the early days of computing. Applications span a broad field of areas ranging from chemistry (Figure 1) to computer vision. Closely related is the problem of detecting symmetries of graphs and of general combinatorial structures. Again this has many application domains, for example, combinatorial optimization, the generation of combinatorial structures, and the computation of normal forms. On the more theoretical side, the problem is of central interest in areas such as logic, algorithmic group theory, and quantum computing.
Figure 1. Nonisomorphic molecular graphs.
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Key Insights
With its many practical and theoretical applications, the graph isomorphism problem remains one of the most important unresolved problems in theoretical computer science.
A recent breakthrough by Laszlo Babai shows that the problem is almost efficiently solvable…theoretically.
We are only starting to explore the of wealth of new ideas the recent advances bring to the field.
Graph isomorphism (GI) gained prominence in the theory community in the 1970s, when it emerged as one of the few natural problems in the complexity class NP that could neither be classified as being hard (NP-complete) nor shown to be solvable with an efficient algorithm (that is, a polynomial-time algorithm). It was mentioned numerous times as an open problem, in particular already in Karp’s seminal 1972 paper23 on NP-completeness as well as in Garey and Johnson’s influential book on computers and intractability.15 Since then, determining the precise computational complexity of GI has been regarded a major open problem in theoretical computer science.
In a recent breakthrough,3 Babai proved that GI is solvable in quasipolynomial time. This means that on n-vertex input graphs, Babai’s algorithm runs in time n p(log n) for some polynomial p(X). This can be interpreted as the problem being almost efficiently solvable—theoretically.
In this paper, we will survey both theoretical and practical aspects of the graph isomorphism problem, paying particular attention to the developments that led to Babai’s result.
Historical development. Graph isomorphism as a computational problem first appears in the chemical documentation literature of the 1950s (for example, Ray and Kirsch35) as the problem of matching a molecular graph (see Figure 1) against a database of such graphs. The earliest computer science reference we are aware of is due to Unger,39 incidentally also in the Communications of the ACM. Maybe the first important step on the theoretical side was Hopcroft and Tarjan’s O(n log n) isomorphism algorithm for planar graphs.22 As the question whether GI is NP-complete gained prominence, it was realized that GI has aspects that distinguish it from most NP-complete problems. In particular, counting the number of isomorphisms between two graphs is not harder than deciding if there is an isomorphism (see Mathon29). Babai et al.8 showed that GI is easy on average with respect to a uniform distribution of input graphs. In fact, this can be extended to most other random graph distributions.
A first wave of substantial progress came in 1979-1980 with Babai2 and Luks’s26 introduction of group theoretic techniques. In his paper, Luks26 showed that isomorphism can be decided in polynomial time on graph classes of bounded degree (that is, the number of edges incident with each vertex is bounded), and Luks laid the foundation for much of the subsequent work on graph isomorphism algorithms by introducing a general divide-and-conquer framework. (We will discuss this framework in some detail.) A combination of Luks’s group theoretic framework with a clever combinatorial trick by Zemlyachenko led to a moderately exponential algorithm for graph isomorphism (see Babai and Luks10). The best bound was , established by Luks in 1983 (see Babai et al.9). This remained the best known bound until Babai’s recent breakthrough.
Around the same time, McKay developed his isomorphism tool Nauty,30 which marks a breakthrough in practical isomorphism testing.
In the mid-1980s, another fascinating facet of the complexity of GI was discovered. Using the newly developed machinery of interactive proof systems, it was shown that the complement of GI has short zero-knowledge proofs16 and this was seen as another indication that GI is not NP-complete. (If GI is NP-complete, then the so-called polynomial hierarchy of complexity classes above NP collapses to its second level, which is regarded as unlikely.) On the other hand, Torán38 proved that GI is hard to solve when the available memory is quite limited (specifically it is hard for nondeterministic logarithmic space under logspace reductions), which gives us at least some complexity theoretic lower bound.
As the group theoretic methods have been introduced in the early 1980s, they have been continually refined. The underlying group theory has progressed (for example, Babai et al.5,9), the complexity of the group theoretic problems has been analyzed in detail (for example, Luks27 and Seress37), and the scope of the methods has been expanded to other structures (for example, Babai et al.7). Another active strand of research has been the design of efficient algorithms for GI restricted to graphs with specific properties (for example, Babai et al.6, Grohe and Marx,19 Lokshtanov et al.,25 Ponomarenko34). More recently, there has also been work on memory restricted algorithms (for example, Datta et al.14).
But no real progress on the general isomorphism problem was made until—out of the blue—Babai published his quasipolynomial-time algorithm in 2015.
After this historical overview, let us get slightly more concrete.
Isomorphisms, automorphisms, and canonical forms. An isomorphism from a graph G = (V, E) to a graph H = (W, F) is a one-to-one mapping π from the vertices of the first graph V onto the vertices of the second graph W that preserves adjacency and nonadjacency, that is, uv ∈ E if and only if π(u) π(v) ∈ F for all pairs uv of vertices in V (Figure 2).
Figure 2. Four isomorphic graphs. The red arrows indicate an isomorphism between the first and the third graph.
An automorphism, or a symmetry, of a graph G is an isomorphism from G to G itself. For example, all n! permutations of the vertex set of a complete graph K n on n vertices are automorphisms. By comparison, an (undirected) path of length n only has two automorphisms, the trivial identity mapping, and the mapping that flips the ends of the path. The collection of all automorphisms of G forms a mathematical structure known as a (permutation) group. As the example of the complete graph shows, automorphism groups can get very large, exponentially large in the number of vertices, but fortunately every permutation group has a generating set linear in the size of the permutation domain (that is, the set of objects being permuted). This allows us to work with automorphism groups efficiently as long as they are represented by sufficiently small generating sets. The problem GI of deciding whether two graphs are isomorphic and the problem of computing a generating set for the automorphism group of a graph (AUT) have the same computational complexity, or more precisely, can be reduced to each other by polynomial-time reductions (see Mathon29).
Another important related problem is the graph canonization problem. A canonical form γ maps each graph G to an isomorphic graph γ(G) in such a way that if graphs G and H are isomorphic then the graphs γ(G) and (H) are identical (not just isomorphic). Observe that a canonical form γ yields an isomorphism test: given G, H, compute γ(G) and γ(H) and check if they are identical. In practical applications, canonical forms are often preferable over isomorphism tests. It is an open problem whether these two problems are actually equivalent (for example, whether the existence of a polynomial-time isomorphism algorithm would yield the existence of a polynomial-time computable canonical form). However, typically for graph classes for which we know a polynomial-time isomorphism algorithm, we also have a polynomial-time canonization algorithm. Sometimes, the extension from isomorphism testing to canonization is straightforward; sometimes it requires extra work (for example, Babai,4 Babai and Luks,10 Schweitzer and Wiebking36).
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Combinatorial Algorithms
To establish that two graphs are isomorphic, we can try to find an isomorphism. To establish that the graphs are nonisomorphic, we can try to find a “certificate” of nonisomorphism. For example, we can count vertices, edges, and triangles in both graphs; if any of these counts differ, the graphs are nonisomorphic. Or we can look at the degrees of the vertices. If there is some d such that the two graphs have a different number of vertices of degree d, the graphs are nonisomorphic.
The Weisfeiler-Leman algorithm provides a systematic approach to generate such certificates of nonisomorphism in an efficient way. Actually, it is a whole family of algorithms, parameterized by a positive integer, the dimension.
Color refinement. We start by describing the 1-dimensional version, which is commonly known as color refinement or naive vertex classification. It is one of the most basic ideas in graph isomorphism testing that has been reinvented several times; the oldest published version that we are aware of can be found in Morgan.32 Color refinement is an important subroutine of almost all practical graph isomorphism tools, and it is also a building block for many theoretical results.
The color refinement algorithm, displayed in Figure 3, iteratively computes a coloring of the vertices of a graph. The actual colors used are irrelevant, what matters is the partition of the vertices into color classes. The final coloring has the property that any two vertices of the same color have the same number of neighbors in each color class. Figure 4 shows an example.
Figure 3. The color refinement algorithm.
Figure 4. Color refinement: a graph, its coloring after 1 refinement round, and the final coloring.
The coloring computed by the algorithm is isomorphism invariant, which means that if we run it on two isomorphic graphs, the resulting colored graphs will still be isomorphic and in particular have the same numbers of nodes of each color. Thus, if we run the algorithm on two graphs and find that they have distinct numbers of vertices of some color, we have produced a certificate of nonisomorphism. If this is the case, we say that color refinement distinguishes the two graphs.
Unfortunately, color refinement does not distinguish all nonisomorphic graphs. Figure 5 shows a simple example. But, remarkably, color refinement does distinguish almost all graphs, in a precise probabilistic sense.8 This, together with its efficiency, is what makes color refinement so useful as a subroutine of practical isomorphism tools.
Figure 5. Two nonisomorphic graphs that are not distinguished by color refinement. Color refinement computes the black-white coloring of the vertices.
The reader may have noticed that color refinement is very similar to other partitioning algorithms, in particular the standard algorithm for minimizing deterministic finite automata. Borrowing ideas from Hopcroft’s DFA minimization algorithm,21 color refinement can be implemented to run in time O((n+m)log n), where n is the number of vertices and m is the number of edges of the input graph.13 Thus, color refinement is indeed very efficient.
Weisfeiler-Leman. We have seen that color refinement is not a complete isomorphism test: it fails to distinguish extremely simple nonisomorphic graphs such as those shown in Figure 5. The k-dimensional Weisfeiler-Leman algorithm (k-WL) is based on the same iterative-refinement idea as color refinement, but is significantly more powerful. Instead of vertices, k-WL colors k-tuples of vertices of a graph. Initially, each k-tuple is “colored” by the isomorphism type of the subgraph it induces. Then in the refinement rounds, the color information is propagated between “adjacent” tuples that only differ in one coordinate (details can be found in Cai et al.11). The 2-dimensional version of the algorithm is due to Weisfeiler and Leman;40 the generalization to higher dimensions is due to Faradẑev, Zemlyachenko, Babai, and Mathon (see Cai et al.11). If implemented using similar ideas as for color refinement, k-WL runs in time O(n k+1 log n).
Higher-dimensional WL is very powerful. Indeed, it is highly nontrivial to find nonisomorphic graphs that are not distinguished by 3-WL. It took a now seminal paper, by Cai et al.11, to prove that for every k, there are nonisomorphic graphs G k, H k that are not distinguished by k-WL. Indeed, these graphs, known as the CFI graphs, have size O(k) and are 3-regular.
It turns out that many natural graph classes do not admit the CFI-graph construction and a low-dimensional WL is a complete isomorphism test. In particular, for all graph classes C that exclude some fixed graph as a minor, there is a constant k such that k-WL distinguishes all nonisomorphic graphs in C.17 This includes the class of planar graphs, for which 3-WL suffices.24
The Weisfeiler-Leman algorithm is remarkably robust. It not only subsumes most combinatorial ideas for graph isomorphism testing but also has a natural characterization in terms of logic.11 Surprisingly, it also corresponds to a natural isomorphism test based on linear programming1 and subsumes various approaches to GI based on algebraic and mathematical programming techniques.
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Group theoretic algorithms
Although most isomorphism algorithms devised over the years are subsumed by the Weisfeiler-Leman algorithm, this is not the case for the group theoretic approach.2, 11 The first application of algorithmic group theory to isomorphism testing was given by Babai.2 Subsequently, Luks26 used a group theoretic approach to devise a polynomial-time isomorphism test for graphs of bounded degree.
As GI and the automorphism group problem AUT are polynomially equivalent (see Mathon29), it suffices to solve the latter. Starting with a suitable group of permutations, we want to compute within it the automorphism group A of interest (technically we want to compute a certain set-stabilizer or the solution to a string isomorphism problem, on which we will not elaborate here). We continually maintain an encasing group Γ ≥ A containing all automorphisms as a subgroup. Our strategy is to iteratively shrink Γ until it agrees with A.
To shrink the group Γ, in case the permutation group Γ has more than one orbit (see Figure 6), we process orbits sequentially.
Figure 6. Basic permutation group concepts.
If the group has only one orbit, we exploit so-called blocks whenever they exist. A block of a permutation group Γ ≤ Sym(Ω) is a subset always mapped to itself or somewhere else entirely, that is, a set B ⊆ Ω of the permutation domain Ω such that for all γ ∈ Γ we have γ(B) = B or γ(B) ∩ B = {}. The set of images of the block {γ(B)|γ ∈ Γ} partitions the domain Ω into blocks of equal size, which together form a so-called block system. The group Γ permutes the blocks of the system and we can consider the induced permutation group Γ’ on the blocks. By choosing B ⊊ Ω inclusion-wise maximal among the blocks, we can ensure that Γ’ does not have any (nontrivial) blocks itself. A group with this property is called primitive. Luks argues that in polynomial time, we can reduce the computation of the automorphism group A to |Γ’| computations, each involving subproblems with significantly smaller orbits, which can then be processed sequentially as mentioned above. In case we started with a primitive group, we use a brute force algorithm, inspecting all permutations in Γ separately.
A crucial observation is now that for graphs of bounded degree, there is a method to guarantee that |Γ’| cannot be too large. Originally Luks presented a more involved argument but a subsequent result by Babai et al.2 directly shows that |Γ’| is polynomially bounded in the permutation domain size. Overall, this bound implies that the entire procedure runs in polynomial time on graphs of bounded degree.
For general graphs, the bottleneck of this procedure occurs when Γ’ is large. In that case, Γ’ is a large primitive group. Such groups are called Cameron groups and a precise classification is known.12, 28 However, this is not a new insight and the fact that Cameron groups form the bottleneck to improving Luks’s method was already known in the 1980s.
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Babai’s Quasipolynomial-Time Algorithm
Attacking exactly this bottleneck, 35 years later, it was Babai who improved the running time of the theoretically fastest general graph isomorphism algorithm. He showed that graph isomorphism can be solved in quasipolynomial time n polylog(n), that is, n(log(n)c) for some constant c. Doing his algorithmic ideas justice is difficult not only because they span 80 pages in his original manuscript but also because the algorithm contains several major, very distinct new ideas that combine smoothly to an overall algorithm. Here, we can thus only sketch the underlying ideas of the various puzzle pieces and how they are combined.
The first step, at quasipolynomial cost, is to reduce the bottleneck of Cameron groups further to what Babai calls Johnson groups. They are groups abstractly isomorphic to a symmetric or an alternating group, but do not necessarily act in their natural action of permuting elements in some ground set, and rather consist of permutations of the t-element subsets of the ground set. An example is the automorphism group of the Petersen graph (see Figure 7).
Figure 7. The Petersen graph is a small graph whose automorphism is a Johnson group. Its nodes correspond to the 2-element subsets of {1,…,5}, with an edge between two nodes if the corresponding subsets have an empty intersection. The automorphism group of the Petersen graph is the symmetric group S5 with its natural action on the 2-element subsets.
As next step, to make Luks’s framework more flexible for his recursive algorithm, Babai does not only maintain an encasing group Γ containing the automorphism group, but also a homomorphism ϕ:Γ → Γ’ into a permutation group Γ’ ≤ Sym(Ω’) over an ideal domain Ω’. This allows the algorithm to make progress by decreasing the size of the ideal domain.
Initially, for Johnson groups, we can choose as ideal domain the abstract ground set mentioned here. This way, the image of ϕ contains almost all permutations, that is, it is the symmetric group or the alternating group on the ideal domain. These two groups are by far the largest primitive groups and therefore called the giants. Accordingly, we speak of a giant homomorphism.
The general strategy of the algorithm is to reduce a problem instance to quasipolynomially many instances that are all smaller than the original instance by at least a constant factor. This is continued until the recursive instances are sufficiently small to be resolved with brute force, leading to an overall quasipolynomial-time algorithm.
If the permutation group induced by the homomorphism ϕ on the ideal domain is intransitive or imprimitive, we can use the strategies of Luks to process orbits sequentially or to consider the actions on blocks, respectively. For this, some nontrivial group theory is required to pull back information from the ideal domain to the original domain.
In case the subgroup A of Γ that we are interested in maps onto the alternating group of the ideal domain, the computation of A is comparatively easy, so let us focus on the case that the image of the subgroup is not a giant.
We then use a local to global approach. We first collect local certificates by testing, for all logarithmic-size subsets T of the ideal domain, whether the homomorphism ϕ applied to the sought-after automorphism group A and restricted to T is a giant homomorphism. We call these sets test sets. A test set is full if the said restriction is a giant homomorphism. As the test set size is only logarithmic and there are only quasipolynomially many such test sets, we can test all test sets for fullness using recursion.
If a test set is full, which certifies high local symmetry, there must be global symmetries certifying this and quite surprisingly such global symmetries can be efficiently constructed. At the core of this statement lies the Unaffected Stabiliser Lemma, a central insight proven by Babai.
If there are a lot of full test sets, the global symmetries allow for efficient recursion. On the other hand, if only few test sets are full, the graph must have a nontrivial structural invariant. Furthermore, we can use the logarithmic-dimensional Weisfeiler-Leman algorithm to construct such a structural invariant in the form of a relational structure of logarithmic arity. This breaks the apparent symmetry.
With the design lemma, we can reduce the relational structure of logarithmic arity to a structure with a binary relation. We obtain a uniprimitive coherent configuration, a particular structure important in algebraic graph theory closely related to the 2-dimensional Weisfeiler-Leman algorithm.
The final puzzle piece is the Split-Or-Johnson combinatorial partitioning algorithm which, from a uniprimitive coherent configuration either produces a split or finds a large canonically embedded Johnson graph, a graph whose automorphism group is a Johnson group. In fact splits, which are invariant partitions of the ideal domain akin to the blocks of a permutation group, can also occur during other parts of the algorithm. They are handled with the techniques similar to the imprimitive case of Luks’s algorithm.
We are left with the case in which a large canonically embedded Johnson graph has been produced. After all, this case had to occur at some point because we know that the resilient Johnson groups exist. But now the Johnson graph is in fact a blessing because we can exploit the well-understood structure of the Johnson graphs to dramatically decrease the size of the ideal domain.
Overall we obtain a quasipolynomial-time algorithm solving the general graph isomorphism problem. Besides the original manuscript, there is also a detailed explanation of the algorithm in the Bourbaki series by Helfgott (see Helfgott et al.20 for an English translation). In fact, Helfgott detected an error in the Split-or-Johnson routine which however was quickly fixed by Babai.
Babai’s algorithm depends on the classification of the finite simple groups, an enormous theorem spanning several hundred journal articles written by numerous authors. Many group theorists prefer to avoid the theorem and indeed Pyber modified Babai’s algorithm to give an alternative analysis that does not depend on the classification.
In further advances, Babai recently extended his result to a canonization algorithm that runs in quasipolynomial time,4 and there is an improvement on Luks’s original result for graphs of maximum degree d testing isomorphism in time n log(d)c).18
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Practical Graph Isomorphism
In practice it is excessive to even run the 2-dimensional Weisfeiler-Leman algorithm, let alone some version of increasing dimension as in Babai’s algorithm. Current isomorphism packages rather use color refinement, that is, the 1-dimensional version. As mentioned, this is already sufficient for almost all graphs. If it turns out not to be sufficient, the algorithms take the route of branching by using the concept of individualization.
Specifically, the individualization-refinement paradigm, which is adopted by virtually all modern competitive isomorphism tools, one by one artificially assigns a different color to all the vertices in a color class. This breaks the symmetry and subsequently color refinement can be potentially applied again to produce a more refined partition of the vertices. In a backtracking manner, the tools continue until a discrete color (a coloring in which all color classes are a singleton) has been reached. The tools use various pruning techniques, such as invariants and pruning with automorphisms, discovered with intricate methods, to drastically improve their performance.
The tools actually compute a canonical form, which also solves the isomorphism problem (as explained earlier). This highly practical method was originally pioneered by McKay with his famous software tool Nauty. There are now various extremely efficient packages such as Bliss, Conauto, Nauty, Saucy, and Traces freely available. Recently many new ideas, responsible for their efficiency, such as the use of the trace for early abortion of color refinement in Traces, have found their way into the tools. We refer the reader to an extensive survey.31 In contrast to Babai’s quasipolynomial-time algorithm, there are, however, graphs on which the running time of all individualization-refinement algorithms scale exponentially.33
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Applications
Graph isomorphism tools can in practice be used to find symmetries of combinatorial objects and as such they have numerous applications in miscellaneous domains. In the context of optimisation, for example, in SAT solving, symmetries are exploited to collapse the search space, as parts equivalent under symmetries only need to be explored once. An alternative way of exploiting symmetries is to add symmetry breaking constraints to the original input again drastically improving performance.
Another application domain exploits canonical labeling to store graph structured data in a database. For example, when molecules are stored in a chemical database, the idea is to store only a canonical representative. To look up a given molecule in the database, we compute its canonical representative and find the result in the database. This way, no isomorphism tests against the elements in the database are required. Other application domains include machine learning, computer graphics, software verification, model checking, and mathematical programming.
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Concluding Remarks
With Babai’s quasipolynomial-time algorithm, we have seen a breakthrough on one of the oldest and best studied algorithmic problems. Undoubtedly, this algorithm and its underlying mathematical framework rank among the most important contributions to theoretical computer science in a long time. We are only starting to explore the potential of the wealth of new ideas they bring to the field.
Current challenges include the group isomorphism problem, one of the core obstacles to even faster graph isomorphism tests. On the practical side, emerging applications in areas such as machine learning demand a better understanding of approximate versions of isomorphism and similarity measures between graphs.
Yet the question whether graph isomorphism is solvable in polynomial time remains open, and we can expect further deep, exciting insights until it will finally be settled.
References
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About the Authors
Authors
Martin Grohe is a professor of computer science at RWTH Aachen University, Aachen, Germany.
Pascal Schweitzer is a professor at TU Kaiserslautern, Germany.
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DOI
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November 2020 Issue
Published: November 1, 2020
Vol. 63 No. 11
Pages: 128-134
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The number of ways in which 4 men and 3 ladies sit at a round table so that no two ladies sit together is
(a) 144
(b) 72
(c) 120
(d) None of these
In this question are men and women indistinguishable? I mean one man can not be different from another because we are classifying them on the criteria of gender.
Another question, how would I incorporate the gap idea into the circular arrangement thing?
I have already found this problem can be modelled using a septagon
combinatorics
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edited Jul 21, 2020 at 5:04
UmbQbify
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asked Jun 4, 2020 at 9:09
Clemens BartholdyClemens Bartholdy
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If indistinguishable then usually not "men and women" are used, but e.g. "blue balls and pink balls". –drhab Commented Jun 4, 2020 at 9:18
uh, that is like just memorizing the question. I am asking how to infer that from the English they have written –Clemens Bartholdy Commented Jun 4, 2020 at 9:19
It is just a comment. Not an answer. –drhab Commented Jun 4, 2020 at 9:26
Humans are not identical. Unless it's told to ignore that. –UmbQbify Commented Jun 4, 2020 at 15:40
I'm arguing that if you group a things based on one property like sex/color then they would be. If you were grouping on personality or something sure , there would be permutations and all –Clemens Bartholdy Commented Jun 4, 2020 at 20:49
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I think all seven people are distinguishable, because in general people are distinguishable. Gender is used for the "not next to one another" restriction, rather than creating distinction.
Also, consider what would happen if men were indistinguishable and women were indistinguishable. There would be exactly one legal seating (or seven, depending on whether you care where around the table the two men who sit next to one another sit (which now that I think about it is another thing we don't know whether it's supposed to matter: do we care who sits closest to the door?)) That's probably not what is asked for here, even without looking at the multiple choices.
But ultimately, this is a game of mind reading, and you have to guess. There is no way to know for absolute certain without asking the problem author.
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answered Jun 4, 2020 at 9:29
ArthurArthur
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In most of the cases people are distinguishable unless mentioned otherwise.
Considering all the people to be distinguishable, we can incorporate the gap method by firstly seating all the men around the circle in 4!4=3!4!4=3! ways. between these 4 men there are 4 gaps created around the circle. we then select 3 of these gaps and permute the women in those to get the total no. of ways.
Ans. = (3!).(4 3).(3!)(3!).(4 3).(3!) = 144
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answered Jul 21, 2020 at 5:19
PranayPranay
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5199 | https://www.youtube.com/watch?v=OKNAqgv7v74 | Find its largest prime factor
letsthinkcritically
73100 subscribers
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Posted: 20 Sep 2021
Math #numbertheory #Factor
In this video we attempt to find the largest prime factor of the number 146419604.
Few interesting facts on divisibility test. I may make a video on these later:
To check whether 3 is a divisor, check whether 3 divides the sum of its digits.
To check whether 5 is a divisor, check whether the last digit is 5 or 0
To check whether 7,11,13 is a divisor, first divide the number by 1001 will help reduce to a smaller number
To check whether 37 is a divisor, first divide the number by 111 will help reduce to a smaller number
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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Transcript:
in this video we are going to find the largest prime factor of the number 146419604 before we move on don't forget to give a like subscribe to my channel and turn on post notifications to find prime factors of large numbers you need a lot of observations such as being able to split this number into sum of powers and try to factorize the sum of powers using algebra so for this the first observation is to notice that it is actually equal to i can split it into one four six four one and then four zeros added by 9604 for the first number is actually equal to at first one four six four one this is in fact 11 to the power four you might have noticed that the sequence from the pascal triangle something that looks like this they are in fact power if you just um combine them into numbers row by row then they are actually powers of 11. so this is 11 to the power four times ten to the power four four zeros and then added by nine zero four going to factorize into four times two four or one and this is in fact 49 squared or in this case i'm going to write this as 7 to the power 4. so now i've written this into sum of four powers and the second term is four times some fourth powers now the next observation is to notice that you can factorize and you can actually factorize expressions of this form so the way to factorize this is to add an extra term which is 4 x squared y squared and you have to subtract it afterwards next you have to subtract that by 4x squared y squared and so for the first three terms is equal to x squared plus 2y squared whole squared and then the second terms as in the last term not the one that's not the second one but the things that are not higher than yellow is 2 x y whole squared now it is in fact difference of squares so it can be written as x squared plus 2y squared plus 2xy or multiplied by x squared plus 2 y squared minus 2xy so that means we've managed to factorize x to the 4 plus 4 times y to the 4. now back to our main problem we can replace x by 110 and replace y by 7. so therefore this can be rewritten as the product while x is one one zero y is seven this whole thing multiplied by something similar is just that we are subtracting two times 1 1 0 times 7 instead now we can compute each bracket separately so that's 1 1 0 squared is 1 2 1 double zero 2 times seven squared is 98 and the third term is one five four zero similar applies to the second bracket but we subtract one five four zero instead so we compute that by hand and the first one will be one three seven three eight multiplied by one one six five eight sorry it's one zero instead one zero six five eight so we've managed to rewrite that big number in the product of two five digit numbers so we're one step closer to our the prime factors now we can further factorize these numbers because it's obvious that we can do that because they are even so the first thing two times six eight six nine and the second number is two times five three two nine so now no more um even factors so we try to factorize them in different ways firstly notice that 5 3 2 9 is actually 73 squared if you manage to remember values of squares then you can quickly settle the green number and what remains will be 6 8 six nine the number six thousand eight hundred and sixty nine is actually a prime so that means this number is in fact our final answer the largest prime factor of the number 146419604 it might be quite strange to end the video in this way so let me end this by briefly explaining how i can deduce that this number 6 869 is actually a prime number without using a calculator so to check whether a number is a prime we usually test all prime factors test all primes that is less than square root of n say if we want to test whether n is a prime so this we have to take a look what's the square root of the number 6 869 so notice that let's take a multiple of 10 say 80 the square is 6 4 double zero so it's larger than 80. now we pick a number um a little bit closer say 83 because this number ends with 9 as well well if we square this number we will get 6889 so it's just a little above the number that we want to test and that means the square root of the number that we want to test is about 82 point something just a wild guess but this is enough because we just want to know a rough estimate of the square root so we know which other problems that we need to test and i've actually listed out the primes already and here are the primes 2 3 5 7 all the way up to 79 i'm not going to test them one by one in this video i'm going to attach some of them actually all the results all the testing results in the comments and slightly explaining some of the special tracks that i have used but this is um all in all how do we test whether a number is prime the most primitive way without using help of any machine |
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